Identify the correct option(s).

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$In {NH}_{4} Cl, N^{'} atom is in s p^{3} d hybridisation$

${({CH}_{3})}_{3} B$

$In S_{8} molecule,$ a total of 16 electrons are left on all the 'S' atoms after bonding

${NH}_{4}^{+} > {NH}_{3} > {NH}_{2}^{-}$ [order of bond angle]

answer is A, B.

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Detailed Solution

The stronger lone pair-lone pair repulsion will cause the bond angle between connected atoms to decrease if an atom possesses both free and bound electron pairs.

${NH}_{2}^{-}$

Has two electron lone pairs and two bond pairs.

${NH}_{3}$

Consists of 3 bond pairs and 1 single pair.

${NH}_{4}^{-}$

Includes 4 bond pairs.

Electron repulsion within a bond pair Electron repulsion between a bond pair and a lone pair as well as lone pair – lone pair electron repulsion
The H-N-H is brought on by increased lone pair – lone pair repulsion among electrons.

To reduce the bond angle. This result is strongest for ${NH}_{2}^{-}$ followed by ${NH}_{3}$

Therefore, the following is the proper order:

${NH}_{4}^{+} > {NH}_{3} > {NH}_{2}^{-}$

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