Class 11th

8.19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + ClO^–(aq) → NO(g) + Cl^-(aq)
(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂^–(aq) + O₂(g) + H⁺

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(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, P₄? is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2?PO₂⁻? with 3.

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water

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The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2?PO₂⁻? with 3.

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

P₄? (s) + 6OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq) + 3OH⁻ (aq)
Subtract 3 hydroxide ions from both sides.

P_4? (s) + 3OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

Ion electron method:
The oxidation half reaction is

P₄? (s) → H₂? PO₂⁻? (aq).
The P atom is balanced.

P₄? (s) → 4H₂? PO₂⁻? (aq)

The oxidation number is balanced by adding 4 electrons on RHS.

P₄? (s) → 4H₂? PO₂⁻? (aq) + 4e⁻

The charge is balanced by adding 8 hydroxide ions on LHS.

P₄? (s)+8OH⁻ (aq)→4H₂? PO₂⁻? (aq)
The O and H atoms are balanced.

The reduction half reaction is P₄? (s)→PH₃? (g).
The oxidation number is balanced by adding 12 electrons on LHS.

P_4? (s)+12e⁻→PH₃? (g)
The charge is balanced by adding 12 hydroxide ions on RHS.

P₄? (s)+12e⁻→PH_3? (g)+12OH⁻
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.

P_4? (s)+3OH⁻ (aq)+3H₂? O (l)→PH₃? (g)+3H₂? PO₂⁻? (aq)

(b) The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction, N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:

−2 +2

N₂H₄ (l)? NO (g)
The N atoms are balanced as:

N₂H₄ (l)?2NO (g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+8e⁻
The O atoms are balanced by adding 6H2O as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+6H₂O (I)+8e⁻. (i)
The reduction half equation is:

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5.26. Assertion: Intermolecular interactions may hold the closely and slowly moving molecules together and thus the solid liquifies.

Reason: Compression brings the molecules in close vicinity and cooling slows down the movement of molecules.

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5.26. (d) Compression brings the molecules in close vicinity and cooling slows down the movement of molecules therefore, intermolecular interactions may hold the closely and slowly moving molecules together and the gas liquifies.

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5.25. Assertion: Liquids and solids are hard to compress.

Reason: Magnitude of the repulsion rises very rapidly as the distance separating the molecules decreases.

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5.25. (a) Other than attractive forces, molecules also exert repulsive forces on one another. When two molecules are brought into close contact with each other, the repulsion between the electron clouds and that between the nuclei of two molecules comes into play. Magnitude of the repulsion rises very rapidly as the distance separating the molecules decreases. This is the reason that liquids and solids are hard to compress. In these states molecules are already in close contact; therefore, they resist further compression; as that would result in the increase of repulsive interactions.

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8.18. Balance the following redox reactions by ion-electron method.
(a) MnO₄^–(aq) +I^–(aq) → MnO₂(s) + I₂ (s) (in basic medium)
(b) MnO₄^–(aq) + SO₂(g) → Mn²⁺(aq) +H₂SO₄^–(in acidic solution)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇^2-(aq) + SO₂ (g) → Cr³⁺(aq) + SO₄^2-(aq) (in acidic solution)

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(a) The balanced half reaction equations are:
Oxidation half equation:

I⁻ (aq) → I₂ (s) - (i)

Reduction half reaction equation:

MnO₄⁻ (aq) → MnO₂ (aq) - (ii)

Balance I atoms and charges in the oxidation half reaction.

2I⁻ (aq) → I₂ (s) + 2e⁻

In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

MnO₄⁻ (aq) + 3e⁻→ MnO₂ (aq)

Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.

MnO4⁻ (aq) + 3e⁻→ MnO₂ (aq)+4OH⁻

To balance O atoms, add 2 water molecules to reactant s

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5.24. Assertion: Viscosity of liquids decreases as the temperature rises.

Reason: Molecules of liquids are held together by attractive intermolecular forces.

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5.24. (b) Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

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Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

5.23. Assertion: The dipole-dipole interactionbetween two HCl moleculesis stronger than the London forces.

Reason: Only partial charges are involved

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5.23. (a) The dipole-dipole interaction between two HCl molecules is stronger than the London forces but is weaker than ion-ion interaction because only partial charges are involved.

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8.17. Consider the reactions:
(a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
(b) H₃PO₂(aq) + 2CuSO₄(aq) + 2H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)
(c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^–(aq→ C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aq) → No change observed
What inference do you draw about the behaviour of Ag⁺and Cu²⁺ from these reactions?

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Reactions (a) and (b) indicate that H₃PO₂ (hypophosphorous acid) is a reducing agent and thus reduces both AgNO₃ and CuSO₄ to Ag and Cu respectively. Conversely, both AgNO₃ and CuSO₄ act as oxidising agent and thus oxidise H₃PO₂to H₃PO₄ (orthophosphoric acid) Reaction (c) suggests that [Ag (NH₃)₂]⁺ oxidises C₆H₅CHO (benzaldehyde) to C₆H₅COO^– (benzoate ion) but reaction (d) indicates that Cu²⁺ ions cannot oxidise C₆H₅CHO to C₆H₅COO^–. Therefore, from the above reactions, we conclude that Ag⁺ ion is a strong deoxidising agent than Cu²⁺ ion.

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8.16. Why does the following reaction occur?

XeO₆^4– (aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g)+ F₂(g) + 3H₂O(l)

What conclusion about the compound Na₄XeO₆ (of which XeO₆^4– is a part) can be drawn from the reaction?

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XeO₆⁴⁻? oxidizes F⁻ and F⁻ reduces XeO₆⁴⁻?
Hence, the given reaction occurs.
The oxidation number of Xe decreases from +8 to +6. The oxidation number of F increases from -1 to 0.
Thus, Na₄? XeO₆? is a stronger oxidising agent than F⁻.

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8.15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

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Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
F₂? + 2Cl⁻ → 2F⁻ + Cl₂?
F₂? + 2Br⁻ → 2F⁻ + Br₂?
F_2? + 2I⁻ → 2F⁻ + I₂?

Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
Cl_2? + Br⁻ → 2Cl⁻ + Br₂?
Cl₂? + I⁻ → 2Cl⁻ + I₂?

Bromine oxidizes iodide ion to iodine.
Br₂? + I⁻ → 2Br⁻ + I₂?

But bromine and chlorine cannot oxidize fluoride to fluorine. Hence, fluorine is the best oxidizing agent amongst the halogens. The decreasing order of the oxidizing power of halogens is F₂? >Cl₂? >Br₂? >I₂?

HI and HBr can reduce sulphuric acid

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Why does the same reductant, thiosulphate react differently with iodine and bromine?

8.14. Consider the reactions:

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