Question

Describe the hybridization of phosphorus in \(\mathrm{PF}_{5}\).

Short answer

The hybridization of phosphorus in \(\mathrm{PF}_5\) is sp3d.

Step-by-step solution

Understand Atomic Orbitals

Before we can derive the correct hybridization, we must understand the atomic orbitals of phosphorus. The electron configuration of phosphorus (\(P\)) is \([Ne] 3s^2 3p^3\). Phosphorus has five valence electrons.

Define Bonding in PF5

In \(\mathrm{PF}_5\), each of the five Fluorine (F) atoms will share a pair of electrons with the Phosphorus (P) atom. Therefore, P needs to have 5 orbitals available for bonding. This exceeds the number of orbitals available in the ground state, therefore, P atom has to excite one of its 3s electrons to the 3d orbital. After exciting one electron, the electron configuration will be: \([Ne] 3s^1 3p^3 3d^1\). Now there are five unpaired electrons, so five hybrid orbitals can be formed.

Determine Hybridization

The hybridization which allows for the formation of five molecular orbitals for bonding is sp3d. This hybridization combines one s orbital, three p orbitals, and one d orbital to form five new sp3d orbitals, each of which can form a bond with one of the 5 fluorine atoms. Hence, the hybridization of phosphorus in \(\mathrm{PF}_5\) is sp3d.