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Bond angle in PH_(4)^(+) is highre than that in PH_(2). Why ? |
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Answer» <P> Solution :P in `PH_(3)` is `sp^(3)`-hybridized. It has three bond pairs and one lone pair around P. DUE to stronger lone pair-bond pair repulsions than bond pair-bond pair repulsions, the tetraheldral angle decreases from `109^(@)-28" to "93.6^(@)`. As a RESULT, `PH_(3)` is pyramidal. However, when it reacts with a proton, it forms `PH_(4)^(+)` ionn which has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and presence of four IDENTICAL bond pair-bond pair interactions, `PH_(4)^(+)` assumes tetrahedral GEOMETRY with a bond angle of `109^(@)-28`. This explains why the bond angle in `PH_(4)^(+)` is higher than in `PH_(3)`.
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