The decreasing order of bond angle in the following molecules is

The decreasing order of bond angle in the following molecules is
  1. $\mathrm{NH}_3>\mathrm{CH}_4>\mathrm{H} 2 \mathrm{O}$
  2. $\mathrm{H}_2 \mathrm{O}>\mathrm{NH}_3>\mathrm{CH}_4$
  3. $\mathrm{CH}_4>\mathrm{H}_2 \mathrm{O}>\mathrm{NH}_3$
  4. $\mathrm{CH}_4>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{O}$

Solution

$\mathrm{CH}_4$ no lone pairs BA is $109^{\circ} 28^{\circ} \mathrm{NH}_3$, One L.P. BA is $107^{\circ}$ and $\mathrm{H}_2 \mathrm{O}$, 2 L.P. $\therefore$ BA is $104^{\circ}$

Asked in: MHT CET 2022 (06 Aug Shift 2)

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