In $PCl_5$, phosphorus is in $sp^3d$ hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
(N/A) In $PCl_5$, the molecule adopts a trigonal bipyramidal geometry.
The three equatorial $P-Cl$ bonds are equivalent, making an angle of $120^{\circ}$ with each other.
The remaining two $P-Cl$ bonds are axial, which experience greater repulsion from the equatorial bond pairs.
Consequently, the axial $P-Cl$ bonds are longer $(240 \ pm)$ than the equatorial $P-Cl$ bonds $(202 \ pm)$, making all five bonds non-equivalent.
The three equatorial $P-Cl$ bonds are equivalent, making an angle of $120^{\circ}$ with each other.
The remaining two $P-Cl$ bonds are axial, which experience greater repulsion from the equatorial bond pairs.
Consequently, the axial $P-Cl$ bonds are longer $(240 \ pm)$ than the equatorial $P-Cl$ bonds $(202 \ pm)$, making all five bonds non-equivalent.
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View Solution$A + 2Cl^{-} \rightarrow B$
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Which of the following statements is $INCORRECT$:
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