STD 12 - p -Block elements - ll — JEE STD 11 Science — Question

Hybridization of $\mathrm{Si}-\mathrm{s} p^{3},$     bond angle $=109\,^o28'$

$\angle Cl-\text{N}-\text{Cl}$ in $NCl_3$ is greater than $\angle \text{Cl}-\text{P}-\text{Cl}$ in $\mathrm{PCl}_{3}$ because partidpation of $s-$ orbital in the hybridization decreases from $NCl_3$ to $PCl_3$

Figure $(2)$

$\angle \text{Cl}-\text{P}-\text{Cl}$ in $\text{PC}{{\text{l}}_{3}}>\angle \text{H}-\text{Sb}-\text{H}$ in $\mathrm{SbH}_{3},$ as in $\mathrm{SbH}_{3}$, bond pairs are formed by overlapping of almost pure $p-$ orbitals.

$\angle \text{H}-\text{Sb}-\text{H}$ in $\text{Sb}{{\text{H}}_{3}}<\angle \text{H}-\text{Te}-\text{H}$ in $\mathrm{H}_{2}Te$ because two lone pairs are present on $Te$ while at $\mathrm{Sb}$ there is one lone pair.