Molecular hydrides are classified as electron deficient, electron precise and electron rich compounds. Explain each type with two examples.
Molecular hydrides are classified according to the relative numbers of electrons and bonds in Lewis structure as follow
(i) Electron deficient hydrides These type of hydrides contain central atom with incomplete octet. These are formed by 13 group elements, e.g., $\mathrm{BH}_3, \mathrm{AlH}_3$, etc. To complele their octet they generally exist in polymeric forms such as $\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_4 \mathrm{H}_{10}$, $\left(\mathrm{AlH}_3\right)_n$ etc. These hydrides act as Lewis acids.
(ii) Electron precise hydrides These hydrides have exact number of electrons required to form normal covalent bonds. These are formed by 14 group elements, e.g., $\mathrm{CH}_4, \mathrm{SiH}_4$, etc. These are tetrahedral in shape.
(iii) Electron rich hydrides These hydrides contain central atom with excess electrons, which are present as ione pairs.
These are formed by 15,16 and 17 group elements, e.g., $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{HF}$, etc. These hydrides act as Lewis bases.
How is heavy water prepared? Compare its physical properties with those of ordinary water.
Heavy water is prepared by prolonged electrolysis of water. Comparison of physical properties of heavy water with those of ordinary water is as follows
| Property | H$$_2$$O | D$$_2$$O |
|---|---|---|
| Molecular mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$ | 18.0151 | 20.0276 |
| Melting point (K) | 273.0 | 276.8 |
| Boiling point (K) | 373.0 | 374.4 |
| Enthalpy of formation $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ | $$-$$285.9 | $$-$$294.6 |
| Enthalpy of vaporisation $-373 \mathrm{~K}^{(}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ | 40.66 | 41.61 |
| Temperature of max. density (K) | 276.98 | 284.2 |
| Density at $298 \mathrm{~K}\left(\mathrm{~g} \mathrm{~cm}^{-3}\right)$ | 1.0000 | 1.1059 |
| Viscosity (centipoise) | 0.8903 | 1.107 |
| Dielectric constant $\left(\mathrm{C}^2 / \mathrm{Nm}^2\right)$ | 78.39 | 78.06 |
| Electrical conductivity at $298 \mathrm{~K}\left(\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right)$ | $$5.7\times10^{-8}$$ | $$-$$ |
| Enthalpy of fusion $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ | 6.01 | $$-$$ |
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.
$$\mathrm{PdS}+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{PbSO}_4+4 \mathrm{H}_2 \mathrm{O}$$