Correct order of bond angle in given species is:

Options

A. `SiO_(4)^(4-) gt PCl_(3) gt NCl_(3) gt SbH_(3) gt H_(2) Te`
B. `SiO_(4)^(4-) gt NCl_(3) gt PCl_(3) gt SbH_(3) gt H_(2)Te`
C. `SiO_(4)^(4-) gt H_(2) Te gt SbH_(3) gt PCl_(3) gt NCl_(3)`
D. `NCl_(3) gt PCl_(3) gt SiO_(4)^(4-) gt SbH_(3) gt H_(2)Te`

Answer

B

Explanation

, Hybridization of `Si-sp^(3)`, bond angle`=109^(@)28'`
`angleCl-N-Cl` ini `NCl_(3)` is greater than `angle Cl-P-Cl` in `PCl_(3)` because participation of s-orbital in hybridization decreases from `NCl_(3)` to `PCl_(3)`

`angleCl-P-Cl` in `PCl_(3) gt angle H-Sb-H` in `SbH_(3)`, as in `SbH_(3)`, bond pairs are formed by overlapping of almost pure p-orbitals.
`angleH-Sb-H` in `SbH_(3)lt angleH-Te-H` in `H_(2)Te` because two lone pairs are present on Te while at Sb there is one lone pair.