Observe the following sets \begin{array}{|l|l|l|}\hline & {|c|}{ Order } & {c|}{ Property } \\\hline i. &…
- i, ii, iv only
- ii, iii only
- ii, iii, iv only
- i, iii, iv only
Solution

In $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ bond angle should be $109.28^{\circ}$ but in ammonia there is a lone pair-bond pair repulsion and in $\mathrm{H}_2 \mathrm{O}$ two lone pair and bond pair repulsion which decrease the bond angle. $\mathrm{SO}_2$ has planar structure and its bond angle should be $120^{\circ}$ but due to lone-pair-bond pair repulsion it decreases. $\therefore \quad \mathrm{SO}_2 \gt \mathrm{NH}_3 \gt \mathrm{H}_2 \mathrm{O}$ (ii)

Dipole moment $\propto$ Electronegativity Oxygen is more electronegative than nitrogen so dipole moment of $\mathrm{H}_2 \mathrm{O}$ is more as compare to $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{~S}$ $\mathrm{H}_2 \mathrm{O} \gt \mathrm{NH}_3 \gt \mathrm{H}_2 \mathrm{~S}$ (iii) More the bond order more will be the bond enthalpy $\begin{array}{|c|c|}\hline \text{Molecule} & \text{Bond-order} \\ \hline \mathrm{N}_2 & 3 \\ \hline \mathrm{O}_2 & 2 \\ \hline \mathrm{H}_2 & 1 \\ \hline\end{array}$ $\therefore$ Bond enthalpy $\mathrm{N}_2 \gt \mathrm{O}_2 \gt \mathrm{H}_2$ (iv) $\begin{array}{|c|c|}\hline \text{Molecule} & \text{Bond order} \\ \hline \mathrm{NO}^{+} & 2.5 \\ \hline \mathrm{O}_2 & 2 \\ \hline \mathrm{O}_2{ }^{2-} & 1 \\ \hline\end{array}$
Bond order $\mathrm{NO}^{+}\gt\mathrm{O}_2\gt\mathrm{O}_2{ }^{2-}$ $\therefore$ (ii), (iii) and (iv) are correct.
Asked in: AP EAMCET 2024 (21 May Shift 1)
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