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Consider the molecules $CH_4$, $NH_3$ and $H_2O$. Which of the given statements is false?
A
The H–O–H bond angle in $H_2O$ is larger than the H–C–H bond angle in $CH_4$.
B
The H–O–H bond angle in $H_2O$ is smaller than the H–N–H bond angle in $NH_3$.
C
The H–C–H bond angle is $CH_4$ is larger than the H–N–H bond angle in $NH_3$.
D
The H–C–H bond angle in $CH_4$, the H–N–H bond angle in $NH_3$ and the H–O–H bond angle in $H_2O$ are all greater than 90°.
Explanation
Bond angle: $CH_4$ > $NH_3$ > $H_2O$.
Detailed Solution
Lone pairs are more diffused than bond pairs. According to VSEPR theory, repulsion order: lp–lp > lp–bp > bp–bp.
Therefore the H–O–H bond angle in $H_2O$ (104.5°) is smaller than the H–C–H bond angle in $CH_4$ (109°28'), so the first statement is false.
Therefore the H–O–H bond angle in $H_2O$ (104.5°) is smaller than the H–C–H bond angle in $CH_4$ (109°28'), so the first statement is false.