Question

$EC{l}_3$ (where $E=B,P,As,Bi$) of these elements are known.
Bond angles are in the following order-

• A. $B>P>As>Bi$
• B. $B>P=As>Bi$
• C. $Bi>As>P>B$
• D. $B>As>Bi>P$

Answer 1

The correct option is A $B>P>As>Bi$

$BC{l}_3$ has no lone pair thus bond angle is ${120}^0$ and $B$ atom is $s{p}^2$ hybridised.
In $PC{l}_3$, $AsC{l}_3$ and $BiC{l}_3$ central atom $E$ is $s{p}^2$ hybridised but also having a lone pair and having (lp-bp) repulsion so the bond angle is less than ${120}^0$, also moving down the group (bp-bp) repulsion decreases thus $P>As>Bi$ and overall order is $B>P>As>Bi$