The decreasing order of bond angles in $\mathrm{BF}_3, \mathrm{NH}_3, \mathrm{PF}_3$ and…
- $\mathrm{I}_3^{-}>\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3$
- $\mathrm{BF}_3>\mathrm{I}_3^{-}>\mathrm{PF}_3>\mathrm{NH}_3$
- $\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3>\mathrm{I}_3^{-}$
- $\mathrm{I}_3^{-}>\mathrm{NH}_3>\mathrm{PF}_3>\mathrm{BF}_3$
Solution
$\text { B is } s p^2 \text {, Bond angle }=120^{\circ}$
$\mathrm{N} \text { is } s p^3 \text { with } 1 ~l p, \text { Bond angle }=107^{\circ}$
$\mathrm{P} \text { is } s p^3 \text { with } 1 ~l p$ , when central atom size $\uparrow$, bond angle $\downarrow, \quad \therefore \mathrm{NH}_3>\mathrm{PF}_3$
$\mathrm{I}$ is $s p^3 d$ (linear), Bond angle $=180^{\circ}$ $\therefore \quad$ Decreasing order of bond angle $\mathrm{I}_3^{-}>\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3$
Asked in: JEE Main 2018 (15 Apr Shift 1 Online)
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