Bond angle in PH_{4}^{+} is more than that of | vedclass

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(A) In $PH_{3}$,the phosphorus atom has one lone pair of electrons. According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the

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lone pair-bond pair repulsion exists in $PH_{3}$

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(A) In $PH_{3}$,the phosphorus atom has one lone pair of electrons. According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to approximately $93.6^{\circ}$,resulting in a pyramidal geometry.In $PH_{4}^{+}$,the phosphorus atom has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and the presence of four identical bond pair-bond pair interactions,$PH_{4}^{+}$ adopts a perfect tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.

Bond angle in $PH_{4}^{+}$ is more than that of $PH_{3}$. This is because

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