Hydrogen

As we know, hydrogen is most abundant element in the universe but the most abundant gas in the troposphere is nitrogen, because hydrogen is lightest element.

Troposphere contains 3 quarters of mass of the entire atmosphere. The air here contain 78% nitrogen and 21% oxygen. The last 1% is made of argon, water vapour and carbon dioxide.

(I) H₂ is a 100% pollution free fuel. So, option (I) is correct.

(II) Molecular weight of H₂(2u).

$ = {1 \over {29}} \times $ molecular weight of butane,

C₄H₁₀ (LPG) [58u].

So, compressed H₂ weighs $\sim$ 30 times more than a petrol tank and option (b) is correct.

(III) NaNi₅, Ti - TiH₂ etc. are used for storage of H₂ in small quantities. Thus, option (c) is correct.

(IV) On combustion values of energy released per gram of liquid dihydrogen (H₂) : 142 kJ g^$-$1, and for LPG : 50 kJ g^$-$1. So, option (d) is incorrect.

In the reaction with potassium periodate (KIO₄) and hydroxylamine (NH₂OH), hydrogen peroxide (H₂O₂) exhibits different roles :

(i) When hydrogen peroxide reacts with potassium periodate (KIO₄), it forms potassium iodate (KIO₃).

The balanced chemical reaction is :

$ \mathrm{H}_2 \mathrm{O}_2 + \underset{\substack{\text{Potassium}\\ \text{periodate}}}{\mathrm{KIO}_4} \longrightarrow \underset{\begin{array}{c}\text{Potassium}\\ \text{iodate}\end{array}}{\mathrm{KIO}_3} + \mathrm{H}_2 \mathrm{O} + \mathrm{O}_2 $

Oxidation state of oxygen in hydrogen peroxide :

(a) In H₂O₂ :

$ \begin{aligned} & 2x + 2 = 0 \\\\ & x = -1 \end{aligned} $

(b) In O₂ :

$ x = 0 $

The oxidation state of oxygen increases from -1 to 0, indicating that H₂O₂ is oxidized and acts as a reducing agent for KIO₄.

(iii) When hydrogen peroxide reacts with hydroxylamine (NH₂OH), it produces nitric acid (HNO₃) and water (H₂O).

$ 3 \mathrm{H}_2 \mathrm{O}_2 + \mathrm{NH}_2 \mathrm{OH} \longrightarrow \mathrm{HNO}_3 + 4 \mathrm{H}_2 \mathrm{O} $

Oxidation state of oxygen :

(a) In H₂O₂ :

$ \begin{aligned} & 2x + 2 = 0 \\\\ & x = -1 \end{aligned} $

(b) In H₂O :

$ 2 + x = 0 $

$ x = -2 $

The oxidation state of oxygen decreases from -1 to -2, meaning H₂O₂ is reduced and acts as an oxidizing agent for NH₂OH.

Reduction is a chemical process by which a substance gains electrons. Hydrogen gas is commonly used as a reducing agent because it can easily donate electrons to other substances. However, the ability of hydrogen to reduce an oxide depends on the reactivity of the metal in the oxide.

Let's consider each option:

Option A: Heated Cupric Oxide (CuO)

Copper(II) oxide can be reduced by hydrogen gas to form metallic copper and water. The reaction is as follows:

$\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}$

Option B: Heated Ferric Oxide (Fe_2O_3)

Ferric oxide can also be reduced by hydrogen gas to form metallic iron and water. The reaction is as follows:

$\text{Fe}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Fe} + 3\text{H}_2\text{O}$

Option C: Heated Stannic Oxide (SnO_2)

Stannic oxide can be reduced by hydrogen gas to form metallic tin and water. The reaction is as follows:

$\text{SnO}_2 + 2\text{H}_2 \rightarrow \text{Sn} + 2\text{H}_2\text{O}$

Option D: Heated Aluminium Oxide (Al_2O_3)

Aluminium oxide cannot be reduced by hydrogen gas. Aluminium has a very high affinity for oxygen, making the reduction of aluminium oxide (Al_2O_3) with hydrogen gas not feasible under normal conditions. This is because aluminium is much more reactive and forms a very stable oxide.

Therefore, the correct answer is Option D: heated aluminium oxide.