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(B) $H_2S$ has a bond angle of approximately $92^{\circ}$. This is because the central atom $S$ uses almost pure $p$-orbitals for bonding with $H$ ato

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$H_2S$

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(B) $H_2S$ has a bond angle of approximately $92^{\circ}$. This is because the central atom $S$ uses almost pure $p$-orbitals for bonding with $H$ atoms,as the electronegativity difference between $S$ and $H$ is very small,leading to minimal hybridization. In contrast,$NH_3$ $(107^{\circ})$,$H_2O$ $(104.5^{\circ})$,and $CH_4$ $(109.5^{\circ})$ exhibit $sp^3$ hybridization with significant bond angles due to the repulsion between electron pairs.

Which one of the following compounds has a bond angle of nearly $90^{\circ}$?

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