Question

Which of the following arrangements correctly represents the decreasing order of bond angles?

• A. $N{H}_3>P{H}_3>As{H}_3$
• B. $N{H}_3>H_{2}O>F_{2}O$
• C. $N{O}_2^{+}>N{O}_2^{-}>N{O}_2$
• D. $C{H}_4>N{H}_3>H_{2}O$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $N{H}_3>P{H}_3>As{H}_3$
B $N{H}_3>H_{2}O>F_{2}O$
D $C{H}_4>N{H}_3>H_{2}O$

All the three hydrides given are having a 15th group element as the central atom bearing three bond pairs and a lone pair on it ($s{p}^3$ hybridised central atom). As we are moving from N to As, the size of the central atom increases and electronegativity decreases. Hence, there will be a decrease in repulsion between the bond pairs and lone-pair and hence decrease in bond angle from $N{H}_3$ to $As{H}_3$.
$N{H}_3$, $H_{2}O$ and $F_{2}O$ are having $s{p}^3$ hybrid central atom with tetrahedral geometry (predicted). But due to the presence of a lone pair on nitrogen in $N{H}_3$, the bond angle reduces to ${107}^{\circ }$ from the ideal tetrahedral angle while it further decreases to ${104.5}^{\circ }$ in the case of $H_{2}O$ due to the presence of two lone pairs on oxygen atom.
Among $H_{2}O$ and $O{F}_2$, the latter is having lesser bond angle as the bond pair is drawn more towards the highly electronegative $F$ atom thus decreasing the repulsion at the centre so that bond angle is slightly reduced (but in the case of $H_{2}O$ the bond pair is drawn more towards the central O-atom).
The order of bond angle is: $N{O}_2^{+}>N{O}_2>N{O}_2^{-}$, this is due to the presence of lone electron in $N{O}_2$ and a lonepair in $N{O}_2^{-}$.
Among $C{H}_4,N{H}_3$ and $H_{2}O$, the order of bond angle is: $C{H}_4>N{H}_3>H_{2}O$. The bond angle decreases because of lone pair-lone pair and lone pair-bond pair repulsions in $N{H}_3$ and $H_{2}O$.