# What is the hybridization of the central phosphorus atom in phosphorus pentafluoride?

Oct 23, 2015

$\text{sp"^3"d}$

#### Explanation:

In order to determine the hybridization of the central phosphorus atom in phosphorus pentafluoride, ${\text{PF}}_{5}$, you must first draw the compound's Lewis structure.

The molecule will have a total of $40$ valence electrons, $5$ from the phosphorus atom, and $7$ from each of the five fluorine atoms.

The phosphorus atom will be the molecule's central atom. It will form single bonds with the five fluorine atoms.

These will account for $10$ of the $40$ valence electrons the molecule has. Each fluorine atom will have three lone pairs of electrons, which will account for the rest of the valence electrons.

The molecule's Lewis structure will look like this

Now you need to focus on the central atom, more specifically on how many regions of electron density surround the central atom - this is known as the steric number

A region of electron density can be a covalent bond - single, double, and triple bonds all count as one region of electron density - or a lone pair of electrons.

Notice that phosphorus is bonded to five fluorine atoms and has no lone pairs of electrons attached, which means that it is surrounded by a total of five regions of electron density, which is equivalent to saying that it has a steric number equal to $5$.

The steric number gives you the number of hybrid orbitals. In this case, a steric number of $5$ means that the phosphorus atom will have $6$ hybrid orbitals

• one s-orbital
• three p-orbitals
• one d-orbital

As a result, the central atom will be $\text{sp"^3"d}$ hybridized.