Question

Are all the five bonds in $PC{l}_5$ molecule equivalent? Justify your answer.

Answer 1

Hybridization of $P$ in $PC{l}_5$:

Electronic configuration of

${}_{15}P\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^3$
$valenceelectrons$
Number of monovalent atoms $=5$
Steric number(n)
$=\text{Number of bond pair}+\text{lone pair}$
$=5+0$
$=5$

For $n=5\to s{p}^{3}d$ hybridization.

Hence, $P\text{in}PC{l}_5$ has $s{p}^{3}d$ hybridization.

$P\text{in}PC{l}_5$ has $s{p}^{3}d$ hybridization with trigonal bipyramidal shape.
Types of bonds in $PC{l}_5$ molecule

$PC{l}_5$ has a trigonal bipyramidal structure and the three equatorials $P-Cl$ bonds are equivalent, while the two axial bonds are different and longer than equatorial bonds.

Hence, in $PC{l}_5$ all five bonds are not equivalent.