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 = $12$.

List-$I$ (Molecule Species)	List-$II$ (Property/Shape)
$A$. $SO_2Cl_2$	$I$. Paramagnetic
$B$. $NO$	$II$. Diamagnetic
$C$. $NO_2^-$	$III$. Tetrahedral
$D$. $I_3^-$	$IV$. Linear

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

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Match List-$I$ with List-$II$ :

List-$I$ (Molecule Species) List-$II$ (Property/Shape)

$A$. $SO_2Cl_2$ $I$. Paramagnetic

$B$. $NO$ $II$. Diamagnetic

$C$. $NO_2^-$ $III$. Tetrahedral

$D$. $I_3^-$ $IV$. Linear

Choose the correct answer from the options given below :

Assertion: $H_2O$ and $XeF_4$ both have a plane of symmetry.
Reason: $H_2O$ is a polar molecule,whereas $XeF_4$ is a non-polar molecule.

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