In the following reaction,how do the Cl-P-Cl | vedclass

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(D) In $PCl_5$,the geometry is trigonal bipyramidal with bond angles of $120^\circ$ (equatorial) and $90^\circ$ (axial).In the product $PCl_4^+$,the p

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Some increase and some decrease

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(D) In $PCl_5$,the geometry is trigonal bipyramidal with bond angles of $120^\circ$ (equatorial) and $90^\circ$ (axial).In the product $PCl_4^+$,the phosphorus atom undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with bond angles of $109.5^\circ$.Comparing the angles: The $120^\circ$ angles decrease to $109.5^\circ$,while the $90^\circ$ angles increase to $109.5^\circ$.Therefore,some angles increase and some decrease.

$I. SnCl_2$	$II. BeF_2$
$III. SO_2$	$IV. NO_2^+$
$V. C_2H_2$

In the following reaction,how do the $Cl-P-Cl$ bond angles change?
$PCl_5 \longrightarrow PCl_4^+ + Cl^-$

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In the following reaction,how do the $Cl-P-Cl$ bond angles change?$PCl_5 \longrightarrow PCl_4^+ + Cl^-$