Sainavle

Thermodynamic Properties And Thermodynamic State Of A System

Thermodynamic properties

The measurable physical quantities by which thermo¬ dynamic state ofa system can be defined completely are called thermodynamic properties or variables ofthe system.

Thermodynamic properties Examples:

The pressure (P), temperature (T), volume (V), composition etc., of a system are the thermodynamic properties or variables of the system because the state of the system can be defined by these variables or properties. The properties or variables required to define the state ofa system are determined by experiment.

• Although a thermodynamic system may have many properties (like— pressure, volume, temperature, composition, density, viscosity, surface tension, etc.), to define a system we need not mention all of them since they are not independent
• If we consider a certain number of properties or variables having certain values to define the state of a system, then the other variables will automatically be fixed. In general, to define the state of a thermodynamic system, four properties or variables are needed. These are pressure, volume, temperature and composition of the system.
• If these variables of a thermodynamic system are fixed then the other variables will also be fixed for that system. For a closed system of fixed composition, the state of the system depends upon pressure (P), temperature (T) and volume (V) of the system.
• If these three variables of the system (P, V, T) are fixed, then other variables (like density, viscosity, internal energy etc.) ofthe system automatically becomes fixed

Class 11 Chemistry Notes For Thermodynamic Properties

Thermodynamic state of a system

A system is said to be in a given thermodynamic state ifthe properties (e.g., pressure, volume, temperature etc.) deter¬ mining its state have definite values.

If the thermodynamic properties or variables of a thermodynamic system remains unchanged with time, then the system is said to be in thermodynamic equilibrium.

A system is said to be in thermodynamic equilibrium if it attains thermal equilibrium, mechanical equilibrium, and chemical equilibrium simultaneously.

• Thermal equilibrium: A system is said to be in thermal equilibrium if the temperature throughout the system is the same and is equal to that of its surroundings.
• Mechanical equilibrium: If no imbalanced force exists within a system and also between the system and its surroundings, the system is said to be in mechanical equilibrium.
• Chemical equilibrium: If the chemical composition throughout a system remains the same with time, the system is said to be in chemical equilibrium

State function of a thermodynamic system definition

A state function of a thermodynamic system is a property whose value depends only on the present state of the system but not on how the system arrived at the present state.

Thermodynamic System Examples:

Pressure (P), volume (V), temperature (T) , internal energy (E or U), enthalpy (H), entropy (S), Gibbs free energy (G) etc., of a thermodynamic system are the state functions because the values of these functions depend only on the present state ofa system, not on how the system arrived at that state.

Class 11 Chemistry Notes For Thermodynamic Properties

Thermodynamic System Change of a state function in a process:

The state of a thermodynamic system at the beginning of a process is called its initial state and the state attained by the system after completion of the process is called its final state. Let X (like P, V,T etc., of a system) be a state function of a thermodynamic system. The values of X at the beginning and at the end of a process are X₁ and X₂ respectively. So, the change in the value of X in the process, ΔX = X₂-X₁

Infinitesimal change in X is represented by dX and finite change in X is represented by AX. For example, the infinitesimal change in pressure (P) of a system is dP and the finite change is ΔP.

If X is a state function of a thermodynamic system, then dX must be a perfect differential as the integration of dX between two states results in a definite value of X

The state function of a system is a path-independent quantity:

A state function of a system depends only on the state of the system. Consequently, the change in any state function ofa system undergoing a process depends only upon the initial and final states of the system in the process, not on the path of the process. Thus the state function of a system is a path-independent quantity

Thermodynamic System Explanation:

Suppose, a system undergoes a process in which its state changes from A (initial state) to B (final state), and because of this, the value of its state function X changes from X_A (value of X at state A ) to X_B (value of X at state B). The process can be carried out by following three different paths as shown

But the change in X, i.e., ΔX= (X_B-X_A) will be the same for all three paths. This is because all the paths have identical initial and final states and consequently X has identical initial and final values for these paths.

Class 11 Chemistry Notes For Thermodynamic Properties

Thermodynamic System Example:

The change in temperature of a system depends only upon the initial and final states of the process. It does not depend on the path followed by the system to reach the final state. So the temperature of a system is a state function. Similarly, the change of other state functions like pressure (P), volume (V), internal energy (U), enthalpy (H), entropy (S), etc., (i.e. ΔP, ΔV, ΔU, ΔH, ΔS etc.) does not depend upon the path ofthe process.

Path-dependent quantity

Two terms commonly used in thermodynamics are heat (q) and work ( w). These are not the properties ofa system. They are not state functions. Heat change or work done involved in a process depends on the path of the process by which the final state of the system is achieved. Thus, heat and ; work are the path-dependent quantities.

In general, capital letters are used to denote the state functions (for example, P, V, T, U, etc.) and small letters are used to denote path functions (for example q , w, etc.). q and w are not state functions. Hence, δq or δw (S = delta) are used instead of dq or dw. Unlike dP or dV, which denotes an infinitesimal change in P or V, δq or δw does not indicate such kind of change in q or w. This is because q and w like P or F are not the properties of a system. δq and δw are generally used to denote the infinitesimal transfer of heat and work, respectively, in a process.

MPBSE Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics Question And Answers

Question 1. Give two examples of path-dependent quantities. Are they properties ofa system?
Answer:

Two path-dependent quantities are heat (q) and work ( w). These are not the properties ofa system.

Question 2. Under what conditions will a system be in thermodynamic equilibrium?
Answer:

A system will be in thermodynamic equilibrium if it simultaneously maintains mechanical equilibrium, thermal equilibrium and chemical equilibrium.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 3. Why is a process occurring in an open container considered to be an isobaric? What is the origin of the internal energy of a system? Why cannot the absolute value of internal energy be determined?
Answer:

A process in an open container takes place under constant atmospheric pressure. Thus, it is an isobaric process.

Read and Learn More Class 11 Chemistry

Question 4. Is the internal energy of a system at 25°C greater or less than its internal energy at 50°C?
Answer:

The internal energy of a system increases with the temperature rise. So, the internal energy ofa system will be greater at 50 °C than that at 25 °C.

Question 5. Under which condition will the pressure-volume work be, \(w=-\int_{V_1}^{V_2} P d V?\) pressure of the gas.
Answer:

⇒ \(w=-\int_{V_1}^{V_2} P_{e x} d V\)

So, in case ofa reversible process, \(w=-\int_{V_1}^{V_2} P d V\)

Question 6. According to the first law of thermodynamics, AU = q + w. Write down the form of this equation for the following processes: Cyclic process Adiabatic process Isothermal expansion of an ideal gas Process occurring in an isolated system.
Answer:

An isolated system does not exchange energy or matter with its surroundings. So, for a process occurring in an isolated system, q = 0 and w = 0. Therefore, ΔU = q + w or, ΔU= 0 + 0 or, ΔU = 0

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 7. The definition of enthalpy shows that for n mol of an ideal gas H = U + nRT.
Answer:

If the enthalpy, internal energy, pressure and volume of ‘n’ mol of an ideal gas at a temperature of TK are, U, P and V respectively, then H = U+PV. For the ‘n’ mol of an ideal gas, PV = nRT. So, H = U+nRT.

Question 8. Prove that for an Ideal gas undergoing an isothermal change, AH = 0.
Answer:

The change in enthalpy of an ideal gas undergoing a process, ΔH = ΔU+nRAT. In an isothermal process, ΔT = O. So, ΔH = A{Again, in an isothermal process of an ideal gas, A U = 0 and hence AH = 0.

Question 9. Under what conditions are

1. ΔU = qv
2. ΔH = qp

Answer:

ΔU = qv; Conditions: Closed system, constant volume, only P-V work is considered

ΔH = qp; Conditions: Closed system, constant pressure, only P-V work is considered.

Question 10. Give an example ofa combustion reaction whose standard enthalpy change is equal to the standard enthalpy of formation ofthe compound formed in the reaction.
Answer:

At 25 C the standard heat of combustion of solid naphthalene [C₁₀H₈(s)] is 5147 kj. mol-1. this means that at 25 c and 1 atm pressure when 1 mol of solid naphthalene is completely burnt in the presence of oxygen the enthalpy change that occurs is 51747kj.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 11. The standard enthalpy of combustion of C_xH_y) at 25°C is Q kj. mol^-1 . Write down the thermochemical equation for the combustion reaction of this compound.
Answer:

The combustion reaction for C(s, graphite) is:

This reaction also C(s, graphite) + O₂(g)→CO₂(g). represents the formation reaction of CO₂(g). Therefore, at 25°C, the standard heat of combustion of C(s, graphite) = the standard heat formation of CO₂(g)

Question 12. Consider the given enthalpy diagram, and > calculate the unknown AH by applying Hess’s law.
Answer:

Following the given diagram, we have

1. A+B→ C + 2D ; ΔH = ?
2. A + B→ E + 2D; ΔH = +27kJ
3. E + 2D →C+ 2D;ΔH = -13kJ

Adding equation 2 and equation 3, we get A + B→C+2D; ΔH = (27- 13)kJ

= 14 kJ

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 13.  For the reaction, A + B →D, AH is -30 kj. Suppose, D is prepared from A and B and then it is again converted into A and B by following the stages D → E → A + B. Calculate the total enthalpy change in these two stages.
Answer:

A + B → D; ΔH = -30 kj ……………………….(1)

The process D→E →A + B comprises the following two steps:

D→E ⋅⋅⋅⋅⋅(2) and E→A + B

Overall reaction: D→A + B The reaction (4) is the opposite ofthe reaction (1).

So, the enthalpy change in reaction (4) is +30 kj.

Hence, the total enthalpy change in steps (2) and (3) is +30 kJ

Question 14. Water remains in equilibrium with its vapour at 100°C and atm. Will the transformation of water into its vapour be spontaneous at this pressure and temperature?
Answer:

No. Since water and its vapour are in equilibrium, neither the forward process (water→vapour) nor the reverse process (vapour →water) is favourable

Question 15. If the process A→B occurs reversibly, then the change in entropy of the system is ΔS₁. When the same process occurs irreversibly, the change in entropy of the system is ΔS₂. Will the value of ΔS₁ be greater than, less than or equal to the value of ΔS₂?
Answer:

Since entropy is a state function, the change in entropy ofa system in a process does not depend upon whether the process is carried out reversibly or irreversibly.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 17. Write the relation between ΔS_sys & AS_surrwhen a process reaches equilibrium. What will be the value of ΔS_univ?
Answer:

For a process at equilibrium

⇒ \(\Delta S_{\text {system }}+\Delta S_{\text {surr }}=0\text { But } \Delta S_{s y g}+\Delta S_{\text {surr }}=\Delta S_{\text {univ }} \text {. So, } \Delta S_{\text {univ }}=0\)

Question 18. Give two examples of spontaneous processes in which the disorderliness of the system decreases.
Answer:

Transformation of water into ice at 1 atm pressure and below (T’C temperature. Condensation of water vapor at1 atm pressure and below 100 C temperature.

Question 19. What do you mean by a perpetual motion machine of the second kind?
Answer:

A machine working in a cyclic process absorbs heat from a single thermal reservoir and completely converts the heat into the equivalent amount of work, is called a perpetual motion machine of a second kind. This type of machine contradicts the second law of thermodynamics, & it is impossible to construct.

Question 20. A certain amount of gas is enclosed in a container with permeable and diathermal walls. Which type of system does the gas belong to?
Answer:

The walls of the container are diathermal. So the system can exchange heat with its surroundings. Again, the walls ofthe container are permeable. So, the system can also exchange matter with its surroundings. Hence, the gas belongs to an open system.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 21. Does the volume of a closed system remain fixed?
Answer:

If the walls of a closed system are non-rigid or movable, then the volume of the system does not remain fixed. For example, a gas enclosed in a cylinder fitted with a movable piston is considered a closed system. Here, the volume of the gas (system) can be increased or decreased by altering the pressure of the gas (system)

Question 22. Give an example of a thermodynamic quantity which is not a state function. Is it a property of a system?
Answer:

Heat is not a state function because heat absorbed or by a system in a process depends upon the path of the realised It is not a property of the system.

Question 23. Give an example of a process which Is simultaneously isothermal and adiabatic.
Answer:

Adiabatic free expansion of an ideal gas (or isothermal free expansion of an ideal gas). The reason is that no exchange of heat occurs between the system and surroundings in this process and the temperature of the tire system remains constant throughout the process.

Question 24. At 25°C, the standard reaction enthalpy for the reaction AB₃(g)→1/2A₂(g)+3/2(g) is. find the standard reaction enthalpy for the reaction.
Answer:

Writing this equation. in reverse manner and multiplying both sides by 2, we get,

A₂(g) + 3B₂(g)→2AB₃(g); -2AH°. So, at 25°C, the standard

Question 25. Mention the standard state of sulphur and iodine at 25
Answer: 

At 25°C, the standard state of sulphur is solid rhombic sulphur [S(rhombic, s)] and that of iodine is solid iodine [1₂(s)].

Question 26. What do you mean by ‘the enthalpy of solidification of water at 0°C and 1 atm pressure = -6.02 kj-mol^-1 .’?
Answer:

This means that 6.02 kj of heat is released when one mole of water completely freezes to ice at 0°C and 1 atm pressure.

Question 27. Why are spontaneous natural processes irreversible?
Answer: ‘

The spontaneous or natural processes are irreversible because the thermodynamic equilibrium of the system is not maintained in such types of processes.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 28. Which of the following will have a greater entropy?

1. 1 mol of H₂ gas (T = 300 K, V = 5ml, )
2. 1 mol of H₂ gas (T = 300 K, V = 10mL).

Answer:

As the entropy of a gas increases with the increase in its volume,

1. The entropy of mol of H₂ (T = 300 K, V = 10 mL)
2. Will be greater than that of l mol of H₂ (T = 300 K, V = 5 mL).

Question 29. For a process, ΔS_sys = -15 J.K^-1 .. For what value of ASsurr will the process be non-spontaneous?
Answer:

The condition of non-spontaneity of a process is \(\Delta S_{s y s}+\Delta S_{s u r r}<0.\) \(\Delta S_{s y s}=-15 \mathrm{~J} \cdot \mathrm{K}^{-1}\), then will be non-spontaneous

Question 30. A gas is allowed to expand against zero external pressure. Explain with reason whether the process is reversible or irreversible.
Answer:

The expansion ofa gas against zero external pressure is an irreversible process. As the opposing pressure is zero, the gas expands rapidly, and it cannot maintain thermodynamic equilibrium during its expansion.

Question 31. In a process, 701 J of heat is absorbed by a system and 394J of work is done by the system. What is the change in internal energy for the process?
Answer:

Given: q = +701J and w = -394 J {-ve sign as the work is done by the system)

Now, ΔU = q + w or, A U = (701- 394)J = +307 J.

So, the change in internal energy of the system = +307 J.

Question 32. The latent heat of the vaporization of water at a normal boiling point is 40.75 kJ. mol^-1 . . Calculate the change in entropy of vaporization.
Answer:

Given:

⇒ \(\Delta H_{\text {vap }}\) = 4075 kJ. mol^-1,

Tb =100C = 375k

⇒ \(\Delta S=\frac{\Delta H_{v a p}}{T_b}=\frac{40.75 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{373}=109.25 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

Question 33. Due mole of ideal gas is expanded isothermally. In this process, which of the quantity (or quantities) among w. q, ΔH , ΔH is(are) zero or >0 or <0?
Answer:

During isothermal expansion, heat is absorbed and work is done by the gas. So q > 0 and w < 0. Again internal energy and enthalpy remain the same during isothermal expansion of an ideal gas. Thus, ΔU = 0 and ΔH = 0 . For the isothermal expansion of mol of ideal gas q>0, w< 0, ΔU = 0, ΔH = 0.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 34. Give examples of two processes by which the internal energy of a gas can be increased.
Answer:

The internal energy of a gas can be increased by increasing the die temperature ofthe gas. If a gas is compressed adiabatically (considering only P-V work) its internal energy increases.

Question 35. Give examples of three processes in which the change in internal energy of the system is zero.
Answer:

The change in internal energy of any cyclic process is zero. The change in internal energy of an ideal gas during isothermal expansion or compression is zero. In adiabatic free expansion of an ideal gas, the change in internal energy is equal to zero.

Question 36. What do you mean by the standard enthalpy of atomisation of chlorine at 25°C = + 121 kj.mol^-1 .?
Answer:

This means that at 25°C and 1 atm pressure, 121 kj of heat is required to produce 1 mol of gaseous Cl-atom from Cl₂(g). Thus, the change in enthalpy for the process,

⇒ \(\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Cl}(\mathrm{g}); \Delta H_{\text {atom }}^0=+121 \mathrm{~kJ}\)

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 37. When does the entropy of the system attain maximum value for a spontaneous or irreversible process occurring in an isolated system? Under this condition, what will be the change in entropy of the system?
Answer:

When a spontaneous or irreversible process occurs in an isolated system, the entropy of the system increases with the progress of the process towards equilibrium. The value of entropy becomes maximum when the process attains equilibrium, and there occurs no further change in the entropy ofthe system. Thus, the value of entropy is maximum at the equilibrium state ofthe process and under this condition, the change in entropy ofthe system is zero.

Question 38. Is the entropy change of a system influenced by the change in temperature? Explain.
Answer:

The entropy of a system is highly dependent on temperature. With the increase or decrease in temperature, the randomness of the constituent particles (atoms, molecules or ions) of a system increases or decreases. Now, the entropy of a system is a measure of the randomness of its constituent particles. Thus, the entropy change ofa system is influenced by the change in temperature.

MPBSE Class 11 Chemistry Chapter 7 Equilibrium Question And Answers

Question 1. A liquid is in equilibrium with its vapor at its boiling point. On average which property of the molecules is equal in two phases?
Answer:

At the boiling point of a liquid in equilibrium with its vapor, the average kinetic energy of the molecules in the two phases is equal.

Question 2. According to Le Chatelier’s principle, what is the effect of adding heat to a solid and liquid in equilibrium?
Answer:

In the equilibrium system solid-liquid, the forward process is endothermic. Therefore, if temperature is increased at equilibrium, then, according to Le Chatelier’s principle, equilibrium will shift to the right, thereby increasing the amount of liquid.

Question 3. Mention two ways by which the equilibrium of the L-given reaction can be shifted to the right.
Answer:

According to Le Chatelier’s principle, the equilibrium of the above reaction can be shifted to the right by the addition of excess reactants [i.e., CH₃COOH(l) or C₂H₅OH(l) ] or by the removal of the products [i.e., CH₃COOC₂H₅(l) or H₂O(l)] from the reaction system at a given temperature, keeping the volume of the reaction system constant.

Read and Learn More Class 11 Chemistry

Question 4. When steam is passed over a red-hot iron, H₂ gas is produced. In this reaction, the yield of H₂(g) is found to increase when the partial pressure of steam is increased. Explain.
Answer:

Reaction:

⇒ \( \mathrm{Fe}(s)+4 \mathrm{H}_{-} \mathrm{O}(\rho) \rightleftharpoons \mathrm{Fe}_{-} \mathrm{O} \cdot(\mathrm{s})+4 \mathrm{H}_{-}(g)\)

A steam is one of the reactants in the above reaction, increasing its partial pressure at equilibrium will shift the equilibrium position to the right. As a result, the yield of the product i.e., H₂(g) will increase.

Class 11 Chemistry Chapter 7 Equilibrium

Question 5. Why does not the equilibrium constant expression for a reaction involving pure solids or liquids contain the concentration terms of the solids or liquids?
Answer:

For a pure solid or liquid, molar concentration is directly proportional to density. Given that density remains constant at a specific temperature, the molar concentration of a pure solid or liquid at that temperature is a constant value, typically regarded as unity (1). Consequently, the equilibrium constant expression for a process involving pure solids or liquids excludes concentration terms for these phases.

Question 6. At constant temperature, the following reaction is at equilibrium in a closed container: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})\) At constant temperature, if the amount of the solid carbon is reduced to half at equilibrium, then what will be the change in the concentration of CO(g)?
Answer:

At a particular temperature, the concentration of any pure solid is independent of its amount. Thus, keeping the temperature constant, if the amount of solid carbon is reduced to half at equilibrium, then its concentration will remain unchanged. So, the concentration of CO(g) will also remain unaffected.

Class 11 Chemistry Chapter 7 Equilibrium

Question 7. The values of the equilibrium constant (if) of a reaction at 25°C & 50°C are 2 ×10^-4 & 2 ×10^-2, respectively. Is the reaction exothermic or endothermic?
Answer:

With the increase in temperature, the value equilibrium constant (K) increases for an endothermic reaction, while it decreases for an exothermic reaction. For the reaction, K(50°C) > K(25°C), indicating it is an exothermic reaction.

Question 8. For a gaseous reaction, K_p > K_c. What will be the effect on equilibrium if pressure is increased at a constant temperature? Will it affect the yields of the products?
Answer:

According to the relation K_p = K_c(RT)^-Δn, if K_p> K_c, then Δn > 0. The positive value of An implies that the reaction occurs with an increase in volume in the forward direction. For such a reaction, if pressure is increased at equilibrium, then according to Le Chatelier’s principle the equilibrium will shift to the left and thus the yield ofthe product will decrease.

Question 9. In the case of the thermal decomposition of H₂(g) to H(g), which conditions of pressure and temperature will be favorable for an increase in the yield of H(g)?
Answer:

Since the formation of H(g) from H₂(g) [H₂(g) 2H(g)] occurs through decomposition, it is an endothermic reaction. Because 2 moles of H(g) are formed from 1 mole of H₂(g), the reaction is associated with a volume increase. So, according to Le Chateliehs principle, the yield of H (g) will increase if the reaction is carried out at a high temperature and low pressure.

Class 11 Chemistry Chapter 7 Equilibrium

Question 10. In the case of the reaction A₂(g) + 4B₂(g), 2AB₄(g), the change in enthalpy (ΔH) is negative. Mention the conditions of pressure and temperature at which the yield of the product, AB₄(g) will decrease.
Answer:

Since ΔH < 0, it is an endothermic reaction. The volume of the reaction system decreases in the forward direction [1 molecule of A₂(g) combines with 4 molecules of B₂(g) to form 2 molecules of AB₄(g)]. Thus; according to Le Chatelier’s principle, under the conditions of high temperature and low pressure, the yield of AB₄(g) will decrease.

Question 11. How will the equilibrium of the reaction, H₂(g) + I₂(g) ⇌ 2HI(g) be affected if the volume of the reaction system at equilibrium is doubled, keeping the temperature constant?
Answer:

Doubling the volume of the reaction system at equilibrium will reduce the total pressure of the system by half. But for the reaction Δn = 0. So, according to Le Chatelier’s principle, the equilibrium ofthe reaction will not be affected by a change in pressure.

Question 12. In the reaction, I₂+I^–→I₃^–, which one acts as a Lewis base?
Answer:

In the reaction between and I₂, the I^– ion donates an electron pair to the I₂ molecule, resulting in the formation of the I₃^– ion [I₂+I^–→I₃^– ]Therefore, the I^– ion acts as a Lewis base in this reaction.

Question 13. The pK_a values of the three weak acids HA, HB, and HC are 4.74, 3.75, and 4.20, respectively. Arrange them in order their of increasing acid strengths.
Answer:

As pK_a = -log₁₀ K_sp, the smaller the value of Ka the larger the value of pKa. So, an acid with a larger pKa will have a smaller Ka. As, pK_a(HB) < pK_a(HC) < pK_a(HA) , pK_a(HB) > Kf_a(HC) > K_a(HA). At a certain temperature, a higher value of Ka for an acid indicates a higher strength of the acid. Therefore, the increasing order of acid strengths of the given acids will be — HA < HC < HB.

Question 14. X and Y are two aqueous solutions of added HA with concentrations of 0.1 M & 0.01M, respectively. In which solution will the degree of ionization of U A be higher
Answer:

According to Ostwald’s dilution law, the degree of ionization of a weak electrolyte increases with the increase in dilution of its aqueous solution. Since, the concentration of solution Y is less than that of X, the degree of ionization of HA will be higher in solution Y.

Class 11 Chemistry Chapter 7 Equilibrium

Question 15. Which one of the following two acids will have a higher concentration of H₃O⁺ ions in their 0.1(M) aqueous solutions HCl and  CH₃COOH?
Answer:

HCl is a strong acid, while CH₆COOH is a weak acid. Thus, HCl ionizes almost completely in aqueous solution, whereas CH₃COOH undergoes partial ionization. As a result, the concentration of H₃0+ ions in 0.1(M) HCl solution is higher than that in 0.1(M) CH₃COOH solution.

Question 16. Show that [OH-]>\(\sqrt{K_w}\) in an alkaline solution.
Answer:

We know, [H₃O⁺] × [OH^– ] = Kw. In pure water, [H₃O⁺] = [OH^–].

This gives \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{K_w}\)

In an alkaline solution, the concentration of OH^– ions is higher than that in pure water.

Therefore, in an alkaline solution, \(\left[\mathrm{OH}^{-}\right]>\sqrt{K_w} \text {. }\).

Question 17. Will the concentration of HgO⁺ ions in pure water at 0°C be more than or less than that at 4°C?
Answer:

Ionization of water is an endothermic t process: [2H₂O(1) H₃O+(aq) + OH^–(aq)]. Hence, with, a temperature rise, the ionic product of water (Kw) increases. Therefore, KM,(4°C) In pure water, [H₃O⁺] = Jÿw- Since,(4°C) > Kw(0°C) , the concentration of H₂O⁺ ions in pure water at 4 °C will be higher than that at 0°C.

Question 18. At a certain temperature, what is the value for the die sum of pH and pOH for an aqueous solution? What will be its value at 25°C?
Answer:

In case of any aqueous solution at a certain temperature pH+POH=pk_w. At 25C, Pk_w= 14. Hence at 25C PH+POH=14.

Class 11 Chemistry Chapter 7 Equilibrium

Question 19. An acid bottle labeled pH – 5 Is this acid a weak acid?
Answer:

The acid may be a weak add or a very dilute strong acid. The pH of an added solution depends upon the die concentration of H₃O⁺ Ions in the solution. So, from the value of pH, it Is not possible to predict whether the acid IN is weak or strong.

Question 20. A, B, and C are three buffer solutions, each of which is composed of a weak acid and its salt. For increasing the pH by 0.02 units, it is found that 1.0, 1.4, and 1.2 millimol of NaOH are required for A, B, and C, respectively. Arrange the solutions in the increasing order of their buffer capacities.
Answer:

The higher the buffer capacity of a buffer solution, the greater the amount of a strong acid or a strong base to be required for increasing the pH of the buffer. It is given that increasing the pH of the buffer by the same amount requires a minimum amount of NaOH for buffer A and a maximum amount of NaOH for buffer B. Therefore, the increasing order of buffer capacity of the given buffers is A < C < B.

Question 21. Of the two bottles, one contains an HCl solution and the other a buffer solution. Each of the bottles b labelled as pH = 5. How can you identify the solutions?
Answer:

Upon measuring the pH of the solutions following the addition of equal drops of NaOH, a significant increase in pH will be observed for one solution, whilst the pH of the second solution remains relatively unchanged. A buffer solution maintains a relatively constant pH, whereas a solution of HCl results in an increase in pH.

Question 22. At a certain temperature, the Ksp of AgCl in water is 1.8 × 10^-10. What will be its Ksp is a 0.1M solution of AgNO₃ at some temperature.
Answer:

At a certain temperature, the solubility of AgCl decreases in the presence of a common ion (Ag⁺), but the solubility product of AgCl remains the same. Therefore, Ksp for AgCl in 0.1(M) aqueous solution of AgNO₃ will be the same as that in water.

Question 23. You are supplied with HCOOH (pK_a =3.74), CH₃COOH (pK_a = 4.74), and NaOH solutions. To prepare a buffer solution of pH =3.8, which acid will you select? Give reason
Answer:

The buffer capacity of a buffer solution consisting of a weak acid and its salt becomes maximum when the pH ofthe buffer solution is equal to the pK_a of the weak acid. Among the given acids, the pK_a of HCOOH is very close to the desired pH of the buffer solution. Hence, one should use HCOOH for preparing the buffer.

Question 24. pH of a buffer solution composed of NH₃ and NH₄Cl is 9.26. Will there be any change in pH if 100 mL of distilled water is added to 100 mL of this buffer solution?
Answer:

For the buffer solution made up of NH₃ and NH₄Cl, the pH of the solutions is given by

⇒  \(p H=14-p K_b-\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_3\right]}\)

If 100 mL of distilled water is added to 100 mL of this buffer solution, no change occurs in the ratio of [NH₄Cl] to [NH₃] and the pH ofthe solution remains the same.

Class 11 Chemistry Chapter 7 Equilibrium

Question 25. Which of the given salts will undergo cationic or anionic or both cationic and anionic hydrolysis? NH₄F, NaCN, AICl₃, Na₂CO₃, NH₄Cl
Answer:

Both NH₄Cl and AlCl₃ are the salts of strong acids and weak bases. In aqueous solution of such salts, cationic hydrolysis takes place. NaCN and Na₂CO₃ are the salts of weak acids and strong bases. In an aqueous solution of such salts, anionic hydrolysis takes place. NH₄F is a salt of weak acid and weak base. In aqueous solution of such salt, both cationic and anionic hydrolysis take place.

Question 26. The values of pure water at 0°C and 25°C are x and y respectively. Is it greater than or less than y?
Answer:

For pure water \(p H=\frac{1}{2} p K_w\) Rise in temperature increases the value of K_w.

So, K_w(0°C) < K_w(25°C) & hence pK_w(0°C) > pK_w(100°C) as pK_w = -log₁₀. As in the case of pure water,

⇒ \(p H=\frac{1}{2} p K_w\) pH of pure at 0 °C will be greater than that at 25 °C. Hence, x > y.

Question 27. pK_w = 12.26 at 100°C. What is the range of pH -scale at this temperature? What will be the pH of a neutral solution at this temperature?
Answer:

At 100 °C, pK_w = 12.26. So, at this temperature, the pH scale ranges from 0 to 12.26. At this temperature,

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{K_w}\)

In a neutral aqueous solution.

Therefore, pH of this solution \(=\frac{1}{2} p K_w=\frac{1}{2} \times 12.26=6.13\)

Question 28. Why does the concentration of OH“ ions in pure water increase with temperature rise? Does this increase make pure water alkaline? Explain.
Answer:

In pure water, [OH^–] = JK_w. K_w increases with temperature, and so does [OH^–]. This increase in [OH^–] does not however mean that pure water becomes alkaline at a higher temperature as pure water always contains an equal number of H₃O⁺ and OH^– ions at any temperature.

Class 11 Chemistry Chapter 7 Equilibrium

Question 29. All Lewis bases are fact bases—explain. Each of HCO₂ and HPO^– can act both as Bronsted acid and hose—why? Write the formula of conjugate base and conjugate acid in each case.
Answer:

A chemical capable of accepting a proton is termed a Bronsted base, whereas a substance that can give a pair of electrons is referred to as a Lewis base. The NH₃ molecule is capable of accepting a proton. Thus, it is a Brønsted base. The NH₃ molecule provides a pair of electrons to establish a coordination bond with a proton. Thus, it also functions as a Lewis base. Consequently, it may be deduced that a Lewis base qualifies as a Bronsted base.

Question 30. Each of HCO₃^– and HPO^– can act both as Bronsted acid and hose—why? Write the formula conjugate base and conjugate acid in each case.
Answer:

According to the Bronsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. Since the given species are capable of accepting and donating a proton, they can act as an acid as well as a base.

Question 31. Of the two solutions of acetic acid with concentrations 0.1(N) and 0.01(N), in which one does acetic acid have a higher degree of dissociation?
Answer:

Acetic acid is a weak acid. Such an acid undergoes partial ionization in water. The degree of ionization of a weak acid in its solution depends upon the concentration of the solution. The higher the concentration of a solution of a weak acid, the smaller the degree of ionization of the acid in that solution. So, the degree of ionization of acetic acid will be higher in 0.01(N) solution.

Question 32. At a certain temperature, the ionization constants of three weak acids HA, HB, and HC are 4.0 × 10^-5 5.2 × 10^-4, and 8.6 × 10^-3, respectively. If the molar concentrations of their solutions are the same, then arrange them in order of their increasing strength.
Answer:

The larger the ionization constant (Ka) of an acid, the greater the extent to which it undergoes ionization in its aqueous solution, and hence the higher the concentration of H30+ ions produced by it in the solution. Alternatively, the larger the value of Ka of an acid, the greater its strength.

The increasing order ofthe given acids concerning their Ka: HA < HB < HC. Consequently, the order of the given acids in terms of increasing strength in their aqueous solutions will be HA < HB < HC.

Class 11 Chemistry Chapter 7 Equilibrium

Question 33. At 25°C, the ionization constant (Ka) of weak acid HA is 10-6. What will the value of the ionization constant (Kb) of its conjugate base (A-) be at that temperature?
Answer:

We know \(K_a \times K_b=10^{-14} \text { [at } 25^{\circ} \mathrm{C} \text { ]. } K_a \text { of } \mathrm{HA}=10^{-6} \text {. }\)

Therefore \(K_b \text { of } \mathrm{A}^{-}=\frac{10^{-14}}{K_a}=\frac{10^{-14}}{10^{-6}}=10^{-8} .\)

Question 34. Give an example of a salt solution whose pH is independent of salt concentration.
Answer:

In the case of a solution of a salt formed from a weak acid and a weak base, the pH of the solution does not depend upon the concentration of the salt. One such salt is CH₃COONH₄.

Question 35. For what kind of solids, is solid vapor equilibrium achieved easily?
Answer:

Solid substances that can be easily converted to vapors on heating under normal pressure (i.e., sublimable substances).

For example: Solid CO₂, camphor, ammonium chloride, naphthalene, etc.

Question 36. At a fixed temperature, a liquid is in equilibrium with its vapors in a closed vessel. Which measurable tint quantity for the liquid gets fixed at equilibrium?
Answer:

When a liquid remains in equilibrium with its vapor in a closed vessel at a particular temperature, the vapor pressure of the liquid is found to acquire a fixed value.

Class 11 Chemistry Chapter 7 Equilibrium

Question 37. At 0°C and 1 atm pressure, why is the equilibrium established between water and ice regarded as dynamic?
Answer:

At 0°C and under 1 atm pressure, ‘water ice’ equilibrium is said to be dynamic since at equilibrium the rate of melting of ice is equal to the rate of freezing of water.

Question 38. Give two examples of chemical reactions for each of the following cases:

1. K_p > K_c
2. K_p<K_c
3.  K_p = K_c

Answer:

We know, K_p = K_c(RT)^Δn

Kp will be greater than K_c if Δn > 0. Reactions for

Question 39. The following reaction is carried out in a closed vessel at a fixed temperature: A(g) 2B(g). The concentrations of A(g) and B (g) in the course ofthe reaction are as follows When does the reaction attain equilibrium? What are the equilibrium concentrations of A and B?
Answer:

From the time of 60 minutes, the concentrations of reactant and product are found to remain the same with time. Therefore, the reaction has arrived at an equilibrium state at 60 minutes.

At the state of equilibrium, [A] = 0.58 mol- L^-1 and [B] = 0.84 mol- L^-1

Question 40. Here are a few salts. Whose aqueous solution(s) at 25°C has (have) a pH greater than 7, less than 7, or equal to 7? (NH₄)₂SO₄, CH₃COONH₄, K₂CO₃, NaNO₃
Answer:

pH < 7 : (NH₄)₂SO₄

pH> 7 : K₂CO₂₃

pH = 7 : CH₃COONH₄, NaNO₃

Class 11 Chemistry Hydrogen Question And Answers

Question 1. Name the isotopes of hydrogen and state their mass ratio.
Answer:

The three isotopes of hydrogen are:

Protium (H or H),

Deuterium (4H or D) and

Tritium (4H or T).

Their mass ratio is protium: deuterium: tritium =1:2:3.

Question 2. What is the source of solar energy?
Answer:

The main source of solar energy is the given nuclear fusion reaction: 4H —> 2He + 2+1e° (positron) + Energy

Question 3. Although Fe is placed above hydrogen in the electrochemical series, dihydrogen is not obtained by its reaction with nitric acid. Explain with reasons.
Answer:

HNO₃ being a strong oxidizing agent oxidizes dihydrogen into the water and itself gets reduced to nitrogen dioxide;\(\mathrm{Fe}+6 \mathrm{HNO}_3 \rightarrow \mathrm{Fe}\left(\mathrm{NO}_3\right)_3+3 \mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{O}\)

Question 4. Give an example and formula of a compound which on electrolysis liberates dihydrogen at the anode.
Answer:

Calcium hydride (CaH₂) on electrolysis, liberates dihydrogen at the anode.

Solutions For Class 11 Chemistry Hydrogen

Question 5. How can one prepare H2 gas from water by using a reducing agent?
Answer:

Reaction between metals such as Na or. JC (strong reducing agents) and water produce hydrogen gas.

⇒ \(2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \uparrow\)

Question 6. Name two compounds, in one of which hydrogen is in +1 and in the other in -1 oxidation state.
Answer:

In HCl, hydrogen is in a +1 oxidation state and in NaH it is in a -1 oxidation state.

Question 7. Holli dihydrogen and carbon monoxide burn in the air with blue flame. How will you distinguish between them?
Answer:

Dihydrogen burns with a blue flame in the air to form water vapor which turns white anhydrous CuSO₄ into hydrated copper sulfate (CuSO₄-5H₂O). However, carbon monoxide on combustion forms CO₂ which does not bring about any change in CuSO₄.

Question 8. What characteristics do you expect from an electron-deficient and an electron-rich hydride with respect to their structures?
Answer:

Electron-deficient hydrides function as electron acceptors, serving as Lewis acids, while electron-rich hydrides operate as electron donors, acting as Lewis bases. B₂H₆ functions as a Lewis acid, whereas NH₃ acts as a Lewis base.

Question 9. Why the boiling point of HF is higher than that of other hydrogen halides?
Answer:

Due to the formation of strong intermolecular hydrogen bonding, the boiling point of HF is higher than that of other hydrogen halides.

Solutions For Class 11 Chemistry Hydrogen

Question 10. How can you separate H2 or D2 from He?
Answer:

Palladium, heated to a high temperature, is cooled in a helium environment mixed with hydrogen or deuterium. As a result, substantial quantities of H₂ or D₂ are absorbed by palladium, whereas He is not. Upon heating palladium, occluded H₂ or D₂ gels are released as free hydrogen or deuterium.

Question 11. Why ionic or salt-like hydrides are used to dry organic solvents?
Answer:

Ionic or salt-like hydrides are used to dry organic solvents because they readily react with water to form the corresponding metal hydroxide along with the evolution of H₂O as- The solvent is then separated from the metallic hydroxide by distillation.

Question 12. Why concentration of D20 increase when electrolysis of water is carried out for a long period of time?
Answer:

Electrolysis of H20 occurs at a faster rate than D₂O because the bond dissociation energy of the O—H bond is greater than that of the O—D bond. So, electrolysis of ordinary water for a prolonged period of time results in an increase in the concentration of D₂O.

Question 13. How would you prepare deuterium peroxide (D₂O₂)?
Answer:

Deuterium peroxide (D₂O₂) can be prepared by the reaction between barium peroxide (BaO₂) and deuterosulphuric acid (D₂SO4).

⇒ \(\mathrm{BaO}_2+\mathrm{D}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+\mathrm{D}_2 \mathrm{O}_2\)

Question 14. How will you prepare deuteroammonia (ND3) from N2?
Answer: Magnesium burns in nitrogen to produce magnesium nitride which further reacts with D₂O to produce ND₃ (deuteroammonia)

⇒ \(\begin{gathered} 3 \mathrm{Mg}+\mathrm{N}_2 \rightarrow \mathrm{Mg}_3 \mathrm{~N}_2 \\ \mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{D}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg}(\mathrm{OD})_2+2 \mathrm{ND}_3 \end{gathered}\)

Question 15. How will you prove that hypophosphorous acid (H3PO₂) is a monobasic acid?
Answer:

When H₃PO₂ is treated with D₂O, only one of its hydrogen atoms is replaced by D. So, it can be said that only one H-atom remains attached to O-atom in hypophosphorous acid (H₃PO₂). Therefore, it is a monobasic acid.

Solutions For Class 11 Chemistry Hydrogen

Question 16. Sodium chloride is less soluble in heavy water than ordinary water—why?
Answer:

As the dielectric constant of D₂O is less than that of H₂O, NaCl (sodium chloride) is less soluble in heavy water than ordinary water. ⁺

Question 17. Explain why the water obtained after passing hard water through cation exchange resins is acidic.
Answer:

When hard water traverses an organic ion exchange resin, the resultant water is acidic due to the exchange of all metal ions in the water with H⁺ ions from the resin. Consequently, the resultant water is devoid of cations and possesses a high concentration of H⁺ ions. The water converts blue litmus paper to crimson.

Question 18. A sugar solution prepared in distilled water is passed successively through cation and anion exchange resins. What will be the taste of the collected water and why?
Answer:

Ion-exchange resins cannot remove sugar(non-electrolyte) from water. Therefore, when a sugar solution is passed successively through cation and anion exchange resins, after being collected tastes sweet.

Question 19. The hardness of the water in a tube well is 300 ppm. What do you mean by this statement?
Answer:

The statement means that in million parts by mass of the sample of water from the tube, well contains salts causing its hardness which are equivalent to 300 parts by mass of calcium carbonate.

Question 20. Will the water obtained by passing hard water through anion exchange resin, form lather with soap? Why?
Answer:

As the sample of water is not free from Ca²⁺ and Mg²⁺ ions, it will not form a lather with soap easily.

Question 21. A sample of water contains MgS04 and urea. How can they be eliminated easily?
Answer: They can be eliminated by a simple distillation method.

Solutions For Class 11 Chemistry Hydrogen

Question 22. It is better to preserve H202 in a polythene bottle than in a glass bottle—why?
Answer:

The decomposition of H₂O₂ is accelerated by the presence of glass, sunlight, and basic substances. So, H₂O₂ is preserved in polythene bottles rather than glass bottles.

Question 23. What do you understand by the expression ’30 volume H₂O₂ solution’?
Answer:

’30 volume H₂O₂ solution’ means that 1 mL of that solution yields 30 mL of oxygen at STP as a result of its complete decomposition.

Question 24. What do you mean by 20% H₂O₂ solution?
Answer:

20% H₂O₂ solution means that momT. of that solution contains 20g of H₂O₂.

Question 25. Calculate the percentage strength of 6.588 volume H₂O₂.
Answer:

Percentage strength of solution \(=\frac{\text { volume strength } \times 34}{11.2 \times 10}\)

Class 11 Chemistry Some P Block Elements Long Questions And Answers

Question 1. Unlike diamond, graphite is a good conductor of electricity —explain.
Answer:

In a diamond crystal, each carbon atom is sp³-hybridized, meaning that all electrons in the valence shell of the carbon atom engage in the formation of sigma bonds with four adjacent carbon atoms.

• Consequently, there are no free electrons remaining in the lattice, rendering the diamond incapable of conducting electricity.
• Conversely, each carbon atom in graphite, being sp²-hybridized, utilizes three of its valence electrons to establish three σ-bonds with three adjacent carbon atoms.
• The fourth electron in the 2pz orbital, oriented perpendicularly to the carbon layer, establishes an n-bond through lateral overlap with the 2pz orbital of any of the three neighboring carbon atoms.
• The electrons of the π-bonds are delocalized throughout the entire crystal via resonance. Graphite exhibits excellent electrical conductivity owing to the existence of mobile electrons.

Read and Learn More Class 11 Chemistry

Question 2. Diamond is extremely hard but graphite is soft and slippery —explain with reason.
Answer:

In diamond crystal, each sp³-hybridized carbon atom is tetrahedrally bonded to four other carbon atoms via covalent connections.

• A multitude of tetrahedral units interconnected to create a three-dimensional extensive array of molecules with robust bonds extending in all directions. Due to its three-dimensional network of robust covalent connections, diamond exhibits exceptional hardness.
• Conversely, in graphite crystals, each sp²-hybridized carbon atom is bonded to three other carbon atoms, creating a network of planar hexagons.
• These two-dimensional layers are arranged in parallel planes, positioned horizontally above one another at a distance of 3.35 Å.
• Due to the weak van der Waals forces binding these layers, one layer can effortlessly glide over another with minimal pressure applied. This elucidates the reasons behind graphite’s softness and lubricity.

Class 11 Chemistry Some P Block Elements

Question 3. Carbon monoxide possesses both oxidising and reducing properties—why?
Answer:

The oxidation slate of carbon in carbon monoxide is +2. It stands in between its highest oxidation state (+4) and lowest oxidation state (-4). As a result, in chemical reactions, the carbon atom in CO may increase its oxidation number from +2 to +4 or it may decrease its oxidation number from +2 to -4. When the oxidation number increases, CO is oxidised and plays the role of a reducing agent. For example, at high temperatures, CO reduces black CuO to red metallic Cu.

On the other hand, in case of a decrease in the oxidation number of carbon, carbon monoxide undergoes reduction, i.e., it acts as an oxidising agent. For instance, in the following reaction, carbon monoxide oxidises hydrogen-producing water and itself gets reduced to methane

Question 4. How will you convert a mixture of CO and CO₂ completely into

1. CO₂ and
2. CO?

Answer:

When a mixture of CO₂ and CO is passed over strongly heated CuO and kept in a combustion tube, CO₂ remains unchanged while CO gets oxidised to CO₂. As a consequence, only CO₂ comes out of the combustion tube. In this way, the mixture is completely converted into CO₂

On the other hand, when a mixture containing CO and CO₂ is passed over white-hot coke, CO remains unchanged but CO₂ gets reduced to CO. In this way, the mixture is completely converted into CO.

Class 11 Chemistry Some P Block Elements

Question 5. Mention 3 similarities like B and Si.
Answer:

Three similarities between boron and silicon are

1. Both boron and silicon react with caustic soda to form H₂ gas.

2B + 6NaOH → + 3H₂

Si + 2NaOH + H₂O →  Na₂SiO₃ + 2H₂

2. Halides of both boron and silicon undergo hydrolysis to form weak acids

BCl₃ + 3H₂O→H₃BO₃+ 3HCl

SiCl₄ + 4H₂O→ H₄SiO₄ + 4HCl

3. Halides of both boron and silicon form complex compounds

BF₃ +HF →  HBF₄; SiF₄ + 2HF → H₂SiF₆

Question 6. When boron trichloride reacts with water, it only B3 forms [B(OH)₄]^– whereas aluminium trichloride forms [AI(H₂O)₄]³⁺ in acidified aqueous solution. State the hybridisation of boron and aluminium in these species and explain your answer
Answer:

Boron trichloride (BCl₃ ) undergoes hydrolysis to form orthoboric acid [B(OH)₃] at first As the atomic size of B is small and its electronegativity is high, B(OH)₃ polarises HO molecule by accepting an OH^– ion thereby forming [B(OH)₄]^– and releasing a proton.

BCl₃ + 3H₂O→B(OH)₃ + 3HCl

B(OH)₃ + H₂O→[B(OH)₄]^–+ H⁺

There are no vacant d -orbitals in B as it lies in the second period and has only one s – and three p – orbitals. So, B can have four pairs of electrons in its valence shell, i.e., its maximum coordination number is 4.

For this reason, B(OH)₃ accepts one OH” ion to form [B(OH)₄]- in which B-atom is sp³ -hybridised. On the other hand, AlCl₃ undergoes hydrolysis in acidic medium to form [Al(H₂O)g]³⁺

AlCl₃ + 6H₂O →[Al(H₂O)₆]₃+ + 3Cl(aq)^–

In an acidic medium, the concentration of OH^– ions is lower than H+ ions. Thus, Al³⁺ ions coordinate with H₂O molecules and not with OH^– ions. Due to the availability of vacant d -d-orbitals in Al³⁺ ions, it can expand its coordination number from 4 to 6. Thus, it can form the complex ion, [Al(H₂O)₆]³⁺ in which hybridisation state of Al is sp³d²

Class 11 Chemistry Some P Block Elements

Question 7. Give reasons for the following:

1. Diamond-tipped tools are used for drilling and cutting purposes.
2. Graphite is used as a lubricant.
3. Silicones are water-repelling in nature.
4. CO gets absorbed by ammoniacal cuprous chloride to form a complex but CO₂ does not

Answer:

1. Diamond has the highest thermal conductivity among all known substances. Because of its high thermal conductivity, diamond-tipped tools do not overheat. Diamond is extremely hard as well. For these reasons, diamond-tipped tools are extensively used for drilling and cutting purposes.

2. Graphite has a layered structure in which any two successive layers are held together by weak van der Waals forces of attraction and thus one layer can smoothly slide over the other. For this reason, graphite acts as a lubricant.

3. Silicones are water-repelling in nature because they are surrounded by non-polar alkyl groups.

4. Carbon monoxide acts as a Lewis base due to the presence of a lone pair of electrons on carbon :C=0: and forms a soluble complex with ammoniacal cuprous chloride (CuCl).

CuCl + NH₃ + CO→ [Cu(CO)NH₃]⁺Cl^–

On the other hand, carbon dioxide  does not have a lone pair ofelectrons on carbon and cannot form complexes.

Question 8. What happens when

1. A mixture of sand and sodium carbonate is melted on heating. 0 At 200°C and under high pressure, carbon monoxide is passed through caustic soda solution and the product is heated to 300°C.
2. At high temperatures, metallic calcium is made to react with carbon and the product obtained is treated with water.
3. Potassium ferrocyanide is heated in the presence of concentrated H₂SO₄ and the gas thus obtained is passed over finely divided nickel powder at 50°C.
4. Silicon is heated with methyl chloride at high temperatures in the presence of copper.
5. SiO₂ is treated with HF.

Answer:

1. When a mixture of sand and sodium carbonate is melted by tremendous heat, sodium silicate and carbon dioxide are obtained.

SiO₂ + Na₂ CO₃ →  Na₂SiO₃ + CO₂↑

2. When CO is passed through NaOH solution at 200°C under high pressure, sodium formate is produced.

NaOH + CO→ HCOONa

When sodium formate is heated at 300°C, sodium oxalate and hydrogen gas are formed.

3. Metallic calcium reacts with carbon at high temperatures to form calcium carbide. Calcium carbide on treatment with water produces acetylene and calcium hydroxide

4. When potassium ferrocyanide is heated with concentrated sulphuric acid, potassium sulphate, ammonium sulphate, ferrous sulphate and carbon monoxide are formed. When the resulting CO gas is passed over finely divided nickel at 50°C, nickel tetracarbonyl (a coordinated complex) is formed

Ni + 4CO →(50°C) Ni(CO)₄

5. When silicon is heated with methyl chloride at high temperature in the presence of copper, a mixture of mono-, di- and trimethylchlorosilane along with a small amount of tetramethylsilane is obtained

SiO₂ reacts with HF to form silicon tetrafluoride. The initially formed SiF₄ dissolves in HF to form hydrofluorosilicic acid.

SiO₂ + 4HF→SiF₄ + 2H₂O ; SiF₄ + 2HF → H₂SiF₆

Class 11 Chemistry Some P Block Elements

Question 9. Starting from boric acid how can you prepare—

1. Boric anhydride
2. Boron trichloride
3. Boron trifluoride
4. Meta and tetraboric acid
5. Boron hydride
6. Ethyl borate

Answer:

1. Boric anhydride is prepared by heating boric acid at high temperatures

2. Boric acid is first converted to boric anhydride, which on heating with Cl₂ forms boron trichloride.

B₂O₃ + 3C + 3Cl₂→2BCl₃ + 3CO

3. Boric acid, on heating with CaF₂ and cone. H2S04 forms boron trifluoride.

CaF₂ + H₂SO₄→CaSO₄ + H₂F₂

2H₃BO₃ + 3H₂F₂ → 2BF₃ + 6H₂O

4. Orthoboric acid on heating at 100°C and 160°C forms metaboric acid and tetraboric acid respectively.

Metaboric Acid:

Tetraboric Acid:

5. Boric acid is first converted to boric anhydride which on heating with Mg-powder forms magnesium boride. On reacting dilute HCl with magnesium boride, boron hydrides are formed.

Ethyl borate is formed by heating a mixture of boric acid and ethanol in the presence of concentrated H₂SO₄.

3C₂H₅OH + H₃BO₃ → (C₂H₅)₃BO₃ + 3H₂O

Class 11 Chemistry Some P Block Elements

Question 10. Consider the compounds, BCI₃ and CCl₄. How will they behave with water? Justify.
Answer:

BCl₃ is an electron-deficient molecule because the core boron atom possesses only six electrons in its valence shell. Consequently, it can take a pair of electrons provided by water and undergo hydrolysis to produce boric acid (H₃BO₃) and HCl.

The central atom of the molecule, designated as atom B, possesses six valence electrons. Shell

Conversely, the carbon atom in CCl₄ possesses 8 electrons in its valence shell and lacks unoccupied d-orbitals to expand its octet. CCl₄ cannot take a pair of electrons from H2O; therefore, it does not undergo hydrolysis.

Question 1. HPO₄^2- can act both as a Bronsted base and as a Bronsted acid. Write the equation of equilibrium established by HPO₄^2- as an acid and a base in an aqueous solution. Also, write the expressions of K_a & K_b in two cases.  What are the conjugate acid and base of HS^–?
Answer:

HPO₄^2-  can denate and accept protons in aqueous solution. Thus it can serve both as an acid and base.

⇒ \(\begin{aligned}
& \mathrm{HPO}_4^{2-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{PO}_4^{3-}(a q)+\mathrm{H}_3 \mathrm{O}_2^{+}(a q) \\
& \mathrm{HPO}_4^{2-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

⇒ \(K_a=\frac{\left[\mathrm{PO}_4^{3-}\right] \times\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HPO}_4^{2-}\right]} ; K_b=\frac{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}\)

HS^– (aq) + H₂O(l) ⇌  H₂S(aq) + OH^– (aq)

Hence, the conjugate acid of HS^–  is H₂S.

HS^–(aq) + H₂O(l) ⇌  S^2-(aq) + H₃O+(aq)

Therefore, the conjugate base of HS^– is S^2--.

Read and Learn More Class 11 Chemistry

Question 2. An aqueous solution of sodium bisulfate is acidic, whereas an aqueous solution of sodium bicarbonate is basic—Explain.
Answer:

Since sodium bisulfate (NaHSO₄) is a salt of strong acid (H₂SO₄) and strong base (NaOH), it is not hydrolyzed in an aqueous solution. NaHS04 in its solution dissociates completely to form Na⁺ and HSO₄ ions and HSO₄ ions so formed get ionized to form H₃O+ and SO₂^-4 ions. As a result, the aqueous solution of NaHSO₄ becomes acidic.

⇒ \(\begin{gathered}
\mathrm{NaHSO}_4(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HSO}_4^{-}(a q) \\
\mathrm{HSO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)
\end{gathered}\)

NaHCO₃ in its solution dissociates completely to form Na+ and HCO₃ ions [NaHCO₃(aq)-Na+(aq) + HCO₃ (aq) ]. HCO₃ can act both as an acid and a base in aqueous solution.

⇒ \(\begin{aligned}
& \mathrm{HCO}_3^{-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{CO}_3^{2-}(a q) \\
& \mathrm{HCO}_3^{-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Class 11 Chemistry Solutions For Equilibrium

Question 3. Calculate the formation constant of [Ag(NH₃)₂]⁺
Answer:

⇒  \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; K_1=3.5 \times 10^3\)

⇒ \(\begin{aligned}
{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; } \\
K_1=1.7 \times 10^3 \ldots
\end{aligned}\)

Equation (3) represents the formation reaction of [Ag(NH₃)₂]2+. Therefore, the equilibrium constant for the reaction represented by equation (3) will be the formation constant for [Ag(NH₃)₂]2+. As equation (3) is obtained by adding equations (1) and (2), the formation constant (fcy) for [Ag(NH₃)₂]2+ equals ky x k2.

Thus, k_y = k_y X k₂ = (3.5 × 10³) × (1.7 ×  10³) = 5.95 ×  10⁶

Question 4. The first and second dissociation constants of an acid H₂A are 1 × 10^-5 and 5 × 10^-10  respectively. Calculate the value of the overall dissociation constant.
Answer:

⇒ \(\begin{aligned}
\mathrm{H}_2 \mathrm{~A}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HA}^{-}(a q) ; \\
K_1=1 \times 10^{-5} \ldots(1)
\end{aligned}\)

⇒ \(\begin{array}{r}
\mathrm{HA}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}^{2-}(a q) ; \\
K_2=5 \times 10^{-10} \ldots
\end{array}\)

The overall dissociation reaction is the sum of the. reactions (1) and (2). H₂A(aq) + 2H₂O(Z) 2H30+(aq) + A3-(aq) Thus, the overall disputation constant K = K₁ × K₂ =  1 × 10^-5 and 5 × 10^-10= 5 ×10^-15

Question 5. a -D-glucose Beta -D glucose, the equilibrium constant for this is 1.8. Calculate the percentage of a -D glucose at equilibrium.
Answer:

Suppose, the initial concentration of a -D glucose is a mol L^-1 and its degree of conversion to p -D glucose at equilibrium is x mol-L^-1

Class 11 Chemistry Solutions For Equilibrium

Therefore, the concentration of a -D glucose and D glucose at equilibrium will be as follows—

⇒ \(\begin{array}{ccc}
\begin{array}{c}
\text { Initial concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a & 0 \\
\begin{array}{c}
\text { Equilibrium concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a-a x & a x
\end{array}\)

Equilibrium constant for this process, \(K=\frac{[\beta-\mathrm{D}-\text { glucose }]}{[\alpha-\mathrm{D}-\text { glucose }]}\)

⇒ \(1.8=\frac{a x}{a(1-x)}=\frac{x}{1-x}\) or, \(x=\frac{1.8}{2.8}=0.6428\)

Thus, the percentage of a -D-glucose at equilibrium is \(\frac{a(1-0.6428)}{a} \times 100=35.72 \%\)

Question 6. Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 x 10-4 M Na₂CO₃ solution. At what concentration of Ba2+ will a precipitate begin to form? ( Ksp for BaCO₃ = 5.1 × 10^-9 )
Answer:

Ba(NO₃)₂ reacts with Na₂CO₃ to form BaCO₃. BaCO₃ is a sparingly soluble compound that forms the following equilibrium in its saturated solution.

⇒ \(\mathrm{BaCO}_3(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{CO}^{2-}(a q)\)

For \(\mathrm{BaCO}_3, K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]\)

In the solution \(\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M}\)

BaCO₃ will precipitate when [Ba²⁺][CO₂^–] → Ksp, i.e., when

⇒ \(\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]>5.1 \times 10^{-9}\left(\text { as } K_{s p}\left[\mathrm{BaCO}_3\right]=5.1 \times 10^{-9}\right)\) \(\text { If }\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M} \text {, then }\left[\mathrm{Ba}^{2+}\right]>\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}\)

Or, [Ba²⁺] > 5.1 × 10^-5M

Thus, the precipitation of BaCO₃ will start when [Ba2+] in the solution is grater than 5.1 × 10^-5 M.

Class 11 Chemistry Solutions For Equilibrium

Question 7. 2.5 m l of \(\frac{2}{5} \mathrm{M}\) weak monoacdic base Kb = 1 x 10-125at 25°C ) is titrated with \(\frac{2}{15}\) in water at 25°C. Calculate the concentration of H + at the equivalence point. (kw = 1 × 10-14)
Answer:

Ana. 2.5 mL of \(\frac{2}{15}\) M monobasic acid \(\equiv \frac{2}{5} \times 2.5 \equiv 1\) mmol of the base.

Suppose, VmL of \(\frac{2}{15} \mathrm{M}\) HC1 is required for the neutralisation. \(V \mathrm{~mL} \text { of } \frac{2}{15} \mathrm{M} \mathrm{HCl} \equiv \frac{2}{15} \times V \mathrm{mmol} \mathrm{HCl} .\)

Therefore \(\) mmol HCL will neutralise 1mmol of the base Hence \(\frac{2 \times V}{15}=1 \text { or, } V=7.5 \mathrm{~mL}\)

The total volume of the solution after neutralization = (2.5 + 7.5) mL = 10 mL.

The number of mmol of the salt formed in the neutralization =1 mmol.

So, the concentration of the salt in the final solution \((C)=\frac{1}{10}=0.1 \mathrm{M}\) As the resulting salt is formed from a weak base and a strong acid, it undergoes hydrolysis. The pH of the solution of such a salt is given by the pH
\(=7-\frac{1}{2} p K_b-\frac{1}{2} \log C\) \(\text { As } K_b=10^{-12}, p K_b=-\log _{10}\left(10^{-12}\right)=12\)

Therefore \(p H=7-\frac{1}{2} \times 12-\frac{1}{2} \log (0.1)=7-6+0.5=1.5\) and the concentration of H+(aq) ions at the equivalence point is [H⁺]

= 10-PH M = 10-1-5 M = 0.316 M.

Question 1. In which of the following two processes, the change in entropy of the system will be negative?

• Fusion of Ice.
• Condensation of water vapor

Answer: The change in entropy of the system is negative during the condensation of water vapor. In this process \(\Delta S_{\text {sys }}=S_{\text {water }}-S_{\text {water vapour }}<0\)

As the molecules in water vapor have more freedom of motion than they have in the water, the molecular randomness is higher in water vapor than in water. Thus, \(S_{\text {water }}<S_{\text {water vapour }}\) Consequently, the value of AS becomes negative.

Read and Learn More Class 11 Chemistry

Question 2. The change in internal energy in different steps of the process A→ B → C→ D are given: A→Bx.kJ-mol-1; B→ C, -y kj-mol-1; C-D,z kj-mol-1. What will the value of AH be for the change A → D?
Answer:

⇒ \(A →{\Delta U_1} B→{\Delta U_2} C →{\Delta U_3} D\)

Δ Ux = UR- UA = x kj.mol-1 ;

ΔU2 = Hc.— UB = -ykjmol-1;

ΔU3 = UD-uc = zkj.mol-1

∴ For the change A D

A U = UD-UA =(UD-UC) + (UC-UB) + (UB-UA)

=(z-y + x) kj-mol-1.

Class 11 Chemistry Solutions For Chemical Thermodynamics

Question 3. Change in enthalpy in different steps of the process A→B→C→A are given: A→B, x kj-mol-1; C→A, y kj-mol-1. Find the value of AH for step B→C.
Answer:

The initial and final states (A) are the same in the given process. So it is a cyclic process. Since H is a state function, AH = 0 for the process. Thus in this process \(\begin{aligned}
& \Delta H=\left(H_B-H_A\right)+\left(H_C-H_B\right) \Psi\left(H_A-H_C\right)=0 \\
& \left.x+\left(H_C-H_B\right)+y=0 \text { or }\left(H_C\right)_B H_B\right)=-(x+y) \mathrm{kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

So change in enthalpy for BC =- (x + y) kj-mol-1.

Question 4. Why are the standard reaction enthalpies of the following two reactions different?
Answer:

⇒ \(\begin{aligned}
& \mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-393.5 \mathrm{~kJ} \\
& \mathrm{C}(\text { diamond }, s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-395.4 \mathrm{~kJ}
\end{aligned}\)
Anszwer: Graphite and diamond are two allotropes ofsolid carbon. Different allotropic forms have different enthalpies. As the given U reactions involve different allotropes, the standard reaction enthalpies ofthese reactions are different.

Question 5. At 25°C, the standard reaction enthalpy for the reaction is -221.0 kj. Does this enthalpy change indicate the standard enthalpy of the formation of CO(g)? If not, then what would be the value of the enthalpy of j formation of CO(g) at 25°C?
Answer:

In the given reaction 2 mol CO(g) is formed from its stable constituent elements. Sd,’ definition, AH0 of this reaction does not represent standard enthalpy of formation of CO(g). 2C(graphite,s) + O₂(g) 2CO(g) ; AH0 = -221.0 kj

Dividing both sides by 2, we get

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g) ; \Delta H^0=-110.5 \mathrm{~kJ} \cdots[1]\)

In reaction [1], 1 mol of CO(g) is formed from its stable constituent elements. Thus, in this reaction AH0 = standard enthalpy of formation of CO(g).

So, at 25°C the standard enthalpy of formation of CO(g) =-110.5 kj-mol-1.

Class 11 Chemistry Solutions For Chemical Thermodynamics

Question 6. Determine the standard reaction enthalpy for the reaction: \(\mathrm{A}_2 \mathrm{~B}_3(s)+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(s)+3 \mathrm{CB}_2(g)\) Given: \(2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow \mathrm{A}_2 \mathrm{~B}_3(s); \Delta H^0=-x \mathrm{~kJ}\)
Answer:  Reversing equation 1 we get \(\mathrm{A}_2 \mathrm{~B}_3(s) \rightarrow 2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) ; \quad \Delta H^0=+x \mathrm{~kJ}\)

Multiplying the equation by 3, we get

⇒  \(3 \mathrm{CB}(g)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow 3 \mathrm{CB}_2(g) ; \Delta H^0=-3 y \mathrm{~kJ}\)

Adding equations [3] and [4], we get

⇒ \(\mathrm{A}_2 \mathrm{~B}_3(\mathrm{~s})+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(\mathrm{~s})+3 \mathrm{CB}_2(g) ; \Delta H^0=(x-3 y) \mathrm{kJ}\)

Therefore, the standard reaction enthalpy for the given reaction =(x- 3y)kJ.

Question 7. In which ofthe following reactions does AH0 at 25 °C indicate the standard enthalpy of formation \(\left(\Delta \boldsymbol{H}_f^0\right)\) of the compound formed in each of the reactions?
Answer:

Here, 2 mol of NH₃(g) is produced from the stable constituent elements, H2(g) and N2(g). Hence according to the definition, at 25°C the standard reaction enthalpy of this reaction is not equal to the standard enthalpy of formation of NH3(g).

The standard state of oxygen at 25°C is O₂(g). So, the given equation does not represent the formation reaction of NO(g). Consequently, at 25°C the standard reaction enthalpy of this reaction is not equal to die standard enthalpy offormation of NO(g).

Here, 1 mol of solid NaCl is formed from its stable constituent elements. Hence at 25°C, the die standard reaction enthalpy of this reaction is equal to the die standard enthalpy of formation of NaCl(s).

Question 8. Discuss the change in die degree of randomness for the following cases— Combustion of kerosene, Sublimation of dry ice, Extraction of salt from seawater, Condensation of water vapour, Crystallisation of a solid from its aqueous solution,
Answer: Combustion of kerosene: Kerosene is a mixture of liquid hydrocarbons. The combustion of kerosene produces C02 and water vapour. As in the reaction, a liquid converts into a gaseous mixture, the molecular randomness ofthe system increases.

Sublimation of dry ice: In dry ice, (solid carbon dioxide), the molecules of C02 exist in an orderly state. When this solid sublimes, the gaseous molecules formed move randomly, i.e., the degree of randomness ofthe molecules increases.

Extraction (Crystallisation) of salt from seawater: In seawater, the attractive forces between Na+ and Cl- ions are weak as the distance between these ions is large. Hence, these ions are virtually free to move randomly. NaCl extracted from the seawater is in a solid state with a crystal structure in which Na+ and Cl- are arranged in a definite order. Hence, the extraction of salt from seawater is associated with a decrease in the degree of randomness.

Condensation of water vapor: In water vapor, the intermolecular forces of attraction between H2O molecules are weak, so the molecules remain in a state of randomness. However, water obtained by condensation of water vapor has less freedom of motion and hence less degree of randomness because of the stronger intermolecular forces of attraction compared to water vapour. Hence, in the case of condensation of water vapour, the degree of randomness decreases.

Crystallization of a solid from its Aqueous solutions:

In an aqueous solution, the solute molecules exist in a state of random motion. When the solute is crystallised from its solution, the solute molecules in the die crystal remain at fixed positions and become almost motionless. So, the crystallization of a solid from its solution causes a decrease in randomness.

⇒ \(\mathrm{Cr}^{3+}+6 \mathrm{H}_2 \mathrm{O}(a q) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)\): In this process, one Cr3+ ion combines with six water molecules to form a single complex ion, [Cr(H20)6]3+. Consequently, the number of particles in the system reduces (from 7 to 1 for each combination), thereby decreasing the degree of randomness.

⇒ \(\mathrm{NH}_4 \mathrm{NO}_2(s) \xrightarrow{\Delta} \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}):\) On decomposition, 1 formula unit of solid ammonium nitrite produces 1 molecule of N2(g) and 2 molecules of water vapour i.e., H20(g). In this process, a solid (in which the molecules are orderly arranged) converts into gaseous substances. Furthermore, the number of molecules also increases. Naturally, this process increases the randomness ofthe system.

Class 11 Chemistry Solutions For Chemical Thermodynamics

Question 9. Mention if the entropy of the system increases or decreases In each of the following cases: Bolling of water, Sublimation of solid iodine,

⇒ \(\begin{aligned}
& 2 \mathrm{O}_3(g) \rightarrow 3 \mathrm{O}_2(g) \\
& \mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_3(g)+\mathrm{HCl}(g) \\
& \mathrm{Hg}(l) \rightarrow \mathrm{Hg}(g) \quad \text { (6) } \mathrm{I}_2(g) \rightarrow \mathrm{I}_2(s) \\
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \text { (8) } 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \\
& \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2(g) \\
& \mathrm{PCl}_5(g) \rightarrow \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
& \text { Haemoglobin }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \text { Oxyhaemoglobin } \\
& \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightarrow \mathrm{Ni}(\mathrm{CO})_4(g) \\
& 4 \mathrm{Fe}(s)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \\
& \mathrm{C}(\text { diamond }) \rightarrow \mathrm{C} \text { (graphite) } \\
&
\end{aligned}\)

Answer: The change in entropy in a phase transition or a reaction depends on the following factors.

For a given amount of substance, the entropy of the substance in different physical states varies in the order: of S_gas > S_liquid > S_soild where S is the molar entropy. Because of this, the entropy of the system increases in the phase changes such as liquid → vapour, solid → vapour, while it decreases in the phase changes such as liquid → solid, vapour → liquid and vapour → solid.

In a reaction, indie reactants are solids or liquids or in a dissolved state in solution and they convert into gaseous products, then the entropy of the die system increases, (b) If the number of gaseous particles (atoms or molecules) in a reaction increases or decrease, then the entropy of the system increaser or decreases.

Increases (liquid — vapour transition), Increases (solid-gas transition), Increases (the number of gaseous particles increases) Increases (gaseous substances are formed from a solid reactant) Increases (liquid — gas transition) Decreases (gas-solid transition) Increases (the number of gaseous particles increases).

Decreases (the number of gaseous molecules decreases) Increases (gaseous substance is formed through the reaction of the reactants, one of which is solid and the other is in a dissolved state in solution) Increases (the number of gaseous molecules increases) Decreases (the number of gaseous molecules decreases)

Class 11 Chemistry Solutions For Chemical Thermodynamics

Decreases (the number of gaseous molecules decreases)

Decreases (the number of gaseous molecules decreases) (g) Increases (the crystal structure of diamond is more compact than that of graphite. As a result, the molecular randomness in graphite is more than that in diamond.

Question 10. At 25°C and 1 atm pressure, for the reaction 3O₂(g) → 20₂(g); H = 286kJ and AS = -137.2 J.K-1. Is this reaction spontaneous? Does the spontaneity of this reaction depend on temperature? Is the reverse reaction spontaneous? If so, then why? Does the spontaneity of the reverse reaction depend on temperature?
Answer: No. (2) For the given reaction ΔH > 0 and ΔS < 0. So, according to the equation ΔG = ΔH-TΔS, AG will be positive at any temperature. Hence, the spontaneity of this reaction is independent of temperature.

The reverse reaction is spontaneous.

In the reverse reaction [20₃(g)→ 3O₂(g)] ; AH = —286 kj and AS = +137.2 J K- . So, according to the equation, AG = AH- TAS, AG is negative.

The spontaneity of the reverse reaction is also independent of temperature as AG is negative.

H<0 and S> 0 at any temperature.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Question 1. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer: In nature, there are three isotopes of hydrogen. These are protium \(\left({ }_1^1 \mathrm{H}\right) \text {, deuterium }\left[{ }_1^2 \mathrm{H} \text { or } \mathrm{D}\right] \text { and tritium }\left[{ }_1^3 \mathrm{H} \text { or } \mathrm{T}\right] \text {. }\)

The mass ratio of \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H} \text { is } 1: 2: 3 \text {. }\)

Question 2. Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer: A hydrogen atom has only one electron and thus, it has one electron less than the stable configuration of the nearest noble gas helium. Thus, to achieve a stable configuration it shares its single electron with the electron of another H -atom to form a stable diatomic molecule (H₂ ).

Read and Learn More Class 11 Chemistry

Question 3. Complete the following reactions:

1. \(\mathrm{H}_2(g)+\mathbf{M}_m \mathrm{O}_{\mathrm{o}}(s) \stackrel{\Delta}{\longrightarrow}\)
2. \(\mathrm{CO}(g)+\mathrm{H}_2(g) \frac{\Delta}{\text { catalyst }}\)
3. \(\mathrm{C}_3 \mathrm{H}_8(g)+3 \mathrm{H}_2 \mathrm{O}(g) \underset{\text { catalyst }}{\longrightarrow}\)
4. \(\mathrm{Zn}(s)+2 \mathrm{NaOH}(a q) \stackrel{\Delta}{\longrightarrow}\)

Answer:

1. \(o\mathrm{H}_2(g)+\mathrm{M}_m\mathrm{O}_o(s)\stackrel{\Delta}{\longrightarrow}m\mathrm{M}(s)+o\mathrm{H}_2\mathrm{O}(l)\)
2. \(\mathrm{CO}(g)+2 \mathrm{H}_2(g) \underset{\text { catalyst }}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH}(l)\)
3. \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \stackrel{\mathrm{Ni}, 1270 \mathrm{~K}}{\longrightarrow} 3 \mathrm{CO}(g)+7 \mathrm{H}_2(g)\)
4. \(\mathrm{Zn}(s)+2 \mathrm{NaOH}(a q) \stackrel{\Delta}{\longrightarrow} \underset{\text { Sodium zincate }}{\mathrm{Na}_2 \mathrm{ZnO}_2(a q)+\mathrm{H}_2(g)}\)

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 4. Discuss the consequences of high enthalpy of the H —H bond in terms of the chemical reactivity of dihydrogen.
Answer: H —H bond length is very small (74 pm) because the H-atom has a very small atomic size. Consequently, H—H bond dissociation enthalpy is very high (435.9 kJ-mol^-1) which makes hydrogen completely inert at ordinary temperature. However, at higher temperatures, the H—H bond undergoes dissociation in the presence of a catalyst to form hydrides with metals and non-metals.

Question 5. What characteristics do you expect from an electron-deficient hydride concerning its structure and chemical reactions?
Answer:

1. There is an insufficient number of electrons in the valence shell of the central atom. So, in this type of hydride, the valence shell of the central atom does not have a complete octet configuration. For example, in BH₃, the valence shell of B has six electrons and the hydrides are trigonal planar shape.
2. Due to electron deficiency, this type of hydrides act as Lewis acids, i.e., they accept electron pairs. For example,\(\mathrm{H}_3 \mathrm{~B}+\mathrm{NH}_3 \rightarrow \mathrm{H}_3 \mathrm{~B} \leftarrow: \mathrm{NH}_3\)
3. To compensate for the electron deficiency, the hydrides form dimers, trimers, and polymers and attain stability. for example B₂H₆,B₄H₁₀,(AIH₃)n etc.
4. Electron-deficient hydrides are extremely reactive. They easily react with both metals and non-metals.

Question 6. Do you expect the carbon hydrides of the (C_nH_2n+2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer: The hydrides of carbon of the type C_nH_2n+2 electron-precise hydrides, i.e., they have an exact number of electrons in the valence shell of the central atom so as to write their conventional Lewis structures. Therefore, they do not have a tendency to either gain or lose electrons and consequently, they do not act as Lewis acids or bases.

Question 7. What do you understand by the term “nonstoichiometric hydrides”? Do you expect this type of hydride to be formed by alkali metals? Justify your answer.
Answer:

Hydrides of d and f-block elements that are low in hydrogen, characterized by a fractional metal-to-hydrogen ratio, are termed non-stoichiometric hydrides.

• The alkali metals, due to their strong reducing properties, donate their lone pair of electrons to the hydrogen atom, resulting in the formation of hydride ions.
• As a hydride ion H- is generated through the complete transfer of an electron, the ratio of metal to hydrogen remains constant in these hydrides, resulting in compositions that reflect a straightforward whole-number ratio. Consequently, the alkali metals only produce stoichiometric hydrides.

Chemical Properties Of Hydrogen

Question 8. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:

• Certain transition metals, such as palladium (Pd) and platinum (Pt), adsorb significant quantities of hydrogen atoms on their surfaces, resulting in the formation of hydrides.
• The incorporation of hydrogen atoms causes the metal lattice to expand, resulting in decreased stability.
• Consequently, upon heating, these metal hydrides breakdown to liberate dihydrogen and revert to a finely split metallic state. The dihydrogen produced in this method can serve as a fuel.
• Consequently, transition metals or their alloys can be utilized for the storage and delivery of hydrogen as a fuel.

Question 9. Among NH₃, and H₂O midIIF, which would you expect to have the highest magnitude of hydrogen bonding & why?
Answer:

The strength of a hydrogen bond depends upon the atomic size mid the electronegativity of the atom to which the H-atom Is covalently linked. Smaller size and higher electronegativity of the other atom favor the formation of stronger H -bonding and that Is due to increase In magnitude of bond polarity. Among N, F, and O atoms, V has the lowest atomic size and the highest electronegativity. Hence the II — F bond is maximum polar and as a result, it will have the highest magnitude of H-bonding.

Question 10. Saline hydrides are known to react with water violently producing fire. Can CO₂, a well-known fire extinguisher, be used in this case? Explain.
Answer: The saline hydrides (For example, NaH, CaH₂, etc.), react with water violently to yield the corresponding metal hydroxides with the evolution of H₂ gas. The liberated H₂ gas undergoes spontaneous combustion causing fire and this is because of the highly exothermic nature of the combustion reaction.

⇒ \(\mathrm{NaH}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{H}_2(g) ;\)

⇒ \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

In this case, CO₂ cannot used as a fire extinguisher because its gels are reduced by the hot metal hydride to form formate ions.

⇒ \(\stackrel{-1}{\mathrm{NaH}}+\stackrel{+4}{\mathrm{CO}_2} \rightarrow \stackrel{+1+2}{\mathrm{H} C O O N a}\)

Question 11. Arrange the following:

1. CaH₂, BeH₂, and TiH₂ in order of increasing electrical conductance.
2. LiH, NaH, and CsH in order of increasing ionic character.
3. If —H, D—D, and F—F in order of increasing bond dissociation enthalpy.
4. NaH, MgH₂, and H₂O in order of Increasing and reducing properties.

Answer:

1. Being a covalent hydride BeH2 does not conduct electricity at all. Being an Ionic hydride CaH₂ conducts electricity in the fused state while TiH₂, being a metallic hydride, conducts electricity at room temperature, ‘t hus, the order of increasing electrical conductance is: BeH₂ < CaH₂ < TiH₂.
2. The electronegativity of the alkali metals decreases down the group from Li to Cs. Therefore, the ionic character of their hydrides also increases in the same order, l.e., LiH < NaH < CsH.
3. The bond dissociation enthalpy of the: F—F: bond is the lowest (242.6 kJ • mol^-1) and this is due to the high concentration of electron density around each F atom in the form of three unshared pairs which significant repulsive interactions. Again, because of the marginally smaller size of D as compared to H, the bond dissociation enthalpy of the D—D bond (443.35 kJ-mol^-1) is slightly higher than that of the H —H bond (435.88 kJ-mol-1). Hence, the bond dissociation enthalpy increases in the order: of F —F < H —H < D —D.
4. NaH, being an ionic hydride, is a more powerful reducing agent than the covalent hydrides MgH₂ and H₂O. MgH₂ is a stronger reducing agent than H20 because the bond dissociation enthalpy of the Mg—H bond is much lower than that of the O —H bond. Therefore, the reducing property increases in the order: of H₂O < MgH₂ < NaH.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 12. Compare the structures of H20 and H202.
Answer: The oxygen atom in water is sp³ -sp³-hybridized. The two O —H bonds are sp3-s sigma bonds. The H —O —H bond angle is 104.5°. This value is a little less than the tetrahedral angle (109°28/) because of stronger lone pair-lone pair and lone pair-bond pair repulsions than bond pair-bond pair repulsion. Thus, water is a bent molecule. Each oxygen atom in H₂O₂ is also sp3 hybridised. The O —0 bond is a sp³-sp³ sigma bond and the two O —H bonds are sp3-s sigma bonds. The two O —H bonds are, however, present in different planes. In the gas phase, the dihedral angle between the two planes (i.e., the planes containing H —O —O system) is 111.5°. So, the molecule has an open-booklike structure.

Question 13. What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer: Self-ionization of water is called auto-protolysis. Self-ionization of water can be expressed by the given equation

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \text { Acid-1 Base-2 } \quad \text { Acid-2 } \quad \text { Base-1 } \\ & \end{aligned}\)

Water exhibits amphoteric properties because of auto-protolysis. Thus water reacts with both acids and bases. It usually acts as a base in the presence of an acid stronger than it and acts as an acid in the presence of a base stronger than it. For example,

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{NH}_3(a q) \longrightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \begin{array}{llll} \text { Acid-1 } & \text { Base-2 } & \text { Acid-2 } & \text { Base-1 } \end{array} \\ & \end{aligned}\)

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{HCl}(a q) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\ & \text { Base-1 } \quad \text { Acid-2 } \quad \text { Acid-1 } \quad \text { Base-2 } \\ & \end{aligned}\)

Question 14. Consider the reaction of water with F2 and suggest, In terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

⇒ \( 2 \mathrm{~F}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{~F}^{-}(a q) Oxidant Reductant\)

⇒ \( 3 \mathrm{~F}_2(g)+3 \mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{O}_3(g)+6 \mathrm{H}^{+}(a q)+6 \mathrm{~F}^{-}(a q) Oxidant Reductant\)

In these reactions, water acts as a reductant and itself gets oxidized to oxygen or ozone. In this case, highly electronegative fluorine acts as an oxidant and gets reduced to F-.

Question 15. Complete the following chemical reactions.

1. \(\mathrm{PbS}(s)+\mathrm{H}_2 \mathrm{O}_2(a q) \rightarrow\)
2. \(\mathrm{MnO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}_2(a q) \rightarrow\)
3. \(\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow\)
4. \(\mathrm{AlCl}_3(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow\)
5. \(\mathrm{Ca}_3 \mathrm{~N}_2(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow\)

Classify the above into [a] hydrolysis, [b] redox, and [c] hydration reactions.

Answer:

⇒ \(\mathrm{PbS}(s)+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow \mathrm{PbSO}_4(s)+4 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\begin{array}{r}
2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{H}_2 \mathrm{O}_2(l)+6 \mathrm{H}^{+}(a q) \longrightarrow \\
2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l)+5 \mathrm{O}_2(g)
\end{array}\)

⇒ \(\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(g) \longrightarrow \mathrm{Ca}(\mathrm{OH})_2(a q)\)

⇒ \(\begin{array}{r}
\mathrm{AlCl}_3(g)+6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \\
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)}
\end{array}\)

⇒ \(\begin{array}{r}
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow} \\
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{OH})\right]^{2+}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)}
\end{array}\)

⇒ \(\mathrm{Ca}_3 \mathrm{~N}_2(s)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2(a q)+2 \mathrm{NH}_3(a q)\)

Reactions 1 and 2 are redox reactions. Reactions 3 and 4 are hydrolysis reactions. Reaction 5 is the hydration reaction.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 16. Is demineralized or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer: Although very pure, demineralized or distilled water is not useful for drinking purposes and this is because it does not contain even useful minerals. To make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralized or distilled water.

Question 17. Describe the usefulness of water in biosphere and biological systems.
Answer:

Water is essential for the survival of various life forms on Earth. It comprises around 65-70% of the body mass of flora and fauna.

• Due to its elevated specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant relative to other liquids, water is essential to the biosphere.
• The significant heat of vaporization of water enables it to regulate body temperature. Water also contributes indirectly to climate regulation.
• Water facilitates the transfer of minerals, nutrients, and enzymes throughout the body and influences metabolic processes.
• Water is an essential element in the process of photosynthesis. Consequently, rivers play a crucial function in the ecosystem.

Question 18. Knowing the properties of H20 and D20, do you think that D20 can be used for drinking purposes?
Answer: Drinking heavy water (D₂O) is harmful to the human beings. Heavy water being highly hygroscopic, absorbs water from different parts of the body. As a result, the body cells may get destroyed. Apart from this, heavy water slows down different biochemical reactions such as mitosis, cell division, etc.

Question 19. What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer: Hydrolysis refers to the reaction of salt with water to form acidic or basic solutions.

For example:

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { (basic solution) }}{2\left[\mathrm{Na}^{+}+\mathrm{OH}^{-}\right]+\mathrm{H}_2 \mathrm{CO}_3 ;}\)

⇒ \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { (acidic solution) }}{\left[\mathrm{H}^{+}+\mathrm{Cl}^{-}\right]+\mathrm{NH}_4 \mathrm{OH}}\)

Hydration, on the other hand, refers to the addition of H20 to ions or molecules to form hydrated ions or hydrated salts. For example:

⇒ \(\underset{\text { Salt }}{\mathrm{NaCl}(s)+\mathrm{H}_2 \mathrm{O}} \rightarrow \underbrace{\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)}_{\text {Hydrated ions }}\)

⇒ \(\begin{array}{cc} \mathrm{CuSO}_4(s)+5 \mathrm{H}_2 \mathrm{O}(l) & \rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(s) \\ \text { Anhydrous salt } & \text { Hydrated salt } \\ \text { (colourless) } & \text { (blue) } \end{array}\)

Question 20. How can saline hydrides remove traces of water from organic compounds?
Answer: Saline hydrides (i.e., NaH, CaH₂, etc.) react with water forming their corresponding metal hydroxides with the liberation of dihydrogen. Thus, traces of water present in organic solvent can be easily removed by distilling them over saline hydrides when dihydrogen escapes into the atmosphere, metal hydroxides are left in the distillation flask, and the dry organic solvent is distilled over.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 21. What do you expect the nature of hydrides to be if formed by elements of atomic numbers 15, 19, 23, and 44 with dihydrogen? Compare their behavior towards water.
Answer:
The element with Z = 15 is non-metal phosphorus and hence it forms the covalent hydride PH₃.

The element with Z = 19 is the alkali metal potassium and hence it forms the saline or ionic hydride K+HT.

The element with Z = 23 is the transition metal vanadium of group-3 and hence it forms the interstitial hydride (VHj 6).

The element with Z = 44 is the transition metal ruthenium (Ru) belonging to group 8. It does not form anhydride (hydride gap). Only the ionic hydride, KH reacts with water evolving dihydrogen.

⇒ \(\mathrm{KH}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{H}_2(g)\)

Question 22. Do you expect different products in solution when aluminum (III) chloride and potassium chloride are treated separately with

Normal water

Acidified water and

Alkaline water? Write equations wherever necessary.

Answer: In normal water: KC1, being a salt of a strong acid and strong base, does not undergo hydrolysis in normal water. It simply dissociates to form K+(aq) and Cl~(aq) ions.

⇒ \(\mathrm{KCl}(s) \stackrel{\text { water }}{\longrightarrow} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

On the other hand, A1C13, being a salt of a weak base Al(OH)3 and a strong acid HC1, undergoes hydrolysis giving acidic solution.

⇒ \(\mathrm{AlCl}_3(s)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Al}(\mathrm{OH})_3(s)+3 \mathrm{H}^{+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

In acidified water: The H+ ions react with Al(OH)3 to form Al3+(aq) ions and H20. Therefore, in acidic water, A1C13 exists as A13+(aq) and Cl~(aq) ions.

⇒ \(\mathrm{AlCl}_3(s) \stackrel{\text { acidified } \mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Al}^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

⇒ \(\mathrm{KCl}(s) \stackrel{\text { acidified water }}{\longrightarrow} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

In alkaline water: A1(0H)3 reacts with OH ions to form soluble tetrahydroxoaluminate complex or meta-aluminateion (A10ÿ2).

KC1 does not react and dissociate to give K+(aq) and Cl~(aq) ions.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Question 1. What is ‘hydrogenite’? Mention its use.

Answer: A mixture of silicon, caustic soda, and slaked lime is called hydrogenite. Dihydrogen can be prepared by heating hydrogenate with water.

⇒ \(\mathrm{Si}+2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{SiO}_3+\mathrm{H}_2 \uparrow\)

∴ \(\mathrm{Si}+\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CaSiO}_3+2 \mathrm{H}_2 \uparrow\)

Question 2. What is denoted by [Hg04]+ ion? Explain

Answer: High charge density and high hydration energy of the H⁺ ion (proton) cause the H⁺ ion to remain solvated to form a hydroxonium ion or hydronium ion (H30+). Hydronium ion again forms hydrogen bonds with three water molecules and remains solvated. Therefore, in water, a proton forms [H₃O(H₂O)₃]⁺ or [H₉O₄]⁺ ion.

Question 3. Pure para-hydrogen is available but not pure ortho hydrogen. Explain.

Answer: At ordinary temperature, ordinary hydrogen is a mixture of 75% of ortho and 25% of para-isomer. With a decrease in temperature, the amount of ortho-hydrogen decreases while that of para-hydrogen increases. At 20K, pure para-hydrogen is obtained. As para-hydrogen is more stable, it is found in the pure form. However, if the temperature is increased, the amount of ortho-isomer of dihydrogen increases but at 400K or above, the ratio of ortho-and para-isomer is fixed (3: 1). Therefore, pure para-hydrogen is available but not pure ortho-hydrogen.

Question 4. How many hydrogen-bonded water molecule(s) are associated with CuS0₄-5H₂0?

Answer: Only one molecule of water, which remains outside the brackets (coordination sphere) is linked by a hydrogen bond to SO as shown below. The remaining four water molecules are associated to Cu²⁺ ion by coordination bonds.

Question 5. Write down the reaction between H₂OZ and hydrazine (NH₂NH₂) in the presence of Cu(II) catalyst. Mention the use of this reaction.

Answer: A highly concentrated solution (about 40%) of H₂0₂ (called high test peroxide) oxidizes hydrazine (N₂H₄) in the presence of Cu(II) into nitrogen gas, itself being oxidized to water (steam).

⇒ \(\stackrel{-2}{\mathrm{~N}_2} \mathrm{H}_4(l)+2 \mathrm{H}_2{ }_{-1}^{-1}(l) \underset{\text { catalyst }}{\stackrel{\mathrm{Cu}(\mathrm{II})}{\longrightarrow}} \stackrel{0}{\mathrm{~N}}(\mathrm{~g})+4 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}}(g)+\text { heat }\)

The reaction is highly exothermic and is accompanied by a large increase in volume which in turn generates high pressure. Due to this, the reaction is employed for propelling rockets.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Hydrogen Multiple Choice Questions And Answers

Question 1. The normality of’30 volume’ of H₂O₂ is—

1. 2.678 (N)
2. 5.336 (N)
3. 8.034 (N)
4. 6.685 (N)

Answer: 2. 5.336 (N)

Volume strength = 5.6 x normality

or, 30= 5.6 x normality

⇒ \(\text { or, normality }=\frac{30}{5.6} \mathrm{~N}=5.357 \mathrm{~N}\)

The normality of 30 volumes of H₂0₂ is 5.357N.

Question 2. A commercial sample of H₂0₂ is labeled as 10V. Its % strength is nearly

1. 3
2. 6
3. 9
4. 12

Answer: 1. 3

10V (10 volume) H₂0₂ means that lmL of H₂0₂ solution will produce lOmL of 02 at STP. Now, the decomposition of H₂O₂ is as below:

⇒ \(\begin{array}{lr} 2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow & 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ (2 \times 34) \mathrm{g} & 22400 \mathrm{~mL} \\ =68 \mathrm{~g} & \text { (at STP) } \end{array}\)

At STP, lOmL of 0₂ is obtained from lmL of H₂0₂

⇒ \(22400 \mathrm{~mL} \text { of } \mathrm{O}_2 \text { is obtained from } \frac{22400}{10} \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2\)

= 2240 mL of H₂0₂

2240mL of H₂0₂ solution contains 68g of H₂0₂

∴ \(100 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains } \frac{68 \times 100}{2240}=3.03 \mathrm{~g}\) of H202 = 3g of H202

The percent strength of 10V H₂O₂ is 3

Class 11 Chemistry Chemical Properties Of Hydrogen Question 3. In aqueous alkaline solution, two-electron reduction of HO₂ gives

1. HO
2. H₂O
3. 0₂
4. O₂–

Answer: 1. Ho

Question 4. Which statement is not correct for ortho- and para-hydrogen

1. They have different boiling points
2. Ortho-form is more stable than para-form
3. They differ in their nuclear spin
4. The ratio of ortho to para-hydrogen changes with a change in temperature

Answer: 2. Ortho-form is more stable than para-form

At normal or high temperatures, ortho-hydrogen is more stable than para-hydrogen but at very low temperatures para-hydrogen is more stable than orthohydrogen.

Question 5. At room temperature, the reaction between water and fluorine produces—

1. HF and H₂0₂
2. HF,0₂ and F₂0₂
3. F©, 0₂ and H®
4. HOF and HF

Answer: 3. F©, 0₂ and H®

Question 6. Which one of the following statements about water is false

1. Water can act both as an acid and as a base
2. There is extensive intramolecular hydrogen bonding in the condensed phase
3. Ice formed by heavy water sinks in normal water
4. Water is oxidized to oxygen during photosynthesis

Answer: 2. There is extensive intramolecular hydrogen bonding in the condensed phase

Water molecules are associated with the formation of intermolecular hydrogen bonding.

Question 7. Hydrogen peroxide oxidises [Fe(CN)₆]₄– to [Fe(CN)₆]₃– in acidic medium but reduces [Fe(CN)₆]₃~ to [Fe(CN)₆]₄– in alkaline medium. The other products formed are, respectively

1. H₂0 and (H₂0 + 0₂)
2. H₂0 and (H₂0 + OH”)
3. (H₂0 + 0₂) and H₂0
4. (H₂0 + 0₂) and (H₂0 + OH-)

Answer: 1. H₂0 and (H₂0 + 0₂)

Acidic medium:

Question 8. Which of the following statements about hydrogen is incorrect

1. Hydrogen has three isotopes of which tritium is the most common
2. Hydrogen never acts as a cation in ionic salts
3. Hydronium ion, H₃0 exists freely in solution
4. Dihydrogen does not act as a reducing agent

Answer: 4. Dihydrogen does not act as a reducing agent

Among three isotopes of hydrogenprotlum (H1), deuterium H₂), and tritium (H₃), normal hydrogen is protium. Among these three, tritium is radioactive and unstable for which it is negligible in amount. In an aqueous solution, metal is reduced by dihydrogen. Again, metallic oxide is also reduced by dihydrogen to free metal.

Question 9. In ice, the oxygen atom is surrounded

1. Tetrahedrallyby 4 hydrogen atoms
2. Octahedrallyby 2 oxygen and 4hydrogen atoms
3. Tetrahedrallyby 2 hydrogen 2 oxygen atoms
4. Octahedrallyby 6 hydrogen atoms.

Answer: 1. X-ray studies have shown that in ice, four hydrogen atoms tetrahedrally surround each oxygen atom.

Class 11 Chemistry Chemical Properties Of Hydrogen MCQs Question 10. Predict the product of the reaction of I₂ with H₂0₂ in a basic medium.

1. I-
2. I_2O3
3. I0₃
4. I₃

Answer: 1. I-

Question 11. Strength of H₂0₂ is 15.18 g.L-1 , then it is equal to—

1. 1 volume
2. 10 volume
3. 5 volume
4. 7 volume

Answer: 3. 5 Volume

⇒ \(\text { Volume strength }=\frac{5.6 \times \text { strength in } \mathrm{g} \cdot \mathrm{L}^{-1}}{\text { equivalent mass of } \mathrm{H}_2 \mathrm{O}_2}\)

∴ \(=\frac{5.6 \times 15.18}{17}=5 \text { volumes }\)

Question 12. Which of the following reactions increases the production of dihydrogen from synthesis gas—

1. \(\mathrm{CH}_4(g)+\mathrm{H}_2 \mathrm{O}(g)\stackrel{\mathrm{1270K}}{\mathrm{Ni}} \mathrm{CO}(g)+\mathrm{H}_2(g)\)
2. \(\mathrm{C}(g)+\mathrm{H}_2 \mathrm{O}(g) \stackrel{1270 \mathrm{~K}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{H}_2(g)\)
3. \(\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \underset{\text { Catalyst }}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(g)}+\mathrm{H}_{2(g)}\)
4. \(\mathrm{C}_2 \mathrm{H}_6(g)+2 \mathrm{H}_2 \mathrm{O}(g) \underset{\mathrm{Ni}}{\stackrel{1270 \mathrm{~K}}{\longrightarrow}} 2 \mathrm{CO}+5 \mathrm{H}_2\)

Answer: 3. \(\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \underset{\text { Catalyst }}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(g)}+\mathrm{H}_{2(g)}\)

The production of dihydrogen can be increased by reacting carbon monoxide of syngas with steam in the presence ofiron chromate as catalyst.

⇒ \(\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2\mathrm{O}_{(\mathrm{g})} \underset{\text { Catalyst}}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}\)

Question 13. Which of the following reactions produces hydrogen

1. Mg + H₂0
2. H₂S₂O₈ + H₂O
3. Ba0₂ + HCl
4. Na₂0₂ + 2HC1

Answer: 1. Mg + H₂0

Alkali and alkaline earth metals react with water to produce hydrogen gas and metal hydroxides. This occurs due to high electropositive character ofthe metals.

⇒ \(\mathrm{Mg}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow\mathrm{Mg}(\mathrm{OH})_2+\mathrm{H}_2\)

Question 14. H₂0₂ can be obtained when following reacts with H₂S0₄ except with

1. Ba0₂
2. Pb0₂
3. Na₂0₂
4. SrOz

Answer: 2. Pb0₂

H₂0₂ is prepared by the reaction of peroxide with H₂S0₄.Pb0₂ is a dioxide. Hence, it does not give H₂0₂ with dilute H₂S0₄.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Very Short-Type Questions

Question 1. Which is the lightest gas known?
Answer: Dihydrogen

Question 2. Which isotope of hydrogen is radioactive?
Answer: Tritium (jH)

Question 3. Give examples of an ionic, a covalent, and a metallic hydride.
Answer: CaH₂ (ionic); NH₃ (covalent) & CrH (metallic)

Question 4. What is hydrolith?
Answer: Calcium hydride (CaH₂)

Chemical Properties Of Hydrogen

Question 5. Name the two nuclear spin isomers of dihydrogen.
Answer: ortho- hydrogen and

Question 6. Give an example of an electron-deficient hydride in which three centre-two electron bonds are present.
Answer: para- hydrogen; 6. B₂Hg

Question 7. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer: Carbon monoxide;

Question 8. How will you prove that a colorless liquid is water?
Answer: 1. White anhydrous CuS04 becomes blue in contact with water

Question 9. What is the unit for expressing the degree of hardness of water?
Answer: ppm (parts per million);

Question 10. Write the names of two chemical substances that are used for removing dissolved oxygen from water meant for the boiler.
Answer: Hydrazine (NH₂NH₂) and sodium sulphite (Na₂S0₃);

Question 11. Why is heavy water used in atomic reactors?
Answer: It is used as a moderator

Question 12. Name a solid and a liquid absorbent of water.
Answer: Cone. H₂S0₄ and P₂0₅;

Question 13. Which chemical is commercially known as ‘perhydrol’?
Answer: H₂0₂ solution;

Question 14. What is called ‘hyper lol or horizon’?
Answer: 2-ethylanthraquinol;

Chemical Properties Of Hydrogen Question 15. What is the die volume strength of a molar solution of H₂0₂?
Answer: A compound of hydrogen peroxide and urea is called hyper lol (NH₂C0NH₂-H₂0₂);

Question 16. Which organic reagent is used for the manufacture of H₂O₂?
Answer: 2-ethylanthraquinol;

Question 17. 10 volume of H₂0₂ = x(N)H₂0₂ .What is the value of x?
Answer: 17. X = 56; 18.

Question 18. What are how water molecules are bonded to the anhydrous salt to form hydrates?
Answer: Coordinate bond and H bond

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen Short-Type Questions

Question 1. How can dihydrogen be obtained from nitric add?
Answer: By the reaction of Mg with HN0₃(2%)

Question 2. Concentrated H₂S0₄ cannot be used for drying H₂ gas why?
Answer: Cone. H₂S0₄ on absorbing H₂0 from moist H₂ produces so much heat that hydrogen catches fire

Chemical Properties Of Hydrogen Question 3. What type of hydrides can be formed by each of the following elements: Li, Zr, P, Hf, N, Ca?
Answer: Ionic, metallic, covalent, metallic, covalent, and ionic respectively.

Question 4. What are the different types of bonds formed by hydrogen in its compounds?
Answer: Ionic, covalent and H-bond

Question 5. In the preparation of H₂0₂, MnOz or Pb0₂ cannot be used instead of Ba0₂ —why?
Answer: Ba0₂ contains peroxo linkage ( —O —O — ), but Mn0₂(0=Mn=0) or Pb0₂(0=Pb=0) does not contain peroxo linkage.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen Numerical Problems

Question 1. 1L of a hard water sample contains 1 mg CaCl₂ and 1 mg MgCl₂. Estimate the degree of hardness of this sample of water.
Answer: Molecular mass of CaCl₂ =111

Nowlllg of CaCl₂= lOOg of CaC0₃

⇒ \(1 \mathrm{mg} \mathrm{CaCl}_2 \equiv \frac{100 \times 0.001}{111} \mathrm{~g}^2 \text { of } \mathrm{CaCO}_3\)

= 9 X 10^-4g of CaC0₂ = 0.9mg of CaC0₃

The molecular mass of MgCl₂ = 95

Now, 95g of MgCl₂=100g of CaC0₃

⇒ \(1 \mathrm{mg} \text { of } \mathrm{MgCl}_2 \equiv \frac{100}{95} \mathrm{mg} \text { of } \mathrm{CaCO}_3=1.05 \mathrm{mg} \text { of }\)CaC03

Equivalent amount of CaC03 corresponding to CaCl₂

and MgCl₂ presentin or 103g of hard water

= (0.90 + 1.05)mg = 1.95mg [1L water = 103g = 106mg water]

The mass of the equivalent amount of CaC03 corresponding to CaCl2 and MgCl2 present in 106mg of water is 1.95 mg. Hence, the degree of hardness of the given sample is 1.95 ppm.

Hence, the hardness of that sample of water is 75 ppm.

Chemical Properties Of Hydrogen

Question 2. Determine the strength of ’30 volume’ H₂0₂ in normality.
Answer: Volume strength = 5.6 x normality

or, 30 = 5.6 x normality

⇒ \(\text { or, normality }=\frac{30}{5.6} \mathrm{~N}=5.35 \mathrm{~N}\)

The normality of 30 volume of H₂0₂ is 5.35N

Question 3. Determine the volume (in liter) of 0₂ obtained at STP when 0.1 liter of 2(M)H₂O₂ solution is decomposed.
Answer:

Now, 1L1(M) 1I₂0₂ = 34g of H₂0₂

1L₂(M) H₂0₂ → (2 X 34)g of H₂0₂

0.1L 2(M) H₂0₂ = (2 X 34 X 0.1 )g of H₂0₂ = 6.8g of H₂O₂

Again at STP, 68g of M₂0₂ produces 22.4L of 0₂.

⇒ \(6.8 \mathrm{~g} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { produces } \frac{22.4}{68} \times 6.8 \mathrm{~L} \text { of } \mathrm{O}_2\)

= 21241, of 0₂

Question 4. When 100 ml of tube-well water is titrated using methyl orange as an indicator, it requires 15 ml of 0.01 (N) HCl. Estimate the hardness of that sample of water.
Answer: 15 mL 0.01 (N) HCI = 1 mL 0.15(N) HCI

⇒ \(1000 \mathrm{~mL} 1(\mathrm{~N}) \mathrm{HCl} \equiv \frac{100}{2} \mathrm{~g} \mathrm{CaCO}_3\)

⇒ \(1 \mathrm{~mL} 0.15(\mathrm{~N}) \mathrm{HCl}=50 \times \frac{1}{1000} \times 0.15 \mathrm{~g} \text { of } \mathrm{CaCO}_3\)

= 7.5x 10^-3g of CaCO₃

Therefore, 100ml, or 100g of that sample of water contains some hardness-producing substance which is equivalent to 7.5 x 10^-3g of CaC0₃. 106g of water contains the hardness producing substance equivalent to\(\frac{7.5 \times 10^{-3} \times 10^6}{100}=75 \mathrm{~g} \text { of } \mathrm{CaCO}_3\)

Chemical Properties Of Hydrogen

Question 5. Calculate the amount of H₂0₂ present in 600 mL of 10-volume H₂0₂ solution.
Answer: 10 volume H₂0₂ means that 1 mL of H₂O₂ solution will produce 10 mL of 0₂ at STP

⇒ \(\begin{gathered} 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ \left(2^{\prime} 34\right) \mathrm{g}=68 \mathrm{~g} \quad 22400 \mathrm{~mL} \text { (at STP) } \end{gathered}\)

At STP, 10 mL 0₂ is obtained from 1 mL of lOvol H₂0₂ solution.

⇒ \(22400 \mathrm{~mL} \text { of } \mathrm{O}_2 \text { is obtained from } \frac{22400}{10} \mathrm{~mL} \text { of } 10 \mathrm{vol}\)

H₂0₂ solution =2240 mL of H₂0₂

2240 mL of H₂0₂ solution contains 68g of H₂0₂

⇒ \(100 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains } \frac{68 \times 100}{2240}=3.03 \mathrm{~g} \text { of }\)H2O2.

⇒ \(\text { So, } 600 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains }=\frac{3.03 \times 600}{100} \mathrm{~g} \text { of }\)

H₂0₂ = 18.18g of H₂0₂ =18.2g of H₂0₂

Chemical Properties Of Hydrogen

Question 6. An excess of acidic KI solution is added to 25 mL of a H₂0₂ solution when iodine is liberated. 20 mL of 0.1(N) sodium thio-sulfate solution is required to titrate the liberated iodine. Calculate the percentage strength, volume strength, and strength in normality of the H₂0₂ solution.
Answer: 25 x ;e(N) = 20 x 0.1(N)

⇒ \(x=\frac{20\times0.1}{25}=\frac{2}{25}=0.08(\mathrm{~N})\)

Amount of H₂0₂ in 25mL0.08(N) II₂0₂ solution

1000 mL 1(N) H₂0₂ solution = 17g H₂0₂

⇒ \(\begin{aligned} & 25 \mathrm{~mL} 0.08(\mathrm{~N}) \mathrm{H}_2 \mathrm{O}_2 \text { solution } \equiv \frac{17 \times 25 \times 0.08}{1000}=0.034 \mathrm{~g} \\ & \mathrm{H}_2 \mathrm{O}_2 \end{aligned}\)

⇒ \(100 \mathrm{ml} 0.08(\mathrm{~N}) \mathrm{H}_2 \mathrm{O}_2 \text { solution contains }=\frac{0.034 \times 100}{25}\)

= 0.136g H₂O₂

% strength of H₂0₂ solution = 0.136

68g H₂0₂ gives → 22.4 L 0₂ at STP

⇒ \(0.034 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2 \text { solution } \rightarrow \frac{22.4 \times 0.034}{68} \text { Litre } \mathrm{O}_2 \text { at STP }\)

= 0.0112 Litre 0₂ at STP

= 11.2 mL 0₂ at STP

25 mL H₂O₂ solution gives 11.2 mL 0₂ at STP.

⇒ \(\text { ImL } \mathrm{H}_2 \mathrm{O}_2 \text { solution gives } \frac{11.2}{25}=0.448 \mathrm{~mL} \mathrm{O}_2 \text { at STP. }\)

Volume strength of the given H₂0₂ solution = 0.448

Question 1. Why is the charge of an electron considered the smallest measurable unit of electricity?
Answer:

American scientist, R. A. Millikan determined the charge of an electron by his famous classical oil drop experiment value was found to be 4.8 x 10_1° esu or 1.602 x 10-19 C. No other particles carry a lesser negative charge.

This is the smallest measurable quantity of charge. The electric charge carried by positively or negatively charged particles is integral multiples of this minimum possible quantity.

Hence, the charge of an electron is considered the smallest unit of electricity.

Question 2. Estate two differences between cathode & anode rays.
Answer:

Cathode rays are composed of negatively charged particles whereas positively charged particles constitute anode rays.

The ratio of charge and mass (e/m ) of the constituent of cathode rays is always constant and it does not depend on the nature of the gas in the discharge tube and the cathode used.

On the other hand, the ratio of charge and mass (e/m ) of the particle constituting the anode rays is not definite. It assumes different values if different gases are used in the discharge tube.

Read and Learn More Class 11 Chemistry

Question 3. Calculate the mass and charge of 1 mol electrons.
Answer:

Charge carried by 1 electron = 1.602 x 10-19 coulomb i.e., the charge associated with 1 mol electron.

=1.602 ×10^-19 × 6.022 × 10²³ coulomb

= 9.647 × 10⁴ coulomb

Mass of an electron = 9.108 ×  10^-28 g

Mass of1mol electrons = 9:108 × 10^-28 ×  6.022×10²³

= 5.485 × 10²⁴ g

Question 4. Compare the values of e/m of electron and proton.
Answer:

If the mass and charge of an electron are mx and ex respectively and the mass and charge of a proton are m2 and e2 respectively, then.

⇒ \(\frac{(\mathrm{e} / \mathrm{m}) \text { of electron }}{(\mathrm{e} / \mathrm{m}) \text { of proton }}-\frac{e_1 / m_1}{e_2 / m_2}=\frac{e_1}{\mathrm{e}_2} \times \frac{m_2}{m_1}=\frac{m_2}{m_1}=1837\)

[As both electron and proton carry a unit charge and a proton is 1837 times heavier than an electron.]

e/m ofelectron = 1837 x e/m of proton.

Structure of Atom Class 11 Questions and Answers

Question 5. If an electron is promoted from the first orbit to the
the third orbit of a hydrogen atom, by how many times
will the radius of the orbit be increased?
Answer:

The radius of the orbit of H -atom, rn \(=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

Radius of the first orbit (r1) \(=\frac{1^2 \times h^2}{4 \pi^2 m e^2}\)

and radius of the third orbit (r3) \(=\frac{3^2 \times h^2}{4 \pi^2 m e^2}\)

∴ \(\frac{r_3}{r_1}=\frac{3^2 \times h^2}{4 \pi^2 m e^2} \times \frac{4 \pi^2 m e^2}{h^2}=9 \text { i.e., } r_3=9 \times r_1\)

∴ If the electron moves from the 1st orbit to the 3rd orbit, the radius of the orbit will be increased 9 times

Question 6. The ionization energy for the H-atom in the ground state is x1-atom-1. Show that the energy required for the process, He+(g) — He2- + e, is 4x J-atom-1.
Answer:

The energy of the electron in the n-th orbit of an H-like particle is

⇒ \(\text { [where } \left.K=\frac{2 \pi^2 m e^4}{h^2}=\text { constant }\right]\)

⇒ \(\text { Now, } \begin{aligned}
I . E_{\mathrm{H}} & =E_{\propto}-E_1=0-\left(-K \times \frac{1^2}{1^2}\right)=K \\
& =x \mathrm{~J} \cdot \text { atom }^{-1}
\end{aligned}\)

The energy required for the given process is the ionization energy of He+.

Since He+ is a hydrogen-like species, so

⇒ \(\begin{aligned}
L . E_{\mathrm{He}^{+}} & =E_\alpha-E_1=0-\left(-K \times \frac{2^2}{1^2}\right) \quad\left[\text { for } \mathrm{He}^{+}, z=2\right] \\
& =4 K=4 x \mathrm{~J} \cdot \mathrm{atom}^{-1}
\end{aligned}\)

Question 7. What does the statement “electronic orbits at stationary state” mean? Does an electron remain stationary in a stationary orbit? Explain.
Answer: According to Bohr’s theory of the hydrogen atom, electrons revolve around the nucleus in some fixed orbits and during its motion, an electron does not lose energy.

For this reason, these orbits are known as “electronic orbits at stationary states”.

When an electron stays in such an orbit, it does not remain stationary at all. Had it been so, the electron, being attracted by the nucleus would have fallen on the nucleus.

The electron always remains in motion so as to overcome the influence of the nuclear attractive force.

Structure of Atom Class 11 Questions and Answers

Question 8. If the radius of the first Bohr orbit is x, then find the de Broglie wavelength of the electron in the third orbit.
Answer: The radius of the the n -th orbit \(r_n \propto n^2\)

∴ r3 = 9x r1 = 9x

Since [r1 =x]

Now, the circumference of the n-th orbit is an integral multiple of the wavelength associated with the motion of the electron, i.e.

2nrn = n x A

∴ 2nr3 = 3 x A

Or, \(\lambda=\frac{2 \pi r_3}{3}=\frac{2 \pi}{3} \times 9 x=6 \pi x\)

Question 9. A photon of wavelength A collides with an electron. After the collision, the wavelength of the photon changes to A’. Calculate the energy of the scattered electron
Answer: Energy ofthe photon before collision \(=h v=\frac{h c}{\lambda}\)

Energy ofthe photon after collision \(=h v^{\prime}=\frac{h c}{\lambda^{\prime}}\)

The difference in energy is imparted to the electron, which suffers collision. Hence energy ofthe scattered electron.

⇒ \(=\frac{h c}{\lambda}-\frac{h c}{\lambda^{\prime}}=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)

Question 10. Find the condition under which the de Broglie wave¬ length of a moving electron becomes twice that of a moving proton. Given that a proton is 1836 times heavier than an electron.
Answer: \(\lambda_e=\frac{h}{m_e v_e} \text { and } \lambda_p=\frac{h}{m_p v_p}\)

⇒ \(\text { Since } \lambda_e=2 \times \lambda_p\)

∴ \(\frac{h}{m_e v_e}=2 \times \frac{h}{m_p v_p}\)

or, \(2 \times m_e v_e=m_p v_p\)

Or, \(v_e=\frac{m_p v_p}{2 m_e}=\frac{v_p}{2} \times \frac{m_p}{m_e}=\frac{v_p}{2} \times 1836=918 v_p\)

i.e., vg = 9lSvp, this is the necessary condition.

Structure of Atom Class 11 Questions and Answers

Question 11. If the kinetic energy of an electron increases by nine times, the wavelength of the de Broglie wave associated with it will increase by how many times
Answer: The de Broglie wavelength associated with a moving electron ofmass m and kinetic energy E is given by

⇒ \(\lambda=\frac{h}{\sqrt{2 E m}}\)

⇒ \(\text { Now } \lambda_1=\frac{h}{\sqrt{2 E m}}, \lambda_2=\frac{h}{\sqrt{2 \times 9 E \times m}}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}\)

∴ \(\frac{\lambda_2}{\lambda_1}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}+\frac{h}{\sqrt{2 E m}}\)

or, \(\lambda_2=\frac{1}{3} \lambda_1\)

So, de Broglie wave associated with the electron will increase by \(\frac{1}{3}\) times.

Question 12. Mention three differences between wave and partide.
Answer: Differences between wave and partied

Question 13. Discuss the limitation of Bohr’s theory based on Heisenberg’s uncertainty principle.
Answer:

• Bohr’s theory posits that negatively charged particles (electrons) within an atom orbit the nucleus in precisely defined paths with predetermined radii. To counterbalance the nuclear attraction force, electrons must exhibit a specific velocity.
• According to Heisenberg’s uncertainty principle, it is impossible to simultaneously ascertain the precise position and momentum (or velocity) of a minuscule particle such as an electron. Consequently, Bohr’s model is in conflict with Heisenberg’s uncertainty principle.

Question 14. What is the difference between the notations L & Z
Answer:

Represents the second Bohr orbit for which n = 2. On the other hand, l denotes azimuthal quantum number which may have values 0, 1, 2, etc.

Shapes of various subshells present within the same principal shell, the number of sub-shells in a ‘certain shell, and their relative energies, etc. depend on the values of’ l ’.

Structure of Atom Class 11 Questions and Answers

For example, in the case of s, p, d, and /-subshells, the values of l are 0, 1, 2, and 3 respectively.

Question 15. How many quantum numbers are required to identify an orbital? Explain it with the help of a specific example.
Answer:

To identify an orbital, three quantum numbers need to be mentioned.

These are the principal quantum number (n), azimuthal quantum number (Z), and magnetic quantum number (m).

For example, in order to denote 2px, 2py and 2Pz > the respective values of the said quantum numbers are—

1. 2px: n = 2, l = 1, m = -1
2. 2py: n = 2, l = 1, m = 0
3. 2pz: n = 2, l = 1, m = +1

Incidentally, it may be noted that the values of all four quantum numbers need to be mentioned to specify a particular electron in any atom.

Question 16. Give the electronic configuration of 24Cr3+. Find the no. of unpaired electrons present in its ion.
Answer: 24Cr: ls22s22p63s23p63d54s1

Question 17. Write down the electronic configurations of Ni and Ni2+ (atomic number of Ni = 28). How many odd electrons do each of them contain?
Answer: 28Ni: ls22s22p63s23p63d84s2

∴ Number of odd electrons = 2;

Electronic configuration of Ni2+: ls22s22p63s23p63d84

∴ A number of odd electrons present in Ni2+ ion 2.

Structure of Atom Class 11 Questions and Answers

Question 18. Which orbit of the Be3+ ion has the same radius as that of the ground state of hydrogen atoms?
Answer: For hydrogen-like atoms or ions, the radius of the n-th orbit is given by, rn \(r_n=0.529 \times \frac{n^2}{Z}\)

In the ground state of H-atom, n =1 (1st orbit) and Z(at. no.) =1

∴ Radius of 1st orbit of H-atom \(r_1=0.529 \times \frac{1^2}{1}=0.529 \)

Let the n -th orbit of Be3+ ion has the same radius as that of the ground state of H-atom.

Now, rn \(r_n\left(\mathrm{Be}^{3+} \text { ion }\right)=0.529 \times \frac{n^2}{4}\) \(=0.529 \times \frac{n^2}{4}\)

Thus, \(0.529 \times \frac{n^2}{4}=0.529\)

or, n2 = 4 or, n = 2

Hence, second orbit of the Be3+ ion has the same radius as that of the ground state of the H-atom.

Question 19. Explain why Fe3+ is more stable than Fe2+ ion.
Answer: Atomic number of Fe-atom =26 and its electronic configuration: ls22s22p63s23p63rf64s2

Electronic config. of Fe2+: lsa2s22p63s23p63d6

and electronic config. of Fe3+ : ls22s22p63s23p63rf5

We know that the electronic configurations of halt-filled or fully-filled subshells are more stable compared to others.

Now, the 3d -subshell of Fe3+ ion is exactly half-filled, and hence Fe3+ is more stable than Fe2.

Structure of Atom Class 11 Questions and Answers

Question 20. The electronic configuration of the electrons in the outer shells of Cu & Cr are 3d104s1 & 3ds4s1 respectively instead of being 3d94s2 and 3d44s2 explain why.
Answer: Half-filled and fully-filled subshells have greater stability. In the case of Cu, the 3d104s1 configuration is more stable than 3d94s2 because in 3d104s1, the 3d -subshell is completely filled, and the 4s -subshell is half-filled.

Similarly, in the case of Cr, the 3d54sx configuration contains half-filled 3d -and 4s -subshells, and hence this will be more stable than the 3d44s2 configuration.

Question 21. All atoms having an even number of electrons always contain paired electrons—is the statement true? If so, then by what principle or rule the above statement can be explained?
Answer:

All the electrons, present in atoms with an even number of electrons, do not necessarily get paired. This can be seen from the electronic configuration of carbon (atomic number 6): ls²2s²2p

In spite of having an even number of electrons (total 6 electrons) the 2p -orbitals of carbon atoms have 2 unpaired or odd electrons. This electronic configuration can be explained by Hund’s rule.

According to this rule—the pairing of electrons in the orbital of a particular subshell (degenerate orbitals) does not occur until all the orbitals of that subshell are singly filled up and the singly occupied orbitals must have electrons with parallel spins.

Question 22. Find the total number of orbitals present in the principal quantum number ‘3’ of an atom. Write the symbols of different types of orbitals and also indicate the number of orbitals in each type
Answer:

S. Total number of orbitals present for the principal quantum number 3 is given by, 32 = 9.

Now n – 3 may have the values 0, 1, and 2 so the types of orbitals present in the given shell tire s. p and d.

Again the no. of -orbitals =(2/ + l) =(2×0) + l =1

Number of p -orbitals = 2/ +1 =(2xl) +1=3

Number of d -orbitals = 2/+l=(2×2)+l=5