The correct order of bond angles in the following compounds/species is :

1. $\mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \stackrel{+}{\mathrm{NH}_4} < \mathrm{CO}_2$
2. $\mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4} < \mathrm{NH}_3 < \mathrm{CO}_2$
3. $\mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4}=\mathrm{NH}_3 < \mathrm{CO}_2$
4. $\mathrm{CO}_2 < \mathrm{NH}_3 < \mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4}$

Solution

According to VSEPR theory, bond angle decreases with increase in lone pair, for molecules of same hybridisation. With increase in s-character, bond angle increase. $\therefore$ order of bond angle $=$

Asked in: NEET 2022 (Phase 2)

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