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NCERT Exemplar Problems Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

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NCERT Exemplar Problems Class 11 Chemistry Chapter  12 Organic Chemistry: Some Basic Principles and Techniques

Multiple Choice Questions
Single Correct Answer Type
Q1. Which of the following^ the correct IUPAC name?
(a) 3-Ethyl-4,4-dimethylheptane
(b) 4,4-Dimethyl-3-ethylheptane
(c) 5-Ethyl-4,4-dimethylheptane
(d) 4,4-Bis(methyl)-3-ethylheptane
Sol: (a) While writing IUPAC name, the alkyl groups are written in alphabetical order. Thus lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-1

Q2. The IUPAC name for
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-2

(a) 1-Hydroxypentane-l, 4-dione
(b) 1,4-Dioxopentanol
(c) l-Carboxybutan-3-one
(d) 4-Oxopentanoic acid

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-3

When more than one functional group lie in the main chain, nomenclature is done according to that functional group which has higher priority. Carboxylic acid (-COOH) has more priority than the keto group (>C = O).
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-4

(a) 1 -Chloro-2-nitro-4-methylbenzene
(b) l-Chloro-4-methyl-2-nitrobenzene
(c) 2-Chloro-1 -nitro-5-methylbenzene
(d) m-Nitro-p-chlorotoluene
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-5

Q4. Electronegativity of carbon atoms depends upon their state of hybridization. In which of the following compounds, the carbon marked with asterisk is most electronegative?
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-6

Sol: (c) Electronegativity increases as the state of hybridization changes from sp3 to sp2 and sp2 to sp. Thus, sp hybridized carbon has the highest electronegativity.

Q5. In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Sol: (c) Alkyl halides do not show functional isomerism. Alcohols and ethers, aldehydes and ketones, cyanides and isocyanides are functional isomers.

Q6. The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is
(a) distillation
(b) crystallisation
(c) distillation under reduced pressure
(d) steam distillation.
Sol: (d) Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.

Q7. During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check  the ink used at two different places. According to you, which technique can give the best results?
(a) Column chromatography (b) Solvent extraction
(c) Distillation (d) Thin layer chromatography
Sol:(d) Thin layer chromatography (TLC) is used to separate the pigments present in ink.

Q8. The principle involved in paper chromatography is
(a) adsorption (b) partition
(c) solubility (d) volatility
Sol:(b) In paper chromatography, separation of the components of a mixture depends upon their partitioning between water held in the stationary phase (i.e. adsorbent paper) and the liquid present in the mobile phase.

Q9. What is the correct order of decreasing stability of the following cations?

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-7
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-8

(a) 2-Ethyl-3-methylpentane
(b) 3,4-Dimethylhexane
(c) 2-sec-Butylbutane
(d) 2,3-Dimethylbutane

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-9

Q11. In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-10
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-11

Q12. Ionic species are stabilized by the dispersal of charge. Which of the following carboxylate ions is the most stable?

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-12

Sol: (d) In all the given carboxylate ions, the negative charge is dispersed which stabilizes these ions. Here, the negative charge is dispersed by two factors, i.e., +R-effect of the carboxylate ion (conjugation) and -I-effect of the halogens.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-13

It is evident from the above structures that +R-effect is common in all the four ions. Therefore, overall dispersal of negative charge depends upon the number of halogen atoms and electronegativity. Since F has the highest electronegativity and two F-atoms are present in option (d), thus, dispersal of negative charge is maximum in option (d).

Q13. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
H3C-HC = CH2 + H+→ ?

(a) 2°Carbanion                                    
(b) 1° Carbocation
(c) 2° Carbocation
(d) l°Carbanion

Sol: (c) When the electrophile attacks CH3 – CH = CH2, delocalisation of electrons can take place in two possible ways
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-14
As 2° carbocation is more stable than 1° carbocation, the first addition is more feasible.

Q14. Covalent bonds can undergo fission in two different ways. The correct representation involving the heterolytic fission of CH3 – Br is

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-15

Sol: (b) Arrow denotes the direction of movement of electrons

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-16
Since, Br is more electronegative than carbon, hence heterolytic fission occurs in such a way that CH3 gets the positive charge and Br gets the negative charge. Thus, option (b) is correct.

Q15. The addition of HC1 to an alkene proceeds in two steps. The first step is the attack of H+ ion to >C = C< portion which can be shown as

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-17

Sol: (b) Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, n electrons of the double bond attack the proton.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-18

More than One Correct Answer Type
Q16. Which of the following compounds contain all the carbon atoms in the same hybridization state?
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-19
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-20

Q17. In which of the following representations given below spatial arrangement of group/atom is different from that given in structure ‘A’?

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-21

Sol: (a, c, d) Different groups are present towards and away from the observer.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-22
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-23
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-24

(a), (c) and (d) have different (anticlockwise) spatial arrangements of atoms or groups of atoms than that given in structure A (clockwise).
Q18. Electrophiles are electron seeking species. Which of the following groups contains only electrophiles?
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-25
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-26
Q19. Which of the above pairs are position isomers?
(a) I and II
(b) II and III
(c) II and IV
(d) III and IV

Sol: (b) II and III are position isomers as they differ in the position of -C- group.

Q20. Two or more compounds having same the molecular formula but different functional groups are called functional isomers. Which of die following pairs are not functional group isomers?
(a) II and III (b) II and IV
(c) I and IV (d) I and II
Sol: (a, c) (a) II and III have the same functional group.
(c) I and IV have the same functional group.

Q21. Nucleophile is a species that should have
(a) a pair of electrons to donate
(b) positive charge
(c) negative charge
(d) electron deficient species
Sol: (a, c) Nucleophile (nucleus-loving) is a chemical species that donates an, electron pair to an electrophile (electron-loving). Hence, a nucleophile should
have either a negative charge or an electron pair to donate. Thus, options (a) r and (c) are correct.

Q22. Hyperconjugation involves delocalization of .
(a) electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsatUrated system.
(b) electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.
(c) π-electrons of carbon-carbon bond.
(d) lone pair of electrons.

Sol: (a, b) Hyperconjugation, also known as sigma-pi conjugation, is the delocalization of sigma electrons. Presence of -H with respect to double bond, triple bond or carbon containing positive charge (in carbonium ion) or unpaired electron (in free radical) is a condition required for hyperconjugation.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-27
Short Answer Type Questions
Note: Consider structures I to VII and answer Questions 23-26.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-28

Q23. Which of the above compounds form pairs of metamers?
Sol: V and VI, VI and VII and V and VII form pairs of metamers because they have different alkyl chains on either side of ethereal oxygen (-O-).

Q24. Identify the pairs of compounds which are functional group isomers.
Sol: Compounds I to IV, i.e., alcohols, and V to VII, i.e., ethers, are functional group isomers with molecular formula C4H10O and different functional groups (-OH in I to IV and -O- in V to VII).
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; HI and V, IH and VI; IH and VII; IV and V, IV and VI, IV and VH are functional group isomers.

Q25. Identify the pairs of compounds that represent position isomerism.
Sol:  I and II, III and IV, VI and VII are position isomers due to different positions of -OH group and -O- group.

Q26. Identify the pairs of compounds that represent chain isomerism.
Sol: I and III, I and IV, II and III, II and IV.

Q27. For testing halogens in an organic compound with AgN03 solution, sodium extract (Lassaigne’s extract) is acidified with dilute HN03. What will happen if a student acidifies the extract with dilute H2S04 in place of dilute HN03?
Sol: On adding dilute H2S04 for testing halogens in an organic compound with AgN03, white precipitate of Ag2S04 is formed. This will interfere with the test of chlorine and this Ag2S04 may be mistaken for white precipitate of chlorine as AgCl. Hence, dilute HN03 should be used instead of dilute H2S04.

Q28. What is the hybridization of each carbon in H7C = C = CH7?
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-29

Q29. Explain how is the electronegativity of carbon atoms related to their state of hybridization in an organic compound?
Sol: Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
sp3 – 25% s-character, 75% P-character
sp2 – 33% s-character, 67% P-character                                             –
sp – 50% s-character, 50% P-character
Hence, the order of electronegativity is sp3 < sp2 < sp

Q30. Show the polarization of carbon-magnesium bond in the following structure.
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-30

Q31. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-31

Sol: They show position isomerism.

Q32. Which of the following selected chains is correct to name the given compound according to IUPAC system?

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-32

Sol: The 4 carbon chain is correct*according to the IUPAC system since it contains both the functional groups. The other three carbon chains are incorrect since none of them contains both the functional groups.

Q33. In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.
Sol: No. DNA and RNA have nitrogen in the heterocyclic rings which cannot be removed as ammonia.

Kjeldahl method cannot be used to estimate nitrogen present in rings, azo and nitro groups as nitrogen present in these cannot be converted into ammonium sulphate.

Q34.If a liquid compound decomposes at its boiling point, which method(s) will you choose for its purification? It is known that the compound is stable at low pressure, steam volatile and insoluble in water.
Sol: Steam distillation can be used for its purification. This method is applied to separate substances which are steam volatile and immiscible with water. Note: Answer Questions 35-38 on the basis of information given below: “Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom, involvement of neighbouring groups in hyperconjugation and resonance.”

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-33

Q36. Which of the following ions is more stable? Use resonance to explain your Answer 

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-34
Sol: Structure (A) is more stable due to resonance. Structure (B) is non-planar and hence it does not undergo resonance. Double bond is more stable within the ring in comparison to outside the ring.

Q37. The structure of triphenylmethyl cation is given here. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation
Sol: Triphenylmethyl cation is a tertiary carbocation which can show nine possible canonical structures and hence is very stable.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-35

Q38. Write the structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations. in order of increasing stability.
Sol: 2-Methylbutane has four possible carbocations

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-36
Stability of carbocation increases in the order 1° < 2° < 3°. Out of I and IV, IV is more stable than I because in IV, CH3 group is at a-carbon and in I, it is at β-carbon and +I-effect decreases with distance

Q39. Three students, Manish, Ramesh and Rajni, were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (LE) independently by the fusion of the compound with sodium metal. Then they added solid FeS04 and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained Prussian blue colour bit Ramesh got red colour. Ramesh repeated the test with the same Lassaigne’s extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation? Also, write the chemical equations to explain the formation of compounds of different colours.

Sol: In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of

ferri ferrocyanide.

6NaCN + FeS04→ Na4[Fe(CN)6] + Na2S04

3Na4[Fe(CN)6] + 2Fe2(S04)3 → Fe4[Fe(CN)6]3 + 6Na2S04
In compounds containing nitrogen and sulphur together, the sodium metal should be in slight excess otherwise in Lassaigne’s test, sodium thiocyanate (NaCNS) is formed which gives red colour with Fe3+ ions and decomposes as follows:

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-37

On the basis of above results, it is clear that Ramesh used less sodium and hence NaSCN was formed in the Lassaigne’s extract which gave red colouration due to Fe(SCN)3 formation while Manish and Rajni used excess sodium and hence, NaCN was formed in the Lassaigne’s extract which gave Prussian blue colour of Fe4[Fe4(CN)6].

Q40. Name the compounds whose line formulae are given below:

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-38

Q41. Write structural formulae for compounds named as
(a) 1-Bromoheptane
(b) 5-Bromoheptanoic acid

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-39

Q42. Draw the resonance structures of the following compounds:
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-40
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-41

Q43. Identify the most stable species in the following set of ions giving reasons:
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-42
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-43

Q44. Give three points of differences between inductive effect and resonance effect.
Sol: 

Inductive effect Resonance effect
(1) This effect involves

displacement of σ-electrons.

The resonance involves displacement of π-electrons or lone pair of electrons.
(2) It operates in saturated compounds. It operates only in unsaturated conjugated systems.
(3) This effect moves up to three •          carbon atoms and becomes

negligible from fourth carbon atom onwards.

This effect moves all along the length of the conjugated system.
(4) This effect causes slight drift of σ-electrons towards the more electronegative atom and hence only partial charges (δ + and δ -) are developed. This results in complete transfer of electrons and hence full +ve and -ve charges are developed.

Q45. Which of the following compounds will not exist as resonance hybrid? Give reason for your answer.

(a) CH3OH
(b) R-CONH2
(c) CH3CH = CHCH2NH2
Sol: (a) CH3OH does not contain -electrons, hence, it cannot exist as resonance hybrid.
(b) Due to the presence of -electrons in C = O bond and lone pair of electrons on N, amide can be represented by the following resonating structures.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-44
(c) CH3CH = CHCH2NH2: Since lone pair of electrons on nitrogen atom is not conjugated with the -electrons, therefore, resonance is not possible.

Q46. Why does S03 act as an electrophile?
Sol: Three highly electronegative oxygen atoms are attached to sulphur atom. This makes sulphur atom electron deficient. Due to resonance, sulphur also acquires positive charge. Both these factors make S03 an electrophile.

ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-45

Q47. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-46
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-47
Q48. By mistake, an alcohol (boiling point 97°C) was mixed with a hydrocarbon (boiling point 68°C). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
Sol: Simple distillation can be used because the two compounds have a difference of more than 20° in their boiling points and therefore, both the liquids can be distilled without any decomposition. ‘

Q49. Which of the two structures (A) and (B) given below is more stabilized by resonance? Explain.
ncert-exemplar-problems-class-11-chemistry-chapter-12-organic-chemistry-some-basic-principles-48
Matching Column Type Questions
In the following questions more than one correlation is possible between options of column I and II. Make as many correlations as you can.
Q50. Match the type of mixture of compounds in Column I with the technique of separation/purification given in column II.

Column I Column II
(a) Two solids which have different solubilities in a solvent and which do not undergo reaction when dissolved in it. (1) Steam distillation
(b) Liquid that decomposes at its boiling point (2) Fractional distillation
(c) Steam volatile liquid (3) Simple distillation
(d) Two liquids which have boiling points close to each other (4) Distillation under reduced pressure
(e) Two liquids with large difference in boiling points. (5) Crystallisation

Sol: (a → 5), (b → 4), (c→1), (d → 2), (e →3)

Q51. Match the terms mentioned in Column I with the terms in Column II.

Column I Column II
(a) Carbocation (1) Cyclohexane and 1-hexene
(b) Nucleophile (2) Conjugation of electrons of C – H σbond with empty p-orbital present at adjacent positively charged carbon.
(c) Hyperconjugation (3) sp2 hybridised carbon with empty p-orbital
(d) Isomers (4) Ethyne
(e) sp hybridization (5) Species that can receive a pair of electrons
(f) Electrophile (6) Species that can supply a pair of electrons

Sol: (a →  3); (b →  6); (c→  2); (d→  1); (e →  4); (f →  5)

Column I Column II Explanation
(a) Carbocation sp2-hybridised carbon with empty p-orbital H3C+ is carbocation. Loss of e makes its p-orbitals empty (sp2– hybridised carbon)
(b) Nucleophile Species that can supply a pair of electrons Nucleus loving, i.e., having negative charge or excess of electrons
(c) Hyperconjugation Conjugation of electrons of C – H σbond with empty p-orbital present at adjacent positively charged carbon
(d) Isomers Cyclohexane and 1-hexene Same molecular formula but different structures
(e) sp hybridization Ethyne HC ≡ CH (sp hybridization)
(f) Electrophile Species that receive a pair of electrons Electron loving, i.e., positive charge or lack of electrons


Q52. Match Column I with Column II.

Column I Column II
(a) Dumas method (1) AgN03
(b) Kjeldahl method (2) Silica gel
(c) Carius method (3) Nitrogen gel
(d) Chromatography (4) Free radicals
(e) Homolysis (5) Ammonium sulphate

 

Sol: (a → 3); (b → 5); (c→ 1); (d →2); (e → 4)

Column I Column II Explanation
(a) Dumas method Nitrogen gel Used for N containing compounds
(b) Kjeldahl method Ammonium

sulphate

Nitrogen converts to ammonium sulphate
(c) Carius method AgN03 Compound is heated in the presence of AgN03
(d) Chromatography Silica gel Adsorbent used is silica gel
(e) Homolysis              ‘ Free radicals Free radicals are formed by homolytic fission

 

NCERT Exemplar Problems Class 11 Chemistry

 

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques appeared first on Learn CBSE.

Food Where Does It Come From – CBSE Notes for Class 6 Science

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Food Where Does It Come From – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Let us see what Mrs. Iyer and Mrs. Kapoor have prepared. Have they prepared the same kind of food? List out the six food items that you see on their table.
• Need for food
• Food from plants
• Food from animals
• What do animals eat
food-come-cbse-notes-class-6-science-1
We get all these food from both plants and animals. Thus, plant parts and animal products are our sources of food. 1. vada , 2. idli , 3. chapathi , 4. chicken curry , 5. Fish curry , 6. Rice .

Food From Plants:

Green plants are known as producers because they prepare their own food. They use light, air (carbon dioxide), water, and chlorophyll (present in their leaves) to prepare their food by the process of photosynthesis.
Different plant parts serve as sources of food for us. Fruits, vegetables, cereals, and pulses that we eat are obtained from different parts of a plant.
Roots Roots of plants like carrot, radish, turnip, sweet potato, and beetroot are eaten (Fig. 1.1).
food-come-cbse-notes-class-6-science-2
Stems Stems of certain plants are eaten. For example, the stem of sugarcane plant is eaten and is also used to make sugar. The stem and flower of banana plant is cooked and eaten in different parts of India. Certain plants have underground stems that we eat. Examples are potato, onion, garlic, and ginger (Fig. 1.2).
food-come-cbse-notes-class-6-science-3
Leaves Leaves of plants like lettuce, spinach, cabbage, coriander, mint, and basil are eaten (Fig. 1.3).
food-come-cbse-notes-class-6-science-4
Flowers Flowers of certain plants like cauliflower, broccoli, and banana are also eaten (Fig. 1.4).
food-come-cbse-notes-class-6-science-5
Seeds Pulses like mung bean, kidney bean, chickpea, and cereals (wheat, maize, and rice) that we eat are seeds of plants (Fig. 1.5).
food-come-cbse-notes-class-6-science-6
Wheat grains are ground to make flour (atta) which is used to make chapattis. Cumin seeds, pepper, and cardamom that we eat as spices are also seeds of different plants.
Sprouted seeds (or sprouts) of mung bean and chickpea (Bengal gram) are very nutritious. Sprouting involves soaking seeds and draining the water and then leaving them till they germinate. Sprouts can be eaten raw as salads or cooked.
Fruits and vegetables Plants also provide us fruits and vegetables (Fig. 1.6).
food-come-cbse-notes-class-6-science-7
Like different plant parts, animal too serve as sources of food. Let us learn about the main food products obtained from animals.

Food From Animal:

Animal products like meat, egg, honey, milk, cheese, butter, and curd are eaten by human beings.
Meat Meat of animals like goat, chicken, fish, and prawns is commonly eaten (Fig. 1.7).
food-come-cbse-notes-class-6-science-8
Egg Hen’s egg is the most common bird egg eaten in the world. It is a rich source of proteins and vitamins. Some people also eat eggs of goose and duck.
Honey: Honey is a sweet liquid made by bees from the nectar of flowers (Fig. 1.8).
food-come-cbse-notes-class-6-science-9
Honey is collected from beehives. It is used in cooking and also has medicinal value.
Milk: Milk is obtained from animals like cow, buffalo, and goat. Fig’1-8 Honev It is a very nutritious food item and is a rich source of proteins. Milk also contains calcium, which is required for proper bone growth and nerve function (Fig. 1.9).
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Products made from milk are known as dairy products. Some common dairy products are discussed below.
Paneer (cottage cheese): Common methods of making paneer include adding lemon juice or vinegar to milk. Then, the liquid portion of milk is drained off and the solid part forms paneer. This process is called curdling.
Cheese: Cheese is made from curdled milk of cow, goat, sheep, or buffalo (Fig. 1.10).
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Cream: Cream is made by collecting the top fatty layer of the milk.
Butter: Butter is made by churning fresh cream.
Ghee: Ghee is made by gently heating butter and removing the solid matter.
Curd Common methods of making curd include adding a small sample of curd in warm milk. The microorganisms (bacteria) present in the curd sample turn the milk into curd.
Unlike green plants, animals cannot make their own food. They depend on plants and other animals for food.
People living in deserts also drink camel’s milk. In ice-cold places, people mostly have yak’s milk.
Word help
Microorganisms Tiny organisms that can be seen only with the help of a microscope
Let’s Remember
Write two examples for each of the following.
1. Roots that we eat
2. Stems that we eat
3. Leaves that we eat
4. Flowers that we eat
5. Seeds that we eat

What Do Animals Eat:

Different animals have different feeding habits. Based on their feeding habits, animals can be divided into three groups: herbivores, carnivores, and omnivores.
Herbivores
Herbivorous animals (Fig. 1.11) or herbivores (herbi, plant; vore, eater) are those that eat only plants and plant products. Cow, deer, horse, giraffe, squirrel, and butterfly are examples of herbivores.
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Special Characteristics of Herbivores
• Herbivores like cow, horse, and goat have wide, blunt teeth. Such teeth are suitable for pulling plants off the ground and grinding them.
• Herbivores like cow and camel have the ability to bring back previously swallowed food to the mouth for chewing it the second time. This helps them to absorb most of the nutrients from hard-to-digest food like grass.
• Squirrels have a pair of broad, sharp-edged front teeth (incisors) in each jaw They use these teeth to gnaw food items like nuts.
• Herbivores like butterfly and hummingbird do not need to worry about chewing
their food. They have mouth-parts shaped like a straw to suck nectar from flowers.
Carnivores
Carnivorous animals (Fig. 1.12) or carnivores (carni, meat; vore, eater) are those that only eat the flesh of other animals. Lion, tiger, jackal, vulture, owl, eagle, snake, and spider are examples of carnivores.
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Special Characteristics of Carnivores
• Carnivores like lion and tiger have sharp and pointed front teeth (canines). They also have sharp claws and powerful jaws which help them to tear flesh.
• Carnivorous birds like eagle have curved, pointed beaks that allow them to tear flesh.
• Carnivores like chameleon and frog have a long, sticky tongue that they use to catch insects.
• Carnivorous fish like shark have several small, sharp teeth that help them bite off chunks of flesh.
Omnivores
Omnivorous animals (Fig. 1.13) or omnivores (omni, all; vore, eater) are those that eat both plants and flesh of other animals. Bear, raccoon, crow, and human beings are examples of omnivores.
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Special Characteristics of Omnivores
• Omnivores like bear and human beings have different types of teeth that help them to eat both plants and flesh of other animals.
• Omnivorous birds like crow have sharp and pointed beak to help them eat a variety of food.
Scavengers and Decomposers
Instead of hunting live animals, some birds and animals eat the flesh of other animals that are already dead.
Vulture is one such bird. These animals or birds are called scavengers. Some other organisms feed on and destroy (or decompose) dead plants and animals. Fungi
and bacteria are examples of such organisms (Fig. Fjg 114 Fung.1.14).
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These organisms are called decomposers. Together with scavengers, decomposers play a very important role in nature. Without these organisms, our planet would be covered with dead plants and animals.

Key Words:
Herbivore An animal that eats only plants and plant products is called a herbivore.
Carnivore An animal that eats only the flesh of other animals is called a carnivore.
Omnivore An animal that eats both plants and flesh of other animals is called an omnivore.
Scavenger An animal that eats only the flesh of animals that are already dead is called a scavenger.
Decomposer An organism that feeds on and decompose dead animals and plants is called a decomposer.

Summary:
• Different plant parts like root, stem, leaf, flower, and fruit serve as sources of food.
• Animal products like meat, egg, honey, milk, curd, cheese, butter, and ghee are eaten by human beings.
• Herbivores have wide blunt teeth that help them to grind and chew plants.
• Carnivores have sharp teeth and claws that help them to tear flesh.

The post Food Where Does It Come From – CBSE Notes for Class 6 Science appeared first on Learn CBSE.

Components of Food – CBSE Notes for Class 6 Science

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Components of Food – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Food is essential for all animals, including human beings. The food that we eat contains different components.
Look at the picture of food items given below. Write the names of components of food that you think are present in the food items. Write your answers in the spaces provided.
components-food-cbse-notes-class-6-science-1
Let us learn more about the various components of food and their importance. 1.Vitamins , 2.proteins, 3.proteins, 4.carbohydrates.

Components Of Food:

The food that we eat consists of different components or nutrients. Nutrients are substances that are needed by our body for proper growth and healthy body function. There are six main components present in food: carbohydrates, fats, proteins, vitamins, minerals, and roughage or dietary fibres. These nutrients fulfill different needs of the body.
Carbohydrates:
Carbohydrates provide energy to your body, which keeps it going throughout the day. There are two major types of carbohydrates in food: sugar and starch.
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Sugars: Sugar is also called simple carbohydrate. Fruits, honey, and table sugar are some sources of sugar.
Starch: Starch is also called complex carbohydrate. Plants store energy in the form of starch. Rice, wheat, corn, potato, and bread are some sources of starch. When we eat plant products, containing sugar and starch, our digestive system breaks them down into glucose. This glucose, which is the simplest form of sugar is then absorbed into the blood and provides us energy.
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Proteins:
Proteins are needed by our body for muscle¬building and repairing worn-out tissues. Our muscles, organs, and even blood are made up of mostly proteins. If we do not eat proteins, our body will not be able to repair damaged cells, or build new ones. Proteins in our diet come from both animal and plant sources (Fig.2.2).
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Meat, fish, egg, and milk are some animal sources of proteins. Pulses, soyabeans, grams, and nuts are some plant sources of proteins.
Vitamins:
Vitamins are needed for the proper functioning of our body. They help in keeping our eyes, bones, teeth, and gums healthy. There are 13 vitamins, each of which has a specific function. Vitamins are of two types: fat-soluble and water-soluble.
Fat-soluble vitamins Vitamins A, D, E, and K are fat-soluble vitamins. These are stored in the fat tissues of our body and are used only when the body needs them.
Our body prepares vitamin D in the presence of sunlight.
Water-soluble vitamins Vitamins Bl, B2, B3, B6, B12, and folic acid (together known as vitamin B complex) and vitamin C are water-soluble vitamins. Since water-soluble vitamins are not stored in the body, these need to be regularly supplied through food items like citrus fruits, spinach, and other green leafy vegetables, etc. Lack of vitamins in the body can cause deficiency diseases.
Minerals:
Just like vitamins, minerals also help our body to stay healthy. Minerals perform important functions like formation of bones, teeth, and blood cells and helps in maintaining a normal heartbeat. Minerals are of two types: macrominerals and trace minerals (Fig. 2.3).
Macrominerals {macro: large) are needed by the body in larger amounts as compared to trace minerals. Calcium, magnesium, sodium, and potassium are examples of macrominerals.
Trace minerals are needed by the body in very small amounts. Iron, zinc, copper, and iodine are examples of trace minerals.
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Roughage or Dietary Fibres:
The portion of the plant food that do not provide any nutrients to our body but help in maintaining a healthy digestive system is called roughage or dietary fibres (Fig. 2.4).
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Roughage are of two types: soluble and insoluble. Soluble roughage are soluble in water whereas insoluble roughage are not. Apple, strawberry, peach, and rice are examples of food items rich in soluble roughage that help in blood circulation. Whole grain, carrot, cabbage, turnip, and cauliflower are examples of food items rich in insoluble roughage. Lack of insoluble roughage in the diet causes the stool to become hard and difficult to pass. This condition is called constipation.
Water:
Almost 70% of our body weight is water. Water is needed by our body for good health.
• It helps to transport substances inside our body.
• It helps our body to absorb nutrients from food.
• It helps to regulate our body temperature.
• It is needed for various chemical reactions that take place inside our body during digestion, excretion, etc.
We get water not only from the liquids we drink but also from the food we eat. Milk, fruits, vegetables, and juices are good sources of water.

Balanced Diet:

Our diet must contain adequate amount of different nutrients for our body to function properly.
A diet that contains adequate amount of different nutrients required for the healthy functioning of our body is called a balanced diet.
A balanced diet must include food items from the following four food groups.
• Milk group: includes milk and milk products
• Meat group: includes meat (chicken, fish, lamb, etc.) and meat substitutes (beans, peas, nuts, and seeds)
• Fruit and vegetable group: includes fruits and vegetables
• Grain group: includes breads and cereals.

Deficiency Diseases:

Lack of carbohydrates, proteins, vitamins, or minerals in the diet can cause diseases. Diseases that are caused due to the lack of nutrients in the diet are called deficiency diseases. Deficiency diseases cannot be transmitted from one person to another.
Deficiency of Carbohydrates:
Carbohydrates are the main energy sources. Lack of carbohydrates in the diet results in lack of energy and stamina. A labourer who does hard manual work needs more carbohydrates in his diet than a person who does his work sitting in his office.
Deficiency of Proteins:
Growing children need more proteins in their diet. Lack of proteins in the diet weakens muscles.
Deficiency of proteins leads to a disease called kwashiorkor (fig. 2.5). Deficiency of proteins along with carbohydrate deficiency is called Protein Energy Malnutrition (PEM). It leads to marasmus. These diseases are more common in children of rural areas.
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A child suffering from kwashiorkor has some or all of the following symptoms: large pot-like belly, stunted growth, swelling of face and limbs (especially the feet), skin diseases, mental retardation, and diarrhoea. If the treatment is started in time, improving protein intake may correct this disease.
Marasmus is more common among infants and children under 5 years of age. A child suffering from marasmus becomes very thin, shows slow body growth, lack of energy, loss of appetite, weak legs, mental retardation, poor muscle development, etc.
To prevent these diseases, the Government of India has started programmes like the Integrated Child Development Scheme (ICDS) and the Mid-day Meal scheme in schools.
Deficiency of Vitamins:
Table 2.1 lists some important vitamins and their sources, functions, deficiency diseases, and symptoms.

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Some vitamins are very sensitive to heat and light. For example, vitamin C is easily
destroyed during cooking. Therefore, vitamin C-rich food items should be eaten raw.
Scurvy was common among sailors in ancient times. Due to lack of cure, several sailors died of this disease during long voyages. In the 18th century, James Lind found that eating citrus fruits reduced the occurrence of scurvy in sailors.
Deficiency of Minerals:
Table 2.2 lists some important minerals and their sources, functions, deficiency diseases, and symptoms. Minerals also assist in certain chemical reactions in the body. Cooking does not destroy them.
Table 2.2 Minerals-Functions, deficiency diseases, symptoms, and sources
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Deficiency of Water in the Body:
Water is very essential for proper functioning of our body. Excess loss of water from the body leads to a condition called dehydration.
Dehydration causes loss of salts and leads to weakness in the body. Oral Rehydrating Solution (ORS) can be given to the patient to recover from dehydration. It is available free of cost at primary health centres. It can also be made at home by mixing 8 teaspoons of sugar and 1 teaspoon of salt in 1 litre of clean, drinking water.

Key Words :

Saturated fats Fats that are normally solid at room temperature are called saturated fats.
Unsaturated fats Fats that are normally liquid at room temperature are called unsaturated fats.
Fat-soluble vitamins Vitamins that are stored in the fat tissue and used only when the body needs them are called fat-soluble vitamins.
Water-soluble vitamins Vitamins that are not stored in the body and need to be regularly supplied through food are called water- soluble vitamins.
Macrominerals Minerals that are needed by the body in larger amounts are called macrominerals.
Trace minerals Minerals that are needed by the body in very small amounts are called trace minerals.
Dietary fibre The portion of plant food that does not provide any nutrients to our body but help in maintaining a healthy digestive system is called dietary fibre.
Balanced diet A diet that contains adequate amount of different components of food required for healthy functioning of the body is called a balanced diet.
Deficiency diseases Diseases that are caused due to the lack of nutrients in the diet are called deficiency diseases.

Summary :

• Carbohydrates, proteins, fats, vitamins, minerals, and dietary fibres are the main components of food.
• Carbohydrates and fats provide energy to the body.
• Proteins are needed for muscle-building and for repairing worn-out tissues.
• Vitamins and minerals are needed for the normal functioning of our body.
• A balanced diet should include food items from four basic food groups.
• Deficiency of carbohydrates causes lack of energy and stamina.
• Deficiency of proteins causes kwashiorkor whereas combined deficiency of proteins and carbohydrates causes marasmus.
• Deficiency of vitamins can cause night blindness, beriberi, anaemia, scurvy, and rickets.
• Deficiency of water can cause dehydration.
• Deficiency of minerals can cause osteoporosis, rickets, anaemia, and goitre.

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Separation of Substances – CBSE Notes for Class 6 Science

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Separation of Substances – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Methods Of Separation:

Different methods are used for separating different substances that are mixed together. Let us learn about some common methods that are used.
Threshing:
Grains or seeds of plants like rice and wheat serve as sources of food. The flour (atta) that is used for making chapattis is made from wheat grains. After these crops have been harvested or cut, the grains need to be separated from the stalks (the dried stems). This is done by threshing.
The process of beating harvested crops to separate the grains from the stalks is called threshing. It is done manually (by hand) or with the help of machines. Manual threshing is done by holding a pile of crop and beating it on a rock or a hard surface (Fig. 3.1). This loosens and separates the grain from the stalk. Sometimes, threshing is also done by crushing the harvested stalks using bullocks.
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Threshing is also done with the help of machines like the combine harvester (Fig. 3.2). Threshed grains may still contain seed coverings and tiny pieces of leaves or stem (collectively called chaff). These are separated by winnowing.
Winnowing:
The method used to separate chaff from the grain by wind or blowing air is called winnowing.
The mixture of chaff and grain is taken in a winnowing basket (Fig. 3.3). The farmer stands at a higher level and lets the mixture fall to the ground.
The grain, being heavier, falls almost vertically whereas the lighter chaff is carried away by the wind and forms a separate heap away from the grain.
The separated chaff is used as fodder for cattle. The direction of the wind plays an important role in the process of winnowing.
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Hand-picking:
Rice, wheat, pulses, etc., that we buy from the market may contain impurities (unwanted or harmful particles) in the form of small stones, unwanted grains, etc. Often, these impurities look very different from the food item and can be spotted easily. The method of separation used in such a case is hand-picking (Fig. 3.4). This method is preferred when
• the quantity of the mixture is small,
• the unwanted substance is present in smaller quantities, and
• the size, shape, or colour of the unwanted substance is different from that of the useful one.
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Sieving:
If the components of a mixture are of different sizes, they can be separated by sieving (Fig. 3.5). The smaller component passes through the pores of the sieve whereas the larger component (stones or husk) is left behind in it. This method is used in some homes to separate wheat bran (the bigger particles) from flour.
However, sieving wheat flour is not advisable as wheat bran, which is removed during sieving, is very rich in nutrients and is also a rich is better to remove visible impurities by hand picking.
The process of sieving is also used to separate pebbles and stones from sand at construction sites. The stones and pebbles present in the mixture remain in the sieve and the fine sand particles pass through the holes of the sieve.
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Sedimentation and Decantation:
Have you seen pulses being washed in your home? When pulses are kept in a bowl of water, they settle down as they are heavy. However, dirt, insects, tiny pieces of straw, and other lighter impurities float at the top. The water, which contains these impurities, is then poured out and discarded. This process involves two methods: sedimentation and decantation.
The process of separating insoluble solids, suspended in a liquid, by allowing them to settle down is called sedimentation.
The solid particles that settle down during sedimentation are called sediments.
The process of pouring out the clear upper liquid without disturbing the sediments is called decantation.
The liquid above the sediments is called a supernatant.
A mixture of sand and water can also be separated by sedimentation and decantation.
The mixture is left undisturbed for some time.
Sand, being heavier, settles down and water is poured out into a separate container.
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Filtration:
The process by which two substances (an insoluble solid and a liquid) are separated by passing the mixture through a filtering device is called filtration.
Filtration is commonly used in our homes. For example, after preparing tea, we filter out the tea leaves using a strainer. Filtration is also done to remove pulp from fresh fruit juice. Water may also contain solid impurities, which can be removed by filtration.
During filtration, the insoluble solid is retained in the filtering device whereas the liquid passes through it. It is important that the particles of the insoluble solid are bigger than the holes in the filtering device for them to be retained in it. A filter paper is a filtering device that has very fine pores in it.
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Condensation:
The process in which gas changes into liquid is called condensation. Condensation is the opposite of evaporation. In nature, water vapour in the air condenses to form its liquid form, the dew. Condensation takes place only when water vapour hits a cold surface.

Solution And Solubility:

When some salt is added to water and stirred, the salt disappears.
This is because the salt has dissolved in the water.
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Dissolving is a change where substances mix completely with the liquid they have been added to.
Not all substances dissolve in water. Only some substances,Salt dissolves in water.
like salt and sugar, dissolve in water and are known as soluble substances. Substances like chalk and sand do not dissolve in water and are known as insoluble substances.
The substance that dissolves is called the solute and the substance in which the solute dissolves is called the solvent. The resulting mixture is called the solution. Thus, solute + solvent = solution.
E.g., sugar + water = sugar solution.
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If we keep adding spoonfuls of sugar to water and stir the solution each time, what will happen after some time? We will notice some grains of sugar at the bottom of the solution. This shows that no more sugar can be dissolved. We say that the solution has become saturated (Fig. 3.7).
A saturated solution is the solution in which no more of the solute can be dissolved.
But what if we heat the solution? Can we then dissolve that ‘extra’ sugar present in the saturated solution?
Yes, we can increase the solubility of a solute by heating the solution. Solubility is the ability of a substance to get dissolved in a given liquid. The quantity of a substance that can dissolve in hot water is much more as compared to that in cold water.
There are some other factors that increase the solubility of a solute.
Stirring We can observe this by taking two glasses of water and adding a spoonful of sugar to each glass. Then we keep one glass undisturbed and stir the other. Sugar dissolves faster when the solution is stirred.
Solute in powdered form We can observe this by taking two glasses of water and adding a whole sugar cube in one glass and powdered or crushed sugar cube in the other. Sugar in the powdered form dissolves first.
Different substances dissolve in different amounts of water while making a saturated solution.

Key Words :

Threshing The process of beating harvested crops to separate seeds from the stalks is called threshing.
Winnowing The method used to separate chaff from the grain by wind or blowing air is called winnowing.
Sedimentation The process of separating insoluble solids suspended in a liquid by allowing them to settle down is called sedimentation.
Decantation The process of pouring out the clear upper liquid without disturbing the sediments is called decantation.
Filtration The process by which an insoluble solid is separated from a liquid by passing the mixture through a filtering device is called filtration.
Saturated solution A solution that can dissolve no more of the solute is called a saturated solution.

Summary :

• Threshing is done either manually or by using machines to separate seeds or grains from the stalks.
• Winnowing involves separating the chaff from the grain by letting the mixture fall to the ground from a height when the wind is blowing.
• Hand-picking involves manually removing small stones, insects, etc. from the grains.
• Sedimentation and decantation are used to separate an insoluble solid from a liquid.
• Insoluble solid impurities present in water can be removed by filtration.
• Common salt can be separated from seawater by evaporation.
• Solubility of a solute can be increased by heating the mixture or it can also be increased by adding the solute in the powdered form.

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Fibre to Fabric – CBSE Notes for Class 6 Science

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Fibre to Fabric – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Clothes are made of different materials. We get these materials from both plants and animals.
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Identify the materials given below as plants or animal product. Write P for the plant products and A for animal products.
Let us learn about how the story of clothing started, the different materials used to make clothes, and how they are made. Answers: Cotton socks, jute rope, silk cloth, lather shoes.

History Of Clothing:

About 30,000 years ago, people started using animal skins for clothing. It is believed that wool was used as early as 6000 years ago.
Domestication of silkworms to produce silk occurred around 3000 BC in China. In India, cotton came into widespread use around 3000 BC. These fabrics were not stitched. They were just wrapped around the body. Even today, sari, dhoti, and turban are unstitched pieces of cloth.
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Fiber And Fabric:

Clothes are made mostly from fibres. Fibres are thin strands of thread, that are woven to make fabric, for example, cotton fabric, silk fabric, etc. The fabric is stitched to make clothes. For example, cotton fabric can be stitched into a cotton frock or a cotton kurta. There are two main processes of making fabric from fibre – weaving and knitting.
Weaving: Weaving involves making fabric by arranging two sets of yarn. It is done using a machine called loom, which can be hand-operated (Fig. 4.1) or power- operated. The pattern in which two sets of threads are arranged in a piece of woven cloth is called a weave (Fig. 4.2).
Knitting: Knitting involves making fabric by forming a series of connected loops of yarn by using knitting needles or machines. Sweaters are made from wool strands by knitting.
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Natural And Synthetic Fibres:

Fibres used to make fabric may be natural or synthetic. Fibres that are obtained from plants or animals are called natural fibres. Examples are cotton, jute, wool, and silk. Fibres that are made by man from chemical substances are called synthetic fibres. Examples are nylon, rayon, polyester, and acrylic. Let us learn more about plant fibres.
Plant Fibres:
Cotton (Fig. 4.3), jute, coir, silk cotton, hemp, and flax are examples of plant fibres. Denim, used to make jeans, is made from cotton.
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Cotton:
The cotton plant is a shrub. It grows well in black soil and warm climate. It needs moderate rainfall. Cotton is a soft fibre that grows around the seeds of the cotton plant. A variety of textile products are made from cotton. In India, ‘lchadi’, a coarse hand-woven cloth, is made from cotton.
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Jute:
Jute is a fibre obtained from the bark of the jute plant (Fig. 4.6). It can be grown in different soil types, ranging from clayey to sandy soil. It grows best in loamy soil (mixture of sand, silt, and clay), sandy soil, and clayey soil. It grows well in regions where it rains a lot. Almost 80% of the world’s high-quality jute comes from Bangladesh. Bangladesh, India, China, Nepal, and Thailand are the main producers of jute.
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Other Useful Plant Fibres:
There are other important plant fibres as well.
Coir: Coir is the fibre obtained from the outer covering or the husk of the coconut. Usually coconuts are left in water for a few months. The husk is then separated from the nut and beaten with wooden mallets to get the fibre. The fibre thus obtained is spun and dyed and is ready for weaving. Coir is used to make several household products like rope and floor covering and also as a stuffing in mattresses and pillows.
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Silk cotton: Silk cotton is another plant fibre that is commonly used as a stuffing in pillow, sleeping bag, and life jacket. This fibre is obtained from the silk cotton tree, also called kapok.
The fruits of the kapok tree contain fibres that are light and fluffy (like cotton). When the fruit ripens, it bursts open, releasing the fibres.
Hemp: Hemp fibres are obtained from the stem of the hemp plant. Hemp fibres are used in the production of ropes, carpets, nets, clothes, and paper.
Flax Fibres obtained from the stem of the flax plant are woven to make a fabric called linen. Flax fibres are also used in the production of rope and high-quality paper.

Key Words :

Fabric The material made by weaving the threads from fibres is called fabric.
Weaving Weaving involves the making of fabric from yarn.
Ginning The process of separating the cotton fibres from its seeds is called ginning.
Spinning The process of making yarn from fibres is called spinning.
Retting The process of rotting the stems of the plants in water to remove the sticky substance and separate fibres is called retting.

Summary :

• Clothing materials are obtained from both plants and animals.
• Fibres are woven to make fabrics and fabrics are stitched to make clothes.
• Fibres may be natural or synthetic.
• Cotton, jute, coir, silk cotton, hemp, and flax are some plant fibres.

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NCERT Exemplar Problems Class 11 Chemistry Chapter 13 Hydrocarbons

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NCERT Exemplar Problems Class 11 Chemistry Chapter  13 Hydrocarbons

Multiple Choice Questions
Single Correct Answer Type
Q1. Arrange the following ia decreasing order of their boiling points.
(A) n-Butane
(B) 2-Methylbutane
(C) n-Pentane
(D) 2,2-Dimethylpropane
(a) A > B > C > D
(b) B > C > D > A
(c) D>C>B>A
(d) C >B>D > A
Sol: (d) As the number of carbon atom increases, boiling point increases. Boiling point decreases with branching.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-1

Q2. Arrange the halogens F2, Cl2, Br2 and I2 in order of their increasing reactivity with alkanes.
(a) I2 < Br2 < Cl2 < F2
(b) Br2 < Cl2 < F2 < I2
(c) F2 < Cl2 < Br2 < I2                               

(d) Br2 < I2 < Cl2 < F2
Sol:
(a) The reactivity order of halogens with alkanes is I2 < Br2 < Cl2 < F2

Q3. The increasing order of reduction of alkyl halides with zinc and dilute HCl is
(a) R-C1<R-I<R-Br
(b) R-Cl<R-Br<R-I
(c) R -1 < R – Br < R – Cl
(d) R-Br<R-I<R-Cl
Sol:(b) The reactivity of reduction bf alkyl halides with Zn/HCl increases as the strength of the C – X bond decreases, i.e., R – Cl < R – Bf < R -I.

Q4. The correct IUPAC name of the following alkane is
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-2

(a) 3,6-Diethyl-2-methyloctane
(b) 5-Isopropyl-3-ethyloctane
(c) 3-Ethyl-5-isopropyloctane
(d) 3-Isopropyl-6-ethyloctane

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-3

Q5. The addition of HBr to 1 -butene gives a mixture of products (A), (B) and (C).
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-4

(C) CH3 – CH2 – CH2 – CH2 – Br
The mixture consists of
(a) (A) and (B) as major and (C) as minor products
(b) (B) as major, (A) and (C) as minor products
(c) (B) as minor, (A) andj(C) as major products
(d) (A) and (B) as minor and (C) as major products.
Sol: (a) The alkene is unsymmetrical, hence will follow Markovnikov rule to give major product.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-5
Since I contains a chiral carbon, it exists in two enantiomers (A and B) which are mirror image of each other.
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-6

Q6. Which of the following will nofshow geometrical isomerism?
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-7

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-8

Q7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
(a) HC1 > HBr > HI
(b) HBr>HI>HCl
(c) HI > HBr > HCl
(d) HCl>HI>HBr
Sol: (c) The decreasing order of reactivity of hydrogen halides with propene is HI > HBr > HC1. As the size of halogen increases, the strength of H – X bond decreases and hence, reactivity increases.

Q8. Arrange the following carbanions in order of their decreasing stability.
(A) H3C-C ≡ C                                        
(B) H-C ≡ C
(C) H3C – CH2

(a) A>B>C
(b) B>A>C
(c) C>B>A
(d) C>A>B

Sol: (b) The order of decreasing stability of carbanions is:

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-9
sp-hybridised carbon atom is more electronegative than sp3-hybridised carbon atom and hence, can accommodate the negative charge more effectively. – CH3 group has +1 effect, therefore, it intensifies the negative charge and, hence, destabilises the carbanion CH3 C = C.

Q9. Arrange the following alkyl halides in decreasing order of the rate of β -elimination reaction with alcoholic KOH.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-10
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-11
is the order of rate of β -elimination with alcoholic KOH.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-12

More the number of β-substituents (alkyl groups), more stable alkene will be formed on β -elimination and more will be the reactivity. Thus, the decreasing order of the rate of β -elimination reaction with alcoholic KOH is: A > C > B.

Q10. Which of the following reactions of methane is incomplete combustion?
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-13

Sol: (c) Dining incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Thus,

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-14

More than One Correct Answef Type
Q11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions?

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-15
Reactions in which methane does not undergo complete combustion to give carbon dioxide and water or incomplete combustion to give carbon and water are controlled oxidation reactions.

Q12. Which of the following alkenes on ozonolysis gives a mixture of ketones only?
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-16

Sol: (c, d) Alkenes which have two substituents on each carbon atom of the double bond give a mixture of ketones on ozonolysis. Thus, options (c) and (d) give mixture of ketones.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-17

Q13. Which are the correct IUPAC names of the following compound?
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-18
(a) 5-Buty 1-4-i sopropyldecane
(b) 5-Ethyl-4-propyldecane
(c) 5-sec-Butyl-4-iso-propyldecane
(d) 4-( 1 -Methylethyl)-5-( 1 -methylpropyl)decane

Sol: (c,d)

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-19

Q14. Which are the correct IUPAC names of the following compound
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-20

(a) 5-(2′,2′-Dimethylpropyl)decane
(b) 4-Butyl-2,2-dimethylnonane
(c) 2,2-Dimethyl-4-pentyloctane
(d) 5-neo-Pentyldecane
Sol: (a,d)
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-21

Q15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring ;
(a) deactivates the ring by inductive effect
(b) deactivates the ring by resonance
(c) increases the charge density at ortho and para-positions relative to meta-position by resonance
(d) directs the incoming electrophile to meta-position by increasing the
charge density relative to ortho and para-positions. ‘
Sol: (a, c) For an electrophilic substitution reaction, the presence of halogen atom in the benzene ring deactivates the ring by inductive effect and increases the charge density at ortho- and para-position relative to meta-position by resonance. –
When chlorine is attached to benzene ring, chlorine being more electronegative pulls the electrons because of its -1-effect. The electron cloud of benzene becomes less dense. Thus, chlorine makes the benzene ring in aryl halide somewhat deactivated. But due to resonance, the electron density on ortho- and para-positions is greater than on meta-position.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-22

Q16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group ________ ‘
(a) deactivates the ring by inductive effect
(b) activates the ring by inductive effect
(c) decreases the charge density at ortho- and para-positions of the ring relative to meta-position by resonance
(d) increases the charge density at meta-position relative to the ortho and para-positions of the ring by resonance
Sol: (a, c) Nitro group by virtue of-I-effect withdraws electrons from the ring and increases the charge and destabilizes carbocation.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-23

In ortho, para-attack of electrophile on nitrobenzene, we are getting two structures (A) and (B) in which positive charge is appearing on the carbon atom directly attached to the nitro group.
As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.

Q17. Which of the following are correct?
(a) CH3 – O – CH+2 is more stable than CH3 – CH+2
(b) (CH3)2CH+ is less stable than CH3 – CH2 – CH+2
(c) CH2 = CH – CH+2 is more stable than CH3 – CH2 – CH+2
(d) CH2 = CH+ is more stable than CH3 – CH+2
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-24

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-25

Q18. Four structures are given in options (a) to (d). Examine them and select the aromatic structures.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-26

Sol: (a, c) In both these options, rings are planar and follow (4n + 2)π-electrons rule
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-27

Cyclooctatetraene is non-planar and has 8π-electrons. It is not aromatic. Cyclopropenyl anion is planar but has 4 π -electrons. It is not aromatic.

Q19. The molecules having dipole moment are________ .
(a) 2,2-Dimethylpropane                      
(b) trans-Pent-2-ene
(c) cw-Hex-3-ene            
(d) 2,2,3,3-Tetramethylbutane
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-28
Since, the +1 effect of CH2CH3 group is higher than that of CH3 group, therefore, the dipole moments of C-CH3 and C-CH2CH3 bonds are unequal. Although these two dipoles oppose each other, yet they do not exactly cancel out each other and hence trans-2-pentene has small but finite dipole moment.
In cis-hex-3-ene, although the dipole moments of the two C – CH2CH3 bond are equal, but they are inclined to each other at an angle of 60° and hence have a finite dipole moment.

Short Answer Type Questions
Q20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reaction? Explain.
Sol: Due to the presence of a π -electron cloud above and below the plane of alkenes and arenes, these are electron rich molecules and, therefore, provide sites for the attack of electrophiles. Hence, they undergo electrophilic reactions. Alkenes undergo electrophilic addition reactions because they are unsaturated molecules. For example,

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-29
Arenes, on the other hand, cannot undergo electrophilic addition reactions. This is because benzene has a large resonance energy of 150.4 kJ mol-1. During electrophilic addition reactions, two new σ-bonds are formed but the aromatic character of benzene gets destroyed and, therefore, resonance energy of benzene ring is lost. Hence, electrophilic addition reactions of arenes are not energetically favourable. Arenes, in contrast, undergo electrophilic substitution reactions in which σ C – H bond is broken and new σ C – X bond is formed: The aromatic character of benzene ring is not destroyed and benzene retains its resonance energy. Hence, arenes undergo electrophilic substitution reactions.

Q21. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will butene formed on the reduction of but-2-yne show geometrical isomerism?
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-30

Thus, but-2-ene is capable of showing geometrical isomerism.

Q22. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
Sol: Ethane contains carbon-carbon sigma (σ) bond. Electron distribution of the sigma molecular orbital is symmetrical around the intemuclear axis of the C – C bond which is not disturbed due to rotation about its axis. This permits free rotation around aC-C single bond. However, rotation around a C – C single bond is not completely free. It is hindered by a small energy barrier due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. The energy difference between the two extreme forms is of the order of 12.5 kJ mol-1, which is very small. It has not been possible to separate and isolate different conformational isomers of ethane.

Q23. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why? .
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-31
In staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon-hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability.

Q24. The intermediate carbocation formed in the reactions of HI, HBr and HC1 with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol-1,363.7 kJ mol1 and 296.8 kJ mol-1 What will be the other of reactivity of these halogen acids?
Sol: The bond dissociation enthalpy decreases in the order HC1 > HBr > HI, therefore, the order of reactivity is in the reverse order i.e., HI > HBr > HCl.
Q25. What will be the product obtained as a result of the following reaction and why?

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-32
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-33

Propyl chloride forms CH3 – CH2 – CH+2 with anhydrous A1C13 which is less stable. This rearranges to a more stable carbocation as:
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-34

Q26. How will you convert benzene into
(i) p-nitrobromobenzene (ii) m-nitrobromobenzene
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-35
Q27. Arrange the following set of compounds in the order of their decreasing . relative reactivity with an electrophile. Give reason.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-36

Sol: The methoxy group (-OCH3) is electron releasing group. It increases the electron density in beniene nucleus due to

resonance effect (+R-effect). Hence, it makes anisole more reactive than benzene towards the electrophile.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-37

In case of alkyl halides, the electron density increases at ortho and para positions due to +R effect. However, the halogen atom also withdraws electrons from the ring because of its -I effect. Since the -I effect is stronger than the +R effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult.
-N02 group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong -R-effect and strong -I-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-38

Q28. Despite their -I effect, halogens are o- andp-direction in haloarenes. Explain.
Sol: In case of aryl halides, halogens are little deactivating because of their strong -I effect. Therefore, overall electron density on the benzene ring decreases. In other words, halogens are deactivating due to -I effect. However, because of the +R-effect, i.e., participation of lone pairs of electrons on the halogen atom with the π-electrons of the benzene ring, the electron density increases more at o- and p-positions than at m-positions.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-39
As a result, halogens are o-, p-directing. The combined result of +R-effect and -I-effect of halogens is that halogens are deactivating but o, p-directing.

Q29. Why does the presence of a nitro group-make the benzene ring less reactive in comparison to the unsubstituted benzene ring? Explain.
Sol: Nitro group is an electron withdrawing group (-R and -I effects). It deactivates the ring by decreasing nucleophilicity for further substitution.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-40

Q30. Suggest a route for the preparation of nitrobenzene starting from acetylene.
Sol:
Acetylene when passed through red hot iron tube at 500°C undergoes cyclic polymerisation to give benzene which upon nitration gives nitrobenzene.

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-41

Q31. Predict the major product(s) of the following reactions and explain their formation.
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-42
Sol: Addition of HBr to unsymmetrical alkenes follows Markonikov rule. It states that negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
Mechanism: Hydrogen bromide provides an electrophile, H+, which attacks the double bond to form carbocation as shown below:
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-43
Addition reaction of HBr to unsymmetrical alkenes in the presence of peroxide follows anti-Markovnikov rule.
Mechanism: Peroxide effect proceeds via free radical chain mechanism as given below:

ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-44
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-45

Q32. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-46
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-47

Q33. The relative reactivity of 1°, 2° and 3° hydrogens towards chlorination is 1: 3.8 : 5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.
ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-48
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ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-50
Q34. Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane and 2-iodopropane.
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ncert-exemplar-problems-class-11-chemistry-chapter-13-hydrocarbons-52

NCERT Exemplar Problems Class 11 Chemistry

 

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NCERT Exemplar Problems Class 11 Chemistry Chapter 14 Environmental Chemistry

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NCERT Exemplar Problems Class 11 Chemistry Chapter  14 Environmental Chemistry

 

Multiple Choice Questions
Single Correct Answer Type
Q1. Which of the following gases is not a greenhouse gas?
(a) CO
(b) 03                        
(c) CH4                    
(d) H20 vapour
Sol:
(a) The gases which absorb solar energy near the earth’s surface and then radiate it back to the earth are called greenhouse gases.

Q2. Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.
(a) N02
(b) 03
(c) S02                                                       

(d) Unsaturated hydrocarbon

Sol: (c) The smog which is formed in the presence of sunlight is called photochemical smog. This occurs in the months of summer when N02 and hydrocarbons are present in large amounts in the atmosphere. Concentration of 03, PAN, aldehydes and ketones increases up in the atmosphere. S02 is not responsible for photochemical smog.

Q3. Which of the following statements is not true about classical smog?
(a) Its main components are produced by the action of sunlight on emissions of automobiles and factories.
(b) Produced in cold and humid climate.
(c) It contains compounds of reducing nature.
(d) It contains smoke, fog and sulphur dioxide.
Sol: (a) Classical smog is initiated by a mixture of S02, particulates and high humidity in the atmosphere in cold conditions. A fog of H2S04 droplets formed condenses on the particulates to form the smog. It is of reducing nature. The gases released by automobiles and factories are not responsible for classical smog.

Q4. Biochemical Oxygen Demand, /BOD) is a measure of organic material present in water. BOD value less than 5 ppm indicates a water sample to be
Sol: (a) The total amount of oxygen consumed by micro-organisms (bacteria) in decomposing organic matter present in certain volume of a sample of water is called Biochemical Oxygen Demand (BOD) of the water.
Water is considered to be pure if it has BOD less than 5 ppm, whereas highly polluted water has BOD more than 17 ppm. Thus, the water having BOD less than 5 ppm is rich in dissolved oxygen.

Q5. Which .of the following statements is wrong? . ‘
(a) Ozone is not responsible for greenhouse effect.
(b) Ozone can oxidize sulphur dioxide present in the atmosphere to sulphur trioxide.
(c) Ozone hole is thinning of ozone layer present in stratosphere.
(d) Ozone is produced in upper stratosphere by the action of UV rays on oxygen .

Sol: (a) Ozone is also one of the greenhouse gases.

Q6. Sewage containing organic waste should not be disposed in water bodies because it causes major water pollution. Fishes in such a polluted water die because of
(a) large number of mosquitoes.
(b) increase in the amount of dissolved oxygen. .
(c) decrease in the amount of dissolved oxygen in water.
(d) clogging of gills by mud.
Sol: (c) Organic waste consumes oxygen and therefore, dissolved oxygen in water decreases and fish in such polluted water die.

Q7. Which of the following statements about photochemical smog is wrong?
(a) It has high concentration of oxidizing agents.
(b) It has low concentration of oxidizing agent.
(c) It can be controlled by controlling the release of N02, hydrocarbons, ozone etc.
(d) Plantation of some plants like pinus helps in controlling photochemical
Sol: (b) Photochemical smog is oxidizing in nature because it contains large concentration of oxidizing agents N02 and 03.

Q8. The gaseous envelope around the earth is known as atmosphere. The lowest layer of this is extended up to 10 km from sea level. This layer is .
(a) Stratosphere
(b) Troposphere
(c) Mesosphere
(d) Hydrosphere
Sol: (b) The atmosphere is divided into four major regions:
(i) Troposphere (ii) Stratosphere . (iii) Mesosphere and (iv) Thermosphere Troposphere is the lowest region of the atmosphere.

Q9. Dinitrogen and dioxygen are main constituents of air but these do not react with each other to form oxides of nitrogen because____________ .
(a) the reaction is endothermic and requires very high temperature.
(b) the reaction can be initiated only in presence of a catalyst.
(c) oxides of nitrogen are unstable.
(d) N2 and 02 are unreactive,
Sol: (a) Major compounds of atmosphere are dinitrogen, dioxygen and water vapour.
N2 = 78.08%, 02 = 20.95%
Both dinitrogen and dioxygen do not react with each other as nitrogen is an inactive gas. The triple bond in N2 is very stable and its dissociation energy is very high. Both react with each other at very high temperature.

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-1

Q10. The pollutants which come directly in the air from source are called primary pollutants. Primary’pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants?
(a) CO
(b) Hydrocarbon
(c) Peroxyacetyl nitrate                       
(d) NO
Sol: (c) Hydrocarbons present in atmosphere combine with oxygen atom produced by the photolysis of N02 to form highly reactive intermediate called free  radical. Free radical initiates a series of reactions.Peroxyacetyl nitrate is formed, which Gan be said as secondary pollutant.
ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-2

Q11. Which of the following statements is correct?
(a) Ozone hole is a hole formed in stratosphere from which ozone oozes out.
(b) Ozone hole is a hole formed in the troposphere from which ozone oozes out.                                                                         .
(c) Ozone hole is thinning of ozone layer of stratosphere at some places.
(d) Ozone hole means vanishing of ozone layer around the earth completely.
Sol: (c) Two types of compounds have been found to be the most responsible for depleting the ozone layer. These are:
(i) NO and (ii) Chlorofluorocarbons.

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-3

Q12. Which of the following practices will not come under green chemistry?
(a) If possible, making use of soap made of vegetable oils instead of using synthetic detergents.
(b) Using H202 for bleaching purpose instead of using chlorine based bleaching agents.
(c) Using bicycle for traveling small distances instead of using petrol/diesel based vehicles.
(d) Using plastic cans for neatly storing substances.
Sol: (d) Plastic is non-biodegradable polymer. Hence, it does not come under green chemistry. Green chemistry includes processes which lead to minimum pollution and less harm to the environment.

More than One Correct Answer Type

Q13. Which of the following conditions shows the polluted environment?
(a) pH of rain water is 5.6.
(b) Amount of carbon dioxide in the atmosphere is 0.03%.
(c) Biochemical oxygen demand is 10 ppm.
(d) Eutrophication
Sol: (c, d) Polluted water may contain nutrients for the growth of algae, which covers the water surface and reduces the oxygen concentration in water. This leads to anaerobic condition, accumulation of obnoxious decay and animal death. This is process of eutrophication.

The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of sample of water is called Biochemical Oxygen Demand. Clean water would have BOD value of 5 ppm whereas highly polluted could have BOD value of 17 ppm or more.

Normally rain water has pH of 6 due to H+ ion formed by reaction of rain water with carbon dioxide in the atmosphere. When the pH of the rain water drops below 5.6, it is called acid rain.

Q14. Phosphate containing fertilizers cause water pollution. Addition of such compounds in water bodies causes
(a) enhanced growth of algae
(b) decrease in amount of dissolved oxygen in water
(c) deposition of calcium phosphate
(d) increase in fish population
Sol:
(a, b) Fertilizers containing phosphate present in water help in the excessive growth of aquatic plants and algae. Micro-organisms which decompose these plants consume oxygen. As a result, the amount of dissolved oxygen in the water decreases.

Q15. The acids present in acid rain are________ .
(a) Peroxyacetylnitrate                        
(b) H2C03
(c) HN03                                                  

(d) H2S04

Sol: (b, c, d) C02 is slightly soluble in water forming carbonic acid.
 Co2 + H2O → H2Co3
The oxides of nitrogen undergo oxidation reaction followed by reaction with water vapours to form nitric acid.

2NO + 02→ 2N02
2N02 + H20 → HN03 + HNO2

The oxidation of S02 into S03 occurs in the presence of dust particles or metal ions. The S03 then reacts with water vapours to form H2S04.

 ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-4

 

Q16. The consequences of global warming may be .
(a) increase in average temperature of the earth
(b) melting of Himalayan Glaciers –
(c) increased biochemical oxygen demand
(d) eutrophication
Sol: (a, b) The rate at which solar radiation is reaching the earth is constant but the amount of C02 in the air is increasing. Consequently, the heat radiated back to the earth will increase and the temperature of the earth surface will also increase.
This increase in temperature will disturb the thermal balance on the earth and could cause glacier  and ice caps to melt.

Short Answer Type Questions
Q17. Greenhouse effect leads to global warming. Which substances are responsible for greenhouse effect?
Sol: The heating of earth due to trapping of radiation is called greenhouse effect. The gases such as C02, CH4, N20, CFC13, CF2C12,03 etc. trap these radiations and are called greenhouse gases.

Q18. Acid rain is known to contain some acids. Name these acids. From where do they come in rain?
Sol: Acid rain contains acids such as HN03, H2S04 and H2C03 (along with small amount of HC1).
HN03 is formed by the oxidation of NO present in air to N02 and N03 and subsequent dissolution in water. H2S04 is formed by the oxidation of S02 present in air to S03 and subsequent dissolution in water.
H2C03 is formed by the dissolution of C02 of the air in water.

Q19. Ozone is a toxic gas and is a strong oxidizing agent, even then its presence in the stratosphere is very important. Explain what would happen if ozone from this region is completely removed.
Sol: Ozone layer acts as a protective umbrella and does not allow the harmful UV radiations to reach the earth’s surface. If ozone is completely removed from the stratosphere, the UV radiations will fall directly on the humans causing skin cancer and on the plants affecting plant proteins.
Q20. Dissolved oxygen in water is v£ry important for aquatic life. What processes are responsible for the reduction of dissolved oxygen in water?
Sol: The discharge of human sewage and organic waste from pulp and paper
industry and presence of leaves, grass, trash etc. in water due to run off result ‘ in phytoplankton growth. The microorganisms which decompose this organic matter need oxygen. Hence, the amount of oxygen in water of lakes etc. decreases.

Q21. On the basis of chemical reactions involved, explain how do chlorofluoro- carbons cause thinning of ozone layer in stratosphere.
Sol: Chlorofluorocarbons are stable compounds. They move to stratosphere by random diffusion. These undergo decomposition in the presence of sunlight to release Cl atoms. These Cl atoms cause catalytic chemical reactions and cause significant depletion of ozone layer as shown below:

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-5
Since the free radicals use ozone and convert it to oxygen, they cause thinning of ozone layer in stratosphere.

Q22. What could be the harmful effects of improper management of industrial and domestic solid waste in a city?
Sol: If domestic waste in a city is not properly managed, it may find its way into . sewers or may be eaten up by the cattle. The non-biodegradable waste like polythene bags, metal scrap etc. choke the sewers. The polythene bags, if swallowed by the cattle, can result into their death. Similarly, if industrial waste is not properly managed, it will cause pollution of the air, soil and water.

Q23.During an educational trip, a student of Botany saw a beautiful lake in a , village. She collected many plants from that area. She noticed that villagers were washing clothes around the lake and at some places, waste material from houses was destroying its beauty. After few years, she visited the same r lake again. She was surprised to find that the lake was covered with algae, stinking smell was coming out and its water had become unusable. Can you explain the reason for this condition of the lake?
Sol: Disposing of waste material and washing clothes in lake water makes the water rich in nutrients like phosphate. It enhances algae growth. Such profuse ‘ algal growth covers the water surface which reduces oxygen concentration in water. This leads to anaerobic conditions with accumulation of dead and decaying water animals, thus, leaving the water with stinking smell and making it unusable.

Q24. What are biodegradable and non-biodegradable pollutants? ,
Sol: Biodegradable pollutants are those which can be decomposed by bacteria. For example, dust particles, sewage, cow dung etc. Non-biodegradable pollutants are those which cannot be decomposed by bacteria. For example, plastic materials, mercury, aluminium, DDT, etc.

Q25. What are the sources of dissolved oxygen in water?
Sol: Sources of dissolved oxygen in water are (i) Photosynthesis (ii) Natural aeration (iii) Mechanical aeration.

Q26. What is the importance of measuring BOD of a water body?
Sol: BOD is the measure of level of pollution caused by organic biodegradable material in terms of how much oxygen will be required to break down the organic material biologically. Clean water would have BOD values less than 5 ppm while highly polluted water could have a BOD value of 17 ppm or more.

Q27. Why does water covered with excessive algal growth become polluted?
Sol: Presence of excessive algal growth shows that water contains a lot of
phosphate due to inflow of fertilizers, etc from the surroundings. Hence, such a sample of water is polluted.

Q28. A factory was started near a village. Suddenly villagers started feeling the presence of irritating vapours in the village and cases of headache, chest pain, cough, dryness of throat and breathing problems increased. Villagers blamed the emissions from the chimney of the factory for such problems. Explain what could have happened. Give chemical reactions for the support of your explanation.
Sol:
The symptoms observed in the villagers show that oxides of nitrogen and sulphur must be coming out of the chimney. This is due to combustion of fossil fuels like coal, oil, natural gas, gasoline, etc. to produce high temperatures at which oxidation of atmospheric nitrogen takes place forming NO and N02:
ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-6
S02 is produced due to combustion of sulphur containing coal and fuel oil or roasting of sulphide ores like iron pyrites (FeS2), copper pyrites (CuFeS2), etc.

Cu2S + 02→2Cu + S02

Q29. Oxidation of sulphur trioxide in the absence of a catalyst is a slow process but this oxidation occurs easily in the atmosphere. Explain how does this happen. Give chemical reactions for the conversion of S02 into S03.
Sol: The presence of particulate matter in polluted air catalyses the oxidation of S02 to S03. The reaction is also promoted by ozone and hydrogen peroxide.

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-7
Q30. From where does ozone come in the photochemical smog?
Sol: When fossil fuels are burnt, nitric oxide and hydrocarbons from unburnt fuels are produced. In sunlight, nitric oxide is converted to nitrogen dioxide. N02  absorbs energy from sunlight and breaks up into NO and free oxygen atoms which are very reactive and combine with 02 to form 03, which reacts with NO to form N02 and 02.

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Q31. How is ozone produced in stratosphere?
Sol: The formation of ozone in the stratosphere takes place in two steps. In the first step, ultraviolet radiation coming from the sun have sufficient energy to split dioxygen into two oxygen atoms. In the second step, the oxygen atoms react with more of dioxygen to form ozone.

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Q32. Ozone is a gas heavier than air. Why does ozone layer not settle down near the earth?
Sol: In stratosphere, the formation of 03 gas goes on continuously, but 03 is also decomposed by UV radiation between 240-360 nm.

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-10

The O-atom reacts with second 03 molecule
03 + 0 -+ 202
Net reaction 203 —> 302
Thus, the reaction forms a delicate balance in which the rate of 03 decomposition matches the rate of 03 formation, i.e., a dynamic equilibrium exists and maintains a constant concentration of 03.

Q33. Sometime ago formation of polar stratospheric clouds was reported over Antarctica. Why were these formed? What happens when such clouds break up by warmth of sunlight?
Sol: In summer season, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms forming chlorine sinks, preventing much ozone depletion, whereas in winter, special type of clouds called polar stratospheric clouds are formed over Antarctica. These polar stratospheric clouds provide surface on which chlorine nitrate gets hydrolysed to form hypochlorous acid. It also reacts with hydrogen chloride to give molecular chlorine.

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When sunlight returns to the Antarctica in spring, the sun’s warmth breaks up the clouds and HOC1, Cl2 are photolysed by sunlight.

ncert-exemplar-problems-class-11-chemistry-chapter-14-environmental-chemistry-12

The chlorine radicals thus formed initiate the chain reaction for ozone depletion.

Q34. A person was using water supplied by Municipality. Due to shortage of water, he started using underground water. He felt laxative effect. What could be the cause?
Sol: The laxative effect is observed only when the concentration of sulphates in water is greater than 500 ppm. Sulphate is harmless at moderate concentration but concentration above 500 ppm produces laxative effects and hypertension.

Matching Column Type Questions

Match the terms given in Column I with the compounds given in Column

Column 1 Column II
(a) Acid rain (1) CHC12-CHF2
(b) Photochemical smog (2) CO
(c) Combination with haemoglobin (3) co2
(d) Depletion of ozone layer (4) so2
(5) Unsaturated hydrocarbons

Sol: (a →3,4); (b → 4, 5); (c → 2); (d → 1)
(a) Acid rain is caused due to oxides of carbon, sulphur and nitrogen.
(b) Photochemical smog is formed by unburnt fuel (unsaturated hydrocarbons). *
(c) Carbon monoxide with haemoglobin is poisonous.
(d) Chlorofluorocarbons (CHC12 – CHF2) cause ozone depletion.

Q36. Match the pollutant(s) in Column I with the effect(s) in Column II.

Column I Column 11
(a) Oxides of sulphur (1) Global warming                        .
(b) Nitrogen dioxide , (2) Damage to kidney
(c) Carbon dioxide (3) ‘Blue baby’ syndrome
(d) Nitrate in drinking water (4) Respiratory diseases
(e) Lead (5) Red haze in traffic and congested areas

Sol: (a → 4); (b → 5); (c →1); (d →3); (e → 2)

(a) Low concentration of sulphur dioxide causes respiratory disease, e.g., asthma, bronchitis etc.
(b) The irritant red haze in traffic and congested places is due to oxides of nitrogen.
(c) The increased amount of C02 in air is mainly responsible for global warming.
(d) Excess’nitrate in drinking water cause methemoglobinemia (blue baby syndrome)
(e) Lead can damage kidney, liver, reproductive system etc.

Q37. Match the activity given in Column I with the type of pollution created by it given in Column II.

Column I Column II
Releasing gases to the atmosphere after burning waste material containing Sulphur. 0) Water pollution
Using           carbamates  as pesticides. (2) Photochemical smog, damage to plant life, corrosion to building material, induce breathing problems, water pollution
Using synthetic detergents for washing clothes. (3) Damaging ozone layer
Releasing gases produced by automobiles and factories in the atmosphere. (4) May cause nerve diseases in human                                     .
Using chlorofluorocarbon compounds for cleaning computer parts. (5) Classical smog, acid rain, water pollution, induce breathing problems, damage to buildings, corrosion of metals

Sol: (a → 5); (b → 4); (c→ 1); (d → 2); (e→ 3)

Q38. Match the pollutants given in Column I with their effects given in Column II.
(a) Sulphur dioxide causes classical smog, acid rain, water pollution, induces breathing problems, causes damage to buildings, corrosion of metals.
(b) Using carbamates as pesticides can cause nerve diseases in humans
(c) Using synthetic detergents for washing clothes causes water pollution.
(d) Unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories cause photochemical smog, damage to plant life, corrosion to building material, induce breathing problems, water pollution.
(e) Chlorofluorocarbons are believed to be the main reason for ozone layer depletion.

Column I Column II
(a) Phosphate fertilizers in water (1) BOD level of water increases
(b) Methane in air (2) Acid rain
(c) Synthetic detergents in water (3) Global warming
(d) Nitrogen oxides in air (4) Eutrophication

Sol: (a → 1,4); (b → 3); (c → 1); (d → 2) ‘
(a) Phosphate fertilizers increase growth of algae, increasing BOD level and causing eutrophication.
(b) Methane oxidises to C02 which causes global warming.
(c) Use of synthetic detergents increases BOD level of water.
(d) Nitrogen oxides present in air combine with water forming nitric acid.

Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q39. Assertion (A): Greenhouse effect was observed in houses used to grow plants and these are made of green glass.
Reason (R): Greenhouse name has been given because glass houses are made of green glass.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (c) There is no scientific relation between greenhouse effect and the given assertion or reason.

Q40. Assertion (A): The pH of acid rain is less than 5.6.
Reason (R): Carbon dioxide present in the atmosphere dissolves in rain water and forms carbonic acid.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (b) In acid rain, pH is less than 5.6. Carbon dioxide dissolves in water to form weak acid.

 H2 + C02→H2C03

Q41. Assertion (A): Photochemical smog is oxidizing in nature.
Reason (R): Photochemical smog contains N02 and 03, which are formed during the sequence of reactions.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (a) Photochemical smog contains N02 and 03; both are oxidizing agents.

Q42. Assertion (A): Carbon dioxide is one of the important greenhouse gases. Reason (R): It is largely produced by respiratory function of animals and plants.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: C02 is produced by respiration of plants and animals; it is a greenhouse gas.

Q43. Assertion (A): Ozone is destroyed by solar radiation in upper stratosphere.
Reason (R): Thinning of the ozone layer allows excessive UV radiations to reach the surface of earth.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (d) Solar radiations never destroy ozone layer.

Q44. Assertion (A): Excessive use of chlorinated synthetic pesticides causes soil and water pollution.
Reason (R): Such pesticides are non-biodegradable.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (a) Chlorine containinginsecticides and pesticides are non biodegradable and they pollute soil and water.

Q45. Assertion (A): If BOD level of water in a reservoir is less than 5 ppm it is highly polluted.
Reason (R): High biological oxygen demand means low activity of bacteria in water.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Sol: (c) High BOD means high activity of Bacteria in water.

NCERT Exemplar Problems Class 11 Chemistry

 

 

 

 


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Sorting Materials Into Groups – CBSE Notes for Class 6 Science

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Sorting Materials Into Groups – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Grouping On The Basic Of Common Properties:

Objects are made of different materials. One material can be used to make different objects. This is possible because different types of materials have different properties. We have to choose materials with the right properties based on what we want to use it for. For example, a chalk made of wood or plastic would be of no use because it cannot be used to write on the blackboard.
Materials have different properties like roughness, lustre, transparency, solubility, flotation, attraction towards a magnet, conduction of heat, and conduction of electricity. Let us study these properties one by one.
Roughness:
Materials can be rough or smooth. Rough materials have bumps or ridges on their surface, which can be felt by touching them. Smooth materials lack these bumps. Examples of rough materials are rocks, sandpaper, and bark of a tree. A glass sheet, flower petals, and surface of an apple are some examples of smooth surfaces (Fig. 5.1).
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Lustre:
Lustre is the shine of a material. All metals in pure state are shiny and said to possess lustre. This property of metals is widely used for making jewellery and other decorative articles. Materials like gold, silver, and bronze have lustre (Fig. 5.2).
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Transparency:
Different materials allow different amounts of light to pass through them depending on a property called transparency. Based on transparency, materials can be of three types: transparent, translucent, and opaque.
Materials that allow all the light to pass through them are called transparent materials (Fig. 5.3). Glass, water, acrylic sheet, and cellophane paper are transparent. Shopkeepers generally prefer to keep items like toffee, biscuit, sweet, etc. in transparent jars so that we can see them easily. Windows are also usually made of glass so that light can pass through and light our rooms.
Materials that allow some light to pass through them are called translucent materials. Oiled paper and coloured glass are translucent materials. Materials that do not allow light to pass through them are called opaque materials. Wood, metal, leaf, stone, and cardboard are opaque materials.
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State:
All substances are made up of matter. Matter exists in three states – solid, liquid, and gas.
Table 5.1 Grouping based on the states of matter
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Solubility:
Different materials have different solubility in water. Based on their solubility, materials can be soluble, insoluble, miscible, or immiscible.
Solid materials that dissolve in water are said to be soluble in water. For example, common salt and sugar. Solid materials that do not dissolve in water are said to be insoluble in water. For example, sand, wood, stone, chalk powder, and wax. Liquids that dissolve in water are said to be miscible in water. For example, alcohol, vinegar, lemon juice, honey, and glycerine. Liquids that do not dissolve in water are said to be immiscible in water. For example, kerosene, coconut oil, and diesel.
Some gases dissolve in water (e.g., carbon dioxide and oxygen). Oxygen dissolved in water is essential for the survival of aquatic organisms. Soft drinks have carbon dioxide dissolved in them. Gases like nitrogen, hydrogen, and helium are insoluble in water.
Flotation:
Certain materials float on water whereas others sink. This property of a material to float on water is called flotation. Generally, materials like wood, leaf, and feather float on water whereas rock and metal sink.
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Attraction towards a magnet:
Materials that are attracted to a magnet are called magnetic materials. This property is called magnetism. Objects made of iron are attracted to a magnet. In addition to iron, nickel and cobalt are also attracted to a magnet.
Conduction of heat:
If you observe the utensils in your kitchen, you will notice that though most of them are made of metals, their handles are made of wood or hard plastic.
sorting-materials-groups-cbse-notes-class-6-science-6
Why aren’t the handles made of metal as well? This is because metals get heated whereas materials like plastic and wood do not (Fig. 5.4). It would be difficult to hold the handles made of metal while cooking.
Materials that allow heat to flow through them are called conductors of heat whereas those that do not allow heat to flow through them are called insulators of heat.
Generally, metals are conductors of heat whereas non-metals like wood, plastic, glass, bamboo, air, and paper are insulators of heat.
Conduction of electricity:
We get electricity in our homes through cables and wires. An electric cable consists of a number of metal wires with or without a plastic covering (Fig. 5.5). The metal wires conduct or transmit electricity whereas the plastic covering do not. Materials that conduct electricity are called conductors. Materials that do not conduct electricity are called insulators. For example, metals are conductors of electricity; wood, air, and plastic are insulators.
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Key Words :

Transparent Materials that allow light to pass through them are called transparent materials.
Translucent Materials that allow some light to pass through them are called translucent materials.
Opaque Materials that do not allow light to pass through them are called opaque materials.
Miscible Liquids that are soluble in water are said to be miscible in water.
Immiscible Liquids that are insoluble in water are said to be immiscible in water.
Magnetic materials Materials that are attracted by a magnet are called magnetic materials.
Conductors Materials that conduct heat or electricity are called conductors.
Insulators Materials that do not conduct heat or electricity are called insulators.

Summary :

• Materials have different properties that make them useful for making different objects.
• An object can be made from different materials.
• Different objects can be made from the same material.
• Materials can be rough or smooth, transparent or opaque, soluble in water or insoluble, can float on water or sink; can be a conductor or insulators of heat or electricity.

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Changes Around Us – CBSE Notes for Class 6 Science

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Changes Around Us – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Reversible And Irreversible Changes:

Changes that occur around us can be broadly categorized as reversible or irreversible depending on whether or not they can be reversed.
Reversible Changes:
Changes that can he reversed are called reversible changes.
What happens to an ice cream if you do not finish it quickly? It melts. Can you change the molten ice cream back into a solid? Yes! Just keep it in the freezer. Molten ice cream can be changed back to its solid form. Thus, melting is a reversible change. Melting of butter and chocolate are also reversible changes (Fig. 6.1).
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What about changes like condensation, freezing, and evaporation of materials? If you take out some ice cubes from the freezer and keep them outside, the ice cubes will absorb heat from the surrounding and melt. When this water (molten ice) is heated for some time, it starts boiling (liquid starts to evaporate) and steam escapes from the container [Fig. 6.2(a)]. Now, if you hold a lid over the container, the steam will again liquify or condense into small droplets of water on coming in contact with the cold lid [Fig. 6.2(b)]. This water can be cooled down further and then kept in the freezer to form ice again [Fig. 6.2(c)]. Thus, the three physical states of water are reversible and can be changed from one state to another by heating or cooling.
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Irreversible Changes:
Changes that cannot be reversed are called irreversible changes.
There are a large number of irreversible changes that take place around us. These result in a new material being produced, which may or may not be useful. Some examples of irreversible changes are given below.
• Ripening of fruits is an irreversible change because it is not possible to get back the raw fruits from ripened or mature ones.
• Blooming of flowers is an irreversible change because flowers cannot change back into buds.
• Milk gets spoiled when not refrigerated, particularly in summer. This is called curdling or souring of milk, which is an irreversible change. Curdling of milk is also done by adding lemon juice to milk for making cottage cheese or paneer.
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• Burning of paper is an irreversible change. A new substance called ash is left or formed after a paper has been burnt. This new substance differs from the paper in its appearance and properties.
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• Cooking of food is an irreversible change because we cannot get back the ingredients in their original form after cooking them. For example, after a cake is baked using flour, egg, milk, chocolate, etc., we cannot get back the ingredients (Fig. 6.3).
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Burning of a candle is often cited as an example of physical change because what we see immediately is melting of wax that solidifies on cooling. However, when a candle burns, the wax is undergoing two changes at the same time: first it melts, and then it burns. What burns is actually melted wax. The melted wax burns on the wick – the wick itself isn’t burning, it is just the wax on it.

Physical And Chemical Changes:

Changes in which no new substances are formed are called physical changes. For example, breaking of a glass (Fig. 6.4), freezing of water, tearing of paper, etc.
Changes in which new substances with different properties are formed are called chemical changes. Cooking of food, burning of substances are chemical changes as entirely new substances are formed. Burning of a candle wax releases carbon dioxide and water vapour (new substances).
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Expansion And Contraction Of Materials:

Some materials expand on heating and some contract on cooling. Heating makes the particles (that form the material) expand or become loose. Cooling makes the particles (that form the material) contract or become tight.
The amount of expansion differs in solids, liquids, Fig- 6-4 Physical change
and gases. Gases expand the most while solids expand the least.
Table 6.1 shows some examples of expansion.
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Cooling does the opposite of heating. Cooling causes a material to contract. Solids contract the least while gases contract the most. Table 6.2 lists some examples of contraction.
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Applications of Expansion and Contraction:
Expansion by heating can be used in several everyday activities.
• The jammed metal lid of a jam jar can be opened by heating. The jar is inverted and just the lid is dipped in hot water. After some time, the lid can be opened easily as the lid gets slightly expanded.
The fact that materials expand on heating is used in thermometers. In many thermometers, mercury is used. When the bulb of the thermometer comes in contact with a hot object, the mercury expands and its level rises in the glass tube, indicating the temperature.
Why the electric lines are never hung tautly between the poles? Wires in the outside
environment are subjected to extreme weather conditions ranging from acute hot to cold temperatures. A taut wire on contraction in winters can snap.

Key Words :

Reversible change: A change that can be reversed is called a reversible change.
Irreversible change: A change that cannot be reversed is called an irreversible change.
Physical change: A change where no new substances are formed is called a physical change.
Chemical change: A change where new substances with different properties are formed is called a chemical change.

Summary :

• Changes happen around us and also within us.
• Some changes are reversible, while some others are irreversible.
• Some changes are physical changes; some are chemical changes.
• Heating causes expansion in a material.
• Cooling causes contraction in a material.
• Gases expand the most and solids expand the least.
• Gases contract the most and solids contract the least.

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Getting to Know Plants – CBSE Notes for Class 6 Science

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Getting to Know Plants – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Root Systems :

There are two main types of root systems: tap root and fibrous root system.
Tap Root System:
In the tap root system (Fig. 8.1), a single root (called the primary root) comes out from the seed after germination. Tap roots are also called true roots.
getting-know-plants-cbse-notes-class-6-science-1
Later, smaller roots called lateral roots branch out from this primary root. Mango, neem, pine, sheesham, pea, carrot, radish, turnip, and beetroot are examples of plants in which tap roots are found.
Fibrous Root System:
Fibrous roots (Fig. 8.2), which grow from the base of the stem have a bushy appearance. These roots are thin and almost equal in size. Grass, maize, wheat, onion, sugarcane, and rice are examples of plants with fibrous roots.
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Functions of Roots:
Some functions of roots are given below:
Anchoring the plant Roots help to anchor the plant firmly into the ground.
Absorption of water and nutrients from the soil They help plants to absorb water and nutrients from the soil, which are essential for their survival.
Desert plants have relatively longer roots because they penetrate deep into the soil in search of water.
Preventing soil erosion They help to bind the soil particles together, thereby preventing them from being carried away by water or wind.
Sometimes roots are modified to perform various other functions like reproduction, nutrition, etc.
Root Modifications:
Roots of some plants are modified to perform additional functions. Let us study some of these modifications and their functions (Fig. 8.3).
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Shoot System:

All parts of a plant that are above the ground form the shoot system. It includes stem, leaf, flower, fruit, etc.
Stem:
The stem is a very important part of the plant.
Functions of a Stem:
• It holds leaves in position and helps them to spread out as the stem and its branches grow. This ensures that they get enough light for photosynthesis.
• It bears flowers, buds, leaves, and fruits.
• It conducts water and mineral salts from the roots to the leaves.
Similarly, it carries the food manufactured by the leaves to other parts of the plant.
• Green stem has chlorophyll and can carry out photosynthesis.
• It has nodes from which leaves arise.
The space between two nodes is called an internode (Fig. 8.4).
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Stem modifications:
Stems of certain plants are modified to perform special functions.
Some of the modifications and their functions are given below.
For storage of water Stems of plants like cactus and jade swell up to store water in them.
To manufacture food Stems of some plants become leaf-like and flattened like that of a cactus and perform photosynthesis.
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For protection Stems may be modified as thorns, like in bougainvillea or may be in the form of hard and sharp prickles, as in rose (Fig. 8.5), to protect the plant from being eaten by animals.
For support Stems of some climbers like grapes and passion flower are modified to form special structures called tendrils (Fig. 8.6). These help the climber plants like, which have weak stems, attach themselves to others for support.
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For storage of food Potato, onion, and ginger are modified stems that store food. There are three kinds of underground stems: tubers (e.g., potato), rhizomes (e.g., ginger), and bulbs (e.g., onion and garlic) (Fig. 8.7).
For multiplication of the plant Rhizomes, bulbs, and tubers also help in the multiplication of plant. Stem cuttings of some plants like rose, jasmine, and hibiscus grow into new plants.
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Leaves:
Leaves are known as food factories of the plant. They arise from at the nodes of the stems and have a characteristic shape and size. Let us study its different parts (Fig. 8.8).
getting-know-plants-cbse-notes-class-6-science-9
The arrangement of veins in a leaf is termed as venation. Venation is of two types: parallel and reticulate. If the veins run parallel to one another from the base to the tip of the leaf, the leaf is said to have parallel venation, e.g., banana and onion (Fig. 8.9). If the veins are arranged in a net-like pattern on both sides of the midrib the leaf is said to have reticulate venation, e.g., peepal and mango (Fig. 8.10).
getting-know-plants-cbse-notes-class-6-science-10
Functions of a leaf:
A leaf performs various important functions for the plant.
* It is usually green due to the presence of a green pigment called chlorophyll. A leaf prepares food for the plants. The process of making food by the plant using carbon dioxide, water, chlorophyll, and light is called photosynthesis.
getting-know-plants-cbse-notes-class-6-science-11
Plants store food in the leaves, fruits, and stems in the form of starch.
• Plants breathe with the help of their leaves. Leaves of most plants have tiny openings called stomata (singular: stoma) (Fig. 8.11) under their surface.
The exchange of gases takes place through the stomata (Fig. 8.12).
• Leaves also lose water through the stomata. The loss of water through the stomata is called transpiration. Transpiration helps the plant in the following ways:
• It helps in cooling the leaves, just as loss of water during sweating helps in keeping our bodies cool.
• During transpiration, more water is ‘pulled’ upwards from the roots to compensate for the lost water. This water brings along important nutrients from the roots, which are required by the leaf. Thus, transpiration helps in the transport of nutrients within the plant.
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Leaf modifications:
• Leaves of some plants are modified to form special structures called tendrils. Tendrils help plants to attach themselves to a support. Plants having tendrils are generally climbers.
• For protection, leaves of certain plants get modified to form spines. Spines also reduce the amount of water lost from the plant.
Flowers, Fruits, and Seeds :
A flower is the reproductive organ of a plant. Figure 8.13 shows the parts of a flower.
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Pollination :
For a flower to develop into a fruit and form seeds, pollen grains must be transferred from its anthers to the stigma.
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The transfer of pollen grains from an anther to a stigma is called pollination (Fig. 8.14).
Many flowers are brightly coloured and have a sweet smell to attract insects like bees (Fig. 8.15). When the insect sits on the flower, the pollen grains stick to its body and may get rubbed off when it sits on another flower. This helps in pollination.
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After pollination, the ovules change into seeds. As seeds (Fig. 8.16) form, the ovary develops into a fruit. Figure 8.17 shows the structure of a bean seed. Under suitable conditions, i.e., availability of sufficient water, air, and warmth, a seed becomes a baby plant.
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Keywords :

Node: Part of the stem from where the leaves arise is called a node.
Internode: The distance between two nodes is called the internode.
Venation: The arrangement of veins in a leaf is called venation.
Transpiration: The loss of water through the stomata is called transpiration.
Stamen: Male part of a flower is called the stamen.
Carpel: Female part of a flower is called carpel.
Pollination: Transfer of pollen grains from the anther to the stigma is called pollination.
Seed coat: The outer covering of a seed is called the seed coat.

Summary :

• The parts that remain under the ground (roots) form the root system.
• The parts above the ground (stem, leaves, flowers, and fruits) form the shoot system.
• There are two main types of root systems: tap root system and fibrous root system.
• Roots may be modified for support, storage of food, propagation, etc.
• Stems may be modified for support, protection, photosynthesis, food storage, and reproduction.
• Leaves are green because they have a green pigment called chlorophyll. Green leaves manufactureb their food with the help of carbon dioxide, water, and light.
• Flower is the reproductive organ of a plant. Flowers have petals, sepals, stamens, and carpel.
• The pollen grains need to be transferred to the stigma from the anther of the flower for pollination.
• The seed has an embryo, which develops into a new plant under suitable conditions.

The post Getting to Know Plants – CBSE Notes for Class 6 Science appeared first on Learn CBSE.

Fun with Magnets – CBSE Notes for Class 6 Science

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Fun with Magnets – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Look at the pictures shown below. Put a V’ mark against the object that you think would stick to a magnet and ‘x’ against the objects that would not stick to a magnet.
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Let us now learn more about magnets. Answers : 1. False, 2. True, 3.True, 4.False, 5.False, 6.True.

Discovery Of Magnets:

According to a legend, the first magnet was discovered by a Greek shepherd named Magnes. It is said that the nails in his shoes and the iron tip of his staff got stuck to a large black rock on which he was standing. Greeks named this strange type of rock ‘magnetite’. The Chinese also knew about magnets. Ancient Chinese sailors used magnets for navigation.

Magnets :

Magnets are made of materials that attract objects made of certain substances like iron, cobalt, and nickel.
Magnets come in various shapes and sizes (Fig. 12.1). They can be found as horseshoe, ring, cylindrical, or bar shape.
Not all objects are attracted to magnets. Objects that are attracted by a magnet are said to be magnetic, e.g., iron and nickel. Objects that are not attracted by a magnet, are said to be non-magnetic, e.g., wood and plastic.
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Poles Of a Magnet :

When magnetic materials (like iron filings) are brought close to a magnet, they do not stick evenly to all parts of the magnet. They stick more on certain parts of the magnet. These are called the poles of the magnet. Magnetic forces are the strongest at the poles. For example, the two ends of a bar magnet are its poles.
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There are two types of poles in every magnet, irrespective of its shape. These are, by convention, called the North Pole (N) and the South Pole (S) (Fig. 12.2). The two poles cannot exist independently. That is, they always come in pairs.
If we break a bar magnet in the middle, we would get two pieces, each having a North Pole and a South Pole. We could go on breaking the magnet into smaller pieces, and everytime we would get both the poles in each piece.

Using a Magnet To Find Directions :

Today, we use magnets for various purposes. In ancient times, the primary use of a magnet was to find directions.
If a magnet is allowed to move freely, it comes to rest in a direction very close to the Earth’s North-South direction. This property of a magnet was used to find directions on the surface of the Earth by travellers. An instrument with a magnet that is used to find directions is called magnetic compass.
It has a small magnetic needle at its centre. This needle can rotate freely and always points in the Earth’s North-South direction. Different directions (north, south, east, and west) are marked on the compass. Figure 12.3 shows how one can find directions using a magnetic compass.
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Do you know why a freely suspended magnet always points in the Earth’s north-south direction? It behaves like it is under the influence of another magnet. But where is this other magnet? It is the Earth itself. This alignment happens because of the influence of the Earth, which itself acts like a giant bar magnet (Fig. 12.4). It influences all the magnets (within its region of influence) to align themselves along its North-South direction.

Attraction And Repulsion :

When two magnets are brought close to each other, they are either pulled towards each other, or pushed away from each other. When the magnets are pulled towards each other, they are said to attract each other. When they are pushed away from each other, they are said to repel each other. Whether the magnets attract or repel depends on which poles of the magnets are facing each other.
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When like poles of the magnets (N-N or S-S) are brought close to each other, they repel. This is called repulsion.
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When unlike poles of the magnets (N-S or S-N) are brought close to each other, they attract. This is called attraction.
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Types Of Magnets :

There are two types of magnets: temporary and permanent. Magnets that retain their magnetic properties only for a short period of time are called temporary magnets. Magnets that retain their magnetic properties for a long period of time are called permanent magnets.
Temporary magnets are usually made of iron, cobalt, or nickel. These materials behave like magnets only when they are near a strong magnet. They quickly lose their magnetic property if the influence of the strong magnet is removed.
Permanent magnets are made from mixtures of iron, cobalt, or nickel with other materials. These make strong magnets and retain their magnetic properties for a long time.

Care Of Magnets :

A magnet can lose its properties due to the following activities.
• Dropping from a height
• Hitting with a hammer
• Applying heat
• Improper storage can also cause loss of magnetic properties.
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Bar magnets should be stored in pairs, with Dropping from a height unlike poles alongside each other. A horseshoe magnet should be stored with a piece of soft iron kept across its poles.

Uses Of Magnets :

Magnets have several uses:
1. Credit cards, ATM cards, and identity cards have a strip of magnetic material that stores information.
2. Television and computer monitors use magnets.
3. Computer hard discs and audio and video cassettes have magnetic material that store information.
4. Magnets are used in picking up substances made of iron from scrapyard.
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Key Words :

Magnet: An object that attracts substances like iron, cobalt, and nickel is called magnet.
Magnetic compass An instrument with a magnet that is used to find directions is called magnetic compass.
Temporary magnets: Magnets that retain their magnetic properties only for a short period of time are called temporary magnets.
Permanent magnets: Magnets that retain their magnetic properties for a long period of time are called permanent magnets.

Summary :

• Only magnetic materials are attracted by magnets.
• Every magnet has two poles: the North Pole and the South Pole.
• Magnetic forces are the strongest at the poles of the magnet.
• A freely suspended magnet will come to rest in the Earth’s North-South direction.
• Like poles of two magnets repel each other.
• Unlike poles of two magnets attract each other.
• Magnets can lose their properties if they are dropped from a height, hit with a hammer, heated, or stored in an improper manner.

The post Fun with Magnets – CBSE Notes for Class 6 Science appeared first on Learn CBSE.

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