CBSE Previous Year Question Papers Class 12 Chemistry 2017 Outside Delhi

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.