Swami Vivekananda Scholarship: Swami Vivekananda scholarship which is also known as Bikash Bhavan scholarship which is merit-cum-means which is for economically backward and meticulous students studying in West Bengal at a higher secondary level, postgraduate level, undergraduate level, above. This scholarship was revamped in the year 2016 and now it comes up with a significantly increased number of seats and different awards. This scholarship is awarded by the West Bengal government where the sanctioned amount has been increased from Rs. 45 crores to Rs. 2000 crores.

This scholarship was announced for assisting the meritorious students of the West Bengal state that belongs to the economically backward class of the society. Also, the Swami Vivekananda scholarship is helpful to students that are studying in class 11, 12 undergraduate levels, postgraduate level, doctorate level.

Swami Vivekananda Scholarship Important Dates

Some of the important dates under the Swami Vivekananda scholarship are as below:

Swami Vivekananda Scholarship – Eligibility Criteria

The key criteria that one needs to follow under the Swami Vivekananda scholarship are as follows

• The students applying for the scholarship must be a resident of West Bengal.
• The student must be pursuing studies at the higher secondary level or above at recognized Institutes in West Bengal.
• The family of the student applying should not be more than Rs. 2.5 lakhs per annum.

Besides the above criteria, there are some course-related specific eligibility criteria that students should follow. They are

How To Apply for Swami Vivekananda Scholarship?

The applications related to the scholarship are only accepted through the official website. The link for application will depend on the category of the student. Also, no hard copies of applications under any circumstances will be acceptable under this scheme. Below is the step by step guide for applying under the Swami Vivekananda scholarship:

• Go to the official website and register yourself based on your category.
• After registering successfully on the official website, login to the website using your registered password and ID.
• After logging into the account, fill all the required details that are marked as mandatory. Makes sure that the information added is correct as any discrepancy will lead to rejection.
• After this, you need to upload the required documents online. Make sure that the documents uploaded are correct documents in specific size and format.
• Once you have provided the Institute verification form properly to the head of the Institution, you can also submit the application form online. The form will be considered complete if you have uploaded the attested form including the verification.

Important Documents for Swami Vivekananda Scholarship

Below is the list of important documents required for Swami Vivekananda scholarship

• Mark sheet of the last exam passed
• Admit card of secondary examination
• Income certificate
• The first page of your bank passbook with all the details specifically mentioned
• Ration card/ Aadhar card/ voter ID
• Income affidavit
• Recent passport size photograph with size not more 10 to 50 KB
• Institute verification form duly signed and authenticated by the head of Institution

Swami Vivekananda Scholarship – Selection Process

Swami Vivekananda scholarship is a merit-cum-means based scholarship. So, it takes student’s merit as well as financial needs into consideration for the final selection process. Below are the steps through selection takes place

The applications submitted online are differentiated based on the categories are initially arranged into descending order. This is done on the basis of marks that have been obtained in the qualifying exam and also on the basis of income criteria.

Once the merit list is prepared, the scholarship amount is distributed to students in their given accounts upon successful completion of documents verification.

Swami Vivekananda Scholarship Amount

The amount to be received under Swami Vivekananda scholarship scheme depends majorly on the course or the level of study that one is currently pursuing. Generally, this amount will vary from Rs. 1000 per month to Rs. 5000 per month. Below is the detailed information related to the eligible amount for the students:

Swami Vivekananda Scholarship – Renewal

Every year the recipients of the Swami Vivekananda scholarship will have to renew their scholarship. This renewal process is applicable to some particular level of studies and the renewal cases are subjected to the decent performances in academics. The minimum score to obtain during renewal from higher secondary level to UG as well as the diploma is 60% marks. However, for getting a renewal at the postgraduate level, one must obtain a minimum of 50% marks. Below is the step by step procedure for applying for renewal process:

• Go to the official website of Swami Vivekananda scholarship
• There is a link for renewal application, click on this link.
• Login to the website using your official ID and password.
• Fill all the required details along with uploading of supporting documents.
• Submit the renewal application once all details are filled.

The important documents required for renewal process are the marksheets for 2nd-year renewal is 1st and 2nd semester’s marksheet. Also, marksheet of 1st and 2nd semester is required for 2nd-year renewal at PG level.

FAQ’s

Question 1.
Who is eligible for Swami Vivekananda scholarship?

Answer:
The student must be a resident of West Bengal. Besides this, there are certain eligibility criteria that one has to meet.

Question 2.
If I have filled the scholarship form in class 11 and I have received the amount then do I need to fill the form for class 12 as well?

Answer:
One needs to renew the Swami Vivekananda scholarship every year. But there certain criteria that are to be met first before applying for the renewal process.

Question 3.
For whom is Swami Vivekananda scholarship useful?

Answer:
Swami Vivekananda scholarship is helpful to students that are studying in class 11, 12,  undergraduate level, postgraduate level, and doctorate level.

Students Can Also Check

• GOKDOM Scholarship – GOKDOM
• Central Sector Scholarship – CSSS

The post Swami Vivekananda Scholarship 2019 | Apply Online, Application, Eligibility appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Chemistry 2018

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

Question 1.
The analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason. [1]

Question 2.
CO(g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? [1]
Answer:
CO(g) and H₂(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.

Question 3.
Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂]. [1]
Answer:
Coordination number: 6;
Oxidation state: +2

Question 4.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? [1]
Answer:
Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction condition, hydrolysis proceeds by nucleophilic substitution mechanism and the benzylic carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.

Question 5.
Write the IUPAC name of the following: [1]

Answer:
The IUPAC name would be 3, 3- Dimethyl-pentane-2-ol.

Question 6.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^-1) in 250 g of water.
(K_f of water = 1.86 K kg mol^-1) [2]
Answer:
Molality (m) of a given solution of Glucose:
m = [(60/180) g mol^-1/250 g] × 1000 = 1.33 mol kg^-1
Now, depression in freezing point is given by, ΔT_f = K_fm
Putting the given values,
ΔT_f = K_fm = 1.86 × 1.33 = 2.5
So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.

Question 7.
For the reaction [2]
2N₂O₅(g) → 4NO₂(g) + O₂(g)
the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1. Calculate the rate of disappearance of N₂O₅(g).
Answer:
Rate of reaction for the given reaction can be given as,
Rate = 1/2 {-Δ[N₂O₅]/Δt} or {-Δ[N₂O₅]/Δt} = 1/2 {[NO₂/ Δt]}
So, the rate of disappearance of N₂O₅ would be half of the rate of production of NO₂ (given 2.8 × 10^-3 Ms^-1).
So, the rate of disappearance of N₂O₅ is 1.4 × 10^-3 Ms^-1.

Question 8.
Among the hydrides of Group-15 elements, which have the [2]
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character?

Question 9.
How do you convert the following? [2]
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
OR
Account for the following:
(a) Aromatic carboxylic acids do not undergo Friedel- Crafts reaction.
(b) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.
Answer:
(a) Conversion of ethanol to Propanone:


OR
(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.

(b) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than the benzoic acid. Being an electron-withdrawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

Question 10.
Complete and balance the following chemical equations: [2]
(a) Fe²⁺ + MnO^–₄ + H⁺ →
(b) MnO^–₄ + H₂O + I^– →
Answer:
(a) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(b) 2MnO^–₄ + H₂O + I^– → 2MnO₂ + 2OH^– + IO^–₃

Question 11.
Give reasons for the following: [3]
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.
(c) Elevation of the boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.
Answer:
(a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.

(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at the higher temperatures.

(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K⁺ and Cl^–, compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.

Question 12.
An element ‘X’ (At. mass = 40 g mol^-1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 10²³ mol^-1) [3]

Question 13.
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^-1 mol^-1) [3]
Answer:
Rate constant for a first-order reaction is given by,

Question 14.
What happens when [3]
(a) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(b) persistent dialysis of a colloidal solution is carried out?
(c) an emulsion is centrifuged?
Answer:
(a) When FeCl₃ is added to a freshly prepared precipitate of Fe(OH)₃, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.

(c) Emulsions are centrifuged to separate them into constituent liquids.

Question 15.
Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process. [3]
Answer:
Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O₂) from which the metal is obtained later by replacement method (using Zinc).
The reactions involved are:
4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[AU(CN)₂]^– (aq) + 4OH^– (aq)
2 [AU(CN)₂]^– (aq) + Zn(s) → 2Au (s) + [Zn(CN)₄]^2- (aq)

Question 16.
Give reasons:
(a) E⁰ value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(b) Iron has a higher enthalpy of atomization than that of copper.
(c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. [3]
Answer:
(a) Mn²⁺ has a d⁵ configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn³⁺, On the contrary, Fe³⁺ attains a half-filled orbital configuration when Fe²⁺ gets oxidized to Fe³⁺. Hence, the E⁰ value for Mn³⁺/ Mn²⁺ couple has more positive E⁰ value.

(b) Fe has a 3d⁶4s² outer electronic configuration whereas Cu has 3d¹⁰4s¹ configuration. Now, more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.

(c) Sc³⁺ has a 3d⁰ configuration whereas Ti³⁺ has a 3d¹ configuration. As there are no electrons in d orbital for Sc³⁺ ion, there is no transition of electrons by absorption of energy and hence no emission in visible range imparting colour to the Sc³⁺ ion.

Question 17.
(a) Identify the chiral molecule in the following pair: [3]

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1 methylcyclohexane with alcoholic KOH.
Answer:
(a) The molecule (i) is a chiral molecule.

(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fitting reaction.

(c) In the 1-Bromo-1-methylcyclohexane, all β-hydrogen atoms are equivalent. Thus dehydrohalogenation takes place, in the reaction of this compound with KOH.

Question 18.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollen’s test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn (Hg)/conc. HCl, give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? [3]
Answer:
(a) Compound A and C give positive Tollen’s test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C₄H₈O the structure of compound would be CH₃COCH₂CH₃ (Butanone).

Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:

(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.

Question 19.
Write the structures of the main products in the following reactions: [3]

Answer:
(i) Sodium borohydride doesn’t reduce esters, so the product would be,

Question 20.

1. Why is bithional added to soap? [3]
2. What is the tincture of iodine? Write its one use.
3. Among the following, which one acts as a food preservative?
   Aspartame, Aspirin, Sodium Benzoate, Paracetamol

Answer:

1. Bithional is added to soaps to impart antiseptic properties to soap.
2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.
3. Sodium benzoate is used as a food preservative.

Question 21.
Define the following with an example of each: [3]
(a) Polysaccharides
(b) Denatured protein
(c) Essential amino acids
OR
(a) Write the product when D-glucose reacts with conc. HNO₃.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins.
Answer:
(a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.

(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.

(c) Essential amino acids: The amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan
OR
(a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.

(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.

(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intra. molecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).

Question 22.
(a) Write the formula of the following coordination compound: Iron (III) hexacyanoferrate (II)
(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅ Cl]SO₄?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]^3-.
(Atomic number of Co = 27) [3]
Answer:
(a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe₄[Fe(CN)₆]₃
(b) [CO(NH₃)₅Cl]SO₄ will show Ionisation isomerism and the possible isomers are [CO(NH₃)₅Cl]SO₄ and [Co (NH₃)₅SO₄]Cl
(c) Electronic configuration of Co³⁺ ion is,

Electronic configuration of sp³d² hybridized (as F is a weak field ligand) orbitals of Co³⁺, with six pairs of electrons from six F ions.

There are 4 impaired electrons in [CoF₆]₃.

Question 23.
Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags. [4]
Answer the following:
(a) Write the values (at least two) shown by Shyam.
(b) Write one structural difference between low-density polythene and high-density polythene.
(c) Why did Shyam refuse to accept the items in polythene bags?
(d) What is a biodegradable polymer? Give an example.
Answer:
(b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.
(c) Shyam refused to take the items in polythene bags as polythene is non-biodegradable neither recyclable,
(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by certain microorganisms and thus are environment-friendly.
For example Poly β -hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Question 24.
(a) Give reasons: [5]
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
OR
(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
(b) Arrange the following in the decreasing order of their reducing character: HF, HCl, HBr, HI
(c) Complete the following reaction:
XeF₄ + SbF₅ →
Answer:
(a) (i) In H₃PO₃ (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These type of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H₃PO₄ (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H₃PO₄.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed-and not FCl₃ because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S₈ molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O₂) and it does not have enough intermolecular attraction and thus exists in gaseous form.

(a) (i) The brown gas A is NO₂ or nitrogen dioxide. On cooling, it dimerises to N₂O₄ and solidifies as a colourless solid.

(iii) Compound A, that is, NO₂ contains the odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N₂O₄ molecule with even number of electrons (thus colourless) and have better intermolecular forces to get solidified. Thus, it changes to solid on cooling.
(b) Decreasing order of reducing character: HI > HBr > HCl > HF
(c) XeF₄ + SbF₅ → [XeF₃]⁺ + [SbF₆]^–

Question 25.
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: [5]
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H₂(g) (1 bar) | Pt(s)
(Given: E° Sn²⁺/Sn = – 0.14V)
(b) Give reasons:
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH₃COOH decreases on dilution.
OR
(a) For the reaction
2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M), ΔG° = -43600 J at 25°C.
Calculate the e.m.f. of the cell. [log 10^-n = -n]
(b) Define fuel cell and write its two advantages.
Answer:
(a) The half cell reactions can be written as;

The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O₂ but on account of overpotential of oxygen, Cl^– gets oxidized preferably, liberating Cl₂ gas.
(ii) The conductivity of CH₃COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.
OR


(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.
Advantages of fuel cells are:

• Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
• Fuel cells are pollution-free.

Question 26.
(a) Write the reactions involved in the following: [5]
(i) Hofmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
(a) Write the structures of the main products of the following reactions:

(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.
(c) Arrange the following in the increasing order of their pK_b values:
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃
Answer:
(a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH₃CONH₂) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

(b) (i) (CH₃)₂ NH is more basic than (CH₃)₃N in aqueous solutions because in (CH₃)₃N the lone pair of electrons on the nitrogen atom is responsible for its basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to (CH₃)₂NH.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.

(b) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’t undergo this reaction.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Chemistry 2018 appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Physics 2019 Delhi

CBSE Previous Year Question Papers Class 12 physics 2019 Outside Delhi Set-I

Section-A

Question 1.
Draw the pattern of electric field lines, when a point charge – Q is kept near an uncharged conducting plate. [1]
Answer:
The positive charge will be induced on the uncharged conducting plate, kept near it. So, the lines of forces will start from metal plate and end -Q

Question 2.
How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant ?
Answer:
The mobility of electrons is given by the

As it’s independent of the applied potential difference, so it will not change if the applied potential difference will be doubled.

Question 3.
Define the term “threshold frequency” in the context of photoelectric emission.
OR
Define the term ‘Intensity’ in photon picture of electromagnetic radiation.
Answer:
The minimum frequency of the radiation incident on a metal surface below which there is no photoelectric emission is called threshold frequency. It is a characteristic property of a photosensitive material.
OR
The amount of light energy or photon energy incident per meter square per second is called intensity of radiation. SI unit of intensity of radiation is W/Sr

Question 4.
What is the speed of light in a denser medium of polarising angle 30°
Answer:

Question 5.
In sky wave mode of propagation, why is the frequency range of transmitting signals restricted to less than 30 MHz ?
OR
On what factors does the range of coverage in ground wave propagation depend ?

Section-B

Question 6.
Two bulbs are rated (P₁, V) and (P₂, V). If they are connected
(i) in series and
(ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P₁and P₂.
Answer:
(i) In series combination :

(ii) In parallel combination

Question 7.
Calculate the radius of curvature of an equiconcave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D ? [2]
OR
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index 4√2 / 5.
Answer:
In an equiconcave lens, radius of curvature of both surfaces are equal



Question 8.
An a-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic Field B, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. [2]
Answer:

Question 9.
State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. [2]
OR
Calculate the orbital period of the electron in the first excited state of hydrogen atom.
Answer :
According to Bohr’s quantization of angular momentum, the stationary orbits are those in which angular momentum of electron


Question 10.
Why a signal transmitted from a TV tower cannot be received beyond a certain distance? Write the expression for the optimum separation between the receiving and the transmitting antenna. [2]

Question 11.
Why is wave theory of electromagnetic radiation not able to explain photo electric effect ? How does photon picture resolve this problem ? [2]
Answer:
Wave theory cannot explain the following laws of photoelectric effect.

• The instantaneous emission of photo electrons.
• Existence of threshold frequency for metal surface.
• K.E. of emitted electrons is independent of intensity of light and depends on frequency.

The concept of photon explained that energy is not only emitted and absorbed in discrete energy quanta, but also it propagates through space in definite quanta with the speed of light. It can explain all the above photoelectric effect, which wave theory cannot explain.

Question 12.
Plot a graph showing variation of de Broglie wavelength (λ) associated with a charged particle of a mass m, versus 1/√v, where V is the potential difference through which the particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle ? [2]
Answer:

Section-C

Question 13.
(a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole. [3]
Answer:



Question 14.
Using Kirchhoff’s rules, calculate the current through the 40 Q and 20 Q resistors in the following circuit: [3]

What is end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5Ω and S respectively.

When the resistance S is shunted with an equal resistance, the new balance length found to be 1.5 l₁ where l₁ is the initial balancing length. Calculate the value of S.




Question 15.
(a) Identify the part of the electromagnetic
spectrum used in
(i) radar and
(ii) eye surgery.
Write their frequency range.
(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. [3]
Answer:
(a)(i) Radar—Microwaves are used in radar. The frequency range is from 3xl0:i Hz to lxlO9 Hz.
(ii) Eye Surgery — Infrared waves are used. The frequency range is from 4xl014Hz to 3xlOu Hz.
(b) Energy density in oscillating electric field is

Question 16.
Define the term wave front. Using Huygens’s wave theory, verify the law of reflection. [3]
OR
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wave front is propagating from a denser to a rarer medium.
Answer:
A wave front is the continuous locus of vibrating particles which are in the same state of vibration or phase.
Laws of Reflection from Huygens principle

Consider a plane wave front AB incident on a plane reflecting surface PQ, Let v be the velocity of the wave. At time t – 0 one end of the wave front just touches the reflecting surface at B. Draw normal NB to PQ. When the wave front strikes the reflecting surface, then due to the presence of it, it cannot advance further. When wave front strikes at B, secondary wavelets starts emitting from B. The secondary wavelets will travel a distance AD = vt during the time the other end A of the wave front AB reaches the surface PQ at Q. To find the reflected wave front, B as a center and AD as radius draw an arc, which represent the secondary wavelets originating from B. As the incident wave front AB advances, the secondary wavelets will touch CD simultaneously.
According to Huygens’s principle CD represents the reflected wave front corresponding to incident wave front AB. BD is common triangles BAD and CBD and BC = AD = vt. Therefore, two triangles are congruent.
So, ∠ABD = ∠CDB
i. e., ∠i = ∠r
Thus, angle of incidence is equal to the angle of reflection. This is the second law of reflection. Also, the incident wave front AB, reflecting surface PQ and the reflected wave front CD are perpendicular to the plane of the paper. So, the incident ray, reflected ray and the normal at the point of incidence, all lie in same plane. This is first law of reflection.
Thus, Huygens’s principle explains both the law of reflection.
OR
The refractive index of a medium is defined as the ratio of speed of light in vacuum to that of the speed of the light in medium.
i.e n= c/v
Where, n = refractive index of medium when light ray passes from vaccum into a medium.
c = velocity of light in vacuum
v = velocity of light in the medium
Proof of Snell’s law of refraction

when a wave front travels from one medium to other, it deviates from its path. In travelling from one medium to other, the frequency of wave remains same and speed and wavelength changes. Let, XY be a surface separating two media T and ‘2’. Let the speed of waves of v₁ and v₂.

Suppose, a plane wave front AB in first medium is incident obliquely on the boundary surface XY and its end touches the surface at A at time t = 0, while the other end B reaches the surface at point B’ after time-interval ‘f Clearly, BB₁ = v₁t.
In the same time, wavelets starts from A and reaches A’ in time ‘f with velocity v₂. Therefore, AA₁ = v₂t According to Huygen’s principle, A₁B₁ is the new position of the wave front in second medium. A¹B¹ is a refracted wave front.
Let, the incident wave front (AB) and reflected wave front (A¹B¹) makes angle i and r with surface XY.
In ΔAB¹B,

Hence, ratio of sine of angle of incidence and the sine of angle of refraction for a given pair of media is constant. This is Snell’s law of refraction.

Question 17.
(a) Define mutual inductance and write its S.I. unit.
(b) A square loop of side carrying a current I2 is kept at distance x from an infinitely long straight wire carrying a current I₁ as shown in the figure. Obtain the expression for the resultant force acting on the loop. [3]

Answer:
(a) Mutual inductance : It is a property of two coils due to which each coil opposes any change of current flowing in the other. The mutually induced e.m.f, in one coil produces the opposition in other coil. Its SI unit is henry (H).
(b)


Question 18.
(a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.
(b) Explain the significance of a radial magnetic field when a current carrying coil is kept in it. [3]
Answer : Torque on current carrying loop in magnetic field :


Consider a coil PQRS placed in a magnetic field. Let 0 be the angle between the plane of the coil and direction of B. When current flow through coil each side experiences a force. The forces on vertical side will constitute a couple.
Moment of torque = One of forces × Perpendicular distance between lines of action of force

Where, is magnetic moment of loop.
(b)A magnetic field, in which the plane of the coil in all positions remains parallel to the direction of magnetic field is called radial magnetic field. In a radial magnetic field, magnetic torque remains maximum for all the positions of the coil.

Question 19.
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.8 x 10⁶ m and the radius of lunar orbit is 3.8 x 106 m. [3]
Answer:
Astronomical telescope in near points adjustment:


Question 20.
(a) State Gauss’ law for magnetism. Explain its significance.
(b) Write the four important properties of the magnetic field lines due to a bar magnet. [3]
OR
Write three points of differences between para, dia, and ferromagnetic materials, giving one example for each.
Answer:
(a) Gauss law for magnetism : If a closed surface is imagined in a magnetic field, the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero.
Gauss law signifies that magnetic mono poles does not exist.
(b)
1. In a bar magnet, each lines of force, starts from a north pole and reaches the south pole externally and then goes from south pole to a north pole internally. Thus, magnetic line of force forms a closed loop.
2. No two lines of force will never intersect each other.
3. In a uniform field, the lines are parallel and equidistant from each other.
4. The lines of force are crowded near the poles.
OR

Note : Any three difference can be written in exam.

Question 21.
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10000 disintegrations and 5,000 disintegrations after 20 hr. and 30 hr. respectively from start. Calculate the half life and initial number of nuclei at t = 0. [3]
Answer:
Decay constant’ of a radioactive sample is defined as the ratio of its instantaneous rate of disintegration to the number of atoms present at that time.



Question 22.
(a) Three photo diodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm ?
(b) Why photo diodes are required to operate in reverse bias? Explain. [3]
Answer:
(a) Energy corresponding to wavelength 600 nm

The photon energy (E = 2.06 eV) is greater than the band gap for diode D₂ only. Hence, diode D₁ and D₃ will not be able to detect the given wavelength.
(b)A photo diode is operated reverse bias because in reverse bias it is easier to observe change in current with change in light intensity,

Question 23.
(a) Describe briefly the functions of the three segments of n-p-n transistor.
(b) Draw the circuit arrangement for studying the output characteristics of n-p-n transistor in CE configuration. Explain how the output characteristics is obtained. [3]
OR
Draw the circuit diagram of a full wave rectifier and explain its working. Also, give the input and output wave forms.
Answer:

For full wave rectifier, we use two junction diodes as shown in figure.
During first half cycle of input a.c. signal the terminal Si is positive relative to S and S₂ is negative, then diode Di is in forward biased and diode D₂ is reverse biased. Therefore, current flows in D₁ not in D₂. In next half cycle, Si is negative and S₂ is positive relative to S. Then D₁ is in reverse biased and D₂ is in forward biased. Therefore, current flows in D₂ not in Di. Thus, for input a.c. signal the output current is a continuous series of unidirectional pulse.
The input and output wave forms are shown in the figure.

Question 24.
(a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of A & B.
(b) A message signal of frequency 20 kHz and peak voltage 10 V is used to modulate a carrier of frequency 2 MHz and peak voltage of 15 V. Calculate the modulation index. Why the modulation index is generally kept less than one ?

Section-D

Question 25.
(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.
(b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit ?
(c) When an inductor is connected to 200 V dc voltage, a current of 1 A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why ? Also, calculate the self inductance of the inductor. [5]
OR
(a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
(b) A small town with a demand of 1200 kW of electric power at 220V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Q per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.
Answer:
(a) Consider an alternating e.m.f. is connected in series with an inductor, resistance R and capacitance C. Let, E and I be the instantaneous values of e.m.f. and current in the LCR circuit V_L, V_C and V_R be the instantaneous values of voltage across inductor, capacitor and resistor respectively. Then,




(b) At resonance
X_L = X_c
iX_L = iX_{c
VL= VC}
The voltage across inductance and capacitance are equal and have a phase difference of 180° at resonance.
(c) Since the reactance of an inductor is zero for d.c. circuit. But the inductor offers resistance to an a.c. circuit. Therefore, the current decreases for the same inductor when it is connected with an a.c. source.
When inductor is connected in a.c. circuit:

OR
(a) Step-down transformer:
It is a device used for converting high alternating voltage at low current into low alternating voltage at high current and vice-versa.
The device works on the principle of mutual induction i.e., if the current or magnetic flux linked with a coil changes then an e.m.f. is induced in the other coil.
In step-down transfomer Np > Ns and transformation ratio is less than 1.

1. Copper losses: Due to resistance of winding’s in primary and secondary coils, some electrical energy is converted into heat energy.
2. Flux losses : Some of the flux produced in primary coil is not linked up with secondary coils.
3. Hysteresis losses : When the iron core is subjected to a cycle of magnetization the core
gets heated up due to hysteresis known as hysteresis loss.
4. Iron losses : The varying magnetic flux produces eddy current in the iron core, which leads to the wastage of energy in the form of heat.
(b) Length of wire line = 20 x 2 = 40 km
Resistance of wire line, r = 40 x 0.5 = 20 Ω
Power to be supplied = 1200 kW = 1200 × 10³ W
Voltage at which power supplied = 4000 V

Question 26.
(a) Describe any two characteristic features
which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double slit experiment.
(b)In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. [5]
OR
(a) Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal length +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

Answer:
(a)

A sources of monochromatic light illuminates two narrow slits S₁ and S₂.The two illuminated slits act as the two coherent sources. The two slits is very close to each other and at equal distance from source. The wave front S₁ and S₂ spread in all direction and superpose and produces dark and bright fringe on screen. Let the displacement of waves from Si and S₂ at point P on screen at time t is




OR
(a) Conditions for total internal refraction :
1. The ray must travel from a denser medium into a rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for the pair of media.
Relation between critical angle and refractive index:

When a ray of light travels from denser to rarer medium, the ray bends away from the normal. When the angle of incidence is equal to the critical angle then the refracted ray grazes the surface of separation represented by ray and



Question 27.
(a) Describe briefly the process of transferring
the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. [5]
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(b) The identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
Answer:
(a) When the plates of the parallel plate capacitor is connected to a battery. Then the first insulated metal plate gets, the positive charge till its potential become maximum. Then, the charge will leak to surroundings. So, the negative charge will be induced on the nearer face of the second plate and the positive charge will be induced on its farther plate.

Consider a capacitor of capacitance C. Initial charge and potential difference be zero. Let, a charge Q be given in small steps. Let at any instant when charge on capacitor be q, the potential difference between its plates,

Now work done in giving an additional charge dq is,

OR
(a) Consider an electric dipole of charges -q and +q separated by a distance 2a and placed in a free space. Let P be a point on equitorial line of dipole at a distance r from the centre of a dipole.




This is the required expression.
(b) Let the two charges of + q each placed at point A and B at a distance 2 m apart in air.

Suppose, the third charge Q (unknown magnitude and charge) is placed at a point O, on the line joining the other two charges, such that OA= x and OB 2-x.
For the system to be in equilibrium, net force on each 3 charges must be zero.
If we assume that charge Q placed at O is positive, the force on it at O may be zero. But the force on charge q at point A or B will not be zero. It is because, the forces on a charge q due to the other two charges will act in same direction. If charge Q is negative, then the forces on q due to other two charges will act in opposite direction.
Hence, Q will be negative in nature. For charge (-Q) to be in equilibrium Force on charge (-q) due to charge (+q) at point A should be equal and opposite to charge (+Q) at B

Therefore, for the system to be in equilibrium a charge – Q is placed at a mid point between the two charges of + q each.

CBSE Previous Year Question Papers Class 12 physics 2019 Outside Delhi Set-II

Note : Except for the following questions, all the remaining questions have been asked in previous set.

Section-A

Question 1.
When unpolarised light is incident on the interface separating the rarer medium and the denser medium. Brewster angle is found to be 60°. Determine the refractive index of the denser medium. [1]
Answer:

Question 2.
When a potential difference is applied across the ends of a conductor, how is the drift velocity of the electrons related to the relaxation time? [1]
Answer:


Question 3.
Draw the equipotential surfaces due to an isolated point charge. [1]
Answer:

Section-B

Question 4.
Explain with the help of Einstein’s photoelectric equation any two observed features in photo-electric effect which cannot be explained by wave theory. [2]
Answer:
Features of photoelectric equation which can not be explained by wave theory :
(a) The wave theory could not explain the instantaneous process of photoelectric effect.
(b) ‘Maximum kinetic energy’ of the emitted photo electrons is independent of intensity of incident light.

Question 5.
A deuteron and an alpha particle having same momentum are in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of the radii of the circular paths described by them. [2]
Answer:
Radius of circular path

Question 6.
(a) Plot a graph showing variation of de Broglie wavelength (X) associated with a charged particle of mass m, versus √v, where V is the accelerating potential.
(b) An electron, a proton and an alpha particle
have the kinetic energy. Which one has the shortest wavelength ? [2]
Answer:

Section-C

Question 7.
(a) State the underlying principle of a moving
coil galvanometer.
(b) Give two reasons to explain why a galvanometer cannot as such be used to measure the value of the current in a given circuit
(c) Define the terms:
(i) voltage sensitivity and
(ii) current sensitivity of a galvanometer. [3]
Answer:
(a) The Principle : When a current flows through the conductor coil, a torque acts on it due to the external radial magnetic field. Counter torque due to suspension balances coil after appropriate deflection due to current in the circuit.
(b) A galvanometer can be used as such to measure current due to following two reasons.

• A galvanometer has a finite large resistance and is connected in series in the circuit, so it will increase the resistance of circuit and hence change the value of current in the circuit.
• A galvanometer is a very sensitive device, it gives a full scale deflection for the current of the order of micro ampere, hence if connected as such it will not measure current of the order of ampere.

(c) (i) Voltage sensitivity : It is defined, as the deflection produced in the galvanometer when a unit voltage is applied across it.
(ii) Current Sensitivity: The ratio of deflection produced by the coil Φ to the current in the coil is called the current sensitivity. It is the deflection of the meter per unit current.

Question 8.
(a) Draw equipotential surfaces corresponding to the electric field that uniformly increases in magnitude along with the z-directions.
(b) Two charges -q and + q are located at point (0, 0, – a) and (0, 0, a). What is the electrostatic potential at the points (0, 0, ± z) and (x, y, 0) ? [3]
Answer:
(a)


Question 9.
(a) Write the relation between half life and average life of a radioactive nucleus.
(b) In a given sample two isotopes A and B are initially present in the ratio of 1 : 2. Their half lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio of 2 :1 ? [3]

Question 10.
(a) Define the term ‘self inductance’ of a coil. Write its S.I. unit.
(b) A rectangular loop of sides a and b carrying current I₂ is kept at a distance ‘a’ from an infinitely long straight wire carrying current I₁ as shown in the figure. Obtain an expression for the resultant force acting on the loop. [3]

Answer:
(a) Self-inductance : Self-inductance of a coil is numerically equal to the amount of magnetic flux linked with the coil when and current flows through the coil. The S.I. unit of Self-inductance is henry (H) or weber per Ampere lH = 1 wb/A
(b) The force on the side AB of rectangle is attractive as current is flowing in the same direction and on side CD will be repulsive the current is flowing in opposite direction with respect to straight conductor. The resultant magnetic force. On sides AD and BC is zero. The side AB is the straight wire. So, the net force will be attractive and rectangular loop will move towards the straight wire.
Now, force between AB and straight wire

Force between CD and straight wire

CBSE Previous Year Question Papers Class 12 Physics 2019 Outside Delhi Set-III

Note : Except for the following questions, all the remaining questions have been asked in previous set.

Section-A

Question 1.
Distinguish between unpolarized and linearly polarized light. [1]
Answer:
Unpolarized light : The light having vibration of electric field vector in all possible directions perpendicular to the direction of wave propagation the light is known as unpolarized light.

Linearly polarized light : The light having vibrations of electric field vector in only one direction perpendicular to the direction of propagation of light is as plane or linearly polarized light

Question 2.
How is the drift velocity in a conductor affected with the rise in temperature ? [1]
Answer:
With the rise in temperature, the collision of electrons occurs more frequently, so relaxation time decreases and hence drift velocity increases.

Question 3.
Draw the pattern of electric field lines when a point charge + q is kept near an uncharged conducting plate. [1]
Answer :
The lines of force start from + Q and terminates at metal plate inducing negative charge on it.

Section -B

Question 4.
(a) Define the terms,
(i) threshold frequency and
(ii) stopping potential in photoelectric effect. [2]
Answer:
(a) (i) Threshold frequency : The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. It is denoted by V₀
(ii) Stopping potential : The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is denoted by V₀ (or Vs).

Question 5.
Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom. [2]
Answer:
According to Bohr’s postulate,


Question 6.
A charged particle q is moving in the presence of a magnetic field B which is inclined to an angle 30° with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path. [2]
Answer:

When a charged particle enter in a magnetic field making angle at 30°. Then velocity component is resolved into 2 components, v cos θ (along the magnetic field) and v sin θ (normal to the magnetic field). As the charged particle moves along XY- plane due to velocity component v sin θ, it also advances linearly due to the velocity component v cos θ. As a result, the charged particle will move in a helical path as shown in figure.

Section-C

Question 7.
(a) Explain briefly how Rutherford scattering of a-particle by a target nucleus can provide information on the size of the nucleus.
(b) Show that density of nucleus is independent of its mass number A.  [3]
Answer:
(a) In Rutherford’s scattering experiment of a-particle, it was observed that the fast and heavy a-particles could be deflected through 180°. But only very small number of particles i.e., 1 in about 8,000 a-particles are deflected through 180° that too from centre only. So by this it was assumed that the size of central part i.e., nucleus is about of
the size of the atom and whole positive charge is concentrated in it.
(b) Consider an atom whose mass number is A and R be the radius of the nucleus. If we neglect the mass of orbital electrons, then mass of the nucleus of the atom of mass number A = A a.m.u.

Question 8.
State the underlying principle of a cyclotron. Explain its working with the help of a schematic diagram. Obtain the expression for cyclotron frequency.  [3]
Answer:
(a) Principle : It is based on a principle that a positive ion can acquire sufficiently large energy with a comparatively smaller alternating potential difference by making it to cross the same electric field again and again by making use of a strong magnetic field.

Working: The positive ions are produced from the source at the centre are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of a.c source are so chosen that as the ions comes out of a dee, it changes its polarity and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. This phenomenon is continued till the ion reaches at the periphery of the dees where an deflecting plate deflects the accelerated ion on the target to be bombarded. Expression for cyclotron frequency Suppose a position ion with charge q moving with a velocity v, then


Question 9.
Two infinitely long straight wire A₁ and A₂ carrying currents I and 21 flowing in the same direction are kept ‘d’ distance apart. Where should a third straight wire A₃ carrying current 1.5 I be placed between A₁ and A₂ so that it experiences no net force due to A₁ and A₂ ? Does the net force acting on A₃ depend on the current flowing throught it ? [3]
Answer:


No, at a same distance the force on the wire A₃ is independent of the direction of the current. As if current is in opposite direction then F₁ and F₂ will be in opposite direction, but will be in equilibrium.

Question 10.
(a) Draw the equipotential surfaces due to an electric dipole.
(b) Derive an expression for the electric field due to a dipole of dipole moment if at a point on its perpendicular bisector. [3]
Answer:
(a)

The Magnitudes of the electric field due to the two charges +q and -q given by.

The directions of E_+q and E_-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
.’. Total electric field

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Physics 2019 Delhi appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Maths 2018

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

Section – A

Question 1.
Find the value of tan^-1 – cot^-1 () [1]
Solution:
We have,

Question 2.
If the matrix A = is skew-symmetric, find the values of’a’ and ‘b’. [1]
Solution:

Question 3.
Find the magnitude of each of the two vectors , having the same magnitude such that the angle between them is 60° and their scalar product is . [1]
Solution:
Let be two such vectors.

Question 4.
If a * b denotes the larger of ‘a’ and ‘b’ and if a o b = (a * b) + 3, then write the value of (5) o (10), where * and o are binary operations.** [1]

Section – B

Question 5.
Prove that: [2]

Solution:
R.H.S = sin^-1(3x – 4x³)
Putting x = sin θ in R.H.S, we get
R.H.S = sin^-1(3 sin θ – 4 sin³ θ)
= sin^-1(sin 3θ)

Question 6.
Given A = , compute A^-1 and show that 2A^-1 = 9I – A. [2]
Solution:

Question 7.
Differentiate with respect to x. [2]
Solution:

Question 8.
The total cost C(x) associated with the pro-duction of x units of an item is given by C(x) = 0.005x³ – 0.02x² + 30s + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. [2]
Solution :
Cost function is given as
C(s) = 0.005x³ – 0.02x² + 30s + 5000
Marginal cost (MC) =
= 0.005(3x²) – 0.02 (2x) + 30
= 0.015x² – 0.04s+ 30
When x = 3,MC = 0.015(3)² – 0.04(3) + 30
= 0.135 – 0.12 + 30
= 30.015

Question 9.
Evaluate:

Solution:
We have,

Question 10.
Find the differential equation representing the family of curves y = ae^{bx + 5}, where a and b are arbitrary constants. [2]
Solution:
Given curve is
y = ae^{bx + 5} …(i)
Differentiating (i) w.r.t. s, we get

Question 11.
If θ is the angle between two vectors and , find sin θ. [2]
Solution:

Question 12.
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. [2]
Solution :
The sample space has 36 Outcomes. Let A be event that the sum of observations is 8.
∴ A = {(2, 6), (3, 5), (5, 3), (4, 4), (6, 2)}

Section – C

Question 13.
Using properties of determinants, prove that: [4]

Solution:

Question 14.
If (x² + y²)² = xy, find . [4]
Solution:
We have,
(x² + y²)² = xy …(i)

OR
If x = a(2θ – sin 2θ) and y = a(1 – cos 2θ), find when θ = .
Solution:
Given, x = a (2θ – sin 2θ)
and y = a (1 – cos 2θ)
Differentiating x* and y w.r.t. θ, we get

Question 15.
If y = sin(sin x), prove that: [4]

Solution:
Given, y = sin (sin x)
Differentiating y w.r.t. x, we get
= cos (sin x) • cos x
Differentiating again w.r.t. x, we get
= cos (sin x) (- sin x) + cos x [-sin (sin x)-cos x]

Question 16.
Find the equations of the tangent and the normal to the curve 16x² + 9y² = 145 at the point (x₁, y₁), where x₁ = 2 and y₁ > 0. [4]
Solution:
Given curve is
16x² + 9y² = 145 ….(i)
Since (x₁, y₁) lies on equation (i),



OR
Find the intervals in which the function

(a) strictly increasing,
(b) strictly decreasing.
Solution:
Given function is

Question 17.
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question ? [4]
Solution:
Let the length, breadth and height of the open tank be x, x and y units respectively.
Then, Volume (V) = x²y …(i)


Area is minimum, thus cost is minimum when x = 2y.
i.e., depth of tank is half of the width.
Value : Any relevant value.

Question 18.
Find: [4]

Solution:

Question 19.
Find the particular solution of the differential equation e^x tan ydx + (2 – ex) sec² ydy = 0, given that y = when x = 0. [4]
Solution:
Given differential equation is,

OR
Find the particular solution of the differential equation + 2y tan x = sin x, given than y = 0 when x = .
Solution:
Given differential equation is,

Hence, particular solution is
y sec² x = sec x – 2
or y = cos x – 2 cos²x

Question 20.
Let Find a vector d which is perpendicular to both . [4]
Solution:
Given,

Question 21.
Find the shortest distance between the lines and [4]
Solution:
Given lines are

Question 22.
Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’, is obtained. If she obtained exactly one “tail7, what is the probability that she threw 3, 4, 5 or 6 with the dice? [4]
Solution :
Let E₁ be the event that girl gets 1 or 2 on the roll and E₂ be the event that girl gets 3, 4, 5, or 6 on the roll of a die.

Let A be the event that she gets exactly one tail. If she tossed coin 3 times and gets exactly one tail then possible outcomes are HTH, HHT, THH

=

Question 23.
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X. [4]
Solution:
First five positive integers are 1, 2, 3, 4, 5. We select two positive numbers in 5 × 4 = 20 ways. Out of these, two numbers are selected at random. Let X denote larger of the two selected numbers. Then, X can have values 2, 3, 4 or 5.
P(X = 2) – P (larger no. is 2) = {(1, 2) and (2, 1)}


Therefore, mean and variance are 4 and 1 respectively.

Section – D

Question 24.
Let A = {x ϵ Z : 0 ≤ x ≤ 12}. Show that R = {(a, b): a, b ϵ A, | a – b | is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]. [6]
Solution:
Given, R = {(a, b): a, b ϵ A, | a – b | is divisible by 4}
Reflexivity: For any a ϵ A
| a – a | = 0, which is divisible by 4
(a, a) ϵ R
So, R is reflexive.

Symmetry: Let (a, b) ϵ R
⇒ | a – b | is divisible by 4
⇒ | b – a | is divisible by 4 [ ∵ | a – b | = | b – a | ]
⇒ (b, a) ϵ R
So, R is symmetric.

Transitive : Let (a, b) ϵ R and (b, c) ϵ R
⇒ | a – b | is divisible by 4
⇒ | a – b | = 4k
∴ a – b = ± 4k, k ϵ Z …(i)
Also, |b – c| is divisible by 4
⇒ |b – c| = 4m
∴ b – c = ±4m, m ϵ Z …(ii)
Adding equations (i) and (ii)
a – b + b – c = ±4 (k + m)
⇒ a – c = ±4 (k + m)
|a – c | is divisible by 4,
⇒ (a, c) ϵ R
So, R is symmetric.
⇒ R is reflexive, symmetric and transitive.
∴ R is an equivalence relation.
Let x be an element of R such that (x, 1) ϵ R
Then | x – 1| is divisible by 4
x – 1 = 0, 4, 8, 12,
⇒ -x = 1, 5, 9 (∵ x ≤ 12)
∴ Set of all elements of A which are related to 1 are {1, 5, 9}.
Equivalence class of 2 i.e.
[2] = {(a, 2): a ϵ A, |a – 2| is divisible by 4}
⇒ | a – 2| = 4k(k is whole number, k ≤ 3)
⇒ a = 2, 6, 10
Therefore, equivalence class [2] is {2, 6, 10]
OR
Show that the function f : R → R defined by f(x) = , ∀x ϵ R is neither one-one nor onto. Also,if g: R → R is defined as g(x) = 2x – 1, find fog(x).
Solution:

Question 25.
If A = , find A^-1. Use it to solve the system of equation: [6]
2x – 3y + 5z = 11
3x + 2y – 4z = -5
x + y – 2z = -3
Solution:



OR
Using elementary row transformations, find the inverse of the matrix:

Solution:
We know that, A = IA

Question 26
Using integration, find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x² + y²= 32. [6]
Solution:
Given curve is

Question 27.
Evaluate: [6]

Solution:


OR
Evaluate as the limit of the sum.
Solution:

Question 28.
Find the distance of the point (-1, -5, -10) from the point of intersection of the line and the plane . [6]
Solution:
Equation of line is

Question 29.
A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand- operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of ₹1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and solve it graphically and find the maximum profit [6]
Solution:
Let the number of packets of screw ‘A’ manufactured in a day be x and that of screw B be y.

Plotting the points on the graph, we get the feasible region OABC as shown (Shaded).


Hence, profit will be maximum if company produces 30 packets of screw A and 20 packets of screw B and maximum profit = ₹ 41.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Maths 2018 appeared first on Learn CBSE.

Important Questions for Class 12 Physics Chapter 13 Nuclei Class 12 Important Questions

Nuclei Class 12 Important Questions Very Short Answer Type

Question 1.
An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?
(Delhi 2008)
Answer:

Question 2.
State the reason, why heavy water is generally used as a moderator in a nuclear reactor. (Delhi 2008)
Answer:
Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction :

Here d is deutron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross-section for neutron absorption.

Question 3.
Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. (Delhi 2008)
Answer:
Control rod or cadmium rod.

Question 4.
State tzvo characteristic properties of nuclear force. (All India 2008)
Answer:
(i) Nuclear forces are the strongest force in nature.
(ii) They are saturated forces.
(iii) They are charge independent.

Question 5.
Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities? (Delhi 2009)
Answer:

Question 6.
Two nuclei have mass number in the ratio 1 : 3. What is the ratio of their nuclear densities? (Delhi 2009)
Answer:
Since nuclear density is independent of the mass number, the ratio of nuclear densities will be 1:1.

Question 7.
Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? (Delhi 2009)
Answer:
Nuclear density is independent of mass number, so the ratio will be 1 : 1.

Question 8.
Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii? (All India 2009)
Answer:

Question 9.
Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii? (All India 2009)
Answer:

Question 10.
Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii? (All India 2009)
Answer:

Question 11.
Write any two characteristic properties of nuclear force. (All India 2009)
Answer:
1. Nuclear forces are strongest forces in nature.
2. Nuclear forces are charge independent.

Question 12.
What is the relationship between decay constant and mean life of a radioactive nucleus? (Comptt All India 2009)

Nuclei Class 12 Important Questions Very Short Answer Type VSA-I

Question 1.
An electron and alpha particle have the same de- Broglie wavelength associated with them. How are their kinetic energies related to each other? (Delhi 2008)
Answer:

Question 2.
State the reason, why heavy water is generally used as a moderator in a nuclear reactor. (Delhi 2008)
Answer:
Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction :

Here d is deutron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross-section for neutron absorption.

Question 3.
Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. (Delhi 2008)
Answer:
Control rod or cadmium rod.

Question 4.
State two characteristic properties of nuclear force. (All India 2008)
Answer:
(i) Nuclear forces are the strongest force in nature.
(ii) They are saturated forces.
(iii) They are charge independent.

Question 5.
Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities? (Delhi 2008)
Answer:

Question 6.
Two nuclei have mass number in the ratio 1 : 3. What is the ratio of their nuclear densities? (Delhi 2008)
Answer:
Since nuclear density is independent of the mass number, the ratio of nuclear densities will be 1:1.

Question 7.
Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? (Delhi 2008)
Answer:
Nuclear density is independent of mass number, so the ratio will be 1 : 1.

Question 8.
Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii? (All India 2008)
Answer:

Question 9.
Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii? (All India 2008)
Answer:

Question 10.
Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii?
Answer:

Question 11.
Write any two characteristic properties of nuclear force. (All India 2008)
Answer:
1. Nuclear forces are strongest forces in nature.
2. Nuclear forces are charge independent.

Question 12.
What is the relationship between decay constant and mean life of a radioactive nucleus? (Comptt. All India 2012)
Answer:
Relationship between decay constant and mean life of a radioactive nucleus is

Question 13.
Write the relationship between the size and the atomic mass number of a nucleus. (Comptt. All India 2012)
Answer:
Relationship between the size and the atomic mass number of a nucleus is

Question 14.
Define the activity of a given radioactive substance. Write its S.I. unit. (All India 2012)
Answer:
The activity of a radioactive substance is defined as the rate of disintegration of the substance. The SI unit for activity is becquerel (Bq).

Question 15.
How is the radius of a nucleus related to its mass number A? (Comptt. All India 2012)
Answer:

Question 16.
Why is it found experimentally difficult to detect neutrinos in nuclear P-decay? (All India 2012)
Answer:
It is found experimentally difficult to detect neutrinos in nuclear P-decay, because of two reasons :
(i) mass of neutrino is extremely small;
(ii) its charge is negligibly small.
Also, neutrinos interact very weakly with matter.

Question 17.
Name and define, the SI unit for the ‘activity’, of a given sample of radioactive nuclei. (Comptt. All India 2012)
Answer:
(i) becquerel is the SI unit of ‘activity’ of a nuclear sample.
(ii) One becquerel activity corresponds to ‘one decay/disintegration per second’.

Nuclei Class 12 Important Questions Short Answer Type (SA-I)

Question 18.

Answer:

Question 19.
Calculate the energy released in MeV in the following nuclear reaction:
(All India 2012)
Answer:

Question 20.
A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme:

The mass number and atomic number of A are 190 and 75 respectively. What are these numbers for A₄? (Delhi 2016)
Answer:

So, the Mass number of A₄ → 69
and Atomic number of A₄ → 172

Question 21.
A radio active nucleus ‘A’ undergoes a series of decays according to the following scheme:

The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A₄?
Answer:
The series can be shown as below:

So, the Mass number of A₄ is 182
and Atomic number of A₄ is 72

Question 22.
(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.
(b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. (All India 2016)
Answer:
(a) When nucleons approach each other to form a nucleus, they strongly attract each other. Their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in’ potential energy results in the decrease in the mass of the nucleons inside the nucleus.

Question 23.
A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV. (Delhi 2016)
Answer:

∴ Gain in binding energy for nucleon = 8.5 – 7.6 = 0.9 MeV
Hence total gain in binding energy per nucleus fission = 240 × 0.9 = 216 MeV

Question 24.
Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. (All India 2016)
Answer:
Two important conclusions :
(i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large-size nucleus.

(ii) Graph explains that force is attractive for distances larger than 0.8 fin and repulsive for distances less than 0.8 fm.

Question 25.
Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged? (All India 2016)
Answer:
(a) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short ranged.
If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk. The property that a given nucleon influences only nucleons close to it, is referred to as saturation property of the nuclear force.

(b) Nuclear force is short-ranged for a sufficiently large nucleus. A nucleon is under the influence of only some of its neighbours, which come within the range of the nuclear force. If a nucleon can have maximum of P neighbours within the range of nuclear force, its binding energy would be proportional to ‘P’ Thus on increasing ‘A’ by adding nucleons binding energy will remain constant.

Question 26.
Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release of energy in the processes of nuclear fission and nuclear fusion can be explained. (All India 2011)
Answer:
1. Nuclear fission : Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.

2. Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

Question 27.
Complete the following nuclear reactions :


(Comptt. Delhi 2011)
Answer:

Question 28.
If both the number of protons and neutrons in a nuclear reaction is conserved, in what way is mass converted into energy (or vice verse)? Explain giving one example. (Comptt. Delhi 2011)
Answer:
Explanation for release of energy in a nuclear reaction : Since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of the nuclear reaction.
But total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in binding energy causes a release of energy in the reaction.
Examples :

Question 29.

(Delhi 2016)
Answer:

Question 30.
Calculate the energy in fusion reaction :

(Delhi 2016)
Answer:

Question 31.
If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction? Explain. (Comptt. All India 2016)
Answer:
The number of protons and neutrons in a nuclear reaction are conserved but the total mass is not conserved.
The total mass of the free protons and neutrons is more than their total mass within the nucleus. The lost mass (= ∆m) known as ‘mass defect’, gets converted into energy as per the relation E = (∆m)c² (c is the velocity of light)

Question 32.
Write two characteristic features of nuclear force.
(b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. (Comptt. All India 2017)
Answer:

(a) Nuclear forces. The strong forces of attraction which hold together the nucleons (neutrons and protons) in the tiny nucleus of an atom are called nuclear forces.
Important properties (characteristics):
1. Nuclear forces are independent of charge (These act between a pair of neutrons, between a pair of protons and between a proton and a neutron).
2. Nuclear forces are the strongest forces in nature.
3. Nuclear forces are very short range forces.
4. Nuclear forces are non-central forces.
5. Nuclear forces are dependent on spin.

(b) A plot of the potential energy between two nucleons as a function of distance is shown in the diagram.

Important conclusions from the graph :
(i) The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. This happens only because the nuclear force is much stronger than the coulomb force. The gravitational force is much weaker than even Coulomb force.
(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometers. This leads to saturation of forces in a medium or a large-sized nucleus, which is the reason for the constancy of the binding energy per nucleon.
(iii) The nuclear force between neutron- neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge.

Question 33.
Write the relation between half life and decay constant of a radioactive nucleus. (Comptt. All India 2017)
Answer:

Question 34.
Two radioactive samples, X, Y have the same number of atoms at t = 0. Their half lives are 3h and 4h respectively. Compare the rates of disintegration of the two nuclei after 12 hours. (Comptt All India 2017)
Answer:

Question 35.
Distinguish between nuclear fission and fusion. Explain how the energy is released in both the processes. (Comptt. All India 2017)
Answer:
In nuclear fission a heavy nucleus breaks up into smaller nuclei accompanied by release of energy; whereas in nuclear fusion two light nuclei combine to form a heavier nucleus accompanied by release of energy.
In both the cases, some mass (= mass defect) gets converted into energy as per the relation :

Nuclei Class 12 Important Questions Short Answer Type SA-II

Question 36.
Draw a plot showing the variation of binding energy per nucleon versus the mass number (A). Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion. (All India 2009)
Answer:
1. Nuclear fission : Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.

2. Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

Question 37.
Define the activity of a radionuclide. Write its S.I. unit. Give a plot of the activity of a radioactive species versus time.
How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8^th of its initial value? (All India 2009)
Answer:
Activity: It is defined as the total decay rate of a sample of one or more radionuclide.
Its S.I. unit is bequerel
I bequerel = 1 decay per second

Question 38.
(i) Define ‘activity’ of a radioactive material and write its S.I. unit.
(ii) Plot a graph showing variation of activity of a given radioactive sample with time.
(iii) The sequence of stepwise decay of a radioactive nucleus is

If the atomic number and mass number of D₂, are 71 and 176 respectively, what are their corresponding values for D?
Answer:
Activity: It is defined as the total decay rate of a sample of one or more radionuclide.
Its S.I. unit is bequerel
I bequerel = 1 decay per second

Question 39.
(a) Write symbolically the P“ decay process of .
(b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life. (All India 2009)
Answer:

(b) Suppose a radioactive sample contains N0 nuclei at time t = 0. After time this number reduces to N. Furthermore, suppose dN nuclei disintegrates in time t to t + dt. As dt is small so the life of each of the dN nuclei can be approximately taken equal to t
∴ Total life of dN nuclei = tdN

Question 40.
State the law of radioactive decay.
Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T_1/2. Depict in the plot the number of undecayed nuclei at
(i) t = 3 T_1/2 and
(ii) t = 5_1/2.(DeIhi 2009)
Answer:
The number of nuclei undergoing decay per unit time, at any instant is proportional to number of nuclei in the sample at that instant. The given figure shows a graph between the number of undecayed nuclei as a function of time.

Question 41.
(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
(ii) Show that the density of nucleus over a wide range of nuclei is constant- independent of mass number A. (Delhi 2009)
Answer:
(i) Saturation is the Short range nature of nuclear forces
(ii) Let A be the mass number and R be the radius of a nucleus
If m is the average mass of a nucleon, then
Mass of nucleus = mA

Clearly, nuclear density is independent of mass number A or the size of the nucleus.

Question 42.
Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is
(i) attractive and
(ii) repulsive. Write any two characteristic features of nuclear forces.
Answer:
The graph indicates that the attractive force between the two nucleons is strongest at a separation r₀ = 1 fm. For a separation greater than the force is attractive and for separation less than r₀, the force is strongly repulsive.

Two characteristic features of nuclear forces :
1. Strongest interaction
2. Short-range force
3. Charge independent character (any two)

Question 43.
Answer the following, giving reasons:
(i) Why is the binding energy per nucleon found to be constant for nuclei in the range of mass number (A) lying between 30 and 170?
(ii) When a heavy nucleus with mass number A = 240 breaks into two nuclei, A = 120, energy is released in’ the process.
(iii) In β-decay, the experimental detection of neutrinos (or antineutrinos) is found to be extremely difficult. (Comptt. All India 2011)
Answer:
(i) Nuclear forces are short ranged. For a particular nucleon inside a sufficiently large nucleus will be under the influence of some of its neighbours which come within the range of the nuclear force. The property that a given nucleon influences only nucleons close to it is also referred to as saturation property of the nuclear force.
(ii) The binding energy per nucleon of the parent nucleus is less than those of the two daughter nuclei. It is this increased binding energy that gets released in this process.
(iii) Neutrinos are chargeless and massless particles, whose interaction with other particles is almost negligible. Hence, they can pass through very large quantity of matter with-out getting detected.

Question 44.
(a) In a typical nuclear reaction, e.g.

although number of nucleons is conserved, yet energy is released. How? Explain.
(b) Show that nuclear density in a given nucleus is independent of mass number A. (Delhi 2011)
Answer:
(a) In all types of nuclear reactions, the law of conservation of number of nucleons is followed. But during the reaction, the mass of the final product is found to be slightly less than the sum of the masses of the reactant components. This difference in mass of a nucleus and its constituents is called mass defect. So, as per mass energy relation E = (∆M)c², energy is released. In the given reaction the sum of the masses of two deutrons is more than the mass of helium and neutron. Energy equivalent of mass defect is released.

Question 45.

though the conserved on both sides of the reaction, yet the energy is released. How? Explain.
(b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the regions where potential energy is
(i) positive and
(ii) negative. (Delhi 2011)
Answer:
(a) Since the total binding energy of nuclei on the left side of the reaction is not the same as the total binding energy of nucleus on the right hand side, this difference of binding energy appears as the energy released.

For separation (r) ≤ 0.8 fermi
Force is repulsive
For r > 0.8 fermi force will be attractive.

Question 46.
(a) The number of nuclei of a given radioactive sample at time t = 0 and t = T are N₀ and N₀/n respectively. Obtain an expression for the half-life (T_1/2) of the nucleus in terms of n and T.
(b) Write the basic nuclear process underlying β-decay of a given radioactive nucleus. (Comptt. Delhi 2013)
Answer:
(a) Decay constant. It is the reciprocal of the time interval in which the number of active nuclei in a radioactive sample reduces to 1/e times of its initial nuclei.
Half life period. It is the time during which half the total number of atoms in radioactive elements (N₀) disintegrates. It is denoted by t_1/2.

(b) β-decay. It is the phenomenon of emission of an electron from a radioactive nucleus. In Beta-minus decay, a neutron transforms into a proton within the nucleus. According to

The emission of β-particle from an atom will change it into a new atom whose atomic number is increased by one without changing its mass number.

Question 47.

(Delhi 2014)
Answer:
(a) Radioactive decay law. It states that “the number of atoms disintegrated per second at any instant is directly proportional to the number of radioactive atoms actually present at that time.” Let N₀ be the total number of atoms present at time t = 0 (initially)
and N be the total number of atoms present at time t, then
According to radioactive decay law, the rate of disintegration at any time t is directly proportional to the number of atoms present at that time t.

(ii) Since new nucleus has the same mass number, hence it would be an Isobar.

Question 48.
Write the relation for binding energy (BE) (in MeV) of a nucleus of , atomic number (Z) and mass number (A) in terms of the masses of its constituents – neutrons and protons.
(b) Draw a plot of BE/A versus mass number A for 2 ≤ A ≤ 170. Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei. (Comptt. Delhi 2014)
Answer:
(a) Mass defect. The difference between the sum of the masses of neutrons and protons forming a nucleus and mass of the nucleons is called mass defect.

(b)

Conclusions :
(i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.

(b) (i) When we move from heavy nuclei region to middle region, we find that there will be a gain in overall binding energy and hence release of energy. This indicates that energy can be released when a heavy nucleus breaks into two roughly equal fragments/nuclear fission.
(ii) Similarly, when we move from lighter nuclei to heavier nuclei, we find that there will be gain in overall binding energy and hence release of energy. This indicates that energy can be released when two lighter
nuclei fuse together to form heavy nucleus/nuclear fusion.
(c) In Beta decay a neutron breaks into a proton, electron and neutrino as

Detection of neutrinos is difficult because they are chargeless and have either no or low mass.

Question 49.
(a) Define the term ‘activity of a sample of radioactive nucleus. Write its S.I. unit.

(Comptt. All India 2014)
Answer:
(a) The activity of a radioactive nucleus equals its decay rate (or number of nuclei decaying per unit time)

Question 50.
(i) Write the relation between ‘average life’ and ‘half-life’ of a radioactive nucleus.

(Comptt. All India 2014)
Answer:

Question 51.
(i) Define the term ‘mass defect’ of a nucleus. How is it related with its binding energy?
(ii) Determine the Q-value of the following reaction:

(Comptt. All India 2014)
Answer:
(i) (a) The mass defect of a nucleus equals the difference between the total mass of its constituents and the mass of the nucleus itself.

Question 52.
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.
Calculate the energy release in MeV in the deuterium-tritium fusion reaction :

(Delhi 2015)
Answer:
(a) The breaking of heavy nucleus into smaller fragments is called nuclear fission; while the joining of lighter nuclei to form a heavy nucleus is called nuclear fusion.
(b) Binding energy per nucleon of the daugher nuclei, in both processess, is more than that of the parent nuclei. The difference in binding energy is released in the form of energy. In both processes some mass gets converted into energy.
(c) Energy released

Question 53.
(a) Write three characteristic properties of nuclear force.
(b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph. (All India 2015)
Answer:
(a) Nuclear forces. The strong forces of attraction which hold together the nucleons (neutrons and protons) in the tiny nucleus of an atom are called nuclear forces.
Important properties (characteristics):
1. Nuclear forces are independent of charge (These act between a pair of neutrons, between a pair of protons and between a proton and a neutron).
2. Nuclear forces are the strongest forces in nature.
3. Nuclear forces are very short range forces.
4. Nuclear forces are non-central forces.
5. Nuclear forces are dependent on spin.

(b) A plot of the potential energy between two nucleons as a function of distance is shown in the diagram.

Important conclusions from the graph :
(i) The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. This happens only because the nuclear force is much stronger than the coulomb force. The gravitational force is much weaker than even Coulomb force.
(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometers. This leads to saturation of forces in a medium or a large-sized nucleus, which is the reason for the constancy of the binding energy per nucleon.
(iii) The nuclear force between neutron- neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge.

Question 54.
Complete the following nuclear reactions :

(c) Why is it found experimentally difficult to detect neutrinos? (Comptt. All India 2017)
Answer:

It is found experimentally difficult to detect neutrinos in nuclear P-decay, because of two reasons :
(i) mass of neutrino is extremely small;
(ii) its charge is negligibly small.
Also, neutrinos interact very weakly with matter.

Question 55.
(a) Write the basic nuclear process involved in the emission of β⁺ in a symbolic form, by a radioactive nucleus.
(b) In the reactions given below :

Find the values of x, y, and z and a, b and c. (All India 2017)
Answer:

Question 56.
Obtain the relation for a sample of radioactive material having decay constant λ, where N is the number of nuclei present at constant λ. Hence obtain the relation between decay constant λ and half life of the sample. (Comptt. Delhi 2017)
Answer:
(i)
(a) Radioactive decay law. It states that “the number of atoms disintegrated per second at any instant is directly proportional to the number of radioactive atoms actually present at that time.” Let N0 be the total number of atoms present at time t = 0 (initially)
and N be the total number of atoms present at time t, then
According to radioactive decay law, the rate of disintegration at any time t is directly proportional to the number of atoms present at that time t.

(ii) Relation between X and
After one half life, Number of nuclei becomes

Question 57.
(i) A radioactive nucleus ‘A’ undergoes a series of decays as given below:

The mass number and atomic number of A₂ are 176 and 71 respectively.
Determine the mass and atomic numbers of A₄ and A.

(Delhi 2017)
Answer:

Question 58.
(a) Draw a graph showing the variation of binding energy per nucleon (BE/A) vs mass number A for the nuclei in 20 ≤ A ≤ 170.
(b) A nucleus of mass number 240 and having binding energy/nucleon 7.6 MeV splits into two fragments Y, Z of mass numbers 110 and 130 respectively. If the binding energy/ nucleon of Y, Z is equal to 8.5 MeV each, calculate the energy released in the nuclear reaction. (Comptt. All India 2017)
Answer:

Conclusions :
(i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.

Nuclei Class 12 Important Questions Long Short Answer Type

Question 59.
(a) Define the term ‘activity’ of a given sample of radionuclide. Write the expression for the law of radioactive decay in terms of the activity of a given sample.
(b) A radioactive isotope has a half life of T years. How long will it take the activity to reduce to 3.125% of its original value?
(c) When a nucleus (X) undergoes β-decay, and transforms to the nucleus (Y), does the pair (X, Y) form isotopes, isobars or isotones? Justify your answer. (Comptt. Delhi 2012)
Answer:
(a) The activity of a radioactive source is measured by the rate of disintegration of the source. It is denoted by ‘A’

Question 60.
(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two important conclusions that can be drawn regarding the nature of nuclear force.
(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.
(c) Write the basic nuclear process of neutron undergoing β-decay. Why is the detection of neutrinos found very difficult?
Answer:

Conclusions :
(i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.

(b) (i) When we move from heavy nuclei region to middle region, we find that there will be a gain in overall binding energy and hence release of energy. This indicates that energy can be released when a heavy nucleus breaks into two roughly equal fragments/nuclear fission.
(ii) Similarly, when we move from lighter nuclei to heavier nuclei, we find that there will be gain in overall binding energy and hence release of energy. This indicates that energy can be released when two lighter
nuclei fuse together to form heavy nucleus/nuclear fusion.
(c) In Beta decay a neutron breaks into a proton, electron and neutrino as

Detection of neutrinos is difficult because they are chargeless and have either no or low mass.

Important Questions for Class 12 Physics

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NCERT Solutions for Class 9 Social Science History Chapter 1 The French Revolution

These Solutions are part of NCERT Solutions for Class 9 Social Science. Here we have given NCERT Solutions for Class 9 Social Science History Chapter 1 The French Revolution. Free PDF download of NCERT solutions for Class 9 Social Science (India and the Contemporary World – I) Chapter 1 – The French Revolution. All questions are explained by the expert Social Science teacher and as per NCERT (CBSE) guidelines.

Question 1.
Describe the circumstances leading to the outbreak of revolutionary protest in France?
Answer:
The following circumstances led to the outbreak of revolutionary protest in France:

1. Louis XVI was an autocratic ruler who could not compromise with his luxurious life. He also lacked farsightedness.
2. When he ascended the throne the royal treasury was empty. Long years of war had drained the financial resources of France. Added to this was the cost of maintaining an extravagant court at the immense palace of Versailles.
3. Under Louis XVI France helped the thirteen American colonies to gain their independence from Britain the war added more than a billion livres to a dept credit, now began to charge 10% interest on loans. So the French government was obliged to spend an increasing percentage of its budget on interest payments alone.
4. The state finally increased taxes to meet its regular expenses su?h as the cost of maintaining an army, running government offices and universities.
5. The French society was divided into three estates but only members of the first two estates i.e,, the clergy and the nobles were exempted to pay taxes. They belonged to privileged class. Thus the burden of financing activities of the state through taxes was borne by the third estate only.
6. The middle class that emerged in the 18th century France was educated and enlightened. They refuted the theory of divine rights of the kings and absolute monarchy. They believed that a person’s social position must depend on his merit. They had access to the various ideas of equality and freedom proposed by the philosophers like John Locke, Jean Jacques Rousseau, Montesquieu etc. Their ideas got popularised among the common mass as a result of intensive discussions and debates in saloons and coffee houses and through books and newspapers.
7. The French administration was extremely corrupt. It did not give weightage to the French Common man.
   The state finally increased taxes to meet its regular expenses such as the cost of maintaining an army, running government offices and universities.

Question 2.
Which groups of French society benefited from the revolution? Which groups were forced to relinquish power? Which sections of society would have been disappointed with the outcome of the revolution?
Answer:

1. The wealthy class of the third estate which came to be known as the new middle class of France benefited the most from the revolution. This group comprised of big businessmen, petty officers, lawyers, teachers, doctors and traders. Previously, these people had to pay state taxes and they did not enjoy equal status. But after the revolution they began to be treated equally with the upper sections of the society.
2. With the abolition of feudal system of obligation and taxes, the clergy and the nobility came on the same level with the middle class. They were forced to give up their privileges. Their executive powers were also taken away from them.
3. The poorer sections of the society, i.e. small peasants, landless labourers, servants, daily wage earners would have been disappointed with the outcome of the revolution. Women also would have been highly discontented.

Question 3.
Describe the legacy of the French Revolution for the peoples of the world during the nineteenth and the twentieth centuries.
Answer:
The French Revolution proved to be the most important event in the history of the world.

1. The ideas of liberty and democratic rights were the most important legacy of the French Revolution. These ideas became an umpiring force for the political movements in the world in the 19th and 20th centuries.
2. The ideas of liberty, equality and fraternity spread from France to the rest of Europe, where feudal system was finally abolished.
3. Colonised people reworked on the idea of freedom from bondage into their movements to Create a sovereign nation state.
4. The idea of Nationalism that emerged after the French Revolution started becoming mass movements all over the world. Now people began to question the absolute power.
5. The impact of the French Revolution would be seen in India too. Tipu Sultan and Raja Rammohan Roy got deeply influenced by the ideas of the revolution.
   In the end, we can say that after the French Revolution people all over the world became aware of their rights.

Question 4.
Draw up a list of democratic rights we enjoy today whose origins could be traced to the French Revolution.
Answer:
Some of the democratic rights which we enjoy today can be traced to the French Revolution are as follows:

1. Right to equality including equality before law, prohibition of discrimination and equality of opportunity in matters of employment.
2. Right to freedom of speech and expression including right to practice any profession or occupation.
3. Right against exploitation.
4. Right to life.
5. Right to vote.

Question 5.
Would you agree with the view that the message of universal rights was beset with contradictions? Explain.
Answer:

1. The message of universal rights was definitely beset with contradictions. Many ideals of the “Declaration of Rights of Man and Citizen” were not at all clear. They had dubious meanings.
2. The French Revolution could not bring economic equality and it is the fact that unless there is economic equality, real equality cannot be received at any sphere. The Declaration of Rights of Man and Citizen laid stress on equality but large section of the society was denied to it. The right to vote and elect their representatives did not solve the poor man’s problem.
3. Women were still regarded as passive citizens. They did not have any political rights such as right to vote and hold political offices like men. Hence, their struggle for equal political rights continued.
4. France continued to hold and expand colonies. Thus, its image as a liberator could not last for a long time.
5. Slavery existed in France till the first half of the 19th century.

Question 6.
How would you explain the rise of Napoleon?
Answer:

1. The political instability of the Directory paved the way for the rise of Napoleon Bonaparte Napoleon had achieved glorious victories in wars. This made France realise that only a military dictator like Napoleon would restore a stable government.
2. In 1804, he crowned himself the emperor of France. He set out to conquer neighbouring European countries, dispossessing dynasties and creating kingdoms where he placed members of his family. Napoleon viewed himself as a moderniser of Europe. He introduced many laws such as the protection of private property and a uniform system of weight and measures provided by the decimal system.
   But his rise to power did not last for a long time. He was finally defeated at Waterloo in 1815.

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Maharashtra Board  Class 10 Solutions for Marathi कुमारभारती – हत्तीचा दुष्टान्त

Maharashtra Board SSC Class 10 Solutions for Marathi कुमारभारती

Maharashtra Board Class 9 Maths Solutions

The post Maharashtra Board Class 10 Solutions for Marathi कुमारभारती – हत्तीचा दुष्टान्त appeared first on Learn CBSE.

Maharashtra Board  Class 10 Solutions for Marathi कुमारभारती – स्त्रीपुरुष तुलना

Maharashtra Board SSC Class 10 Solutions for Marathi कुमारभारती

Maharashtra Board Class 10 Maths Solutions

The post Maharashtra Board Class 10 Solutions for Marathi कुमारभारती – स्त्रीपुरुष तुलना appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Maths 2017 Outside Delhi

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

CBSE Previous Year Question Papers Class 12 Maths 2017 Outside Delhi Set I

Section – A

Question 1.
If for any 2 × 2 square matrix A, A(adj A) = then write the value of | A |. [1]
Solution:
We have,

Question 2.
Determine the value of ‘k’ for which the following function is continuous at x = 3 [1]

Solution:
Given,

Question 3.
Find: . [1]
Solution:

Question 4.
Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20. [1]
Solution:
Since, the direction ratios of the normal to the given planes are proportional.
i.e.,
Thus, the given planes are parallel.
Now, let P(x₁, y₁, z₁) be any point on 2x – y + 2z – 5 = 0
Then, 2x₁ – y₁ + 2z₁ – 5 = 0
or 5x₁ – 2.5y₁ + 5z₁ – 12.5 = 0 …(i)
The length of the perpendicular from P(x₁, y₁, Z₁) to 5x – 2.5y + 5z – 20 = 0,

Therefore, the distance between the given planes is 1 unit.

Section – B

Question 5.
If A is a skew-symmetric matrix of order 3, then prove that det A = 0. [2]
Solution :
Given, A is a skew-symmetric matrix of order 3.
So, A^T = -A
Now, |A^T| = |-A |
|A^T| = (-1 )³|A| [1 |kA | = kⁿ | A | where n is order of A]
|A | = – | A| [ | A^T | = | A | ]
⇒ |A| + |A| =0
∴ 2|A| = 0 or |A| = 0.
i.e., det A = 0 Hence Proved.

Question 6.
Find the value of c in Rolle’s theorem for the function f(x) = x² – 3x in [-, 0] [2]
Solution:
We know that the polynomial function f(x) = x² – 3x is everywhere continuous and differentiable.
So, f(x) is continuous on [-,0].
Also, f(x) is differentiable on(-,0)
Now, f(-) = (-)³ – 3 (-)

Question 7.
The volume of a cube is increasing at the rate of 9 cm3/s. How fast is its surface area increasing when the length of an edge is 10 cm ? [2]
Solution:
Let, the side of the cube be a cm then, volume of cube (V) = a³
Differentiating V w.r.t.t, we get

Question 8.
Show that the function f(x) = x³ – 3x² + 6x – 100 is increasing on R. [2]
Solution:
We have,
f(x) = x³ – 3x² + 6x – 100
then, f'(x) = 3x² – 6x + 6
= 3(x² – 2x + 1) + 3
= 3 (x – 1)² + 3 > 0 for all x ϵ R.
Hence, the function f(x) is increasing on R.
Hence Proved

Question 9.
The x-coordinate of a point on the line joining the points P( 2, 2, 1) and Q (5, 1, – 2) is 4. Find its z-coordinate. [2]
Solution :
Given, the points P(2, 2, 1) and Q (5, 1, – 2) of a line.
Then, equation of line PQ,

Question 10.
A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find if A and B are independent events. [2]
Solution:
Since, A be the event of number obtained is even then, A = {2, 4, 6}
and B be the event of number obtained is red

Hence, the events A and B are not independent events.

Question 11.
Two tailors, A and B earn ₹ 300 and ₹ 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP. [2]
Solution:
Let x and y be the number of days for which the tailors A and B work respectively.
Total cost per day = ₹ (300x + 400y)
Let Z denote the total cost in rupees, then,
Z = 300₹ + 400y
Since in one day 6 shirts are stitched by tailor A and 10 shirts are stitched by tailor B and it is desired to produce atleast 60 shirts.
∴ 6₹ + 10y ≥ 60
It is given that 4 pairs of trousers are stitched by each tailor A and B per day to produce atleast 32 pairs of trousers.
∴ 4₹ + 4y ≥ 32
Finally, the no. of shirts and pair of trousers cannot be negative.
∴ x ≥ 0, y ≥ 0
Thus, mathematical formulation of the given LPP is as follows:
Minimize Z = 300x + 400y
Subject to constraints:
6x + 10y ≥ 60
4x + 4y ≥ 32
x ≥ 0, y ≥ 0

Question 12
Find: . [2]
Solution:
We have,

Section – C

Question 13.
If , then find the value of x.
Solution:
We have,

Question 14.
Using properties of determinants, prove that [1]

Solution:

OR
Find matrix A such that

Solution:
We have,

Question 15.
If x^y + y^x = a^b, then find . [4]
Solution:
We have, x^y + y^x = a^b

OR
If e^y(x + 1) = 1, then show that .
Solution:
We have,

Question 16.
Find: [4]

Solution:

Question 17.
Evaluate: [4]

Solution:


OR
Evaluate :
Solution:

Question 18.
Solve the differential equation [4]
Solution:
We have,

Question 19.
Show that the points A, B, C with position vectors respectively, are the vertices of a right-angled triangle. Hence, find the area of the triangle. [4]
Solution:
Given, the position vectors of the points A, B and C are respectively.

Question 20.
Find the value of X, if four points with position vectors and are coplanar. [4]
Solution:

Question 21.
There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X. [4]
Solution:
Given, X denote the sum of the numbers on the two drawn cards.
Then, X can take values 4, 6, 8,10,12 and sample space (S) = {(1, 3), (1, 5), (1, 7), (3, 1), (3, 5), (3, 7), (5, 1), (5, 3), (5, 7), (7, 1), (7, 3), (7, 5)]
So, the probability distribution of X is as given below:

Question 22.
Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he was found to have an A grade. What is the probability that the student has 100% attendance ? Is regularity required only in school ? Justify your answer. [4]
Solution:
Consider the following events:
A : the student has grade A.
E₁ : the student has 100% attendance.
E₂ : the student is irregular.
Then, probability of the students having 100% attendance:
P(E₁) = 30% = 0.3
Similarly, P(E₂) = 70% = 0.7
Now, by previous year report, the probability of the students having grade A who have 100% attendance:
P(A/E₁) = 70% = 0.7
and the probability of the students having grade A who are irregular :
P(A/E₂) = 10% = 0.1
Then, the probability of the student having 100% attendance who already has attain A grade = P(E₁/A)
By Bayes’ theorem,

No, regularity is required in school as well as in life.
It helps to be disciplined in every aspect of life. Or, when you work regularly, inspiration strikes regularly.

Question 23.
Maximise Z = x + 2y
subject to the constraints :
x + 2y ≥ 100 2x – y ≤ 0
2x + y ≤ 200 x, y ≥ 0
Solve the above LPP graphically. [4]
Solution:
Given,
Maximise Z = x + 2y
Subject to the constraints :
x + 2y ≥ 100 2x – y ≤ 0
2x + y ≤ 200 x, y ≥ 0
Converting the inequations into equations we obtain the lines
x + 2y = 100,
2x – y = 0,
2 x + y = 200.
Then, x + 2y = 100

Plotting these points on the graph, we get the shaded feasible region i.e., ADFEA.

Section – D

Question 24.
Determine the product: and use it to solve the system equations x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1. [6]
Solution:

Question 25.
Consider f : given by . Show that f is bijective. Find the inverse of f and hence find f^-1(0) and x such that f^-1(x) = 2. [6]
Solution:
For one-one:





OR
Let A = Q × Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) ϵ A. Determine, whether * is commutative and associative. Then, with respect to * on A.**
(i) Find the identity element in A
(ii) Find the invertible elements of A.

Question 26.
Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. [6]
Solution:
Let the length and breadth of the cuboid of square base be x and height be y.



Hence, it is a cube since the length, breadth and height of a cube are equal. Hence Proved.

Question 27.
Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4). [6]
Solution:
We have, A(4, 1), B (6, 6) and C(8, 4) as the vertices of a triangle ABC.



OR
Find the area enclosed between the parabola 4y = 3x² and the straight line 3x – 2y + 12 = 0.
Solution :
Given, the equation 4y = 3x² and 3x – 2y + 12 = 0.

Question 28.
Find the particular solution of the differential equation (x – y) = (x + 2y), given that y = 0 when x = 1. [6]
Solution:
We have,

Question 29.
Find the coordinates of the point where the line through the points (3, -4, -5) and (2, – 3, 1), crosses the plane determined by the points (1, 2, 3),(4, 2, -3) and (0, 4, 3). [6]
Solution:
Equation of the plane determined by the points (1, 2, 3), (4, 2, – 3) and (0, 4, 3) is


OR
A variable plane which remains at a constant distance 3p from the origin cuts, the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is

Solution:
Let the coordinates of A, B, C are (a, 0,0), (0, b, 0) and (0,0, c)

CBSE Previous Year Question Papers Class 12 Maths 2017 Outside Delhi Set II

Note : Except for the following questions, all the remaining questions have been asked in previous set.

Section – B

Question 12.
The length x, of a rectangle is decreasing at the rate of 5 cm/minute and the width y, is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle. [2]
Solution:

Section – C

Question 20.
Find: . [4]
Solution:

Question 21.
Solve the following linear programming problem graphically:
Maximise Z = 34x + 45y
under the following constraints
x + y ≤ 300
2x + 3y ≤ 70
x ≥ 0, y ≥ 0 [4]
Solution:
We have,
Maximise Z = 34x + 45y
Subject to the constraints :
x + y ≤ 300
2x + 3y ≤ 70
x ≥ 0, y ≥ 0
Converting the given inequalities into equations, we obtain the following equations :
x + y = 300
2x + 3y = 70
Then, x + y = 300


Plotting these points on the graph, we get the shaded feasible region i.e., OCDO.

Clearly, the maximum value of Z is 1190 at (35, 0).

Question 22.
Find the value of x such that the points A (3, 2, 1), B (4, x, 5), C(4, 2, – 2) and D (6, 5, – 1) are coplanar. [4]
Solution:
Given, the points A(3, 2, 1), B (4, x, 5), C (4, 2, – 2) and D (6, 5, -1).

Question 23.
Find the general solution of the differential equation:
y dx – (x + 2y²) dy = 0 [4]
Solution:
We have,

Section – D

Question 28.
AB is the diameter of a circle and C is any point on the circle. Show that the area of triangle ABC is maximum, when it is an isosceles triangle. [6]
Solution:
Let r be the radius of the circle then,
AB = 2 r ( AB is diameter)

Let, BC = x units
We know that angle subtended by diameter in a circle is right angle
∴ ∠C = 90°

Question 29.
If A = , find A^-1. Hence using A^-1 solve the system of equations 2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3.
Solution:
We have,

CBSE Previous Year Question Papers Class 12 Maths 2017 Outside Delhi Set III

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Section – B

Question 12.
The volume of a sphere is increasing at the rate of 8 cm³/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12 cm. [2]
Solution:
We have,

Section – C

Question 20.
Solve the following linear programming problem graphically: [4]
Maximise Z = 7x + 10y
subject to the constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
x ≥ 10
x ≥ 0, y ≥ 0
Solution:
We have,
Maximise Z = 7x + 10y
Subject to the constraints :
4x + 6y ≤ 240
6x + 3y ≤ 240
x ≥ 10
x ≥ 0, y ≥ 0
Converting the given inequalities into equations, we obtain the following equations :
4x + 6y = 240
6x + 3y = 240
x = 10
Then, 4x + 6y = 240


Plotting these points on the graph, we get the shaded feasible region i.e., DEFGD.

Question 21.
Find: [4]
Solution:

Question 22.
If then express in the form of , where is parallel to is perpendicular to . [4]
Solution:
We have,

Question 23.
Find the general solution of the differential equation – y = sin x. [4]
Solution:
We have,

Section – D

Question 29.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. [6]
Solution:
Let ABCD be a window of rectangular form surmounted by a semicircle with diameter AB.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Maths 2017 Outside Delhi appeared first on Learn CBSE.

Important Questions for Class 12 Chemistry Chapter 8 The d- and f-Block Elements Class 12 Important Questions

The d- and f-Block Elements Class 12 Important Questions Very Short Answer Type

Question 1.
What is meant by ‘lanthanoid contraction’? (Delhi 2011)
Answer:
The steady decrease in the ionic radius from La³⁺ to Lu³⁺ is termed as lanthanoid contraction.

Question 2.
Why do transition elements show variable oxidation states? (Comptt. Delhi 2014)
Answer:
The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons, i.e. (ns) and (n -1) d electrons have approximate equal energies.

Question 3.
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. (Delhi 2017)
Answer:
Permanganate ion, i.e., MnO₄^– with oxidation number +7.

Question 4.
What happens when (NH₄)₂Cr₂O₇ is heated? (Delhi 2017)
Answer:

Question 5.
Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number. (Delhi 2017)
Answer:
Cr₂O₇^2- (dichromate ion) in which oxidation state of Cr is +6 which equal to its group number 6.

The d- and f-Block Elements Class 12 Important Questions Short Answer Type -I [SA – I]

Question 6.
Explain the following observations :
(i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions. (Delhi 2010)
Answer:
(i) From titanium to copper the atomic size of elements decreases and mass increases as a result of which density increases.
(ii) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d- orbitals and variable oxidation states.

Question 7.
Explain the following observations :
(i) Transition elements generally form coloured compounds.
(ii) Zinc is not regarded as a transition element. (Delhi 2010)
Answer:
(i) Because of presence of unpaired d electrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours. This is how transition elements form coloured compounds.
(ii) Zinc in its common oxidation state of +2 has completely filled d-orbitals. Hence considered as non-transition element.

Question 8.
Assign reasons for the following :
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibit greater range of oxidation states than lanthanoids. (Delhi 2011)
Answer:
(i) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu
(ii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 9.
Assign reasons for each of the following :
(i) Transition metals generally form coloured compounds.
(ii) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements. (Delhi 2011)
Answer:
(i) Because presence of unpaired d electrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours.
(ii) Manganese exhibits highest oxidation of +7 among 3d series of transition elements because all the oxidation states are exhibited from +2 to +7 by Mn and no other element of this series shows this highest state of oxidation.

Question 10.
How would you account for the following :
(i) Cr²⁺ is reducing in nature while with the same d-orbital configuration (d⁴) Mn³⁺ is an oxidising agent.
(ii) In a transition series of metals, the metal
which exhibits the greatest number of oxidation states occurs in the middle of the series. (All India 2011)
Answer:
(i) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half filled t₂g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.
(ii) Due to presence of more unpaired electrons and use of all 4s and 3d electrons in the middle of series.

Question 11.
Complete the following chemical equations : (All India 2011)
(i) MnO₄ (aq) + S₂O₃^2- (aq) + H₂O (1) →
(ii) Cr₂O₇^2- (aq) + Fe₂₊ (aq) + H₊ (aq) →
Answer:

Question 12.
State reasons for the following :
(i) Cu (I) ion is not stable in an aqueous solution.
(ii) Unlike Cr³⁺, Mn²⁺, Fe³⁺ and the subsequent other M²⁺ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species. (All India 2011)
Answer:
(i) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu
(ii) Because high enthalpies of atomisation of 4d and 5d series and high ionization enthalpies, the M.P. and B.P. of heavier transition elements are greater than those of first transition series which is due to stronger intermetallic bonding. Hence 4d and 5d series metals generally do not form stable cationic species.

Question 13.
Explain giving a suitable reason for each of the following :
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series. (All India 2011)
Answer:
(i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d- orbitals and variable oxidation states.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series as these have their electrons of outer most shell at greater distance from the nucleus, as compared to atoms of 3d transition metals.

Question 14.
Explain giving reasons :
(i) Transition metals and their compounds generally exhibit a paramagnetic behaviour.
(ii) The chemistry of actinoids is not so smooth as that of lanthanoids. (All India 2011)
Answer:
(i) Because of presence of unpaired electrons in their d-subshell in atomic and ionic state.
(ii) Lanthanoids show limited number of oxidation state, viz. +2, +3 and +4 (out of which +3 is most common) because of large energy gap between 4f and 5d subshells. Actinoids also show stable +3 oxidation state but show a number of oxidation states i.e. +4, +5 and + 6, +7 due to small energy difference between 5f, 6d and 7s subshells.

Question 15.
Complete the following chemical equations : (Delhi 2012)

Answer:

Question 16.
Explain the following :
(a) The enthalpies of atomization of transition metals are quite high.
(b) The transition metals and many of their compounds act as good catalysts. (Comptt. Delhi 2012)
Answer:
(a) In transition elements, there are large number of unpaired electrons in their atoms, thus they have a stronger inter atomic interaction and thereby stronger bonding between the atoms. Due to this they have high enthalpies of atomization.
(b) Because of the availability of d-orbitals, they can easily form intermediate products which are activated. The sizes of transition metal atoms and ions are also favourable for transition complex formation with the reactants.

Question 17.
Complete the following chemical reaction equations : (All India 2012)

Answer:

Question 18.
(a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequency and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
SC³⁺, V³⁺, Ti⁴⁺, Mn²⁺.
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (Delhi 2013)
Answer:
(a) Copper exhibits + 1 oxidation state more frequently i.e., Cu⁺¹ because of its electronic configuration 3d¹⁰4s¹. It can easily lose 4s¹ electron to give stable 3d¹⁰ configuration.

(b) SC³⁺ = 4S⁰ 3d³⁺ = no unpaired electron
V³⁺ = 3d² 4s⁰ = 2 unpaired electron
Ti⁴⁺ = 3d⁰ 4s⁰ = no unpaired electron
Mn²⁺ = 3d⁵ 4s⁰ = 5 unpaired electron
Thus V³⁺ and Mn²⁺ are coloured in their aqueous solution due to presence of unpaired electron.

Question 19.
What is Lanthanoid contraction? What are its two consequences? (Comptt. Delhi 2013)
Answer:
Lanthanoid contraction: The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.

Consequences :
(i) Similarity in properties: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.

Question 20.
Assign a reason for each of the following observations:
(i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points.
(ii) The ionization enthalpies (first and second) in the first series of the transition elements are found to vary irregularly. (Comptt. Delhi 2014)
Answer:
(i) Because of stronger metallic bonding and high enthalpies of atomization.
(ii) Due to irregulaties in the electronic configuration there is irregularities in the enthalpies of atomisation. Hence there is irregular variation in I.E.

Question 21.
What is lanthanoid contraction? Write a conse-quence of lanthanoid contraction. (Comptt. Delhi 2014)
Answer:
Lanthanoid contraction: The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.
Consequences :
(i) Similarity in properties: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.

Question 22.
How would you account for the following?
(i) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(ii) The oxidising power of the following three oxoions in the series follows the order: (Comptt. Delhi 2014)

Answer:
(i) The highest oxidation state of a metal is exhibited in its oxide or fluoride due to its high electronegativity, low ionisation energy and small size.
(ii) La³⁺ (Z = 57) and Lu³⁺ (Z = 71) do not show any colour in solutions.

Question 23.
Assign reason for each of the following :
(i) Transition elements exhibit paramagnetic behaviour.
(ii) Co²⁺. is easily oxidised in the presence of a strong ligand. (Comptt. Delhi 2014)
Answer:
(i) Because of presence of unpaired electrons in their d-subshell in atomic and ionic state.
(ii) Co²⁺ ion is easily oxidised to Co³⁺ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d²sp³).

Question 24.
Describe the general trends in the following properties of the first series (3d) of the transition elements :
(i) Number of oxidation states exhibited
(ii) Formation of oxometal ions (Comptt. Delhi 2014)
Answer:
(i) The number of oxidation states increases upto middle of series i.e. unto +7 and then decreases.
(ii) Oxometal ions are polyatomic ions with oxygen.
Example : VO₂⁺, VO⁺², T₁O²⁺

Question 25.
Assign reasons for the following :
(i) Copper(I) ion is not known to exist in aqueous solutions.
(it) Both O₂ and F₂ stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine. (Comptt. All India 2014)
Answer:
(i) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu
(ii) The ability of O₂ to stabilize higher oxidation states exceeds that of fluorine because oxygen can form multiple bonds with metals.

Question 26.
Assign reasons for the following :
(i) Transition metals and many of their compounds act as good catalysts.
(ii) Transition metals generally form coloured compounds. (Comptt. All India 2014)
Answer:
(i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Because of presence of unpaired electrons in d-orbital, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours.

Question 27.
What are the transition elements? Write two characteristics of the transition elements. (Delhi 2015)
Answer:
Elements which have partially filled d-orbital in its ground states or any one of its oxidation states are called transition elements.

• They show variable oxidation states.
• They form coloured ions.
• They form complex compounds.

Question 28.
Write one similarity and one difference between the chemistry of lanthanoids and that of actinoids. (All India 2015)
Answer:
Similarity : Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference: Actinoids show wide range of oxidation states but lanthanoids do not.

Question 29.
What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution. (Comptt. Delhi 2015)
Answer:
Disproportionation: In a disproportionation reaction an element undergoes self-oxidation as well as self-reduction forming two different compounds.

Question 30.
Suggest reasons for the following features of transition metal chemistry :
(i) The transition metals and their compounds are usually paramagnetic.
(ii) The transition metals exhibit variable oxidation states. (Comptt. Delhi 2015)
Answer:
(i) The transition metals and their compounds are usually paramagnetic because of the presence of unpaired electrons in their d-orbitals.
(ii) The transition metals exhibit variable oxidation states because of very close energies of incompletely filled (n – l)d orbitals and ns orbitals due to which both can participate in bonding.

Question 31.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid? Write the ionic equations for the reactions. (Comptt. All India 2015)
Answer:
Potassium Permangante (KMnO₄) is prepared from pyrolusite ore (MnO₂). The ore (MnO₂) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidising agent like KNO₃ to give dark green potassium manganate (K₂MnO₄). K₂MnO₄ disproportionates in a neutral or acidic solution to give potassium permanganate.

Question 32.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with
(i) iodine (ii) H₂S. (Comptt. All India 2015)
Answer:
Potassium dichromate (K₂Cr₂O₇) acts as a strong oxidising agent in acidic medium using H₂SO₄.
K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]
Ionic reactions :

Question 33.
When chromite ore FeCr₂O₄ is fused with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KC1 forms an orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C). (Delhi 2016)
Answer:
The chromite ore FeCr₂O₄ on fusion with NaOH in presence of air, forms a yellow coloured compound (A) i.e. Sodium chromate.

Sodium dichromate (B) on reaction with KCl forms orange coloured compound Potassium dichromate (C).
Na₂Cr₂O₇ + 2KCl → 2NaCl + K₂Cr₂O₇ (C)
(i) Thus (A) → Sodium chromate Na₂CrO₄
(A) → Sodium dichromate Na₂Cr₂O₇
(B) → Potassium dichromate K₂Cr₂O₇
(ii) (C) is used as a strong oxidising agent in acidic medium in volumetric analysis.

Question 34.
Complete the following chemical equations: (Delhi 2016)

Answer:

Question 35.
Give reasons :
(i) Transition metals show variable oxidation, states.
(ii) Actinoids show wide range of oxidation states. (Comptt. Delhi 2016)
Answer:
(i) Because in transitional elements ns and (n – 1 )d electrons have approximate equal energies.
(ii) Due to comparable energies of 5f, 6d and 7s orbitals.

Question 36.
Give reasons :
(i) Zn is not regarded as a transition element.
(ii) Cr²⁺ is a strong reducing agent. (Comptt All India 2016)
Answer:
(i) Zinc Atomic no. 30 have EC 3d¹⁰, 4s². It has completely filled ‘d’ orbitals.
(ii) Cr⁺² (3d⁴) after loosing one electron forms Cr⁺³ (d³). It has configuration. Hence more stable. Hence Cr⁺² acts as reducing agent.
Or
Cr⁺³ is more stable than Cr⁺², therefore Cr⁺² looses one electron. Hence acting as strong reducing agent.

Question 37.
What happens when (NH₄)₂Cr₂O₇ is heated? (Delhi 2017)
Answer:

Question 38.
Explain the following observations :
(i) Copper atom has completely filled d orbitals (3d¹⁰) in its ground state, yet it is regarded as a transition element.
(ii) Cr²⁺ is a stronger reducing agent than Fe²⁺ in aqueous solutions. (Comptt. All India 2017)
Answer:
(i) Copper atom has completely filled d orbitals (3d¹⁰) in its ground state, yet it is regarded as a transition element due to incompletely filled d-orbital in its ionic states i.e. Cu²⁺ (3d⁹).
(ii) The highest oxidation state for Cr is +6, therefore it can loose 3 more electrons, whereas Fe needs to loose only 1 electron to achieve its highest oxidation state of +3. Thus, Cr³⁺ is more reducing than Fe²⁺.

Question 39.
Explain the following observations :
(a) Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state, yet it is regarded as a transition element.
(b) E⁰ value for Mn³⁺ /Mn²⁺ couple is much more positive than Cr³⁺/Cr²⁺. (Comptt. All India 2017)
Answer:
(a) Because silver has incomplete d-orbital (4d⁹) in its +2 oxidation state, hence it is a transition element.
(b) The large positive E° value for Mn³⁺/Mn²⁺ shows that Mn²⁺ is much more stable than Mn⁺³ due to stable half filled configuration (3d⁵). Therefore the 3^rd ionisation energy of Mn will be very high and Mn³⁺ is unstable and can be easily reduced to Mn²⁺. E° value for Cr³⁺ | Cr²⁺ is positive but small i.e. Cr³⁺ can also be reduced to Cr²⁺ but less easily. Thus Cr³⁺ is more stable than Mn³⁺.

Question 40.
Explain the following observations :
(i) Zn²⁺ salts are colourless.
(ii) Copper has exceptionally positive value. (Comptt. All India 2017)
Answer:
(i) Zn²⁺ salts are colourless due to absence of unpaired electrons in its ground state and ionic state i.e. Zn²⁺ = [Ar] 3d¹⁰4s⁰4p⁰
(ii) The E⁰_M²⁺/M for any metal is related to the sum of the enthalpy changes taking place in the following steps :
M(g) + Δ_aH → M(g) (Δ_aH = enthalpy of atomization)
M(g) + Δ_iH → M²⁺(g) (Δ_iH = ionization of atomization)
M²⁺(g) + aq → M²⁺(g) + Δ_hydH (Δ_hydH = hydration atomization)
Copper has high enthalpy of atomization (i.e. energy absorbed and low enthalpy of hydration (i.e. energy released). Hence E⁰_M²⁺/M for copper is positive. The high energy required to transform Cu(s) to Cu₂₊(aq) is not balanced by its hydration enthalpy.

The d- and f-Block Elements Class 12 Important Questions Short Answer Type -II [SA – II]

Question 41.
How would you account for the following :
(i) Many of the transition elements and their compounds can act as good catalysts.
(ii) The metallic radii of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second series.
(iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids. (All India 2009)
Answer:
(i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Due to lanthanoid contraction in second series after lanthanum, the atomic radii of elements of second and third series become almost same and hence show similarities in properties.
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 42.
How would you account for the following?
(i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
(ii) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ couple or Fe²⁺/Fe²⁺ couple.
(iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride. (Delhi 2010)
Answer:
(i) Due to lanthanoid contraction in second series after lanthanum, the atomic radii of elements of second and third series become almost same and hence show similarities in properties.
(ii) The electronic configuration of Mn²⁺ ion is more symmetrical as compared to that of Cr²⁺ ion. So 3^rd ionisation potential of Mn²⁺ is much higher. As a result E° value of Mn³⁺/ Mn²⁺ couple is much more positive than for Cr³⁺/Cr²⁺ couple.
(iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride due to its high electronegativity.

Question 43.
Complete the following chemical equations : (Delhi 2011)

Answer:

Question 44.
How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
(iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical. (Delhi 2012)
Answer:
(i) The transition metals form a large number of interstitial compounds in which small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty spaces in the crystal lattices of transition metals.
(ii) Because of lanthanoid contraction.
(iii) This is due to comparable energies of 5f, 6d, 7s orbitals in actinoids.

Question 45.
How would you account for the following?
(i) With the same d-orbital configuration (d⁴) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidizing agent.
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions. (Delhi 2012)
Answer:
(i) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half filled t₂g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.
(ii) Due to comparable energies of 5f 6d and 7s orbitals of actinoids, these show larger number of oxidation states than corresponding members of lanthanoids.
(iii) Due to the presence of unpaired electrons in d-orbital, transition metal exhibits colours in aqueous solution or due to d-d transition.

Question 46.
Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
(ii) There occurs much more frequent metal- metal bonding in compounds of heavy transition metals (i.e. 3^rd series).
(iii) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation. (Delhi 2012)
Answer:
(i) In transition elements, there are large number of unpaired electrons in their atoms, thus they have a stronger inter atomic interaction and thereby stronger bonding between the atoms. Due to this they have high enthalpies of atomization.
(ii) Because of high enthalpy of atomisation of 3^rd series, there occurs much more frequent metal-metal bonding in compounds of heavy transition metals.
(iii) The 3d orbital in Mn²⁺ is half-filled and is more stable compared to Fe²⁺ has 6 electrons in the 3d orbital. Mn²⁺ prefer to lose an electron or get oxidised whereas Fe²⁺ will readily loose one electron or get oxidised. Therefore, Mn²⁺ is much more resistant than Fe²⁺ towards oxidation.

Question 47.
How would you account for the following :
(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, + 2 and + 4 ions are also obtained.
(ii) The E°_M²⁺/M for copper is positive (0.34 V).
Copper is the only metal in the first series of transition elements showing this behaviour.
(iii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series. (All India 2012)
Answer:
(i) Among lanthanoids, Ln (III) compounds are predominant. However +2 and +4 ions in solution or in solid compounds are also obtained. This is because they have empty, half-filled and completely filled 4/ sub-shell respectively which show extra stability.
(ii) E°_M²⁺/M for any metal is related to the sum of enthalpy of atomisation, ionization and hydration enthalpy. Copper has high enthalpy of atomisation and low enthalpy of hydration. Hence E°_Cu²⁺/Cu is positive. The high energy required to transform Cu_(s) to Cu²⁺_(aq) is not balanced by its hydration enthalpy.
(iii) Because of lanthanoid contraction the metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series.

Question 48.
Explain the following observations :
(i) Many of the transition elements are known to form interstitial compounds
(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
(iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. (All India 2012)
Answer:
(i) The transition metals form a large number of interstitial compounds in which small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty spaces in the crystal lattices of transition metals.
(ii) From titanium to copper the atomic size of elements decreases and mass increases as a result of which density increases.
(iii) This is due to comparable energies of 5f 6d and 7s orbitals in actinoids.

Question 49.
Explain each of the following observations :
(i) With the same d-orbital configuration (d⁴), Cr²⁺ is a reducing agent while Mn³⁺ is an oxidising agent.
(ii) Actinoids exhibit a much larger number of oxidation states than the lanthanoids.
(iii) There is hardly any increase in atomic size with increasing atomic numbers in a series of transition metals. (All India 2012)
Answer:
(i) Cr⁺² is reducing agent as its configuration changes from d⁴ to d³ which is having half filled t₂g level whereas Mn⁺³ to Mn⁺² results in stable half filled d⁵ configuration hence it is oxidising.
(ii) Due to comparable energies of 5f, 6d and 7s orbitals actinoids exhibit a much larger number of oxidation states than the lanthanoids.
(iii) Along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the penultimate d-subshell increases the screening effect which counter balances the effect of increased nuclear charge. Thus, atomic radius does not change much.

Question 50.
Give reasons :
(a) There is a gradual decrease in the size of atoms with increasing atomic number in the series of lanthanoids.
(b) Sc (21) is a transition element but Ca (20) is not.
(c) The Fe²⁺ is much more easily oxidised to Fe²⁺ than Mn²⁺ to Mn³⁺. (Comptt. Delhi 2012)
Answer:
(a) This is due to lanthanoid contraction.
(b) Sc (21) is regarded as a transition element due to the presence of incomplete d- subshell (3d¹4s²) but Ca (20) does not have any d-subshell.
(c) Fe⁺²[Ar] 3d⁶ can be easily oxidised to Fe⁺³ [Ar] 3d⁵, which is half filled and is more stable but Mn⁺² is d⁵ in configuration which is again half filled configuration and Mn⁺³ is d⁴ in configuration.

Question 51.
(a) What is lanthanoid contraction? Mention its main consequences.
(b) Write the balanced ionic equation for the reaction between ferrous sulphate and acidified potassium permanganate solution. (Comptt. Delhi 2012)
Answer:
(a) Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.

Consequences :
(i) Similarity in properties : Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.

(b) Balanced ionic equation :
MnO₄^– + 5Fe⁺² + 8H⁺ → Mn⁺² + 5Fe⁺³ + 4H₂O

Question 52.
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst. (Delhi 2013)
Answer:
(i) Because the energy difference between (n-1) d-orbitals and ns-orbitals is very less. Since there is very little energy difference between these orbitals, both energy levels can be used for bond formation. Thus transition elements exhibit variable oxidation states.
(ii) Zr and Hf have almost identical radii due to lanthanoid contraction which is due to weak shielding of d-electrons.
(iii) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.

Question 53.
Complete the following chemical equations: (Delhi 2013)

Answer:

Question 54.
Write balanced chemical equations for the two reactions showing oxidizing nature of potassium permanganate. (Comptt. Delhi 2013)
Answer:
Reactions showing oxidising nature of KMnO₄
(i) 2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄+ 2MnSO₄ + 2H₂SO₄
(ii) 2KMnO₄ + 3H₂SO₄ + 5H₂O → K₂SO₄ + 2MnSO₄ + 3H₂O + 5S

Question 55.
Give reasons for the following :
(i) Transition metals exhibit a wide range of oxidation states.
(ii) Cobalt(II) is very stable in aqueous solutions but gets easily oxidised in the presence of strong ligands.
(iii) Actinoids exhibit a greater range of oxidation states than lanthanoids. (Comptt. All India 2014)
Answer:
(i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals as ns, and (n – 1) d electrons have very less energy difference.
(ii) Co²⁺ ion is easily oxidised to Co³⁺ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d²sp³).
(iii) Actinoids because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 56.
Assign reasons for the following :
(i) Cu(I) ion is not known to exist in aqueous solutions.
(ii) Transition metals are much harder than the alkali metals.
(iii) From element to element actinoid contraction is greater than the lanthanoid contraction. (Comptt. All India 2014)
Answer:
(i) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu
(ii) In transitional elements, in addition to metallic bonding there is extra covalent bonding due to presence of unpaired electrons in their ‘d’ orbitals, hence they are much harder.
(iii) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are buried deep inside the atom.

Question 57.
(a) How would you account for the following :
(i) Actinoid contraction is greater than lanthanoid contraction.
(ii) Transition metals form coloured compounds.
Complete the following equation :
2MnO₄^– + 6H⁺ + 5NO₂^– → (Delhi 2015)
Answer:
(a) (i) Actinoid contraction is greater than lanthanoid because 5f electrons (in actinoids) have a poorer shielding effect than 4f electrons (in lanthanoids).
(ii) Transition metals from coloured compounds because they have unpaired electrons in the d-orbital.
(b) 2MnO₄^– + 6H⁺+ 5NO₂ → 2Mn⁺² + 5NO₃^– + 3H₂O

Question 58.
(a) Account for the following :
(i) Cu+ is unstable in an aqueous solution.
(ii) Transition metals form complex compounds.
(b) Complete the following equation :
Cr₂O₇^2- + 8H⁺ + 3NO^–₂ → (All India 2015)
Answer:
(a) (i) Cu⁺ is unstable in an aqueous solution because Cu+ undergoes disproportionation reaction as follows:
2Cu⁺ →Cu²⁺+ Cu
(ii) Transition metals form complex compounds due to small size of metal, higher nuclear (ionic) charge and availability of vacant or incompletely filled d-orbitals.
(b) Cr₂O₇^2- + 8H⁺ + 3NO^–₂ → 2Cr³⁺ + 3NO^–₃ + 4H₂O

Question 59.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction? (Comptt. Delhi 2015)
Answer:
Lanthanoid contraction: The overall decrease in atomic and ionic radii with increasing atomic number from La to Lu due to imperfect shielding of 4f-orbital is known as lanthanoid contraction. Cause : As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons added in f-subshell causes imperfect shielding which is unable to counterbalance the effect of the increased nuclear charge. Hence the contraction in size occurs.

Consequences :
(i) Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoids becomes very difficult.
(ii) There is similarity in size of elements belonging to same group of second and third transition series.
Example: Zr and Hf are known as chemical twins due to their almost identical radii.

Question 60.
What is meant by ‘’disproportionation’? Give one example of disproportionation reaction in aqueous solutions. (Comptt. All India 2015)
Answer:
Disproportionation: When in a reaction, the oxidation of an element in a compound increases in one of the products and decreases in the other product, it is said to undergo disproportionation of oxidation state. In other words, the reaction in which an element undergoes self-oxidation and self-reduction simultaneously.
Example : In acidic solution Mn (VI) in
changes to Mn (VII) in the product and to Mn (IV) in the product MnO₂.

Question 61.
Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations. (Delhi 2016)
Answer:
(i) Because oxygen stabilizes the highest oxidation state (+7 of Mn) even more than fluorine i.e., +4 since oxygen has the ability to form multiple bonds with metal atoms.
(ii) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons.
(iii) This happens because the energy difference between 5f, 6d and 7s subshells of the actinoids is very small and hence electrons can be accomodated in any of them.

Question 62.
In the 3d series (Sc = 21 to Zn = 30) :
(i) Which element shows maximum number of oxidation states?
(ii) Which element shows only +3 oxidation state?
(iii) Which element has the lowest enthalpy of atomization? (Comptt. Delhi 2016)
Answer:
(i) Mn
(ii) Sc
(iii) Zn

Question 63.
Describe the preparation of potassium permangnate. How does the acidified permanganate solution react with oxalic acid? (Comptt. All India 2016)
Answer:
Pyrollusite ore is digested in KOH in the presence of oxygen
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄^2- + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O
Reaction with oxalic acid in Acidic medium
5C₂O₄^2- + 2MnO₄^– + 16H⁺ → 2Mn⁺² + 8H₂O + 10CO₂

Question 64.
Define lanthanoid contraction. Write its two consequences. (Comptt. All India 2016)
Answer:
Steady and regular decrease in atomic or ionic size with increasing atomic no. in Lanthanoid series.
Consequences :

• Due to lanthanoid contraction elements of 4d and 5d have similar size, hence occurs together.
• Basic character of trivalent hydroxide decreases.

Question 65.
A mixed oxide of iron and chromium is fused with sodium carbonate in free access of air to form a yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B), which is a strong oxidizing agent. Identify compound (A) and (B). Write chemical reactions involved. (Comptt. All India 2017)
Answer:
The mixed oxide of iron and chromium is chromite or chrome ion i.e. FeO.Cr₂O₃

Question 66.
(a) Give reasons for the following:
(i) Compounds of transition elements are generally coloured.
(ii) MnO is basic while Mn₂O₇ is acidic.
(iii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. (Comptt. All India 2017)
Answer:
(a) (i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d- orbitals and variable oxidation states. The colour of transition metal ions is due to d-d transition. When electrons jump from one orbital to another in their partially filled d-orbitals, another light is emitted due to which the compounds of transition elements seem to be coloured.
(ii) MnO is basic while Mn₂O₇ is acidic because the basic nature decreases as the oxidation state or number of oxygen atoms increases i.e. MnO (+4) and Mn₂O₇ (+7)
(b) Divalent ion with atomic number 26 is Fe²⁺

The d- and f-Block Elements Class 12 Important Questions Long Answer Type [LA]

Question 67.
(a) Complete the following chemical reaction equations :

(b) Explain the following observations about the transition/inner transition elements :
(i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29).
(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3rd series).
(iii) The members in the actinoid series exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series. (Delhi & All India 2009)
Answer:
(a)

(b) (i) As we move along a transition series from left to right, the atomic radii decrease due to increase in nuclear charge. Hence, the atomic volume decreases. At the same time, atomic mass increases from Ti to Cu, therefore density increases.
(ii) It is due to their low ionization energies and variable oxidation state.
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.
Or
Because 5f electrons are less burned than 4 ‘f’ electrons. Hence 5f electrons are also taking part in chemical bonding. Hence actinoids are showing large number of oxidation states.

Question 68.
(a) Complete the following chemical equations for reactions :

(b) Give an explanation for each of the following observations :
(i) The gradual decrease ‘n’ size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction).
(ii) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
(iii) With the same d-orbital configuration d⁴, Cr²⁺ ion is a reducing agent but Mn³⁺ ion is an oxidising agent. (Delhi 2009)
Answer:
(a)

(b) (i) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are buried deep inside the atom.
(ii) Due to presence of more unpaired electrons and use of all 4s and 3d electrons in the middle of series.
(iii) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half filled t₂g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.

Question 69.
(a) Complete the following chemical reaction equations :
(i) Fe²⁺(aq) + + H⁺(aq) →
(ii) + I^–(aq) +H⁺(aq) →
(b) Explain the following observations :
(i) Transition elements are known to form many interstitial compounds.
(ii) With the same d⁴ d-orbital configuration Cr²⁺ ion is reducing while Mn³⁺ ion is oxidizing.
(iii) The enthalpies of atomization of the transition elements are quite high. (Delhi 2009)
Answer:
(a) Fe²⁺(aq) + + H⁺(aq) → Mn²⁺ + 4H₂O + 5Fe³⁺
(ii) + I^–(aq) +H⁺(aq) → 2Cr³⁺ + 7H₂O + 3I₂

(b) (i) The transition metals form a large number of interstitial compounds in which small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty spaces in the crystal lattices of transition metals.

(ii) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half filled t₂g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.

(iii) Enthalpy of atomization is the amount of heat required to break the metal lattice to get free atoms. As transition metals contain a large number of unpaired electrons, they have strong interatomic attractions (metallic bonds). Hence they have high enthalpies of atomization.

Question 70.
(a) Complete the following chemical reaction equations :
(i) (aq) + H₂ S(g) + H⁺(aq) →
(ii) MnO₂(s) + KOH(aq) + O₂ →
(b) Explain the following observations :
(i) Transition metals form compounds which are usually coloured.
(ii) Transition metals exhibit variable oxidation states.
(iii) The actinoids exhibit a greater range of oxidation states than the lanthanoids. (Delhi 2009)
Answer:
(a)

(b) (i) Colour is due to the presence of unpaired electrons in their d-subshells.
(ii) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals. In transitional elements ns and (n – 1)d electrons have approximate equal energies hence in addition to ns electrons, (n – 1)d electrons are also taking part in chemical bonding.
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 71.
(a) Complete the following chemical reaction equations :

(b) Explain the following observations :
(i) In general the atomic radii of transition elements decrease with atomic number in a given series.
(ii) The E°_M²⁺/M, for copper is positive (+ 0.34 V). It is the only metal in the first series of transition elements showing this type of behaviour.
(iii) The E° value for Mn³⁺ | Mn²⁺ couple is much more positive than for Cr³⁺ | Cr²⁺ or Fe³⁺ | Fe²⁺ couple. (Delhi 2009)
Answer:
(a) (i) (aq) + 6I^–(aq) + 14H⁺(aq) → 2Cr³⁺ + 7H₂O + 3I₂
(ii) 5Fe²⁺ + + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

(b) (i) Because of increase in effective nuclear charge and weak shielding effect of d-electrons, the atomic radii decreases.

(ii) The E°_M²⁺/M for any metal is related to the sum of the enthalpy changes taking place in the following steps :
M(g) + Δ_aH → M(g) (Δ_aH = enthalpy of atomization)
M(g) + Δ_iH → M²⁺(g) (Δ_iH = ionization enthalpy)
M²⁺(g) + aq → M²⁺(aq) + Δ_hydH (Δ_hydH = hydration enthalpy)
Copper has high enthalpy of atomization (i.e. energy absorbed and low enthalpy of hydration (i.e. energy released). Hence E°_M²⁺/M for copper is positive. The high energy required to transform Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

(iii) The large positive E° value for Mn³⁺ | Mn²⁺ shows that Mn²⁺ is much more stable than Mn³⁺ due to stable half filled configuration (3d⁵). Therefore the 3^rd ionization energy of Mn will be very high and Mn³⁺ is unstable and can be easily reduced to Mn²⁺. E° value for Fe³⁺ | Fe²⁺ is positive but small i.e. Fe³⁺ can also be reduced to Fe²⁺ but less easily. Thus Fe³⁺ is more stable than Mn³⁺.

Question 72.
(a) What is meant by the term lanthanoid contraction? What is it due to and what consequences does it have on the chemistry of elements following lanthanoids in the periodic table?
(b) Explain the following observations :
(i) Cu⁺ ion is unstable in aqueous solutions.
(ii) Although Co²⁺ ion appears to be stable, it is easily oxidised to Co³⁺ ion in the presence of a strong ligand.
(in) The E°_Mn²⁺/Mn value for manganese is much more than expected from the trend for other elements in the series. (Delhi 2009)
Answer:
(a) Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction.
Cause: As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons added in f-subshell causes imperfect shielding which is unable to counterbalance the effect of the increased nuclear charge. Hence the contraction in size occurs.
Consequences :
(i) Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoids becomes very difficult.
(ii) There is similarity in size of elements belonging to same group of second and third transition series.
Example: Zr and Hf are known as chemical twins due to their similar radii.

(b) (i) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu
(ii) Co²⁺ ion is easily oxidised to Co³⁺ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d²sp³). Co³⁺ can accomodate more no. of electrons from the ligand and offer better stability to the resulting complex. Hence Co²⁺ oxidises to Co³⁺.
(iii) The value of E° for Mn is more negative than expected from the general trend due to greater stability of half filled d-subshell (d⁵) in Mn²⁺.

Question 73.
(a) Complete the following chemical equations :
(i) (aq) + H₂S (g) + H⁺ (aq) →
(ii) Cu²⁺ (aq) + I^–(aq) →
(b) How would you account for the following :
(i) The oxidising power of oxoanions are in the order
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺. (All India 2010)
Answer:
(a) (i) latex]\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}[/latex] + 3H₂S + 8H⁺ → 2Cr³⁺ + 7H₂O + 3S
(ii) 2Cu²⁺ (aq) + 4I^– (aq) → Cu₂I₂ + I₂

(b) (i) The ions in which the central metal atom is present in the highest oxidation state will have the highest oxidising power. In V is present in the +5 oxidation state, while in Cr₂O₇^2- ion, Cr is present in the +6 oxidation state. Similarly in the MnO₄^– ion, Mn is present in the +7 oxidation state. Thus as the oxidation state of the central metal atom increases in the order V < Cr < Mn, therefore the oxidising power of the oxoanions increases in the order
VO₂⁺ < Cr₂O₇^2- < MnO₄^–
(ii) Because third electron has to be removed from stable half filled 3d-orbitals (Z = 25 has 3d⁵ 4s²).
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺ because E°_{Cr³⁺/Cr²⁺} is negative (-0.41 V)
whereas E°_{Fe³⁺/Fe²⁺} is positive (0.77 V). Thus Cr²⁺ is easily oxidised to Cr³⁺ but Fe²⁺ cannot be easily oxidized to Fe³⁺.

Question 74.
(a) Complete the following chemical equations :

(b) Explain the following observations :
(i) La³⁺ (Z = 57) and Lu³⁺ (Z = 71) do not show any colour in solutions.
(ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
(iii) Cu⁺ ion is not known in aqueous solutions. (All India 2010)
Answer:

It has maximum five unpaired electrons, so maximum paramagnetism
(iii) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hyd (hydration enthalpy) for Cu²⁺(aq) is much more negative than that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu⁺ → Cu²⁺ + Cu

Question 75.
(a) Describe the preparation of potassium dichromate from chromite ore. What is the effect of change of pH on dichromate ion?
(b) How is the variability in oxidation states of transition elements different from that of non-transition elements? Illustrate with examples. (Comptt. All India 2012)
Answer:
(a) Potassium dichromate is prepared from chromate by reacting chromite ore with Na₂CO₃
4 FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ .
The yellow solution of sodium chromate is filtered off and acidified with H₂SO₄ to give orange sodium dichromate
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + H₂O + 2Na
Sodium dichromate is then treated with KCl to give potassium dichromate as orange crystals.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution.
2CrO₄^-2 + 2H⁺ → Cr₂O₇-2 + H₂O
Cr₂O₇^-2 + 2OH^–→ 2CrO₄^-2 + H₂O

(b) In transition elements, the successive oxidation state differs by unity, e.g, Mn shows all the oxidation states from +2 to +7. On the other hand, non-transition metals exhibit variable oxidation states which differ by two units, e.g. Pb(II), Pb(IV), Sn(II), Sn(IV) etc.

Question 76.
(a) Describe the preparation of potassium permanganate from pyrolusite ore. What happens when acidified potassium permanganate solution reacts with ferrous sulphate solution? Write balanced chemical equations.
(b) Account for the following :
(i) Mn²⁺compounds are more stable than Fe²⁺ compounds towards oxidation to their +3 state.
(ii) Cr²⁺ is reducing and Mn³⁺ oxidizing when both have d⁴ configuration. (Comptt. All India 2012)
Answer:
(a) KMnO₄ is prepared by fusion of MnO₂ with an alkali metal hydoxide and an oxidising agent like KNO₃. This produces the dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give KMnO₄
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄^-2 + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O
Fe⁺²(green) is converted to Fe⁺³(yellow)
5Fe⁺² + MnO₄^– + 8H⁺ → Mn⁺²+ 4H₂O + 5Fe⁺³

(b) (i) The electronic configuration of Mn⁺² is [Ar] 3d⁵ i.e. all the five d-orbitals are singly occupied. Thus, this configuration is stable and resists further oxidation i.e. loss of further electrons requires high energy. On the other hand, the electronic configuration of Fe⁺² is [Ar]3d⁶. Hence it loses one electron and achieves the stable configuration i.e. oxidation of Fe⁺² to Fe⁺³ is easily achieved.(ii) Cr⁺² is reducing in nature as its configuration changes from d⁴ to d³ which is a stable configuration having half filled t₂g orbitals. On the other hand, Mn⁺³ is oxidising in nature as the configuration changes from d⁴ to d⁵ (a stable configuration having half filled t₂g and e_g orbitals).

Question 77.
(a) Give reasons for the following :
(i) Mn³⁺ is a good oxidising agent.
(ii) E°_M²⁺/M values are not regular for first row transition metals (3d series).
(iii) Although ‘F is more electronegative than ‘O’, the highest Mn fluoride is MnF₄, whereas the highest oxide is Mn₂O₇.
(b) Complete the following equations : (All India 2013)

Answer:
(a) (i) Mn³⁺ has electronic configuration 3d⁴4s⁰.
On reduction it gains one electron to become 3d⁵4s⁰ which is half filled stable configuration. Hence it is a good oxidising agent.
(ii) E°_M²⁺/M values are not regular for first row transition metals due to abnormalities and irregularities in their ionization enthalpies (IE₁ + IE₁) and sublimation enthalpies.
(iii) Because oxygen stabilizes the highest oxidation state even more than fluorine and has ability to form multiple bonds with metal atoms.

Question 78.
(a) Why do transition elements show variable oxidation states?
(i) Name the element showing maximum number of oxidation states among the first series of transition metals from Se (Z = 21) to Zn (Z = 30).
(ii) Name the element which shows only +3 oxidation state.
(b) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals. (All India 2013)
Answer:
(a) Because the energy difference between (n-1) d-orbitals and ns-orbitals is very less. Since there is very little energy difference between these orbitals, both energy levels can be used for bond formation. Thus transition elements exhibit variable oxidation states.
(i) It is Manganese (Mn) which shows oxidation states from +2 to +7
(ii) Scandium (Sc) shows only +3 oxidation state.

(b)Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.

Consequences :
(i) Similarity in properties : Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.
Important alloy: Mischmetal alloy which contains 95% lanthanoids and 5% Fe.

Question 79.
Give reasons :
(i) Zirconium (Z = 40) and Hafnium (Z = 72) have almost similar atomic radii.
(ii) d-block elements exhibit more oxidation states than f-block elements.
(iii) The enthalpies of atomization of the transition metals are high.
(iv) The variation in oxidation states of transition metals is of different type from that of the non-transition metals.
(v) Orange solution of potassium dichromate
turns yellow on adding sodium hydroxide to it. (Comptt. All India 2013)
Answer:
(i) Zirconium and Hafnium have almost similar atomic radii due to the effect of lanthanoid contraction.
(ii) d-block elements exhibit more oxidation states due to small energy gap between ns and (n – 1)d subshell while f – block elements show less oxidation state due to large energy gap between ns and (n -2)f subshell.
(iii) Enthalpy of atomization is the amount of heat required to break the metal lattice to get free atoms. As transition metals contain a large number of impaired electrons, they have strong interatomic attractions (metallic bonds). Hence they have high enthalpies of atomization.
(iv) In transition elements, the successive oxidation state differs by unity, e.g, Mn shows all the oxidation states from +2 to +7. On the other hand non-transition metals exhibit variable oxidation states which differ by two units, e.g. Pb(II), Pb(IV), Sn(II), Sn(IV) etc.
(v) The orange coloured potassium dichromate solution when treated with basic NaOH solution, is converted to chromate which gets a faint colour like yellow.
The reaction is given as:
K₂Cr₂O₇ + 2NaOH → K₂CrO₄ + Na₂CrO₄ + H₂O.
This formation of chromate (CrO₄^–) ion converts the colour of solution to yellow.

Question 80.
(a) Describe the preparation of potassium permanganate from pyrolusite ore. Write balanced chemical equation for one reaction to show the oxidizing nature of potassium permanganate.
(b) What is lanthanoid contraction and what is it due to? Write two consequences of lanthanoid contraction. (Comptt. All India 2013)
Answer:
(a) Preparation of KMnO₄ from pyrolusite ore occurs in two steps :
(i) Conversion of Mn02 into potassium manganate :

(ii) Oxidation of potassium manganate to potassium permanganate : The obtained potassium manganate bubbled with carbon dioxide or chlorine or ozonised oxygen.

(b) Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.

Consequences :
(i) Similarity in properties: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference: Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.

Question 81.
(a) How do you prepare :
(i) K₂MnO₄ from MnO₂?
(ii) Na₂Cr₂O₇ from Na₂CrO₄?
(b) Account for the following :
(i) Mn²⁺ is more stable than Fe²⁺ towards oxidation to +3 state.
(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
(iii) Actinoid elements show wide range of oxidation states. (Delhi 2014)
Answer:
(a) (i) K₂MnO₄ from MnO₂ (Pyrolusite) :
Finely powdered pyrolusite is fused with KOH in the presence of air to give green coloured potassium manganate.

(b) (i) The 3d orbital in Mn²⁺ is half-filled and is more stable compared to Fe²⁺ has 6 electrons in the 3d orbital. Mn²⁺ prefer to lose an electron or get oxidised whereas Fe²⁺ will readily loose one electron or get oxidised. Therefore, Mn²⁺ is much more resistant than Fe²⁺ towards oxidation.
(ii) As there are no unpaired electrons in zinc, it is soft and has low melting point and low enthalpy of atomization.
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 82.
(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive E⁰(M²⁺/M) value and why?
(iii) Out of Cr³⁺ and Mn³⁺, which is a stronger oxidizing agent and why?
(iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation :
MnO₄^– + 8H⁺ + 5e^– → (Delhi 2014)
Answer:
(i) Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
(ii) Copper has positive E⁰(Cu²⁺/Cu) value because of its high enthalpy of atomization and low enthalpy of hydration. The high energy required to oxidise Cu to Cu²⁺ is not balanced by its hydration energy.
(iii) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half filled t₂g orbitals. Therefore Cr²⁺ is reducing agent, it gets oxidized to Cr³⁺. While Mn²⁺ has stable half filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.
(iv) Eutropium is well known to exhibit +2 oxidation state.
(v) MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

Question 83.
(a) Complete the following equations :
(i) Cr₂O₇^2- + 2OH^– →
(ii) MnO₄^– + 4H⁺ + 3e^– →
(b) Account for the following :
(i) Zn is not considered as a transition element.
(ii) Transition metals form a large number of complexes.
(iii) The E⁰ value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ couple. (All India 2014)
Answer:
(a) (i) Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O
(ii) MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^–

(b) (i) Zinc in its common oxidation state of +2 has completely filled d-orbitals. Hence considered as non-transition elements.
(ii) Because of smaller size of their ions, high ionic charge and availability of vacant d-orbitals, transition metals from a large number of complexes.
(iii) The large positive E° value for Mn³⁺ | Mn²⁺ shows that Mn²⁺ is much more stable than Mn³⁺ due to stable half filled configuration (3d⁵). Therefore the 3^rd ionization energy of Mn will be very high and Mn³⁺ is unstable and can be easily reduced to Mn²⁺. E° value for Fe³⁺ | Fe²⁺ is positive but small i.e. Fe³⁺ can also be reduced to Fe²⁺ but less easily. Thus Fe³⁺ is more stable than Mn³⁺.

Question 84.
(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation :
MnO₄^– + 8H⁺ + 5e^– →
(iv) Out of Mn³⁺ and Cr³⁺, which is more paramagnetic and why? (Atomic nos. : Mn = 25, Cr = 24)(All India 2014)
Answer:
(i) Difference between lanthanoids and actinoids :

• The general electronic configuration of lanthanoids is [Xe]⁵⁴ 4f^1-14 5d^0-1 6s² whereas that of actinoids is [Rn⁸⁶ 5f^1-14 6d^0-1 7s². Thus lanthanoids belong to 4/ series whereas actinoids belong to 5/ series.
• Actinoids are radioactive while lanthanoids are not radioactive.
• The ionisation enthalpies of the early actinoids are lower than those of the early lanthanoids. Actinoids show the oxidation state from +3 (most common) to +7, while lanthanoids show the oxidation state from +3 up to + 7.
• The first few members of the lanthanoids series are quite reactive. Therefore they combine with H₂ on gentle heating while the actinoids are highly reactive especially in the finely divided state, therefore they combine with most of the non-metals at moderate temperature.

(ii) Cerium
(iii) MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O
(iv) Mn³⁺ is more paramagnetic due to presence of 4 unpaired electrons than Cr³⁺ having 3 unpaired electrons.

Question 85.
(a) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Cr²⁺ is a strong reducing agent.
(iii) Cu²⁺ salts are coloured while Zn²⁺ salts are white.
(b) Complete the following equations: (All India 2016)

Answer:
(a) (i) Because oxygen stabilizes the highest oxidation state (+7 of Mn) even more than fluorine i.e., +4 since oxygen has the ability to form multiple bonds with metal atoms.
(ii) Cr²⁺ exists in the d⁴ system and is easily oxidized to Cr³⁺ by loosing one electron which has the stable d³/t₂g orbital configuration. So, Cr²⁺ is a strong reducing agent.
(iii) Cu²⁺ has the configuration 3d⁹ with one unpaired electron which gets excited in the visible region to impart its colour while Zn²⁺ has 3d¹⁰ configuration without any unpaired electron so no d – d transition possible and hence colourless.

Question 86.
The elements of 3d transition series are given as:
Se Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(i) Write the element which shows maximum number of oxidation states. Give reason.
(ii) Which element has the highest m.p.?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why? (All India)
Answer:
(i) Mn shows, maximum number of oxidation states upto +7. It has the maximum number of unpaired electrons.
(ii) Cr has the highest melting point.
(iii) Sc shows only +3 oxidation state.
(iv) Mn is a strong oxidizing agent in +3 oxidation state because after reduction it attains +2 oxidation state in which it has the most stable half-filled (d⁵) configuration.

Question 87.
(a) Account for the following:
(i) Transition metals form large number of complex compounds.
(ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.
(iii) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. (Delhi 2017)
Answer:
(a) (i) Transition metals form large number of complexes because:

• small size of metal ion;
• high ionic charge; and
• availability of empty d-orbitals.

(ii) The lowest oxide of transition metal is basic because of low oxidation state some of valence electrons are not involved in bonding and acts as base by donating electron. However in higher oxides due to high oxidation state, it cannot donate electrons but can accept electrons due to high effective nuclear charge. Hence, they are acidic in nature.

(iii) Mn²⁺ exists in half-filled d⁵ state which is very stable while Mn³⁺ is d⁴ which is not so stable. Mn³⁺ can be easily reduced to Mn²⁺. Conversion from d⁴ to d⁵ will be quick and have negative ΔG value. Hence, because of the stability factor the E° value is high for this process. While Cr³⁺ is d³ is half-filled (t_2g3) is stable in nature and Cr²⁺ is d⁴, has one extra electron which it would like to donate to attain the stable half-filled (t_2g3) configuration. Hence for the process Cr³⁺ to Cr²⁺, the value of E° is less.

(b) Similarity: Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference: Actinoids show wide rage of oxidation states but lanthanoids do not.

Question 88.
(a) (i) How is the variability in oxidation
states of transition metals different from that of the p-block elements?
(ii) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?
(iii) Orange colour of Cr₂O₇^2- ion changes to yellow when treated with an alkali. Why?
(b) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons. (Delhi 2017)
Answer:
(a) (i) In transition elements, the oxidation
states can vary from +1 to highest oxidation state by removing all its valence electrons and the oxidation states differ by 1, e.g., Fe²⁺ and Fe³⁺ while in p-block elements, the oxidation states differ by 2, e.g., +2 and +4 or +3 and +5, etc.

(ii) Cu²⁺(aq) is much more stable than Cu⁺(aq). This is because although second ionization enthalpy of copper is large but Δ_hydH for Cu²⁺(aq) is much more negative that for Cu⁺(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows:
2Cu⁺ → Cu²⁺ + Cu

(iii) The orange coloured potassium dichromate solution when treated with basic NaOH solution, is converted to chromate which gets a faint colour like yellow.
The reaction is given as:
K₂Cr₂O₇ + 2NaOH → K₂CrO₄ + Na₂CrO₄ + H₂O.
This formation of chromate (CrO₄^–) ion converts the colour of solution to yellow.

(b)

• Lanthanoids show limited oxidation states, i.e., +2, +3, +4 out of which +3 is most common which is due to large energy gap between 4f and 5d subshells while actinoids show large number of oxidation states due to small energy gap between 5f, 6d and 7s subshells.
• Actinoid contraction is greater than lanthanoid contraction due to poor shielding of 5f electrons.
• Actinoids are radioactive in nature.

Question 89.
(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd and Hg are soft metals.
(iii) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. (All India 2017)
Answer:
(a) (i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals. In transitional elements ns, and (n – 1) d electrons have a approximate equal energies hence in addition to ns electrons, (n – 1) d electrons are also taking part in chemical bonding.

(ii) Zn, Cd and Hg are soft metals because they do not exhibit covalency due to completely filled d-orbitals. Absence of unpaired d electrons causes weak metallic bonding.

(iii) Mn²⁺ exists in half-filled d⁵ state which is very stable while Mn³⁺ is d⁴ which is not so stable. Mn³⁺ can be easily reduced to Mn²⁺. Conversion from d⁴ to d⁵ will be quick and have negative ΔG value. Hence, because of the stability factor the E° value is high for this process. While Cr³⁺ is d³ is half-filled (t_2g3) is stable in nature and Cr²⁺ is d⁴, has one extra electron which it would like to donate to attain the stable half-filled (t_2g3) configuration. Hence for the process Cr³⁺ to Cr²⁺, the value of E° is less.

(b) Similarity: Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference: Actinoids show wide rage of oxidation states but lanthanoids do not.

Question 90.
(a) Following are the transition metal ions of 3d series:
Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V= 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equations: (All India 2017)

Answer:
(a) (i) Cr³⁺ is most stable because of its small size and t³_2g configuration.
(ii) Mn³⁺ is a strong oxidising agent because after gaining one electron it is converted into Mn²⁺ which has stable d⁵ configuration.
(iii) Ti⁴⁺ is colourless due to d° configuration, i.e., no unpaired electrons.

(b) (i) 2MnO₄^– + 16H⁺ + 5S^2- → 2Mn²⁺ + 8H₂O + 5S

Question 91.
When chromite ore is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (A) is obtained. On acidifying the yellow solution with sulphuric acid, compound (B) is crystallised out. When compound (B) is treated with KC1, orange crystals of compound (C) crystallise out. Identify (A), (B) and (C) and write the reactions involved. (Comptt. Delhi 2017)
Answer:
Fusion of chromite ore with sodium carbonate :

Question 92.
(a) (i) Which transition element in 3d series has positive E⁰_M²⁺/M and why?
(ii) Name a member of lanthanoid series which is well know to exhibit +4 oxidation state and why?
(b) Account for the following :
(i) The highest oxidation state is exhibited in oxoanions of transition metals.
(ii) HCl is not used to acidify KMnO₄ solution.
(iii) Transition metals have high enthalpy of atomisation. (Comptt. Delhi 2017)
Answer:
(a) (i) Copper has positive E⁰_M²⁺/M value because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(ii) Cerium shows +4 oxidation state because it acquires stable empty orbital configuration and therefore Ce⁺⁴ is also used as a good analytical reagent and good oxidising agent.

(b) (i) The highest oxidation state shown in oxoanions of transition metals is (Cr shows +6) and it is due to the ability of oxygen to form multiple bonds with the metal atoms.
(ii) HCl is not used to the acidify KMnO₄ solution because KMnO₄ is a very strong oxidizing agent and it can oxidize HCl to liberate chlorine gas.
(iii) In transition elements, there are large number of unpaired electrons in their atoms, thus they have a stronger inter atomic interaction and thereby stronger bonding between the atoms. Due to this they have high enthalpies of atomization.

Important Questions for Class 12 Chemistry

The post Important Questions for Class 12 Chemistry Chapter 8 The d- and f-Block Elements Class 12 Important Questions appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi Set I

Question 1.
Write the formula of the compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄. [1]

Question 2.
Write the IUPAC name of the following compound: [1]

Answer:

Question 3.
What is the effect of adding a catalyst on
(a) The activation energy (E_a), and
(b) Gibbs energy (ΔG) of a reaction?
Answer:
On adding a catalyst
(a) Activation energy of the reaction decreases.
(b) Gibbs energy doesn’t change.

Question 4.

Answer:

Question 5.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. [1]
Answer:
When a liquid is dispersed in solid, ‘gel’ colloid is formed. Examples Jelly, butter, cheese, curd, etc.

Question 6.
(a) Arrange the following compounds in the increasing order of their acid strength: [2]
p-cresol, p-nitrophenol, phenol
(b) Write the mechanism (using curved arrow notation) of the following reaction:

OR
Write the structures of the products when Butan-2-ol reacts with the following:
(a) CrO₃
(b) SOCl₂
Answer:

Question 7.
Calculate the number of unit cells in 8.1g of aluminum if it crystallizes in a face-centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol^-1) [2]

Question 8.
Draw the structures of the following:
(a) H₂SO₃
(b) HClO₃
Answer:

Question 9.
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. [2]
Answer:
Electrolytic cells are generally used is hearing aids. At the cathode, reduction of metal takes place and at the anode, oxidation of metal takes place.
Cathode: M + e^– → M⁺
Anode: M⁺ → M + e^–

Question 10.
Using IUPAC norms write the formulae for the following:
(a) Sodium dicyanidoaurate (I)
(b) Tetraamminechloridonitrito-N-platinum (IV) sulfate [2]
Answer:
(a) Sodium dicyanoaurate (I)
Na [Au (CN)₂]
(b) Tetraammine chloridonitrito-N-platinum (IV)
Sulphate [Pt(NH₃)₄(Cl) (NO₂)]SO₄

Question 11.
(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example. [3]

Question 12.
(a) The cell in which the following reaction occurs:
2Fe³⁺ (aq) + 2I^– (aq) → 2Fe²⁺ (aq) + I₂ (s)
has E_cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given: 1F = 96,500 C mol^-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours?
(Given: 1 F = 96,500 C mol^-1) [3]
Answer:
(a) ΔG = -nFE_cell = -2 × 96500 × 0.236 = -45.548 kJ/mol
(b) According to Faraday’s first law the amount of metal deposited (W).
W = i × t = 0.5 × 7200 = 3600C
1F = 96500 C mol^-1
That is e^– flows from 96500 C = 1 mol
e^– flows from 3600 C = = 0.037 mol.
No. of electrons = 0.037 × 6.023 × 10²³
= 0.2246 × 10²³
= 22.46 × 10²¹ electrons

Question 13.
(a) What type of isomerism is shown by the complex [CO(NH₃)₅ (SCN)]²⁺?
(b) Why is [NiCl₄]^2- paramagnetic while [Ni(CN)₄]^2- is diamagnetic?
(Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed? [3]
Answer:
(a) Linkage isomerism
(b) [NiCl₄]^2-, Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸
Cl^– is a weak field ligand.

2 electrons are impaired in [NiCl₄]^2- which provides paramagnetism to the complex.
[Ni(CN)₄]^2-
Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸
CN^– is a strong, field ligand

no electron is unpaired in [Ni (CN)₄]^2- That’s why the complex is diamagnetic.
(c) In the tetrahedral complex, CFSE is very low and it is difficult for the tetrahedral complexes to exceed the pairing energy. Usually, electrons prefer to move to higher energy orbitals for pairing. Thus they usually form high spin complexes.
(CFSE) tetrahedral = (CFSE)octahedral

Question 14.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Homogeneous catalysis and Heterogeneous catalysis [3]
OR
(a) Write the dispersed phase and dispersion medium of milk.
(b) Write one similarity between physisorption and chemisorption.
(c) Write the chemical method by which Fe(OH)₃ sol is prepared from FeCl₃.
Answer:
(a) Multimolecular colloids are the colloids in which the dispersed phase consists of aggregates of atoms or molecules with molecular size less than 1 nm whereas associated colloids are the substances that are dissolved in a medium, behave as normal electrolytes at low concentration but as colloids at higher concentration.

(b) Coagulation is the process of precipitation of a colloidal solution by the addition of an excess of an electrolyte whereas peptization is the process responsible for the formation of stable dispersion of colloidal particles in the dispersion medium.

(c) Homogeneous catalysis is the one in which the phases of the reactants and the catalysts are the same whereas in heterogeneous catalysis the phases of the reactants and the catalysts are not the same.
OR
(a) Milk
Dispersed phase – Liquid
Dispersion medium – Liquid
(b) Both physisorption and chemisorption depends on the surface area. Both increases with an increase in the surface area.
(c) Fe(OH)3 sol is prepared from FeCl3 by hydrolysis method.

Question 15.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [3]
Answer:
For first order reaction,

Question 16.
The following compounds are given to you: 2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(a) Write the compound which is most reactive towards SN₂ reaction.
(b) Write the compound which is optically active.
(c) Write the compound which is most reactive towards (3-elimination reaction. [3]
Answer:
(a) 1-Bromo pentane > 2-Bromo pentane > 2-Bromo-2-methyl pentane (Reactivity towards, SN₂ reaction)
(b) 2-Bromo pentane
CH₃CH₂CH₂CHBrCH₃

This compound is most reactive towards β-elimination.

Question 17.
Write the principle of the following:

1. Zone refining
2. Froth floatation process
3. Chromatography [3]

Answer:

1. Zone refining
   • This process is used for the metals which are required in very high purity like silicon, germanium, boron, gallium, etc.
   • This method is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.
   • In this method, impure metal is cast into a thin bar.
2. Froth floatation process
   • This method is based on the principle that difference in the wetting properties of the ore and gangue particles with water and oil.
   • This method is used for the extraction of those metals in which the ore particles are preferentially wetted by oil and gangue by water.
   • This method has been used for the concentration of sulphide ores like PbS, ZnS, CuFeS_2, etc.
3. Chromatography
   • This is a modem method of purification based on the difference in the adsorbing capacities of the metal and its impurities on a suitable adsorbent.
   • This technique is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 18.
Write the structures of compounds A, B and C in the following reactions:

Answer:

Question 19.
Write the structures of the monomers used for getting the following polymers:
(a) Nylon-6,6
(b) Melamine-formaldehyde polymer
(c) Buna-S [3]
Answer:
(a) Monomers of Nylon-6, 6

(b) Monomers of Melamine-formaldehyde polymer

Question 20.
Define the following: [3]
(a) Anionic detergents
(b) Limited spectrum antibiotics
(c) Antiseptics
Answer:
(a) Anionic detergents: These detergents contain an anionic hydrophilic group. These are manufactured from the long chain of alcohols. These long chain alcohols are treated with cone. H₂SO₄ to form alkyl hydrogen sulphates of high molecular mass and then are neutralized with alkali to form salts.

(b) Limited Spectrum Antibiotics: The antibiotics which are effective against single organism or disease are called limited spectrum antibiotics, example-streptomycin.

(c) Antiseptics: The chemical substances that are used to either kill or prevent the growth of micro-organisms are called antiseptics. These are not harmful to living tissues and can be safely applied on wounds, cuts, ulcers, etc., for example, Soframycin.

Question 21.
Give reasons for the following:
(a) Red phosphorus is less reactive than white phosphorus.
(b) Electron gain enthalpies of halogens are largely negative.
(c) N₂O₅ is more acidic than N₂O₃. [3]
Answer:
(b) Electron gain enthalpies of halogens are largely negative in their respective periods.
This is due to the fact that the atoms of these elements have only one electron less than the stable noble gas (ns²np⁶) configuration. Therefore, they have maximum tendency to accept an additional electron.

Question 22.
Give reasons for the following:
(a) Acetylation of aniline reduces its activation effect.
(b) CH₃NH₂ is more basic than C₆H₅NH₂.
(c) Although-NH₂ is olp directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. [3]
Answer:
(a)

In acetanilide, the oxygen atom of the group withdraws electrons from the NH₂ group as shown below:

As a result, the electron pair on nitrogen gets displaced to the carboxyl group. Therefore, the unshared pair of electron on nitrogen is less available for a donation of the electron to the aromatic ring.

(b) In aniline, lone pair of e~ present on ‘N’ is in conjugation with the benzene ring and become less available for protonation because of resonance.

This conjugation of lone pair of e- is not present in case of methylamine and lone pair of e^– of ‘N’ are fully available for protonation. That’s why the basicity order of aniline and methylamine is:

The reason for the formation of a large amount of m-nitroaniline is that under strongly acidic conditions, aniline gets protonated to anilinium ion (-NH₃ group). This is a deactivating group and is meta-directing in nature.

Question 23.
After watching a program on TV about the presence of carcinogens (cancer causing agents) Potassium bromate and potassium iodate in bread and other bakery products, Rupali a Class XII student decided to make others aware about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers and other bakery products to the students. The principal took immediate action and instructed the canteen contractor to replace the bakery products with some protein and vitamin rich food like fruits, salads, sprouts, etc. The decision was welcomed by the parents and the students.

After reading the above passage, answer the following questions:
(a) What are the values (at least two) displayed by Rupali?
(b) Which polysaccharide component of carbohydrates is commonly present in bread?
(c) Write the two types of secondary structures of proteins.
(d) Give two examples of water soluble vitamins. [4]
Answer:
(b) Starch
(c) 1. α-helix structure.
2. β-pleated sheet structure.
(d) Vitamin B and Vitamin C

Question 24.
(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd, and Hg are soft metals.
(iii) E⁰ value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [5]
OR
(a) Following are the transition metal ions of 3d series: Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equation:
(i) 2MnO₄ + 16H⁺ + 5S^2- →
(ii) KMnO₄ →
Answer
(a) (i) Transition metal ions show variable oxidation states due to the participation of (n-1) d electrons in addition to outer ns-electrons because the energies of ns and (n-1)d subshells are almost equal. As a result of which the electrons of (n-1)d and ns subshell both part in bond formation.

(ii) Zn, Cd, and Hg are soft metals because of their completely filled 3d, 4d, and 5d orbitals respectively Due to completely filled d-orbitals these metals are reluctant to form Zn-Zn, Cd-Cd, and Hg-Hg bonds.

(iii) A highly positive value of E⁰ for Mn³⁺/Mn²⁺ shows that Mn²⁺ (d⁵) is particularly stable. While low value of E⁰ for Cr³⁺/Cr²⁺ shows that Cr²⁺ (d⁴) is less stable than Cr³⁺ (d³)

(b) Similarity: In lanthanoids and actinoids both the added electron enters the antepenultimate shell 4f and 5f respectively.
Difference: Lanthanoids show a common oxidation state of +3 while actinoids show different oxidation states other than +3.
OR
(a) Ti⁴⁺ = 1s²2s²2p⁶3s²3p⁶
V²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
Mn³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴
Cr³⁺ = 1s² 2s² 2p⁶ 3s² 3d³
(i) Ti⁴⁺ is most stable in an aqueous solution because of full filled valence shell (3s² 3p⁶) onfiguration (noble gas configuration).
(ii) Mn³⁺ is the strong agent as it oxidises other species it will reduce itself by taking an electron and will stabilise its configuration to 3d⁵.
(iii) Ti⁴⁺ is colourless due to absence of impaired electrons (3s² 3p⁶)

Question 25.
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given:
The molar mass of sucrose = 342g mol^-1
The molar mass of glucose = 180 g mol^-1
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass [5]
OR
(a) 30 g of urea (M = 60g mol-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
Answer:
(a) T0 (freezing point of water) = 273.15K
Ts (freezing point of sucrose solution) = 269.15K
Weight of sucrose in solution = 10 g
Weight of glucose in solution = 10 g
The molar mass of sucrose = 342 g mol^-1
The molar mass of glucose = 180 g mol^-1
The freezing point of glucose = x
Depression in freezing point


So, the freezing point of glucose solution = 265.55 K.

(b) (i) Molality: It is the number of moles of the solute dissolved per 1000 g of the solvent. It is denoted by m.

(ii) Abnormal molar mass: Those solute that dissociates or associate in solution, show an abnormal molar mass in solution, for example, Molar mass of ethanoic acid is greater than normal molar mass.

The molar mass of KCl in solution is reduced than normal molar mass.
KCl → K⁺ + Cl^–
OR
(a) W_B = 30 g
M_B = 60 g mol^-1
W_A = 846 g
M_A = 18 g mol^-1
P⁰ = 23.8mm Hg
P_s = x
Relative lowering of vapour pressure

So, the vapour pressure of water for this solution = 23.597 mm Hg
(b)

Question 26.
(a) Write the product(s) in the following reactions:

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one
(ii) Benzoic acid and Phenol
OR
(a) Write the reactions involved in the following:
(i) Etard reaction
(ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to Benzaldehyde
(ii) Acetophenone to Benzoic acid
(iii) Ethanoic acid to 2-Hydroxyethanoic acid
Answer:


(b) (i) Butanal and Butan-2-one

Thus Butanal gives silver mirror test with Tollen’s reagent whereas Butan-2-one does not.
(ii) Benzoic acid and phenol

Thus, Benzoic acid gives sodium benzoate on reaction with sodium bicarbonate whereas phenol gives no reaction with sodium bicarbonate.
OR
(a) (i) Etard reaction: The oxidation of toluene to benzaldehyde with chromyl chloride (CrO₂Cl₂) dissolved in CCl₄ or CS₂.

(ii) Stephen reaction: The partial reduction of alkyl or aryl cyanides to the corresponding aldehydes with a suspension of anhydrous SnCl₂ in ether saturated with HCl at room temperature followed by hydrolysis.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question 3.
Write the IUPAC name of the following compound. [1]

Answer:

Question 5.

Answer:

Question 6.
Using IUPAC norms write the formulae for the following:
(a) Tris (ethane-1, 2-diamine) chromium (III) chloride.
(b) Potassium tetrahydroxozincate (II). [2]
Answer:
(a) [Cr(en)₃]Cl₃
(b) K₂[Zn(OH)₄]

Question 7.
Draw the structures of the following:
(a) H₂S₂O₈
(b) ClF₃
Answer:

Question 8.
Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. [2]
Answer:
The lead storage battery is commonly used in inverters.
Reactions taking place at the anode

Question 11.
(a) Write the principle of vapour phase refining.
(b) Write the role of dilute NaCN in the extraction of silver.
(c) What is the role of collectors in the froth floatation process? Give an example of a collector. [2]
Answer:
(a) Vapour phase refining: This method is based on the principle that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation-C.

(b) NaCN is used to leach the silver ore in the presence of air. Pure silver is obtained by replacement in the process of extraction of silver.

(c) In the froth floatation process, collectors enhance the non-wettability of the mineral particles. Example of collectors is pine oil, eucalyptus oil, fatty acids, etc.

Question 16.
Define the following:
(a) Narrow spectrum antibiotics.
(b) Antacids [3]
Answer:
(a) Narrow spectrum antibiotics: The antibiotics which are effective mainly against gram-positive or gram-negative bacteria are called narrow-spectrum antibiotics Example: Penicillin.

(b) Antacids: The chemical substances which neutralize excess acids in the gastric juices and gives relief from acid indigestion, acidity, heart bums, and gastric ulcers are called antacids. Example Sodium hydrogen- carbonate (baking soda) in water.

Question 17.
Write the structures of the monomers used for getting the following polymers:
(a) Polyvinyl chloride (PVC)
(b) Buna-N [3]
Answer:
(a) A monomer of polyvinyl chloride (PVC)
Vinyl chloride CH₂ = CH — Cl

(b) Monomer of Buna-N

Question 22.
(a) Based on the nature of intermolecular forces, classify the following solids: Benzene, Silver
(b) AgCl shows Frenkel defect while NaCl does not. Give reason.
(c) What type of semiconductor is formed when Ge is doped with Al? [3]

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question 1.

Answer:

Question 3.
Write the formula of the compound of iodine which is obtained when cone. HNO₃ oxidises I₂. [1]
Answer:
Iodic acid, HIO₃ is obtained on the oxidation of I₂ by HNO₃.

Question 4.
What type of colloid is formed when a gas is dispersed in a liquid? Give an example. [1]
Answer:
‘Foam’ colloid is formed when a gas is dispersed in a liquid. For example, whipped cream or soda water.

Question 5.
Write the IUPAC name of the following compound:

Answer:

Question 6.
Draw the structures of the following: [2]
(a) XeF₄
(b) BrF₅
Answer:
(a) XeF₄

(b) BrF₅

Question 7.
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. [2]
Answer:
Dry cells are used in transistors.
At anode
Zn → Zn2+ + 2e-
At cathode
2NH⁺₄ (aq) + 2MnO₂(s) + 2e^– → 2MnO(OH) + 2NH₃

Question 9.
Using IUPAC norms write the formulae for the following:
(a) Potassium trioxalatoaluminate (III).
(b) Dichloridobis (ethane-1, 2-diamine) cobalt (III) [2]
Answer:
(a) K₃[Al(Ox)₃]
(b) [CoCl₂(en)₂]⁺

Question 14.
(a) Based on the nature of intermolecular forces, classify the following solids: Sodium sulphate, Hydrogen
(b) What happens when CdCl₂ is doped with AgCl?
(c) Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances? [3]

Question 15.
(a) Write the principle of electrolytic refining.
(b) Why does copper obtained in the extraction from copper pyrites have a blistered appearance?
(c) What is the role of depressants in the froth floatation process? [3]
Answer:
(a) Electrolytic refining: This method is based on the principle of electrolysis. In this method, impure metal is made to act as anode and a strip of the same metal in pure form is used as a cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metals.

(b) In the extraction of copper from CuFeS₂, SO₂, N₂ and O₂ escape from the metal. As the metal solidifies, the dissolved gases escape producing blisters on the metal surface, which provides blister appearance to copper.

(c) Depressants are used to prevent certain types of particles from forming the froth with air bubbles.
For example, NaCN can be used as a depressant in the separation of ZnS and PbS.

Question 19.
Define the following:
(a) Cationic detergents
(b) Broad-spectrum antibiotics
(c) Tranquilizers [3]
Answer:
(a) Cationic detergents: These are the quaternary ammonium salts of amines with acetates, chlorides or bromides as anions. The cationic part possesses a long hydrocarbon chain with a positive charge on the nitrogen atom. Example. Cetyl trimethyl ammonium chloride.

(b) Broad-spectrum antibiotics: Antibiotics which kills or inhibit a wide range of gram-positive and gram-negative bacteria are called broad-spectrum antibiotics. Example Chloramphenicol.

(c) Tranquilizers: The chemical substances used for the treatment of stress, fatigue, mild and severe mental diseases are called tranquilizers. Example: Phenelzine (Nardil).

Question 20.
Write the structures of the monomers used for getting the following polymers:
(a) Teflon
(c) Neoprene [3]
Answer:

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Physics 2018

Section-A

Question 1.
A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency ? [1]
Answer:
Frequency of revolution of a particle

OR
Since mass of electron is less than that of proton, therefore, its frequency of revolution will be higher than that of proton

Question 2.
Name the electromagnetic radiations used for
(a) water purification, and
(b) eye surgery.  [1]
Answer:
(a) Water purification : Ultraviolet radiation.
(b) Eye surgery : Ultraviolet radiation/laser

Question 3.
Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. [1]
Answer :
Graph for photoelectric current (I) versus applied potential for radiations of same frequency and varying intensity.

Question 4.
Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two the parent or the daughter nucleus would have higher binding energy per nucleon ? [1]
Answer:
When lighter nuclei combine to form a heavier nucleus, binding energy per nucleon increases and energy is released. Thus, the daughter nucleus would have higher binding energy per nucleon.

Question 5.
Which mode of propagation is used by short wave broadcast services [1]

Section-B

Question 6.
Two electric bulbs P and Q have their resistances in the ratio of 1 : 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. [2]
Answer :
Let resistances of bulbs P and Q be R and 2R respectively. As they are connected in series, so current through each bulb is same. Let the current be I.

Question 7.
A10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Q as shown in the figure. Find the value of current in the circuit. [2]

OR
In a potentiometer arrangement for deter-mining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Q is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.


Question 8.
(a) Why are infra-red waves often called heat waves ? Explain.
(b) What do you understand by the statement, “Electromagnetic waves trans-port momentum” ? [1]
Answer:
(a) Infra-red waves are called heat waves because they raise the temperature of the object on which they fall and hence increase their thermal motion. They also affect the photographic plate and are readily absorbed by most of the materials.
(b) Electromagnetic waves transport momentum. This means that when an electromagnetic wave travels through space with energy U and speed c, then it transports linear momentum p = U/C. If a surface absorbs the waves completely, then momentum ‘p’ is delivered to the surface. If the surface reflects the wave, then momentum delivered by both incident and reflected wave adds on to give ‘2_p’ momentum.

Question 9.
If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? [2]


Since, the energy of incident radiation is greater than the work function of sodium and potassium, but less than that of calcium and molybdenum, therefore, photoelectric emission will take place in sodium and potassium.

Question 10.
A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%. [2]

Section-C

Question 11.
Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the
(a) resultant electric force on a charge Q and
(b) potential energy of this system. [3]
R
(a) Three point charges q – 4_q and 2_q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

b) Find out the amount of the work done to separate the charges at infinite distance.
Answer :
(a) Force on charge Q at B due to charge q at A






Question 12.
(a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.
Answer:
(a) Conductivity of a metallic wire is defined as its ability to allow electric charges or heat to pass through it. Numerically, conductivity of a material is reciprocal of its resistivity.
SI unit : ohm^-1 m^-1 or mho m^-1 or Siemen m^-1

(b) Consider a potential difference V be applied across a conductor of length l and cross section A.
Electric field inside the conductor, E = v/l.
Due to the external field the free electrons inside the conductor drift with velocity V_d.
Let, number of electrons per unit volume = n,
charge on an electron = e
Total electrons in length, l = nAl  And,
total charge, q = neAl




Question 13.
A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate
(a) the work done in turning the magnet to align its magnetic moment
(i) normal to the magnetic field,
(ii) opposite to the magnetic field, and
(b) the torque on the magnet in the final orientation in case (ii).  [3]
Answer:

Question 14.
(a) An iron ring of relative permeability μ has winding’s of insulated copper wire of n turns per meter. When the current in the winding’s is I, find the expression for the magnetic field in the ring.
(b) The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. [3]

Question 15.
(a) Show using a proper diagram how unpolarised light can be linearly polarised by reflection from a transparent glass surface.
(b) The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index 3/2
placed in water of refractive index 4/3 Will this ray suffer total internal reflection on striking the face AC ?




Question 16.
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light ? [3]
Answer :
(a) The resultant intensity in Young’s experiment is given by

When slit is not covered, then I₀ is the intensity from each slit.
Maximum intensity (I_max) occurs when Φ = 0°.
Minimum intensity (I_min) occurs when (Φ ) = 180°.
If one slit is covered with glass to reduce its intensity by 50%, then

(b) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpen¬dicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Further, the shortest visible wave, blue, produces a bright fringe first.

Question 17.
A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y. [3]

Answer:
Given, refractive index of lens, μg = 1.5. The distance of the needle from the lens in the first case = The focal length of the combination of convex lens and planoconcave lens formed by the liquid, f= x And, the distance measured in the second case = Focal length of the convex lens, f₁ = y If the focal length of planoconcave lens formed by the liquid be f₂, then


Question 18.
(a) State Bohr’s postulate to define stable- orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. [3]
Answer :
(a) Bohr’s postulate for stable orbits in hydrogen atom : An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of h/2π, where h is Planck’s constant.
If n is the principal quantum number of orbit, then an electron can revolve only in . certain orbits or definite radii. These are called stable orbits.

de Broglie explanation of stability of orbits:
According to de Broglie, orbiting electron around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of . circular orbit contains a whole number of wavelengths i.e., for orbit circumference of n^th orbit as 2nπr_n

OR

Question 19.
(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.
(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ? [3]
Answer:
(a) Plot of binding energy per nucleon mass number :

1. When we move from the heavy nuclei region to the middle region of the plot, we find that there will be a gain in the overall binding energy and hence results in release of energy. This indicates that energy can be released when a heavy nucleus (A ~ 240) breaks into two roughly equal fragments. This process is called nuclear fission.
2. Similarly, when we move from lighter nuclei to heavier nuclei, we again find that there will be gain in the overall binding energy and hence release of energy takes place. This indicates that energy can be released when two or more lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

Question 20.
(a) A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw the labelled circuit diagram she would use and explain how it works.
(b) Give the truth table and circuit symbol for NAND gate. [3]
Answer :
(a) Full wave rectifier :


Explanation : In positive half cycle of AC, end A becomes positive and D1 becomes forward biased and D2 is reverse biased, so conducts and D2 doesn’t. So conventional current flows through D1, RL and upper half of secondary winding. Similarly, during negative half cycle of AC, diode D2 becomes forward biased and D1 is reverse biased, current flows through D2, RL and lower half of secondary winding. Thus, current flows in same direction in both half cycles of input AC voltage.

Question 21.
Draw the typical input and output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine (a) the input resistance (r ), and (b) current amplification factor (β). [3]

Question 22.
(a) Give three reasons why modulation of a message signal is necessary for long distance transmission.
(b) Show graphically an audio signal, a carrier wave and an amplitude modulated wave. [3]

Section-D

Question 23.
The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage. [4]
(a) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.
(b) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
(c) Write two values each shown by the teachers and Geeta.
Answer :
(a) The device used to change alternating voltage to a higher or lower value is a transformer.
Causes of power dissipation in this device are :
1. Core losses due to eddy currents and hysteresis loop due to alternating flux.
2. Copper losses due to resistance of winding in primary and secondary coils.
3. Loss of power due to leakage of magnetic flux in coil.
(b) The loss of power in the transmission lines is I²R, where I is the strength of current and R is the resistance of the wires. To reduce the power loss a.c. is transmitted over long distances at extremely high voltages. This reduces I in the same ratio. Therefore, I²R becomes negligibly low. For the same reason, at the generating stations, the voltage is stepped up to transmit it over long distances to minimize power loss. Therefore a.c. is used because stepping up is not possible for direct current.

Section-E

Question 24.
(a) Define electric flux. Is it a scalar or a vector quantity ? A point charge q is at a distance of dl2 directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

(b) If the point charge is now moved to a distance ‘d ‘ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected. [5]
OR
(a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r₁to r₂( r₂> r₁).

Answer:
(a) Electric flux : Electric flux through an area is defined as the product of electric field strength E and area dS perpendicular to the field. It represents the field lines crossing the area. It is a scalar quantity. Imagine a cube of edge d, enclosing the charge. The square surface is one of the six faces of this cube. According to Gauss’ theorem in electrostatics,

Total electric flux through the cube =
This is the total flux through all six surtace
∴ Electric flux through the square surface = 

(b) On moving the charge to distance d from the center of square and making side of square 2d, does not change the flux at all because flux is independent of side of square or distance of charge in this case.
OR
(a) Electric field due to a straight uniformly charged infinite line of charge density λ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts :
(i) curved surface S₁
(ii) flat surface S₂ and
(iii) flat surface S₃.
By symmetry, the electric field has the same magnitude E at each point of curved surface Sj and is directed radially outward. We consider small elements of surfaces S₁, S₂ and S₃.


b) Graph showing variation of E with perpendicular distance from line of charge : The electric field is inversely proportional to distance V from line of charge.


Question 25
(a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ω in a magnetic field , directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/ hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10^-4 T and the angle of dip is 30°. [5]
OR
A device X is connected across an ac source of voltage V = V₀ sin ωt . The current through X is given as I = I₀ sin
(a) Identify the device X and write the expression for its reactance.
(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.
(c) How does the reactance of the device X vary with frequency of the ac ? Show this variation graphically.
(d) Draw the phasor diagram for the device X.
Answer :
(a) Principle of ac generator :
The ac generator is based on the principle of electromagnetic induction. When closed coil is rotated in a uniform field with its axis perpendicular to field, then magnetic flux changes and emf is induced.
Working :
When the armature coil rotates, the magnetic flux linked with it changes and produces induced current. If initially, coil PQRS is in vertical position and rotated clockwise, then PQ moves down and SR moves up. By Fleming’s right hand rule, induced current flows from Q to P and S to R which is the first half rotation of coil. Brush B₁ is positive terminal and B₂is negative. In second half rotation, PQ moves up and SR moves down. So induced current reverses and the alternating current is produced in this manner by the generator.




Question 26.
(a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
(b) Obtain the mirror formula and write the expression for the linear magnification.
(c) Explain two advantages of a reflecting telescope over a refracting telescope. [5]
OR
(a) Define a wave front. Using Huygens’ principle, verify the laws of reflection at a plane surface.
(b) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? Explain.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the obstacle. Explain why ?
Answer :
(a) Concave mirror produces real, inverted and magnified image for object placed between F and C :

(b) Derivation for mirror formula and magnification: Consider an object AB be placed in front of a concave mirror beyond center of curvature C.



(c) Advantages of reflecting telescope over a refracting telescope are :
1. Reflecting telescope is free from chromatic and spherical aberrations unlike refracting telescope. Thus image formed is sharp and bright.
2. It has a larger light gathering power so that a bright image of even far off object is obtained.
3. Resolving power of reflecting telescope is large.
OR
(a) Wave front is defined as the continuous locus of all the particles of a medium which are vibrating in the same phase. Verification of laws of reflection using Huygens’s principle : Let XY be a reflecting surface at which a wave front is being incident obliquely. Let v be the speed of the wave front and at time t = 0, the wave front touches the surface XY at A. After time t, point B of wave front reaches the point B’ of the surface.

According to Huygens’s principle each point of wave front acts as a source of secondary waves. When the point A of wave front strikes the reflecting surface, then due to presence of reflecting surface, it cannot advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. As the wave front AB advances further, its points A₁, A₂, A₃ etc. strike the reflecting surface successively and send spherical secondary wavelets in the first medium.

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Physics 2018 appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Delhi

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Delhi Set I

Section – A

Question 1.
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. [1]
Answer:
Manganese belongs to group number 7 and its oxidation state in KMnO₄ is +7
i.e., KMnO₄
1 + x + 4 (-2) = 0
⇒ 1 + x – 8 = 0
⇒ x = 7
Thus, the formula of the oxo-anion is KMnO₄.

Question 2.
Write IUPAC name of the following compound: [1]
(CH₃CH₂)₂NCH₃
Answer:
N-Ethyl-N-methylhexanamine.

Question 3.
For a reaction R → P, half-life (t_1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction? [1]
Answer:
Since half-life is independent of the initial concentration of the reactants. Thus it is a first-order Reaction. The formula for the half-life of the first-order reaction.
t_1/2 =

Question 4.
Write the structure of l-Bromo-4-chlorobut-2- ene. [1]
Answer:
BrCH₂CH = CHCH₂Cl

Question 5.
Write one similarity between physisorption and Chemisorption. [1]
Answer:
Physisorption and chemisorption both are the surface phenomenon and both increases the surface area during the process of adsorption.

Section – B

Question 6.
Complete the following reactions:
(i) NH₃ + 3Cl₂(excess) →
(ii) XeF₆ + 2H₂O →
OR
What happens when
(i) (NH₄)2Cr₂O₇ is heated?
(ii) H₃PO₃ is heated?
Write the equations. [2]
Answer:
(i) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl.
(ii) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
OR
(i) (NH₄)2Cr₂O₇(s) → Cr₂O₃(s) + N₂(g) + 4H₂O(g)
(ii) 4H₃PO₃ → 3H₃PO₄ + PH₃.

Question 7.
Define the following terms:
(i) Colligative properties
(ii) Molality (m) [2]
Answer:
(i) Colligative properties are those which depends on a number of moles of solute irrespective of their Nature.
(ii) Molality is defined as the number of moles of solute dissolved per kg of the solvent. It is independent of temperature.

Question 8.
Draw the structures of the following:
(i) H₂S₂O₇
(ii) XeF₆
Answer:

Question 9.
Calculate the degree of dissociation (α) of acetic acid if it’s molar conductivity (∧_m) is 39.05 S cm² mol^-1.
Given: ∧⁰ (H⁺) = 349.6 S cm² mol^-1 and ∧⁰ (CH₃COO^–)= 40.9 S cm² mol^-1. [2]
Answer:
Given: Molar conductivity (∧m) for acetic acid = 39.05 S cm² mol^-1.
∧⁰ (H⁺) = 349.6 S cm² mol^-1
∧⁰ (CH₃COO^–) = 40.95 S cm² mol^-1
We know that:

Thus, the degree of dissociation of acetic acid is 0.1.

Question 10.
Write the equations involved in the following reactions:
(i) Wolff-Kishner reduction
(ii) Etard reaction. [2]
Answer:

Section – C

Question 11.
A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15K.
[Given: (Molar mass of sucrose = 342 g mol^-1), (Molar mass of glucose = 180 g mol^-1)] [3]
Answer:
Given: Freezing point of = 269.15 K 10% solution of glucose
The freezing point of pure water = 273.15 K
The molar mass of sucrose = 342 g mol^-1.
The molar mass of glucose = 180 g mol^-1.
We know that:

Question 12.
(a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO₃ for 15 minutes.
Given: Molar mass of Ag = 108 g mol^-1, 1F = 96500 C mol^-1)
(b) Define fuel cell [3]
Answer:
(a) Given:
Current = 2 amperes
Time = 15 minutes
Molar mass of Ag = 108 g mol^-1
1F = 96500 C mol^-1
Amount of metal deposited (m) = ZQ
Q = It = 2 × 15 × 60 = 1800 C
Silver deposited Ag⁺ + 2e^– → Ag(s)
1 mole of electron or 1 × 96500 C of current deposit silver = 108 g
1800 C of current will deposit =
Amount of Ag deposited = 2.01g

(b) Fuel cell is the cell which converts the energy of combustion of fuels directly into electrical energy.

Question 13.
(i) What type of isomerism is shown by the complex [Co(NH₃)₆]Cr(CN)₆]?
(ii) Why a solution of [Ni(H₂O)₆]²⁺ is green while a solution of [Ni(CN)₄]^2- is colourless? (At no. of Ni = 28)
(iii) Write the IUPAC name of the following complex: [3]
[CO(NH₃)₅(CO₃)]Cl.
Answer :
(i) Both shows coordination isomerism because both cationic and anionic entities and isomers differ in the distribution of ligands in the coordination entity of cationic and anionic part.

(ii) In [Ni(H₂O)₆]²⁺ Ni is in +2 oxidation state with electronic configuration 3d⁸. In the presence of weak ligand H₂O the two unpaired electrons do not pair up and hence the complex has two impaired electrons. Therefore, it is coloured and shows d-d transitions which absorb red light and emits green complimentary light.
In case of [Ni(CN)₄]^2- Ni also shows +2 oxidation state but CN ligand is strong ligand and two unpaired electrons undergo pairing, to no d-d transitions takes place and it shows no colour.

(iii) Pentaamminecarbonatocobalt(III) chloride.

Question 14.
Write one difference in each of the following:
(i) Lyophobic sol and Lyophilic sol.
(ii) Solution and Colloid
(iii) Homogeneous catalysis and Heterogeneous catalysis. [3]
Answer:
(i) Lyophobic colloidal sols are not hydrated and have a weak affinity with the dispersion medium whereas lyophilic colloidal sols are heavily hydrated and have a strong affinity with the dispersion medium.

(ii) The solution is a homogeneous mixture of solute and solvent whereas colloid is the heterogeneous mixture of the dispersed phase and the dispersion medium.

(iii) Homogeneous catalysis is the catalysis in which the reactants and the catalysts are in the same phase whereas in the heterogeneous catalysis the reactants and the catalysts are in the different phases.

Question 15.
Following data are obtained for reaction:
N₂O₅ → 2NO₂ + O₂

(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010, log 4 = 0.6021) [3]
Answer:

Question 16.
Following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane

1. Write the compound which is most reactive towards SN₂ reaction.
2. Write the compound which is optically active.
3. Write the compound which is most reactive towards β-elimination reaction. [3]

Answer:

1. 1-Bromopentane is most reactive towards SN2 reaction as it follows the order 1° > 2° > 3°.
2. 2-Bromopentane is optically active.
3. 2-Bromo-2-methyl butane is most reactive towards β-elimination reaction.

Question 17.

1. Write the principle of the method used for the refining of germanium.
2. Out of PbS and PbCO₃ (ores of lead), which one is concentrated by froth floatation process preferably?
3. What is the significance of leaching in the extraction of aluminium? [3]

Answer:

1. Zone refining method is used for the refining of germanium and it is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.
2. PbS, Sulphide ore has more tendency to stick to the oil which comes on the surface being lighter and easily skimmed off so PbS is concentrated by froth floatation method.
3. Leaching of alumina is done to remove the impurities like SiO₂ by using NaOH solution and pure alumina is obtained.

Question 18.
Write structures of compounds A, B and C in each of the following reactions: [3]

OR
Do the following conversions in not more than two steps:
(a) Benzoic acid to Benzaldehyde
(b) Ethylbenzene to Benzoic acid
(c) Propanone to Propene
Answer:

Question 19.
Write the structure of the monomers used for getting the following polymers:
(i) Dacron [3]
Answer:
(i) Monomers of Dacron:

Question 21.
Give reasons:
(i) Thermal stability decreases from H₂O to H₂Te.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion.
(iii) Nitrogen does not form pentahalide. [3]
Answer:
(i) As we move down in a group atomic radius increases as a result bond length increases. Larger the bond length lesser will be the bond dissociation enthalpy. So thermal stability decreases from O to Te.
(ii) Fluoride ion is the smallest ion in the group and it has high charge density and charges size ratio. That is why it has a high hydration enthalpy.

Section – E

Question 24.
(a) Account for the following:
(i) Transition metals form a large number of complex compounds.
(ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.
(iii) E⁰ value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compare to Cr³⁺/Cr²⁺.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [5]
OR
(a) (i) How is the variability in oxidation states of transition metals different from that of the p-block elements?
(ii) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?
(iii) The orange colour of Cr₂O₇^2- ion changes to yellow when treated with an alkali. Why?
(b) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons.
Answer:
(i) Transition metals form a large number of complexes due to:

• The small size of atoms and ions of transition metals.
• High nuclear charge.
• Presence of incompletely filled d-orbitals.

(ii) As the oxidation state increases the size of ion goes on decreasing thus the covalent character increases as a result of this amphoteric and acidic strength increases. While in case of lower oxides of transition metals ionic size increases and thus basic character increases.

(iii) Because Mn²⁺ has 3d⁵ as a stable oxidation state which is half-filled and stable. Mn has very high third ionization energy for change from d⁵ to d⁴ but in case of Cr³⁺, 3d³ is more stable due to completely half-filled t_2g orbitals (crystal field spitting theory) and that is why Mn³⁺ /Mn²⁺ is highly positive as compared to Cr³⁺ /Cr²⁺.

(b) Both Lanthanoids and Actinoids have the t³ oxidation state and both show contraction or irregular electronic configuration while the major difference between the lanthanoids and actinoids is actinoids are radioactive while lanthanoids are not radioactive in nature.
OR
(a) (i) In p block elements the difference in oxidation state is 2 and in transition elements, the difference is 1.
(ii) Cu⁺ is unstable in aq. solution because it undergoes disproportion reaction and has low hydration enthalpy.
(iii) In alkaline medium dichromate ions Cr₂O₇^2- changes to chromate ion CrO₄^2-, which is yellow in colour due to which the colour changes when treated with an alkali.

(b) Chemistry of actinoids is complicated as compared to lanthanoids due to the following reasons:

• They show multiple oxidation states namely +5, +6 and +7 oxidation states respectively which permits the formation of higher oxidation states through the removal of the periphery electrons.
• They are radioactive and have a strong propensity to form complex reactions because of its unstable isotopes, some actinoids are formed naturally by radioactive decay.

Question 25.
(a) An element has atomic mass 93 g mol^-1 and density 11.5 g cm^-3. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(b) Write any two differences between amorphous solids and crystalline solids. [5]
OR
(a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in an f.c.c. structure.
(Atomic mass of Al = 27 g mol^-1)

(b) Give reasons:

• In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.
• Silicon on doping with phosphorous forms n-type semiconductor.
• Ferrimagnetic substances show better magnetism than antiferromagnetic substances.

Question 26.
(a) Write the product(s) in the following reactions:

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanol and Phenol
(ii) Propanol and 2-methyl propane-2-ol [5]
OR
(a) Write the formula of reagents used in the following reactions:
(i) Bromination of phenol to 2, 4, 6-tribromophenol
(ii) Hydroboration of propene and then oxidation to propanol.

(b) Arrange the following compound groups in the increasing order of their property indicated:
(i) p-nitrophenol, ethanol, phenol (acidic character)
(ii) Propanol, propane, Propanal (boiling point)

(c) Write the mechanism (using curved arrow notation) of the following reaction:

Answer:

(b) (i) Ethanol and phenol: When neutral ferric chloride is added to both the compounds phenol gives violet coloured complex whereas ethanol does not give this complex when treated with ferric chloride solution.
(ii) Propanol and 2-methyl propane-2-ol: When both the solutions were treated with anhydrous ZnCl₂ and conc. HCl (Luca’s test) the 2-methyl propane-2-ol gives the turbidity immediately whereas propanol does not give the turbidity immediately.
OR
(a) (i) Aq-Br₂
(ii) B₂H₆ and then H₂O₂ and OH-

(b) (i) Ethanol < Phenol < p-Nitrophenol.
(ii) Propane < Propanal < Propanol

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – A

Question 1.
Write the structure of 2, 4-dinitrochlorobenzene. [1]
Answer:

Question 4.
Write IUPAC name of the following compound:
CH₃NHCH(CH₃)₂ [1]
Answer:
N-methyl propane-2-amine.

Question 5.
Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number. [1]
Answer:
Chromium belongs to group number 6 and its oxidation state in K₂Cr₂O₇ is +6
i.e., K₂Cr₂O₇
1 × 2 + 2x + (-2 × 7) = 0
⇒ 2 + 2x – 14 = 0
⇒ 2x – 12 = 0
⇒ 2x = 12
⇒ x = 6
Thus, the formula of the oxo-anion is K₂Cr₂O₇

Section – B

Question 7.
Draw the structures of the following: H₃PO₂ [2]
Answer:

Question 8.
Define the following terms:
(i) Ideal solution
(ii) Molarity (M) [2]
Answer:
(i) Ideal solutions are those solutions that obey Raoult’s law over the entire range of concentration.
Example: Benzene and toluene, n-heptane and n-hexane.

(ii) Molarity is defined as the number of moles of solute dissolved per liter of solution.

Question 9.
Complete the following reactions: [2]
(i) Cl₂ + H₂O →
(ii) XeF₆ + 3H₂O →
OR
What happens when
(i) conc. H₂SO₄ is added to Cu?
(ii) SO₃ is passed through water?
Write the equations.
Answer:
(i) Cl₂ + H₂O → 2HCl + [O]
(ii) XeF₆ + 3H₂O → XeO₃ + 6HF
OR
(i) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
(ii) SO₃ + H₂O → H₂SO₄

Question 10.
Write the reactions involved in the following:
(i) Hell-Volhard-Zelinsky reaction
(ii) Decarboxylation reaction [2]
Answer:
(i) Hell-Volhard Zelinsky reaction:

(ii) Decarboxylation reaction:

Section – C

Question 13.
Write the principles of the following methods:
(i) Vapour phase refining [3]
Answer:
(i) Vapour phase refining: It is based on the principle that the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal.

Question 15.
Define the following:
(iii) Disinfectants [3]
Answer:
(iii) Disinfectants: These are the substances that are applied to non-living objects to destroy microorganisms that are present on the objects.
Example: 1-2% Phenol solution.

CBSE Previous Year Question Papers Class 12 Chemistry 2017 Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – A

Question 4.
Write the structure of 3-Bromo-2-methylprop-1-ene. [1]
Answer:
BrCH₂(CH₃)C = CH₂

Question 5.
Write IUPAC name of the following compound:
(CH₃)₂N-CH₂CH₃) [1]
Answer:
N, N-dimethylethanolamine

Section – B

Question 6.
Write the reactions involved in the following reactions:
(i) Clemmensen reduction
(ii) Cannizzaro reaction [2]
Answer:
(i) Clemmensen reduction:

(ii) Cannizzaro reaction:

Question 7.
Draw the structures of the following:
(i) H₄P₂O₇
(ii) XeOF₄
Answer:

Question 8.
Define the following terms:
(ii) van’t Hoff factor (i) [2]
Answer:
(ii) van’t Hoff factor(i): It is defined as the extent of dissociation or association or the ratio of the observed colligative property to the calculated colligative property.

Question 10.
Complete the following chemical equations: [2]
(i) F₂ + 2Cl^– →
(ii) 2XeF₂ + 2H₂O →
OR
What happens when
(i) HQ is added to MnO₂?
(ii) PCl₅ is heated?
write the equations involved.
Answer:
(i) F₂ + 2Cl^– → 2F^– + Cl₂
(ii) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
OR
(i) MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
(ii) PCl₅ → PCl₃ + Cl₂

Section – C

Question 14.
Write the structures of the monomers used for getting the following polymers:
(i) Nylon-6
Answer:
(i) Monomers of Nylon-6:

Question 19.
Write one difference between each of the following:
(i) Multimolecular colloid and Macromolecular colloid
(ii) Sol and Gel
(iii) O/W emulsion and W/O emulsion [3]
Answer:
(i) In multimolecular colloids, a large number of atoms or smaller molecules of a substance aggregates together to form species having a size in the colloidal range.
Example: Sulphur sol whereas in macro-molecular colloids the colloidal particles are large molecules having colloidal dimensions.
Example: Starch.

(ii) In sol the dispersing phase is solid and dispersing medium is liquid;
Example: paint, gold sol etc., whereas in Gel the dispersing phase is liquid and dispersing medium is solid;
Example: Jelly, butter etc.

(iii) In O/W emulsion, oil is the dispersed phase while water is the dispersion medium
Example: milk, vanishing cream etc whereas in W/O emulsion water is the dispersed phase while oil is the dispersion medium.
Example: Cold cream, butter etc.

Question 20.
(i) What type of isomerism is shown by the complex [Co(en)₃]Cl₃?
(ii) Write the hybridisation and magnetic character of [Co(C₂O₄)₃]^3- (At. no. of Co = 27)
(iii) Write IUP AC name of the following Complex [Cr(NH₃)₃Cl₃]. [3]
Answer:
(i) Since the given coordinate compound does not have a plane of symmetry and the ligand attached is bidentate ligand so it will show optical isomerism.

(ii) [Co(C₂O₄)₃]^3- Co is in +3 oxidation state with an electronic configuration of 3d⁶ Oxalate is a strong field ligand so the pairing of electrons takes place.

(iii) Triamminetrichloridochromium(III).

CBSE Previous Year Question Papers

The post CBSE Previous Year Question Papers Class 12 Chemistry 2017 Delhi appeared first on Learn CBSE.

CBSE Previous Year Question Papers Class 12 Maths 2017 Delhi

Time allowed: 3 hours
Maximum marks : 100

General Instructions:

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.
• All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted. You may ask for logarithmic tables, if required.

CBSE Previous Year Question Papers Class 12 Maths 2017 Delhi Set I

Section – A

Question 1.
If A is a 3 × 3 invertible matrix, then what will be the value of k if det (A^-1) = (det A)^k. [1]
Solution:
Given, A is an invertible matrix.
∴ A. A^-1 = I
⇒ det (A.A^-1) = det (I)
⇒ det (A).det(A^-1) = 1 [ ∵ det(I) = 1]
⇒ det (A), (det A)^k = 1 [∵ det(A^-1)=(det A)^k]

on comparing both sides, we get
k = -1

Question 2.
Determine the value of the constant ‘k’ so that the function f(x) = is continuous at x = 0. [1]
Solution:
Given, that the function is continuous

Question 3.
Evaluate: . [1]
Solution:

Question 4.
If a line makes angles 90° and 60° respectively with the positive directions of X and Y-axes, find the angle which it makes with the positive direction of Z-axis. [1]
Solution:
We know that:
l² + m² + n² = 1 …(i)
and l = cos α, m = cos β, n = cos γ
Given, α = 90°, β = 60°
∴ cos α = cos 90° = 0 and cos β = cos 60° =

Section – B

Question 5.
Show that all the diagonal elements of a skew symmetric matrix are zero. [2]
Solution:

Hence, all the diagonal elements of a skew, symmetric matrix are zero (as diagonal elements are : a₁₁; a₂₂, …….. a_nn). Hence Proved.

Question 6.
Find at x = 1, y = if sin² y + cos xy = K. [2]
Solution:
Given, sin² y + cos xy = K
Differentiating both sides w.r.t x, we get

Question 7.
The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of increase of its surface area, when the radius is 2 cm. [2]
Solution:
Let V be the volume and r be the radius of sphere at any time t.

∴ Rate of increase of surface area of the sphere is 3 square centimetre per second.

Question 8.
Show that the function f(x) = 4x³ – 18x² + 27x – 7 is always increasing on R. [2]
Solution:
Given, f(x)= 4x³ – 18x² + 27x – 7
Differentiating f(x) w.r.t. x, we get
f'(x) = 12x² – 36x + 27 = 3(4x² – 12x + 9)
= 3(2x – 3)² for any x ϵ R
3 > 0 and (2x – 3)² ≥ 0
∴ f'(x) ≥ o
⇒ The fimction is always increasing on R.

Question 9.
Find the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line 5x – 25 = 14 – 7y = 35z. [2]
Solution:
Given line is 5x – 25 = 14 – 7y = 35z

∴ Vector equation of the line which passes through the point A (1, 2, – 1) and its direction ratio are proportional to 7, – 5, 1 is

Question 10.
Prove that if E and F are independent events, then the events E and F’ are also independent. [2]
Solution:
Since E and F are independent events :
P(E ∩ F) = P (E) . P (F) …(i)
P(E ∩ F’) = P (E) – P (E ∩ F)
= P (E) – P (E) P (F) [From (i)]
= P(E)(1 – P(F))
= P (E) P (F’)
∴ E and F’ are also independent. Hence Proved.

Question 11.
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300. Formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit ? It is being given that at least one of each must be produced. [2]
Solution:
Let the manufacturer produces x pieces of necklaces and y pieces of bracelets.
Since total number of necklaces and bracelets that can be handle per day are 24.
so, x + y ≤ 24 …(i)
To make bracelet one needs one hour and half an hour is need to make necklace and maximum time available is 16 hours
so, x + y ≤ 16 ….(ii)
Now, let Z be the profit and we have to maximize it, so our LPP will be
Maximize Z = 100x + 300y
Subject to constraints :
x + y ≤ 24
x + y ≤ 16
or x + 2y ≤ 32
and x, y ≥ 1

Question 12.
Find: . [2]
Solution:

Section – C

Question 13.
Prove that [4]

Solution:

Question 14.
Using properties of determinants, prove that: [4]

Solution:

OR

find the matrix D such that CD – AB = 0.
Solution:
Let D be the matrix of order 2 × 2,

Question 15.
Differentiate the function (sin x)^x + sin^-1 with respect to x.
Solution:


OR
IF x^myⁿ = (x + y)^m+n, prove that = 0.
Solution:

Question 16.
Find [4]

Solution:

Question 17.
Evaluate: [4]

Solution:


OR
Evaluate:
Solution:

Question 18.
Prove that x² – y² = c(x² +y²)² is the general solution of the differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy, where c is a parameter. [4]
Solution:
Given differential equation is,

Question 19.
Let then:
(a) Let c₁ = 1 and c₂ = 2, find c₃ which makes coplanar.
(b) If c₂ = – 1 and c₃ = 1, show that no value of c₁ can make coplanar. [4]
Solution:

Question 20.
If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to . Also, find the angle which makes with . [4]
Solution:

Question 21.
The random variable X can take only the values 0, 1, 2, 3. Given that P (X = 0) = P (X = 1) = p and P(X = 2) = P(X = 3) such that Σp_ix_i² = 2Σp_ix_i , find the value of p. [4]
Solution:
Given, P(X = 0) = P(X = 1) = p and P(X = 2) = P (X = 3)
Let P (X = 2) = P(X = 3) = k