Chemistry NCERT Exemplar Solutions Class 11th Chapter Nine

This is a short answer type question as classified in NCERT Exemplar

(i) PBS (s)+4H₂O₂ (aq) → PbSO₄ (s)+4H₂O (l)

(ii) CO (g)+2H₂ (g) → CH₃OH (l)

This is a short answer type question as classified in NCERT Exemplar

Ice has a highly ordered 3D hydrogen bonded structure. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm.

This is a short answer type question as classified in NCERT Exemplar

NaH belongs to Saline hydrides and B₂H₆ and H₂O belong to molecular hydrides.

This is a short answer type question as classified in NCERT Exemplar

These are formed by many d-block and/block elements. However, the metals of group 7, 8 and 9 do not form hydride. Unlike saline hydrides, they are almost always non- stoichiometric, being deficient in hydrogen. For example, LaH 2.87, YbH 2.55, TiH 1.5-1.8, ZrH 1.3-1.75. Dihydrogen forms molecular compounds with most of the jp-block elements.

Most familiar examples are CH₄, NH₃, H₂O and HF.

This is a short answer type question as classified in NCERT Exemplar

Water-gas shift reaction: The water-gas shift reaction is a chemical reaction in which carbon monoxide reacts with steam in the presence of a catalyst to form carbon dioxide and hydrogen.

CO+H₂+H₂O → CO₂+2H₂

This is a long answer type question as classified in NCERT Exemplar

It is lithium hydride (LiH) because it has significant covalent character due to the smallest alkali metal, LiH is very stable. It is almost nonreactive towards oxygen and chlorine. It reacts with Al2Cl2 to form lithium aluminium hydride.

8LiH+ Al₂Cl₆ → 2LiAlH₄ → LiCl

This is a long answer type question as classified in NCERT Exemplar

Since, a colourless liquid 'A' contains only hydrogen and oxygen and decomposes slowly on exposure to light but is stabilised by addition of urea, therefore, liquid A may be hydrogen peroxide. A is H₂O₂. 

Structure of H₂O₂ is given below.

This is a long answer type question as classified in NCERT Exemplar

Mass of H₂O₂ = 68 g.

Mol. Mass of H₂O₂=34gmol^-1

∴1L of M solution of H₂O₂ will contain H₂O₂=34*5 g

∴2L of 5 M solution will contain H₂O₂=34*5*2=340g

or 200 mL of 5 solution will contain H₂O₂ = $\frac{340}{2000}$ $*$  200 = 34 g

Now 2H₂O₂ → 2H₂O+O₂

           64 g  32g

Now 68 g of H₂O₂ on decomposition will give O₂=32g

∴34g of H₂O₂ on decomposition will give O₂ = $\frac{34}{68}$ $*$    34 =16 g

(i) Industrial preparation: H₂O₂ is prepared by the auto-oxidation of 2- alkylanthraquinols

2-ethylanthraquinol ↔ H₂O₂⁺ oxidised product

2Fe₂+ (aq)+2H+ (aq)+H₂O₂ → Fe3 (aq)+2H₂O (l)

PbS (s)+4H₂O₂ (l) → PbSO4 (s)+4H₂O (l)

(ii) Reducing action of hydrogen peroxide 

2MnO₄^-+6H⁺+5H₂O₂→ 2Mn²⁺+8H₂O+5H₂

HOCl+H₂O₂ → H₃O⁺+Cl^-+O₂

Oxidising action of hydrogen peroxide 

2Fe²⁺+H₂O₂→ 2Fe³⁺+2OH^-

Mn²⁺+H₂O₂ → Mn⁴⁺+2OH^-

Acidic properties of H₂O₂

I₂+H₂O₂+2OH^-→ 2I^-+2H₂O+H₂

2MnO₄^-+3H₂O₂→ 2MnO₂+3O₂+2H₂O+2OH^-