The total number of right angles angle ClPCl | vedclass

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(A) $1$. $PCl_5$ has a trigonal bipyramidal geometry ($sp^3d$ hybridization). The axial $Cl-P-Cl$ bonds are at $180^{\circ}$,and the equatorial $Cl-P-

Vedclass · Accepted Answer

$6, 0, 12$

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(A) $1$. $PCl_5$ has a trigonal bipyramidal geometry ($sp^3d$ hybridization). The axial $Cl-P-Cl$ bonds are at $180^{\circ}$,and the equatorial $Cl-P-Cl$ bonds are at $120^{\circ}$. The axial bonds are at $90^{\circ}$ to the three equatorial bonds. Thus,there are $3 + 3 = 6$ right angles.$2$. $PCl_4^+$ has a tetrahedral geometry ($sp^3$ hybridization). All bond angles are $109.5^{\circ}$. Therefore,the number of $90^{\circ}$ angles is $0$.$3$. $PCl_6^-$ has an octahedral geometry ($sp^3d^2$ hybridization). In an octahedron,each equatorial $Cl$ atom is at $90^{\circ}$ to four other $Cl$ atoms (two axial and two equatorial). With $4$ equatorial positions,this gives $4 	imes 2 = 8$ angles. Additionally,the two axial positions are at $90^{\circ}$ to the four equatorial positions,adding $4$ more. Total right angles = $12$.

Element	Electronegativity Value
$W$	$2.7$
$X$	$2.1$
$Y$	$0.8$
$Z$	$3.4$

The total number of right angles $\angle ClPCl$ present in $PCl_5$,$PCl_4^+$,and $PCl_6^-$ are . . . . . . , . . . . . . , . . . . . . respectively.

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$X$ $2.1$
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