CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

$\text{ B is }s{p}^2\text{, Bond angle }=120^{\circ }$
$\mathrm{N}\text{ is }s{p}^3\text{ with }1lp,\text{ Bond angle }=107^{\circ }$
$\mathrm{P}\text{ is }s{p}^3\text{ with }1lp$ , when central atom size $\uparrow$, bond angle $\downarrow ,\therefore {\mathrm{NH}}_3>{\mathrm{PF}}_3$
$\mathrm{I}$ is $s{p}^{3}d$ (linear), Bond angle $=180^{\circ }$ $\therefore$ Decreasing order of bond angle ${\mathrm{I}}_3^{-}>{\mathrm{BF}}_3>{\mathrm{NH}}_3>{\mathrm{PF}}_3$