Question

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

Answer 1

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

• Diatomic molecule `=>` No hybridization at all! (EX: `"HCl"`)
• 2 electron groups `=>` `sp` hybridization (EX: `"CO"_2`)
• 3 electron groups `=>` `sp^2` hybridization (EX: `"BF"_3`)
• 4 electron groups `=>` `sp^3` hybridization (EX: `"CH"_4`)
• 5 electron groups `=>` `sp^3d` hybridization (EX: `"PF"_5`)
• 6 electron groups `=>` `sp^3d^2` hybridization (EX: `"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.

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HYPERVALENCY OF PHOSPHORUS

Since phosphorus (`"P"`, atomic number `15`) is on the third period of the periodic table, it has access to orbitals of principal quantum number `n = \mathbf(3)`.

That means it can use its `3d` orbitals in addition to its typical `3s` and `3p` 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 `"PF"_6^(-)` requires that all six `"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 `2p_z` orbitals is not the same as the overlap between a `2p_z` and a `3d_(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 `"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 `3s`, `3p`, and `3d` orbitals of phosphorus to generate `sp^3d^2` orbitals.

Then, what you will have is six sets of `stackrel("P")overbrace(sp^3d^2)-stackrel("F")overbrace(p_y)` orbital overlaps. These overlaps generate the `"P"-"F"` bonds.

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