The H-N-H angle in ammonia is 107.6^{circ} wh | vedclass

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(A) Key point: As the percentage of $s$-character in a bond increases, the bond angle also increases. According to Bent's Rule, the lone pair occupies

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(A) Key point: As the percentage of $s$-character in a bond increases, the bond angle also increases. According to Bent's Rule, the lone pair occupies an orbital with more $s$-character as the electronegativity of the central atom increases. In $NH_3$, the $N$ atom is more electronegative than $P$ in $PH_3$. Since the bond angle in $NH_3$ $(107.6^{\circ})$ is greater than in $PH_3$ $(93.5^{\circ})$, the bonding orbitals in $NH_3$ have more $s$-character. Consequently, the lone pair orbital in $NH_3$ has more $s$-character and less $p$-character compared to the lone pair in $PH_3$. Thus, the $p$-character of the lone pair on ammonia is less than that of phosphine. Hence, option $(A)$ is the correct answer.

$S_8$	$A$
$P_4$	$B$
$S_6$	$C$
$O_3$	$D$

The $H-N-H$ angle in ammonia is $107.6^{\circ}$ while the $H-P-H$ angle in phosphine is $93.5^{\circ}.$ Relative to phosphine, the $p$-character of the lone-pair on ammonia is expected to be

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