Arrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.
A$\mathrm{CH_4} > \mathrm{H_2O} > \mathrm{NH_3}$
B$\mathrm{NH_3} > \mathrm{CH_4} > \mathrm{H_2O}$
C$\mathrm{H_2O} > \mathrm{NH_3} > \mathrm{CH_4}$
D$\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$
Answer & Solution
Correct answer: D. $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$
All three central atoms are $sp^3$ hybridised, so the base bond angle is $109.5°$. Lone pairs compress that angle (the lone-pair-to-bond-pair repulsion is stronger than bond-pair-to-bond-pair).
Approximate bond angles:
- $\mathrm{CH_4}$ (0 lone pairs): $109.5°$
- $\mathrm{NH_3}$ (1 lone pair): $107°$
- $\mathrm{H_2O}$ (2 lone pairs): $104.5°$
More lone pairs means more compression, so the order is $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$.
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