Question

What is the geometry and hybridization of `"AuCl"_4^-`?

Answer 1

Warning! Long Answer. The `"AuCl"_4^(-)` ion is square planar and `"dsp"^2` hybridized.

Explanation:

CRYSTAL FIELD THEORY

Crystal field theory postulates that

• `"d"` orbitals pointing directly at an axis are most destabilized by electrostatic interactions with a ligand.
• `"d"` orbitals pointing away from an axis are least destabilized by electrostatic interactions with a ligand.

(From www.slideshare.net)

For example, in an octahedral field, the diagram above shows that:

• the `"d"_(z_2)` and `"d"_(x^2-y^2)` orbitals are most destabilized.
• the `"d"_text(xy), "d"_text(xz)`, and `"d"_text(yz)` orbitals are least destabilized.

In a square planar complex, the four ligands are only in the `xy` plane, so any orbital in the `xy` plane has a higher energy level.

The absence of ligands along the `z`-axis relative to an octahedral field stabilizes the `"d"_(z^2)`, `d_(xz)`, and `d_(yz)` levels, and leaves the `"d"_(x^2-y^2)` level the most destabilized.

When you work it out, there turns out to be four different energy levels in a square planar field:

The `"d"_(x^2-y^2)` level is at such a high level that it remains unoccupied in almost all `"d"^8` complexes.

`bb("AuCl"_4^-)`

(a) The electron configuration of `"Au"` is `"[Xe] 6s"^1 "4f"^14 "5d"^10`.

(b) The electron configuration of `"Au"^"3+"` is `"[Xe] 6s"^0 "4f"^14 "5d"^8`.

(c) As the 4 `"Cl"^(-)` ions, which are weak-field ligands, approach the `"Au"^"3+"` ion in preparation for bonding, the `"5d"` electrons pair up since repulsions are minimal and the crystal-field splitting energy is relatively small.

(d) The `"Au"^"3+"` ion can then combine its vacant `"5d"_(x^2-y^2), 6s", "6p_x"` and `"6p_y` orbitals to create four new equivalent, hybridized `"dsp"^2` orbitals.

These orbitals can accept a lone pair from each of the chloride ions and form `"Au-Cl"` bonds.

The bonds point to the corners of a square, forming a square planar molecular geometry.

(From www.chemtube3d.com)