Question

The incorrect order of bond angles is:

• A. $H_{2}O>H_{2}S>H_{2}Se>H_{2}Te$
• B. $C_2{H}_2>C_2{H}_4>C{H}_4>N{H}_3$
• C. $S{F}_6<N{H}_3<H_{2}O<O{F}_2$
• D. $Cl{O}_2>H_{2}O>H_{2}S>S{F}_6$

Answer 1

The correct option is C $S{F}_6<N{H}_3<H_{2}O<O{F}_2$

a) According to VSEPR, as electronegativity of the central atom decreases, bond angle decreases. So, bond angle of $H_{2}O>H_{2}S>H_{2}Se>H_{2}Te$.

b) $C_2{H}_2$ is $sp$ hybridized, $C_2{H}_4$ is $s{p}^2$ hybridized, $C{H}_4$ and $N{H}_3$ are $s{p}^3$ hybridized. We know that, as percentage $s$ character increases, bond angle increases. $C{H}_4$ and $N{H}_3$ are $s{p}^3$ hybridized, but $N{H}_3$ has a lone pair. So, due to the lone pair-bond pair repulsion, bond angle in $N{H}_3$ is less than that in $C{H}_4$.
Hence, the order of bond angles is: $C_2{H}_2>C_2{H}_4>C{H}_4>N{H}_3$.

c) $N{H}_3,H_{2}O$ and $O{F}_2$ have $s{p}^3$ hybridization. In $N{H}_3$, $N$ has only one lone pair. But in $H_{2}O$ and $O{F}_2$, oxygen has two lone pairs. So, the bond angle of $N{H}_3$ is greater than in $H_{2}O$ and $O{F}_2$ due to less lone pair-bond pair repulsions. In $H_{2}O$ and $O{F}_2$, both have two lone pairs of electrons. Fluorine being highly electronegative than hydrogen, the bond pair of electrons are drawn more towards $F$ in $O{F}_2$. Whereas in $H_{2}O$, it is drawn towards $O$. So, in $O{F}_2$ the bond pairs being displaced away from the central atom have very little tendency to open up the angle. But in $H_{2}O$, this opening up is more as the bond pair electrons are closer to each other. So, the bond angle of $O{F}_2$ is less than that in $H_{2}O$. $S{F}_6$ has $s{p}^3{d}^2$ hybridization and hence it has ${90}^{\circ }$ bond angle.

d) $Cl{O}_2$ bond angle is highest due to its $s{p}^2$ hybridization, $H_{2}O$ and $H_{2}S$ are $s{p}^3$ hybridized and $S{F}_6$ is $s{p}^3{d}^2$ hybridized. Hence, the correct bond angle is
$Cl{O}_2>H_{2}O>H_{2}S>S{F}_6$