How will you concentrate $\mathrm{H}_2 \mathrm{O}_2$ ? Show differences between structures of $\mathrm{H}_2 \mathrm{O}_2$ and $\mathrm{H}_2 \mathrm{O}$ by drawing their spatial structures. Also mention three important uses of $\mathrm{H}_2 \mathrm{O}_2$.
(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) Give a method for the manufacture of hydrogen peroxide and explain the reactions involved therein.
(ii) Illustrate oxidising, reducing and acidic properties of hydrogen peroxide with equations.
(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) What mass of hydrogen peroxide will be present in 2 L of a 5 molar solution?
(ii) Calculate the mass of oxygen which will be liberated by the decomposition of 200 mL of this solution.
(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 $\frac{32 \times 34}{68}=16 \mathrm{~g} \mathrm{O}_2$
A colourless liquid ' $A$ ' contains H and 0 elements only. It decomposes slowly on exposure to light. It is stabilised by mixing urea to store in the presence of light.
(i) Suggest possible structure of $A$.
(ii) Write chemical equations for its decomposition reaction in light.
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.
(i) The structure of H$$_2$$O$$_2$$ is
(a) $\mathrm{H}_2 \mathrm{O}_2$ structure in gas phase, dihedral angle is $111.5^{\circ}$.
(b) $\mathrm{H}_2 \mathrm{O}_2$ structure in solid phase at 110 K , dihedral angle is $90.2^{\circ}$.
(ii) $2 \mathrm{H}_2 \mathrm{O}_2 \xrightarrow[\text { Sunlight }]{\text { hv }} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$
An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $\mathrm{Al}_2 \mathrm{Cl}_6$.
It is LiH because it has significant covalent character due to the smallest alkali metal, Li. LiH is very stable. It is almost unreactive towards oxygen and chlorine.
It reacts with $\mathrm{Al}_2 \mathrm{Cl}_6$ to form lithium aluminium hydride.
$$8 \mathrm{LiH}+\mathrm{Al}_2 \mathrm{Cl}_6 \longrightarrow 2 \mathrm{LiAlH}_4+6 \mathrm{LiCl}$$