Write one chemical reaction for the preparation of $\mathrm{D}_2 \mathrm{O}_2$.
The one chemical reaction for the preparation of $\mathrm{D}_2 \mathrm{O}_2$ is by the action of $\mathrm{D}_2 \mathrm{SO}_4$ dissolved in water over $\mathrm{BaO}_2$.
$$\mathrm{BaO}_2+\mathrm{D}_2 \mathrm{SO}_4 \longrightarrow \mathrm{BaSO}_4+\mathrm{D}_2 \mathrm{O}_2$$
Calculate the strength of 5 volumes $\mathrm{H}_2 \mathrm{O}_2$ solution.
By definition, 5 volumes $\mathrm{H}_2 \mathrm{O}_2$ solution means that 1 L of this $\mathrm{H}_2 \mathrm{O}_2$ solution on decomposition produces 5 L of $\mathrm{O}_2$ at STP.
$$\begin{aligned} 2 \mathrm{H}_2 \mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ 2 \times 34 \mathrm{~g} & \longrightarrow 22.7 \mathrm{~L} \text { at STP } \end{aligned}$$
If $22.7 \mathrm{LO}_2$ at STP will be obtained from $\mathrm{H}_2 \mathrm{O}_2=68 \mathrm{~g}$
$\therefore 5 \mathrm{~L}$ of $\mathrm{O}_2$ at STP will be obtained from $\mathrm{H}_2 \mathrm{O}_2=\frac{68 \times 5}{22.7} \mathrm{~g}=14.98=15 \mathrm{~g}$
$\therefore$ Strength of $\mathrm{H}_2 \mathrm{O}_2$ in 5 volume $\mathrm{H}_2 \mathrm{O}_2$ solution $=15 \mathrm{~g} \mathrm{~L}^{-1}$.
$\Rightarrow \quad$ Percentage strength of $\mathrm{H}_2 \mathrm{O}_2$ solution $=\frac{15}{1000} \times 100=1.5 \%$
Therefore, strength of $\mathrm{H}_2 \mathrm{O}_2$ in 5 volume $\mathrm{H}_2 \mathrm{O}_2$ solution $=15 \mathrm{~g} / \mathrm{L}=1.5 \% \mathrm{H}_2 \mathrm{O}_2$ solution.
(i) Draw the gas phase and solid phase structure of $\mathrm{H}_2 \mathrm{O}_2$.
(ii) $\mathrm{H}_2 \mathrm{O}_2$ is a better oxidising agent than water. Explain.
(i) $\mathrm{H}_2 \mathrm{O}_2$ has a non-planar structure. The molecular dimensions in the gas phase and solid phase are given below
(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) $\mathrm{H}_2 \mathrm{O}_2$ is better oxidising agent than water as discussed below
(a) $\mathrm{H}_2 \mathrm{O}_2$ oxidises an acidified solution of KI to give $\mathrm{I}_2$ which gives blue colour with starch solution but $\mathrm{H}_2 \mathrm{O}$ does not.
$$\begin{aligned} & 2 \mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2 \\ & \mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2 \end{aligned}$$
(b) $\mathrm{H}_2 \mathrm{O}_2$ turns black PbS to white $\mathrm{PbSO}_4$ but $\mathrm{H}_2 \mathrm{O}$ does not.
$$\mathrm{PdS}+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{PbSO}_4+4 \mathrm{H}_2 \mathrm{O}$$
Melting point, enthalpy of vaporisation and viscosity data of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{D}_2 \mathrm{O}$ is given below
| H$$_2$$O | D$$_2$$O | |
|---|---|---|
| Melting point/K | 373.0 | 374.4 |
| Enthalpy of vaporisation at (373 K)/kJ mol$$^{-1}$$ | 40.66 | 41.61 |
| Viscosity/centipoise | 0.8903 | 1.107 |
On the basis of this data explain in which of these liquids intermolecular forces are stronger?
Given that,
| H$$_2$$O | D$$_2$$O | |
|---|---|---|
| Melting point/K | 373.0 | 374.4 |
| Enthalpy of vaporisation at (373 K)/kJ mol$$^{-1}$$ | 40.66 | 41.61 |
| Viscosity/centipoise | 0.8903 | 1.107 |
From this data, it is concluded that the values of melting point, enthalpy of vaporisation and viscosity depend upon the intermolecular forces of attraction. Since, their values are higher for $\mathrm{D}_2 \mathrm{O}$ as compared to those of $\mathrm{H}_2 \mathrm{O}$, therefore, intermolecular forces of attraction are stronger in $\mathrm{D}_2 \mathrm{O}$ than in $\mathrm{H}_2 \mathrm{O}$.
Dihydrogen reacts with dioxygen $\left(\mathrm{O}_2\right)$ to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen? Justify your answer.
The isotope of hydrogen which contains one proton and one neutron is deuterium (D). Therefore, when dideuterium reacts with dioxygen, heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$ is produced.
$\underset{\text { Dideuterium }}{2 \mathrm{D}_2(g)}+\underset{\text { Dioxygen }}{\mathrm{O}_2(g)} \xrightarrow{\text { Heat }} \underset{\begin{array}{c}\text { Deuterium oxide } \\ \text { (Heavy water) }\end{array}}{2 \mathrm{D}_2 \mathrm{O}}$
The reactivity of $H_2$ and $D_2$ towards oxygen will be different. Since, the $D-D$ bond is stronger than $\mathrm{H}-\mathrm{H}$ bond, therefore, $\mathrm{H}_2$ is more reactive than $\mathrm{D}_2$.