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Bond angle in PH_(4^(+) is higher than in PH_3 Why? |
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Answer» Solution :Phosphorous in both `PH_3` and `PH_4^(+)` is `sp^3` hybridised . Due to the absence of lone, pair -bond pair repulsion and presence of FOUR identical bond pair- bond pair interactions, `PH_4^+` assumes tetrahedral geometry with a bond angle of `109^@ 28.` But `PH_3` has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from `109^@ 28.` to `93.6^@` . As a result `PH_3` is PYRAMIDAL. `PH_3` -Pyramidal with bond angle of `93.6^@` `PH_4^(+4)` - Tetrahedral with bond angle of `109^@ 28.`. |
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