The correct increasing order of adjacent bond angle among BF_3 ,P…

MCQ

The correct increasing order of adjacent bond angle among $BF_3 ,PF_3$ and $ClF_3$

A
$BF_3 < PF_3 < ClF_3$
B
$PF_3 < BF_3 < ClF_3$
✓
$ClF_3 < PF_3 < BF_3$
D
$BF_3 = PF_3 = ClF_3$

✓

Answer

Correct option: C.

$ClF_3 < PF_3 < BF_3$

c
The bond angle in $BF _3$ is $120^{\circ}$ since $B$ is $sp ^2$ - hybridized. In $PF _3, P$ is $sp ^3$ - hybridized but the bond angle is little less that $109^{\circ} .28$ (but far greater than $90^{\circ}$ ) due to bond pair - Ione pair repulsion. In $CIF _3, Cl$ is $sp ^3 d$ hybridized. It has $T$ - shape and hence the bond angle is around $90^{\circ}$. Therefore the correct order of increasing bond angle is $ClF _3\,< \,PF _3\,<\, BF _3$

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