Atomic hydrogen is highly unstable. Since, the electronic configuration of atomic hydrogen is $1 s^1$, it needs one more electron to complete its configuration and gain stability. Therefore, atomic hydrogen is very reactive and combines with almost all the elements.

It, however, reacts in three different ways i.e.,

(i) by loss of its single electron to form $\mathrm{H}^{+}$,

(ii) by gain of one electron to form $\mathrm{H}^{-}$ and

(iii) by sharing its electron with other atoms to form single covalent bonds. In contrast, the bond dissociation energy form $\mathrm{H}-\mathrm{H}$ bond is very high ( $435.88 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ). As a result, molecular hydrogen is almost inert at room temperature and hence reacts only with a few elements.

(i) $\mathrm{D}_2 \mathrm{O}$ can be prepared by prolonged electrolysis of water.

(ii) Physical properties

(a) $\mathrm{D}_2 \mathrm{O}$ is colourless, odourless, tasteless liquid. It has maximum density $-1.1073 \mathrm{~g} \mathrm{~mL}^{-1}$ at $11.6^{\circ} \mathrm{C}$ (Maximum density of water at $4^{\circ} \mathrm{C}$ ).

(b) Solubility of salts in heavy water is less than in ordinary water because it is more viscous than ordinary water.

(c) Nearly, all physical constants of $\mathrm{D}_2 \mathrm{O}$ are higher than $\mathrm{H}_2 \mathrm{O}$. It is due to the greater nuclear mass of deuterium atom than H -atom and stronger H -bonding in $\mathrm{D}_2 \mathrm{O}$ than $\mathrm{H}_2 \mathrm{O}$.

(iii) Exchange reactions of hydrogen with deuterium

$\begin{aligned} \mathrm{NaOH}+\mathrm{D}_2 \mathrm{O} & \longrightarrow \mathrm{NaOD}+\mathrm{HOD} \\ \mathrm{HCl}+\mathrm{D}_2 \mathrm{O} & \longrightarrow \mathrm{DCl}+\mathrm{HOD} \\ \mathrm{NH}_4 \mathrm{Cl}+\mathrm{D}_2 \mathrm{O} & \longrightarrow \mathrm{NH}_3 \mathrm{DCl}+\mathrm{HOD}\end{aligned}$

(i) Industrially, $\mathrm{H}_2 \mathrm{O}_2$ is prepared by the auto-oxidation of 2-alkylanthraquinols.

2-ethylanthraquinol $\underset{\mathrm{H}_2 / \mathrm{Pd}}{\stackrel{\mathrm{O}_2 /(\text { air }}{\rightleftarrows}} \mathrm{H}_2 \mathrm{O}_2+$ Oxidised product

In this case, $1 \% \mathrm{H}_2 \mathrm{O}_2$ is formed. It is extracted with water and concentrated to $\sim 30 \%$ (by mass) by distillation under reduced pressure.

It can be further concentrated to $\sim 85 \%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $\mathrm{H}_2 \mathrm{O}_2$.

(ii) $\mathrm{H}_2 \mathrm{O}_2$ has a non-planar structure.

The molecular dimensions in the gas phase and solid phase are shown below

In the gas phase, $\mathrm{H}_2 \mathrm{O}$ is a bent molecule with a bond angle of $104.5^{\circ}$ and $\mathrm{O}-\mathrm{H}$ bond length of 95.7 pm as shown below

(iii) Following are the three important uses of $\mathrm{H}_2 \mathrm{O}_2$

(a) In daily life, it is used as a hair bleach and as a mild disinfectant. As an antiseptic it is sold in the market as perhydrol.

(b) It is used in the synthesis of hydroquinone, tartaric acid and certain food products and pharmaceuticals (cephalosporin) etc.

(c) It is employed in the industries as a bleaching agent for textiles, paper pulp, leather, oils, fats etc.

(i) $\mathrm{H}_2 \mathrm{O}_2$ is industrially manufactured by the auto-oxidation of 2alkylanthraquinols

$$\text{2-ethylanthraquinol}\mathrm{\mathrel{\mathop{\rightleftharpoons} \limits_{{H_2}/Pd}^{{O_2}/(air)}} {H_2}{O_2}}+ \text{Oxidided product}$$

In this case, $1 \% \mathrm{H}_2 \mathrm{O}_2$ is formed. It is extracted with water and concentrated to $\sim 30 \%$ (by mass) by distillation under reduced pressure. It can be further concentrated to $\sim 85 \%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $\mathrm{H}_2 \mathrm{O}_2$.

(ii) $\mathrm{H}_2 \mathrm{O}_2$ acts as an oxidising as well as reducing agent in both acidic and alkaline media. Following reactions are described below

(a) Oxidising action in acidic medium

$$\begin{aligned} & 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \\ & \mathrm{PbS}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{PbSO}_4(\mathrm{~s})+4 \mathrm{H}_2 \mathrm{O}(l) \end{aligned}$$

(b) Reducing action in acidic medium

$$\begin{aligned} 2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_2 \mathrm{O}_2 & \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2 \\ \mathrm{HOCl}+\mathrm{H}_2 \mathrm{O}_2 & \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{O}_2 \end{aligned}$$

(c) Oxidising action in basic medium

$$\begin{gathered} 2 \mathrm{Fe}^{2+}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{OH}^{-} \\ \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{Mn}^{4+}+2 \mathrm{OH}^{-} \end{gathered}$$

(d) Reducing action in basic medium

$$\begin{aligned} & \mathrm{I}_2+\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{I}^{-}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ & 2 \mathrm{MnO}_4+3 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{MnO}_2+3 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{H}^{-} \end{aligned}$$

(i) Molar mass of $\mathrm{H}_2 \mathrm{O}_2=34 \mathrm{~g} \mathrm{~mol}^{-1}$

1 L of 5 M solution of $\mathrm{H}_2 \mathrm{O}_2$ will contain $34 \times 5 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2$

2 L of 5 M solution of $\mathrm{H}_2 \mathrm{O}_2$ will contain $34 \times 5 \times 2=340 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2$

Mass of $\mathrm{H}_2 \mathrm{O}_2$ present in 2 L of 5 molar solution $=340 \mathrm{~g}$

(ii) 0.2 L (or 200 mL ) of 5 M solution will contain

$$\frac{340 \times 0.2}{2}=34 \mathrm{~g} \mathrm{~H}_2 \mathrm{O}_2$$

$$\mathop {2{H_2}{O_2}}\limits_{2 \times 34 = 68\,g} \to 2{H_2}O + \mathop {{O_2}}\limits_{2 \times 16 = 32\,g} $$

68 g H$$_2$$O$$_2$$ on decomposition will give 32 g O$$_2$$

$\therefore 34 \mathrm{~g} \mathrm{~H}_2 \mathrm{O}_2$ on decomposition will give $\frac{32 \times 34}{68}=16 \mathrm{~g} \mathrm{O}_2$