Question

The decreasing order of bond angles in $B{F}_3,N{H}_3,P{F}_3$ and $I_3^{-}$ is:

• A. $B{F}_3>I_3^{-}>P{F}_3>N{H}_3$
• B. $I_3^{-}>N{H}_3>P{F}_3>B{F}_3$
• C. $B{F}_3>N{H}_3>P{F}_3>I_3^{-}$
• D. $I_3^{-}>B{F}_3>N{H}_3>P{F}_3$

Answer 1

The correct option is D $I_3^{-}>B{F}_3>N{H}_3>P{F}_3$

$lp-lp>l_p-bp>bp-bp$

Due to $Dp-lp$ repulsion in $P{F}_3,F-P-F$ cmgle will be less than that of $B{F}_3$

$\therefore B{F}_3<P{F}_3$

$B{F}_3>N{H}_3$, as the geometry of $B{F}_3$ is triagonal planar while that $N{H}_3$ is tetrahedral

As the Fluorine atom pulls the lone pair of electron on P atom, $lp-bp$ repulsion is more pre dominant. Thus $N{H}_3>P{F}_3$

thus, $B{F}_3>N{H}_3>P{F}_3$

$I_3^{-}$ ion linear. This bond angle is ${180}^o$.

$\therefore$ is $I_3^{-}>B{F}_3>N{H}_3>P{F}_3$