ncert solutions class 11 chemistry

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NCERT Exemplar Class 11 Chemistry Chapter 14 Environmental Chemistry

NCERT Exemplar Class 11 Chemistry Chapter 14 Environmental Chemistry are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar 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.

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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.
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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.

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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.

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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.

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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.

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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 1Column 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 IColumn 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 IColumn 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 IColumn 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 Class 11 Chemistry Chapter 13 Hydrocarbons

NCERT Exemplar Class 11 Chemistry Chapter 13 Hydrocarbons are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar 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

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

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Q5. The addition of HBr to 1 -butene gives a mixture of products (A), (B) and (C).

(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.

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Since I contains a chiral carbon, it exists in two enantiomers (A and B) which are mirror image of each other.
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Q6. Which of the following will nofshow geometrical isomerism?

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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
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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?

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,

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

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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?

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.

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Q13. Which are the correct IUPAC names of the following compound?

(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

(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)
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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.

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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.

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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

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

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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

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?

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? .

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?

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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
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Q27. Arrange the following set of compounds in the order of their decreasing . relative reactivity with an electrophile. Give reason.

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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.

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.
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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.



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.

NCERT Exemplar Class 11 Chemistry Solutions

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

NCERT Exemplar Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar 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.

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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?

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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

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

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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

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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?

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

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

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(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?


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.

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Short Answer Type Questions
Note: Consider structures I to VII and answer Questions 23-26.

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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?

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.

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.”

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

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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.

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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:

Q43. Identify the most stable species in the following set of ions giving reasons:

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

Inductive effectResonance 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.


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.

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 IColumn 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 IColumn 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 IColumn IIExplanation
(a) Carbocationsp2-hybridised carbon with empty p-orbitalH3C+ is carbocation. Loss of e makes its p-orbitals empty (sp2– hybridised carbon)
(b) NucleophileSpecies that can supply a pair of electronsNucleus loving, i.e., having negative charge or excess of electrons
(c) HyperconjugationConjugation of electrons of C – H σbond with empty p-orbital present at adjacent positively charged carbon
(d) IsomersCyclohexane and 1-hexeneSame molecular formula but different structures
(e) sp hybridizationEthyneHC ≡ CH (sp hybridization)
(f) ElectrophileSpecies that receive a pair of electronsElectron loving, i.e., positive charge or lack of electrons


Q52. Match Column I with Column II.

Column IColumn 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 IColumn IIExplanation
(a) Dumas methodNitrogen gelUsed for N containing compounds
(b) Kjeldahl methodAmmonium sulphateNitrogen converts to ammonium sulphate
(c) Carius methodAgN03Compound is heated in the presence of AgN03
(d) ChromatographySilica gelAdsorbent used is silica gel
(e) Homolysis              ‘Free radicalsFree radicals are formed by homolytic fission

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Chemistry Chapter 11 The p-Block Elements

NCERT Exemplar Class 11 Chemistry Chapter 11 The p-Block Elements are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 11 The p-Block Elements.

Multiple Choice Questions
Single Correct Answer Type

Q1. The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is
(a) B
(b) A1
(c) Ga
(d) In
Sol: (c) The melting point of gallium is 30°C and boiling point is 2240°C. Thus, the element exists in liquid state for a wide range of temperature.

Q2. Which of the following is a Lewis acid?
(a) AlCl3
(b) MgCl2                 
(c) CaCl2                 
(d) BaCl2
Sol:
(a) A1C13 is electron-deficient and hence acts as a Lewis acid.

Q3. The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [B(OH)4]- and the geometry of the complex are respectively
(a) sp3, tetrahedral
(b) sp3, square planar                          
(c) sp3d2, octahedral                              
(d) dsp2, square planar

Sol: Boron has die electronic configuration:
ls22s22px12p0y2p°z
In the excited state, 2s-orbital electrons are impaired and one electron is shifted to a p-orbital. Now, hybridisation occurs between one s-and three p-orbitals to give sp3 hybridisation and tetrahedral geometry.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-1

Q4. Which of the following oxides is acidic in nature?
(a) B203
(b) A1203                  
(c) Ga203                   
(d) In203
Sol: (a) B203 is acidic in nature. It reacts with basic oxides to form metal borates. Acidic nature decreases on moving down the group.

Q5. The exhibition of highest co-ordination dumber depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in MF63- ?
(a) B
(b) AI
(c) Ga
(d) In
Sol: (a) The element M in the complex ion MF63-  has a coordination number of six. Since B has only s- and p-orbitals and no d – orbitals, therefore, at the maximum it can show a coordination number of 4. Thus, B cannot form complex of the type MF63-, i.e., option (a) is correct.

Q6. Boric acid is an acid because its molecule
(a) contains replaceable H+   ion
(b) gives up a proton.
(c)accepts OHfrom water releasing proton.
(d) combines with proton from water molecule.
Sol: (c) Because of the small size of boron atom and presence of only six electrons in its valence shell, B(OH)3 accepts a pair of
electrons from OH ion of H20, releasing a proton.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-2

Q7.  Catenation, i.e., linking •of similar atoms depends on size and electronic  configuration of atoms. The tendency of catenation in Group 14 elements follows the order
(a) C > Si > Ge > Sn
(b) C » Si > Ge = Sn
(c) Si > C > Sn > Ge
(d) Ge > Sn > Si > C
Sol:
(b) The decrease in catenation property is linked with M – M bond energy which decreases from carbon to tin.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-3

Q8. Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding f (a) MeSiCl3 (b) Me2SiCl2 (c) Me3SiCl (d) Me4Si
Sol: (c) The chain length of the polymer can be controlled by adding (CH3)3SiCl  which blocks the ends.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-4

Q9. Ionisation enthalpy (∆ tH1 kJ mol-1) for the elements of Group 13 follows the order.
(a) B > A1 > Ga > In > T1
(b) B < A1 < Ga< In <T1
(c) B < A1 > Ga < In < T1                    
(d) B > A1 < Ga > In < T1

Sol: (d) On moving down the group from B to Tl, a regular decreasing trend in the ionisation energy values is not observed.
ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-5

In Ga, there are ten d-electrons in the penultimate shell which screen the nuclear charge less effectively and thus, outer electron is held firmly. As a result, the ionisation energy of both A1 and Ga is nearly the same. The increase in ionisation energy from In to Tl is due to poor screening effect of 14f electrons present in the inner shell.

Q10. In the structure of diborane
(a) all hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane.
(b) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
(c) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
(d) all the atoms are in the same plane.
Sol: (b) Four terminal hydrogen atoms and two boron atoms lie in the same plane and two hydrogen atoms forming bridges lie in a plane perpendicular to the rest of the molecule.
ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-6

Q11. A compound X, of boron reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with lithium aluminium hydride. The compounds X and Y are represented by the formulas.
(a) B2H6,B3N3H6
(b) B203, B3N3H6
(c) BF3, B3N3H6
(d) B3N3H6 , B2H6

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-7

Q12. Quartz is extensively used as a piezoelectric material, it contains
(a) Pb
(b) Si
(c) Ti
(d) Sn
Sol: (b) Quartz is a crystalline form of silica.

Q13. The most commonly used reducing agent is

(a) AlCl3
(b) PbCl2
(c) SnCl4
(d) SnCl2
Sol:
d) +4 oxidation state of Sn is more stable than +2 oxidation state. Therefore, Sn2+ can be easily oxidised to Sn4+ and hence SnCl2 acts a reducing agent.
SnCl2 + 2Cl → SnCl4 + 2e

Q14. Dry ice is(a) Solid NH3 (b) Solid S02 (c) solid C02  (d) solid N2
Sol:
(c) Solid C02 is known as dry ice.

Q15. Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group(s) are present in the mixture?
(a) group 2                                               
(b) groups 2,13 and 14
(c) groups 2 and 13                                
(d) groups 2 and 14
Sol: (b) Cement contains elements of group 2 (Ca), group 13 (Al) and group 14 (Si).

More than One Correct Answer Type

Q16. The reason for small radius of Ga compared to Al is_________ .
(a) poor screening effect of d and f orbitals
(b) increase in nuclear charge
(c) presence of higher orbitals
(d) higher atomic number
Sol: (a, b) The additional 10 d-electrons offer poor screening effect for the outer electrons from the increased fluclear charge in Gallium. Hence, atomic radius of Gallium is less than that of aluminium.

Q17. The linear shape of C02 is due to  ______ .
(a) sp3 hybridisation of carbon
(b) sp hybridisation of carbon
(c) pπ-pπ bonding between carbon and oxygen
(d) sp2 hybridisation of carbon
Sol: (b, c) The linear shape of C02 is due to pπ-pπ bonding between carbon and oxygen and sp hybridisation of carbon.

Q18. Me3SiCl is used during polymerisation of organo silicones because
(a) the chain length of organo silicone polymers can be controlled by adding Me3
(b) Me3SiCl blocks the end terminal of silicone polymer.

(c) Me3SiCl improves the quality and yield of the polymer. –
(d) Me3SiCl acts as a catalyst during polymerization.
Sol: (a, b) The chain length of the polymer can be controlled by adding Me3SiCl which blocks the ends of the silicon polymer.
Q19. Which of the following statements are correct?
(a) Fullerenes have dangling bonds.
(b) Fullerenes are cage-like molecules.
(c) Graphite is thermodynamically most stable allotrope of carbon.
(d) Graphite is slippery and hard and therefore used as a dry lubricant in
Sol:
(b, c) Fullerenes are cage-like (soccer or rugby ball) molecules and graphite is thermodynamically most stable allotrope of carbon. Thus, options (b) and (c) are correct

Q20. Which of the following statements are correct? Answer on the basis of figure.

(a) The two bridged hydrogen atoms and the two boron atoms lie in one plane.
(b) Out of six B – H, bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(c) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds.
(d) The four terminal B – H bonds are two centre-two electron regular bonds.
Sol: (a, b, d) Each of the two boron atoms is in sp3-hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each, of the boron atom overlap with two terminal hydrogen atoms forming two normal B – H σ-bonds. One of the remaining hybrid orbitals (either empty or singly occupied) of one of the boron atoms, 15-orbital of H (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalized orbital covering the three nuclei with a pair of electrons. This is three centre two electron bond. Similar overlapping occurs with the second hydrogen atom (bridging) forming three centre two electrons bond.

Q21. Identify the correct resonance structures of carbon dioxide from the ones given below:

Short Answer Type Questions

Q22. Draw the structure of BC13.NH3 and AlCl3 (dimer).
Sol: In BCl3, the central B atom has six electrons in the valence shell. It is, therefore, an electron deficient molecule and needs two more electrons to ‘ complete its octet. In other words, BCl3 acts as a Lewis acid. NH3, on the other hand, has a lone pair of electrons which it can donate easily. Therefore, NH3 acts as a Lewis base. The Lewis acid (BC13) and the Lewis base (NH3) combine together to form an adduct as shown below:
ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-10
In A1C13, A1 has six electrons in the valence shell. Therefore, it is an electron deficient molecule and needs two more electrons to complete its octet.
Chlorine, on the other hand, has three lone pairs of electrons. Therefore, to complete its octet, the central A1 atom of one molecule accepts a lone pair of electrons from Cl atom of the other molecule forming a dimeric structure as shown below.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-11

Q23. Explain the nature of boric acid as a Lewis acid in water.
Sol:Boric acid is a weak monobasic acid and acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3 + 2H20 →[B(OH)4] + H30+

Q24. Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-12
[B(OH)4] units are present in water. Boron has sp3 hybridisation in [B(OH)4] unit.

Q25. Explain why the following compounds behave as Lewis acids?
(i) BC13
(ii) AICI3
Sol:(i) BCl3: Boron has 6 electrons in its outermost orbital and has a vacant p orbital. Thus, it is an electron deficient compound. Hence, it acts as Lewis acid and accepts a lone pair of electrons.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-13

(ii) AlCl3 is also an electron deficient compound and acts as Lewis acid. It generally forms a dimer to achieve stability.

Q26. Give reasons for the following:
(a) CCl4 is immiscible in water, whereas SiCl4 is easily hydrolysed.
(b) Carbon has a strong tendency for catenation compared to silicon.
Sol: (i) CC14 is a covalent compound while H20 is a polar compound. Therefore, it is insoluble in water. Alternatively, CCl4 is insoluble in water because carbon does not have (/-orbitals to accommodate the electrons donated by oxygen atom of water molecules. As a result, there is no interaction between CC14 and water molecules and hence CC14 is insoluble in water. On the other hand, SiCl4 has d-orbitals to accommodate the lone pair of electrons donated by oxygen atom of water molecules. As a result, there is a strong interaction between SiCl4 and water molecules. Consequently, SiCl4 undergoes hydrolysis by water to form silicic acid.

(b)  The bond dissociation energy decreases rapidly as the atomic size increases. Since the atomic size of carbon is much smaller (77 pm) as compared to that of silicon (118 pm), therefore, carbon-carbon bond dissociation energy is much higher (348 kJ mol-1) than that of silicon-silicon bond (297 kJ mol-1). Hence, because C – C bonds are much stronger as compared to Si-Si bonds, carbon has a much higher tendency for catenation than silicon.

Q27. Explain the following:
(i) C02 is a gas whereas Si02 is a solid.
(b) Silicon forms SiF62- ion whereas corresponding fluoro compound of carbon is not known.
Sol: (a) Because of its small size and good π-overlap with other small atoms, carbon forms strong double bonds with two oxygen atoms to give discrete C02 molecules.                                            –
Silicon atom, on account of large size, does not have good π -overlap with other atoms. It uses its four valence electrons to form four single bonds directed towards the four apices of a tetrahedron (sp3-hybridisation). Each oxygen is linked with two silicon atoms, i.e., a giant three dimensional structure comes into existence which is very stable. Thus, C02 is a gas and Si02 is a solid.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-14

(b) Silicon has 3d-orbitals in the valence shell and thus expands its octet giving sp3d2 hybridisation while d-orbitals are not present in the valence shell of carbon. It can undergo ip3-hybridisation only. Thus, carbon is unable to form CF62- anion.

Q28. The+1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.
Sol: In group 13 and 14, as we move down the group, the tendency of s-electrons of the valence shell to participate in bond formation decreases. This is due to ineffective shielding of s-electrons of the valence shell by the intervening d- and f-electrons. This is called inert pair effect.
Due to this, s-electrons of the valence shell of group 13 and 14 are unable to participate in bonding. Hence, +1 and +2 oxidation states, in group 13 and 14 respectively, become -more stable with increasing atomic number.

Q29. Carbon and silicon both belong to the group 14, but in spite of the stoichiometric similarity, the dioxides (i.e., carbon dioxide and silicon dioxide) differ in their structures. Comment.
Sol: Carbon, the first member of group 14 possesses a pronounced ability to form stable p-p multiple bonds with itself and with other first row elements such as nitrogen and oxygen. In C02, both the oxygen atoms are linked with carbon atom by double bonds.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-15

However, silicon shows its reluctance in forming p-p multiple bonding due to large atomic size. Thus, in Si02, oxygen atoms are linked to silicon atom by single covalent bonds giving three dimensional network.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-16

Q30. If a trivalent atom replaces a few silicon atoms in three dimensional network of silicon dioxide, what would be the type of charge on overall structure?
Sol: If a few tetrahedral Si atoms in a three dimensional network structure of Si02 are replaced by an equal number of trivalent atoms, then one valence electron of each Si atom will become free. As a result, each substitution of Si atom by a trivalent atom introduces one unit negative charge into the three dimensional network structure of Si02. Hence, Si02 becomes negatively charged.

Q31. When BC13 is treated with water, it hydrolyses and forms [B(OH)4]” only whereas A1C13 in acidified aqueous solution forms [A1(H20)6]3+ Explain what is the hybridisation of boron and aluminium in these species?
Sol: BC13 + 3H20→ B(OH)3 + 3HC1
B(0H)3 + H20→[B(OH)4] + H+
B(OH)3 due to its incomplete octet accepts an electron pair (as OH) to give [B(OH)4]. Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridisation of B in [B(OH)4] is sp3.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-17

Q32. Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character, A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn’t proceed. Explain the reason.
Sol: Al being amphoteric dissolves both in acids and alkalies evolving H2 gas which bums with a pop sound.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-18

Q31. Explain the following:

  • Gallium has higher ionisation enthalpy than aluminium.
  • Boron does not exist as B3+
  • Aluminium forms [A1F6]3- ion but boron does not form [BF6]3-
  • PbX2 is more stable than PbX4.
  • Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.
  • Electron gain enthalpy of chlorine is more negative as compared to fluorine.
  • TI(N03)3 acts as an oxidising agent.
  • Carbon shows catenation property but lead does not.
  • BF3 does not hydrolyse.
  • Why does the element silicon, not form a graphite like structure whereas carbon does.

Sol: (i) Due to ineffective shielding of valence electrons by the intervening 3d electrons, the effective nuclear charge on Ga is slightly higher than that on A1 and hence the ∆Hi of gallium is slightly higher than that of Al. Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆

(ii) Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆iH1 + ∆iH2 + ∆iH3), boron does not lose all its valence electrons to form B3+ ions.

(iii) Al has vacant d-orbitals and hence can expand its coordination number from 4 to 6 and hence forms octahedral [A1F6]3- ion in which Al undergoes sp3d2 hybridisation. In contrast, B does not have d-orbitals. Therefore, it can have a maximum coordination number of 4. Therefore, B forms [BF4] (in which B is sp3-hybridised) but not [BF6]3-.

(iv) Due to inert pair effect, +2 oxidation state of Pb is more stable than its +4 oxidation state. Consequently, PbX2 in which the oxidation state of Pb is +2 is more stable than PbX4 in which the oxidation state of Pb is +4.

(v) Inert pair effect is less prominent in Sn than in Pb. Therefore, +2 oxidation of Sn is less stable than its +4 oxidation state. In other words, Sn2+ can easily lose two electrons to form Sn4+ and hence Sn2+ acts as a reducing agent.
Sn2+→Sn4+ + 2e

In contrast, the inert pair effect is’ more prominent in Pb than in Sn. Therefore, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb4+ can easily lose two electrons to form Pb2+ and hence Pb4+ acts as an oxidising agent.

Pb4+ + 2e→-Pb2+

(vi) Due to small size, the electron-electron repulsions in the relatively compact 2p-subshell of F are quite strong and hence the incoming electron is not accepted with the same ease as in case of bigger Cl atom where repulsions are comparatively weak. Thus, electron gain enthalpy of chlorine is more negative as compared to that of fluorine.

(vii) Due ta strong inert pair effect, the +3 oxidation state of T1 is less stable than its +1 oxidation state. Since in T1(N03)3, oxidation state of T1 is +3, therefore, it can easily gain two electrons to form T1N03 in which the oxidation state of T1 is +1. Consequently, T1(N03)3 acts as an oxidising agent.

(viii) Property of catenation depends upon the strength of element-element bond which, in turn, depends upon the size of the element. Since the atomic size of carbon is much smaller than that of lead, therefore, carbon-carbon bond strength is much higher than that of lead-lead bond. Due to stronger C-C than Pb-Pb bonds, carbon has a much higher tendency for catenation than lead.

(ix) Unlike other boron halides, BF3 does not hydrolyse completely. Instead, it hydrolyses incompletely to form boric acid and fluoroboric acid. This is because the HF first formed reacts with H3B03.

ncert-exemplar-problems-class-11-chemistry-chapter-11-the-p-block-elements-19

(x) In graphite, carbon is sp2-hybridised and each carbon is linked to three other carbon atoms by forming hexagonal rings. Each carbon is now left with one unhybridised p-orbital which undergoes sideways overlap to form three p-p double bonds. Thus, graphite has two-dimensional sheet like (layered) structure consisting of a number of benzene rings fused together. Silicon, on the other hand, does not form an analogue of carbon because of the following reason:
Due to bigger size and smaller electronegativity of Si than C, it does not undergo sp2-hybridisation and hence it does not form p-p double bonds needed for graphite like structure. Instead, it prefers to undergo only sp3-hybridisation and hence silicon has diamond like three­dimensional network

Q34. Identify the compound A, X and Z in the following reactions:

Q35. Complete the following chemical equations:

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.
Q36. Match the species given in Column I with the properties mentioned in Column II.

Column IColumn II
(i) BF4(a) Oxidation state of central atom is +4
(ii) A1C13(b) Strong oxidising agent
(iii) SnO(c) Lewis acid
(iv) Pb02(d) Can be further oxidised
(e) Tetrahedral shape

Sol: (i→e); (ii → c); (iii —> d); (iv →a, b)
(i) BF4 : Tetrahedral shape, sp3 hybridisation, regular geometry.
(ii) AlCl3: Octet of A1 not complete, acts as Lewis acid.
(iii) SnO: Sn2+ can show +4 oxidation state.
(iv)Pb02: Oxidation.state of Pb in Pb02 is +4. Due to inert pair effect, Pb4+ is less stable than Pb2+ and hence acts as strong oxidising agent.

Q37. Match the species given in Column I with properties given in Column II.

Column IColumn II
(i) Diborane(a) Used as a flux for soldering metals
(ii) Gallium                     ‘(b) Crystalline form of silica
(iii) Borax(c) Banana bonds
(iv) Aluminosilicate(d) Low melting, high boiling, useful for measuring high temperatures
(v) Quartz(e) Used as catalyst in petrochemical industries

Sol: (i → c); (ii → d); (iii → a); (iv → e); (v → b)

  • BH3 is unstable, forms diborane B2H6 by 3 centre -2 electron bonds, shows banana bonds.
  • Gallium with low melting point and high boiling point makes it useful to measure high temperatures.
  • Borax is used as a flux for soldering metals.
  • Aluminosilicate is used as catalyst in petrochemical industries.
  • Quartz, is a crystalline form of silica.

Q38.  Match the species given in Column I with the hybridisation given in Column II.

Column IColumn II
(i) Boron in [B(OH)4]“(a) sp2
(ii) Aluminium in [A1(H20)6]3+(b) sp3
(iii) Boron in B2H6(c) sp3d2
(iv) Carbon in Buckminsterfullerene
(v) Silicon in SiO44-
(vi) Germanium in [GeCl6]2-

Sol:(i →b); (ii→ c); (iii → b); (iv→ a); (v →b); (vi→ c)

  • Boron in [B(OH)4] is sp3
  • Aluminium in [A1(H20)6]3+ is sp3d2 hybridised
  • Boron in B2H6 is sp3
  • Carbon in Buckminsterfullerene sp2 is hybridised.
  • Silicon in SiO44- is sp3
  • Germanium in [GeCl6]2- is sp3 d2

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): If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.
Reason (R): Aluminium is trivalent while silicon is tetravalent.
(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) In aluminosilicates (anion), some of the silicon atoms are replaced by aluminium. Since, aluminium is trivalent while silicon is tetravalent, hence we get negatively charged ion.

Q40. Assertion (A): Silicones are water repelling in nature.
Reason (R): Silicones are organosilicon polymers, which have (-R2SiO-) as repeating unit.
(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) Silicones are organosilicon polymers and they are hydrophobic in nature. Silicones neither react nor absorb water molecules.

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements

NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 10 The S-Block Elements.

Multiple Choice Questions
Single Correct Answer Type

Q1. The alkali metals are low melting. Which of the following alkali metals is expected to melt if the room temperature rises to 30°C?
(a) Na (b) K (c) Rb (d) Cs
Sol: (d) Among alkali metals, melting point decreases as the strength of metallic bonding decreases with increasing size of the atom. Thus, Cs has the lowest melting point (28.5°C) and will melt at 30°C.

Q2. Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously?
(a) Li . (b) Na (c) K (d) Cs
Sol: (a) Both melting point and heat of reaction of alkali metals with water decrease down the group from Li to Cs. Although the heat of reaction of Li is the highest, but due to its high melting point, even this heat is not sufficient to melt the metal, which exposes greater surface to water for reaction. As a result, Li has the least reactivity but the reactivity increases as the melting point of alkali metals decreases down the group from Li to Cs.

Q3. The reducing power of a metal depends oh various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
(a) Sublimation enthalpy (b) Ionisation enthalpy
(c) Hydration enthalpy (d) Electron-gain enthalpy
Sol: (c) Lithium has highest hydration enthalpy which accounts for its high negative E° value and its high reducing power.

Q4. Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
(a) MgC03

(b)CaC03
(c)SrCQ3                            

(d)BaC03

 Sol: (d) Thermal stability of metal carbonates increases as the electropositive character of the metal or the basicity of the metal hydroxide increases from Be(OH)2 to Ba(OH)2. Thus, BaC03 is the most stable.

Q5. Which of the carbonates given below is unstable in air and is kept in C02 atmosphere to avoid decomposition.
(a) BeCO3
(b) MgC03
(c) CaC03
(d) BaCO3
Sol: (a) Due to least electropositive character or least basicity of Be, BeC03 is less stable and hence decomposes to give BeO and C02.
BeC03→BeO + C02
Since the decomposition reaction is reversible, therefore, to increase the stability of BeC03 or to reverse the above equilibrium, BeC03 is kept in an atmosphere of C02.

Q6. Metals form basic hydroxides. Which of the following metal hydroxides is the least basic?
(a) Mg(OH)2 (b) Ca(OH)2               (c) Sr(OH)2              (d) Ba(OH)2
Sol:
(a) As the ionization enthalpy increases from Mg →Ba, the M – O bond becomes weaker and weaker down the group and hence basicity increases down the group. Thus, Mg(OH)2 is least basic.

Q7. Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is
(a) BeCl2
(b) MgCl2                  
(c) CaCl2                
(d) SrCl2

Sol: Due to small size, high electronegativity and high ionization enthalpy of Be, BeCl2 is covalent and hence most soluble in organic solvents such as ethanol.

Q8. The order of decreasing ionization enthalpy in alkali metals is

(a) Na > Li > K > Rb (b) Rb < Na < K < Li

(c) Li > Na > K > Rb (d) K < Li < Na < Rb

Sol: (c) Ionization enthalpy decreases with increase in Size of the atom in a group. Hence, the order is:

Li > Na > K > Rb.

Q9. The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to
(a) ionic nature of lithium fluoride. . .
(b) high lattice enthalpy. ‘
(c) high hydration enthalpy for lithium ion.
(d) low ionization enthalpy of lithium atom.
Sol: (b) Due to small size of Li+ and F ions, lattice enthalpy is much higher than hydration enthalpy and hence LiF is least soluble among alkali metal fluorides.

Q10. Amphoteric hydroxides react with both alkalies and acids. ‘Which of the following group 2 metal hydroxides is soluble in sodium hydroxide?
(a) Be(OH)2
(b) Mg(OH)2
(c) Ca(OH)2
(d) Ba(OH)2
Sol:
(a) Be(OH)2 reacts with NaOH to give beryllate ion, becoming soluble in it. Be(OH)2 + 20H→[Be(OH)4]2

Q11. In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH4C1 with Ca(OH)2. The by-product obtained in this process is
(a) CaCl2
(b) NaCl  
(c) NaOH
(d) NaHC03
Sol:
(a) Sodium carbonate is synthesised by Solvay or ammonia soda process. The reactions involved are
ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-1

Q12.When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to
(a) ammoniated electron                     
(b) sodium ion
(c) sodium amide                                   
(d) ammoniated sodium ion
Sol: (a)M+(x+y)NH3→ M+(NH3)x + e(NH3)y
The colour of solution (deep blue) is due to the ammoniated electron which absorbs energy in the visible region.

Q13. By adding gypsum to cement
(a) setting time of cement becomes less.
(b) setting time of cement increases.
(c) colour of cement becomes light.
(d) shining surface is obtained.
Sol: (b) Raw materials for cement are limestone, clay and gypsum. Cement is a dirty greyish heavy powder containing calcium aluminates and silicates. Gypsum (CaS04 -2H20) is added to the components to increase the setting time of cement so that it gets sufficiently hardened. Setting of cement is an exothermic process and involves hydration of calcium aluminates and , silicates.

Q14. Dead burnt plaster is

On heating plaster of Paris at 200°C, it forms anhydrous calcium sulphate, i.e., dead plaster which has no setting property as it absorbs water very slowly.
ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-3

Q15. Suspension of slaked lime in water is known as
(a) lime water                                          
(b) quick lime
(c) milk of lime                                       
(d) aqueous solution of slaked lime
Sol: (c) Suspension of slaked lime in water is known as milk of lime.

Q16. Which of the following elements does not form hydride by direct heating with dihydrogen?
(a) Be                     
(b) Mg                     
(c) Sr                        
(d) Ba
Sol:
(a) Due to high ionization enthalpy and small size, Be does not react with hydrogen by direct heating.

Q17. The formula of soda ash is
(a)    NaHCO3.10H2O
(b)Na2C03.2H20
(c) Na2C03.H20
(d) Na2C03   
Sol: (d) On heating washing soda, it loses its water of crystallization. Above 373 K, it becomes completely anhydrous white powder called soda ash.

ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-4

Q18. A substance which gives brick red flame and breaks down on heating to give oxygen and brown gas is
(a)    Magnesium nitrate                       
(b)     Calcium nitrate
(c)     Barium nitrate                               
(d)     Strontium nitrate
Sol: (b) 2Ca(N03)2→2CaO + 4N02 + 02
N02 is a brown gas. Ca2+ imparts brick red colour to the flame.

Q19.Which of the following statements is true about Ca(OH)2?
(a) It is used in the preparation of bleaching powder.
(b) It is a light blue solid.
(c) It does not possess disinfectant property.
(d) It is used in the manufacture of cement.

ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-5

Q20. A chemical A is used for the preparation of washing soda to recover ammonia. When C02 is bubbled through an aqueous solution of A, the solution tons milky. It is used in white washing due to disinfectant nature. What is the chemical formula of A?
(a) Ca(C03)2          
(b) CaO                   
(c) Ca(OH)2
(d) CaC03

Q21. Dehydration of hydrates of halides of.calcium, barium and strontium, i.e., CaCl2.6H20, BaCl2.2H20, SrCl2.2H20, can be achieved by heating. These become wet oh keeping in air. Which of the following statements is correct about these halides?
(a) Act as dehydrating agents.
(b) Can absorb moisture from air.
(c) Tendency to form hydrate decreases from calcium to barium.
(d) All of the above.
Sol: (d) Chlorides of alkaline earth metals are hydrated salts. Due to their hygroscopic nature, they can be used as dehydrating agents to absorb moisture from air.

Extent of hydration decreases from Mg to Ba, i.e., MgCl2.6H20, CaCl2.6H20, BaCl2 2H20, SrCl2.2H20.

Q22. Metallic elements are described by their standard electrode potential, frision enthalpy, atomic size, etc. The alkali metals are characterized by which of the following properties?
(a) High boiling point. ‘
(b) High negative standard electrode potential.
(c) High density.
(d) Large atomic size.
Sol: (b, d) Alkali metals have high negative standard electrode potential and large atomic size.

Q23. Several sodium compounds find use in industries. Which of the following compounds are used for textile industry?
(a) Na2C03            
(b) NaHC03            
(c) NaOH               
(d) NaCl
Sol: (a, c) Na2C03 and NaOH are used in textile industry.         .

Q24. Which of the following compounds are readily soluble in water?
(a) BeS04              
(b) MgS04              
(c) BaS04                
(d) SrS04
Sol:
(a, b) Solubility decreases down the group because hydration enthalpy decreases more rapidly than lattice enthalpy. Thus, BeS04 and.MgS04 are soluble.

Q25. When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water, the sodium ions are exchanged with which of the following ion(s)?
(a) H+ions                                               
(b) Mg2+ions’
(c) Ca2+ ions                                           
(d) SO24_ ions
Sol:
(b, c) Sodium zeolite removes Ca2+ and Mg2+ ions from hard water. Na2Z + M2+ → MZ + 2Na+ (M = Ca, Mg) where, Z = Al2Si208H20

Q26. Identify the correct’ formula of halides of alkaline earth metals from the following.
(a) BaCl2.2H20
(b) BaCl2 .4H20
(c) CaCl2 . 6H20
(d) SrCl2.4H20
Sol:(a, c) Tendency to form halide hydrates gradually decreases down the group. The hydrates are MgCl2.6H20, CaCl2.6H20, SrCl2.6H20 and BaCl2.2H20.

Q27. Choose the correct statements from the folio-wing.
(a) Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
(b) Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of Be2+ overcomes the lattice enthalpy factor.
(c) Beryllium exhibits coordination number more than four.
(d) Beryllium oxide is purely acidic in nature.
Sol: (a, b) Be does not exhibit coordination number more than four and BeO is amphoteric in nature.

Q28. Which of the following are the correct reasons for anomalous behaviour of lithium?
(a) Exceptionally small size of its atom.
(b) Its high polarizing power.
(c) It has high degree of hydration.
(d) Exceptionally low ionization enthalpy.
Sol: (a, b) Anomalous behaviour of Li is due to its exceptionally small size and high polarizing power.

Short Answer Type Questions
Q29. How do you account for the strong reducing power of lithium in aqueous solution? .
Sol: Electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. It mainly depends upon the following three factors
(i) Sublimation enthalpy
(ii) Ionization enthalpy
(iii) Enthalpy of hydration
The sublimation enthalpies of alkali metals are almost similar. Since Li has the smallest size, its enthalpy of hydration is the highest among alkali metals. Although ionization enthalpy of Li is the highest among alkali metals, it is more than compensated by the high enthalpy of hydration. Thus, Li has the most negative standard electrode potential (-3.04 V) and hence Li is the strongest reducing agent in aqueous solution mainly because of its high enthalpy of hydration.
ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-7

Q30. When heated in air, the alkali metals form various oxides. Mention the oxides formed by Li, Na and K.
Sol: The reactivity of alkali metals towards oxygen increases down the group as the atomic size increases. Thus, Li forms only lithium oxide (Li20), sodium forms mainly sodium peroxide (Na202) along with a small amount of sodium oxide while potassium forms only potassium superoxide (K02).

ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-8

This is because of the following two reasons:
(i) As the size of.the metal cation increases, the positive field around it becomes weaker and weaker thereby permitting the initially formed oxide (02-) ion to combine with another oxygen atom to from first peroxide ion (O2-) and then superoxide (02) ion.
ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-9

(ii) Since larger cations stabilize larger anions due to higher lattice energies, therefore, the stability increases from oxide → peroxide→superoxide as the size of the metal cation increases down the group and the size of the anion increases from oxide → peroxide → superoxide.

ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-10

Q32. Lithium resembles magnesium in some of its properties. Mention two such properties and give reasons for this resemblance.
Sol:(i) Both Li and Mg are harder and lighter than other elements in their groups.
(ii) Both form ionic nitrides Li3N and Mg3N2 by heating in an atmosphere of nitrogen.
Li resembles Mg due to similar atomic radii and ionic radii.

Q33. Name an element’ from Group 2 which forms an amphoteric oxide and awater soluble sulphate.                   Sol: Due to small size and somewhat high ionization enthalpy of Be, Be(OH)2 is amphoteric in nature, i.e., it reacts with both acids and bases. Further, due to small size, the hydration enthalpy of Be2+ ions is much higher than the lattice enthalpy of BeS04. As a result, BeS04 is highly soluble in water.

Q34. Discuss the trend of the following:
(i) Thermal stability of carbonates of Group 2 elements.
(ii) The solubility and the nature of oxides, of Group 2 elements.
Sol: (i) All the alkaline earth metals form carbonates (MC03). All these carbonates decompose on heating to give C02 and metal oxide. The thermal stability; of these carbonates increases down the group, i.e., from Be to Ba,
BeC03 < MgC03 < CaC03 < SrC03 < BaC03
BeC03 is unstable to the extent that it is stable only in atmosphere of C02. It however shows reversible decomposition in closed container

BeC03 ⇌BeO + C02

Hence, more is the stability of oxide formed, less will be stability of carbonates. Stability of oxides decreases down the group. Since beryllium oxide is high stable, it makes BeC03 unstable.

(ii) All the alkaline earth metals form oxides of formula MO. The oxides are very stable due to high lattice energy and are used as refractory material. Except BeO (predominantly covalent), all other oxides are ionic and their lattice energy decreases as the size of cation increases.
The oxides are basic and basic nature increases from BeO to BaO (due to increasing ionic nature).
ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-11
BeO dissolves both in acid and alkalies to give salts and is amphoteric.
The oxides of the alkaline earth metals (except BeO and MgO) dissolve in water to form basic hydroxides and evolve a large amount of heat. BeO and MgO possess high lattice energy and thus are insoluble in water.

Q35. Why are BeS04 and MgS04 readily soluble in water while CaS04, SrS04 and BaS04 are insoluble?
Sol: The hydration enthalpies of BeS04 and MgS04 are quite high because of small size of Be2+ and Mg2+ ions. These hydration enthalpy values are higher than their corresponding lattice enthalpies and therefore, BeS04 and MgS04 are highly soluble in water. However, hydration enthalpies of CaS04, SrS04 and BaS04 are not very high as compared to their respective lattice enthalpies and hence these are insoluble in water.

Q36. All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.
Sol: Because of the small size, high electronegativity and high ionization enthalpy, lithium compounds have considerable covalent character while compounds of other alkali metals are ionic in nature. As a result, compounds of lithium are more soluble in organic solvents while those of other alkali metals are more soluble in water.

Q37. In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing (NH4)2C03 with sodium chloride? Explain.
Sol: No. In Solvay process, ammonium hydrogencarbonate is prepared from ammonium carbonate, which then reacts with sodium chloride to form sodium hydrogencarbonate. Due to low solubility of NaHC03, it gets precipitated and decomposes on heating to give Na2C03.
We cannot obtain sodium carbonate directly by treating the solution containing (NH4)2C03 with sodium chloride as both the products formed on reaction, i.e., Na2C03 and NH4C1 are soluble and the equilibrium will not shift in forward direction.
(NH4)2C03 + 2NaCl⇌Na2C03 + 2NH4Cl

ncert-exemplar-problems-class-11-chemistry-chapter-10-the-s-block-elements-12

Q39. Why do beryllium and magnesium not impart colour to the flame in the flame test?

Sol: All alkaline earth metals (except Be and Mg) impart a characteristic colour to the Bunsen flame. The different colours arise due to different energies . required for electronic excitation and de-excitation.
Be and Mg atoms, due to their small size, bind their electrons more strongly (because of .higher effective nuclear charge). Hence, they require high excitation energy and are not excited by the energy of the flame with the result that no flame colour is shown by them.

Q40.What is the structure of BeCl2 molecule in gaseous and solid state? Beryllium chloride has different structures in solid and vapour state. In solid state, it exists in the form of polymeric chain structure in which each Be- atom is surrounded by four chlorine atoms, having two of the chlorine atoms covalently bonded while the other two by coordinate bonds. The resulting bridge structure contains infinite chains.

In vapour state, above 1200 K, it exists as a monomer having linear structure and zero dipole moment. But below 1200 K, it exists as dimer even in the vapour state.

Matching Column Type Questions
In the following questions, more than one option of column I and II may be correlated.
Q41.Match the elements given in Column I with the properties mentioned in Column II.

Column IColumn II
(i) Li(a) Insoluble sulphate
(ii) Na(b) Strongest monoacidic base
(iii) Ca(c) Most negative E° value among alkali metals
(iv) Ba(d) Insoluble oxalate                                     ‘
(e) 6s2 outer electronic configuration

Sol:(i → c); (ii → b); (iii → d); (iv → a, e)

(i) Li – Most negative E° among alkali metals
[Due to very high hydration energy the resulting E° is most negative].
(ii) Na – Strongest monoacidic base
[Alkalies are more acidic than alkaline earth metals. LiOH has covalent character], –
(iii) Ca – insoluble oxalate
[Calcium oxalate is insoluble in water.]
(iv) Ba – Insoluble sulphate
[Hydration energy decreases as size of cation increases].
6s2 outer electronic configuration
56Ba = Is2,2s2,2p,3s2, 3p6, 3d10,4s2,4p6,4d10, 5s2, 5p6,6s2

Q42. Match the compounds given in Column I with their uses mentioned in Column II.

Column IColumn II
(i) CaC03(a) Dentistry, ornamental work
(ii) Ca(OH)2(b) Manufacture of sodium carbonate from caustic soda
(iii) CaO(c) Manufacture-of high quality paper
(iv) CaS04(d) Used in white washing

Sol: (i → c); (ii →d); (iii → b); (iv→a)
(i) CaC03 – Manufacture of high quality paper
(ii) Ca(OH)2 – Used in white washing
(iii) CaO – Manufacture of sodium carbonate from caustic soda
(iv) CaS04 – Dentistry, ornamental work

Q43. Match the elements given in Column I with the colour they impart to the flame given in Column II.

Column IColumn II
(i) Cs(a) Apple green
(ii) Na(b) Violet
(iii) K(c) Brick red
(iv) Ca(d) Yellow
(v) Sr(e) Crimson red
(vi) Ba(f) Blue

Sol: (i → f); (ii →d); (iii → b); (iv → c); (v →e); (vi → a)

Elements with the characteristic flame colour are as follows ‘
(i) Cs – Blue
(ii) Na-Yellow
(iii) K- Violet
(iv) Ca-Brick red
(v) Sr – Crimson red
(vi) Ba – Apple green

Flame colours are produced from the movement of the electrons in the metal ions present in the compounds. These movements of electrons (electronic excitation and de-excitation) requires energy.
Each atom has particular energy gap between ground and excited energy ‘ level. Therefore, each of these movements involves a specific amount of energy emitted as light energy, and each corresponds to a particular colour. As we know energy gap between ground and excited state energy level increases, wavelength of light absorbed decreases and complementary colour is observed.

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.
Q44. Assertion (A): The carbonate of lithium decomposes easily on heating to , form lithium oxide and C02.
Reason (R): Lithium being very small in size polarizes large carbonate ion leading to the formation of more stable Li20 and C02.
(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 hot correct but R is correct.
Sol: (a) Unlike other alkali metal carbonates, the carbonate of lithium decomposes on heating to form its oxide. Its oxide is stablised by polarization.

Q44. Assertion (A): Beryllium carbonate is kept in the atmosphere of carbon dioxide.
Reason (R): Beryllium carbonate is unstable and decomposes to give beryllium oxide and carbon dioxide:
(a) Both A and Rare 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) BeC03 is kept in the atmosphere of C02, otherwise it will decompose to give its oxide and carbon dioxide.

Long Answer Type Questions

Q46. The s-block elements are characterized by their larger atomic sizes, lower ionization enthalpies, invariable +1 oxidation state and solubilities of their oxosalts. In the light of these features, describe the nature of their oxides, halides and oxosalts.
Sol: (i) Nature of oxides – Alkali metals form M20, M202 and M02 types of oxides. The stability of the peroxide or superoxide increases as the size of metal cation increases. This is due to the stabilization of large anions by larger cations.
(ii) Nature of halides – Alkali metal halides have general formula MX. All halides are soluble in water. LiF is very less soluble in water due to its high lattice energy. Their melting points and boiling points follow the trend – fluoride > chloride > bromide > iodide. This is because with increase in size of the halide ion, lattice energy increases.
(iii) Oxosalts – Oxosalts of alkali metals are generally soluble in water and thermally stable. As electropositive character increases down the group, stability of carbonates and bicarbonates increases.

Q47. Present a comparative account of the alkali and alkaline earth metals with respect to the following characteristics:
(a) Tendency to form ionic/covalent compounds
(b) Nature of oxides and their solubility in water
(c) Formation of oxosalts
(d) Thermal stability of oxosalts

Alkali metalsAlkaline earth metals
(i) All alkali metals except Li form ionic compounds.(i) All alkaline earth metals except Be form ionic compounds.
(ii) The solubility of oxides of al­kali metals increases down the group. The basic character of the ox­ides increases down the group(ii) The solubility of oxides of Mg, – Ca, Sr and Ba increases from Mg to Ba. BeO, however, is covalent and insoluble in water. The basic character of oxides in­creases from MgO to BaO. BeO is, however, amphoteric.
(iii) All alkali metals form oxo salts such as carbonates, sul­phates and nitrates.(iii) All alkaline earth metals form oxo salts such as carbonates, sulphates and nitrates.
(iv) Solubility of carbonates and sul­phates increases down the group.(iv) Solubility of carbonates and sul­phates decreases down the group.
(v) Carbonates and sulphates of Li decompose on heating while the stability, of carbonates and sulphates of other metals in­creases down the group.(v) The carbonates and sulphates of alkaline earth metals all decom­pose on heating but the tempera­ture of their decomposition in­creases down the group, i.e., their thermal stability increases.

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen

NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 9 Hydrogen.

Multiple Choice Questions
Single Correct Answer Type

Q1. Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(a) Its tendency to lose an electron to form a cation.
(b) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration.
(c) Its low negative electron gain enthalpy value.
(d) Its small size.
Sol: (b) Halogens have the tendency to gain one electron and acquire inert gas configuration. Hydrogen also accepts one electron and acquires helium configuration.
ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-1

Q2. Why does H+ ion always get associated with other atoms or molecules?
(a) Ionisation enthalpy of hydrogen resembles that of alkali metals.
(b) Its reactivity is similar to halogens.
(c) It resembles both alkali metals and halogens.
(d) Loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to small size it can not exist freely.
Sol: (d) H→H+ +e
H+ has a very small size (~1.5 x 10-3 pm) compared to normal atomic and ionic sizes of 50 to 220 pm. It does not exist freely and is always associated with other atoms or molecules.

Q3. Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is
(a) LiH > NaH > CsH > KH > RbH
(b) LiH < NaH < KH < RbH < CsH
(c) RbH > CsH > NaH > KH > LiH
(d) NaH > CsH > RbH > LiH > KH
Sol: (b) Ionic character increases as the size of the atom increases.
LiH < NaH < KH < RbH < CsH

Q4. Which of the following hydrides is electron-precise hydride?
(a) B2H6
(b) NH3                    
(c) H20                    
(d) CH4
Sol:
(d) CH4 is an electron precise hydride since there are exact number of electrons to form normal covalent bonds.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-2

Q5. Radioactive elements emit a, p and y rays and are characterized by their half lives. The radioactive isotope of hydrogen is
(a) Protium (b) Deuterium (c) Tritium (d) Hydronium
Sol: (c) Nucleides with n/p (neutron-proton) ratio > 1.5 are usually radioactive. For example, tritium (n = 2,p = 1).

Q6. Consider the reactions
(A) H202 + 2HI → I2 + 2H20
(B) HOCl + H2O2 → H30+ + Cl + 02
Which of the following statements is correct about H202 with reference to these reactions? Hydrogen peroxide is   _______      

(a) an oxidizing agent in both (A) and (B)
(b) an oxidizing agent in (A) and reducing agent in (B)
(c) a reducing agent in (A) and oxidizing agent in (B)
(d) a reducing agent in both (A) and (B)

O.N. of oxygen is decreased from -1 (H202) to -2 (H20), therefore, it is reduced and acts as an oxidizing agent.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-4

O.N. of oxygen is increased from -1 (H202) to 0 (02), therefore, it is oxidized and acts as a reducing agent.

Q7. The oxide that gives H202 on treatment with dilute H2S04 is
(a) Pb02
(b) Ba02 -8H20
(c) Mn02
(d) Ti02

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-5

Q8. Which of the following equatibns depicts the oxidizing nature of H202?
(a) 2Mn04 + 6H+ + 5H202 → 2Mn2+ + 8H20 + 502
(b) 2Fe3+ + 2H+ + H202 → 2Fe2+ + 2H20 + 02
(c) 2I + 2H+ + H202 → I2 + 2H20

(d) KI04 + H202 → KI03 + H20 + 02

Q9. Which of the following equations depicts reducing nature of H202?
(a) 2[Fe(CN)6]4- + 2H+ + H202 → 2[Fe(CN)6]3- + 2H20
(b) I2 + H202 + 2OH→ 2I + 2H20 + 02
(c) Mn2+ + H202 → Mn4+ + 20H

(d) PbS + 4H202 → PbS04 + 4H20
Sol: (b) I2 + H202 + 2OH→ 2I + 2H20 + 02
I2 is reduced to I. Thus, H202 acts as a reducing agent.

Q10. Hydrogen peroxide is ._________ .
(a) an oxidizing agent
(b) a reducing agent
(c) both an oxidizing and a reducing agent
(d) neither oxidizing nor reducing agent.
Sol: (c) H202 acts both as oxidizing and reducing agent.

Q11. Which of the following reactions increases production of dihydrogen from synthesis gas?

Sol:(c) To increase the production of H2 from synthesis gas, CO is oxidized to C02 by passing it over steam at 673 K in presence of a catalyst.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-8

Q12. When sodium peroxide is treated with dilute sulphuric acid, we get .
(a) sodium sulphate and water
(b) sodium sulphate and oxygen
(c) sodium sulphate, hydrogen and oxygen
(d) sodium sulphate and hydrogen peroxide.

Q13. Hydrogen peroxide is obtained by‘the electrolysis of _______.
(a) water
(b) sulphuric acid
(c) hydrochloric acid
(d) fused sodium peroxide
Sol:
(b) Hydrogen peroxide is obtained by electrolysis of sulphuric acid.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-10

Q14. Which of the following reactions is an example of use of water gas in the synthesis of other compounds?

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-11

Sol: (d) Water gas is used in synthesis of compounds like methanol.

Q15. Which of the following ions will cause hardness in water sample?

Q16. Which of the following compounds is used for water softening?
(a) Ca3(P04)2
(b) Na3P04           
(c)  Na6P6018                             
(d) Na2HP04
Sol:
(c)Na6P6018
(Sodium hexametaphosphate) commercially known as calgon is used for water softening.
2CaCl2 + Na2[Na4(P03)6] →Na2[Ca2(P03)6] + 4NaCl

Q17. Elements of which of the following group(s) of periodic table do not form hydrides?
(a) Groups 7,8, 9
(b) Group 13
(c) Group 15,16, 17
(d) Group 14
Sol: (a) Group 7, 8, 9 elements do not form hydrides.

Q18. Only one element of forms hydride.
(a) group 6
(b) group 7
(c) group 8
(d) group 9
Sol: (a) Only one element chromium from group 6 forms hydride, (CrH).

More than One Correct Answer Type
Q19. Which of the following statements are not true for hydrogen?
(a) It exists as diatomic molecule.
(b) It has one electron in the outermost shell.
(c) It can lose an electron to form a cation which can freely exist.
(d) It forms a large number of ionic compounds by losing an electron.
Sol. (c, d) It can lose an electron to form a cation which cannot freely exist.
Generally, it does not form ionic compounds by losing an electron but forms a large number of covalent compounds by sharing electron.

Q20. Dihydrogen can be prepared on commercial scale by different methods. In its preparation by the action of steam on hydrocarbons, a mixture of CO and H2 gas is formed. It is known as _____ .
(a)water gas
(b) syn gas
(c) producer gas
(d) industrial gas
Sol:
(a, b) A mixture of CO + H2 is known as water gas or syn gas (synthesis gas).

Q21. Which of the following statement(s) is/are correct in the case of heavy water?
(a) Heavy water is used as a moderator in nuclear reactor.
(b) Heavy water is more effective as solvent than ordinary water.
(c) Heavy water is more associated than ordinary water.
(d) Heavy water has lower boiling point than ordinary water.
Sol: (a, c) Heavy water is used as a moderator in nuclear reactor and is more associated than ordinary water.

Q22.Which of the following statements about hydrogen are correct?
(a) Hydrogen has three isotopes of which protium is the most common.
(b) Hydrogen never acts as cation in ionic salts.
(c) Hydrogen ion, H+, exists freely in solution.
(d) Dihydrogen does not act as a reducing agent.

Q23. Some of the properties of water are described below. Which of them is/are not correct?
(a) Water is known to be universal solvent.
(b) Hydrogen bonding is present to a large extent in liquid water.
(c) There is no hydrogen bonding in the frozen state of water.
(d) Frozen water is heavier than liquid water.
Sol:(c, d) There is extensive hydrogen bonding in ice. Ice is lighter than water due to empty spaces present in tetrahedrons formed by hydrogen bonds.

Q24. Hardness of water may be temporary or permanent. Permanent hardness is due to the presence of
(a) Chlorides of Ca and Mg in water
(b) Sulphate of Ca and Mg in water
(c) Hydrogen carbonates of Ca and Mg in water
(d) Carbonates of alkali metals in water.
Sol: (a, b) Permanent hardness of water is due to presence of following salts: CaCl2; MgCl2; CaS04; MgS04

Q25. Which of the following statements are correct?
(a) Elements of group 15 form electron deficient hydrides.
(b) All elements of group 14 form electron precise hydrides.
(c) Electron precise hydrides have tetrahedral geometries.
(d) Electron rich hydrides can act as Lewis acids.
Sol: (b, c) All elements of group 14 form electron precise hydrides like CH4 which are tetrahedral in geometry.

Q26. Which of the following statements are correct?
(a) Hydrides of group 13 act as Lewis acids.
(b) Hydrides of group 14 are electron deficient hydrides.
(c) Hydrides of group 14 act as Lewis acids.
(d) Hydrides of group 15 act as Lewis bases.
Sol: (a, d) All elements of group 13 will form electron deficient compounds which acts as Lewis acids.
All elements of group 14 will form electron precise compounds.
Electron rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. NH3 has one lone pair, H20 has two lone pairs and HF has three lone pairs, and so these compounds act as Lewis bases.

Q27. Which of the following statements are correct?
(a) Metallic hydrides are deficient of hydrogen.
(b) Metallic hydrides conduct heat and electricity.
(c) Ionic hydrides do not conduct electricity in solid state.
(d) Ionic hydrides are very good conductors of electricity in solid state.
Sol: (a, b, c) Metallic hydrides are non-stoichiometric hydrides. They conduct heat and electricity. Ionic hydrides conduct electricity only in molten or aqueous state.

Short Answer Type Questions
Q28. How can production of hydrogen from water gas be increased by using water gas shift reaction?
Sol: Water gas is produced when superheated steam is passed over red hot coke or coal at 1270 K in the presence of nickel as catalyst.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-14
Pure H2 from water gas cannot be obtained easily because it is difficult to remove CO. Therefore, to increase the production of H2 from water gas, CO is oxidized to C02 by mixing it with more steam and passing the mixture over heated FeCr04 catalyst at 673 K.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-15

The C02 produced is removed by scrubbing with sodium arsenite solution.

Q29. What are metallic/interstitial hydrides? How do they differ from molecular hydrides?
Sol: Metallic hydrides are formed by d- and f-:block elements. Their hydrides conduct heat and electricity. They are non-stoichiometric, being deficient in hydrogen. For example, LaH2.87, ybH2.55etc.

Metallic hydridesMolecular hydrides
(1)These are formed by el­and f-block elements.(1)These are formed by p-block elements and some s-block elements (Be and Mg).
(2)They conduct electricity.(2)They do not conduct electricity.
(3)They are hard and have metallic luster.(3)They are volatile compounds having low melting and boiling points.

Q30. Name the classes of hydrides to which H20, B2H6 and NaH belong.
Sol: H20 – Electron rich covalent hydride/molecular hydride
B2H6 – Electron deficient molecular hydride
NaH – Ionic hydride

Q31. If same mass of liquid water and a piece of ice are taken, then why is the density of ice less than that of liquid water?
Sol: The mass per unit volume (i.e., mass/volume) is called density. Since water expands on freezing, therefore, volume of ice for the same mass of water is more than liquid water. In other words, density of ice is lower than liquid water and hence ice floats on water.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-16

Q33. Give reasons:
(i) Lakes freeze from top towards bottom.
(ii) Ice floats on water.
Sol: (i) During severe winter, the temperature of water in the lake keeps on decreasing. Since cold water is heavier, it keeps on going into the interior of the lake while warm water keeps on coming to the surface of the lake. This process continues till the temperature of entire water of lakes becomes 4°C. Since density of water is maximum at 277 K, any further decrease in the temperature will decrease its density. As a result, the temperature of the surface water keeps on decreasing and it ultimately freezes. Now, any further decrease in the temperature will decrease the temperature of water below 4°C. This process continues and as a result, the lakes keep on freezing from top to bottom.
(ii) Density of ice is less than water due to presence of empty spaces created because of H-bonding between H20 molecules. Hence, ice floats on water.

Q34. What do you understand by the term ‘auto-protolysis of water? What is its significance?
Sol: Autoprotolysis is a reaction in which two same molecules react to give ions with proton transfer. Water undergoes autoproteolysis, i.e., a proton from one molecule is transferred to another molecule.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-17

On account of autoprotolysis, water is amphoteric in nature.

Q35. Discuss briefly de-mineralisation of water by ion exchange resin.
Sol: Demineralised water free from all soluble mineral salts is obtained by passing water successively through a cation exchange and an anion exchange resin.
In cation exchange process, H+ exchanges for Na+, Ca2+, Mg2+ and in the anion exchange process OH exchanges for anions like CH, HCO3, S02-4, etc. H+ and OH released combine to form water.
H+ + OH → H20

Q36. Molecular hydrides are classified as electron deficient, electron precise and electron rich compounds. Explain each type with two examples.
Sol: Covalent or molecular hydrides are classified into three categories:
(i) Electron deficient hydrides: These hydrides do not have sufficient number of electrons to form normal covalent bonds. Examples are the hydrides of group 13 such as B2H6, (AlH3)n etc.
(ii) Electron precise hydrides: These have exact number of electrons to form normal covalent bonds. Examples are the hydrides of group 14 such as CH4, SiH4, etc.
(iii) Electron rich hydrides: These have more number of electrons than normal covalent bonds. The excess electrons are present in the form of lone pairs. Examples are the hydrides of group 15, 16 and 17 such as

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-18

Q37. How is heavy water prepared? Compare its physical properties with those of ordinary water.
Sol: Heavy water can be prepared by exhaustive electrolysis of water. Comparison of physical properties of H20 and D20.

PropertyH2oD2o
(i)Molecular mass (g mol-1)18.01520.027
(ii)Melting point (K)273.0276.8
(iii)Boiling point (K)373.0374.4
(iv)Density (298 K) g cm-31.00001.1059
(v)Enthalpy of vapourisation (kJ mol-1)40.6641.61

Q38. Write one chemical reaction for the preparation of D202.
Sol: D202 can be prepared by the reaction of D2S04 dissolved in water over Ba02.
Ba02 + D2S04→ BaS04 + D202

Q39. Calculate the strength of 5 volume H202
Sol:
5 volume H202 solution means that 1 L of 5 volume H202 on decomposition gives 5L of 02 at NTP.
ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-19

Q40. (i)Draw the gas phase and solid phase structure of H202.
(ii) H202 is a better oxidizing agent than water. Explain.
Sol:
(i) Structure of H202 is slightly different in gas phase and solid phase.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-20
ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-21

Q41. Melting point, enthalpy of vapourisation and viscosity data of H20 and D20 are given below:

H2oD2o
Boiling point / K373.0374.4
Enthalpy of vapourisation at (373 K)/ kJ mol-140.6641.61
Viscosity/centipoise0.89031.107

On the basis of this data, explain in which of these liquids intermolecular forces are stronger?

Sol: The melting point, enthalpy of vapourisation and viscosity values of all these items depend upon the intermolecular forces

of attraction. Since their values are higher for D20 as compared to those of H20, therefore, intermolecular forces of attraction

are stronger in D20 than in H20.

Q42. Dihydrogen reacts with dioxygen (02) to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen? Justify your answer.
Sol:The isotope of hydrogen which contains one proton and one neutron is deuterium (D). Thus, when dideuterium reacts with dioxygen, deuterium oxide, i.e., heavy water is produced.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-22
The reactivity of H2 and D2 towards oxygen will be different. Since the D – D bond is stronger than H – H bond, therefore, H2 is more reactive than D2 towards reaction with oxygen.

Q43. Explain why HC1 is a gas and HF is a liquid.
Sol: Due to greater electronegativity of F over Cl, F forms stronger H-bonds as compared to Cl. As a result, more energy is needed to break the H-bonds in HF than in HC1 and hence the b.p. of HF is higher than that of HCl. Consequently, H-F is liquid while HC1 is a gas at room temperature.

Q44. When the first element of the periodic table is treated with dioxygen, it gives a compound whose solid state floats on its liquid state. This compound has an ability to act as an acid as well as a base. What products will be formed when this compound undergoes autoionisation?
Sol: The first element is hydrogen and its molecular form is dihydrogen (H2). It reacts with oxygen to form water whose solid state is ice which is lighter than water and floats over water.
Water is amphoteric in nature, i.e., it acts as an acid in the presence of strong bases and acts as a base in the presence of strong acids.

ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-23
ncert-exemplar-problems-class-11-chemistry-chapter-9-hydrogen-24

Q45. Rohan heard that instructions were given to the laboratory attendant to store a particular chemical, i.e., keep it in the dark room, add some urea in it, and keep it away from dust. This chemical acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.
(i) Write the name of this compound.
(ii) Explain why such precautions are taken for storing this chemical.
Sol: (i) The name of the compound is H202. It acts as an oxidizing as well as reducing agent in both acidic and basic media.
(ii) H202 is decomposed by light and dust particles. Urea is added as a negative catalyst, i.e., to check its decomposition.

2H202(1)→2H20(1) + 02(g)

Because of the oxidizing properties, H202 is widely used to control pollution by oxidation of harmful cyanides and obnoxious smelling sulphides present in domestic and industrial effluents. It also helps in sewage disposal by supplying 02 for oxidation of organic matter present – in sewage waters.

NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions

NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 8 Redox Reactions.

Multiple Choice Questions
Single Correct Answer Type

Q1. Which of the following is not an example of redox reaction? 
(a) CuO + H2 → Cu + H20                    
(b) Fe202 + 3CO → 2Fe + 3C02
(c).2K + F2 →2KF                                   

(d) BaCl2 + H2S04 →BaS04 + 2FIC1
Sol: (d) BaCl2 + H2S04 —> BaS04 + 2HC1 is not a redox reaction. It is an example of double displacement reactions.

Q2. The more positive the value of E°, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below, find out which of the following is the strongest oxidizing agent. E° values: Fe3+/Fe2+ = +0.77; I2(g)/I = +0.54;
Cu2+/Cu = +0.34; Ag+/Ag = +0.80 V
(a) Fe3+
(b) I2(g)                    
(c) Cu2+                    
(d) Ag+
Sol:
(d) Since Ag+/Ag has highest positive value of E°, therefore, Ag+ is the strongest oxidizing agent with highest tendency to get reduced

Q3. E° values of some redox complexes are given below. On the basis of these values choose the correct option.
E° values: Br2/Br = +1.90; Ag+/Ag(s) = +0.80 Cu2+/Cu(s) = +0.34; I2(s)/I = +0.54 V
(a) Cu will reduce Br
(b) Cu will reduce Ag
(c) Cu will reduce I                               
(d) Cu will reduce Br2

Sol: (d) Copper will reduce Br2, if the E° of the redox reaction, 2Cu + Br2
CuBr2 is +ve.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-1

Since E° of this reaction is +ve, therefore, Cu can reduce Br2 and hence option (d) is correct.

Q4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
E° values: Fe3+/Fe2+ = +0.77; I2/I = +0.54;
Cu2+/Cu = +0.34; Ag+/Ag = +0.80 V
(a) Fe3+ and I
(b) Ag+ and Cu
(c) Fe3+ and Cu
(d) Ag and Fe3+
Sol: (d)
ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-2

Q5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2S2032_ + I2→S4062- + 2I
S2032- + 2Br2 + 5H20 →2S042- + 4Br + 10H+
Which of the following statements justifies the above dual behaviour of thiosulphate?

(a) Bromine is a stronger oxidant than iodine.
(b) Bromine is a weaker oxidant than iodine.
(c) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(d) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

Bromine being stronger oxidizing agent than I2, it oxidises S of S2O2-3 to SO42- whereas I2 oxidises it only into S4O62- ion.

Q6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(a) The oxidation number of hydrogen is always +1.
(b) The algebraic sum of all the oxidation numbers in a compound is zero.
(c) An element in the free or the uncombined state bears oxidation number zero.
(d) In all its compounds, the oxidation number of fluorine is -1.
Sol: (a) Oxidation number of hydrogen is -1 in metal hydrides like NaH.

Q7. In which of the following compounds, an element exhibits two different oxidation states.
(a) NH2OH
(b) NH4NO3
(c) N2H4

(d) N3H

Sol: (b) NH4NO3 is an ionic compound consisting of NH4+ and NO3 ion.
The oxidation number of N in two species is different as shown below:
In NH;,
Let the oxidation state of N in NH4+ be x.
x + 4x(+l) = +l
x = -3
In NO3
Let the oxidation state of N in NO3  be y,
y + 3 x (-2) = -1
y = +5
ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-4

Only in arrangement (a), the O.N. of central atom increases from left to right. Therefore, option (a) is correct.

Q9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
(a) 3d14s2               
(b) 3d2 4s2               
(c) 3d54s1                 
(d) 3d54s2
Sol:
(d) Highest O.N. of any transition element = (n – 1)d electrons +ns electrons. Therefore, larger the number of electrons in the 3d orbitals, higher is the maximum O.N.
(a) 3d14s2= 3; ‘
(b) 3d2 4s2 = 3 + 2 = 5;
(c) 3d54s1=5 + 1=6
(d) 3d54s2 = 5+2 = 7

Q10. Identify disproportionation reaction
(a) CH4 + 202 → C02 + 2H20
(b) CH4 + 4C12 → CC14 + 4HCl
(c) 2F2 + 20H→2F + OF2 + H20
(d) 2N02 + 20H → N02 + NO3 + H20
Sol: (d) Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,
ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-5

Thus, in reaction (d), N is both oxidized as well as reduced since the O.N. of N increases from +4 in N02 to +5 in N03 and decreases from +4 in N02 to +3 in N02.

Q11. Which of the following elements does not show disproportionation tendency?
(a) Cl
(b) Br  
(c) F  
(d) I
Sol: (c) Being the most electronegative element, F can only be reduced and hence it always shows an oxidation number of-1. Further, due to the absence of d-orbitals, it cannot be oxidized and hence it does not show +ve oxidation numbers. In other words, F cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option (c) is correct.

More than One Correct Answer Type

Q12. Which of the following statement(s) is/are not true about the following decomposition reaction?
2KCIO3 →2KC1 + 302
(a) Potassium is undergoing oxidation.

(b) Chlorine is undergoing oxidation.
(c) Oxygen is reduced.
(d) None of the species are undergoing oxidation or reduction.
Sol: (a, b, c, d) Writing the oxidation number of each element above its symbol,

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-6

(a) The O.N. of K does not change, K. undergoes neither reduction nor oxidation. Thus, option (a) is not correct.
(b) The O.N. of chlorine decreases from +5 in KCl03 to -l in KCl, hence, Cl undergoes reduction.
(c) Since, O.N. of oxygen increases from -2 in KC103 to 0 in 02, oxygen is oxidized.
(d) This statement is not correct because Cl is undergoing reduction and O is undergoing oxidation.

Q13. Identify the correct statement(s) in relation to the following reaction:
Zn + 2HCl → ZnCl2 + H2
(a) Zinc is acting as an oxidant.

(b) Chlorine is acting as a reductant.
(c) Hydrogen ion is acting as an oxidant.
(d) Zinc is acting as a reductant.

(a) The O.N. of Zn increases from 0 to +2 (in ZnCl2) and therefore, Zn acts as a reductant and not as an oxidant. Hence, option (a) is not correct.
(b) The O.N. of Cl does not change and therefore, it neither acts as a reductant nor an oxidant. Hence, option (b) is not correct.
(c) The O.N. of H decreases from +1 in H+ to 0 in H2. Therefore, H+ acts an oxidant. This option is correct.
(d) Zinc acts as reductant because its O.N. changes from 0 to +2. This option is correct.

Q14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its/their compounds.
(a) 3s1
(b) 3dl4s2                 
(c)  3d24s2
(d) 3s23p3

Sol:(b, c, d) Elements which have only s-electron in the valence shell do not show more than one oxidation state. Thus, element with 3s1 as outer electronic configuration shows only one oxidation state of +1.
Transition elements, i.e., elements (b, c) having incompletely filled orbitals in the penultimate shell show variable oxidation states. Thus, element with outer electronic configuration as 3d1 4s2 shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d24s2 shows variable oxidation states of +2, +3 and +4.
p-Block elements also show variable oxidation states due to a number of reasons such as involvement of J-orbitals and inert pair effect. For example, element (d) with 3s2 3p3 as (i.e., P) as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbitals.

Q15. Identify the correct statements with reference to the given reaction.
P4 + 30H + 3H20→ PH3 + 3H2 P02
(a) Phosphorus is undergoing reduction only.

(b) Phosphorus is undergoing oxidation only.
(c) Phosphorus is undergoing oxidation as well as reduction.
(d) Hydrogen is undergoing neither oxidation nor reduction
Sol:

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-8
Because O.N. of P increases from 0 (P4) to +1 (H2P02) and decreases from 0 (P4) to -3 (PH3), therefore, P has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because O.N. of H remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.

Q16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(a)   A13-/A1;  E°= -1.66 V
(b)    Fe2+ /Fe;  E°= -0.44 V
(c) Cu2+/ Cu E°=34 V
(d) F2(g)/2F(aq) E°= 2.87 V
Sol:
(a, b) The electrodes having negative electrode potentials are stronger reducing agents than H2 gas and therefore, will act as anodes.

Short Answer Type Questions
Q17. The reaction Cl2(g) + 20H(aq)→ Cl0(aq) + Cl(aq) + H20(l) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidizing action.
Sol:

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-9
In this reaction, O.N. of Cl increases from 0 (in Cl2) to +1 (in CIO) and decreases to -1 (in Cl). Therefore, Cl2 is both oxidized to CIO and reduced to Cl. Since Cl ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), therefore, Cl2 bleaches substances due to oxidizing action of hypochlorite, ClO ion.

Q18. Mn02-4 undergoes disproportionation reaction in acidic medium but Mn04 does not. Give reason.
Sol:  
In Mn04, Mn is in the highest oxidation state of+7 (i.e.. cannot be oxidized further) and hence it cannot undergo disproportionation.
In contrast, the O.N. of Mn in Mn042- is +6. Therefore, it can increase its O.N. to+7 or decrease its O.N. to some lower value. Thus, Mn042- undergoes disproportionation according to the following reaction.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-10

Here, the O.N. of Mn increases from +6 in Mn042- to +7 in Mn04 and decreases to +4 in Mn02. Thus, Mn042- undergoes disproportionation in acidic medium.

Q19. PbO and Pb02 react with HC1 according to following chemical equations:
2PbO + 4HCl → 2PbCl2 + 2H20
Pb02 + 4HC1 → PbCl2 + Cl2 + 2H20
Why do these compounds differ in their  reactivity?

In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acid-base reaction, because PbO is a basic oxide which reacts with HCl acid.
The reaction (ii) is a redox reaction in which Pb02   gets reduced and acts as an oxidizing agent.

Q20. Nitric acid is an oxidizing agent and reacts with PbO but it does not react with Pb02. Explain why?
Sol: Nitric acid is an oxidizing agent and reacts with PbO to give a simple acid- base reaction without any change in oxidation state. In Pb02, Pb is in +4 oxidation state and cannot be oxidized further, hence no reaction takes place between Pb02 and HN03.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-12

Q21. Write balanced chemical equations for the following reactions:
(i) Permanganate ion (Mn04) reacts with sulphur dioxide gas in acidic medium to produce Mn2+ and hydrogensulphate ion. (Balance by ion- electron method)
(ii) Reaction of liquid hydrazine (N2H4) with chlorate ion (C103) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)
(iii) Dichlorine heptaoxide (C1207) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (C102) and oxygen gas. (Balance by ion-electron method)
Sol:
(i) 2Mn04 + 5S02 + 2H20 + H+→5HS04 + 2Mn2+
Balancing by ion-electron method:

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Q22. Calculate the oxidation number of phosphorus in the following species.
(a) HPO32- and (b) P043-
Sol: (a) Let the oxidation number of P inHPO32- be x. So,
+1 + x + (—6) = —2 x = +3
(b) Let the oxidation number of P in P034 be x. So, x + (-8) = -3
x = +5

Q23. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) Na2S203               
(b) Na2S406             
(c) Na2S03
(d) Na2S04
Sol:
(a) Na2S203

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-17

Thus, the oxidation states of two S atoms in Na2S203 are -2 and +6.
(b) Na2S406

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-18

Q24. Balance the following equations by the oxidation number method.
(i) Fe2+ + H+ + Cr2072- →Cr3+ + Fe3+ + H20
(ii) I2 + N03→ N02 +I03
(iii) I2 + S2032- →I + S4062-

(iv) MnO, + C2042-→ Mn2+ + CO2

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Total increase in O.N. = 5×2= 10
Total decrease in O.N. = 1
To equalize O.N. multiply NO3, by 10
I2 + 10no3→ 10no2 + IO3
Balancing atoms other than O and H
I2 + 10no3→ 10NO2 + 2 IO3
Balancing O and H
I2 + lO no3 + 8H+→ 10NO2 + 2 IO3 + 4H20

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-21

To equalize O.N. multiply C02 by 2. –
Mn02 + C202-4 →Mn2+ + 2CO2
Balance H and O by adding 2H20 on right side, and 4H+ on left side of equation.
MnO2 + C202-4 + 4H+ →Mn2+ + 2CO2 + 2H20

Q25. Identify the redox reactions out of the following reactions and identify the oxidizing and reducing agents in them

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-22

Sol: (i) Writing the O.N. of on each atom,
ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-23
Here, O.N. of Cl increases from -1 (in HCl) to 0 (in Cl2). Therefore, Cl is oxidized and hence HCl acts as a reducing agent.
The O.N. of N decreases from +5 (in HN03) to +3 (in NOC1) and therefore, HN03 acts as an oxidizing agent.
Thus, reaction (i) is a redox reaction.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-24
Here O.N. of none of the atoms undergo a change and therefore, this is not a redox reaction.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-25
Here, O.N. of Fe decreases from +3 (in Fe203) to 0 (in Fe) and therefore, Fe203 acts as an oxidizing agent.
O.N. of C increases from +2 (in CO) to +4 (in C02) and therefore, CO acts as a reducing agent. Thus, this is a redox reaction.
ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-26
Here O.N. of none of the atoms undergo a change and therefore, it is not a redox reaction.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-27

Here. O.N. of N increases from -3 (in NH3) to 0 in (N2) and therefore, NH3 acts as a reducing agent. O.N. of O decreases from 0 (in 02) to -2 (in H,0) and therefore, 02 acts as an oxidizing agent.
Thus, reaction (v) is a redox reaction.

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-28
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Dividing the equation into two half reactions:
Oxidation half reaction: I→ I2
Reduction half reaction: Cr2072-→ Cr3+
Balancing oxidation and reduction half reactions separately as:

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-30

(ii) Step-1: Separate the equation into two half reactions.
The oxidation number of various atoms are shown below:

ncert-exemplar-problems-class-11-chemistry-chapter-8-redox-reactions-31

(i) Balance all atoms other than H and O. This step is not needed, because, it is already balanced.
(ii) The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as:
Fe2+→Fe3+ + e
(iii) Charge is already balanced.
(iv) No need to add H or O.
The balanced half equation is:
Fe2+→ Fe3+ + e         …(i)
Consider the second half equation
Cr2072-→ Cr3+

(i) Balance the atoms other than H and O.

Cr2072-→2Cr3+
(ii) The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there are two Cr atoms, add 6e on the left.
Cr2072- + 6e→ 2Cr3+
(iii) Since the reaction takes place in acidic medium add 14H+ on the left to equate the net charge on both sides.
Cr2072-+6e+14H+→2Cr3+

(iv) To balance FI atoms, add 7H2O molecules on the right.

Cr2072- + 6e + 14H+→ 2Cr3+ + 7H20          .. .(ii)

This is the balanced half equation.

Step-3: Now add up the two half equations. Multiply eq. (i) by 6 so that electrons are balanced.

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NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Chemistry Chapter 7 Equilibrium

NCERT Exemplar Class 11 Chemistry Chapter 7 Equilibrium are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 7 Equilibrium.

Multiple Choice Questions
Single Correct Answer Type

Q1. We know that the relationship between Kc and Kp is Kp = Kc(RT)∆n What would be the value of An for the reaction NH4C1(s) ⇆ NH3(g) + HCl(g)?
(a) 1
(b) 0.5                       
(c) 1.5                    
(d) 2
Sol: (d) The relationship between Kp and Kc is
Kp = Kc (RT)n
Where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants)
For the reaction,
NH4C1(s) ⇆ NH3(g) + HCl(g)
∆n = 2 – 0 = 2

Q2. For the reaction H2(g) + I2(g) ⇌2Hl(g), the standard free energy is ∆G° >0  The equilibrium constant (K) would be
(a) K = 0
(b)> 1                
(c) K=1                  
(d)A:<l
Sol: (d) ∆G° = -RTln K. ∆G° > 0 means ∆G° is +ve. This can be so only if In K is -ve, i.e., K< 1.

Q3. Which of the following is not a general characteristic of equilibria involving physical processes?
(a) Equilibrium is possible only in a closed system at a given temperature.
(b) All measurable properties of the system remain constant.
(c) All the physical processes stop at equilibrium.
(d) The opposing processes occur at the same rate and there is dynamic but stable condition.
Sol: (c) All the physical processes like melting of ice and freezing of water, etc., do not stop at equilibrium.

Q4. PC15, PC13 and Cl2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 x 10“3 molL-1, 1.2 x 10-3 mol L 1 and 1.2 x 10-3 mol L-1 The value of Kc for the reaction PCl5(g) ⇌PCl3(g) + Cl2(g) will be
(a) 8 x 103 molL-1
(b) 1.8 x 10-3 molL-1
(c) 1.8 x 10“3 molL-1                             

(d) 0.55 x 104 molL-1
Sol:
(b) PCl5(g)   ⇌ PCl3(g) + Cl2(g)

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-1

Q5. Which of the following statements is incorrect?
(a) In equilibrium mixture of ice and water kept in perfectly insulated flask, mass of ice and water does not change with time.
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.
(c) On addition of catalyst the equilibrium constant value is not affected.
(d) Equilibrium constant for a reaction with negative AH value decreases as the temperature increases.

Sol: (b)Fe3++SCN ⇌ FeSCN2+(Red)
When oxalic acid is added to a solution containing iron nitrate and potassium thiocyanate, oxalic acid reacts with Fe3+ ions to form a stable complex ion [Fe(C204)3]3-, thus, decreasing the concentration of free Fe3+ ions which in mm decreases the intensity of red colour.
Fe3+ + SCN⇌ [Fe(SCN)]2+ (Red)

Q6. When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture, it becomes pink. On the basis of this information mark the correct answer.
[Co(H20)6]2+ (aq) + 4Cl(aq) [CoCl4]2-(aq) + 6H20(l)

(Pink)                                                                                (Blue)
(a)∆H > 0 for the reaction
(b) ∆H < 0 for the reaction
(c) ∆H = 0 for the reaction
(d) The sign of ∆H cannot be predicted on the basis of this information.
Sol:(a) Since the reaction shifts to backward direction on cooling, this means that the backward reaction is exothermic reaction. Therefore, the forward reaction is endothermic reaction and ∆H > 0.

Q7. The pH of neutral water at 25°C is 7.0. As the temperature increases, ionization of water increases, however, the concentration of H+ ions and OH ions are equal. What will be the pH of pure water at 60°C?
(a) Equal to 7.0          

(b) Greater than 7.0
(c) Less than 7.0                                    
(d) Equal to zero
Sol:(c) At 25°C, [H+] = [OH] = 10 7 and Kw = [H+] [OH ] = 10-14. On heating, Kw increases, i.e., [H+] [OH] > 10-14 As [H+] = [OH], [H+]2 > 10-14 or [H+] > 10-7 M or pH < 7.

Q8. The ionization constant of an acid, Ka is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 x 10-5, 3.0 x 10-8 and 1.8 x 10-4 Which of the following orders of pH of 0.1 mol dm-3 solutions of these acids is correct?
(a) acetic acid > hypochlorous acid > formic acid
(b) hypochlorous acid > acetic acid > formic acid
(c) formic acid > hypochlorous acid > acetic acid
(d) formic acid > acetic acid > hypochlorous acid

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-2
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Q10. Acidity of BF3 can be explained on the basis of which of the following concepts?
(a) Arrhenius concept
(b) Bronsted Lowry concept
(c) Lewis concept
(d) Bronsted Lowry as well as Lewis concept
Sol: (c) According to Lewis concept, a positively charged or an electron deficient species acts as Lewis acid. BF3 is an electron deficient compound with B having 6 electrons only.

Q11. Which of the following will produce a buffer solution when mixed in equal volumes?
(a) 1 mol dm-3 NH4OH and 0.1 mol dm-3 HC1
(b) 0.05 mol dm-3 NH4OH and 0.1 mol dm-3 HC1
(c) 1 mol dm-3 NH4OH and 0.05 mol dm-3 HC1
(d) 1 mol dm-3 CH3COONa and 0.1 mol dm-3 NaOH

Sol: (c) In (c), all HC1 will be neutralized and NH4C1 will be formed. Also some NH4OH will be left unneutralized. Thus, the final solution will contain NH4OH and NH4C1 and hence will form a buffer.

Q12. In which of the following solvents is silver chloride most soluble?
(a) 0.1 mol dm-3 AgN03 solution
(b) 0.1 mol dm-3 HC1 solution
(c) H20                                                     
(d) Aqueous ammonia
Sol: (d) Silver chloride forms a soluble complex with aqueous ammonia.
AgCl + 2NH3→ [Ag(NH3)2]Cl

Q13. What will be the value of pH of 0.01 mol dm-3 CH3COOH (K.A -74 x 10-5)?
(a) 3.4                   
(b) 3.6                      
(c) 3.9                       
(d) 3.0

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Q14. Ka for CH3COOH is 1.8 x 10-5 and Kb for NH4OH is 1.8 x 10-5. The pH of ammonium acetate will be
(a) 7.005                                                 
(b) 4.75
(c) 7.0                                                      
(d) Between 6 and 7
Sol: (c) Ammonium acetate is a salt of weak acid and weak base.

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-6

Q15. Which of the following options will be correct for the stage of half completion of the reaction A ⇌B.
(a) ΔG° = 0                                             
(b) ΔG° > 0
(c) ΔG° < 0                                            
(d) ΔG° = -RT In 2
Sol: (a) A ⇌B
ΔG° = -RT In K
At the stage of half completion of reaction [A] = [B],
Therefore, K = 1. Thus, ΔG° = 0.

Q16. On increasing the pressure, in which direction will the gas phase reaction proceed to re-establish equilibrium, is predicted by applying the Le Chatelier’s principle. Consider the reaction,
N2(g) + 3H2(g) ⇌2NH3(g)
Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?
(a) K will remain same
(b) K will decrease
(c) K will increase
(d) K will increase initially and decrease when pressure is very high
Sol:(a)N2(g) + 3H2(g) ⇌2NH3(g)
According to Le Chatelier’s principle, at constant temperature, the equilibrium composition will change but K will remain same.

Q17. What will be the correct order of vapour pressure of water, acetone and ether at 30°C? Given that among these compounds, water has maximum boiling point and ether has minimum boiling point.
(a) Water < Ether < Acetone
(b) Water < Acetone < Ether
(c) Ether < Acetone < Water
(d) Acetone < Ether < Water
Sol: (b) Greater the boiling point, less is the vapour pressure. Hence, the correct order of vapour pressures will be:
water < acetone < ether.

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-7

Q19. In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?
(a) H2(g) + I2(g) ⇌2HI(g) ‘

(b) PCl5(g) ⇌PCl3(g) + Cl2(g)
(c) N2(g) + 3H2(g) ⇌2NH3(g)
(d) The equilibrium will remain unaffected in all the three cases.
Sol: (d) The equilibrium will remain unaffected in all three cases on addition of small amount of inert gas at constant volume.

Q20. For the reaction N204(g) ⇌2N02(g), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?
(a) The reaction is endothermic.
(b) The reaction is exothermic.
(c) If NO2(g) and N204(g) are mixed 400 K at partial pressures 20 bar and 2 bar respectively, more N204(g) will be formed.
(d) The entropy of the system increases.
Sol: (a, c, d)
(a) As the value of K increases with increase of temperature and K = Kf / Kb, this means that kf increases, i.e., forward reaction is favoured.
Hence, reaction is endothermic.
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Q21. At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist in equilibrium. Which of the following term defines this temperature? .
(a) Normal melting point
(b) Equilibrium temperature
(c) Boiling point
(d) Freezing point
Sol: (a, d) These are normal melting point and freezing point since they are measured at atmospheric pressure.

Q22. The ionization of hydrochloric acid in water is given below:
HCl(aq) + H20(l) ⇌H30 + (aq) +Cl(aq)
Label two conjugate acid-base pairs in this ionization.
Sol: HCl (Acid)               Cl (Conjugate base)
H20 (Base)            H30+ (Conjugate acid

Q23. The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionization and how is it affected by concentration of sodium chloride?
Sol:
(i) Sugar being a non-electrolyte does not ionize in water, whereas NaCl ionizes completely in water and produces Na+ and Cl ions which help in the conduction of electricity.
(ii) When concentration of NaCl is increased, more Na+ and Cl ions will be produced. Hence, conductance increases.

Q24. BF3 does not have proton but still acts as an acid and reacts with NH3. Why is it so? What type of bond is formed between the two?
Sol: BF3 is an electron deficient compound. Hence, it acts as Lewis acid. NH3 has a lone pair of electrons. Hence, acts as Lewis base. A coordinate bond is formed between the two.
H3N: →BF3

Q25. Ionization constant of a weak base MOH, is given by the expression

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-9
Values of ionization constant of some weak bases at a particular temperature are given below:

BaseDimethylamineUreaPyridineAmmonia
5.4 x 10-41.3 x 10-141.77 x lO-91.77 xlO-5

Arrange the bases in decreasing order of the extent of their ionization at equilibrium. Which of the above base is the strongest?
Sol: Greater is the ionization constant (Kb) of a base, greater is the ionization of the base. Order of extent of ionization at equilibrium is dimethylamine > ammonia > pyridine > urea. Dimethylamine is the strongest base due to maximum value of Kb.

Q26. Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
OH, RO, ch3coo , cl
Sol:
Conjugate acids of given bases are H20, ROH, CH3COOH, HC1.
Their acidic strength is in the order
HCl > CH3COOH > H20 >ROH Hence, basic strength is in the order RO > OH > CH3COO > Cl

Q27. Arrange the following in increasing order of pH.
KN03(aq), CH3COONa(aq), NH4Cl(aq), C6H5COONH4(aq)
Sol: KN03: salt of strong acid-strong base, solution is neutral, pH = 7. CHjCOONa: salt of weak acid-strong base, solution is basic, pH > 7.
NH4Cl: salt of strong acid-weak base, solution is acidic, pH < 7.
C6H5COONH4: both weak put NH4OH is slightly stronger than C6H5COOH, pH close to 7 but slightly > 7.
Hence, in order of pH, NH4C1 < C6H5COONH4 > KN03 < CH3COONa.

Q28. The value of Kc for the reaction
2HI(g) ⇌H2(g) + I2(g) is 1 x 10-4At a given time, the composition of reaction mixture is [HI] = 2 x 10-5 mol, [H2] = 1 x 10-5 mol and [I2] = 1 x 10-5 mol. In which direction will the reaction proceed?

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Q29. On the basis of the equation pH = -log [H+], the pH of 10-8 mol dm-3 solution of HC1 should be 8. However, it is observed to be less than 7.0. Explain the reason.
Sol: Concentration 10-8 mol dm-3 indicates that the solution is very dilute. So, we cannot neglect the contribution of H30+ ions produced from H20 in the solution. Total [H30+] = 10-8 + 10-7 M. From this we get the value of pH close to 7 but less than 7 because the solution is acidic.
From calculation, it is found that pH of 10-8 mol dm-3 solution of HC1 is equal to 6.96.

Q30. pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution a 100 times?
Sol:
pH = 5 i.e., [H+] = 10-5 mol L-1
On dilution by 100 times [H+] = 10-7 mol L-1 For a very dilute solution,
Total [H+] = [H30+ ions from acid] + [H20+ ions from water]
= 107 + 107
pH = -log[H+] = -log (2 x 10-7) = 7 – log 2
= 7-0.3010 = 6.6990

Q31. A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaS04 in water is 8 x 10-4 mol dm-3, calculate its solubility in 0.01 mol dm-3 of H2S04.

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Q32. pH of 0.08 mol dm3HOC1 solution is 2.85. Calculate its ionization constant.
Sol: pH of HOC1 = 2.85
-pH = log [H+] or -2.85 = log [H+]
=> [H+] = 1.413 x 10-3

Q33. Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.
Sol: pH of solution A = 6
[H+] = 10-6mol L 1
pH of solution B = 4
[H+] = 10-4 molL-1On mixing one litre of each solution Total volume = 1 L + 1 L = 2 L
Total amount of H+ in 2 L solution formed by mixing solutions A and B = 10-6  + 10-4  mol

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-13
Q34. The solubility product of Al(OH)3 is 2.7 x 10-11. Calculate its solubility in gL and also find out pH of this solution. (Atomic mass of A1 = 27 u).

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-14
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Q35. Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Ksp of PbCl2 = 3.2 x 10-8, atomic mass of Pb = 207 u).

Q36. A reaction between ammonia and boron trifluoride is given below:
:NH3 + BF3 →H3N : BF3
Identify the acid and base in this reaction. Which theory explains it? What is the hybridization of B and N in the reactants?

Sol:Although BF3 does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with NH3 by accepting the lone pair of electrons from NH3 and completes its octet. The reaction can be represented by
BF3 + :NH3→BF3 ← :NH3
Lewis electronic theory of acids and bases can explain it. Boron in BF3 is sp2 hybridised, whereas N in NH3 is sp3 hybridised.

Q37. Following data is given for the reaction:
CaC03(s) → CaO(s) + C02(g)
fH° [CaO(s)] = -635.1 kJ mol1
fH° [COz(g)] = -393.5 kJ mol1
fH° [CaC03(s)] = -1206.9 kJ mol1
Predict the effect of temperature on the equilibrium constant of the above reaction.
Sol:
rH° = ∆fH°  [CaO] + ∆fH° [C02] – ∆fH° [CaC03]
= [-635.1] + [-393.5] – [-1206.9] = 178.3 kJ mol-1 Thus, the reaction is endothermic. Hence, according to Le Chatelier’s principle, on increasing the temperature, the equilibrium will proceed in the forward direction.
Matching Column Type Questions
Q38. Match the following equilibria with the corresponding condition

Column IColumn II
(i)Liquid⇌Vapour(a)Saturated solution
(ii)Solid ⇌Liquid(b)Boiling point
(iii)Solid ⇌Vapour(c)Sublimation point
(iv)Solute(s) ⇌Solute (solution)(d)Melting point ‘
(e)Unsaturated solution

Sol: (i) →(b), (ii) → (d), (iii) → (c), (iv) → (a)

(i) Liquid ⇌Vapour equilibrium exists at the boiling point.
(ii) Solid ⇌Liquid equilibrium exists at the melting point.
(iii) Solid ⇌ Vapour equilibrium exists at the sublimation point.
(iv) Solute(s) ⇌Solute (solution) equilibrium exists in saturated solution.

Q39. For the reaction: N2(g) + 3H2(g) ⇌2NH3(g) equilibrium constant,

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-18
Some reactions are written below in Column I and their equilibrium constants in terms of Kc are written in Column II. Match the following reactions with the corresponding equilibrium constant.

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-19

Sol: (i) →(d), (ii) → (c), (iii) → (b)
ForN2(g) + 3H2(g)⇌2NH3(g)

ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-20

Q40. Match standard free energy of the reaction (Column I) with the corresponding equilibrium constant (Column II).
ncert-exemplar-problems-class-11-chemistry-chapter-7-equilibrium-21

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 6 Thermodynamics.

Multiple Choice Questions
Single Correct Answer Type

Q1. Thermodynamics is not concerned about .
(a) energy changes involved in a chemical reaction.
(b) the extent to which a chemical reaction proceeds.
(C) the rate at which a reaction proceeds.
(d) the feasibility of chemical reaction.
Sol: (c) Thermodynamics is not concerned with rate at which a reaction proceeds. The rate of reaction is dealt by kinetics.

Q2. Which of the following statements is correct?
(a) The presence of reacting species in a covered beaker is an example of open system.
(b) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(c) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(d) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.
Sol: (c) For a closed vessel made up of copper, no matter can be exchanged between the system and the surroundings but energy exchange can occur through its walls.

Q3. The state of gas can be described by quoting the relationship between
(a) pressure, volume, temperature
(b) temperature, amount, pressure
(c) amount, volume, temperature
(d) pressure, volume, temperature, amount
Sol: (d) State of a system can be described by state functions or state variables which are pressure, volume, temperature and amount of the gas (PV= nRT).

Q4. The volume of gas is reduced to half from its original volume. The specific
heat will .
(a) reduce to half (b) be doubled
(c) remain constant (d) increase four times
Sol: (c) The specific heat of a substance is the heat required to raise the temperature of 1 gram of a substance by one degree (1 K or 1 °C). It is an intensive property and is independent of the volume of the substance.

Q5. During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-1
Sol: (c) Exothermic reaction for combustion of one mole of butane is represented as:

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-2
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-3

Q7. In an adiabatic process, no transfer-of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-4
Sol: (c) For free expansion w = 0 For adiabatic process q = 0 From first law of thermodynamics,
∆U=q + w
= 0 + 0 = 0
Since there is no change of internal energy, hence temperature will also remain constant, i.e., ∆T = 0

Q8. The pressure-volume work for an ideal gas can be calculated by using the expression

The work can also be calculated from the pV

plot by using the area under curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from Vi to Vf, choose the correct option.
(a) w (reversible) = w (irreversible)
(b) w (reversible) < w (irreversible)
(c) w (reversible) > w (irreversible)
(d) w (reversible) = w (irreversible) + pex. ∆V
Sol: (b) w (reversible) < w (irreversible)
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-6
Area under the curve is greater in irreversible compression than that of reversible compression.

Q9. The entropy change can be calculated by using the expression ∆S = q rev / T.  When water freezes in a glass beaker, choose the correct statement amongst the following:

When water freezes in a glass beaker, choose the correct statement amongst the following:

(a) ∆S(system) decreases but ∆S(surroundings) remains the same.
(b) ∆S(system) increases but ∆S(surroundings) decreases.
(C) ∆S(system) decreases but ∆S(surroundmgs) increases.
(d) ∆S(system) decreases but ∆S(surroundings) also decreases.

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-7

Q10. On the basis of thermochemical equations (i), (ii) and (iii), find out which of the algebraic relationships given in options (a) to (d) is correct.

Q11. Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (a) to (d) is correct?
(i) C(g) + 4H(g) → CH4(g); ∆rH= kJ mol-1
(ii) C(graphite, s) + 2H2(g) → CH4(g); ∆rH = y kJ mol 1
(a) x = y                   (b) x = 2y           (c)x >y     (d)x< y

Sol: (c) x > y because same bonds are formed  in reactions (i) and (ii)  but bonds
between reactant molecules are broken only in reaction (ii). As energy is absorbed when bonds are broken, energy released in reaction (i) is greater than that in reaction (ii).

Q12. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(a) is always negative
(b) is always positive
(c) may be positive or negative
(d) is never negative
Sol:
(c) Heat of formation of a compound may be positive or negative, e.g.,

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-9

Q13. Enthalpy of sublimation of a substance is equal to
(a) enthalpy of fusion + enthalpy of vapourisation
(b) enthalpy of fusion
(c) enthalpy of vapourisation
(d) twice the enthalpy of vapourisation.
Sol: (a) Enthalpy of sublimation of a substance is equal to enthalpy of fusion + enthalpy of vapourisation.
Sublimation is direct conversion of solid to vapour, i.e., solid → vapour
Writing in two steps, we have solid → liquid → vapour.
Solid → liquid requires enthalpy of fusion.
Liquid →vapour requires enthalpy of vapourisation

Q14. Which of the following is not correct?
(a) ∆G is zero for a reversible reaction.
(b) ∆G is positive for a spontaneous reaction
(c) ∆G is negative tor a spontaneous reaction
(d) ∆G is positive for a non-spontaneous reaction.
Sol:(b) ∆G gives a criterion for spontaneity at constant pressure and temperature.
(i) If ∆G is negative (< 0). the process is spontaneous.
(ii) If ∆G is positiv e (> 0). the process is non-spontaneous.
(iii) If ∆G is zero then reaction is at equilibrium.

More than One Correct Answer Type
Q15. Thermodynamics mainly deals with
(a) interrelation of various forms of energy and their transformation front one from to another.
(b) energy changes in the processes which depend only on initial and final states of the microscopic system containing a few molecules.
(c) how and at what rate these energy transformations are carried out.
(d) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.
Sol: (a, d) Thermodynamics deals with interrelation of various forms of energy and their transformation into each other. It also deals with thermal or mechanical equilibrium. However, if does not tell anything about the rate of reaction.

Q16. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system
(a) qP will be negative                               
(b) ∆γHwill be negative
(c) qp will be positive                                
(d) ∆γHwill be positive.
Sol:(a, b) For an exothermic reaction, qp = -ve, ∆γH = -ve

Q17. The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
(a) flow of heat from colder to warmer body.
(b) gas in a container contracting into one comer.
(c) gas expanding to fill the available volume.
(d) burning carbon in oxygen to give carbon dioxide.
Sol:(c, d) Gas expands or diffuses in available space spontaneously, e.g., leakage of cooking gas gives smell of ethyl mercaptan spontaneously. Moreover, burning of carbon to C02 is also spontaneous.

Q18. For an ideal gas. the work of reversible expansion under isothermal condition 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
can be calculated by using expression w = -nRT In Vf / Vi A sample containing
(a) Work done at 600 K is 20 times the work done at 300 K.
(b) Work done at 300 K is twice the work done at 600 K
(c) Work done at 600 K is twice the work done at 300 K.
(d) ∆U= 0 in both cases.

i.e., work done at 600 K is twice the work done at 300 K. Since each case involves isothermal expansion of an ideal gas, there is no change in internal energy, i.e., U = 0.

Q19. Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
2Zn(s) + 02(g) → 2ZnO(s); ∆H=-693.8 kJ mol-1
(i) The enthalpy of two moles ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(iii) 8 kJ mol -1 energy is evolved in the reaction.
(iv) 693.8 kJ mol-1 energy is absorbed in the reaction.

Short Answer Type Questions

Q20. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is
40.79 kJ mol-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?
Sol: Enthalpy of a reaction is the energy change per mole for the process.
18 g of H20 = 1 mole (∆Hvap = 40.79 kJ moE1)
Enthalpy change for vapourising 2 moles of H20 = 2 x 40.79 = 81.58 kJ ∆H°vap = 40.79 kJ mol -1

Q21. One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
Sol: Water has higher enthalpy of vapourisation. (∆Hr)water > (∆Hr)acetone

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics- 12
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-13
Q24. Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆rH and ∆rH1, ∆rH2, ∆rH3 …. represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆rH for overall reaction and ∆rH1, ∆rH2….. etc. for intermediate reactions.
Sol: By Hess’s law ∆rH = ∆rH1+ ∆rH2 + ∆rH3……………………

Q25. The enthalpy of atomisation for the reaction CH4(g) → C(g) + 4H(g) is 1665 kJ mol-1. What is the bond energy of C – H bond?
Sol: CH4 → C + 4H, ∆H= + 1665 kJ mol-1          ,
Bond energy of (C – H) bond = 1665/4 =416.2 kJ mol

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-14
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-15
Q27. Given that ΔH= 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Sol: It is a spontaneous process. Although enthalpy change is zero but randomness or disorder (ΔS) increases and ΔS is positive. Therefore, in equation, ΔG = ΔH – TΔS, the term TΔS will be negative. Hence ΔG will be negative.

Q28. Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
Sol: Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematicalrelation which relates these three parameters is ΔS = q rev/ T
Here, ΔS = change in entropy                                                  ^
qrcv = heat of reversible reaction ‘
T = temperature

Q29. Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Sol: Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are same.

Q30. At 298 K, Kp for the reaction N204(g)⇌ 2N02(g) is 0.98. Predict whether the reaction is spontaneous or not.
Sol: ΔrG° = -RT ln Kp
= -RT ln (0.98)
Since In (0.98) is negative
.’. ΔrG° is positive
=> the reaction is non spontaneous

Q31. A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in the figure. What will be the value of ΔHfor the cycle as a whole?

ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-16

Sol: For a cyclic process, ΔH = 0

Q32. The standard molar entropy of H2O(l) is 70 J K-1 mol-1. Will the standard molar entropy H20(s) be more, or less than 70 J K -1 mol-1?
Sol: The standard molar entropy of H20 (1) is 70 J K-1 mol-1. The solid form of H20 is ice. In ice, molecules of H20 are less random than in liquid water. Thus, molar entropy of H20 (s) < molar entropy of H20 (1). The standard molar entropy of H20 (s) is less than 70 J K 1 mol-1.

Q33. Identify the state functions and path functions out of the following: enthalpy, entropy, heat, temperature, work, free energy.
Sol: State functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work

Q34. The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Sol: Molar enthalpy of vapourisation is more for water due to hydrogen bonding between water molecules.

Q35. Which quantity out of ΔrG and ΔrG° will be zero at equilibrium?
Sol: Gibbs energy for a reaction in which all reactants and products are in standard state. ΔrG° is related to the equilibrium constant of the reaction as follows
ΔrG = ArG° + RT In K
At equilibrium, 0 = ΔrG° + RT In A ({ΔrG = 0) or    ΔrG° =-RT lnK
ΔrG° = 0 when K= 1
For all other values of K, ArG° will be non-zero.

Q36. Predict the change in internal energy for an isolated system at constant volume.
Sol: For an isolated system w = 0, q = 0
Since ΔU= q + w = 0 + 0 = 0, ΔU= 0

Q37. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Sol: At constant volume
q = ΔU + (-w)
-w = pΔ q = AU + pΔV
ΔV = 0 (at constant volume)
Hence, qv = ΔU + 0 = ΔU= change in internal energy At constant pressure, qp = AU + pΔV
Since ΔU + pΔV=ΔH
=> qp = ΔH change in enthalpy
Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.

Q38. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre.
Sol: During free expansion, external pressure is zero, so Work done, w = -pextΔV
= -0(5 – 1) = 0
Since the gas is expanding isothermally, therefore, q = 0
ΔU = q + w =0+0=0

Q39. Heat capacity (CP) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?
Sol: For water, molar heat capacity = 18 x Specific heat or
Cp = 18 x c
But, specific heat,
C = 4.18 J g-1 K-1 Heat capacity,
Cp = 18 x 4.18 JK 1 = 75.24 JK-1

Q40. The difference between Cp and Cv can be derived using the empirical relation H = U + pV. Calculate the difference between Cp and Cv for 10 moles of an ideal gas.
Sol: Given that, Cv = heat capacity at constant volume,
Cp = heat capacity at constant pressure Difference between Cp and Cv is equal to gas constant (R).
.’. Cp – Cv = nR                                (where, n = no. of moles)
= 10 x 8.314 = 83.14J

Q41. If the combustion of 1 g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Sol: Molar enthalpy change for graphite (ΔH)
= enthalpy change for 1 g x molar mass of C = -20.7×12 = -2.48 x 102 kJ mol-1
Since the sign of ΔH = -ve, it is an exothermic reaction.

Q42. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction?
H2(g) + Br2(g)→2HBr(g)
Given that bond energy of H2, Br2 and HBr is 4.35 kJ mol-1,192 kJ mol-1 and 368 kJ mol -1 respectively.
ncert-exemplar-problems-class-11-chemistry-chapter-6-thermodynamics-17

NCERT Exemplar Solutions Class 11 Chemistry

NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter

NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 5 States of Matter.

Multiple Choice Questions
Single Correct Answer Type

Q1. A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude
(a) pressure increases.
(b) temperature decreases.
(c) pressure decreases.
(d) temperature increases.
Sol: (c) At high altitudes, pressure is low. Hence, boiling takes place at lower temperature and therefore, cooking takes more time. In pressure cooker, pressure is high and hence boiling point increases.

Q2. Which of the following properties of water can be used to explain the spherical shape of rain droplets?
(a)Viscosity
(b)Surface tension
(c)Critical phenomena
(d)Pressure
Sol: (b) Due to surface tension, the surface of the water drops is under tension and tends to take spherical shape to reduce the tension.
Q3. A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin.
The plots at different values of pressure are shown in the figure. Which , of the following order of pressure is correct for this gas?
ncert-exemplar-problems-class-11-chemistry-chapter-5-states-of-matter-1

(a)P1 >P2 >P3 >P4
(b) P1 =p2 =p3 =p4
(c) P1 <P2<P3<P4

(d) px <p2=p3<p4

Sol: (c) At a particular temperature pV= constant.
Thus,
P1V1 =P2V2=P3V3=P4V4
As V1> V2 > V3 > V 4, therefore, P1 <P2 < P3 < P4.

Q4. The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon
(a) charge of interacting particles.
(b) mass of interacting particles.
(c) polarisability of interacting particles.
(d) strength of permanent dipoles in the particles.
Sol: (c) London dispersion forces operate only over very short distance. The energy of interaction varies as
1/(distance between two interacting particles)6 Large or more complex are the molecules, greater is the magnitude of London forces. This is obviously due to the fact that the large electron clouds are easily distorted or polarised.
Hence, greater the polarisability of the interacting particles, greater is the magnitude of the interaction energy.

Q5. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is
(a) more than unit electronic charge.
(b) equal to unit electronic charge.
(c) less than unit electronic charge.
(d) double the unit electronic charge.
Sol: (c) Partial charge is a small charge developed by displacement of electrons. It is less than unit electronic charge and is represented as δ+ or δ

Q6. The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?
(a) 0.8 x 105atm                                     
(b) 0.008 Nm-2
(c) 8 x 104 Nm -2                                     

(d) 0.25 atm
Sol: (c) Let the number of moles of dihydrogen and dioxygen be 1 and 4.

ncert-exemplar-problems-class-11-chemistry-chapter-5-states-of-matter-2

Q7. As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(a) Increases
(b) Decreases
(c) Remains same
(d) Becomes half
Sol: (a) At constant volume, as the temperature is increased, pressure also increases.

Q8. Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles.
Following are the critical temperatures of some gases.

Gasesh2He02n2
Critical temperature in Kelvin33.25.3154.3126

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first.
(a) H2, He, 02, N2
(b) He,02,H2,N2
(c) N2,02,He,H2                                       

(d) 02,N2,H2,He
Sol: (d) Higher the critical temperature, more easily is the gas liquefied. Hence, order of liquefaction starting with the gas liquefying first will be: 02, N2, H2, He.

Q9. What is SI unit of viscosity coefficient (η)?
(a) pascal
(b) N s m-2              
(c) km -2 s                
(d) N m-2

ncert-exemplar-problems-class-11-chemistry-chapter-5-states-of-matter-3
Q10. Atmospheric pressures recorded in different cities are as follows:

CitiesShimlaBengaluruDelhiMumbai
p in N/m21.01 x 1051.2 x 1051.02 x 1051.21 x 105

Consider the above data and mark the place at which liquid will boil first.
(a) Shimla
(b) Bengaluru
(c) Delhi
(d) Mumbai
Sol: (a) Shimla has lowest atmospheric pressure, hence liquid will boil first in this city. Boiling of a liquid takes place when the vapour pressure becomes equal to the atmospheric pressure.

Q11. Which curve in figure represents the curve of ideal gas?

ncert-exemplar-problems-class-11-chemistry-chapter-5-states-of-matter-4

(a) only
(b) C and D only
(c)E and F only                                    
(d)A and B only

Sol: (a) For curve B, value of PV is constant and for an ideal gas plot of PV vs P is a straight line.

Q12. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?
(a)Increase
(b)No effect
(c) Decrease
(d) No regular pattern will be followed
Sol: (c) Intermolecular force of liquid decreases with increase in temperature, hence viscosity of liquid also decreases. However, some exceptions are there like liquid proteins and liquid sulphur.

Q13.How does the surface tension of a liquid vary with increase in temperature?
(a) Remains same
(b) Decrease
(c) Increase
(d) No regular pattern is followed
Sol: (b) Surface tension of a liquid decreases with increase in temperature due to less forces of attraction between the molecules.

More than One Correct Answer Type
Q14.With regard to the gaseous state of matter which of the following statements are correct?
(a) Complete order of molecules (b) Complete disorder of molecules
(c) Random motion of molecules (d) Fixed position of molecules
Sol: (b, c) In gaseous state, molecules are in a state of random motion, i.e., it is the state in which molecules are disorderly arranged. Gaseous state has higher entropy than the liquid as well as solid.

Q15. Which of the following figures does not represent 1 mole of dioxygen gas at STP?
(a) 16 grams of gas                                
(b) 22.7 litres of gas
(c) 6.022 x 1023 dioxygen molecules
(d) 11.2 litres of gas
Sol: (a, d) 1 mole of dioxygen represents 32 g of 02, 22.7 L of 02 or 6.022 x 1023 molecules of o2 gas.

Q16. Under which of the following two conditions applied together, a gas deviates most from the ideal behaviour?
(a) Low pressure (b) High pressure
(c) Low temperature (d) High temperature
Sol: (b, c) A gas which obeys the ideal gas equation, p V = nRTunder all conditions of temperature and pressure is called an ‘ideal gas’.
However, there is no gas which obeys the ideal gas equation under all conditions of temperature and pressure. Hence, the concept of ideal gas is only theoretical or hypothetical. The gases are found to obey the gas laws fairly well when the pressure is low or the temperature is high.
Such gases are, therefore, known as ‘real gases’. All gases are real gases. Hence, at high pressure and low temperature, a real gas deviates most from ideal behaviour.

Q17. Which of the following changes decrease the vapour pressure of water kept in a sealed vessel?
(a) Decreasing the quantity of water
(b) Adding salt to water
(c) Decreasing the volume of the vessel to one-half
(d) Decreasing the temperature of water
Sol: (b, d) Vapour pressure does not depend upon the quantity of water or size of the vessel. It decreases on adding salt to water or decreasing the temperature of water.

Short Answer Type Questions

Q18. If 1 gram of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume? CO, H20, CH4, NO

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Q19. Physical properties of ice, water and steam are very different. What is the chemical composition of water in all the three states?
Sol: The chemical composition of water remains the same in all the physical states, i.e., solid, liquid and gas.

Q20. The behaviour of matter in different states is governed by various physical laws. According to you what are the factors that determine the state of matter?
Sol: Pressure, temperature, mass and volume are the factors that determine the
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Q21. Use the information and data given below to answer the questions (a) to (c):
• Stronger intermolecular forces result in higher boiling point.
• Strength of London forces increases with the number of electrons in the molecule.
• Boiling point of HF, HC1, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.
(a) Which type of intermolecular forces are present in the molecules HF, HC1, HBr and HI?
(b) Looking at the trend of boiling points of HC1, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here?
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?
Sol: (a) All the given molecules viz. HF, HC1, HBr and HI have permanent di-poles. Hence, all of them possess dipole-dipole and London forces. HF in addition to dipole-dipole and London forces also has hydrogen bonding.
(b) Electronegativity of Cl, Br and I is in the order: Cl > Br > I. Therefore, polar character and hence dipole-dipole interactions should be in the order HCl > HBr > HI. But boiling points are found to be in the order HCl < HBr < HI. This shows that London forces are predominant. This is because London forces increase as the number of electrons in the molecule increases. In this case, the number of electrons increases from HC1 to HI.
(c) Due to very high electronegativity of F, HF is most polar and also there is hydrogen bonding present in it. Hence, it has the highest boiling point.

Q22. What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Sol: When temperature and pressure of a gas are 273.15 K (or 0°C) and 1 atm
(or 1 bar or 105 pascal), such conditions are called standard temperature and pressure conditions (STP). Under these conditions, the volume occupied by 1 mole of each and every gas is 22.4 L. Hence, the molar volume of N2 and Ar at 273.15 K and 1 atm is 22,4 L.

Q23. A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Sol:  At high temperature and low pressure, the gases behave ideally since the two postulates of kinetic theory of gases are true under these conditions.
(i) The volume of a molecule of a gas is negligible as compared to its complete volume.
(ii) There is negligible force of attraction between the molecules of a gas.

Q24. Two different gases ‘A’ and ‘9’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly, the gas ‘A’ liquefies but gas ‘B’ does not liquefy even on applying high pressure until it is cooled. Explain this phenomenon.
Sol: Gas ‘A’ is at critical temperature and therefore liquefies. Gas ‘B’ is at a temperature higher than critical temperature and therefore, does not liquefy even on applying high pressure.

Q25. Value of universal gas constant (R) is same for all gases. What is its physical significance?
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Q26. One of the assumptions of kinetic theory of gases states that “there is no force of attraction between the molecules of a gas.” How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
Sol: The statement that there is no force of attraction between the molecules of a gas is true at low pressure and high temperature only. If the statement is true under all conditions of temperature and pressure, it will not be possible to liquefy an ideal gas.

Q27. The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:

water, alcohol (C2H5OH) and hexane [CH3(CH2)4CH3)].

Sol: In the above given molecules, only hexane [CH3(CH2)4CH3] is a non-polar molecule in which only London dispersion forces exist. These forces are very weak while both water and ethanol are polar molecules in which dipole- dipole interactions as well as H-bonding exists.
However, since H-bonding interactions are much stronger in water than ethanol, therefore, it possesses stronger intermolecular forces than alcohol and hexane. Hence, the increasing order of surface tension is
Hexane < Alcohol < Water

Q28. Pressure exerted by saturated water vapour is called aqueous tension. What correction term will you apply to the total pressure to obtain pressure of dry gas?
Sol: Whenever a gas is collected over water, it is moist and saturated with water vapour which exerts their own pressure. The pressure due to water vapour is called aqueous tension. Thus, the total pressure of the gas (pmoist gas) is

Pmoist gas = P drygas

Thus P drygas  is given as

P drygas = Pmoist gas – aqueous tension. Hence, the correction term applied to the total pressure of the gas in order to obtain pressure of dry gas is Pmoist gas – aqueous tension.

Q29. Name the energy which arises due to motion of atoms or molecules in a body. How is this energy affected when the temperature is increased?
Sol: The energy generated due to motion of atoms or molecules in a body is thermal energy. It is measured as average kinetic energy of molecules. It increases with increase in temperature.
K. E ∝ T

Q30. Name two intermolecular forces that exist between HF molecules in liquid state.
Sol: (i) Dipole-dipole interactions
(ii) Hydrogen bonding

Q31. One of the assumptions of kinetic theory of gases is that there is no force of attraction between the molecules of a gas.
State and explain the evidence that shows that the assumption is not applicable for real gases.
Sol: Real gases can be liquefied on cooling and compressing which shows that there are forces of attractions between gaseous molecules.

Q32. Compressibility factor, Z, of a gas is given as Z = PV/nRT
(i) What is the value of Z for an ideal gas?
(ii) For real gas what will be the effect on value of Z above Boyle’s temperature?
Sol: (i) For ideal gas, Z = 1.
(ii) For a real gas, above Boyle’s temperature, gas shows positive deviation and hence Z > 1

Q33. The critical temperature (Tc) and critical pressure (Pc) of C02 are 30.98°C and 73 atm respectively. Can C02(g) be liquefied at 32°C and 80 atm pressure?
Sol: C02 gas cannot be liquefied at 32°C by applying a pressure of 80 atm. This is because the temperature is higher than critical temperature of C02.

Q34. For real gases the relation between P, V and T is given by van der Waals equation:

where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas.
‘a’ is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ‘b’. Give reason. 02, C02, H2, He
(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason.CH4, O2, H2
Sol:
(i) As ‘b’ represents molar volume occupied by the gas molecules, greater the size of the molecules, greater is the volume occupied by 1 mole of molecules. The size and hence the value of ‘b’ increase in the order: H2 < He < 02 < C02.
As all the given molecules are non-polar, the magnitude of intermolecular attractions and hence the value of ‘a’ increases with the increase in number of electrons in the molecule, i.e., in the order: CH4 > 02 > H2. (Greater the number of electrons, greater is the size of electron cloud, greater is the polarization of the molecule, greater is the attraction).

Q35. The relation between pressure exerted by an ideal gas (Pideal) and observed pressure (Preal) is given by the equation

If pressure is taken in N m-2, number of moles in mol and volume in m3,Calculate the unit of ‘a’. What will be the unit of ‘a’ when pressure is in atmosphere and volume in dm3?

Q36. Name two phenomena that can be explained on the basis of surface tension.
Sol:(i) Spherical shape of liquid drops.
(ii) Cleansing action of soaps and detergents. ‘
(iii) Capillary action —» Rise and dip of liquid in the column of capillary.

Q37. Viscosity of a liquid arises due to strong intermolecular forces existing between the molecules. Stronger the intermolecular forces, greater is the viscosity. Name the intermolecular forces existing in the following liquids and arrange them in the increasing order of their viscosities. Also give reason for the assigned order in one line.Water, Hexane (CH3CH2CH2CH2CH2CH3), Glycerine (CH2OHCH(OH)CH2OH)

Sol: The increasing order of viscosity is Hexane < Water < Glycerine
In hexane, there are only London forces, whereas in water and glycerine there is hydrogen bonding which is stronger in case of glycerine (because of the presence of three -OH groups) than that in water.

Q38. Explain the effect of increasing the temperature of a liquid on intermolecular forces operating between its particles. What will happen to the viscosity of a liquid if its temperature is increased?
Sol: If the temperature is increased, the intermolecular forces become weak and kinetic energy increases. As the temperature increases the kinetic energy can overcome intermolecular forces, hence viscosity decreases and liquids can flow easily.

Q39. The variation of pressure with volume of the gas at different temperatures can be graphically represented as shown in figure.

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On the basis of this graph answer the following questions.
(i) How will the volume of a gas change if its pressure is increased at constant temperature?
(ii) At a constant pressure, how will the volume of a gas change if the temperature is increased from 200 K to 400 K?
Sol: (i) The volume of a gas will decrease if the pressure on the gas is increased keeping the temperature constant.
(ii) The volume of a gas increases if the temperature is increased keeping the pressure constant.

Q40. Pressure versus volume graphs for a real gas and an ideal gas are shown in the figure.

Answer the following questions on the basis of this graph.
(i) Interpret the behaviour of real gas with respect to ideal gas at low pressure.
(ii) Interpret the behaviour of real gas with respect to ideal gas at high pressure.
(iii) Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.

Sol: (i) At low pressure, the two* curves almost coincide. This shows that at low pressure, the real gases show very small deviation from ideal behaviour.
(ii) At high pressure, the curves are far apart. This shows that real gases show large deviations at high pressure.
(iii) At point A, where the two curves intersect each other, the real gas behaves exactly like ideal gas and the pressure and volume corresponding to this point are P1 and V1

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Matching Column Type Questions
Q41. Match the graph between the following variables with their names.

Column I (Graphs)Column II (Names)
(i) Pressure vs temperature graph at constant molar volume.(a) Isotherms
(ii) Pressure vs volume graph at constant temperature.(b) Constant temperature curve
(iii) Volume vs temperature graph at constant pressure.(c) Isochores
(d) Isobars


Sol: (i) →(c); (ii) → (a); (iii) → (d)
(i) Pressure vs temperature graph at constant volume – Isochores
(ii) Pressure vs volume graph at constant temperature – Isotherms
(iii) Volume vs temperature graph at constant pressure – Isobars

Q42. Match the following gas laws with the equation representing them.

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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.
Q44. Assertion (A): Three states of matter are the result of balance between intermolecular forces and thermal energy of the molecules. .
Reason (R): Intermolecular forces tend to keep the molecules together but thermal energy of molecules tends to keep them apart.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (a) States of matter of a substance depend on the intermolecular forces.

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Intermolecular forces keep the molecules together while thermal energy of molecules tends to keep them apart.

Q45. Assertion (A): At constant temperature, PV vs V  plot for real gases is not a straight line.
Reason (R): At high pressure all gases have Z> 1 but at intermediate pressure most gases have Z < 1.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (b) PV versus Fplot is straight line for ideal gases but not for real gases.
At high pressure, there is positive deviation from ideal behaviour (Z < 1) but at intermediate or moderate pressure, the real gases pass negative deviation from ideal behavior (Z < 1).

Q46. Assertion (A): The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature.
Reason (R): At high altitude atmospheric pressure is high.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (c) The temperature at which the vapour pressure of a liquid becomes equal to the external pressure is called its boiling point. Atmospheric pressure decreases with increase in altitude.

Q47. Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.
Reason (R): Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.
(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) A is true but R is false.
(d) A is false but R is true.
Sol:(a) According to Andrews, a real gas cannot be liquefied‘above critical ‘ temperature, whatever pressure is applied. Intermolecular force of gas molecules is low above critical temperature.

Q48. Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.
Reason (R): The density of liqtiid and gaseous phase is equal to critical temperature.
(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) A is true but R is false.
(d) A is false but R is true.
Sol:(a) At critical state, surface tension of liquid is zero hence liquid passes into gaseous state imperceptibility and continously . At this stage , the liquid and vapour phase becomes equal.

Q49. Assertion (A): Liquids tend to have maximum number of molecules at their surface.
Reason (R): Small liquid drops have spherical shape.
(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) A is true but R is false.
(d) A is false but R is true.
Sol: (d) Liquid tends to acquire minimum surface area due to surface tension. Thus, small liquid drops are spherical. In spherical shape, surface area is minimum.
Long Answer Type Questions
Q50. Isotherms of carbon dioxide at various temperatures are represented in the following figure. Answer the following questions based on this figure.
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(i) In which state will C02 exist between the points a and b at temperature T1
(ii) At what point will Co2 start liquefying when temperature is T1?

(iii) At what point will C02 be completely liquefied when temperature is T2?
(iv) Will condensation take place when the temperature is T3
(v) What portion of the isotherm at T1 represent liquid and gaseous C02 at equilibrium?

Sol: (i) At T1 (between points ‘a’ and ‘b’) C02 exists in gaseous state.
(ii) At point b, C02 starts liquefying at T1
(iii) At temperature T2, at point ‘g’ C02 will be completely liquefied.
(iv) No condensation at T3 since T3 > Tc.
(iv) At T1 liquid and gaseous C02 are at equilibrium between b and c.

Q51. The variation of vapour pressure of different liquids with temperature is shown in figure

(i) Calculate graphically boiling points of liquids A and B.
(ii) If we take liquid C in a closed vessel and heat it continuously, at what temperature will it boil?
(iii) At high altitude, atmospheric pressure is low (say 60 mm Hg). At what temperature liquid D boils?
(iv) Pressure cooker is used for cooking food at hill station. Explain in terms of vapour pressure why is it so?

Sol: (i) Boiling point of liquid A ≃315 K, B ≃ 345 K
(ii) In a closed vessel, liquid C will not boil because pressure inside keeps on increasing.
(iii) Temperature corresponding to 60 mm ≃ 313 K.
(iv) A liquid boils when its vapour pressure becomes equal to atmospheric pressure. At hill station, atmospheric pressure is low. Therefore, liquid boils at a lower temperature and cooking is not perfect. In a pressure cooker the pressure inside increases and the liquid boils at a higher temperature.

Q52. Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid up to critical temperature in a closed vessel? In this situation what will be the state of the substance?
Sol: When a liquid is heated up to its critical temperature in a closed vessel, it does not pass through a two phase region and substances remain in one phase. There is a continuity between a gaseous and liquid state. The term fluid is used for either a liquid or a gas to recognize this continuity. Liquid and gas can be distinguished only when the fluid is below its critical temperature and the surfaces separating them can be seen. At critical temperature, liquid passes into gaseous state continuously and the surface separating the two phases disappears. A gas below the critical temperature can be liquefied by applying pressure.

Q53. Why does sharp glass edge become smooth on heating it up to its melting point in a flame? Explain which property of liquids is responsible for this phenomenon.
Sol: On heating the glass, it melts and takes up rounded shape at the edges which has minimum surface area. This is due to the property of surface tension of liquids.

Q54. Explain the term ‘laminar flow’. Is the velocity of molecules same in all the layers in laminar flow? Explain your answer.
Sol: When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of the upper layers increases as the distance of layers from the fixed layer increases.
This type of flow in which there is a regular gradation of velocity on passing from one layer to the next is called laminar flow.

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In laminar flow, the velocity of molecules is not same in all the layers because every layer offers some resistance or friction to the layer immediately below it.

Q55. Isotherms of carbon dioxide gas are shown in figure. Mark a path for changing . gas into liquid such that only one phase (i.e. either a gas or liquid) exists at any time during the change. Explain how the temperature, volume and pressure should be changed to carry out the change.

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Sol: In isotherm of carbon dioxide, it is possible to change a gas into liquid or a liquid into gas by a process in which always a single phase is present.
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If we move vertically from point A to F by increasing the temperature, then we can reach the point G by compressing the gas at constant temperature along this (isotherm at 31.1°C). Now we can move vertically downwards to D by lowering the temperature. As soon as we cross point H on critical isotherm, we get liquid. If process is carried out at critical temperature, substance always remains in one phase. Hence the path for the change is A → F → G → H → D