Hydrogen

$ \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) \xrightarrow[(1273 \mathrm{~K})]{\mathrm{Ni}} \mathrm{CO}+3 \mathrm{H}_2 $

(Used for industrial preparation of dihydrogen gas)

(D, E) explanation

(D) $\mathrm{H}-\mathrm{H}$ bond strength/ bond dissociation energy/bond energy of $\mathrm{H}_{2}$ can not be lowest because bond formed between hydrogen atoms is due to overlapping of $1 \mathrm{~s}$-$1 \mathrm{~s}$.

(E) Hydrogen can not reduces oxides of highly reactive metal.

Presence of particulate matter in polluted air catalyzes the oxidation of SO₂ to SO₃

i.e., 2SO₂ (g) + O₂ (g) $\to$ 2SO₃ (g)

This reaction can also be promoted by O₃ and H₂O₂, as

SO₂ (g) + O₃ (g) $\to$ SO₃ (g) + O₂ (g)

SO₂ (g) + H₂O₂ (l) $\to$ H₂SO₄ (aq)

In neutral or faintly alkaline solution.

$\mathop {2MnO_4^ - }\limits^{ + 7} + {H_2}O + \mathop {{I^ - }}\limits_{(iodide)}^{ - 1} \to \mathop {2Mn{O_2}}\limits^{ + 4} + 2O{H^ - } + \mathop {IO_3^ - }\limits_{(iodate)}^{ + 5} $

Manganese (Mn) oxidation state change from +7 to +4.

Hydrogen has three isotopes : protium($_1^1$H) deuterium($_1^2$H or D) and tritium ($_1^3$H or T). Of these isotopes, only tritium is radioactive and emits low energy $\beta$^$-$ particles (t_1/2, 12.33 years).

Clark's method used to remove temporary hardness of water

Ca(HCO₃)₂ + Ca(OH)₂ $ \to $ 2CaCO₃ + 2H₂O

Chlorine water on standing loses yellow colour due to formation of HCl and HOCl which in turns provide nascent oxygen.

$\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HCl}+\mathrm{HOCl}$

$\mathrm{HClO} \longrightarrow \mathrm{HCl}+$ $\mathrm{\mathop {[O]}\limits_{Nascent\,oxygen}}$

Thus, production of $\mathrm{HCl}$ and $\mathrm{HOCl}$ are responsible for fading of yellow colour of chlorine water.

$\mathrm{H}_2 \mathrm{O}_2$ can be obtained by $\mathrm{BaO}_2 . 8 \mathrm{H}_2 \mathrm{O}$ as follows:

$\mathop {Ba{O_2}.8{H_2}O(s)}\limits_{Barium\,peroxide} + {H_2}S{O_4}(aq)\buildrel {} \over \longrightarrow BaS{O_4}(s) + \mathop {{H_2}{O_2}(aq)}\limits_{Hydrogen\,peroxide} + 8{H_2}O(l)$

Anhydrous barium oxide is not used here because the precipitate $\mathrm{BaSO}_4$ forms a protective layer on the unreacted barium peroxide and this prevents its further participation in the reaction. Hence, option (a) is the correct.

Heavy water is used for slowing down the speed of neutrons in nuclear reactors, hence used as moderator. Boiling point of heavy water is greater (374.42 K) than that of ordinary water (373 K), hence heavy water is more associated. Dielectric constant of ordinary water is greater than that of heavy water, hence ordinary water is a better solvent.

In H₂O₂ , the O–H groups are not in the same plane. So it has non-planar structure. It has a half-opened book structure in which the two O–H groups lie on the two pages of the book. The angle between two pages of the book is 94^o and H–O–O bond angle is 97^o.