The total number of right angles angle ClPCl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$6, 0, 12$

Vedclass · Answer

(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 = $12$.

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

Similar Questions

The decreasing order of bond dissociation energies of $C-C$, $C-H$, and $H-H$ bonds is:

Born-Haber cycle may be used to calculate:
$I$. $\Delta_{eg}H$
$II$. Mass number
$III$. Electronegativity
$IV$. Lattice energy

Number of $S-S$ bonds in $H_2S_nO_6$ is:

The bond length $(pm)$ of $F_2$, $H_2$, $Cl_2$ and $I_2$, respectively is

Find the correct option regarding the given chemical properties.

Vedclass Test Series

Exam Paper Generator

Online Exam Module