The incorrect order of bond angle is:

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(D) $XeF_2$ has a linear geometry with a bond angle of $180^{\circ}$.$BeF_3^-$,$BF_3$,and $CH_3^+$ are all $sp^2$ hybridized with $0$ lone pairs,resul

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$PH_3 > PCl_3$

Vedclass · Answer

(D) $XeF_2$ has a linear geometry with a bond angle of $180^{\circ}$.$BeF_3^-$,$BF_3$,and $CH_3^+$ are all $sp^2$ hybridized with $0$ lone pairs,resulting in a bond angle of $120^{\circ}$.$CH_4$,$SiCl_4$,and $BF_4^-$ are all $sp^3$ hybridized with $0$ lone pairs,resulting in a bond angle of $109.5^{\circ}$.$CCl_4$ has a bond angle of $109.5^{\circ}$ and $H_2O$ has a bond angle of $104.5^{\circ}$.In $PH_3$ and $PCl_3$,the bond angle depends on the electronegativity of the central atom and the surrounding atoms. $PH_3$ has a bond angle of $\approx 93^{\circ}$ and $PCl_3$ has a bond angle of $\approx 100^{\circ}$.Therefore,the order $PH_3 > PCl_3$ is incorrect.

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