Question
Bond angle in $PH_4^+$ is higher than that in $PH_3.$ Why?
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Answer
In $PH_3,$ P is sp hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with $Sp_{3}$ changed to pyramidal. $PH_3$ combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in $PH_3.$
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