Question

Which of the following molecules has the least bond angle?

• A. $N{H}_3$
• B. $Be{F}_2$
• C. $H_{2}O$
• D. $C{H}_4$

Answer 1

The correct option is C $H_{2}O$

Order of repulsions: $l.p.-l.p.>l.p.-b.p.>b.p-b.p$

In $Be{F}_2$, the steric number of $Be$ is 2. So, it is $sp$ hybridised and the bond angle is ${180}^{\circ }$.

$N{H}_3,H_{2}O,C{H}_4$ are $s{p}^3$ hybridised. Therefore, their bond angles are less than that of $Be{F}_2$.

Also among $N{H}_3,H_{2}O$ and $C{H}_4$, $H_{2}O$ has 2 lone pairs, $N{H}_3$ has one and $C{H}_4$ has none.

Therefore, the bond angle of $H_{2}O$ is the least because it has lone pair-lone pair replusion and two lone pair-bond pair repulsions.