Marksparks

NCERT Class 11 Chemistry Chemical Thermodynamics Long Question And Answers

Question 1. Classify the following systems into open, closed, or isolated:

1. Living cell,
2. A gas is enclosed in a cylinder fitted with a movable piston.
3. The walls ofthe container and the piston are impermeable and thermally insulated.
4. The substances present in a soda water bottle. The chemicals participated in a chemical reaction occurring in a closed glass container.
5. Hot tea is kept in a thermos flask.

Read and learn More NCERT Class 11 Chemistry

Answer:

1. An open system: A living cell exchanges matter and energy with its surroundings to maintain itself.
2. A closed system: The walls of the container and the piston are impermeable and thermally insulated. So, the system cannot exchange matter and heat with its surroundings.  However, as the piston is movable, the system can do work or work can be done on it if the pressure on the die piston is decreased or increased. So, the system can exchange energy in the form of work with its surroundings,
3. A closed system: Here, the components present in the bottle constitute the system. As the bottle is closed, the system cannot exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings.
4. A closed system: Here, the chemicals constitute the system. As the reaction container is closed, the system is unable to exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings,
5. An isolated system: The walls of a thermos flask are made up of insulating materials. Again, the mouth of the flask is closed. So, the system can exchange neither matter nor energy with its surroundings.

Question 2. Identify the following an extensive or intensive property: Enthalpy, internal energy, pressure, viscosity, heat capacity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, number of moles, refractive index, entropy.
Answer:

An extensive property ofa system depends upon the mass ofthe substance present in the system. Its value increases as the amount of substance in the system increases. Enthalpy, internal energy, heat capacity, number of moles, and entropy are extensive properties.

An intensive property of a system is independent of the amount ofthe substance present in the system. Pressure, viscosity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, and refractive index are intensive properties.

Chemical Thermodynamics Long Answer Questions Class 11

Question 3. Thermodynamic state functions are path-independent quantities. Explain with an example.
Answer:

A state function is a path-independent quantity. This means that when a system undergoes a process, the change in any state function depends only on the initial and final states of the system, and not on the path of the process.

To make it clear, let us consider the following process in which the mol of an ideal gas changes its state:

2 atm, 4L, 273K→1 atm, 8L, 273K

We can make this change by either of the following two processes. However, in each of these processes, the change in a state function, viz., P or V is the same.

Question 4. Why is the change in any state function in a cyclic process zero? Is the change in any function both reversible and irreversible cyclic processes zero?
Answer:

1. The value of any state function of a system depends only on the present state of the system. In a cyclic process, the initial and the final states of a system are the same, and so are the values ofa state function at these two states.
2. Hence, the change in a state function will be zero in a cyclic process. The change in any state function depends only on the initial and the final states ofthe system.
3. It does not depend on the path followed to carry out the change. This means that a state function undergoes the same change in a process with the specified initial and final states irrespective of whether the process is carried out reversibly or irreversibly.
4. Now, in a cyclic process, the change in any state function is always zero. Hence, for both reversible and irreversible cyclic processes, the change in any state function will be zero.

Question 5. 1mol of an ideal gas participates in the process as described in the figure.

1. What type is the overall process?
2. Is this an isothermal process?
3. Mention the isobaric and isochoric steps in this process.

Answer:

1. It is a cyclic process because the system returns to its initial state after performing a set of consecutive processes.
2. This is not an isothermal process, because the temperature of the system does not remain constant throughout the process although the initial and the final temperatures are the same.
3. Step BC represents an isobaric process because the pressure ofthe system remains constant in this step. Step CA represents an isochoric process because the volume ofthe system remains constant in this step.

Question 6. Are the following changes reversible or irreversible? Give proper explanations:

1. Melting of ice at 0°C and 1 atm pressure
2. The pressure ofa gas enclosed in a cylinder fitted with a piston is 5 atm.
3. The gas is expanded against an external pressure of 1atm.

Answer:

1. At 0°C and 1 atm pressure ice remains in equilibrium with water. If the temperature ofthe system is increased by an infinitesimal amount, ice melts into the water slowly. Again, if the temperature of the system is decreased by an infinitesimal amount, water freezes into ice slowly.
2. Thus, an infinitesimal increase or decrease in temperature causes a change in the direction of the process. Hence, the melting of ice at 0°C and 1 atm pressure can be considered as a reversible process,
3. An irreversible process: The external pressure is considerably less than the pressure of the gas. So, the gas will expand rapidly without maintaining thermodynamic equilibrium during the process. Hence, this expansion will occur irreversibly.

Question 7. In the process A→ B→ C, the change in internal energy of the system in the steps A→ B and B→ C are -x kJ.mol¯1 and y kj-mol^-1, respectively. What will be the change in the internal energy ofthe system in step C→ A?
Answer:

A→ B→ C

Given, ΔU_{A→ B} = -x KJ. mol^-1 and ΔU_{B→ C} = -x KJ. mol^-1

∴ ΔU_C→A  = -ΔU_A→C = (x-y) kJ. mol^-1

[Since U is a state function, its change in die forward direction of a process is the same as that in the backward direction but opposite in sign].

Question 8. A certain amount of a gas participates in the cyclic process ABCD (follow figure). Calculate the total work done g in the process
Answer:

Pressure volume work, w = -P_ex(V₂-V₁)=-P_exΔV

In step: AB: P_ex = x atm, ΔV= (2y-y)L=yL

∴ w₁ = -P_ex ΔV = -xy L . atm

In steps BC and DA. work done is zero because the system remains constant in these steps. Instep CD: P_ex = 0.5x atm and AV = (y- 2y) L = -y L.

∴ ω₂ = -P_exΔV = —0.5x (-y L) = 0.5.xy} L . atm

So, the total work done in the process, ω = ω₁ + ω₂

= (—xy + 0.5xy) L .atm =-0.5xy L.atm

= -0.5xy x 101.3 J [since 1Latm = 101.3 J] =-50.65xy J

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics

Question 9. A particular amount of gas participates separately in the two processes given below: Process-1 For which process, the work done is maximum?
Answer:

⇒ Process-1:

Step-1: ω₁ = -P_ex-ΔV = -P₁(V₂– V₁)

⇒ Step-2: ω₂ = 0 [Since the volume of  the system is constant]

⇒ Total work, ω = ω₁ + ω₂ = -P₁(V₂– V₁)

⇒ So, |ω| = |P₁(V₂-V₁)|

⇒ Process-2:

Step-1: ω₁ = 0 [∴ the volume of the system is constant]

⇒ Step-2: ω₂ = —P_ex ΔV = -P₂(V₂-V₁)

⇒ Total work, ω’ = ω₁+ ω₂ = -P₂(V₂– V₁)

⇒ So, |ω’| = |P₂(V₂-V₁)| [Since P₂ < P₁ , |ω’| < |ω’| ]

Question 10. For an ideal gas, the isothermal free expansion and adiabatic free expansion are the same processes— Explain For chemical changes, why is the change in enthalpy more useful than the change in internal energy?
Answer:

According to the first law of thermodynamics, ΔU = q+ω. In the free expansion of a gas, w = 0. Again, in an adiabatic process, q = 0 Hence, in an adiabatic free expansion of an ideal gas, the change in internal energy ΔU =q + 0=0.

Also, in an isothermal process, the change in the internal energy of an ideal gas is zero. Again for free expansion of an ideal gas, w = 0. So, in an isothermal free expansion of an ideal gas, ΔU = q + w or, 0 = q+ 0 or q = 0. Therefore, it can be concluded that both processes are the same.

Question 11. The heat required to raise the temperature of 1 mol ofa gas by 1°C is q at constant volume and q’ at constant pressure. Will q be better than, less than or equal to q’? Explain
Answer:

The heat required to raise the temperature of1 mol of gas by 1 ° at constant pressure is greater than that required at constant volume. At constant pressure, the heat absorbed by a gas is used up in two ways.

One part of it is used by the gas for doing external work, and the remaining part is utilised for increasing the temperature of the gas.

At constant volume, the heat absorbed by a gas is completely utilised for increasing the temperature of the gas as no external work (P-Vwork) is possible at constant volume. Therefore, q’ must be greater than q.

Question 12. A 0.5 mol sample of H₂(g) reacts with a 0.5 mol sample of Cl₂(g) to form 1 mol of HCl(g). The decrease in enthalpy for the reaction is 93 kj. Draw an enthalpy diagram for this reaction.

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) ; \Delta H=-93 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The reaction is associated with a decrease in enthalpy. So, it is an exothermic reaction. In such a reaction, the total enthalpy of the product(s) (2Hp) is less than that of the reactant(s)(£flfi). Therefore, in the enthalpy diagram for the reaction, ZHp lies below ZHR.

Question 13. A 1 mol sample of N₂ (g) reacts with 1 mol of O₂(g) to form 2 mol of NO₂(Og), where the increase in enthalpy is 180.6kj. Draw An enthalpy diagram for this reaction.
Answer:

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) ; \Delta H=+180.6 \mathrm{~kJ}\)

In the reaction, the enthalpy increases. So, it is an endothermic reaction. In such a reaction the total enthalpy ofthe product(s)(ΣH_R) is greater than that ofthe reactant(s) (ΣH_p). Therefore, in the enthalpy diagram for the reaction, ΣH_p lies above ΣH_R

ΣH_p

ΣH  =  +180.6 kJ

ΣH_R

Class 11 Chemistry Chemical Thermodynamics Long Answer Solutions

Question 14. Identify the exothermic and endothermic changes:
Answer: 

⇒ \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g)+57.0 \mathrm{~kJ}\)

Answer:

⇒  \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+57.0 \mathrm{~kJ}\).

Heat is realed in this reacton,. so it’s an exothermic reaction.

H₂O(s) + 6.02 kJ→H₂O(l) . Heat is absorbed in this reaction. So, it is an endothermic process.

⇒ \(\mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)-130 \mathrm{~kJ}\)

⇒ \(\text { i.e. } \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g)+130 \mathrm{~kJ} \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)\)

Question 15. Write down the thermochemical equations for the following reactions:

1. A 1mol sample of methane gas reacts with 2 mol of oxygen gas to form lmol of carbon dioxide and 2mol of water. In this reaction, 890.5 kl of heat is produced.
2. A 1mol sample of carbon (graphite) reacts with 1 mol ofoxygen to form 1 mol of carbon dioxide gas. The heat evolved in this reaction is 393.5 kj.
3. 6 mol of carbon dioxide gas reacts with 6 mol of water to form 6 mol of oxygen gas and lmol of glucose. The heat absorbed in this reaction is 2200 kj.

Answer:

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.5 \mathrm{~kJ}\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \( \Delta H=-890.5 \mathrm{~kJ}\)

⇒ \(6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g)-2800 \mathrm{~kJ}\)

⇒ \(\text { i.e., } 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \text {; }\)

⇒  \( \Delta H=-2800 \mathrm{~kJ}\)

Question 16. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l); \dot{2} \Delta \mathrm{H}=-285.8 \mathrm{~kJ}\)  What will be the value of ΔH for the reaction: 2H₂O(l)→ 2H₂(g) + O₂(g)?
Answer:

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \dot{2} \Delta H=-285.8 \mathrm{~kJ}\)

If this equation is written in the reverse manner, we have

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) ; \Delta H=+285.8 \mathrm{~kJ}\)

Multiplying this equation by 2, we have

⇒ \(2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=4571.6 \mathrm{~kJ}\)

Chemical Thermodynamics Long Questions and Answers Class 11

Question 17. Why is ΔH = ΔU for the following two reactions? Explain.

1. NaOH(aq) + HCl(aq)→ NaCl(aq) + H₂O(l)
2. CHΔ(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g)

Answer: This reaction occurs In a solution. For a reaction occurring in a solution, All = AH

For this reaction An (total number of moles of gaseous products – total number of moles of gaseous reactants) = (1 +2)-(l + 2) = 0. So, Δ7 = ΔU according to the relation ΔH = ΔU + ΔT.

Question 18. Give an example of a reaction for each of the following relations between AH and ΔH: 

1. ΔH<ΔH
2. ΔH > ΔH
3. ΔH = AH.

Answer:

In a reaction, if a gaseous substance (either as a reactant or as a product or both) participates, the change in enthalpy in the reaction at constant pressure and temperature is given by ΔH= AU + ΔnRT. K An (total number of moles of gaseous products – total number of moles of gaseous reactants) >0, < 0 or =0, then AH > AH, AH < AH or All = AH respectively.

⇒ \( 2 \mathrm{H}_2 \mathrm{O}(g) \rightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta n=(2+1)-2=+1 \text {. So, } \Delta H>\Delta U\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta n=0-(2+1)=-3 \text {. So, } \Delta H<\Delta U\)

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)\)

⇒ \(\Delta n=2-(1+1)=0 . \text { So, } \Delta H=\Delta U\)

Chemical Thermodynamics Chapter 6 Long Answer Questions

Question 19. In which of the following reactions At 25°C, does the standard enthalpy change correspond to the standard enthalpy of formation of H₂O(Z)? Give reasons.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)\(2 \mathrm{H}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
Answer:

1. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z) because, the standard state of oxygen at 25 °C, is O₂(g) and not O₃(g).
2. The standard enthalpy change in this reaction refers to the standard heat of formation of H₂O(Z) because 1 mol of H₂O(Z) is formed from its stable constituent elements.
3. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z). This is because the stable state of hydrogen at 25 °C, is not H(g).

Question 20. Which one of the given reactions indicates the formation reaction ofthe compound produced in the reaction?

1. S (monoclinic) + O₃(g)→ SO₃(g)
2. C (graphite, s) + 2H₂(g)→CH₄(g)
3. N₂(g) + O₂(g)→2NO(g)

Answer:

In the formation reaction of a compound, 1 mol of the compound is formed from its constituent elements. S(s, monoclinic) +O₂(g)→SO₂(g), at 25 °C this reaction does not represent the formation of S02(g) because, the stable form of sulphur is S(s, rhombic) at 25°C.

2. C(s, graphite) + 2H₂(g)→CH₄(g), at 25 °C reaction represents the formation reaction of CH₄(g) because lmol of CH₄(g) is formed from its stable constituent elements.

3.  N₂(g) + O₂(g)→2NO(g), at 25°C this reaction does not represent the formation reaction of NO(g) because 2 mol of NO(g) are formed in the reaction.

Question 21. The standard heats of combustion of CH₄(g) and C₂H₆(g) are -890 kj.mol^-1 and -1560 kj. mol^-1 respectively. Why is the calorific value of C₂H₆(g) lower than that of CH₄ (g)?
Answer:

The enthalpy of combustion of a compound is always negative. So, the standard enthalpy of combustion of the given compound, ΔHº_C = -Q kj .mol-1. The thermochemical equation for this combustion reaction

⇒ \(\mathrm{C}_x \mathrm{H}_y(l)+\left(x+\frac{y}{4}\right) \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_c^0=-Q \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 22. Identify whether the enthalpy of the initial state is greater than, less than or equal to that of the final state in the following changes: solid;→ liquid; vapour→liquid →vapour
Answer: Solid→Liquid:

It is an endothermic process. In this process

⇒ \(\Delta H>0 \text {, i.e., } H_{\text {liquid }}-H_{\text {solid }}>0 \text { or, } H_{\text {liquid }}>H_{\text {solid }} \text {. }\)

Therefore, the enthalpy of the final state will be greater than that ofthe initial state.

Vapour→Solid: It is an exothermic process. So, in this process

Question 23. What docs Mi signify in each of the following equations?

1. HCl(g) + 5H₂O(l)→HCl(5H₂O); ΔH = -64 kJ
2. HCl(g) + aq→HCl(aq); ΔH = -75 kJ
3. HCl(5H₂O)+20H₂O(l)→HCl(25H₂O); ΔH=-8.1 kJ

Answer:

In this process, 1 mol of HCl dissolves in 5 mol of water, forming a solution of definite concentration. The enthalpy change in such a process is known as the integral heat of the solution. So, ΔH, in the process Indicates the integral heat of the solution.

In this process, 1 mol HCl dissolves in a large amount of water, forming an infinitely dilute solution. The enthalpy change in such a process is called the heat of solution. So, vH, in process 2, indicates the heat of the solution.

This is a dilution process because a solution with a definite concentration is diluted by adding solvent to it. So, ΔH, in this process, indicates the heat of dilution.

Change in enthalpy remains the same whether a reaction is carried out in one step or several steps under similar reaction conditions Explain the rearms.

Question 24. Given (at 25°C and 1 atm pressure):

1. C (s, diamond) + O₂(g)→CO₂(g); ΔH°=-393.5 kj-mol^-1
2. C (s, graphite) +O₂(g)→CO₂(g); ΔH°=-391.6 kj-mol^-1

Find the standard heat of transition from graphite to diamond.
Answer:

⇒ \(\mathrm{C}(s, \text { diamond })+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-391.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

[Subtracting equation 1 From equation We get C(s, graphite)→ C(s, diamond); AH° = 1.9 kj. mol^-1 So, the heat of transition for this process is 1.9 kj.mol^-1

Class 11 Chemical Thermodynamics Long Question and Answers

Question 25. At 25°C, if standard enthalpies of formation o/MX(s), M+(ag) and X-(aq) are -x, y and -zkj-mol^-1 respectively, then what will the heat of reaction before the reaction M+(aq) + X~(aq)-+ MX(s).
Answer:

M+(aq) + X-(aq)→MX(s)

The standard heat of reaction,

⇒ \(\Delta H^0=\Delta H_f^0[\mathrm{MX}(s)]-\Delta H_f^0\left[\mathrm{M}^{+}(a q)\right]-\Delta H_f^0\left[\mathrm{X}^{-}(a q)\right]\)

=\((-x-y+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Question 26. At 25°C, the bond dissociation energy of N₂(g) is 946 kj.mol-1. What does it mean? What would be the standard atomisation enthalpy of N₂(g) at 25°C?
Answer:

At 25°C, the standard bond dissociation energy of N₂(g) is 946 kj.mol-1. This means that the energy required to break 1 mol of N=N bonds completely in the gaseous state to form gaseous nitrogen atoms is 948 kj. At 25 °C, the standard state of nitrogen is N₂(g). Now, the formation of 1 mol of N(g) takes place by the following process \(\frac{1}{2} \mathrm{~N}_2(g) \rightarrow \mathrm{N}(g)\)

Since 1 mol of N(g) is produced from N₂(g) in the process [1], the enthalpy change in this process at 25 will be equal to the standard atomisation enthalpy of nitrogen.

In process [1], change in enthalpy \(=\frac{1}{2}\) x bond dissociation energy of N=N

= \(\frac{1}{2} \times\) 946 = 473 k.mol-1 Therefore, at 25 °C the standard atomisation enthalpy of nitrogen is 473 kJ.mol-1.

Chemical Thermodynamics Long Answer NCERT Solutions

Question 27. A—B bonds present in AB3(g) molecule undergo stepwise dissociation by the following sequence of steps.

1. AB₃(g)→AB₂(g) + B(g) 
2. AB₂(g)→AB(g) + B(g)
3. AB(g)→A(g) + B(g)

Answer:

Bond energy of A — B bond in AB₃(g) molecule = the average bond dissociation energy of three A— B bonds in AB₃(g) molecule. If the standard enthalpy change in step is ΔH° kj.mol-1, then the bond dissociation energy of the A-B bond in AB₃(g) molecule

= \(\frac{x+\Delta H^0+z}{3} \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Now, the bond dissociation energy of the A-B bond = kJ.mol^-1

⇒ \(y=\frac{x+\Delta H^0+z}{3} \quad \text { or, } \Delta H^0=3 y-(x+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Therefore, the standard enthalpy change in step (2)is [3y-(x+z)kJ.mol^-1

Question 28. The following changes are performed on 1 mol of N₂ gas

1. Pressure Is decreased at a constant temperature
2. Volume is decreased at a constant temperature
3. What will be the sign of S_sys In These changes?

Answer:

If the pressure of a gas is decreased at a constant temperature, the volume of the gas increases. At a larger volume of a gas, the gas molecules get a greater chance to move about. As a result, the randomness of the molecules increases, which increases the entropy ofthe gas. Hence, A = +ve.

At constant temperature, the decrease in the volume of a gas reduces the availability of space for the movement of gas molecules. This results in a decrease in the randomness of the molecules; consequently, the entropy ofthe gas decreases. Hence, Δ = -ve

Question 29. The following two reactions occur spontaneously. What will be the signs of AS and AS_surr in these two reactions?

1. A(s)→ B(s) + C(g); ΔH > 0
2. 2X(g)→ X₂(g); ΔH < 0

Answer:

At a given temperature and pressure, the change in entropy of the surroundings, \(\Delta S_{\text {surr }}=-\frac{\Delta H}{T}\)…………………(1)

In this reaction, ΔS_sys = + ve is the gaseous substance produced in the reaction. Since, in this reaction, ΔH > 0; according to equation ΔS_surr = – ve.

In this reaction, Δ S_sys = -ve as the number of gas particles decreases. Again, in this process, AH < 0. So, according to the equation, ΔS_surr > 0

NCERT Chemical Thermodynamics Long Answer Solutions Class 11

Question 30. For a reaction ΔH > 0, and another ΔH < 0. For both the reactions ΔSsys < 0. Which one is likely to occur spontaneously? Which one always occurs nonspontaneous? define

1. Gibbs free energy
2. The standard energy of formation of a substance
3. The standard free energy change in a chemical reaction.

Answer:

For the reaction with ΔH>0, ΔS_surr. is -ve. Again, for this reaction ΔS_sys < 0, and hence ΔS_sys, + ΔS_surr < 0 i.e.  ΔS_univ < 0. So, this reaction will be non-spontaneous.

For the reaction with ΔH < 0,  ΔS_surr is -ve. Again, for this reaction ΔS_surr < 0, Therefore,

ΔS_univ = (ΔS_sys, + ΔS_surr may be Positive or negative depending on the magnitudes of S_sys and S_surr If |ΔS_sys|< |ΔS_surr| then the reaction will be spontaneous.

NCERT Class 11 Chemistry Chemical Thermodynamics Multiple Choice Questions

Question 1. Which of the following statements is true

1. Entropy increases when water vaporizes
2. Randomness decreases in the fusion of ice
3. Randomness increases in the condensation of water vapor
4. Randomness remains unchanged during the vaporization of water

Answer: 1. Entropy increases when water vaporizes

Vaporization of water (water → water vapor) involves an increase in the entropy of the system because the molecular randomness in water vapor is greater than that in water.

Question 2. Identify the correct statement in a chemical reaction

1. The entropy always increases
2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction
3. The enthalpy always decreases
4. Both tire enthalpy and tire entropy remain constant

Answer: 2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction.

For a reaction occurring at a constant temperature and pressure, ΔG<0, ie., ΔH-TΔS<0 the change in entropy (Δs) and (flng with the change in enthalpy (ΔH) determines the spontaneity of a reaction.

Question 3. The condition for the spontaneity of a  process is—

1. Lowering of entropy at constant temperature & pressure
2. Lowering of Gibbs free energy of tire system at constant temperature and pressure
3. Increase in entropy of tire system at constant temperature and pressure
4. Increase in Gibbs free energy of the universe at constant temperature and pressure

Answer: 2. Lowering of Gibbs free energy of tire system at constant temperature and pressure

The condition of spontaneity for a reaction to occur at constant t and p is ag < 0.

Chemical Thermodynamics MCQs Class 11

Question 4. P-v work done by an ideal gaseous system at constant volume is the internal energy of the system)

1. – ΔP/P
2. Zero
3. -VΔp
4. -Δe

Answer: 2. Zero

As the system’s volume remains constant in the process, the system cannot do any external work.

Question 5. Mixing of two different ideal gases under an isothermal reversible condition wall leads to

1. Increase in Gibbs free energy of the system
2. No change in the entropy of the system
3. Increase in entropy of the system
4. Increase in enthalpy of the system

Answer: 3. Increase in entropy of the system

The molecular randomness in a gas mixture is larger than that in the individual gases because a gas mixture contains a larger number of molecules than that in the individual gases. Consequently, the mixing of two gases will lead to an increase in the entropy of the system.

Question 6. For an isothermal expansion of an ideal gas, the correct of the thermodynamic parameters will be

1. ΔU =0,Q = 0,ω≠0 and ΔH≠ 0
2. ΔU ≠0,Q ≠ 0,ω≠0 and ΔH= 0
3. ΔU =0,Q ≠ 0,ω=0 and ΔH≠ 0
4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

Answer: 4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

For an ideal gas undergoing an isothermal process, ‘ah = 0. When a gas undergoes expansion it absorbs heat from its surroundings and performs external work. So, in such a process, q≠0 and w≠0.

We know, Δh = Δu+ Δ(pv). For an ideal gas, pv = nrΔt.

∴ Δh = Δu+Δ(nrt) = Δu+nrΔt

In an isothermal process of an ideal gas, aΔu = 0, at- 0, and hence ΔH = 0.

Question 7. The change in entropy (ds) is defined as

1. \(d s=\frac{\delta q}{t}\)
2. \(d s=\frac{d h}{t}\)
3. \(d s=\frac{\delta q_{r e v}}{t}\)
4. \(d s=\frac{(d h-d g)}{t}\)

Answer: 3. \(d s=\frac{\delta q_{r e v}}{t}\)

If a system undergoing a reversible process at tk absorbs an amount of heat, 6qrev, then the change in entropy of the system in the process, \(d s=\frac{\delta q_{r e v}}{t}.\).

Question 8. Δh for cooling 2 mol ideal monoatomic gas from 225 °c to 125°c at constant pressure will be?

1. 250R
2. -500R
3. 500R
4. -250R

Answer: 2. ΔH=nCp(T₂-T₁)

Question 9. For a spontaneous process, correct statement(s) is (are)

1. \(\left(\Delta G_{s y s}\right)_{T, P}>0\)

2. \(\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>\)

3. latex]\left(\Delta G_{s y s}\right)_{T, P}<0[/latex]

4. \(\left(\Delta U_{s y s}\right)_{T, V}>0\)

Answer: 3. \(\left(\Delta G_{s y s}\right)_{T, P}<0\)

For a spontaneous process at constant pressure and temperature,

ΔS >0 or, ΔS + ΔS_surr> 0

Now, ΔG = -TΔS_univ. For a spontaneous process

ΔS_univ > 0 , and hence ΔG < 0

Question 10. Given: C+O₂→CO₂; ΔH⁰ = -xkj; 2CO+O₂→2CO₂; ΔH⁰=-ykj The heat of formation of carbon monoxide will be

1. Y+2x/2
2. Y+2x
3. 2x-y
4. 2x-y/2

Answer: 1. Y+2x/2

C+O₂→CO₂; ΔH°=-xKJ; 2CO+O₂→2CO₂;ΔH°=-yK

Subtracting equation [2] from the equation obtained by multiplying equation [1] by 2, we have,

2CO+O₂→2CO₂;ΔH⁰=-yK

∴ Heat of formation of co \(=\left(\frac{y-2 x}{2}\right) \mathrm{kj} \cdot \mathrm{mol}^{-1}\)

Question 11. The enthalpy of vaporization of a certain liquid at its boiling point of 35°c is 24.64 kl-mol^-1. The value of change in entropy for the process is—

1. 704 J.k^-1 Mol^-1
2. 80J.k^-1 Mol^-1
3. 24.64j.k^-1. Mol^-1
4. 7.04 j.k^-1. Mol^-1

Answer: 2. 80J.k^-1. Mol^-1

⇒ \(\Delta S_{v a p}=\frac{\Delta H_{v a p}}{T_b}\)

= \(=\frac{24.64 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{(273+35) \mathrm{K}}\)

⇒  \( =80 \mathrm{~j} \cdot \mathrm{k}^{-1} \cdot \mathrm{mol}^{-1}\)

Question 12. ΔH and ΔS of a certain reaction are -400 kj. mol^-1 and -20kj.mol^-1.k^-1 respectively. The temperature below which the reaction is spontaneous is

1. 100K
2. 20°C
3. 20K
4. 120°C

Answer: 3. 20K

For a spontaneous reaction, Δg = ΔH- TΔs <0

⇒ \(T \Delta S>\Delta H\)

Or, \(T>\frac{\Delta H}{\Delta S}\)

Or, \(T>\frac{400}{} \mathrm{~K}_1 \text { or, } T>20 \mathrm{~K}\)

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics MCQs

Question 13. For the reaction x²y⁴(f)→2xy(g) at 300 k the values of a u and as are 2kcal and 20 cal.k-1 respectively. The value of Δg for the reaction is-

1. -3400 Cal
2. 3400 Cal
3. -2800 Cal
4. 2000 Cal

Answer: 3.  ΔH = ΔU+Δn_gRt or, ΔH = 2000 + 2 × 2 × 300

Δn_g = 2-0 = 2, R = 2 cal -k^-1– mol^-1 ]

ΔH = 3200 cal

ΔG = ΔH – TΔS

= 3200- (300 ×  20)

= -2800 cal

Question 14. For the reaction 2SO₂(g) + O(g)⇌ 2SO₃(g) at 300 k, the value of ΔG⁰ 0the equilibrium constant value for the reaction ax that temperature is (r is gas constant)—

1. 10 Atm^-1
2. 10 Atm
3. 10
4. 1

Answer: 3. 10

2SO₂(g) + O₂(g)⇌ 2SO₃(g)

Now, ΔG = -RTlnk

Or, -690.9R = -RTlnk

∴ K = 10¹

Question 15. The condition for the reaction to occur spontaneously is

1. ΔH must be negative
2. ΔS must be negative
3. (ΔH- tΔs) must be negative
4. (Δ H+ tΔs) must be negative

Answer: 3. (ΔH- tΔs) must be negative

At constant temperature and pressure, for a spontaneous process of the reaction ΔG < 0. According to Gibbs’s equation ΔG = ΔH- TΔS. Therefore the condition spontaneity is, ΔH- TΔS < 0.

Question 16. During a reversible adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio \(\frac{c_p}{c_v}\) for the gas is-

1. \(\frac{3}{2}\)
2. \(\frac{7}{2}\)
3. \(\frac{5}{3}\)
4. \(\frac{9}{7}\)

Answer: 1. \(\frac{3}{2}\)

For adiabatic reversible process, \(t p^{\frac{(1-\gamma)}{\gamma}}=\text { constant }\)

Or, \(p t^{\frac{\gamma}{(1-\gamma)}}=\text { constant }\)

Again as mentioned, P ∝ T³

Or, PT^-3= constant

Comparing equations (1) and (2), it may be written as

⇒ \(\frac{\gamma}{1-\gamma}=-3\)

∴ \(\gamma=\frac{3}{2}\)

Question 17. The heat of neutralization of a strong base and a strong add is 13.7 kcal the heat released when 0.6 mol HCl solution is added to 0.25 mol of NaOH is-

1. 3.425kcal
2. 8.22kcal
3. 11.645kcal
4. 13.7kcal

Answer: 1. 3.425kcal

⇒  \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\) ‘

⇒  \(\Delta H^0=-13.7 \mathrm{kcal}\)

⇒ \(0.25 \mathrm{~mol} \mathrm{HCl}(a q)+0.25 \mathrm{~mol} \mathrm{NaOH}(a q)\)

⇒ \(0.25 \mathrm{~mol} \mathrm{NaCl}(a q)+0.25 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta H^0=-13.7 \times 0.25=-3.425 \mathrm{kcal}\)

∴ Amount of heat released = 3.425 kcal

Question 18. The entropy change involved in the isothermal reversible expansion of 2 mol of an ideal gas from a volume of 10 cm³ to a volume of 100 cm3 at 27°cis —

1. 38.3j.mol^-1.k^-1
2. 35.8j.mol^-1.k^-1
3. 32.3j.mol^-1. K^-1
4. 42.3j.mol^-1.k^-1

Answer: 1. 38.3j.mo^-1.k^-1

For isothermal reversible expansion,

⇒ \(\delta s=n r \ln \frac{v_2}{v_1}=2 \times 8.314 \ln \frac{100}{10}=38.3 \mathrm{~j} \cdot \mathrm{mol}^{-1} \cdot k^{-1}\)

Question 19. Among the following expressions which one is incorrect-

1. \(w_{r e v, i s o}=-n R T \ln \frac{V_f}{V_i}\)

2. \(\ln K=-\frac{\Delta H^0-T \Delta S^0}{R T}\)

3. \(K=e^{-\Delta G^0 / R T}\)

4. \(\frac{\Delta G_{\text {sys }}}{\Delta S_{\text {total }}}=-T\)

Answer: 2. \(\ln k=-\frac{\delta h^0-t \delta s^0}{r t}\)

ΔG° = -RTInk and ΔG° = ΔH₀– TΔS₀

⇒  \(\Delta H^0-T \Delta S^0=-R T \ln K \text { and } \ln K=-\left(\frac{\Delta H^0-T \Delta S^0}{R T}\right)\)

Question 20.  A cylinder filled with 0.04mol of ideal gas expands reversibly from 50ml to 375ml at a constant temperature of 37.0°c. As it does so, it absorbs 208 j heat, q and w for the process will be (a = 8.314 j.mol^-1.k^-1 )

1. q = +208 J, w = +208 J
2. q = +208 J, w = -208 J
3. q = -208 J , w = -208 J
4. q = -208 J , w = +208 J

Answer: 2. Q = +208 J, w = -208 J

The process is isothermal and the system is an ideal gas. So, in this process, a u = 0. Given that q = +208J.

∴ Au = q + w or, 0 = 208 + w

∴ W = -208J .

Question 21. For complete combustion of ethanol, \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

The amount of host produced ao measured in a bomb calorimeter, u 1364,47kJ mol 1 at 25°C, assuming ideality the enthalpy of combustion, ΔH_c(  – mol^-1 j for the reaction will be (R= 8.3m J. k^-1. Mol^-1)

1. -1350.50
2. -1366.95
3. -1361.95
4. -1460.50

Answer: 2. -1366.95

In a bomb calorimeter, a reaction occurs under constant volume. Hence, q_v = ΔU. For the given reaction, an = 2-3 = -1 we know, ΔH = ΔU+ΔnRT

∴ ΔH = [- 1364.47 × 8.314 × 10^-3× 298] kj . Mol^-1

=-1366.95 kj- mol^-1

Multiple Choice Questions for Class 11 Chemistry Chemical Thermodynamics

Question 22. The following reaction is performed at 298k. 2No(g) + O₂(g/ v= 2nO₂(g) the standard free energy of formation is 86.6 kJ/mol at 298k, what is the standard free energy of formation of NO₂(g) at 298k (kp – 1.6 × 10¹²) 

1. 8660- \(\frac{\ln \left(1.6 \times 10^{12}\right)}{r(298)}\)

2. 0.5[2 × 86600 -R(298) in(1.6 × 10¹²)

3. R(298)ln(1.6 × 10¹²)- 86600

4. 86600 + R(298) In (1.6 × 10¹²)

Answer: 2. 0.5[2 × 86600 -R(298) in(1.6 × 10¹²)

Given: \(T=298 \mathrm{~K}, \Delta G_f^0(\mathrm{NO})=86.6 \mathrm{~kJ} / \mathrm{mol},\)

⇒ \(\Delta G_f^0\left(\mathrm{NO}_2\right)=?, K_p=1.6 \times 10^{12}\)

⇒ \( 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_0(g) \rightleftharpoons 2 \mathrm{NO}_0(g)\)

∴ \(\Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \Delta G_f^0(\mathrm{NO})-\Delta G_f^0\left(\mathrm{O}_2\right)\)

⇒ \(\text { or, } \Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600\)

⇒ \(Again,\Delta G_r^0=-R T \ln K_p
or, 2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600=-R(298) \ln \left(1.6 \times 10^{12}\right)\)

⇒ \(\text { or, } \Delta G_f^0\left(\mathrm{NO}_2\right)=\frac{2 \times 86600-R(298) \ln \left(1.6 \times 10^{12}\right)}{2}\)

= 0.5[2 × 86600- (298)Ln(1.6× 10¹²)

Question 23. The standard Gibbs energy change at 300k for the reaction 2az±b + c is 2494.2 j. At a given time, the composition of the reaction mixture is [a] \([a]=\frac{1}{2},[b]=2 \text { and }[c]=\frac{1}{2}\) reaction proceeds in the [a = 8.314)/k/mol, c=2.718] 

1. Forward direction because q<kc
2. Reverse direction because q<kc
3. Forward direction because q > kc
4. Reverse direction because q > kc

Answer: 4. Reverse direction because q > kc

⇒ \(2 A \rightleftharpoons B+C\)

Given: T=300K, ΔG° = 2494.2J,

R = 8.314 J. K^-1.mol^-1

Now, ΔG° = -2.303 RTlogK_c

or, 2494.2 = -2.303 × 8.314 × 300 ×  logk_c

or, logK_c = -0.4342

∴ K_c = 0.3679

⇒ \(Q_c=\frac{[B][C]}{[A]^2}=\frac{2 \times \frac{1}{2}}{(1 / 2)^2}\)

= 4

Question 24. The heats of carbon and carbon monoxide combustion are -393.5 and -283.5 kj.mol^-1 respectively. The heat of formation (in kj) of carbon monoxide per mole is

1. 110.5
2. 676.5
3. -676.5
4. -110.5

Answer: 4. -110.5

C(s) + O₂(g) →CO₂(g) , ΔH° = -393.5 kj.mol^-1 ………………(1)

⇒ \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g), \Delta H^0=-283.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  ………………(2)

Subtracting equation (2) from equation (1) we get,

⇒ \(\mathrm{C}(s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)\)

ΔH° =[- 393.5- (-283.5)] kj.mol^-1

=-110.0 kj-mol^-1

Therefore, the heat of the formation of CO(g)

= -110.0 kj-mol^-1

Question 25. Given, C(graphite) + O₂(g) → CO₂(g), Δ_rH⁰= -393.5 kj .mol^-1

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \delta_r H^0=-285.8 \mathrm{~kj} \cdot \mathrm{mol}^{-1} \)

⇒  \(\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{h}_2 \mathrm{O}(l) \rightarrow \mathrm{CH}_4(g)+2 \mathrm{O}_2(\mathrm{~g})\)

Based on the above thermochemical equations, the value of arh° at 298 k for the reaction, c(graphite) + 2h₂(g) → CH₄(g) , will be—

1. -748 Kj.mol^-1
2. -144.0kj.mol^-1
3. +74.8 kj.mol^-1
4. +144.0kj.mol^-1

Answer: 1. -748 Kj.mol^-1

⇒ \(\mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta_r H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  …………………..(1)

⇒ \(\mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CH}_4(g)+2 \mathrm{O}_2(g)\) …………………..(2)

⇒ \(\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta_r H^0=+890.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) …………………..(3)

(1) eqn. +2 x eqn. (2) no. + eqn. (3) no.

⇒  \(\mathrm{C}(\text { graphite })+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g})\)

⇒  \(\Delta_r H^0=[-393.5-2 \times 285.8+890.3] \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

= \(-74.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 26. The combustion of benzene (z) gives CO₂(g) and h₂O(z). Given that the heat of combustion of benzene at constant volume is -3263.9 kj -mol^-1 at 25°c, the heat of combustion (in kj .mol^-1 ) of benzene at constant pressure will be [r = 8.314 j.k^-1.mol^-1 )

1. -4152.6
2. -452.46
3. 3260
4. -3267.6

Answer: 4. -3267.6

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta n=6-\frac{15}{2}=-\frac{3}{2}\)

⇒ \(\Delta H=\Delta U+\Delta n R T\)

⇒ \(\Delta H=\left[-3263.9-\frac{3}{2} \times 8.314 \times 10^{-3} \times 298\right] \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

=\(-3267.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 27. If \(\frac{1}{2} a \rightarrow b, \quad \delta h=+150 \mathrm{~kj} \cdot \mathrm{mol}^{-1} ; \quad 3 b \rightarrow 2 c+d\)   ah=-125 kj mol^-1 ; e + a-*2d , ΔH= + 350 k. Mol^-1 then ah of the reaction b + d→ £ + 2c will be

1. 525Kj.mol^-1
2. -175Kjmol^-1
3. -325Kj.mol^-1
4. 352Kj.mol^-1

Answer: 2. -175Kjmol^-1

½A→ B: ΔH = + 150kJ.mol^-1

3B→2c + D; ΔH = -125 kj. mol^-1

E + A →2D; ΔH = +350 kj . mol^-1

Subtracting equation [3] from (2 x equation (1) + equation (2), we have

B + D→E + 2c; ΔH = 2 × 150- 125- 350

= -175 kJ.

Question 28. Which is the correct option for free expansion of an ideal gas under adiabatic conditions—

1. q=0, ΔT≠0, W=0
2. q≠0, ΔT≠0, W=0
3. q=0, ΔT≠0, W=0
4. q=0, ΔT<0, W≠0

Answer: 3. q=0, ΔT≠0, W=0

For an adiabatic process, q = 0, and for free expansion of a gas, w = 0.

∴ Δu=q + w = 0 + 0 = 0. For an ideal gas undergoing an isothermal process, ah = 0. So, the given process is isothermal and hence at = 0.

Question 29. The enthalpy change for the reaction, 4h(g)→2h2(g) is -869.5 kl. The dissociation energy of the H – Hbond is

1. -434.8 kJ
2. -869.6kJ
3. +434.8 kJ
4. +217.4kJ

Answer: 3. +434.8 kJ

We know, bond formation enthalpy =(-)x bond dissociation enthalpy. Now, bond formation enthalpy for 2 mol h —h bonds = -869 kJ

Bond dissociation enthalpy for h— h bond \(=\frac{1}{2} \times 869=434.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 30. If the enthalpy change for the transition of liquid water to steam is 30 kj.mol-1 at 27°c, the entropy change in ] mol^-1 k^-1 for the process would be

1. 10
2. 1.0
3. 0.1
4. 100

Answer: 4. 100

Question 31. In which of the given reactions, standard reaction entropy change (as0) is positive and standard Gibbs energy change (ag°) decreases sharply with increasing temperature—

1. \(\mathrm{c} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)
2. \(\mathrm{co}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
3. \(\mathrm{mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{mgo}(\mathrm{s})\)
4. \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

Answer: 1. \(\mathrm{C} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)

In the reaction, C(graphite) \(+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\) number of gas molecules increases. As a result, the system’s entropy increases (ΔS° > 0).

We know, ΔG = ΔH – TΔS …………….(1)

for the given reaction ΔH° < 0, as it is a combustion reaction. Since at ΔH° < 0 and ΔS° < 0 according to the reaction (1), ΔG ° decreases with rise in temperature.

Question 32. For the reaction, X₂O₄(l)→2XO₂(g), ΔU=2.1 kcal, as = 20 cal.k^-1 at 300 k. Hence, ΔG is—

1. 2.7 kcal
2. -2.7 kcal
3. 9.3kcal
4. -9.3kcal

Answer: 2. -2.7 kcal

ΔH= ΔU +ΔnRT

For the given reaction, an = 2.

∴ ΔH = 2.1 + 2 × 1.987 × 10^-3 × 300 =3.2922 kcal and

ΔG = ΔH- TΔS= 3.2922 -300 × 20 × 10^-3 =-2.7 kcal

Chemical Thermodynamics Chapter 6 NCERT MCQs

Question 33. The heat of combustion of carbon to CO₂ is -393.5 kj/mol. The heat released upon the formation of 35.2g of CO₂ from carbon and oxygen gas is

1. -315 Kj
2. +315 Kj
3. -630 Kj
4. -3.15kj

Answer: c(s) + O₂(g)→CO₂(g); afh = -393.5 kj.mol^-1

The heat released on formation of 44g CO₂ =-395.5 kj-mol^-1

The heat released by the formation of 35.2g of CO₂

⇒ \(-\frac{393.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}}{44 \mathrm{~g}} \times 35.2 \mathrm{~g}=-315 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 34. For a sample of perfect gas when the pressure is changed  isothermally from p_i to p_f the entropy change is given by

1. \(\Delta S=n R T \ln \left(\frac{p_f}{p_i}\right)\)

2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

3. \(\Delta S=n R T \ln \left(\frac{p_f}{p_i}\right)\)

4. \(\Delta S=R T \ln \left(\frac{p_i}{p_f}\right)\)

Answer: 2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

For the expansion of n mol of an ideal gas, the change in entropy of the reversible isothermal process is

⇒ \(\Delta S=n R \ln \frac{V_f}{V_l}\)

At constant temperature, for n mol of an ideal gas \(\frac{V_f}{V_i}=\frac{p_i}{p_f}\)

Therefore \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

Question 35. Consider the following liquid-vapor equilibrium. Liquid ⇌ vapour which of the following relations is correct

1. \(\frac{d \ln P}{d \mathrm{~T}^2}=-\frac{\Delta H_v}{T^2}\)

2. \(\frac{d \ln P}{d \mathrm{~T}}=\frac{\Delta H_v}{R T^2}\)

3. \(\frac{d \ln G}{d T^2}=\frac{\Delta H_v}{R T^2}\)

4. \(\frac{d \ln P}{d T}=-\frac{\Delta H_v}{R T}\)

Answer: 2. \(\frac{d \ln P}{d \mathrm{~T}}=\frac{\Delta H_v}{R T^2}\)

For liquid-vapor equilibrium, the relationship between equilibrium pressure (P), the heat of vaporization (ΔH_v) and temperature (T) will be

InP = \(-\frac{\Delta H_v}{R T}+Z\)

Z = constant

Differentiating the equation concerning, we get

⇒ \(\frac{d \ln P}{d T} \frac{\Delta H_\nu}{R T^2}\)

Question 36. The correct thermodynamic conditions for die spontaneous reaction at all temperatures is—

1. ΔH < 0 and ΔS > 0
2. ΔH < 0 and ΔS < 0
3. ΔH < 0 and ΔS = 0
4. ΔH > 0 and ΔS < 0

Answer: 1. ΔH < 0 and ΔS > 0

At constant temperature and pressure, the reaction is to be spontaneous if ΔG _< 0. According to the Gibbs free energy ag = ΔH – TΔS, ΔG will be negative at any temperature if ΔH  < 0 and ΔS  > 0

Question 37. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 l to a final volume of 4.50 l. The change in internal energy a u of the gas in joules will be

1. -500 Jk^-1
2. -505 J
3. +505 J
4. 1136.25j

Answer: 2. -505 J

Work done of irreversible process,

w = \(-P_{e x} \Delta V=-2.5(4.5-2.5)=-5 \mathrm{~L} \cdot \mathrm{atm}\)

=-5 × 101.3)J

=-506.5J

Since, it is an insulated system, q = 0.

From the first law ofthermodynamics

Δ U = q+w

Or, ΔU = 0-506.5 J

=-506. 5J

Question 38. For a given reaction, ah = 35.5 kj. mol^-1 and as = 83.6 j.mol^-1. The reaction is spontaneous at (assume that ah and as do not vary with temperature)-

1. T > 425 k
2. All temperatures
3. T> 298 k
4. T< 425 k

Answer: 1. T > 425 k

ΔG = ΔH-TΔS = 35.5 ×  103- T× 83.6

Reaction is to be spontaneous if Δ < 0.

Thus, 35.5 ×10³– T ×  83.6 < 0

Therefore, T × 83.6 > 35.5 × 10³ or, T> 424.64 K

Question 39. The bond dissociation energies of x₂, y₂, and xy are in the ratio of 1 : 0.5: 1 . Ah for the formation of xy is -200 kj.mol^-1. The bond dissociation energy of x² will be

1. 200Kj.mo1^-1
2. 100Kj.mol^-1
3. 800 Kj.mol^-1
4. 400Kj.mol^-1

Answer: 3. 800 Kj.mol^-1

⇒ \(\frac{1}{2} \mathrm{x}_2+\frac{1}{2} \mathrm{y}_2 \rightarrow \mathrm{xy}\)

⇒  \(\Delta \mathrm{H}_{\text {reaction }}=\sum(\mathrm{BE})_{\text {reactant }}-\sum(\mathrm{BE})_{\text {product }}\)

BE = Bond energy

If the bond energy of x² is a kj.mol^-1 then the bond energy of y² and xy are 0.5a and a kj.mol^-1 respectively.

∴ \(-200=\frac{a}{2}+\frac{0.5}{2} a-a=-0.25 a \quad \text { or, } a=800\)

Therefore, bond energy of x₂ = 800kj.mol^-1

Question 40. Which of the following is an intensive property

1. Enthalpy
2. Entropy
3. Specific heat
4. Volume

Answer: 3. Specific heat

Intensive property: specific heat; extensive property-: enthalpy, entropy, volume.

Question 41. Which of the tires following is not a thermodynamic function—

1. Internal energy
2. Work done
3. Enthalpy
4. Entropy

Answer: 2. Workdone

Thermodynamic functions are internal energy, enthalpy, entropy, pressure, volume, temperature, free energy, and number of moles.

Question 42. For the adiabatic process, which is correct-

1. At = 0
2. As = 0
3. Q = 0
4. Qp = 0

Answer: 3. Q = 0

For the adiabatic process, no exchange of heat takes place between the system and surroundings. i.e., Q = 0.

Class 11 Chemistry Chemical Thermodynamics MCQs

Question 43. The enthalpy of formation of CO(g), CO₂(g), N₂O(g), and N₂O₄(g) is -110, -393, +811′ and kj/mol respectively for the reaction N₂O₄(g) + 3CO(g)→N₂O(g) + 3CO₂(g) ahr (kj/mol) is –

1. -212
2. +212
3. +48
4. -48

Answer: 4. -48

N₂O₄(g) + 3CO(g)→ N₂O(g) + 3CO₂(g)

ΔH_reaction =∑ Heat of formation of products heat of formation of reactants

-∑ Heat of formation of reactants

Question 44. The bond dissociation energy of CH₄ is 360 kj/mol and C₂H₆ is 620 kj/mol. Then bond dissociation energy of the C- C bond is—

1. 170 Kj/mol
2. 50Kj/mol
3. 80Kj/mol
4. 220Kj/mol

Answer: 3. 80Kj/mol

Dissociation energy of-methane = 360 kj.mol^-1

∴ Bond energy of c—h bond \(=\frac{360}{4}=90 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

The bond energy of ethane,

1× B.E. (C-C) + 6 × B.E. (C-H) = 620 KJ.mol^-1

Or, B.E.(C—C) + 6 × 90 = 620

Or, B.E. (C—C) + 540 = 620

Or, B.E. (C—C) = 620-540

Or, B.E. (C—C) = 80 KJ.mol^-1

Bond dissociation of c —C bond = 80 kj. mol^-1

Question 45. Which thermodynamic parameter is not a state function-

1. Q at constant pressure
2. Q at constant volume
3. W at adiabatic
4. W at isothermal

Answer: 4. W at isothermal

H and u are state functions but w and q are not state functions.

From the equation, Δh = ΔU+Δpv

At constant pressure, Δh = ΔU+pΔV

At constant volume, ΔH = ΔU+ VΔp

At constant pressure, Δp = 0, ΔH = qp so, it is a state function.

At constant volume, ΔV = 0, ΔU = q_V so, it is a state function.

Work done in any adiabatic process is a state function.

ΔU = q- w

∴ (Δq – 0)

ΔU = -w work done in the isothermal process is not a state function.

W = -q

∴ ΔT = 0, q ≠ 0

Question 46. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Alt < 0 , w < 0
3. Al/> 0, w = 0
4. A17 > 0 , w> 0

Answer: 1. Au< 0, w = 0

For adiabatic conditions, PV^ϒ  = constant

⇒ \(p_1 v_1^\gamma=p_2 v_2^\gamma ; v_2=\frac{1}{2} v_1\)

⇒ \(p_2=p_1\left(\frac{v_1}{v_2}\right)^\gamma \text { [for diatomic gas, } \gamma=1.4 \text { ] }\)

⇒ \(p_2=p_1\left(\frac{v_1 \times 2}{v_1}\right)^{1.4} p_2=p_1(2)^{1.4}=(2)^{1.4} p\)

Question 47. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Au< 0, w <0
3. Au> 0, w = 0
4. Au> 0, w>0

Answer: 1. Au< 0, w = 0

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb.

If the process is rim in a sealed container then no expansion or compression is allowed, so w = 0 and au = q ΔU< 0, w = 0.

Question 48. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25° c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g^-1 °c^-1, and the value of the heat of combustion of benzoic acid is

1. 881.1kcal
2. 771.124kcal
3. 981.1kcal
4. 871.2kcal

Answer: 2. 771.124kcal

Given:

Weight of benzoic acid = 1.89 g;

The temperature of the bomb calorimeter =25°c=298k;

The mass of water (m) = 18.94 kg = 18940 g;

Increase in temperature (ΔT) = 0.632°c and specific heat of water (s) = 0.998 cal- g°.C^-1

We know that heat gained by water or heat liberated by benzoic acid (q) = msΔT

= 18940 ×  0.998 × 0.632 = 11946.14 cal

Since 1.89 g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122 g of acid = \(\frac{11946.14 \times 122}{1.89}\)

= 771126.5 cal

= 771.12 kcal

(Where 122 g is the molecular weight of benzoic acid)

Question 49. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25°c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g^-1.°C^-1, and the value of the heat of combustion of benzoic acid is

1. 881.1kcal
2. 771.124kcal
3. 981. kcal
4. 871.2kcal

Answer: 2. 771.124 kcal

NCERT Class 11 Chemistry Chemical Thermodynamics Multiple Choice

Question 50. What is the entropy change in 2 mol n2, when its temperature is taken from 400 k to 800 k, at constant pressure

1. 30J/k
2. 60J/k
3. 40J/k
4. 20J/k

Answer: 3. 40J/k

⇒ \(\Delta S=n C_P \ln \frac{T_2}{T_1}\)

Or, \(R T \ln \frac{4}{2}=3 R T \ln \frac{x}{2}\)

Question 51. 1 Mole of an ideal gas expands isothermally reversible from 2 liters to 4 liters and 3 moles of the same gas expand from 2 liters to x liter and do the same work, what is ‘x’-

1. (8)^1/3
2. (4)^2/3
3. 2
4. 4

Answer: 2. (4)^2/3

⇒ \(w=n R T \ln \frac{V_2}{V_1} \text { or, } R T \ln \frac{4}{2}=3 R T \ln \frac{x}{2}\)

Or, \(\ln 2=\ln \left(\frac{x}{2}\right)^3 \text { or, }\left(x^3=16\right) \text { or, } x=(16)^{\frac{1}{3}}=4^{\frac{2}{3}}\)

Question 52. Which of the following are extensive properties

1. Volume and enthalpy
2. Volume and temperature
3. Volume and specific heat
4. Pressure and temperature

Answer: 1. Volume and enthalpy

Extensive properties depend upon the quantity of the matter contained in the system,

Example: Volume and enthalpy, etc.

Intensive properties depend only upon the nature of the substance and are independent of the amount of the substance present in the system

Example: Temperature, pressure-specific heat, etc.

Question 53.  H₂O(l) → H+(aq)+OH-(aq); ΔH°=57.32 kj .mol^-1 H₂(g) + ½ (g)→H₂O(Z); ΔH°=-285.8 kj . mol^-1 at 25°C. If Δ⁰H₂[H⁺+(aq)] = 0, then the standard heat of formation (kj.mol^-1) for OH^–(aq) at 25^°C ls

1. -142.9
2. -228.48
3. -343.12
4. -253.71

Answer: 2. -228.48

Question 54. For a reaction at T K, ΔH> 0 and ΔS > 0. If the reaction attains equilibrium at a temperature of T₁K, (assume ΔH and ΔS are independent of temperature) then

1. T<T₁
2. T>T₁
3. T=T₁
4. T>T₁

Answer: 2. T>T₁

Question 55. The change in entropy for 2 mol ideal gas in an isothermal reversible expansion from 10 mL to 100 mL at 27°C is—

1. 26.79 J. K^-1
2. 38.29 J.K^-1
3. 59.07 J-K^-1
4. 46.26 J-K^-1

Answer: 2. 38.29 J.K^-1

Question 56. Which of the statements is true

1. A reaction, in which a 77 < 0 is always spontaneous;
2. A reaction, in which ah > 0 can never occur spontaneously
3. For a spontaneous process in an isolated system,
4. For a spontaneous process in an isolated system,

Answer: 3. For a spontaneous process in an isolated system,

Question 57. For the reaction, CaCO₃(s)→CaO(s) + CO₂(g), ΔH⁰= +179.1 kl-mol-1 and ΔSO = 160.2 If ΔH⁰ and ΔS⁰ are temperature independent, then the temperature above which the reaction will be spontaneous is equal to—

1. 1008 K
2. 1200 K
3. 845 K
4. 1118 K

Answer: 4. 1118 K

Question 58. When a gas (molar mass =28 g-mol^-1) of mass 3.5g is burnt completely in the presence of excess oxygen in a bomb calorimeter) the temperature of the calorimeter increases from 208 K to 298.45 K. The heat of combustion at constant volume for the gas (Given; the heat capacity of the calorimeter 2.5 k K4)

1. 4.5 kJ .mol^-1
2. 8.0 kJ. mol^-1
3. 9.0 KJ .mol^-1
4. 9.5 kJ mol^-1

Answer: 3. 9.0 KJ .mol^-1

Question 59. For the reaction, 2NH₃(g)→ N₂(g)+3H₂(g)-

1. ΔH< 0, ΔS> 0
2. ΔH> 0, ΔS > 0
3. ΔH > 0, ΔV <0
4. ΔH < 0, Δ5 < 0

Answer: 2. AH> 0, AS > 0

Question 60. Which of the following pairs is true for the process C₆H₆(g)[1atm, 80.1°C]→C₆H₆(l)[1 atm, 80.1°C]

1. ΔG< 0, ΔS> 0
2. ΔG< 0, ΔS< 0
3. ΔG = 0, ΔS < 0
4. ΔG = 0, ΔS > 0

Answer: 3. ΔG = 0, ΔS < 0

Question 61. The internal energy change when a system goes from state P to Q is 30kJ> mol^-1 If the system goes from P to Q by a reversible path and returns to state P by an irreversible path, what would be the net change in internal energy

1. 30kj
2. <30kJ
3. zero
4. >30kJ

Answer: 3. Zero

MCQs on Chemical Thermodynamics Class 11 Chemistry

Question 62. If at normal pressure and 100°C the changes in enthalpy and entropy for the process, H₂O(l)→H₂O(g), are ΔH and ΔS respectively, then ΔH-ΔU is—

1. 5.6 kj. mol^-1
2. 6.2 kj – mol^-1
3. 3.1 kj-mol^-1
4. 4.8 kj-mol^-1

Answer: 3. 3.1 kj-mol^-1

Question 63. At 25°C, the standard heat of formation for Br₂(g) is 30.9 kj.mol^-1. At this temperature, the heat of vaporization for Br₂(l) is—

1. <30.9 kj. mol^-1
2. 30.9 kj .mol^-1
3. >30.9 kj. mol^-1
4. Cannot Be Predicted

Answer: 2. 3.1 kj-mol^-1

Question 64. At 25°C, when 0.5 mol of HCl reacts completely with 0.5 mol of NaOH in a dilute solution, 28.65 kj of heat is liberated. If at 25°C Δ _f H⁰[H₂O(l)]=  then Δ _f H⁰OH^– (aq) is—

1. -314.45 kj. mol^-1
2. -285.8 kj.mol^-1
3. -228.5 kj. mol^-1
4. -343.1 kJ. mol^-1

Answer: 2.  -285.8 kj.mol^-1

Question 65. On combustion, C_xH_Y(l) forms CO₂(g) and H₂O(l). At a given temperature and pressure, the value of \(\left(\frac{\Delta H-\Delta U}{R T}\right)\) in this combustion reaction is—

1. \(\frac{x}{5}\)
2. \(\frac{x+y}{3}\)
3. \(\frac{y}{4}\)
4. \(\frac{x-y}{4}\)

Answer: 3. \(\frac{y}{4}\)

Question 66. An ideal gas is compressed isothermally at 25°C from a volume of 10 L to a volume of 6 L. Which of the following is not true for this process—

1. q<0
2. w>0
3. ΔU = 0
4. ΔH > 0.

Answer: 4. ΔH > 0.

Question 67. At 27°C, for the reaction aA(g) + B(g)→2C(g) PΔV = -2.5 kj . The value of ‘a’ is

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 68. When 1 mol of an ideal gas is compressed in a reversible isothermal process at T K, the pressure of the gas changes from 1 atm to 10 atm. In the process, if the work done by the gas is 5.744 kJ, then T is—

1. 400k
2. 300k
3. 420k
4. 520k

Answer: 2. 300k

Question 69. For 0.5 mol of an ideal gas, 15 cal of heat is required to raise its temperature by 10 K at constant volume. The molar heat capacity for the gas at constant pressure is-

1. 3cal.K^-1 . mol^-1
2. 4 cal.k^-1mol^-1
3. 5cal.k^-1 .mol^-1
4. 4.5cal.k^-1.mol^-1

Answer: 3. 5cal.k-1 .mol^-1

Question 70. According to the enthalpy diagram given below, the standard heat of formation (kJ mol^-1) of CO(g) at 25°C is

1. -283.0
2. -110.5
3. +283.0
4. +110.5

Answer: 2. -110.5

Question 71. 1 mol of an ideal gas is enclosed in a cylinder fitted with a frictionless and weightless piston. The gas absorbs x kJ heat and undergoes expansion. If the amount of expansion work done by the gas is x kJ, then the expansion is—

1. Adiabatic
2. Cyclic
3. Isothermal
4. It cannot Be Predicted

Answer: 3. Isothermal

Question 72. Given (at 25°C)

⇒ \(\mathrm{Ca}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CaO}(\mathrm{s}) ; \Delta H^0=-635.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \)

⇒  \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s}) ; \)

⇒  \(\Delta H^0=-65.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The heat of formation (in kj.mol^-1 ) for Ca(OH)₂(s) is –

1. -855.4
2. -673.9
3. -986.6
4. -731.7

Answer: 3. -986.6

Question 73. At 25°C, the standard heats of formation of H₂O(g) , H₂O₂(g), H(g) and O(g) are -241.8, -135.66, 218 and 249.17 k).mol^-1 respectively. The bond energy (in kj.mol-1 ) of O —O bond in H₂O₂(g) molecule is—

1. 179.23
2. 160.19
3. 142.60
4. 157.16

Answer: 3. 142.60

Question 74. Given (at 25°C):

⇒ \(\mathrm{C}(\mathrm{s} \text {, graphite })+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)

⇒ \(\Delta H^0=-110.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{O}(\mathrm{g}) ; \Delta H^0=+249.1 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \)

⇒ \(\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{O}(\mathrm{g}) ; \Delta H^0=1073.24 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The standard enthalpy change for the process, C(s, graphite)→C(g) at 25°C is—

1. +934.64 KJ.Mol^-1
2. 713.64 KJ.Mol^-1
3. 962.64 KJ.Mol^-1
4. 652.64 KJ.Mol^-1

Answer: 2. 713.64 KJ.Mol^-1

Question 75. At 25°C, for the reaction, H+{aq) + OU-(aq)yH₂O(l) , AHO = -57.3 kj . mol^-1. If the ionization enthalpy of HCN in water is 45.2 kj.mol^-1, then the standard heat of reaction (in kj mol^-1 ) for the reaction, HCN(aq) + NaOH(at jr)-NaCN(ag) + H₂O(Z), in dilute aqueous solution is

1. -113.5
2. -12.1
3. -102.5
4. -35.7

Answer: 2. -12.1

Question 76. The temperature of a bomb calorimeter changes from 25°C to 32.7°C when wg of naphthalene mass (molar = 128 g. mol^-1 ) is burnt completely in the calorimeter. If the heat of combustion at constant volume for naphthalene is -5152 kj mo^-1 then w is (heat capacity of the calorimeter = 8.19 kj K^-1)

1. 0.87 g
2. 1.91 g
3. 2.37 g
4. 1.57g

Answer: 4. 1.57g

Question 77. In which of the following processes the change in entropy for the system is zero

1. Irreversible adiabatic processes
2. Reversible adiabatic process
3. A spontaneous process occurring in an isolated system
4. Isothermal expansion of an ideal gas

Answer: 2. Reversible adiabatic process

Question 78. A system undergoes the process: A→ B → C→ D. In this process, the change in a state function (.X) of the system is x. In steps A→B and B→C of the process, if the changes in X are y and z respectively, then the change in X in step D→C is

1. x-y-z
2. x-z+y
3. y+z-x
4. y-z-x

Answer: 3. y+z-x

Question 79. At 27°C, ΔH = + 6 kJ for the reaction A + 2B-3C. In the reaction, if ΔASuniv = 2 J . K^-1 , then ΔS_sys (in J . K^-1 ) is

1. +2
2. +3
3. +20
4. +22

Answer: 4. +22

Chemical Thermodynamics Multiple Choice Questions for Class 11

Question 80. An LPG cylinder contains 14 kg of butane. A family requires 2 X 104 kj of heat for their cooking purpose every day. By how many days will the butane in the cylinder be used up (Given: heat of combustion for butane = -2658 kj-mol^-1 )

1. 15 days
2. 20 days
3. 32 days
4. 40 days

Answer: 3. 32 days

Question 81. For a reaction involving 1 mol of Zn and 1 mol of H₂SO₄ in a bomb calorimeter

1. ΔH> 0 , w> 0
2. ΔU> 0 , w> 0
3. ΔU< 0 , w> 0
4. ΔU< 0 , w> 0

Answer: 4. ΔU< 0 , w> 0

Question 82. Assuming that water vapor is an ideal gas, the internal energy change (ALT) when mol of water is vaporized at bar pressure and 100°C, will be (Given: at bar and 373K, molar enthalpy of vaporization of water is 41 kj. mol^-1, R = 8.3 J.mol^-1 . K^-1 )

1. 4.100 KJ. mol
2. 3.7904 KJ. mol-1
3. 37.904 kj. mol-1
4. 41.00 kj. mol-1

Answer: 3. 37.904 kj. mol-1

Question 83. At 25°C and 1 atm pressure, ΔH and pressure-volume work for the reaction, 2H₂(g) + O₂(g)y₂H₂O(g) are —483.7 kj and 2.47 kj respectively. In this reaction the value U is-

1. -483.7 kJ
2. -481.23 KJ
3. -400.23 Kj
4. -492.6 KJ

Answer: 2. -481.23 KJ

Question 84. An ideal gas’s initial state of 1 mol is (P₁, V₂, T₁ ). The gas is expanded by a reversible isothermal process and also by a reversible adiabatic process separately. If the final volume of the gas is the same in both of the processes, and changes in internal energy in the isothermal and adiabatic processes are ΔU₁ and ΔU₂ respectively, then

1. ΔU₁=ΔU₂
2. ΔU₁<ΔU₂
3. ΔU₁>ΔU₂
4. Cannot Be Predicated

Answer: 3. ΔU₁>ΔU₂

Question 85. At constant pressure, the amount of heat required to raise the temperature of 1 mol of an ideal gas by 10-C is x kj. If the same increase in temperature were carried out at constant volume, then the heat required would be

1. > xKJ
2. <x Kj
3. = x KJ
4. > xKJ

Answer: 1. > xKJ

Question 86. The enthalpy of fusion of ice at 0- C and 1 atm is 6.02 kj – mol^-1. The change in enthalpy (J K^-1)of the surroundings when 9 g of water is frozen at 0°C and 1 atm pressure is

1. +11.02
2. -11.02
3. -20.27
4. +23.09

Answer: 2. -11.02

Question 87. At a given pressure and a temperature of 300 K, Δ_Surr and Δ_sys for a reaction are 8.0 J. K^-1 and 4.0 J . K^-1surr respectively. ΔG for this reaction is-—

1. -3.0 KJ
2. -3.6 KJ
3. 3.0 kJ
4. -4.2 kJ

Answer: 2. -3.6 KJ

Question 88. In a reversible process, if changes in the entropy of the system and its surroundings are ΔS₁ and ΔS₂ respectively, then

1. ΔS₁ +ΔS₂ >0
2. ΔS₁ +ΔS₂<0
3. ΔS₁+ΔS₂= 0
4. ΔS₁ +ΔS₂>0

Answer: 3. ΔS₁+ΔS₂= 0

Question 89. A flask of volume 1 L contains 1 mol of an ideal gas. The flask is connected to an evacuated flask, and as a result, the volume of the gas becomes 10 L. The change in entropy (J. K^-1) of the gas in this process is

1. 9.56
2. 19.14
3. 11.37
4. 14.29

Answer: 2. 19.14

Question 90. The heats of neutralization of four acids A, B, C, and D are 13.7, 9.4, 11.2, and 12.4kcal respectively when they are neutralized against a common base. The weakest add among A, B, C, and D is

1. A
2. B
3. C
4. D

Answer: 2. B

Question 91. In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which is true-

1. ΔU= w≠0, q=0
2. ΔU=0, w=0, q≠0
3. ΔU=0 w=0, q≠0
4. w=0 w≠0, q=0

Answer: 1. ΔU= w≠0, q=0

Question 92. How many calories are required to increase the temperature of 40g of Ar from 40-C to 100°C at a constant volume (R = 2 cal.mol^-1.K^-1)

1. 120
2. 2400
3. 1200
4. 180

Answer: 4. 180

Question 93. Water is supercooled to -4°C. The enthalpy (H) of the supercooled water is

1. Same as ice at -4°c
2. More than ice at -4°c
3. Same as ice at 0°c
4. Less than ice at -4°c

Answer: 4. Less than ice at -4°c

Question 94. The standard entropy of X₂, Y₂, and XY₃ are 60, 40, and 50 J.K-1.mol^-1 respectively. For the reaction\(\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3,(\Delta H=-30 \mathrm{~kJ})\) (AH = -30 kl ) to be at equilibrium, the temperature will be

1. 1250 k
2. 750k
3. 500 k
4. 1000 k

Answer: 2. 750k

Question 95. Two moles of gas of volume 50L and pressure 1 atm are compressed adiabatically and reversibly to 10atm. What is the atomicity of the gas (T₁/T₂= 0.4)

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 96. Given (at 25°C): C(s, graphite)-C(g) ; ΔH⁰ = +713.64 kj. mol^-1

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g}); \Delta H^0=+218 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) C(s, graphite)+3H₂(g)→C₆H₆(g); ΔH°-+82.93kj – mol^-1At 25°C, if the bond energy of C—H and C— C bonds are 418 and 347 kj .mol^-1 respectively, then the C=C bond energy is

1. +679.81 KJ. mol^-1
2. +652.63 kj. mol^-1
3. +808.75 KJ. mol^-1
4. +763.39 kJ. mol^-1

Answer: 2. +652.63 kj. mol^-1

Question 97. Given (at 25 °C):

⇒ \(\mathrm{C}_6 \mathrm{H}_6(g)+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H^0=-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25 °C, if the stand heat of formation of C₃H₈(g) is -103.8 kj. mol-1 , then the standard heat of reaction for the reac3h8ction; C₃H₈(g) + H₂(g)yC₂H₆(g) + CH₄(g) is

1. +98.45 kJ
2. -55.70KJ
3. 62.37 KJ
4. -47.25 KJ

Answer: 2. -55.70KJ

Question 98. At 0°C and normal pressure, the enthalpy of fusion of ice is 334.7 J . g^-1. At this temperature and pressure, if 1 mol of water is converted into 1 mol ice, then the change in entropy of the system will be

1. 16.7 J.K^-1
2. -16.7 J.K^-1
3. 22.06 J.K^-1
4. -22.06 J.K^-1

Answer: 4. -22.06 J.K^-1

Question 99. 5 mol of gas is put through a series of changes as shown graphically in a cyclic process. The process X→Y, Y→Z and Z→X respectively are

1. Isochoric, isobaric, isothermal
2. Isobaric, isochoric, isothermal
3. Isothermal, isobaric, isochoric
4. Isochoric, isothermal, isobaric

Answer: 1. Isochoric, isobaric, isothermal

Question 100. Given \(\mathrm{NH}_3(g)+3 \mathrm{Cl}_2(g) \rightarrow \mathrm{NCl}_3(g)+3 \mathrm{HCl}(g) ;-\Delta H_1\) N₂(g) + 3H₂(g)→2NH₃(g) ; ΔH₂; H₂(g) + Cl₂(g→2HCl(g);ΔH₃ Heat of formation (ΔHf) of NCl₃(g) in terms of ΔH₁ , ΔH₂ and ΔH₃ is-

1. \(-\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

2. \(\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

3.\(\Delta H_1-\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

4. None of the above

Answer: 4. None of the above

Chemical Thermodynamics MCQs Class 11 Solutions

Question 101. When lg of graphite is completely burnt in a bomb calorimeter, the temperature of the bomb and water rises from 25°C to 30.5°C. The heat capacity of the calorimeter is 5.96 kj. °C-1, then the heat of combustion per mole of graphite at constant volume is

1. -357.13 kj.mol^-1
2. -289.71 kj.mol^-1
3. -393.36 kj .mol^-1
4. -307.94 kj-mol^-1

Answer: 3. -393.36 kj .mol^-1

Question 102. The volume of a gas is reduced to half of its original volume. The specific heat will—

1. Reduce to half
2. Double
3. Remain Constant
4. Increase Four times

Answer: 3. Remain Constant

Question 103. For which molar-specific heat is temperature-independent

1. Argon
2. Hydrogen
3. Nitrogen
4. Carbon dioxide

Answer: 1. ARgn

Question 104. Which of the following quantities are state functions

1. q
2. q+w
3. w
4. U+pv

Answer: 1. q

Question 105. A monoatomic ideal gas undergoes the cyclic process, Which of the comments are true for this process

1. For the whole process q = +1.134J
2. For the whole process Δ_sys> 0
3. For the whole process Δ_sys = 0
4. For the whole process q = -2.310 J

Answer: 1. For the whole process q = +1.134J

Question 106. Which of the following comments is true

1. Only for an ideal gas, cp m > cv m
2. For any gas, cp m > cv m
3. For a solid substance, cp m – cv> m
4. For ‘ n ’ mol ofideal gas, cp m- cv m = nr

Answer: 2. For any gas, cp m > cv m

Question 107. When 3g of ethane gas is brunt at 25°C, 156 kl of heat is liberated. If the standard enthalpies of formation for CO₂(g) and H₂O(l) are -393.5 and -285.8 kj.mol^-1respectively, then for ethane gas-

1. Standard heat of combustion = -1560 kJ mol^-1
2. Standard heat of formation =-67.9 kJ. Mol^-1
3. Standard heat of combustion =-832 kJ .mol^-1
4. Standard heat of formation = -84.4 kJ. Mol^-1

Answer: 1. Standard heat of combustion = -1560 kJ. Mol-^-1

Question 108. A reaction is spontaneous at a temperature of 300K, but it is non-spontaneous at a temperature of 400 K. If ΔH and ΔS for the reaction do not depend on temperature, then

1. ΔH> 0
2. ΔH < 0
3. ΔS > 0
4. ΔS<0

Answer: 2. AH < 0

Question 109. The reaction, 3O₂(g)→2O₃(g), is non-spontaneous at any temperature. Hence

1. The reverse reaction is spontaneous at any temperature
2. ΔH < 0 and ΔS < 0 for the reverse reaction
3. ΔH > 0 , ΔS > 0 for the reverse reaction
4. ΔH < 0 and ΔS > 0 for the reverse reaction

Answer: 1. The reverse reaction is spontaneous at any temperature

Question 110. For the isothermal free expansion of ideal gas

1. ΔH =0
2. ΔS < 0
3. ΔS > 0
4. ΔH > 0

Answer: 1. ΔH =0

Question 111. The changes in which of the following quantities are for a cyclic process

1. Enthalpy
2. Work
3. Entropy
4. Internal energy

Answer: 1. Enthalpy

Question 112. Which of the following relations are true for the reaction, PCl₅(g)→PCl₃(g) + Cl₂(g)

1. ΔH< 0
2. ΔH >0
3. ΔS <0
4. ΔS>0

Answer: 2. ΔH >0

Question 113. An ideal gas performs only pressure-volume work in the given cyclic process. In the diagram, AB, BC, and CA are the reversible isothermal, isobaric, and isochoric processes respectively. Identify the correct statements regarding this cycle

1. Total Work Done In This Process ( W) = WA→B + WB→C

2. Changes In Internal Energy In The Step Ab = 0

3.ΔSA→b = ΔSB→C+ΔSC→A

4. If The Total Heat And Work Involved In The Process Are Q And W Respectively, Then q + W = 0

Answer: 2. Changes In Internal Energy In The Step Ab = 0

Question 114. Identify the correct statements

1. Standard state of bromine (25°C, 1 atm) is Br2(g)

2. C (graphite, s)-C (diamond, s); here AH=0

3. Standard enthalpy change for the reaction N₂(g) + O₂(g)→2NO(g) at 25°C and 1 atm is standard enthalpy of formation of NO(g)

4. At a particular temperature and pressure, if ΔH = xkj for the reaction A + 3.B→2C then AH \(-\frac{x}{2} \mathrm{~kJ}\) for the reaction \(C \rightarrow \frac{1}{2} A+\frac{3}{2} B\)

Answer: 2. C (graphite, s)-C (diamond, s); here ΔH=0

Question 115. Which of the following statements is correct

1. In any adiabatic process, ΔSsys = 0
2. In the isothermal expansion of ideal gas, ah = 0
3. An endothermic reaction will be spontaneous if in this reaction ΔSsys > 0
4. Heat capacity is a padi-dependent quantity

Answer: 2. In the isothermal expansion of ideal gas, ah = 0

Question 116. Correct statements are

1. A + B→D, ΔH = x kj.This reaction is performed in the following two steps: 1. A + B→C  2. C→D. If in step 1. ΔH = y kj. then 2. ΔH = (x-y)J in step
2. for a spontaneous process occurring in an isolated system ΔSsys> 0 at equilibrium
3. In a spontaneous chemical reaction at constant temperature and pressure,  ΔG = =TΔ_surr
4. In a chemical reaction ΔH > 0 and ΔS > 0. The reaction attains equilibrium at temperature, Tg. At constant pressure and constant temperature TK the reaction will be spontaneous, if T > T

Answer: 1. A + B-D, ΔH = x kj.This reaction is performed in the following two steps:

1. A + B→C→C
2. C→D.

If in step

1. ΔH = ykj. then
2. ΔH = (x-y)J in step

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 117. Some reactions and their ΔH° values are given below:

⇒ C(graphite,s)\(+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta H^0=a \mathrm{~kJ}\)

⇒  \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=b \mathrm{~kJ}\)

⇒  \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=c \mathrm{~kJ}\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=d \mathrm{~kJ}\)

⇒ \(2 \mathrm{C} \text { (graphite,s) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; \Delta H^0=m \mathrm{~kJ}\)

Which of the following statements is correct

1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1
2. Standard heat of combustion of C₂Hg = (2a + 2b + 3c- m) kj.mol^-1
3. Standard heat of combustion of carbon = a kj.mol^-1
4. Standard of formation of CO₂(g)=(a+b) kj.mol^-1

Answer: 1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1

Question 118. At 25°C, in which of the given reactions do standard enthalpies of reactions indicate standard enthalpies of formation of the products in the respective reactions

1. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{3} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{HI}(\mathrm{g})\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Question 119.

1. CaCO₃(s)→CaO(s)+CO₂(g)
2. \(\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)\)
3. PCl₅(g)⇒ Pcl₃(g) + C1₂(g)
4. N₂(g) + O₂(g)→2NO(g)

For which of these reactions, P-V work is negative

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 120. Which of the given reactions are endothermic

1. Combustion of methane
2. Decomposition of water
3. Dehydrogenation of ethane to ethene
4. Conversion of graphite to diamond

Answer: 2. Decomposition of water

Question 121. At constant volume and 298K, mol of gas is heated and the final temperature is 308 K. Ifheat supplied to the gas is 500 J, then for the overall process

1. w = 0
2. w = -500 J
3. AU = 500 J
4. AU = 0

Answer: 1. w = 0

Question 122. True for spontaneous dissolution of KCl in water are

1. ΔG<0
2. ΔH > 0
3. ΔS_surr < 0
4. ΔH<0

Answer: 1. ΔG<0

Question 123. When a bottle of perfume is opened, odorous molecules mix with air and diffuse gradually throughout the room. The correct facts about the process are

1. ΔS = 0
2. ΔG < 0
3. ΔS> 0
4. ΔS < 0

Answer: 2. ΔG < 0

Question 124. mol of an ideal gas undergoes a cyclic process ABC A represented by the following diagram

Which of the given statements is correct for the process —

1. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)
2. Work done by the gas in the overall process is P₀ V₀
3. Heat absorbed by the gas in path AB is 2P₀ V₀
4. Heat absorbed by the gas in path BC is \(\frac{1}{2} P_0 V_C\)

Answer: 2. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)

Question 125. At 0°C and 10 atm pressure 14g of oxygen is subjected to undergo a reversible. adiabatic expansion to a pressure of 1 atm. Hence in this process

1. The final temperature of the gas is 141.4 K
2. The final temperature of the gas is 217.3 K
3. Work done = 293.2 cal
4. Work done = -286 cal

Answer: 1. Final temperature of the gas is 141.4 K

Question 126. Choose the reactions in which the standard reaction enthalpy (at 25°C) represents the standard formation enthalpy of the product

1. \(\mathrm{H}_2(g)+\frac{1}{3} \mathrm{O}_3(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g)\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Question 127. Which of the following is (are) not a state function?

1. Enthalpy
2. Heat capacity
3. Heat
4. Work done

Answer: Heat and work

Question 128. A thermodynamic state function is a quantity

1. Used to determine heat changes
2. Whose value is independent of the path
3. Used to determine pressure-volume work
4. Whose value depends on temperature only

Answer: 2. Whose value is independent of the path

A thermodynamic state function of a system is a quantity whose value depends only on the present state of the system. Its value does not depend on the path of a process in which the system participates.

Question 129. For the process to occur under adiabatic conditions, the correct condition is 

1. ΔT=0
2. ΔP=0
3. q=0
4. w=0

Answer: 3. q=0

In an adiabatic process, no exchange of heat takes place between the system and its surroundings.

Question 130. Enthalpies of all elements in standard states are

1. Unity
2. Zero
3. <0
4. Different for each element

Answer: 2. Zero

By convention, the enthalpies of all the elements in their standard states are considered to be zero.

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 131. ΔU° for combustion of methane is -XkJ .mol ^-1. The value of ΔH° is

1. ΔU°
2. >ΔU°
3. <ΔU°
4. 0

Answer: 3. <ΔU°

Question 132. An amount of work w is done by the system and a q amount of heat is supplied to the system. By which the following relations the change in internal energy of the system can be expressed-

1. Δ U = q-w
2. Δ U =q+w
3. ΔU=q
4. ΔU=w-q

Answer: 4. ΔG<0

Question 133. At 25°C which of the following has an enthalpy of formation zero

1. HCL(g)
2. O₂(g)
3. O₃(g)
4. NO(g)

Answer: 2. O₂(g)

Question 134. Which one is the correct unit of entropy

1. k-1. mol^-1
2. J.k-1. mol^-1
3. J.mol^-1
4. J-1.k-1. mol^-1

Answer: 2. J.k-1. mol^-1

Question 135. For the reversible reaction A + 2B→ C + Heat, the forward reaction will proceed at

1. Low temperature and low pressure
2. Low pressure
3. High pressure and low temperature
4. High pressure and high temperature

Answer: 3. High pressure and low temperature

For the reversible reaction A + 2 BC + A, the forward reaction will proceed at high pressure and low temperature.

Question 136. Which ofthe following is an example of a closed system

1. A hot water-filled thermos flask
2. An ice water-filled airtight metallic bottle
3. A water-filled stainless steel bowl
4. A hot water-filled glass beaker

Answer: 2. An ice water-filled airtight metallic bottle

An ice-water-filled airtight metallic bottle is an example of a closed system.

Question 137. Which is an intensive property of a system

1. Internal energy
2. Entropy
3. Mass
4. Density

Answer: 4. Density Density is an intensive property.

Question 138. If one monoatomic gas is expanded adiabatically from 2L to 10 L at atm external pressure then the value of a ΔU (in atm. L )is

1. -8
2. 0
3. -66.7
4. 58.2

Answer: 1. -8

q = 0 (since process is adiabatic.) ∴ Δu =w = -pav

=-1(10 -3) atm.L=-8 atm.L

Question 139. The equation of state for ‘ n’ mol of an ideal gas is PV – nRT. fn this equation, the respective number of intensive and extensive properties are

1. 2,3
2. 3,2
3. 1,4
4. 4, 1

Answer: 2. 3,2

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 140. The enthalpy of combustion of methane, graphite, and dihydrogen at 298K are -890.3 kj.mol^-1, -393.5 kj.mol^-1, and -285.8 kj.mol^-1 respectively. Enthalpy of formation of CH₄(g) will be—

1. -74.8KJ. mol^-1
2. —52.27kl.mol^-1
3. +74.8KJ. mol^-1
4. +52.26KJ. mol^-1

Answer: 1.  -74.8KJ. mol^-1

According to the given data

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-890.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-285,8 \mathrm{~kJ} \cdot \text { mol }^{-1}\)

By 1 xeq.(2) + 2xeq.(3)-eq.(1), we get the thermochemical equation involving the formation reaction of CH₄(g)

⇒ \(\mathrm{C}(s \text {, graphite })+\mathrm{O}_2(g)+2 \mathrm{H}_2(g)+\mathrm{O}_2(g)-\mathrm{CH}_4(g)-2 \mathrm{O}_2(g)\)

⇒ \(\mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(g)-2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{C}(s \text {, graphite })+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g}) ; \Delta H^0=-74.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

This equation represents the formation reaction of CH₄(g) . Hence, the enthalpy of the formation of CH₄(g) is -74.8kJ.mol^-1

Question 141. A reaction, A + B→+C + D + q is found to have a positive entropy change. The reaction will be

1. Possible at high temperature
2. Possible only at low temperature
3. Not possible at any temperature
4. Possible at any temperature

Answer: 4. Possible at any temperature

The Thermochemical equation for the reaction indicates that the reaction is exothermic. So, for this reaction, ΔH < 0. It is given that ΔS > 0 for the reaction. So, according to the relation ΔG = ΔH- TΔS, ΔG will be <0 at any temperature. Hence, the reaction is possible at any temperature.

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Short Question And Answers

Question 1. Li⁺ ion is the smallest one among the ions of group- 1 elements. It would, therefore, be expected to have much higher ionic mobility, and hence the solutions of lithium salts would be expected to have higher conductivity than the solutions of cesium salts. However, in reality, the reverse is observed. Explain
Answer: 

Due to high charge density, very small Li⁺ ion gets much more hydrated compared to the large Cs⁺ ion. Thus, the size of the hydrated lithium-ion is much larger than that of the hydrated cesium ion. For this reason, the mobility of Li⁺ ion is much lower than that of Cs⁺ ion and consequently, the solutions of lithium salts have much lower conductivity than the solutions of cesium salts.

Read and Learn More NCERT Class 11 Chemistry Short Answer Questions

Question 2. The E° value for Cl₂/Cl^– is +1.36, for I₂/I^– is + 0.53V, for Ag⁺/Ag is + 0.79 V, for Na⁺/Na is -2.71V and for Li⁺/Li is -3.04V. Arrange the following atoms and ions in decreasing order of their reducing strength: I^–, Ag, Cl^–, Li, Na
Answer:

The lower the value of standard reduction potential, the greater the tendency of the reduced form to be oxidized, i.e., the reduced form will serve as a stronger reductant. Therefore, the decreasing order of reducing strength of the given atoms and ions is

Li > Na > I^–> Ag > Cl^–

Question 3. The alkali metals are obtained not by the electrolysis of the aqueous solutions of their salts but by the electrolysis of their molten salts. Explain.
Answer:

The solutions of alkali metal salts contain metal cations, anions, H⁺ ions, and OH^– ions. The discharge potential of H⁺ ions is lower than that of metal cations. So, on electrolysis of tire solutions of alkali metal salts, hydrogen is discharged at the cathode rather than the metal. However, when the molten salts of alkali metals are electrolysed, the metal cation being the only cation present, gets discharged at the cathode.

S-Block Elements Short Answer Questions Class 11

Question 4. The alkali metals are paramagnetic but their salts are diamagnetic—why?
Answer:

Due to the presence of an unpaired valence electron the alkali metals are paramagnetic. During salt formation, this unpaired electron of the outermost shell leaves the metal atom and becomes attached to a non-metal atom. As a result, the cation and the anion thus obtained contain no unpaired electron. Hence, the alkali metal salts are diamagnetic.

Question 5. Beryllium & magnesium do not give color to flame whereas other alkaline earth metals do so. Why?
Answer: 

Due to their smaller size, valence electrons of Be and Mg are more tightly held by the nucleus. Therefore, they need a large amount of energy for the excitation of their valence electrons to higher energy levels. Since such a large amount of energy is not available from Hansen flame, these two metals do not impart any color to the flame.

Question 6. E° for M²⁺(aq) + e →M(s) (where, M = Ca, Sr or Ba) Is nearly constant. Comment.
Answer:

The value of standard electrode potential (E°) of any M²⁺/M electrode depends upon three factors.

1. Enthalpy of vaporisation
2. Ionization enthalpy, and
3. Enthalpy of hydration.

Since the combined effect of these factors is approximately the same for Ca, Sr, and Ba, their standard electrode potential (E°) values are nearly constant.

Question 7. Both alkaline earth metals and their salts are diamagnetic. Explain.
Answer:

The alkaline earth metals are diamagnetic as all the orbitals are filled with paired electrons. The ions of alkaline earth metals, M²⁺ have stable noble gas configurations in which all the orbitals are doubly occupied. Also, there are no unpaired electrons in the anions. Hence, the salts of alkaline earth metals are also diamagnetic.

Question 8. Beryllium salts can never have more than 4 molecules of water of crystallization, i.e., they can never achieve a coordination number > 4 while other metal ions tend to have a coordination number of 6, for example: [Ca(H₂O₆)²⁺. Explain.
Answer: 

Beryllium does not exhibit coordination numbers more than 4 because, in its valence shell of Be²⁺ ion, there are only four available orbitals (one s and three p) present. The remaining members of the group can have a coordination number of six by using their d -d-orbitals along with s -and p -orbitals.

Question 9. Anhydrous is used as a drying agent — why?
Answer:

Anhydrone or magnesium perchlorate, Mg(ClO₄)₂ is  used as a drying agent because, due to its smaller size and higher charge, Mg has a greater tendency to complexation with water molecules by forming Mg(ClO₄)₂ -6H₂O

Question 10. Li salts are commonly hydrated while other alkali metal salts are usually anhydrous. Explain.
Answer:

Due to the smallest size among all alkali metal ions, the Li⁺ ion can interact with water molecules more easily than the other alkali metal ions. Hence the salts of lithium are commonly hydrated. On the other hand, other alkali metal ions being larger have little tendency to get hydrated. Therefore, their salts are generally anhydrous. For example, lithium chloride crystallizes as LiCl. 2H₂O₂ but sodium chloride crystallizes as NaCl.

Question 11. Although the abundance of Na and K in the earth’s crust are comparable, sodium is nearly 30 times more abundant than potassium in seawater—why?
Answer:

These metals were leached from the aluminosilicate rocks by weathering. The potassium salts having large anions are less soluble than the sodium salts because of higher lattice energy. Moreover, potassium is preferentially absorbed by the plants. For this reason, sodium is more abundant than potassium in seawater.

Question 12. When caustic soda solution is kept in a glass bottle, the inner surface of the bottle becomes opaque. Explain
Answer:

Caustic soda (NaOH) being strongly basic reacts with acidic silica (SiO₂) present in glass to form sodium silicate (Na₂SiO₃)

SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

As a result, the inner surface of the bottle becomes opaque.

Question 13. Why are alkali metals stored in kerosene?
Answer:

When the highly reactive alkali metals are exposed to air,  they readily react with oxygen, moisture, and carbon dioxide of air to form oxides, hydroxides, and carbonates respectively. To prevent these reactions, alkali metals are normally stored in kerosene, an inert liquid.

Question 14. The second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals. Explain.
Answer:

The loss of a second electron from an alkali metal cation (M⁺) causes a loss of stable noble gas configuration while the loss of a second electron from an alkaline earth metal cation leads to the attainment of a very stable noble gas configuration. This explains why the second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals.

Question 15. MgO is used as a refractory material— Explain why
Answer:

Due to greater charge on both the cation (Mg²⁺) and die anion (O^2-), MgO possesses higher lattice energy and for tills, it has a very high melting point and does not decompose on heating. For this reason, it is used as a refractory material

Question 16. BaSO₄ is insoluble in water whereas BeSO₄ is soluble in water—1-explain with reason._
Answer: 

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 17. BeCl₂ fumes in moist air but BaCl₂ does not. Explain
Answer:

BeCl₂ being a salt of a weak base, Be(OH)_2, and a strong acid, HCl undergoes hydrolysis in moist air to form HCl; which fumes in air. BaCl₂ on the other hand, being a salt of a strong base, Ba(OH)_2, and a strong acid, HCl does not undergo hydrolysis to form HCl and hence does not fume in moist air

BeCl₂ + 2H₂O→Be(OH)₂ + 2HCl↑

BaCl₂ +H₂O→ Ba(0H)₂ + 2HCl

NCERT Solutions Class 11 Chemistry Chapter 10 S-Block Elements SAQ

Question 18. Mg₃N₂ when reacts with water, gives off NH₂ but HCl is not evolved when MgCl₂ reacts with water at room temperature. Give reasons.
Answer:

Mg₃N₂ is a salt of the strong base, Mg(OH)₂ and the weak acid, NH₃

Hence it gets hydrolyzed to give NH₃.

Mg₃N₂ + 6H₂O→ 3Mg(OH)₂ + 2NH₃T↑

MgCl₂,  On the other hand, is a salt of the strong base, Mg(OH)_2, and the strong acid, HCl. Hence, it does not undergo hydrolysis.

Question 19. A piece of burning magnesium ribbon continues to bum in sulfur dioxide.
Answer:

A piece of magnesium ribbon continues to bum in SO₂ since it reacts to form MgO and S. This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning.

This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning.

Question 20. Ba²⁺ ions are poisonous, still, they are provided to patients before taking stomach X-rays. Explain
Answer:

A barium meal (suspension of BaSO₄ in water) is generally given to patients when X-ray photographs of the alimentary canal are required. The salt provides a coating of the alimentary canal and hence X-ray photograph can be taken since it is quite transparent to the X-ray otherwise. BaSO₄ is almost insoluble in water and hence it does not pass from the digestive system to the circulatory system and can therefore be safely used for the purpose.

Question 21. Can sodium hydride be dissolved in water? Justify.
Answer:

Sodium hydride cannot be dissolved in water because it gets hydrolyzed with brisk effervescence of hydrogen gas.

NaH + H₂ O→  H₂ + NaOH

Question 22. Why does sodium impart a yellow color in the flame?
Answer:

The ionization enthalpy of Na is relatively low. Therefore when this metal or its salt is heated in Bunsen flame, its valence shell electron is excited to higher energies by absorption of energy. When the excited electron returns to its initial position in the ground state, it liberates energy in the form of light in the yellow region of the electromagnetic spectrum. That’s why sodium imparts a yellow color to the flame.

Question 23. Wind makes lithium exhibit uncommon properties compared to the rest of the alkali metals.
Answer:

The unusual properties of lithium as compared to other alkali metals are since

• Li – atom and L⁺  ion are exceptionally small in size and
• Li⁺ ion has the highest polarising power (i.e.„ charge/size ratio).

Question 24. What is the common oxidation state exhibited by the alkali metals and why?
Answer:

The alkali metals easily lose their valence electrons (ns¹) to acquire a stable octet, (i.e., the stable electronic configuration of the nearest noble gas) and because of this, the common oxidation state exhibited by the alkali metals is +1.

Question 25. What is the difference between baking soda and baking powder?
Answer:

Both baking soda and baking powder are leavening agents. Baking soda is pure sodium bicarbonate. When baking soda is combined with moisture and an acidic ingredient (for example,  Yoghurt, buttermilk) the resulting chemical reaction produces CO₂ gas bubbles that cause baked goods to rise. Baking powder contains NaHCO₃, but it includes an acidifying agent (cream of tartar) already, and also a drying agent (usually starch).

Baking powder is available as single-acting baking powder and as double-acting baking powder. Single-acting baking powders are activated by moisture, so we must bake recipes that include this product immediately after mixing. Double-acting powder reacts in two phases and can stand for a while before baking.

Question 26. Though table salt is not deliquescent it gets wet ! in the rainy season— Explain.
Answer:

Pure NaCl is not deliquescent but table salt contains impurities like MgCl₂ and CaCl₂ These impurities being deliquescent absorb moisture from air in the rainy season. As a result, table salt gets wet.

Question 27. What precautions should be taken while handling beryllium compounds and why?
Answer:

Contact of Be compounds with skin dermatitis, and inhaling dust or smoke of Be-compounds causes a disease called berylliosis which is rather similar to success- Therefore, beryllium compounds should be handled with care.

Question 28. Explain why the elements of group 2 form M²⁺ Ions, but not M3+ ions.
Answer:

Loss of third electron from group-2 metal atoms causes loss of stable noble gas configuration and for this reason group- 2 elements form M²⁺ ions but not M³⁺ ions. fifl Arrange Be(OH)₂, Ba(OH)₂ & Ca(OH)₂in order of increasing solubility in water and explain the order. Answer: Among the alkaline earth metal hydroxides having a common anion, the cationic radius influences the lattice enthalpy. Since the lattice enthalpy decreases much more than the hydration enthalpy with increasing cationic size, the solubility increases on moving down the group.

Question 29. The reaction between marble and dilute H₂SO₄ is not used to prepare CO₂ gas—why?
Answer:

Marble (CaCO₃) reacts with dilute H₂SO₄ to form insoluble CaSO₄ which deposits on the surface of marble and prevents further reaction. So, the evolution of CO₂ ceases after some time. Thus the reaction between marble and dilute H₂SO₄ prepares CO₂ gas.

Question 30. Name the important compound of Li used in organic synthesis. How the compound is prepared?
Answer:

The compound is lithium aluminum hydride (LiAlH₄). It is a useful reducing agent and is used in organic synthesis. It is prepared by the reaction of lithium hydride and aluminum chloride in a dry ether solution.

4LiH + AlCl₃ → LiAlH₄ + 3LiCl

Question 31. What is the oxidation state of K in KO₂ and why is this compound paramagnetic?
Answer:

The superoxide ion is represented as O^–₂. It has one unit of negative charge. Since the compound is neutral, therefore, the oxidation state of K is +1. The structure of Or is 6J0 O^–₂ Since it has one unpaired electron in π∗_2p MO, therefore, the compound is paramagnetic.

Question 32. The crystalline salts of alkaline earth metals contain more water for crystallization than the corresponding alkali metals. Explain.
Answer:

In the salts of alkaline earth metals, the metal ions have a smaller size and higher charge compared to the corresponding metal ions of the alkali metals of the same period. Thus, alkaline earth metals have a greater tendency to get hydrated & form crystalline salts compared to alkali metals. Thus, NaCl is completely anhydrous whereas MgCl₂ exists as MgCl₂-6H₂O.

Question 33. Lithium salts are more stable if the anion present in the salt is small. Explain.
Answer:

Small anions have more ionic character and hence the salts of lithium containing those ions have more lattice enthalpy. Large anions, on the other hand, are highly polarisable, and hence they impart covalent character to the salt. Thus, lithium salts are more stable with small anions than that with large anions.

Question 34. Alkali metals become opaque when they are kept open in the air Why?
Answer:

As the alkali metals are highly reactive, they readily react 20; with oxygen to form oxides. These oxides undergo a reaction with the water vapor present in the air to produce hydroxides. The formed hydroxides immediately react with CO₂ of air to produce carbonate compounds. These carbonate compounds form layers on the surface of alkali metals. Consequently, they become opaque.

Question 35. BaSO₄ is insoluble in water, but BcSO₄ is soluble in water-Explain.
Answer:

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence, it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 36. An aqueous solution of Be(NO₃)₂ is strongly acidic. Explain.
Answer:

In the hydrated ion, [Be(H₂O)₄]²⁺, water molecules are extensively polarised, ultimately leading to the weakening of the O —H bond.

Hydrolysis takes place and the solution becomes distinctly acidic:

(H₂O)₃Be₂ +—OH₂ + H₂O→(H₂O)₃ 3 Be⁺ —OH + H₃O⁺

Question 37. The hydroxides and carbonates of Na and K are readily soluble in water while the corresponding salts of Mg and Ca are sparingly soluble. Explain.
Answer:

Due to smaller size and higher ionic charge, the lattice enthalpies of alkaline earth metals are much higher than those of alkali metals and hence the solubility of alkali metal hydroxides and carbonates is much higher than those of alkaline earth metal hydroxides and carbonates;

Question 38. BeO is insoluble but BeSO₄ is soluble in water. Explain.
Answer:

The higher lattice enthalpy of BeO formed by the combination of a small cation and small anion is more than its hydration enthalpy but the lattice enthalpy of BeSO₄ formed by the combination of a small cation and large anion is less than its hydration enthalpy;

Short Answer Questions for Class 11 Chemistry S-Block Elements

Question 39. BaO is soluble but BaSO₄ is insoluble in water — why?
Answer:

The lattice enthalpy of BaO formed by the combination of a large cation and a small anion is less than its hydration enthalpy but the lattice enthalpy of BaSO₄ formed by the combination of a large cation and a large anion is more than its hydration enthalpy;

Question 40. Lil is more soluble than KIin ethanol. Explain.
Answer:

Due to the much higher polarising power of very small Li⁺ ion, Lil is predominantly covalent but due to the low polarising power of relatively large K⁺ ion, KI is predominantly ionic and for this reason, Lil is more soluble in the organic solvent ethanol;

Question 41. How can fused calcium chloride be prepared? Give two important uses of it.
Answer:

When CaCl₂ 2H₂O is heated above 533K, anhydrous CaCl₂ forms. This melts at 1046K. When the molten salt is cooled, it solidifies as white lumps of crystalline mass which is known as fused calcium chloride;

Question 42. Hydrated magnesium chloride is heated in the presence of ammonium chloride (NH₄Cl).
Answer:

Magnesium chloride when heated with NH₄Cl forms an additional compound (MgCl₂-NH₄Cl₆H₂O) which on heating forms anhydrous MgCl₂

Question 43. Why are alkali metals not found in nature?
Answer:

Alkali metals are highly electropositive and extremely reactive elements. Thus, they easily react with atmospheric oxygen and carbon dioxide. These metals have a high tendency to lose electrons to form cations because of their low ionization potential. For this reason, they readily react with highly electronegative elements or radicals to form compounds. So, alkali metals are not found in a free state in nature.

Question 44. Explain why is sodium less reactive than potassium.
Answer:

Due to its small size, the ionization enthalpy of sodium (496kL-mol^-1) is greater than potassium (419kJ. mol^-1) and the standard electrode potential value of potassium (-2.925V) is more negative than that of sodium (-2.714V). Thus, sodium is less reactive than potassium.

Question 45. Basicity of oxides
Answer:.

Basicity of oxides: The ionization enthalpy of alkali metals is less than that of the corresponding alkaline earth metals, i.e., the electropositive character of alkali metals is greater than that of alkaline earth metals. Thus, the basicity of oxides of alkali metals is more than the oxides of alkaline earth.

S-Block Elements Class 11 Short Answer Solutions

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Warm-Up Exercise Question And Answers

Question 1. Name the lightest and the heaviest metal.
Answer:   Lithium is the lightest (density: 0.53g-cm^-3) and osmium is the heaviest (density: 22.6g-cm^-3 ) metal.

Question 2. Which one between water and pyrene (CCl₄), can be used to extinguish fire caused by metallic sodium?
Answer:  Pyrene (CCl₄) can be used.

Question 3. Which among the alkali metal ions has the lowest mobility In aqueous solution?
Answer: Li⁺ ions have the lowest mobility because they remain highly hydrated in aqueous solution.

Question 4. Which of the alkali metal cations has the highest polarising power?
Answer: Due to its very small size, Li⁺ ion has the highest polarising power among the alkali metal ions

NCERT Class 11 Chemistry S-Block Elements Short Answer Solutions

Question 5. The density of alkali metals is very low Why?
Answer: Due to large atomic size and weak metallic bonding, the densities of alkali metals are very low.

Question 6. Name a radioactive alkali metal and write its atomic number.
Answer:  Francium, atomic number =87

Question 7. Mention the similarity shown in the electronic configurations of the alkali metals.
Answer: All alkali metals have similar valence shell electronic configuration of ns¹

Question 8. Why alkali metals are called s -s-block elements?
Answer: Alkali metals are called s -s-block elements because the last electron enters the ns-orbital

Question 9. What is trona?
Answer:

Trona ( Na₂ CO₃ , NaHCO₃  .  2H₂O) is an important mineral of sodium

Question 10. Arrange lithium, sodium, and potassium according to their abundance in the earth’s crust.
Answer:

According to their abundance in nature, the elements are arranged as lithium < potassium < sodium.

S-Block Elements Chapter 10 Short Answer Questions Class 11

Question 11. Give an example of a double salt formed by an alkali metal and alkaline earth metal.
Answer:

The double salt formed by an alkali metal and an alkaline earth metal is carnallite (KCl-MgCl₂-6H₂O)

Question 12. Arrange LiF, NaF, KF, RbF, and CsF in increasing order of their lattice energies.
Answer:

The increasing order of lattice energies of the given compounds is CsF < RbF < KF < NaF < LiF.

Question 13. Why are alkali metals paramagnetic?
Answer: Due to the presence of unpaired electrons in the valence shell of alkali metals, they are paramagnetic.

Question 14. Which alkali metal is generally used in photoelectric cells?
Answer: Cesium is generally used in photoelectric cells

Question 15. Which alkali metals form superoxides when heated in excess air?
Answer: The alkali metals that form superoxides when they are heated in excess air are potassium (K), rubidium (Rb), and cesium (Cs).

Question 16. Differentiate between Na₂CO₃ and NaHCO₃
Answer:

When NaHCO₃ is heated, it liberates CO₂ which turns lime-water milky. On the other hand, when Na₂CO₃ is heated, it does not undergo decomposition

Question 17. Arrange MCI, MBr, MF, and MI (where M= alkali metal) according to increasing covalent character.
Answer:

The covalent character of metallic chlorides increases with an increase in the size of the anion. Therefore, the order of the metallic chlorides according to increasing covalent character is MF < MCI < MBr < MI.

Question 18. Explain why the peroxides and superoxides of the alkali metals act as strong oxidizing agents.
Answer:

In reaction with water, peroxides produce MOH along with H₂O₂ and superoxides produce MOH and O, along with H₂O₂. H₂O₂ is a strong oxidizing agent. Thus, peroxides and superoxides of the alkali metals act as strong oxidizing agents.

M₂O₂ + 2H₂O → 2MOH + H₂O₂

2MO₂ + 2H₂O→ 2MOH + H₂O₂ + O

Question 19. Give a simple test to distinguish between KNO₃ & LiNO₃.
Answer:

Colourless O₂ gas evolves on heating KNO₃. But heating, LiNO₃ dissociates into colorless gas and brown NO₂ gas.

Question 20. Explain why a solution of Na₂CO₃ is alkaline in nature whereas a solution of Na₂SO₄ is neutral.
Answer:

For Na₂CO₃, Na₂SO₄ is a salt of strong base (NaOH) and strong acid (H₂SO₄). So, the nature of the solution of Na₂SO₄ is neutral.

Question 21. Among the sulfate salts of lithium, sodium, potassium, and rubidium, which salt does not form double salt?
Answer: Lithium sulfate (Li₂, SO₄) does not form any double salt

Question 22. What happens when sodium sulfate is fused with charcoal? Give equation.
Answer:

When sodium sulfate is fused with charcoal, it reduces to sodium sulfide, and carbonÿis oxidized to carbon monoxide:

Na₂SO₄ + 4→ Na₂S + 4CO↑

Question 23. Which alkali metal bicarbonate has no existence in the solid state?
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Question 24. Mention the property for which lithium is used to separate N2 gas from a gas mixture.
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Class 11 Chemistry S-Block Elements SAQ

Question 25. Give a simple test to distinguish between Li₂CO₃ & Na₂CO₃. 
Answer:

On heating, Li₂CO₃ decomposes to CO₂ which turns lime water milky. On the other hand, Na₂ CO₃ does not decompose on heating

Question 26. Write down the name of the alkali metal compound which i

1. Effective in the treatment of manic depressive psychosis
2. Used in baking powder

Answer:

1. Li₂CO₃
2. NaHCO₃

Question 27. Why a standard solution of sodium hydroxide (NaOH) cannot be prepared?
Answer:

Sodium hydroxide, being a hygroscopic substance absorbs moisture from the atmosphere. It also absorbs CO₂ from the air and forms Na₂CO₃. Thus, sodium hydroxide cannot be accurately weighed and so a standard solution of NaOH cannot be prepared.

Question 28. How the group-2 elements are commonly known? Why are they so called?
Answer:

Group-2 elements are commonly known as ‘alkaline earth metals’. These are so-called because their oxides are basic and are found in the earth’s crust.

Question 29. Why group-2 elements are called s -s-block elements?
Answer: Group-2 elements are called s -s-block elements because the last electron enters the s -s-orbital

Question 30. Which group-2 element has a slightly different electronic configuration than the rest of the elements?
Answer:  Beryllium has 2 electrons in its penultimate shell while the rest have 8 electrons in their penultimate shells.

Question 31. Explain why the atomic and ionic radii of Mg are less than those of Na and Ca.
Answer:

The electrons of Mg having a higher nuclear charge are more strongly attracted towards the nucleus. Thus, the atomic and ionic radii of Mg are less than Na. Again on moving down the group (from Mg to Ca), the atomic and ionic radii increase due to the addition of new shells, and the increasing screening effect of joindy overcomes the effect of increasing nuclear charge. Thus, the atomic and ionic radii of Mg are less than Ca.

Question 32. Arrange Mg²⁺, Ba²⁺, Sr²⁺, Be²^+, and Cav according to decreasing order of their hydration enthalpies. Explain your answer.
Answer: The correct order is  → Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

S-Block Elements Chapter 10 NCERT Short Answer Questions

Question 33. Alkaline earth metals predominantly form ionic compounds. However, the first member of the group forms covalent compounds—Explain why.
Answer:

Due to relatively high electronegativity, the first member of each of these groups tends to form covalent compounds.

Question 34. A white residue is obtained when metallic Mg is burnt in air. This residue when heated with water emits an ammoniacal smell. Explain these observations.
Answer:

On burning metallic magnesium in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are formed. So, the white residue obtained is a mixture of MgO & Mg₃N₂.

This residue when heated with water, results in the hydrolysis of Mg₃N₂ which emits an ammoniacal smell.

Question 35. Why calcium is better than sodium in eliminating a small amount of water from alcohol?
Answer:

Both Na and Ca react with water to form the corresponding hydroxide. However, sodium rapidly reacts with alcohol to form sodium ethoxide (NaOC₂H₅) but calcium reacts quite slowly with alcohol

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑

Thus, for eliminating a small amount of water from alcohol, calcium is better than sodium.

Question 36. Why Mg-ribbon continues to burn in the presence of SO₂ gas
Answer:

During the combustion of Mg-ribbon, the amount of heat generated leads to the decomposition of sulfur dioxide into sulfur and oxygen. This oxygen is responsible for the continued burning of Mg-ribbon.

2Mg + SO₂→ 2MgO + S

Question 37. Except Be(OH)₂, all other alkaline earth metal hydroxides are basic and their basic strength increases down the group.
Answer:

Be has high ionization enthalpy, for which Be(OH)₂ is amphoteric.

Question 38. Why is Mg(OH)₂ less basic than NaOH?
Answer:

Due to the larger ionic size and lower ionization enthalpy of Na, the Na — OH bond in NaOH is weaker than the Mg — OH bondin Mg(OH)₂. Thus, NaOH is more basic than Mg(OH)₂

Question 39. Sparingly soluble carbonate salts of alkaline earth metals become easily soluble in water in the presence of CO₂. Why?
Answer:

Sparingly soluble carbonate salts of alkaline earth metals are converted into soluble bicarbonate salts in the presence of CO₂ So, these salts become easily soluble in water.

CaCO₃(s) + CO₂(g) + H₂O(l)→ Ca(HCO₃)₂ (aq)

Short Answer Solutions for S-Block Elements Class 11 Chemistry

Question 40. The anhydrous chloride salt of which alkaline earth metal is used as a drying agent in the laboratory?
Answer:

Anhydrous chloride salt of calcium metal, i.e., calcium chloride (CaCl₂) is used as a drying agent in the laboratory.

Question 41. Why BeCO₃ is kept in an atmosphere of CO₂?
Answer:

BeCO₃ being highly unstable easily decomposes to give off CO₂ when kept in an open atmosphere.

Question 42. Write down some important points of difference between beryllium and magnesium.
Answer:

• Beryllium is harder than magnesium,
• Compounds of Be are largely covalent, whereas most of the compounds of Mg are ionic,
• Be does not exhibit a coordination number of more than 4, while Mg exhibits a coordination number of 6

Question 43. Be usually forms covalent compounds but other elements of group ionic compounds. Why?
Answer:

Be usually forms covalent compounds due to its high ionisation enthalpy and small size. However, due to comparatively low ionization enthalpy and large size, other elements of group-2 form ionic compounds

Question 44. Which compounds of the alkaline earth metals are used as refractory substances?
Answer:

The oxides of alkaline earth metals (MO) have high melting points and so are used as refractory substances.

Question 45. Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris. Explain.
Answer:

Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris because carbon will reduce CaSO₄ to CaS.

CaSO₄ + 4C →  CaS + 4CO ↑

Question 46. The fire caused by sodium in the laboratory cannot be extinguished by spraying water Why?
Answer:

Sodium reacts vigorously with water producing H₂ gas which catches fire by the heat evolved in the reaction. So water cannot be used for extinguishing sodium-fire

Question 47. Why does Li not exist with Na or K in their minerals?
Answer:

Lithium forms independent minerals and does not exist with Na or K because the Li⁺ ion is too small to replace the more abundant Na+ or K⁺ ions in their minerals.

Question 48. How can you prepare propyne from magnesium carbide?
Answer:

When magnesium carbide (MgC₂) is heated, Mg₂C₃ (magnesium allylide) is obtained. This on hydrolysis, produces propyne (CH₃C=CH)

Question 49. Why do halides of Be dissolve in organic solvents while those of Ba do not?
Answer:

Halides of Be are covalent. Hence, they dissolve in organic solvents while Ba is ionic. Hence, they do not dissolve in organic solvents.

Question 50. Write with a balanced equation the reaction for the manufacture of sodium bicarbonate from sodium carbonate.
Answer:

Sodium bicarbonate (NaHCO₃) Is manufactured by passing CO2 through a saturated solution of sodium carbonate (Na₂CO₃).

Na₂CO₃ + CO₂ + H₂O ⇌ 2NaHCO₃↓

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Short Answer Questions

Question 51. Write the balanced equation(s) for the reaction when excess carbon dioxide is passed through brine saturated with ammonia
Answer:

When carbon dioxide is passed through an aqueous solution of NaCI (brine, 28% NaCl solution) saturated with ammonia, sodium bicarbonate is formed.

Question 52. Why is LiCl soluble in organic solvents?
Answer:

As the polarising power of Li⁺ is very high, LiCl is covalent. Hence, it is soluble in an organic solvent.

Question 53. Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods.
Answer:

Alkali and alkaline earth metals are strong, reducing agents and it is difficult to reduce their oxides or chlorides by any other reducing agent.

Question 54. With the help of a drop of an indicator solution, how would you know whether a solution consists of Na₂CO₃ or NaHCO₃?
Answer:

A drop of phenolphthalein will change the colorless solution of Na₂CO₃ to purple but the colorless solution of NaHCO₃ will remain unchanged;

Question 55. The crystalline salts of alkaline earth metals contain more water of crystallization than the corresponding alkali metal salts—why?
Answer:

Due to their smaller size and higher charge, the alkaline earth metal ions have a greater tendency to coordinate with water molecules as compared to alkali metal ions.

Question 56. Beryllium chloride hydrate loses no water over P₄O₁₀ — why?
Answer:

Due to very small size and stronger hydrating tendency of Be²⁺ ion, it is not possible for P₄O₁₀ to abstract water molecules from beryllium chloride hydrate, [Be(H₂O)4]Cl₂;

Question 57. Why are fumes seen when barium halides are kepi In open air?
Answer:

In the open air, barium halide undergoes hydrolysis by water vapor (moisture), and fumes of halogen acid (except HF) evolve.

Question 58. Explain why the compounds of beryllium are much more covalent than the other Gr-2 metal compounds.
Answer:

Due to the very small size and high charge of Be²⁺ ion, it has much higher polarising power and because of this, the compounds of Be are much more covalent than the other group-2 metal compounds.

Question 59. Beryllium compounds are extremely toxic why?
Answer:

Beryllium compounds are extremely toxic because of their very high solubility and their ability to form complexes with enzymes in the body.  Also, beryllium displaces magnesium from some enzymes.

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Short Answer Questions

Question 60. Which calcium salt causes the formation of the kidney? Stones?
Answer:

Calcium oxalate, \(\mathrm{Ca}^2+\stackrel{\ominus}{\mathrm{O}}_2 \mathrm{C}-\mathrm{C} \stackrel{\ominus}{\mathrm{O}}_2\) causes formation of kidney stones

Question 61. Alkali metals are good reducing agents—Why?
Answer: 

The smaller the ionization enthalpy, the greater the reducing strength. Since alkali metals have lower ionization enthalpies, they are good reducing agents.

Question 62. Explain why the alkali metals cannot be obtained by the reduction method.
Answer: Alkali metals are strong reducing agents and it is difficult to reduce their oxide with any other reducing agent

Question 63. Which alkali metal ion has the maximum polarising j power and why?
Answer:  Li⁺ ion has the highest polarising power among all the alkali metal ions as the value of charge to size ratio of the smallest Li+ ion is the highest.

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Multiple Choice Questions

Question 1. NO₂ is not obtained on heating—

1. AgNO₃
2. KNO₃
3. Cu(NO₃)₂
4. Pb(NO₃)₂

Answer: 2. KNO₃

KNO₃ on heating decomposes to form potassium nitrite (KNO₂) and oxygen gas

Question 2. Which one of the following has the lowest ionization

1. ls²2s²2p⁶
2. ls²s²2p⁶3s¹
3. ls²2s²2p⁵
4. ls²2s²2p³

Answer: 2. ls²s²2p⁶3s¹

Since the electronic configuration ls²s²2p⁶3s¹ is that of an alkali metal, its ionization potential value is the lowest.

Question 3. Which of the following represents the composition of carnallite mineral—

1. K₃O-Al₂O₃.6SiO₂
2. KNO₃
3. K₂SO4.MgSO₄-MgCl₂-6H₂O
4. KCl-MgCl₂-6H₂O

Answer: 4. KCl-MgCl₂-6H₂O

Carnallite:  KCl-MgCl₂-6H₂O

Question 4. Chlorine gas reacts with red-hot calcium oxide to give—

1. Bleaching powder and dichlorine monoxide
2. Bleaching powder and water
3. Calcium chloride and chlorine dioxide
4. Calcium chloride and oxygen

Answer: 4. Calcium chloride and oxygen

2CaO + 2Cl₂ → 2CaCl₂ + O₂

S-Block Elements MCQs Class 11

Question 5. The decreasing order of the basic characters of K₂O, BaO, CaO, and MgO is

1. K₂O > BaO > CaO > MgO
2. K₂O > CaO > BaO > MgO
3. MgO > BaO > CaO > K₂O
4. MgO>CaO>BaO>K₂O

Answer:   4. MgO>CaO>BaO>K₂O

The oxides of alkali metals are highly basic. The basic character of the oxides of alkaline earth metals increases on moving down the group.

Question 6. Match the flame colors of the alkaline earth metal salts in the Bunsen burner.

Answer: 4. 1 -A, 2-B,3-C

1. 1 -A, 2-C,3-B
2. 1 -C, 2-A,3-B
3. 1 -B, 2-C,3-A
4. 1 -A, 2-B,3-C

Question 7. The correct order of solubility in water is

1. CaSO₄>BaSO₄>BeSO₄>MgSO₄>SrSO₄
2. BeSO₄ > MgSO₄> CaSO₄> SrSO₄ > BaSO₄
3. BaSO₄>SrSO₄>CaSO₄>MgSO₄>BeSO₄
4. BeSO₄> CaSO₄ > MgSO₄ > SrSO₄> BaSO₄

Answer: 2. BeSO₄ > MgSO₄> CaSO₄> SrSO₄ > BaSO₄

Question 8. When BaCl₂ is added to an aqueous solution, a white precipitate is obtained. The anion among CO^2-₃ SO^2-₃ and SO^2-₄ that was present in the solution can be

1. CO^2-₃ But any of the other two
2. SO^2-₃ But not any of the other two
3. SO^2-₄ But not any of the other two
4. Any of them

Answer: 4. Any of them

BaSO₃, BaCO_3, and BaSO₄ are insoluble in water, thus they are precipitated out in an aqueous solution.

Question 9. Which of the following is least thermally stable

1. MgCO₃
2. CaCO₃
3. SrCO₃
4. BeCO₃

Answer: 4. BeCO₃

Question 10. Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides

1. MgO < K₂O < Al₂O₃ < Na₂O
2. Na₂O<K₂O<MgO<Al₂O₃
3. K₂O<Na₂O<Al₂O₃<MgO
4. Al₂O₃ < MgO < Na₂O < K₂ O

Answer: 4.Al₂O₃ < MgO < Na₂O < K₂ O

The basicity of the metallic oxides increases with the increase in electronegativity values of the metals.

Question 11. Which of the following on thermal decomposition yields a basic as well as an acidic oxide

1. KClO
2. CaCO₃
3. NH₄NO₃
4. NaNO₃

Answer: 2.CaCO₃

CaCO₃ at high temperature decomposes to form CaO and CO₂

Question 12. Which one of the following alkaline earth metal sulfates has its hydration enthalpy greater than its lattice

1. BaSO₄
2. SrSO₄
3. CaSO₄
4. BeSO₄

Answer: 4. BeSO₄

The hydration enthalpy of BeSO₄ is greater than Its lattice energy because of the very small size of Be ²⁺

Question 13. The main oxides formed on combustion of, Na and Kin excess of air respectively are

1. LiO₂, Na₂O and KOH
2. LiO₂, Na₂O₂ and K₂O
3. Li₂O₂, Na₂O₂ and KO
4. Li₂O₂, Na₂O₂ and KO₂

Answer: 4. Li₂O₂, Na₂O₂ and KO₂

4Li + O₂ →2Li₂O

2Na + O→Na₂O₂

K + O→KO

Question 14. Although lithium and magnesium resemble each other in properties due to a diagonal relationship, the following statement is not correct

1. Both of them form a nitride compound
2. When nitrates of both Li and Mg are heated, NO₂ and O₂ are obtained
3. Both of them form basic carbonate salt
4. Both of them form soluble bicarbonate salt

Answer: 3. Both of them form basic carbonate salt

Magnesium forms basic bicarbonate salt [3MgCO₃.Mg(OH)₂ .3H₂O] whereas lithium forms carbonate salt [Li₂CO₃]. I cannot form any basic bicarbonates.

Question 15. Match Column – A Column – B for the compositions of substances and select the correct answer using the code given below

1. 1 -C, 2-D, 3-  A, 4- B
2. 1 -B, 2-C, 3-  D, 4- A
3. 1 -A, 2-B, 3-  C, 4- E
4. 1 -D, 2-C, 3-  A, 4- B

Answer: 2. 1 -B, 2-C, 3-  D, 4- A

Plaster of Paris: CaSO₄– ½H₂O

Epsomite: MgSO₄-7H₂O

Kieserite: MgSO₄-H₂O

Gypsum: CaSO₄-2H₂O

Question 16. Which one of the following is present as an active ingredient in bleaching powder for bleaching action

1. CaOCl₂
2. Ca(OCl)₂
3. CaO₂Cl
4. CaCl₂

Answer: 2.Ca(OCl)₂

Bleaching powder is – a mixture of calcium chlorohypochlorite [Ca(OCl)₂] and basic calcium chloride [CaCl₂-Ca(OH)₂-2H₂O]. The compound present as an active ingredient in bleaching powder for bleaching action is Ca(OCl)₂

Question 17. Which of the following has the lowest melting point

1. CaCl₂
2. CaBr₂
3. Cal₂
4. CaF₂

Answer: 3.Cal₂

With the increase in the size of the halogens (from toI), the covalent character of the corresponding compounds (from CaF₂ to Cal₂) increases. Therefore, the melting points of the compounds decrease.

NCERT Solutions Class 11 Chemistry Chapter 10 S-Block Elements MCQs

Question 18. In the case of replacement reaction, the reaction will be most favorable if M happens to be

1. K
2. Rb
3. Li
4. Na

Answer: 2. Rb

For a particular alkyl halide, the reactivity of the alkali metal fluorides increases gradually from Li to Cs because their ionization enthalpies decrease and their electronegativities increase. Therefore, for the given reaction, if M = Rb, then the reaction becomes most favorable.

Question 19. Which one of the alkali metals forms only the normal oxide M₂O on heating in air

1. Li
2. Na
3. Rb
4. K

Answer: 1. Li

Question 20. The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins follows the order

1. K⁺ < Na+ ⁺ Rb⁺ < Li⁺
2. Na⁺ < Li⁺ < K⁺ < Rb⁺
3. Li⁺ < K⁺ < Na⁺ < Rb⁺
4. Rb⁺ < K+< Na+< Li⁺

Answer: 4. Li⁺ < K⁺ < Na⁺ < Rb⁺ 

The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins decreases with the increase in the size of the alkali metal. The correct order of the sizes of the alkali metals is Li⁺ < Na⁺ < K⁺ < Rb⁺ .4.

The ease of adsorption of the hydrated alkali metal ions on the ion-exchange resins decreases with the increase in the size of the alkali metal. The correct order of the sizes of the alkali metals is: Li⁺ < Na⁺ < K⁺ < Rb⁺

Question 21. Be²⁺is isoelectronic with which of the following ions

1. Na⁺
2. Mg²⁺
3. H⁺
4. Li⁺

Answer: 4. Li⁺

No. ofelectrons in Be = 4

.-. No. of electrons in Be²⁺ = 2

Again, no. of electrons in Li = 3

No. ofelectrons in Li+ = 2

Therefore, Be²⁺ is isoelectronic with Li⁺

Question 22. On heating which of the following releases CO₂ most easily

1. K₂CO₃
2. Na₂CO₃
3. MgCO₃
4. CaCO₃

Answer: 3.

The thermal stability of the given compounds follows the order: of K₂CO₃ > Na₂CO₂ > CaCO₂ > MgCO₂.

Thus, MgCO₃ Can release CO₂ most easily on heating.

Question 23. In contrast with beryllium, which one of the following statements is incorrect

1. It forms Be₂C
2. Its salts rarely hydrolyzed
3. Its hydride is electron deficient and polymeric
4. It is rendered passive by nitric acid

Answer: 2. Its salts rarely hydrolyzed

Beryllium forms covalent compounds, thus hydrolysis of beryllium compounds occurs readily.

Question 24. The suspension of slaked lime in water is

1. Quicklime
2. Milk of lime
3. Aqueous solution of slaked lime
4. Lime water

Answer: 2. Milk of lime

Suspension of slaked lime Is known as ‘milk ofIimei

Question 25. Which of the following statements is false

1. Mg ²⁺ ions form a complex with ATP
2. Ca²⁺ ions are important in blood clotting
3. Ca²⁺ ions are not important in maintaining the regular beating of the heart
4. Mg²⁺ ions are important in the green parts of plants

Answer: 3. Ca²⁺ ions are not important in maintaining the regular beating of the heart

Ca²⁺ plays an important role in the regular beating of the heart

Question 26. The product obtained as a result of a reaction of nitrogen

1. Ca(CN)₂
2. CaCN₃
3. Ca₂CN
4. CaCN

Answer: The given options are not correct

Question 27. Which one of the following takes up CO₂ and releases

1. K₂O
2. CaO
3. KO₂
4. KOH

Answer: 3.KO₂

2KO₂ + CO₂ → K₂ CO₃ + \(\frac{3}{2}\) O₂

Question 28. Ionic mobility of which of the following alkali metal ions is lowest when the aqueous solution of these salts is put under an electric field

1. K
2. Rb
3. Li
4. Na

Answer: 3. Li

Question 29. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is ls²2s²2p³ the simplest formula for this compound is

1. Mg₃X₂
2. Mg₂X₃
3. Mg₂X
4. Mg X₂

Answer: 1. Mg₃X₂

Electronic configuration of X: ls²2s²2p³

Valency of X = 3

Here, the valency of Mg = 2

Thus the compound will be Mg₃X₂

Multiple Choice Questions for Class 11 Chemistry S-Block Elements

Question 30. Which of the following oxides is the most acidic in nature

1. CaO
2. MgO
3. BaO
4. BeO

Answer: 4. BeO

On moving down the group basic character of the gr-2 On moving down the group basic character of the gr-2

BeO < MgO < CaO < SrO < BaO

Thus, BeO is acidic among all the given oxides.

Question 31. Among CaH₂, BeH₂, BaH₂ the order of ionic character

1. BaH₂< BeH₂ < CaH₂
2. BeH₂ < CaH₂ < BaH₂
3. BeH₂<BaH₂<CaH₂
4.  CaH₂ < BeH₂, < BaH₂

Answer: 2.BeH₂ < CaH₂ < BaH₂

Question 32. Which of the following is not hygroscopic

1. CsCl
2. MgCl₂
3. CaCl₂
4. LiCI

Answer: 1. CsCl is not hygroscopic while MgCl,  CaCI_4, and LiCl are hygroscopic

Question 33. Which is the correct order of solubility in water—

1. Ba(OH)₂ <Mg(OH)₂
2. BaCO₃ >CaCO₃
3. CaSO₄ <MgSO₄
4. Ca(OH)2^Mg(OH)₂

Answer: 3. CaSO₄ <MgSO₄

Question 34. Bleaching powder does not contain—

1. CaCl₂
2. Ca(OH)₂
3. Ca(OCl)₂
4. Ca(ClO₃ )₂

Answer: 4.Ca(ClO₃)₂

Bleaching powder is a mixture of calcium hypochlorite, Ca(OCI)_2, and the basic chloride CaCl₂, H₂O with some slaked lime, Ca(OH)₂.

Question 35. Which of the following statements is incorrect—

1. Li⁺ has a minimum degree of hydration
2. The oxidation state of K in KO₂ is +1
3. Na is used to make a Na/Pb alloy
4. MgSO₄ is readily soluble in water

Answer: 1.  Li⁺ has a minimum degree of hydration

The hydration enthalpies of alkali metal ions decrease with an increase in ionic sizes. Hence, the orderis Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺. Therefore Li⁺ has a maximum degree of hydration

Question 36. Which of the following has the highest hydration energy

1. MgCl₂
2. CaCl₂
3. BaCl₂
4. SrCl₂

Answer: 1. MgCl₂

Smaller-sized and highly charged metal ions have higher hydration energy. Therefore, the order of hydration energy will be

Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

Question 37. Which is the correct order of solubility in water

1. Ba(OH)₂< Mg(OH)₂
2. BaCO₃ >CaCO₃
3. CaSO₄ < MgSO₄
4. Ca(OH)₂ ≅ Mg(OH)₂

Answer: 3.CaSO₄ < MgSO₄

Question 38. What is the product of the reaction between the dilute solution of Ba(OH)₂ and H₂O₂ + ClO₂ 

1. HOCl
2. Ba(OCl)₂
3. Ba(ClO₃)₂
4. Ba(ClO₂)₂

Answer: 4.Ba(ClO₂)₂ 

Ba(OH)₂  +H₂ O₂  + 2ClO₂  → Ba(ClO₂ )₂  + 2H₂ O + O₂ 

Question 39. Which of the following compounds transforms baking soda into baking powder

1. KHCO₃
2. NaHCO₃
3. KHC₄H₄O₆
4. KCl

Answer: 3.KHC₄H₄O₃

Question 40. Which of the following pair of compounds does not undergo any chemical change on heating

1. MgCO₃,KHCO₃
2. CS₂CO₃,KNO₂
3. Na₂CO₃,NaNO₃
4. Li₂CO₃, KNO₂

Answer: 2.  CS₂CO₃, KNO₂

S-Block Elements Chapter 10 NCERT MCQs

Question 41. Ca(OH)₂ (anhydrous)  CaCl₂ +B In this reaction, B is

1. CaOCl₂
2. Ca(ClO₃)₂
3. Ca(OH)₂
4. Ca(ClO₂)₂

Answer: 2. Ca(ClO₃)₂

Question 42. The blue color of potassium solution in liquid ammonia is due to the presence of

1. Solvated electron
2. Potassium amide
3. Impurities present in potassium
4. Potassium oxide

Answer: 1. Solvated electron

Question 43. The alkali metal that does not participate in the reaction MI + I₂ → MI₃ is

1. Na
2. K
3. Rb
4. Cs

Answer: 1. Na

Question 44. The lithium compound which is soluble in water is

1. Li₂CO₃
2. LiNO₃
3. LiF
4. Li₃PO₄

Answer: 2. LiNO₃

Question 45. An alkaline earth metal when heated along with nitrogen forms X. On hydrolysis, X forms an insoluble basic compound and a gas which turns CuSO₄ solution deep blue. The metal is

1. Be
2. Ca
3. Mg
4. Ba

Answer: 3. Mg

Question 46. Which of the following compounds exists in only an aqueous solution

1. Li₃PO₄
2. NaHCO₃
3. [Li(NH₃)₄]I
4. LiHCO₃

Answer: 4.LiHCO₃

Question 47. Which of the following acids does not liberate CO₂ on reacting with sodium carbonate

1. Dilute HCl
2. Dilute H₃BO₃
3. Dilute H₃PO₄
4. Dilute H₂PO₄

Answer: 2. Dilute H₃BO₃

Question 48. Hydrated BeCl₂ acts as a/an

1. Lewis base
2. Arrhenius base
3. Arrhenius acid
4. Lewis acid

Answer: 4. Lewis acid

Question 49. Which of the following compounds or pair of compounds is responsible for turning yellow sodium peroxide white in the presence of air

1. Na
2. NaOH and Na₂CO
3. H₂O₂
4. NaOH and H₂O₂

Answer: 2.NaOH and Na₂CO

Question 50. The chemical formula ofmicrocosmic saltis

1. NaHPO₄.2H₂O
2. NH₄₃HPO₄.2H₂O
3. Na(NH₄)HPO₄.4H₂O
4. (NH₄)3HPO₄.4H₂O

Answer: 3.Na(NH₄)HPO₄.4H₂O

Question 51. The hydride which does not form as a result of a direct reaction between the metal and hydrogen is

1. CaH₂
2. MgH₂
3. BeH₂
4. NaH

Answer: 3.BeH₂

Question 52. An amphoteric oxide dissolves in HCl to form a salt. The salt does not impart any characteristic color to the flame but fumes in moist air. The oxide is

1. BaO₂
2. MgO
3. BeO
4. CaO

Answer: 3. BeO

Question 53. The difference in the number of water molecules in gypsum and Plaster of Paris is—

1. 5/2
2. 2
3. 1/2
4. 3/2

Answer: 4. 3/2

Question 54. Compound A imparts a brick-red color to the flame and decomposes on heating to produce oxygen and a brown gas. A is

1. Mg(NO₃)₂
2. Ba(NO₃)2
3. Ca(NO₃)₂
4. Sr(NO₃)₂

Answer: 2. Ba(NO₃)₂

Class 11 Chemistry S-Block Elements MCQs

Question 55. The sodium salt of an unknown anion when heated with MgCl₂ forms a white precipitate. The anion is

1. SO^2-₄
2. CO^2-₃
3. HCO^–₃
4. NO^–₃

Answer: 3.HCO^–₃

Question 56. The salt which is added to table salt to keep it dry and free-flowing

1. KCl
2. Ca(PO₄)₂
3. KI
4. Na₃PO₄

Answer: 2.Ca(PO₄)₂

Question 57. The compound which acts as an oxidizing as well as a reducing agent

1. NaNO₃
2. Na₂O
3. Na₂O₂
4. KNO₃

Answer: 3.

Question 58. The compound which is used to extinguish fire caused by combustion of alkali metals is—

1. CC1₄
2. Sand
3. Water
4. Kerosene

Answer: 1.  CC1₄

Question 59. An aqueous solution of sodium sulfate is electrolyzed using inert electrodes. The products formed at the cathode and anode respectively are

1. O₂,H₂
2. H₂,O₂
3. O₂, Na
4. O₂,SO₂

Answer: 1.O₂, H₂

Question 60. Sodium when heated at 300°C in air forms X which absorbs CO₂ to form Na₂CO₃ and a compound Y. The compound Y is 

1. H₂
2. O₂
3. H₂O₂
4. O₃

Answer: 2.O₂

Question 61. Excess Na⁺ ions in the human body cause

1. Diabetes
2. Anaemia
3. Low blood pressure
4. High blood pressure

Answer: 4. High blood pressure

Question 62. The metal carbide which on hydrolysis produces allylene or propyne is

1. Be
2. Ca
3. Al
4. Mg

Answer: 4. Mg

Question 63. A metal M readily forms water-soluble sulfate MSO₄, water-insoluble hydroxide M(OH)_2, and oxide MO. The oxide remains inert on heating. The hydroxide is soluble in NaOH. M is

1. Be
2. Mg
3. Ca
4. Sr

Answer: 1. Be

Question 64. When compound A is heated, it produces a colorless gas, and the residue obtained is dissolved in water to form compound B. The compound C is formed when excess CO₂ is passed through the aqueous solution of B. The compound C can be separated in a solid state from its solution. C in its solid state when heated forms the compound A. Ais

1. CaCO₃
2. Na₂CO₃
3. K₂CO₃
4. CaSO₄.2H₂O

Answer: 1.CaCO₃

Question 65. When metals A and B are heated in air, A only forms oxide but B forms both oxide and nitride. A and B are

1. Cs, K
2. Mg, Ca
3. Li, Na
4. K, Mg

Answer: 4. K, Mg

Question 66. The alkali metal which emits a light of the longest wavelength in the flame testis

1. Na
2. K
3. Cs
4. Li

Answer: 2. K

NCERT Class 11 Chemistry S-Block Elements Multiple Choice Questions

Question 67. The compound which does not form a double salt is

1. Li₂SO₄
2. Na₂SO₄
3. K₂SO₄
4. Rb₂SO₄

Answer: 1. Li₂SO₄

Question 68. The decreasing order of stability of the chloride salts of alkali metals is

1. LiCl > KCl > NaCl > CsCl
2. CsCl > KCl > NaCl > LiCl
3. NaCl > KCl > LiCl > CsCl
4. KCl > CsCl > NaCl > LiCl

Answer: 4. KCl > CsCl > NaCl > LiCl

Question 69. The affinity of sodium for water is used for drying

1. Alcohol
2. Ammonia
3. Benzene
4. Phenol

Answer: 3. Benzene

Question 70. The ions present in an anhydrous mixture of potassium fluoride and hydrofluoric acid are

1. K⁺,H⁺ F^–
2. (KF)⁺+ (HF)^–
3. KH⁺, F^–
4. K⁺,HF₂^–

Answer: 4. K⁺,HF₂^–

Question 71. The correct order of covalent character in the following compounds is

1. LiCl < NaCl < BeCl₂
2. BeCl₂ > LiCl > NaCl
3. NaCl < LiCl < BeCl₂
4. BeCl₂ < NaCl < LiCl

Answer: 3. NaCl < LiCl < BeCl₂

Question 72. A chemical compound A is used for the recovery of ammonia during the preparation of washing soda. When CO₂ is passed through the aqueous solution of A, the solution turns turbid. A is used for whitewashing because of its disinfecting properties. The chemical formula of A is

1. Ca(HCO₃)₂
2. CaO
3. Ca(OH)₂
4. CaCO₃

Answer: 3. Ca(OH)₂

Question 73. The compound used for drying neutral or basic gases is—

1. Calcium carbonate
2. Sodium carbonate
3. Sodium bicarbonate
4. Calcium oxide

Answer: 4. Calcium oxide

Question 74. Which of the following compounds does not contain calcium carbonate

1. Dolomite
2. Marble statue
3. Burnt gypsum
4. Snail shell

Answer: 3. Burnt gypsum

Question 75. The compound which on thermal decomposition produces a basic and an acidic oxide is

1. KClO₃
2. Na₂CO₃
3. NaNO₃
4. CaCO₃

Answer: 4.CaCO₃

Question 76. Which of the following oxides does not react with water

1. BeO
2. CaO
3. MgO
4. SrO

Answer: 1. BeO

Question 77. Which of the following carbonates is soluble in water

1. SrCO₃
2. BaCO₃
3. Al₂(CO₃)₃
4. Rb₃CO₃

Answer: 3.Al₂(CO₃)₃

Question 78. The compound whose aqueous solution is called ‘baryta water’ is

1. BaSO₄
2. BaO
3. BaCO₃
4. Ba(OH)₂

Answer: 4. Ba(OH)₂

Question 79. The alkali metal that emits a light of the shortest wavelength in the flame test is

1. Na
2. K
3. Cs
4. Li

Answer: 3. Cs

Question 80. The correct order of ionic mobility of the following ions in their aqueous solution is

1. Na⁺> K⁺> Rb⁺> Cs⁺
2. K⁺> Na⁺ > Rb⁺> Cs⁺
3. Cs⁺ > Rb⁺ > K⁺ > Na⁺
4. Rb⁺ > K⁺ > Cs⁺ > Na⁺

Answer: 3. Cs⁺ > Rb⁺ > K⁺ > Na⁺

Question 81. Which of the following compounds is not used for storing or immersing metallic sodium

1. Benzene
2. Kerosene
3. Ethanol
4. Toluene

Answer: 3. Ethanol

Question 82. Which of the following compounds is paramagnetic

1. KO₂
2. SiO₂
3. TiO₂
4. BaO₂

Answer: 1.KO₂

MCQs on S-Block Elements Class 11 Chemistry

Question 83. KO₂ is used in oxygen cylinders that are used for submarines and spacecraft because—

1. It increases the amount of oxygen by absorbing CO₂
2. It eliminates water vapor
3. It absorbs CO₂
4. It forms O₃

Answer: 1. It increases the amount of oxygen by absorbing CO₂

Question 84. Which of the following compounds is the most stable

1. LiF
2. LiCl
3. LiBr
4. Lil

Answer: 1. LiF

Question 85. The alkali metal for which the photoelectric effect is maximum is

1. Cs
2. Na
3. K
4. Li

Answer: 1. Cs

Question 86. The melting points of alkali metals are low. Which of the alkali metals melts when room temperature becomes more than 30 °C 

1. K
2. Na
3. Cs
4. Rb

Answer: 3. Cs

Question 87. Which of the following alkali metal hydroxides is the most basic

1. CsOH
2. KOH
3. Li OH
4. RbOH

Answer: 1.CsOH

Question 88. The metal whose carbonate salt is the most stable is

1. Na
2. Mg
3. A1
4. Si

Answer: 1. Na

Question 89. Which of the following compounds is used for manufacturing soft soaps

1. KOH
2. NaOH
3. LiOH
4. Mg(OH)₂

Answer: 1.

Question 90. The atomic number of a radioactive alkali metal is

1. 55
2. 87
3. 19
4. 37

Answer: 2. 87

Question 91. Which of the following is used for the manufacture of high? temperature thermometers

1. An alloy of Li and Na
2. An alloy of Na and Cs
3. An alloy of Na and K
4. An alloy of K and Rb

Answer: 3. An alloy of Na and K

Question 92. The mixture of MgCl₂ and MgO is known as

1. Sorel’s cement
2. Portland cement
3. Alum
4. Magnesium oxychloride

Answer: 1. Sorel’s cement

Question 93. Which of the following alkaline earth metals does not form its corresponding hydride by directly reacting with hydrogen

1. Mg
2. Sr
3. Be
4. Ba

Answer: 3. Be

Question 94. Which of the following alkaline earth metal hydroxides is soluble in NaOH solution

1. Ba(OH)₂
2. Ca(OH)₂
3. Mg(OH)₂
4. Be(OH)₂

Answer: 4. Be(OH)₂

Question 95. The chloride salt which is soluble in ethanol is

1. BeCl₂
2. CaCl₂
3. SrCl₂
4. MgCl₂

Answer: 1.BeCl₂

S-Block Elements Chapter 10 MCQs Class 11

Question 96. Which of the following is produced when one mole of magnesium nitride reacts with an excess of water

1. One mole of ammonia
2. One mole of nitric acid
3. Two moles of ammonia
4. Two moles of nitric acid

Answer: 3. Two moles of ammonia

Question 97. Which of the following methods is used for the preparation of calcium

1. Reduction of CaO by carbon
2. Reduction of CaO by hydrogen
3. Electrolysis of a mixture of anhydrous CaCl₂ and KCl
4. Electrolysis of molten Ca(OH)₂

Answer: 3. Electrolysis of a mixture of anhydrous CaCl₂ and KCl

Question 98. Which alkaline earth metal ion plays a vital role in the contraction of muscles

1. Ba²⁺
2. Sr²⁺
3. Mg²⁺
4. Ca²⁺

Answer: 4. Ca²⁺

Question 99. Which of the following correctly indicates the formula of halides of alkaline earth metals

1. BeCl₂.2H₂O
2. BeCl₂.4H₂O
3. CaCl₂.6H₂O
4. SrCl₂.4H₂O

Answer: 1.BeCl₂.2H₂O

Question 100. The compounds of sodium which are used in the textile Industry are

1. Na₂CO₃
2. NaHCO₃
3. NaOH
4. NaCl

Answer: 1. Na₂CO₃

Question 101. Which of die following pairs of elements have similar properties

1. Be, Cs
2. K, Cs
3. Sr, Rb
4. Be, Al

Answer: 2. K, Cs

Question 102. The chlorides which are soluble in pyridine are

1. LiCl
2. CsCl
3. NaCl
4. BeCl₂

Answer: 1 and 4

Question 103. The gases in which magnesium burns are

1. CO₂
2. N₂O
3. N₂
4. SO₂

Answer: 1,2,3 and 4

Question 104. Which of the following oxides have rock salt structure with coordination number 6:6 

1. MgO
2. CaO
3. SrO
4. B₂O₃

Answer: 1,2 and 3

Question 105. Which of the following pairs of compounds cannot exist in aqueous solution

1. NaH₂PO₄ and Na₂HCO₃
2. Na₂CO₃ and NaHCO₃
3. NaOH and NaH₂ PO₄
4. NaHCO₃ and NaOH

Answer: 3 and 4

Question 106. The compounds which on heating do not form oxides are

1. NaNO₃
2. CsOH
3. LiOH
4. SrCO₃

Answer: 1 and 2

Question 107. Which of the following pairs of elements will give superoxides and peroxides respectively when heated with excess air

1. K, Br
2. Nay Rb
3. K, Rb d
4. Na, Ba

Answer: 3 and 4

Question 108. Which of the following do not respond to flame test

1. Be
2. Mg
3. KO₂
4. Sr

Answer: 1 and 2

Question 109. Which of the following compounds are not paramagnetic in nature

1. K₂O₂
2. NO₂
3. KO₂
4. K₂O

Answer: 1 and 4

Question 110. Which of the following is incorrect

1. Soda ash: Na₂CO₃
2. Pearl ash: Cu₂CO₃
3. Bone ash: K₂CO₃
4. Baking soda: NaHCO₃

Answer: 2 and 3

Question 111. Which ions of water are replaced by sodium ions when hard water is passed through zeolite (hydrated sodium aluminum silicate)

1. H⁺
2. Mg⁺
3. Ca²⁺
4. SOl₄^2-

Answer: 2 and 3

Question 112. The compounds which are soluble in organic solvents are

1. CaCl₂
2. BaCl₂
3. BeCl₂
4. AlCl₃

Answer: 3 and 4

Question 113. The compounds formed when potassium superoxide reacts with water are

1. KOH
2. H₂O₂
3. K₂O₂
4. O₂

Answer: 1, 2 and 4

Question 114. Which of the following hydrated salts undergo hydrolysis on heating

1. BaCl₂.2H₂O
2. MgCl₂.6H₂O
3. SrCl₂.2H₂O
4. CaCl₂.6H₂O

Answer: 2 and 4

Question 115. Which of the following compounds are extensively used as drying agents

1. Anhydrous CaCl₂
2. Mg(ClO₄)2
3. BeC
4. Ca(OH)₂

Answer: 1 and 2

S-Block Elements Multiple Choice Questions for Class 11 Chemistry

Question 116. Each of the following compounds reacts with water but which of these liberates the same gas

1. Na
2. Na₂O₂
3. KO₂
4. NaH

Answer: 1 and 4

Question 117. The polarisability of LiCl is higher than that of NaCl. Concerning this, which of the following statements is true

1. The melting point of LiCl is less than that of NaCl
2. LiCl is sparingly soluble in organic solvents
3. LiCl dissociates to a greater extent in water than NaCl
4. The conductivity of molten LiCl is less than that of NaCl

Answer: 1 and 4

Question 118. Which of the following statements is correct

1. Electronegativity of alkali metals decreases with an increase in atomic number
2. Lithium is the lightest metal
3. Alkali metals are strong reducing agents –
4. The electronegativity of alkali metals ranges from 1.0 to 0.7

Answer: 1, 3, 4

Question 119. Which of the following statements are incorrect about the hydrates of alkali metals

1. Conduct electricity in their molten states
2. These compounds act as oxidizing agents
3. These compounds dissolve in water to liberate
4. These compounds are covalent

Answer: 2 and 4

Question 120. Which of the following statements are correct about the ionic solids KI and CaO lattice enthalpy of CaO is greater than that of K₃

1. I am soluble in benzene
2. The melting point of CaO is high
3. The melting point of KI is high

Answer: 1, 2, 3

Question 121. The compounds which do not form NO₂ on undergoing thermal decomposition are

1. LiNO₃
2. NaNO₃
3. KNO₃
4. RbNO₃

Answer: 2, 3, and 4

Question 122. Metals are identified by their standard reduction potential, enthalpy of fusion, and atomic size. The alkali metals are identified by their

1. High boiling point
2. High negative standard reduction potential
3. High density
4. Greater atomic size

Answer: 2 and 4

Question 123. Which of the following sulfates easily dissolve in water

1. BeSO₄
2. MgSO₄
3. BaSO₄
4. SrSO₄

Answer: 1 and 2

Question 124. The properties of beryllium nitride which are different from the nitrides of other alkaline earth metals are

1. Its volatility
2. Its covalent nature
3. Unable to undergo hydrolysis
4. Its ionic nature

Answer: 1 and 2

Question 125. Which of the following options is correct for RbO₂

1. It is a peroxide
2. It is diamagnetic
3. It is a superoxide
4. It is paramagnetic

Answer: 3 and 4

Question 126. Which of the following statements is correct for the alkaline earth metals

1. Hydration enthalpy of Sr²⁺is less than that of Ba²⁺
2. CaCO₄ undergoes decomposition at a higher temperature than BaCO₃
3. Ba(OH)₂ is a stronger base than Mg(OH)₂
4. SrSO₄  is more soluble in water than CaSO₄

Answer: 2 and 3

Question 127.  Which of the following statements are correct about the alloy formed by sodium and potassium

1. It is used in Lassaigne’s test
2. It is liquid at ordinary temperature
3. It is used in specially designed thermometers
4. It is used as a coolant in nuclear reactors

Answer: 2 and 3

Question 128. Which of the following statements is incorrect

1. BeCl₂ molecule is linear in the gaseous state
2. Calcium hydride is known as hydrolith
3. Carbides of both beryllium and calcium react with water to form acetylene
4. Oxides of both Be and Ca are amphoteric

Answer: 3 and 4

Question 129. Which of the following is less stable thermally

1. LiF
2. KCl
3. RbF
4. CsF

Answer: 2. KCl

The thermal stability of a compound increases with the increasing value of enthalpy of formation. Among the given compounds the value of enthalpy of formation is minimum (-428kJ.mol^-1) for KCl. Hence, it has the lowest thermal stability among the given compounds.

Question 130. Which of the following pairs is responsible for developing an electric potential across the membrane of living cells

1. Ca²⁺and Na⁺
2. Na⁺ and K⁺
3. K⁺ and Ba²⁺
4. Mg²⁺ and Ca²⁺

Answer: 3. K⁺ and Ba²⁺

NCERT Class 11 Chemistry Chapter 10 S-Block Elements MCQs

Question 131. Which one of the following chlorides is a soluble organic solvent

1. CaCl₂
2. NaCl
3. MgCl
4. BeCl₂

Answer: 4.MgCl

Question 132. Which of the following alkaline earth metal carbonates is thermally least stable

1. BaCO₃
2. CaCO₃
3. SrCO₃
4. BeCO₃

Answer: 4.SrCO₃

Question 133. The cause of the different colors of the flame in the flame test is

1. Lowionisation potential
2. Low melting point
3. Malleability
4. The presence of one electron in the outermost orbit

Answer: 1. Lowionisation potential

Question 134. Which of the following alkaline earth metal sulfate is most soluble in water

1. CaSO₄
2. SrSO₄
3. BaSO₄
4. MgSO₄

Answer: 4. BaSO₄

Question 135. What will be the order of reducing powder of the following elements Li, Na,  K, Rb, Cs

1. Cs > Rb > K > Na > Li
2. Rb > Cs > K > Na > Li
3. K > Rb > Cs > Na > Li
4. Na > Li> K >Cs > Rb

Answer: 1. Cs > Rb > K > Na > Li

The order of reducing free states is:

Cs > Rb > K > Na > Li

Question 136. Which one of the following elements shows a diagonal relationship with magnesium

1. Na
2. Li
3. Be
4. Ca

Answer: 2. Li

Question 137. Sodium is preserved in which of the following liquids

1. Water
2. Ethanol
3. Kerosene oil
4. Methanol

Answer: 3. Kerosene oil

Question 138. Which of the alkali metals has having least melting point?

1. Na
2. K
3. Rb
4. Cs

Answer:  4. Cs

As the atomic size of the metal increases, the strength of metallic bonding decreases and consequently, the melting point decreases. Since the size of the cesium atom is the largest, it has the least melting point.

NCERT Class 11 Chemistry Chapter 10 S-Block Elements MCQs

Question 139. Which one of the following alkali metals gives hydrated salts? 

1. Li
2. Na
3. K
4. Cs

Answer: 1.  Li

 Li; Explanation:

Among all the alkali metal ions, Li+ is the smallest and thus it has the highest charge density. As a result, Li⁺ attracts water molecules more strongly than any other alkali metal cation and forms hydrated salts.

Question 140. Which one of the alkaline earth metal carbonates is thermally the most stable? 

1. MgCO₃
2. CaCO₃
3. SrCO₃
4. BaCO₃
   

Answer:  4. BaCO₃

BaCO₃; Explanation:

Among all the alkaline earth metals, the Ba²⁺ ion is the largest. Again, CO₃^2- ion is also quite large. Thus, among the given carbonates, Ba²⁺ and CO₃^2-  ions are most tightly packed in the crystal lattice of BaCO₃. Consequently, the lattice enthalpy of BaCO₃ is the highest and so, is most thermally stable.

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Long Question And Answers

Question 1. How can anhydrous magnesium chloride be prepared from magnesium chloride hexahydrate?
Answer:

Anhydrous magnesium chloride cannot be prepared by simply heating MgCl₂-6H₂O because it gets hydrolyzed by its water of crystallization.

However, when hydrated MgCl₂ is heated at 650K in the presence of HCl, its hydrolysis is prevented; and it loses its water of crystallisation to form anhydrous MgCl₂

Read and learn More NCERT Class 11 Chemistry

However, when hydrated MgCl₂ is heated at 650K in the presence of HCI, its hydrolysis is prevented, and it loses its water of crystallization to form anhydrous MgCl₂.

S-Block Elements Class 11 Long Question and Answers

Question 2. Which out of BeCl₂ and CaCl₂ would give an acidic solution when dissolved in water?
Answer: 

Being a covalent compound and a good Lewis acid, BeCl₂ forms a hydrated salt, Be(H₂O₂)4Cl₂. The hydrated ion undergoes hydrolysis in solution producing H₃O⁺. This occurs because the Be — O bond is very strong and so in the hydrated ion this weakens the O —H bond. Hence, there is a strong tendency to lose protons. For this reason, the aqueous solution of BeCl₂ is acidic.

[Be(H₂O)₄]²⁺ + H₂O ⇌ [Be(H₂O)₃OH]⁺ + H₃O⁺

On the other hand, CaCl₂ is a strongly ionic compound and does not behave as a Lewis acid (the size of Ca is relatively large and its octet is filled up). Moreover, since it is a salt of strong acid and strong base, it does not undergo hydrolysis and therefore, its aqueous solution is neutral.

Question 3 Explain The below Observation

1.

2.

Answer:

1. The smaller Li⁺ ion exerts strong polarising power and distorts the electron cloud of the nearby oxygen atom of the OH^– ion. This results in the formation of a strong Li —O bond and the weakening of the O —H bond. This ultimately facilitates the decomposition of LiOH into Li₂O and H₂O. The polarising power of the large Na+ ion is much lower and thus, NaOH remains unaffected by heating.

2. The smaller Li⁺ ion exerts a strong polarising power on highly polarisable H^– ion and as a result, the two atoms remain strongly attached by a covalent bond. On the other hand, due to the low polarising power of Na⁺ ion, NaH is essentially ionic & so it dissociates on heating to yield metallic sodium and dihydrogen.

Question 4. Discuss the roles of Na₂O₂, KO_2, and LiOH in the purification of air.
Answer:

Sodium peroxide (Na₂O₂) is used to purify the air in submarines and confined spaces as it removes carbon dioxide (CO₂) and produces O₂.

Potassium superoxide (KO₂) is used to purify the air in space capsules, submarines, and breathing masks because it can absorb carbon dioxide (CO₂) thereby removing it and producing O₂ Both functions are important life support systems.

Lithium hydroxide (LiOH) is used for the absorption of carbon dioxide (CO₂) in space capsules and submarines.

Question 5. Lithium forms monoxide while sodium forms) peroxide in the presence of excess oxygen why
Answer:

Larger cations can be stabilized by larger anions because if both the ions are comparable in size, the coordination number will be high and this gives rise to a high lattice enthalpy. Lithium-ion, Li⁺ as well as oxide ion, O^2-, have small ionic radii and high charge densities.

Hence these small ions combine and form a very stable lattice of lithium monoxide (Li₂O). Similarly, the formation of sodium peroxide (Na₂O₂) can be explained based on the stable lattice formed by the packing of relatively large Na⁺ ion and peroxide ion O₂

Question 6. A freshly cut piece of sodium metal appears shiny but its metallic lustre soon gets tarnished when exposed to air. Give reason
Answer:

When a freshly cut piece of metallic sodium is exposed to moist air, it readily reacts with oxygen to form sodium monoxide (Na₂O). The resultant sodium monoxide and also the metal itself readily react with the moisture of the air to form sodium hydroxide (NaOH).

In the subsequent step, both Na₂O and NaOH combine with CO₂ of air to form sodium carbonate (Na₂CO₃). Thus, a coating of sodium carbonate is formed on the surface of the metal and as a result of this, the metallic luster is tarnished.

4Na + O₂→ 2Na₂O; Na₂O + H₂O →2NaOH

2Na + 2H₂O→2NaOH + H₂↑; Na₂O+ CO₂→Na₂CO₃

2NaOH + CO₂ →Na₂CO₃ + H₂O

NCERT Solutions Class 11 Chemistry Chapter 10 Long Question Answers

Question 7. What happens when each of the following compounds is heated? 

1. Li₂CO₃ 
2. Na₂CO₃
3. LiNO₃
4. KNO₃

Answer:

1. Lithium carbonate (Li₂CO₃ ) decomposes readily on heating to give lithium monoxide (Li₂O) and carbon dioxide (CO₂).

2. Na₂CO₃ does not decompose on heating.

3. Lithium nitrate (LiNO₃) decomposes on heating to form lithium monoxide (Li₂O), nitrogen dioxide (NO₂), and dioxygen (O₂).

4.  Potassium nitrate (KNO₃) decomposes on heating to give potassium nitrite (KNO₂) and dioxygen

Question 8. What happens when

1. HCl gas is passed through a concentrated solution of common NaCl-containing impurities like Na₂SO₄, CaSO₄, CaCl₂, MgCl_2, etc.
2. Caustic soda beads are exposed to air for a long time.
3. How will you convert Na₂CO₃ into NaHCO₃ and vice versa?

Answer:

1. When HCl gas is passed through a concentrated solution of common NaCl with impurities, crystals of pure NaCl separate out because of the common ion effect,  Caustic soda (NaOH) is a deliquescent substance and becomes wet on exposure to air.

On long exposure, the solid beads dissolve in the absorbed water Moist caustic soda then absorbs carbon dioxide (CO₂) from the air to form sodium carbonate (Na₂CO₃) which forms a coating over the surface of the material. As sodium carbonate (Na₂CO₃) is not a deliquescent substance the wet sodium hydroxide becomes dry again.

2NaOH + CO₃→ Na₂CO₃+ H₂O

2.  Sodium bicarbonate (NaHCO₃) can be obtained by passing carbon dioxide (CO₂) through a saturated solution of sodium carbonate. Sodium bicarbonate, being less soluble gets separated from the solution as a white crystalline substance.

Na₂ CO₃ + CO₂+ H₂O→2NaHCO₃

When sodium bicarbonate (NaHCO₃) is heated, it decomposes to give sodium carbonate (Na₂CO₃) and carbon dioxide (CO₂)

Question 9. Why are the group-2 metals harder and have higher melting and boiling points than group-1 metals?
Answer:

The magnitude of the cohesive energy determines the hardness as well as melting and boiling points of the metals and it depends on the number of electrons involved in metallic bonding. In the case of group 1 metals, one electron per atom (valence electron, ns² ) is involved in metallic bonding while in group 2 metals, two electrons per atom (valence electrons, ns²) are involved in metallic bonding. Moreover, atoms of group 2 metals are smaller in size than those of group 1 metals.

Consequently, stronger metallic bonding exists in group 2 metals which results in higher cohesive energy and close packing of the atoms. This accounts for the greater hardness and higher melting and boiling points of group 2 metals as compared to group 1 metals.

Question 10.

1. Give some common tests used for the detection of calcium compounds. 

2. A white solid When heated liberates a colorless gas that does not support combustion. The residue is dissolved in water to form (B) which can be used for whitewashing. When excess CO₂ gas is passed through the solution of (B), it gives a compound (C) which on heating forms (A). Identify (A), (B) and (C). Give the reactions
Answer:

The following tests may be performed for the detection of Ca -compounds:

1. Calcium salts give a brick-red color in the flame test,
2. When ammonium oxalate solution is added to a solution of a calcium salt, a white precipitate of crystalline calcium oxalate is obtained which is insoluble in acetic acid but soluble in mineral acids.
3. In addition to a solution of sodium carbonate to a neutral (or ammoniacal) solution of a Ca-salt, white calcium carbonate is precipitated, which is soluble in acids.
4. The observations suggest that the compound (A) is limestone, i.e., CaCO₃, the compound (B) is calcium hydroxide & the compound (C) is calcium bicarbonate.

The corresponding reactions are as:

Long Answer Questions Class 11 Chemistry Chapter 10 S-Block Elements

Question 11. Compare the alkali metals and alkaline earth metals concerning their

1. Basicity of oxides
2. Solubility of hydroxides and
3. Solubility of nitrates.

Answer:

1. Basicity of oxides:

The ionization enthalpy of alkali metals is less than the corresponding alkaline earth metals. So the alkali metal oxides are more basic than the corresponding alkaline earth metal oxides.

2. Solubility of hydroxides:

Due to of small size and higher ionic charge, the lattice enthalpies of alkaline earth metal hydroxides are much higher than those of alkali metal hydroxides and hence, the solubility of alkali metal hydroxides is much higher than that of alkaline earth metal hydroxides.

3. Solubility of nitrates:

Nitrates of alkali and alkaline earth metals are soluble in water. However, the solubility of alkali metal nitrates increases down the group because their lattice enthalpies decrease more rapidly than their hydration enthalpies. Nitrates of alkaline earth metals follow the reverse trend i.e., their solubility decreases down the group and this is because their hydration enthalpies decrease more rapidly than their lattice enthalpies.

Question 12. What are the properties that make oxides of MgO and  BeO useful for lining furnaces?
Answer:

The given properties make MgO and BeO useful for lining furnaces

• They have high melting points [melting point of Beo is 2500°C (approx) and MgO is 2800°C (approx)].
• They are very good conductors of heat.
• They have very low vapor pressure.
• They are chemically inert.
• They are insulators

Chapter 10 S-Block Elements Long Answer Class 11 NCERT Solutions

Question 13. The halides of alkali metals are soluble in water except for LiF. Why?
Answer:

The solubility of a salt in water depends on its lattice enthalpy as well as its hydration enthalpy. A salt dissolves in water when its hydration enthalpy exceeds its lattice enthalpy value. The lattice enthalpy of LiF is very high and its hydration enthalpy value does not exceed this value. As a result, LiF is insoluble in water.

However, the other halides of alkali metals possess higher hydration enthalpy values compared to their corresponding lattice enthalpies. Hence, they are soluble in water.

Question 14. Why Is LiCO₃ decomposed at a lower temperature whereas Na₃ CO₃ at a higher temperature
Answer: 

Li+ ion being smaller in size forms a more stable lattice with the smaller oxide ion (O^2-) than the larger carbonate ion (CO₃ ) and consequently, Li₂CO₃  decomposes into Li₂ O at a much lower temperature. The high polarising power of very small Li⁺ ions also facilitates the decomposition of Li₂CO₃.

On the other hand, the larger Na+ ion forms a more stable lattice with the larger CO₃ ion than with the smaller O^2- ion. Therefore, Na₂ CO₃ is quite stable and decomposes only at very high temperatures.

Question 15. Compare the solubility and thermal stability of the given compounds of the alkali metals with those of the alkaline earth metals. 

1. Nitrates
2. Carbonates Sulphates.

Answer:

1. Nitrates:

The nitrates of alkali metals as well as alkaline earth metals are highly soluble in wOn heating, nitrates of both alkali and alkaline earth metals undergo decomposition.

Nitrates of alkali metals decompose to form metallic nitrite and oxygen. On the other hand, nitrates of alkaline earth metals decompose to form the corresponding oxides with the evolution of NO₂ and O_2.

Due to the diagonal relationship, lithium nitrates behave similarly to magnesium nitrate.

2. Carbonates:

1. Except for Li₂CO₃, other carbonates of alkali metals readily dissolve in water. However, carbonates of alkaline earth metals are practically insoluble in water. Their solubilities decrease on moving down the group. So, BeCO₃ is sparingly soluble in water while BaCO₃ is insoluble in water.

2. Carbonates of alkali metals except Li₂CO₃ are stable and do not decompose on heating, but carbonates of alkaline earth metals decompose on heating to give metal oxide and carbon dioxide.

The thermal stability of the carbonates increases down the group. Like MgCOg, Li₂CO₃ decomposes on heating.

3. Sulphates:

1. Except Li₂SO₄, the remaining sulfates of alkali metals are water-soluble. The sulfates of alkaline earth metals are relatively less soluble in water than the corresponding sulfates of alkali metals. Further, their solubilities decrease down the group,

2. The sulfates of alkali metals are stable compounds and do not decompose on heating. On the other hand, alkaline earth metals dissociate on heating to give metal oxides and sulfur trioxide.

The thermal stability of the sulfates increases down the group. Li₂SO₄ dissociates on heating just like MgSO_4.

S-Block Elements Chapter 10 NCERT Long Answer Solutions

Question 16. Starting with sodium chloride how would you proceed to prepare

1. Sodium metal
2. Sodium peroxide

Answer:.

1. Metallic sodium can be obtained by the electrolysis of a mixture of sodium chloride (40%) and calcium chloride (60 %) in a fused state. The function of calcium chloride is to lower the reaction temperature from 807°C (m.p. of NaCl) to about 577°C. The molten sodium metal thus obtained is liberated at the cathode

Overall reaction: Nacl → Na⁺ + Cl^–

At cathode: Na⁺ + e→ Na;

At anode: Cl^– → Cl + e; Cl + Cl → Cl₂ ↑

2.  Sodium chloride is first converted to sodium by electrolytic reduction. The metal is then heated more than air. The initially formed sodium oxide reacts with excess O₂ to form Na₂O₂.

Question 17. What happens when  

1. Magnesium is burnt in the air 
2. Calcium nitrate is heated?

Answer:

1. When magnesium bums in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are obtained as products

2. On heating calcium nitrate, it decomposes to form CaO, NO_2, and O₂ reaction

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Detailed Answers

Question 18. The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:

Alkali metals form monovalent cations (such as Na⁺, and K⁺) while alkaline earth metals form divalent cations (such as Mg²⁺, and Ca²⁺). Due to the increase in charge of the cations, the lattice energy of the corresponding salt increases. For this reason, hydroxides and carbonates of sodium and potassium have lower lattice enthalpy values than the hydroxides and carbonates of magnesium and calcium.

As hydration enthalpies of the hydroxides and carbonates of sodium and potassium are greater than their lattice enthalpies, these salts readily dissolve in water. However, in the case of the hydroxides and carbonates of calcium and magnesium, the lattice enthalpy values are greater than that of hydration enthalpy and consequently, these salts are less soluble in water.

Question 19. Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:

As Li⁺ is the smallest ion among all the alkali metal ions, it can polarise water molecules more easily than the other alkali metal ions. So, numerous water molecules get attached to lithium salts as water of crystallization. However, this is not observed in the case of other alkali metal ions. Thus, lithium salts are commonly hydrated.

Example: LiCl-2H₂O  and those of the other alkali ions are usually anhydrous.

NCERT Class 11 S-Block Elements Long Q&A

Question 20. Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone?
Answer:

The lattice enthalpy of ionic LiF (formed by small Li⁺ion and F^– ion) is higher than its hydration enthalpy. On the other hand, the lattice enthalpy of LiCl containing small Li⁺ ion and large Cl^– ion is considerably lower than its hydration enthalpy.

Thus, LiF is almost insoluble in water while LiCl is soluble. Furthermore, Li⁺ ions can polarise bigger Cl^– ions more easily than smaller F^– ions. As a result, LiCl has more covalent character than LiF and so, it is also soluble in the organic solvent acetone.

Question 21. What happens when

1. Sodium metal is dropped in water?
2. Sodium metal is heated in a free supply of air
3. Sodium peroxide dissolves in water?

Answer:

Sodium hydroxide is formed. H₂ gas is evolved which catches fire due to the exothermicity of the reaction.

2Na(s) + 2H₂O(l)→ 2NaOH(aq) + H₂(g)

2. Sodium peroxide is formed by heating sodium metal in a free supply of air.

2Na(s) + O₂ (g)→ 2Na₂O₂(s)

3. H₂O₂ is formed when sodium peroxide dissolves in water

Na₂O₂(s) + 2H₂O(l)→2NaOH(aq) + H₂O₂(Z)

Question 22. Comment on each of the given observations:

1. Lithium is the only alkali metal to form a nitride directly.
2.  M²⁺+(aq) + 2e → M(s) (where M = Ca, Sr, or Ba) is nearly constant.

Answer:

1. The lattice energy of lithium nitride (Li₃N) which consists of a small cation (Li⁺) and a small anion (N₃-) is much higher and this energy compensates for the high bond dissociation energy of the N=N bond and the energy to form N₃– ion. Larger alkali metal ions cannot compensate for these energy requirements. Hence, Li⁺ is the only alkali metal that forms nitride directly.

2. M²⁺ (aq) + 2e → M(s) Where M = Ca, Sr, Ba

Standard electrode potential, \(E_{\mathrm{M}^{2+} / \mathrm{M}}^0\) depends on 3

1. Enthalpy of vaporisation
2. Ionization enthalpy and
3. Enthalpy of hydration.

As the combined effect of these factors is almost the same for Ca, Sr, and Ba, their E° values are nearly constant.

Question 23. State as to why

1.  A solution of Na₂CO₃ is alkaline?
2. Alkali metals are prepared by electrolysis of their fused chlorides.
3. Sodium is found to be more useful than potassium.?

Answer:

1. Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH. In an aqueous solution, it ionizes to form Na⁺ and CO^2-₃ – ions.

Na₂CO₃ ⇌  2Na⁺  (aq) + CO^2-₃ (aq)

The formed CO^2-₃ ions hydrolyze in an aqueous solution to produce acetic acid and OH^– ion.

CO^2-₃ (aq) + 2H₂O(Z) ⇌ H₂CO₃ ⇌+ 2OH^– ( aq)

As H₂CO₃ is a weak acid, it remains mostly unionized. Consequently, the concentration of OH^– ions increases in the solution thereby making the solution alkaline.

2.

• As alkali metals are strong reducing agents, they cannot be extracted by chemical reduction from their oxides or other compounds,
• As alkali metals are highly electropositive, these metals cannot be displaced from their salts with the help of other elements,
• Alkali metals cannot be obtained even by the electrolysis of aqueous solutions of their salts.

In this case, H₂, instead of the alkali metal is liberated at the cathode because the discharge potential of alkali metals is higher than that of hydrogen.

Thus, to prepare alkali metals, electrolysis of their fused chlorides is carried out. For example,

NaCl →Na⁺ + Cl^–

During electrolysis, at the cathode,

2Na⁺ + 2e →2Na; and at the anode, 2Cl→ Cl₂ + 2e

3. Sodium is found to be more useful because Na is not as reactive as K. For this reason, reactions of sodium with different substances can be controlled and usage of sodium is far more safe than potassium. Thus, sodium is more useful than potassium

Class 11 Chemistry Chapter 10 S-Block Elements Long Answer Solutions

Question 24. Write balanced equations for reactions between 

1. Na₂O₂ & Water,
2. KO₂ & water,
3. Na₂ O& CO₂

Answer: The balanced equations of the given reactions are.

1.

2. 2K O₂ (S) + 2H₂ O(l) → 2KOH(aq) + H₂ O₂ (aq) + O₂ (g)

3. Na₂ O + CO₂→ Na₂ CO₂

Question 25. How would you explain the given observations? 

1. BeO is almost insoluble but BeSO₄ is soluble in water.
2. BaO is soluble but BaSO4 is insoluble in water.
3. Lil is more soluble than KI in ethanol.

Answer:

1. O^2- is smaller in size than SO₄^2-. Consequently, a small Be²⁺ ion is tightly packed with a small O^2- ion, and thus, the lattice enthalpy of BeO is greater than its hydration enthalpy. So BeO is insoluble in water. On the other hand, a small Be²⁺ ion is loosely packed with a large SO^2-. Ion and thus, the lattice enthalpy of BeSO₄ is less than its hydration enthalpy. So, BeSO₄ is soluble in water.

2. Large Be²⁺ ion is tightly packed with large SO₄^2-.ion and thus, the lattice enthalpy of BaSO₄ is greater than its hydration enthalpy. So, BaSO₄ is insoluble in water On the contrary, a large Ba²⁺ ion is loosely packed with a small Oion, and thus lattice enthalpy of BaO is less than its hydration enthalpy. So, BaO is water soluble.

KI is predominantly ionic. On the other hand, due to the high polarising power of very small Li+ ions, Lil is predominantly covalent. For this reason, Lil is more soluble than KI in the organic solvent ethanol.

Long Questions S-Block Elements Class 11 NCERT

Question 27. How will you distinguish between:

1. Mg and Ca
2. Na₂SO₄ and BaSO₄
3. Na₂CO₃ and NaHCO₃,
4. LiNO₃ and KNO₃

Answer:

1. Calcium, when heated, imparts brick brick-red color to the flame but magnesium does not.

2. Sodium sulfate (Na₂SO₄) is soluble in water but barium sulfate (BaSO₄) is insoluble.

3. Na₂CO₃ is stable to heat but NaHCO₃ decomposes on heating to produce CO₂ gas which turns limewater milky

4. When LiNO₃ is heated, it decomposes to yield reddish-brown vapors of NO₂. However, when KNO₃ is heated, it decomposes to yield colorless O₂ gas

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Very Short Question And Answers

Question 1. Explain why sodium carbonate is used in fire extinguishers.
Answer: In fire extinguishers, Na₂CO₃ is used as it reacts with dil. H₂SO₄ to produce CO₂

Question 2. Why docs sodium hydrogen carbonate called baking soda?
Answer: NaHCO₃ is an important ingredient in baking powder. Thus, it is also called baking soda.

Question 3. Name a bivalent element whose oxide is soluble in excess NaOH solution.
Answer:  Beryllium; BeO + 2NaOH→ Na₂BeO₂ + H₂O

Question 4. Which of the alkali metals exhibits abnormal behavior?
Answer: Lithium

S-Block Elements Very Short Answer Questions Class 11

Read and Learn More NCERT Class 11 Chemistry Very Short Answer Questions

Question 5. Which alkali metal is most abundant in the earth’s crust?
Answer: Sodium (Na)

Question 6. What type of crystal lattice is formed by the alkali metals?
Answer:  Alkali metals form a body-centered cubic lattice

Question 7. Name an alkali metal that is used in Lassaigne’s test for the detection of nitrogen, sulfur, and halogen in the organic compounds.
Answer: Sodium

Question 8. Write the general electronic configuration of group-2 elements.
Answer: Electronic configuration: [Inert gas] ns²

Question 9. Which element of group 2 shows abnormal behavior?
Answer: Beryllium (Be) shows abnormal behavior

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Very Short Answer Questions

Question 10. Why CaF₂ is considered the most important fluoride salt among all the fluoride salts of alkaline earth metals?
Answer:  CaF₂ is the most important source for the preparation of fluorine

Question 11. Which out of LIF of Lil is more covalent?
Answer:  Lil is more covalent.

Question 12. Which alkali metal carbonate easily liberates CO₂ on heating?
Answer: Li₂CO₃ easily liberates CO₂ on heating.

Question 13. Which alkali metal chloride imparts violet color to the bunsen burner flame?
Answer: Potassium chloride (KCl)

Question 14. Which alkali metal hydride is used as a source of hydrogen for filling up meteorological balloons?
Answer: Lithium hydride (LiH).

Question 15. Name the products formed in the following reaction. Explain Lil + KF→? + ?
Answer: LiF and KI. The larger K⁺ ion stabilizes the larger I^– ion while the smaller Li⁺ ion stabilizes the smaller F- ion.

Question 16.  What is water glass?
Answer: Sodium silicate (NaOSiO₃).

Question 17. Which alkali metal combines with nitrogen to form the corresponding nitride?
Answer: Lithium (Li).

Question 13. What are the raw materials used for the manufacture of sodium carbonate by the Solvay process?
Answer: The raw materials are NaCl, CaCO and NH₃.

Question 18. Which out of the Na₂CO₃ solution & NaHCO₃ solution, changes the color of phenolphthalein into pink?
Answer: Na₂CO₃ solution.

Question 19. What is the main ingredient of baking powder?
Answer: Sodium bicarbonate (NaHCO₃).

Question 20. Which compound is used to treat the flue gases from coal and oil-fired power stations and to remove SO₂ and H₂SO₄ responsible for acid rain?
Answer: Na₂CO₃.

Question 21. What are the ingredients of the fusion mixture that is used in dry tests in inorganic analysis?
Answer: K₂CO₃ and Na₂CO₃

Question 22. Which alkaline earth metal is the most abundant in the earth’s crust?
Answer: Calcium (Ca).

Question 23. Which alkaline earth metal is radioactive?
Answer: Radium (Ra).

Very Short Answer Questions for Class 11 Chemistry S-Block Elements

Question 24. Name the metal of group-2 which is used to prepare Grignard reagent.
Answer: Magnesium (Mg).

Question 25. Name the element of group 2 that resembles lithium in characteristics.
Answer: Magnesium (Mg).

Question 26. Which alkaline earth metals do not impart any color to the flame of a Bunsen burner?
Answer: Be anil Mg.

Question 27. Name an alkaline earth metal compound that can be used as a portable source of hydrogen.
Answer: Calcium hydride (CaH₂).

Question 28. Which Gr-2 metal forms covalent compounds?
Answer: Beryllium (Be).

Question 29. Which Gr-2 metal burns readily when exposed to air?
Answer: Barium (B a).

Question 30. Which alkaline earth metal reacts with alkali to form hydrogen gas?
Answer: Beryllium (Be).

Question 31. What is the composition of the alloy, electron?
Answer: 95% of Mg+ and 5% of Zn.

S-Block Elements Class 11 Very Short Answer Questions

Question 32. Which reagent is used to analyze Ca²⁺ and Mg²⁺ ions quantitatively?
Answer: EDTA (Ethylenediaminetetraacetic acid).

Question 33. What is the medicinal name of the aqueous solution of Mg(OH)₂?
Answer: Milk of magnesia.

Question 34. What is anhydrous?
Answer: Magnesium perchlorate, Mg(ClO₄)₂ is known as anhydrous.

S-Block Elements Chapter 10 Short Answer Solutions Class 11

Question 35. How will you distinguish between BeSO₄ & BaSO₄?
Answer: BeSO₄ is water soluble but BaSO₄ is insoluble in water.

Question 36. Which out of MgCO₃, SrCO_3, and BaCO₄, possesses the highest thermal stability?
Answer: BaCO₄ has the highest thermal stability.

Question 37. Distinguish between Be(OH)₂ and Ba(OH)₂
Answer: Be(OH)₂ is soluble in caustic soda solution while Ba(OH)2 is insoluble in it.

Question 38. Why is BeCl₂ soluble in organic solvents?
Answer: BeCl₂ is covalent.

Question 39. Which Gr-2 metal carbonate is unstable in the air?
Answer: BeCO₃.

Question 40. Which alkaline earth metal sulfate is useful in diagnosing stomach ulcers by X-ray?
Answer: BaSO₄ (in ‘barium meal’ X-ray).

Question 41. BeO is covalent but still, it has a much higher melting point— mention the reason.
Answer: This is due to its polymeric structure.

Question 42. What is the difference between lime water and milk of lime?
Answer: Mg²⁺

Question 43. Write a reaction by which BeCl₂ can be prepared.
Answer:

NCERT Class 11 Chemistry S-Block Elements VSAQ Solutions

Question 44. What is the composition of soda lime used for the preparation of hydrocarbons in the laboratory?
Answer: NaOH and CEO
.
Question 45. ED Name two acidic oxides which react similarly with calcium hydroxide [Ca(OH)₂].
Answer: CO₂ and SO₂.

Question 46. Which alkaline earth metal oxide is used as a flux in metallurgy to remove siliceous impurities?
Answer: Calcium oxide (CaO)

Question 47. What is the commercial name of the disinfectant powder obtained when Cl2 reacts with slightly moist slaked lime at 40°C?
Answer: Bleaching powder.

Question 48. What is Plaster of Paris?
Answer: Hemihydrate of calcium sulphate, (CaSO₄)₂-H₂O is called Plaster of Paris.

Question 49. What type of impurities in gypsum should be avoided in preparing Plaster of Paris from it?
Answer: Carbonaceous impurities are to be avoided.

Question 50. Why KNO₃ is used instead of NaNO₃ in gunpowder?
Answer: NaNO₃ is deliquescent. Hence, KNO₃ is preferred over NaNO₃ for the preparation of gunpowder.

Question 51. What do you mean by ‘black ash
Answer: A mixture of sodium carbonate (Na₂CO₃) and calcium sulfide (CaS) is called black

Question 52. What happens when magnesium is heated with | acetylene at 875K?
Answer: Magnesium on heating with acetylene at 875K forms magnesium carbide (Mg₃C₂)

Question 53. Write the composition of gunpowder.
Answer: Gunpowder is an explosive mixture containing KNO₃ along with charcoal and sulfur

Class 11 Chemistry Chapter 10 S-Block Elements Short Answer Questions

Question 54. Why is Na₂S₂O₃ used in photography?
Answer: In photography, Na₂S₂O₃ is used to dissolve the unexposed AgBr.

Question 55. The affinity of sodium towards water is used in drying benzene. Explain.
Answer: Sodium does not react with benzene. Hence it can be used effectively for drying benzene.

Question 56.  Name a pair of elements that exhibits a diagonal relationship.
Answer:  Lithium (Li) and magnesium (Mg) CO₂

Question 57. Name an alkaline earth metal.
Answer: Calcium (ca)

Question 58. Write the balanced equation for the reaction when water is added to calcium carbide.
Answer: CaC₂ + 2M₂O →MC=CH + Ca(OH)₂

Question 59. Which alkali metal ion has the highest polarising power?
Answer: Li⁺

Question 60. What are the common oxidation states exhibited by group-1 and group-2- 2 metals respectively?
Answer: +2 And +1 respectively

Question 61. Name the alkali metals which form superoxides when heated with excess oxygen.
Answer: K, Rb, and cs

Question 62. Which one is the lightest and which one is the heaviest of all the metals?
Answer: Lithium (Li) and Osmium(os) respectively

Question 63. What is the composition of the white powder obtained when metallic magnesium is burnt in the air?
Answer: A mixture of MgO and Mg₃N₂

Question 64. Name two metals of group 2 which do not impart any color to the flame.
Answer: Mg and Be

Question 65. Which alkali metal cannot be stored in kerosene?
Answer: Li

Question 66. Which alkali metal is used as a scavenger in metallurgy to remove O₂ and N₂ gases?
Answer: Li

Question 67. Which Gr-2 metal carbide reacts with water to produce methane?
Answer: Be₂ C

S-Block Elements Chapter 10 NCERT Very Short Answer Questions

Question 68. In an aqueous solution, which alkali metal ion has the lowest mobility?
Answer: Li⁺

Question 69. Which is the most abundant alkaline earth metal in the earth’s crust?
Answer: Calcium

Question 70. Name one group-2 metal.
Answer: Chlorophyll

Question 71. Which alkali metal ion forms a stable complex with 18- crown-6
Answer: Potassium(K)

Question 72. Which alkaline earth metal is largely used as a lightweight construction metal?
Answer: Mg

Question 73. Which alkaline earth metal forms an organometallic compound known as Grignard reagent?
Answer: Mg

NCERT Solutions Class 11 Chemistry Chapter 10 S-Block Elements VSAQ

Question 74. Which alkali and alkaline earth metals are radioactive?
Answer: Fr and Ra

Question 75.  The salts of which alkali metals are commonly hydrated?
Answer: Li

Question 76. Which alkali metal acts as the strongest reducing agent in aqueous solution?
Answer: Li

Question 77. Which is the least stable alkali metal carbonate?
Answer: Li₂CO₃

Question 78. What is baryta water?
Answer: Ba(OH ₂ solution

Question 79. Which alkaline earth metal hydroxide is most soluble in water?
Answer: Ba(OH)₂

Question 80. Which alkaline earth metal imparts crimson color to flame?
Answer: Srcl₂

Question 81. BeO is covalent and still has high melting points — why?
Answer: It is polymeric

Question 82. What is anhydrous?
Answer: Magnesium perchlorate Mg(ClO₄)₂

Question 83. Which alkaline earth metal carbonate can be kept only in an atmosphere of CO₂?
Answer: BeCO₃

Question 84. Which alkaline earth metal chloride is used as a desiccant in the laboratory?
Answer: CaCl₂

Very Short Answer Solutions for S-Block Elements Class 11 Chemistry

Question 85. The basic strength of which alkali metal hydroxide is the highest?
Answer: CsOH

Question 86. Which alkaline earth metal hydroxide is amphoteric?
Answer: Be(OH)₂

Question 87. How will you distinguish between Ba(OH)₂ and Be(OH)₂?
Answer: Be(OH)₂ dissolves in alkali but Ba(OH)₂ does not

Question 88. What are the raw materials used for the manufacture of washing soda by the Solvay process?
Answer: NaCl, CaCO₃ and NH₃

Question 89. What is soda ash?
Answer: . Anhydrous Na₂CO₃

Question 90. Which alkaline earth metal hydroxide and alkali metal carbonate are used for softening hard water?
Answer: Ca(OH)₂ and Na₂CO₃;

Question 91. What is the formula of Plaster of Paris?
Answer: 2CaSO₄ .H₂O

Question 92. What is dead burnt plaster?
Answer: Anhydrous CaSO₄;

Question 93. Which alkaline earth metal carbonate is used as an ingredient of chewing gum?
Answer: CaCO₃

Question 94. What is a fusion mixture?
Answer: A mixture of Na₂CO₃ and K₂CO₃

Question 95. What is the most abundant source of sodium chloride?
Answer: Seawater

Question 96. For which of its chief properties is Plaster of Pariswide used?
Answer: On mixing with water, it forms a plastic mass which sets into a hard mass within 5 to 15 minutes;

Question 97. What is used for making blackboard chalks?
Answer: CaCO₃

Question 98. Which compound is generally used for the detection of CO₂ in the laboratory?
Answer: Lime water [Ca(OH)₂]

Question 99. What is the diagonal relationship? Give one example
Answer:

See ‘General Discussion on s -block elements

Very Short Answer Questions for Class 11 Chemistry Chapter 10 S-Block Elements

Question 100. Which of the alkaline earth metal hydroxides are amphotericin in nature?
Answer:

Beryllium hydroxide [Be(OH)₂] is amphoteric in nature

Class 11 Chemistry S-Block Elements VSAQ

Question 101. What is hydrolysis?
Answer:

Calcium hydride [CaH₂] is known as hydrolytic

Question 102. Find out the oxidation state of sodium in Na₂O₂.
Answer:

Let the oxidation state of Na be x. The oxidation state of oxygen in the peroxides is -1

2x + 2(-1) = 0; 2x = 2,x= +1.

Question 103. Which two alkaline earth metals cannot be identified by flame test?
Answer:

Beryllium (Be) and magnesium (Mg) cannot be identified by flame test

Question 104. Magnesium occurs in nature largely as MgCO₃but beryllium never occurs as BeCO₃. Explain.
Answer:

The strong polarising power of Be²⁺ ion makes BeCO₃ unstable;

Question 105. Out of Na and Mg which one has a higher second ionization enthalpy? Why?
Answer:

The second ionization enthalpy of Na is higher than that of Mg;

Question 106. The chloride of a metal is soluble in an organic solvent. The chloride can be CaCl₂, NaCl, MgCl₂, BeCl₂
Answer: BeCl₂

Question 107. Why does common salt become wet in the rainy season?
Answer:

Due to the absorption of aerial moisture by deliquescent impurities like MgCl₂, CaCl_2, etc;

S-Block Elements Chapter 10 Very Short Answer Q&A Class 11

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Fill In The Blanks

Question 1. _____________ has no d-orbital in its valence shell.
Answer: Lithium

Question 2. _____________ reacts with nitrogen to give nitrides.
Answer: Lithium

Question 3. _____________ imparts a golden-yellow color to the flame.
Answer: Sodium

Question 4. Potassium, in reaction with dioxygen, produces _____________
Answer: KO₂

Question 5. The outermost electronic configuration of the radioactive alkali metal is _____________
Answer: 7s¹

Question 6. The bicarbonate salt_____________ of does not exist in solid state.
Answer: Lithium

Chapter 10 S-Block Elements VSAQ Class 11 NCERT Solutions

Question 7. Ionic conductance of Li+ ion in aqueous solution is lowest _____________ is highest
Answer: Hydration

Question 8. _____________ion has maximum polarising power
Answer: Li⁺

Question 9. _____________ is the most abundant alkali metal in Earth’s
Answer: Sodium

Question 10. The basicity of the alkali metal hydroxides _____________ down the group.
Answer: Increases

Question 11._____________ is used as a source of oxygen in submarines, space shuttles, and oxygen masks.
Answer: KO₂

Question 12. The alkali metals combine with mercury to give_____________
Answer: Amalgams

Question 13. The alkali metals exist as_____________ lattices having cordination number 8.
Answer: Body Center cubic

Question 14. Lil _____________ is more soluble than KI in ethanol.
Answer: More

Question 15. K₂CO₃ cannot be produced by the Solvay process because _____________ does not get precipitated in the aqueous solution.
Answer: KHCO₃

Question 16. _____________ and _____________ do not respond to flame test.
Answer: Be,  Mg

Question 17. _____________ ion exhibits maximum does not respond tendency to complexes.
Answer: Be²⁺

Question 18. Only _____________ can displace hydrogen from dilute HNO₃
Answer: Mg

NCERT Class 11 Chemistry Chapter 10 S-Block Elements VSAQ

Question 19. Common salt gets wet due to the presence of _____________ as impurity.
Answer: MgCl₂

Question 20. Hydration enthalpy of Mg²⁺ is _____________ Ca²⁺. as than that of
Answer: Greater

Question 21. Hydrolysis of calcium carbide produces _____________
Answer: Acetylene

Question 22. The commercial name of _____________is hydrolith
Answer: CaH₂,

Question 23. BeCO₃ is stable only in an atmosphere of _____________
Answer: CO₂

NCERT Class 11 S-Block Elements Short Answer Questions

Question 24. Lime water is a transparent aqueous solution of _____________
Answer: Ca(OH)₂

Question 25. The second ionization enthalpy of the alkaline earth metals is _____________ than their first ionization enthalpy.
Answer: Greater

Question 26. _____________ is used to prepare Grignard reagents
Answer: Greater

NCERT Class 11 Chemistry Chapter 10 S-Block Elements VSAQ

Question 27. The melting point of the alkaline earth metals is _____________ than that of alkali metals
Answer: Mg

Question 28. Between Ca and Na, _____________ is used to dehydrate alcohols
Answer: Ca

Question 29. Among the alkaline earth metals, _____________ is abundant in the earth’s crust.
Answer: Ca

Question 30. Temperature of the mixture of _____________  -54°C. Is most and ice is about
Answer: CaCl²⁺

NCERT Solutions For Class 11 Chemistry Chapter 11 Some P Block Elements Multiple Choice Questions

Question 1. Which of the following metals has the largest abundance in the earth’s crust

1. Aluminium
2. Magnesium
3. Calcium
4. Sodium

Answer: 1. Aluminium

Question 2. For BCl₃, AlCl₃ and GaCl3 the increasing order of ionic character is

1. BCl₃ < AlCl₃< GaCl₃
2. GaCl₃ < AlCl₃ < BCl₃
3. BCl₃< GaCl₃< AlCl₃
4. AlCl₃ < BCl₃ < GaCl₃

Answer: 3.  BCl₃< GaCl₃< AlCl₃

The ionic character of a compound depends on the polarising power of the cation present in it. The more the polarising power of the cation, the less the ionic character of the compound containing the cation. The polarising power of the cations B⁺³, Al⁺³ and Ga⁺³ is Al⁺³ < Ga⁺³ < B⁺³. Hence, the increasing order of ionic character will be BCl₃ < GaCl₃< AlCl₃

Question 3. In borax, the number of B- O- B link and B- OH bonds present are respectively

1. Five and four
2. Four and five
3. Three and four
4. Five and five

Answer: 1. Five and four

Some P-Block Elements Class 11 Multiple Choice Questions

Question 4. In diborane, the number of electrons that account for bonding in the bridges is

1. Six
2. Two
3. Eight
4. Four

Answer: 4. Four

Two bridging bonds are present in diborane. In the bridges, each H-atom is bonded to two B-atoms by sharing of only one pair of electrons i.e., 3c-2e bonds. Thus, the total number ofelectrons for two bridges is four.

Question 5. The main reason that SiCl4 is easily hydrolysed as compared to CCl₄ is that

1. SiCl bond is weaker than C — Cl bond
2. SiCl₄ can form hydrogen bonds
3. SiCl₄ is covalent
4. Si can extend its coordination number beyond four

Answer: 4. Si can extend its coordination number beyond four

In SiCl₄ , Si has a vacant d -orbital. Therefore, it can expand its coordination through coordinate bonding to get hydrolysed

Question 6. Which ofthe following ions cannot be formed by boron

1. BF^3-₆
2. BH^–₄
3. B(OH)^–₄
4. BO^–₂

Answer: 1. BF^3-₆

Boron cannot expand its covalency beyond four due to the unavailability of π-orbitals. Thus it cannot form BF^3-_6.

Question 7. Which of the following exists as covalent crystals in the solid state

1. Phosphorus
2. Iodine
3. Silicon
4. Sulphuric

Answer: 3. Silicon

Silicon exists as covalent crystals in a solid state

Question 8. Which of the following are Lewis acids

1. PH₆ and SiCl₄
2. BCl₃ and AlCl₃
3. PH₃ and BC
4. AlCl₃ and SiCl₄

Answer: 2. BCl₃ and AlCl₃

Both BCl₃ and AlCl₃ have incomplete octets around the central atom. Thus they can act as Lewis acids.

Question 9. Which of the following oxides is amphoteric

1. SnO₂
2. CaO
3. SiO₂
4. CO₂

Answer: 1. SnO₂

SnO₂ reacts with both acids as well as with bases. Hence, it is an amphoteric oxide

SnO₂ + 4HCl→SnCl₂+ 2H₂O

SnO₂ + 2NaOH→Na₂SnO₃+ H₂O

NCERT Solutions Class 11 Chemistry Chapter 11 P-Block MCQs

Question 10. Which of the following statements is incorrect—

1. Pure sodium dissolves in liquid ammonia to give a blue solution
2. NaOH reacts with glass to give sodium silicate
3. Aluminium reacts with excess NaOH to give Al(OH)₃
4. NaHCO₃ on heating gives Na₂CO₃

Answer: 3. Aluminium reacts with excess NaOH to give Al(OH)₃

Aluminium reacts with NaOH to form sodium metaaluminate (NaAlO₂).

Question 11. Name the two types ofthe structure of silicate in which one oxygen atom of [SiO₄]^4- is shared

1. Linear chain silicate
2. Sheet silicate
3. Borosilicate
4. Three-dimensional

Answer: 3.  Pyrosilicate

Question 12. Which of these is least likely to act as a Lewis base

1. BF₃
2. PF
3. CO
4. F

Answer: 1. BF₃

BF₃ is a Lewis acid

Question 13. Which ofthe following is electron-deficient

1. (BH₃)₂
2. PH₃c
3. (CH₃)₂
4. (SiH₃)₂

Answer: 1. (BH₃)₂

Boron hydrides are electron-deficient compounds

Question 14. Number of carbon atoms per unit cell of diamond unit cell

1. 1
2. 4
3. 8
4. 6

Answer: 2. 4

Diamond exists as a face-centred cubic unit cell. Thus, at the centre of each face one atom is present i.e., ½ × 6 = 3 atoms. Also, ½ × 8 = 1 atom is present at the comers. Hence, the total no. of atoms per unit cell = 3 +1 = 4 atoms.

Question 15. . Which of these is not a monomer for a high molecular mass silicone polymer 

1. PhSiCl₃
2. MeSiCl₃
3. Me₂SiCl₂
4. Me₃SiCl

Answer: 4. Me₃SiCl

Polymerisation involves the repeating of monomeric units. If Me3 SiCl units are joined, then a dimer is formed rather than a polymer.

P-Block Elements MCQs Class 11 Chemistry

Question 16. Which ofthe following structures is similar to graphite

1. B₂H₆
2. BN
3. B
4. B₄C

Answer: 2. BN

Question 17. The basic structural unit of silicate is

1. SiO^2-₄
2. SiO^–
3. SiO^4-₄
4. SiO^2-₃

Answer: 3.SiO^4-₄

Question 18. The stability of the +1 oxidation state among Al, Ga, In and Tl increases in the sequence

1. Ga<In<Al<Tl
2. Al < Ga < In < Tl
3. Tl < In < Ga < A1
4. In <T1 < Ga < Al

Answer: 2. Al < Ga < In < Tl

Question 19. AlF₃ is soluble In HF only In the presence of HF. It is due to the formation of

1. K₃(AlF₆)
2. AlH₃
3. K₃(AlF₃H)
4. K₃(AlF₃H₃)

Answer: 1. K₃(AlF₆)

The maximum coordination number shown by AF₆^3- is 6 and thus it can form A1F|- ion.

Question 20. Boric acid is an acid because its molecule

1. Gives up a proton
2. Accepts OH from water releasing a proton
3. Combines with a proton from the water molecule
4. Contains replaceable H -ion

Answer: 2. Accepts OH from water releasing a proton

H₃BO₃ is a Lewis acid and it takes up OH ion giving up H ion in the solution.

Question 21. It is because of the inability of ns² electrons of the valence shell to participate in bonding that

1. Sn²⁺ is oxidising while Pb⁴⁺ is reducing
2. Sn²⁺ and Pb⁴⁺ are both oxidising and reducing
3. Sn⁴⁺ is reducing while Pb⁴⁺ is oxidising
4. Sn²⁺ is reducing while Pb⁴⁺ is oxidising

Answer: 4.  Sn²⁺ is reducing while Pb⁴⁺ is oxidising

The inert pair effect increases down the group. Hence Sn²⁺ acts as a reducing agent whereas Pb⁴⁺ is an oxidising agent

Question 22. Which one of die following elements is unable to form MF₆^3- ion

1. In
2. Ga
3. B
4. Al

Answer: 3.  B

Boron does not have any vacant orbital. Therefore it cannot expand its coordination more than 4. Thus it is unable to form a compound ofthe type MF₆^3-

Question 23. The correct order of atomic radii in group 13 elements is

1. B < Ga < Al < In < Tl
2. B < Al < In < Ga < Tl
3. B < Ga < Al < Tl < In
4. B < Al < Ga < In < Tl

Answer: 1. B < Ga < Al < In < Tl

Due to the weak screening effect of d -d-orbital, the atomic radius of Ga is less than that of AI.

Class 11 Chemistry Chapter 11 Some P-Block Elements MCQs

Question 24. The wrong statement about fullerene is

1. It has 5-membered carbon ring
2. It has 6-membered carbon ring
3. It has sp² hybridisation
4. It has 5-membered rings more than 6-membered rings

Answer: 4. It has 5-membered rings more than 6-membered rings

Fullerene consists of 12 five-membered rings and 20 six-membered rings. So it has five-membered rings less than six-membered rings

Question 25. Iodine oxidises sodium borohydride to give

1. B₂H₆
2. Sodium hydride
3. HI
4. I^–₃

Answer: 1. B₂H₆

The oxidation of sodium borohydride with iodine in diglyme gives diborane.

Question 26. Which material is used as a neutron moderator

1. Graphite
2. Boron
3. Cadmium
4. Uranium

Answer: 1. Graphite

Neutron moderators slow down the speed of neutrons by collisions. They do not absorb neutrons. For

Example:  Water and graphite

Question 27. For silicon which is not correct

1. It is a type of silicate
2. It is hydrophilic
3. It is thermally unstable
4. The repeating unit is R₂SiO

Answer: 3. Silicon is hydrophobic

Question 28. Which ofthe following is not sp² -hybridised

1. Graphite
2. Graphene
3. Fullerene
4. Dry ice

Answer: 4. Dry ice

Solid CO₂ is dry ice in which carbon atom undergoes sp -hybridization

Question 29. The pair of amphoteric hydroxides is

1. Be(OH)₂, Al(OH)₃
2. Al(OH)₃,LiOH
3. B(OH)₃,Be(OH)₂
4. Be(OH)₂, Mg(OH)₂

Answer: 1. Be(OH)₂, Al(OH)₃

Al and Be show similar properties due to their diagonal relationship

Question 30. Which ofthe following reactions do not take place

BF₃ + F^–→BF₄^– …………..(1)

BF₃ + 3F^–→BF₄^3-…………..(2)

AlF₃ + 3F^–→AlF^3-₆ ………….(3)

1. Only 1
2. Only 2
3. Only 3
4. Only 1 and 3

Answer: 2. Only 2

BF₃ forms complex halides of the type BF^–₄ in which the B atom has coordination number 4. It cannot extend its coordination number beyond 4 due to the unavailability of d -d-orbitals in its configuration. Hence, BFg ion (sp³d³ hybridisation) is not formed. On the other hand, AI can extend its coordination number beyond 4 due to the availability of d -d-orbitals in its configuration.

Question 31. Select the correct options from the following

1. Graphene is an atomic layer of graphite.
2. Graphene is an atomic layer composed of sp² – hybridised carbon.
3. Chemical bonds in graphite are similar in strength to that of diamond.
4. All of these.

Answer: 4. All of these.

Question 32. Among the following substituted silanes, the one which will give rise to cross-linked silicone polymer on hydrolysis is

1. R₃SiCl
2. R₃SiCl
3. RSiCl₃
4. R₃SiCl

Answer: 3. RSiCl₃

Question 33. Hydride of boron occurs as B2H6 but B2C16 does not exist This is because

1. pπ-dπ back bonding is possible in B₂H₆ but not in R₂SiCl₂
2. Boron and hydrogen have almost equal values of electronegativity
3. Boron and chlorine have almost equal atomic sizes
4. Small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.

Answer: 1. pπ-dπ back bonding is possible in B₂H₆ but not in R₂SiCl₂

NCERT Class 11 Chemistry Some P-Block Elements Multiple Choice Q&A

Question 34. Which of the given compounds does not react with dilute HC1 at high temperature

1. SnSO₄
2. PbSO₄
3. BiOCl
4. CdSO₄

Answer: 2. PbSO₄

PbSO₄ belongs to group I so, it is insoluble in HCl. Acidity increases with the addition of ethylene glycol. It also exhibits a 3D arrangement due to hydrogen bonding.

Question 35. Boric acid is a weak acid, but in the presence of which ofthe following compounds, it behaves as a stronger acid

1. Glycerol
2. Acetic acid
3. Ethanol
4. Ethylene

Answer: 1. Glycerol

Question 36. The structure of diborane (B₂H₆) contains—

1. Four 2c-2e bonds and four 3c-2e bonds
2. Two 2c-2e bonds and two 3c-3e bonds
3. Two 2c-2e bonds and four 3c-2e bonds
4. Four 2c-2e bonds and two 3c-2e bonds

Answer: 4. Four 2c-2e bonds and two 3c-2e bonds

Question 37. Which of the following elements is used in high-temperature thermometry

1. Al
2. Ga
3. Hg
4. In

Answer: 2. Ga

Question 38. An important ingredient of Pyrex glass is—

1. Zn
2. Pb
3. B
4. Fe

Answer: 3. B

Question 39. Which ofthe following is the purest allotrope of carbon

1. Diamond
2. Fullerene
3. Graphite
4. Charcoal

Answer: 2. Fullerene

Question 40. The number of isomers possible for disubstituted borazine, B₃N₃H₄X₂ is-

1. 3
2. 4
3. 5
4. 6

Answer: 2. 4

Question 41. Pentaborane-9 (B₅H₉) is an example of

1. Arachno-borane
2. Pseudo-borane
3. Nido-borane
4. Close-borane

Answer: 1. Arachno-borane

Question 42. So when reacts with A forms B. A & B respectively are

1. HF, H₂SiF₄
2. HF, H₂SiF₆
3. HCl, H₂SiCl6
4. HI, H₂SiI₆

Answer: 2. HF, H₂SiF₆

Question 43. Boric acid is a

1. Monobasic and weak Lewis acid
2. Monobasic and weak Bronsted acid
3. Monobasic and strong Lewis acid
4. Tribasic and weak Bronsted acid

Answer: 1. Monobasic and weak Lewis acid

Multiple Choice Questions Some P-Block Elements Class 11 NCERT

Question 44. Which ofthe following does not exist in a free state

1. BF₃
2. BCl₃
3. BBr₃
4. BH₃

Answer: 4. BH₃

Question 45. The correct order of decreasing Lewis acid character is

1. BCl₃ > AlCl₃> GaCl₃ > InCl₃
2. AlCl₃ > BCl₃> InCl₃> GaCl₃
3. AlCl₃ > GaCl₃> BCl₃ > InCl₃
4. InCl₃ > GaCl₃ > AlCl₃ > BCl₃

Answer: 1. BCl₃ > AlCl₃> GaCl₃ > InCl₃

Question 46. Which of the following is present in the chain structure of silicate

1. (Si₂O^2-₅)_n
2. (Si₂O^2-₃)_n
3. Si₂O^4-₄
4. Si₂O^6-₇

Answer: 2. (Si₂O^2-₃)_n

Question 47. A metal, M forms chlorides in +2 and +4 oxidation states. Which ofthe following statements about these chlorides is correct

1. MCl₂ is more volatile than MCl₄
2. MCl₂ is more ionic than MCl₄
3. MCl₂ is more soluble in any. ethanol than MCl₄
4. MCl₂ is more easily hydrolysed than MCl₄