Question #dd0d4

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Question #dd0d4

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Answer 1 · 2017-03-13T22:04:59

The hybridization of $P$ $"P"$ in ${PCl}_{5}$ $"PCl"_5$ is ${sp}^{3} d$ $"sp"^3"d"$ .

Explanation:

The Lewis structure of ${PCl}_{5}$ $"PCl"_5$ is

Lewis structure

According to VSEPR theory, a molecule with five bonding pairs must have both a trigonal bipyramidal electron geometry and molecular geometry.

The $P$ $"P"$ atom needs five orbitals to form the five $P-Cl$ $"P-Cl"$ bonds.

It has a $3s$ $"3s"$ and three $3p$ $"3p"$ orbitals, so it must use one of its $3d$ $"3d"$ orbitals to form the fifth bond.

These orbitals are hybridized to form five ${sp}^{3} d$ $"sp"^3"d"$ orbitals that (surprise!) just happen to point in the right directions to form a trigonal bipyramid.

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