The decreasing order of bond angles in   BF 3 ,   NH 3 ,   PF 3 and I 3 - is:

1. ${\mathrm{I}}_3^{-}>\mathrm{ }{\mathrm{BF}}_3>\mathrm{ }{\mathrm{NH}}_3>\mathrm{ }{\mathrm{PF}}_3$
2. ${\mathrm{BF}}_3>\mathrm{ }{\mathrm{I}}_3^{-}>\mathrm{ }{\mathrm{PF}}_3>\mathrm{ }{\mathrm{NH}}_3$
3. ${\mathrm{BF}}_3>\mathrm{ }{\mathrm{NH}}_3>\mathrm{ }{\mathrm{PF}}_3>{\mathrm{I}}_3^{-}$
4. ${\mathrm{I}}_3^{-}>\mathrm{ }{\mathrm{NH}}_3>\mathrm{ }{\mathrm{PF}}_3>\mathrm{ }{\mathrm{BF}}_3$

Solution

Bond angle $={120}^o, s{p}^2$
Bond angle $={107}^o, s{p}^3$ (1 lone pair)
Bond angle ≈ 92^o
When central atom size increases, bond angle decreases  
$s{p}^3, N{H}_3>P{F}_3$
Bond angle $={180}^o, s{p}^{3}d$

Asked in: JEE Main 2018 (15 Apr)

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