The Great Stone Face – II – CBSE Notes for Class 8 English Honeydew

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

Many years passed. Ernest was now a middle aged man. Age brought white hairs upon his head and wrinkles across his face. But it also made him very wise. He became famous in the valley. Learned men from cities came to see him and talk with him.
When Earnest had been growing old, a new poet who was previously the native of that valley arrived there. He had spoken high of the Great Stone Face also in his poem. His songs also reached Ernest’s ears. It appeared to him that his face had the likeness of the Stone Face.
The poet too had heard of Ernest’s wisdom and wished to meet him. One summer day he arrived at Ernest’s door. He sought a night’s shelter. Ernest readily agreed. The two talked together. The poet found his host very wise, gentle, kind and hospitable. Ernest looked into the poet’s eyes and features. He compared the poet’s face with that of the Stone Face. When the poet asked him why he was looking sad he told that all through his life he had awaited the fulfilment of a prophecy and when he read his poems he became convinced that the poet was the real Stone Faced man.
The poet, however, claimed that he did not bear the likeness to the stone face. It was true that he had high dreams in his mind but sometimes he himself had no belief in those dreams. The eyes of both were wet with tears. In the evening, together they went to a meeting place. Ernest spoke out his thoughts. His words had power because they had depth. They were ttye words of life, a life of good deeds and selfless love. His face took on a grand expression. The poet cried out that Ernest was the real likeness of that Stone Face. The people agreed with him. The prophecy was fulfilled, the poet thought so. But Ernest still kept hoping that some wiser and better man than himself would appear looking very much like the Great Stone Face.

The post The Great Stone Face – II – CBSE Notes for Class 8 English Honeydew appeared first on Learn CBSE.

How The Camel Got His Hump – CBSE Notes for Class 8 English It So Happened

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

The world had just begun. The animals had started working for humans. The Horse, the Dog, and the Ox pulled loads and served man. But there was only one animal Camel that refused to work. It lived in the desert and ate thorny plants, When anybody spoke to him, he said ‘Humph!’.
One Monday morning the Horse came to the Camel. He told the desert animal to work like them. So did the Dog and the Ox. But they had no success. Then the three complained to Man against the Camel. But Man grew angry and asked them to work double to compensate Camel’s work. So the animals were not happy. They held a panchayat. The Camel laughed at them and went away.
Soon the Djinn or the Spirit in charge of deserts came there. He listened to their complaint and agreed with them. He decided to punish such an idle animal. He found the Camel looking at his own reflection in the pool of water. He had made the three other animals do extra work since Monday morning. When the Djinn aked him to clarify, the Camel only said ‘Humph’! He showed no willingness to work.
The Djinn put a curse on the Camel. Soon the Camel’s back puffed up into a big hump. It was Thursday. The Camel was told to work for three days without eating anything, because his food was stored in his hump. But the hump on his back made it difficult for him to carry’ the load. The Djinn said that it would serve as a storehouse of food. He also said that it would disappear when the Camel would learn to behave properly.
The Camel went away to join the Three. From that day he has been wearing a hump but there was no escape from w’ork. He has, however, still not learned to behave.

The post How The Camel Got His Hump – CBSE Notes for Class 8 English It So Happened appeared first on Learn CBSE.

Children at Work – CBSE Notes for Class 8 English It So Happened

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

Children are made to work and earn in different ways. Some of them help their family make a livelihood. Many run away from unhappy homes and they need to support themselves. They don’t go to any school.
Velu was an eleven year old boy. He ran away from home because his father was addicted to drinking. He used to beat him and his sister and snatched whatever they earned.
The Kanyakumari Express pulled in at Chennai Station. Velu got down and sat on a bench. He had never before seen so many people with their luggage. The noise was terrible. He was feeling tired, miserable and hungry. In his bundle he carried a shirt, a towel and a comb. He had neither money nor ticket. He tried to sleep on the floor near the door. He heard the rough voice of a girl around his own age. The girl was carrying a big bag. She was picking up plastic cups and putting them into the sack. Her name was Jaya. She sat down next to him. She said she could find some food for him to eat. She stood up to go. Velu had nowhere else to go. So he ran after her.
The two got to the road. Smoke and dust made Velu’s head spin. The girl dragged him to the other side of the street. She took him on to a bridge. Velu peeped over the railing. The road under him ran to the city. Velu saw big buildings in the distance. Jaya warned him not to get caught otherwise he would be in jail. His bare feet burned on the hot tar road as he walked.
They finally stopped in front of a big building. A marriage was going to be performed inside. Behind the hall was a garbage heap. The girl picked up a crushed banana and a vada which Velu ate unwillingly. He felt better. Then the girl went to home to drop her full bag and get another one. Velu also joined her.
Jaya and Velu walked along the roads and came to a bridge across a dirty drain. They came to a row of huts built out of wood, plastic and sheets. It was Jaya’s village. She dumped her sack there and gave him a pair of old shoes, a sack and stick. He wondered what work she wanted him to do. She told him that they were rag pickers. They collected things from rubbish bins-paper, plastic and glass. It was sold to Jaggu, a junk dealer. Velu had no option. He decided to work as a rag picker until he found a better job.

The post Children at Work – CBSE Notes for Class 8 English It So Happened appeared first on Learn CBSE.

The Selfish Giant – CBSE Notes for Class 8 English It So Happened

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

The Giant’s garden was very beautiful. Children loved to play in it. Every afternoon, as they were coming from school, they used to enter the large lovely garden with flowers and peach-trees.
But the Giant was selfish. One day when he returned home after seven years of ‘ stay with his friend, he saw the children playing in the garden. He shouted at them
angrily and they ran away. The Giant decided not to allow anybody else to play there. So he built a high boundary wall around it. He also put up a notice that outsiders entering the garden will be punished.
The poor children had no other place to go. They didn’t like to play on the dusty road full of stones.
Then the spring came all over the country. Only in the garden of the selfish Giant it was still winter. No flowers bloomed and no birds sang. The only visitors were the snow and the frost which painted all the trees silver. The north wind roared there all day. It also asked the hailstones to come.
The Giant wondered why the spring passed by his garden. It was always winter there.
One morning he heard some lovely music of a linneta song bird outside his window. He jumped out of bed and looked out. He saw a wonderful sight. The children had crept in through a hole in the wa^l. They were sitting on the branches of trees which gladly welcomed them with flowers. Only in one corner it was still winter. A small boy was wandering all around the tree and crying. He was too small to climb up.
The Giant’s heart melted. He went out in the garden. The children ran away on seeing him. The garden became winter again. But when he put the small boy up into the tree, it broke into flowers. The boy threw his arms round his neck and kissed him. Other children noticed the change in the Giant’s behaviour. So they too came back. The Giant knocked down the wall and started playing with them. He wanted to see the small boy whom he had put up into the tree. But he had gone away.
Years went by and he grew old and weak. He realised that the children were the most beautiful flowers of all.
One winter morning he saw a tree covered with lovely white flowers and silver fruit. Under it stood the little boy. He came running, close to the boy. He noticed two nail wounds each on the boy’s arms and on the little feet. He declared angrily that he would kill cruel man who had injured him. At this the child told him that those were the wounds of Love. It further told that he had come to take him to his garden, which was paradise. In the afternoon the Giant was found lying dead under the tree.

The post The Selfish Giant – CBSE Notes for Class 8 English It So Happened appeared first on Learn CBSE.

The Treasure Within – CBSE Notes for Class 8 English It So Happened

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

Every child is capable of doing big things. He can become a hero in the area of his interest. What is needed most is to discover his talent or aptitude. Hafeez is a living example of such a hero.
Hafeez Contractor is today India’s leading architect. But he was unhappy and careless at school. He loved doing things and hated mechanical learning. He used to have terrible dreams about appearing for a maths examination in which he knew nothing at all. When he reached the third standard, he lost interest in studies. He took interest in games and playing practical jokes on others. He passed all the school exams by using unfair means.
But, later, one sentence spoken to him by his Principal changed his life. He was in the 11th standard then. The Principal said that Hafeez was a good student, but had little interest in studies. He (the principal had taken care of him till that day but now he could no longer take care of him He should study hard. Only sports won’t take him far. He took his final examination. He could never get more than second class.
No doubt he was pulled up by his teachers at school. He received a caning every week for not doing homework or for his bad behaviour. Caning did hurt him badly, but he forgot about it soon when h$ was in playground. Every Saturday he went to see a movie. He was the leader of a gang of boys and took part in going fights. He used to open his textbooks only on the day just before exams.
How then he could become an architect? Hafeez told the interviewer that he wanted to join the army or the police force. But his mother told him to do his graduation first. So he joined Jaihind College in Bombay. There he studied French for seven years without learning anything. He later learnt this language from his cousin, an architect’s wife.
It so happened that Hafeez saw somebody drawing a window detail. He told the man that the window won’t open. He had a bet and he won it. His cousin’s husband asked him to design a house and he did it. He then asked Hafeez to join architecture. And Hafeez got first position in the entrance test in that college of architecture. He thinks that all his qualities grew from what he used to play and do during school.
He admitted frankly that he was bad at languages. He was also poor in maths. Gifted boys are said to be poor in learning. Class toppers often do very ordinarily later on. He himself could sketch on the spot. He judged the taste of a client from his way of talking and eating. Hafeez always hated maths. But now as an architect he loves that subject.

The post The Treasure Within – CBSE Notes for Class 8 English It So Happened appeared first on Learn CBSE.

NCERT Exemplar Problems 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

Sol: (c) Exothermic reaction for combustion of one mole of butane is represented as:

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.

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 V_i to V_f, choose the correct option.
(a) w (reversible) = w (irreversible)
(b) w (reversible) < w (irreversible)
(c) w (reversible) > w (irreversible)
(d) w (reversible) = w (irreversible) + p_ex. ∆V
Sol: (b) w (reversible) < w (irreversible)

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.

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.,

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) q_P will be negative                               
(b) ∆_γHwill be negative
(c) q_p will be positive                                
(d) ∆_γHwill be positive.
Sol:(a, b) For an exothermic reaction, q_p = -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 V_f / V_i 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 (∆H_vap = 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. (∆H_r)_water > (∆H_r)_acetone

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

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                                                  ^
q_rcv = 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, K_p for the reaction N₂0₄(g)⇌ 2N0₂(g) is 0.98. Predict whether the reaction is spontaneous or not.
Sol: Δ_rG° = -RT ln K_p
= -RT ln (0.98)
Since In (0.98) is negative
.’. Δ_rG° 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?

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 Δ_rG and Δ_rG° will be zero at equilibrium?
Sol: Gibbs energy for a reaction in which all reactants and products are in standard state. Δ_rG° is related to the equilibrium constant of the reaction as follows
Δ_rG = A_rG° + RT In K
At equilibrium, 0 = Δ_rG° + RT In A^– ({Δ_rG = 0) or    Δ_rG° =-RT lnK
Δ_rG° = 0 when K= 1
For all other values of K, A_rG° 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, q_v = ΔU + 0 = ΔU= change in internal energy At constant pressure, q_p = AU + pΔV
Since ΔU + pΔV=ΔH
=> q_p = 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 = -p_extΔV
= -0(5 – 1) = 0
Since the gas is expanding isothermally, therefore, q = 0
ΔU = q + w =0+0=0

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

Q40. The difference between C_p and C_v can be derived using the empirical relation H = U + pV. Calculate the difference between C_p and C_v for 10 moles of an ideal gas.
Sol: Given that, C_v = heat capacity at constant volume,
C_p = heat capacity at constant pressure Difference between C_p and C_v is equal to gas constant (R).
.’. C_p – C_v = 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

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 6 Thermodynamics appeared first on Learn CBSE.

Princess September – CBSE Notes for Class 8 English It So Happened

CBSE NotesCBSE Notes Class 8 EnglishNCERT Solutions English

Summary:

The King and Queen of Siam had many daughters. In order to avoid confusion, the King called the princesses by the name of the months like January, February, March etc. The last daughter was named September.
The King of Siam had a peculiar habit. Instead of accepting birthday gifts, he used to give everyone gifts. One year on his birthday, he gave each one of the princesses a green parrot in a golden cage. The princesses taught the bird to say; “God save the king”; some of the parrots would say, “Pretty polly” in seven oriental languages.
One day Princess September found her parrot lying dead. She burst into tears. The Queen called her crying nonsense and put her to bed without any supper. Princess September was woken up by the song of a little bird. He (bird) flew into her room and sang beautifully. Princess September forgot crying. She agreed to have it in place of her dead parrot.
When the princess awoke the next day, the little song bird was still there and he said ‘Good morning!’ He ate rice out of her hand. Everybody was surprised to see that. Princess September decided to show her bird to her eight sisters. For each of them the bird sang a different song. The parrots could only say “God Save the King”. The King and the queen were also surprised and delighted to hear the bird’s song.
The King had got tired of hearing those parrots say “Pretty Polly”.
All the eight princesses were worried now. The parrots also looked sad. Only Princess September ran in the palace singing like a lark.
The jealous sisters took a nasty decision. They pointed out to their youngest sister that she should put her bird into a cage lest he should fly away for good.
The little bird came late that evening. September’s worry was about his safety from hawks and the hunter’s traps. So as he came, she pushed him into a cage and shut the door. The bird was taken aback. He hoped that he would be set free in the morning. But he cried in vain for liberty. As a result, he lost his singing talent. He refused to eat anything.
The next day September found the bird lying with his eyes closed. He looked as if he were dead. Seeing his condition, she agreed to let him leave the golden cage. And the bird promised to return to her every day and sing to her. Her sisters made fun of her. They said he would never return. But he did, at last. And he sat on September’s shoulder and ate out of his hand and sang her the beautiful songs he had learned.
September kept her window open all day and night for the bird to fly in and out. Fresh air from the open window made the princess extremely beautiful. She was married to the king of Cambodia. But her sisters who kept the windows shut, grew ugly as well as unpleasant. So they were married away with the King’s ministers.

The post Princess September – CBSE Notes for Class 8 English It So Happened appeared first on Learn CBSE.

NCERT Exemplar Problems Class 11 Chemistry Chapter 7 Equilibrium

Multiple Choice Questions

Single Correct Answer Type

Q1. We know that the relationship between K_c and K_p is K_p = K_c(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 K_p and K_c is
K_p = K_c (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 K_c 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)

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 K_w = [H⁺] [OH ^–] = 10^-14. On heating, K_w 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, K_a is the measure of strength of an acid. The K_a 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

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

Q14. K_a for CH₃COOH is 1.8 x 10^-5 and K_b 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.

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.

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 = K_f / K_b, this means that k_f increases, i.e., forward reaction is favoured.
Hence, reaction is endothermic.

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

Values of ionization constant of some weak bases at a particular temperature are given below:

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 (K_b) 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 K_b.

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 K_c for the reaction
2HI(g) ⇌H₂(g) + I₂(g) is 1 x 10^-4. At 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?

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]
= 10^–7 + 10^–7
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 (Q_sp) 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.

Q32. pH of 0.08 mol dm^–3HOC1 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^-1
On 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

Q34. The solubility product of Al(OH)₃ is 2.7 x 10^-11. Calculate its solubility in g^–L and also find out pH of this solution. (Atomic mass of A1 = 27 u).

Q35. Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (K_sp 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 mol^–1
∆_fH° [CO_z(g)] = -393.5 kJ mol^–1
∆_fH° [CaC0₃(s)] = -1206.9 kJ mol^–1
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

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,

Some reactions are written below in Column I and their equilibrium constants in terms of K_c are written in Column II. Match the following reactions with the corresponding equilibrium constant.

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

Q40. Match standard free energy of the reaction (Column I) with the corresponding equilibrium constant (Column II).

NCERT Exemplar Problems Class 11 Chemistry

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 7 Equilibrium appeared first on Learn CBSE.

NCERT Exemplar Problems 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.

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)

Q5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2S₂0₃^2_ + I₂→S₄0₆^2- + 2I^–
S₂0₃^2- + 2Br₂ + 5H₂0 →2S0₄^2- + 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 S₂O^2-₃ to SO₄^2- whereas I₂ oxidises it only into S₄O₆^2- 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

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 + NO^–₃ + 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,

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 N0^–₃ and decreases from +4 in N0₂ to +3 in N0^–₂.

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,

(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) 3d^l4s²                 
(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 P0^–₂
(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:

Because O.N. of P increases from 0 (P₄) to +1 (H₂P0^–₂) 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)   A1^3-/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:

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. Mn0^2-₄ undergoes disproportionation reaction in acidic medium but Mn0^–₄ does not. Give reason.
Sol:  In Mn0^–₄, 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 Mn0₄^2- is +6. Therefore, it can increase its O.N. to+7 or decrease its O.N. to some lower value. Thus, Mn0₄^2- undergoes disproportionation according to the following reaction.

Here, the O.N. of Mn increases from +6 in Mn0₄^2- to +7 in Mn0^–₄ and decreases to +4 in Mn0₂. Thus, Mn0₄^2- 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.

Q21. Write balanced chemical equations for the following reactions:
(i) Permanganate ion (Mn0^–₄) 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 (C10^–₃) 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 (C10^–₂) and oxygen gas. (Balance by ion-electron method)
Sol: (i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺→5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

Q22. Calculate the oxidation number of phosphorus in the following species.
(a) HPO₃^2- and (b) P0₄^3-
Sol: (a) Let the oxidation number of P inHPO₃^2- 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

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

Q24. Balance the following equations by the oxidation number method.
(i) Fe²⁺ + H⁺ + Cr₂0₇^2- →Cr³⁺ + Fe³⁺ + H₂0
(ii) I₂ + N0^–₃→ N0₂ +I0₃
(iii) I₂ + S₂0₃^2- →I^– + S₄0₆^2- ‘
(iv) MnO, + C₂0₄^2-→ Mn²⁺ + CO₂

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

To equalize O.N. multiply C0₂ by 2. –
Mn0₂ + C₂0^2-₄ →Mn²⁺ + 2CO₂
Balance H and O by adding 2H₂0 on right side, and 4H⁺ on left side of equation.
MnO₂ + C₂0^2-₄ + 4H⁺ →Mn²⁺ + 2CO₂ + 2H₂0

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

Sol: (i) Writing the O.N. of on each atom,

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.

Here O.N. of none of the atoms undergo a change and therefore, this is not a redox reaction.

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.

Here O.N. of none of the atoms undergo a change and therefore, it is not a redox reaction.

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.

Dividing the equation into two half reactions:
Oxidation half reaction: I^–→ I₂
Reduction half reaction: Cr₂0₇^2-→ Cr³⁺
Balancing oxidation and reduction half reactions separately as:

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

(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
Cr₂0₇^2-→ Cr³⁺

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

Cr₂0₇^2-→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.
Cr₂0₇^2- + 6e ^–→ 2Cr³⁺
(iii) Since the reaction takes place in acidic medium add 14H⁺ on the left to equate the net charge on both sides.
Cr₂0₇^2-+6e^–+14H⁺→2Cr³⁺

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

Cr₂0₇^2- + 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.

NCERT Exemplar Problems Class 11 Chemistry

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 8 Redox Reactions appeared first on Learn CBSE.

Real Numbers – CBSE Notes for Class 10 Maths

CBSE NotesCBSE Notes Class 10 MathsNCERT Solutions Maths

The post Real Numbers – CBSE Notes for Class 10 Maths appeared first on Learn CBSE.

Polynomials – CBSE Notes for Class 10 Maths

CBSE NotesCBSE Notes Class 10 MathsNCERT Solutions Maths

The post Polynomials – CBSE Notes for Class 10 Maths appeared first on Learn CBSE.

हम पंछी उन्मुक्त गगन के – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post हम पंछी उन्मुक्त गगन के – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

दादी माँ – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post दादी माँ – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

हिमालय की बेटियां – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post हिमालय की बेटियां – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

कठपुतली – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post कठपुतली – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

मीठाईवाला – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post मीठाईवाला – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

रक्त और हमारा शरीर – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post रक्त और हमारा शरीर – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.

NCERT Exemplar Problems 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.

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.

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.

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

Q8. Which of the following equatibns depicts the oxidizing nature of H₂0₂?
(a) 2Mn0^–₄ + 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.

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.

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

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.

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.

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, LaH_2.87, ybH_2.55etc.

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.

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.

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, S0^2-₄, 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

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.

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.

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.

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

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.

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.

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 Problems Class 11 Chemistry

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 9 Hydrogen appeared first on Learn CBSE.

NCERT Exemplar Problems 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

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.

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.

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.

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.

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).

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 (0^2-) ion to combine with another oxygen atom to from first peroxide ion (O^2-) and then superoxide (0^–₂) ion.

(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.

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).

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

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.

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.

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.

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

NCERT Exemplar Problems Class 11 Chemistry

The post NCERT Exemplar Problems Class 11 Chemistry Chapter 10 The S-Block Elements appeared first on Learn CBSE.

रहीम की दोहे – CBSE Notes for Class 7 Hindi

CBSE NotesCBSE Notes Class 7 HindiNCERT Solutions Hindi

The post रहीम की दोहे – CBSE Notes for Class 7 Hindi appeared first on Learn CBSE.