What is the geometry and hybridization of "PF"_6^(-)? What orbitals are used to hybridize?

Jun 1, 2016

In general, the typical pattern for basic/simple molecules is:

• Diatomic molecule $\implies$ No hybridization at all! (EX: $\text{HCl}$)
• 2 electron groups $\implies$ $s p$ hybridization (EX: ${\text{CO}}_{2}$)
• 3 electron groups $\implies$ $s {p}^{2}$ hybridization (EX: ${\text{BF}}_{3}$)
• 4 electron groups $\implies$ $s {p}^{3}$ hybridization (EX: ${\text{CH}}_{4}$)
• 5 electron groups $\implies$ $s {p}^{3} d$ hybridization (EX: ${\text{PF}}_{5}$)
• 6 electron groups $\implies$ $s {p}^{3} {d}^{2}$ hybridization (EX: ${\text{SF}}_{6}$)

You can count these up and see that the number of orbitals used in the hybridization equal the number of electron groups around the central atom.

HYPERVALENCY OF PHOSPHORUS

Since phosphorus ($\text{P}$, atomic number $15$) is on the third period of the periodic table, it has access to orbitals of principal quantum number $n = \setminus m a t h b f \left(3\right)$.

That means it can use its $3 d$ orbitals in addition to its typical $3 s$ and $3 p$ valence orbitals.

This generates an octahedral molecular and electron geometry. You can see the final shape of this at the bottom.

PRINCIPLES OF ORBITAL HYBRIDIZATION

Orbital hybridization in ${\text{PF}}_{6}^{-}$ requires that all six $\text{P"-"F}$ bonds are identical; not necessarily in bond angle, per se, but in the orbitals used to construct the bond.

(The ideal bond angles are a separate phenomenon based on electron repulsions.)

For example, an overlap between two $2 {p}_{z}$ orbitals is not the same as the overlap between a $2 {p}_{z}$ and a $3 {d}_{{z}^{2}}$ orbital, even though they are both possible. They are both head-on (sigma/$\sigma$) overlaps, but they are not the same bonds.

The problem is that they should be the same bonds for a molecule making six identical $\text{P"-"F}$ bonds. Otherwise, the bond lengths will not all be the same, when they should be.

GENERATING THE HYBRIDIZED ORBITALS

To make the bonds more uniform, hybridization must occur between the $3 s$, $3 p$, and $3 d$ orbitals of phosphorus to generate $s {p}^{3} {d}^{2}$ orbitals.

Then, what you will have is six sets of $\stackrel{\text{P")overbrace(sp^3d^2)-stackrel("F}}{\overbrace{{p}_{y}}}$ orbital overlaps. These overlaps generate the $\text{P"-"F}$ bonds.

You can see the adjacent bond angles are all ${90}^{\circ}$ in an ideal octahedral geometry, because that is the largest angle that the atoms can be separated while still making six identical $\text{P"-"F}$ bonds.