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

May 26, 2017

Warning! Long Answer. The ${\text{AuCl}}_{4}^{-}$ ion is square planar and ${\text{dsp}}^{2}$ hybridized.

#### Explanation:

CRYSTAL FIELD THEORY

Crystal field theory postulates that

• $\text{d}$ orbitals pointing directly at an axis are most destabilized by electrostatic interactions with a ligand.
• $\text{d}$ 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 ${\text{d}}_{{z}_{2}}$ and ${\text{d}}_{{x}^{2} - {y}^{2}}$ orbitals are most destabilized.
• the ${\text{d"_text(xy), "d}}_{\textrm{x z}}$, and ${\text{d}}_{\textrm{y z}}$ orbitals are least destabilized.

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

The absence of ligands along the $z$-axis relative to an octahedral field stabilizes the ${\text{d}}_{{z}^{2}}$, ${d}_{x z}$, and ${d}_{y z}$ levels, and leaves the ${\text{d}}_{{x}^{2} - {y}^{2}}$ level the most destabilized.

When you work it out, there turns out to be four different energy levels in a square planar field: The ${\text{d}}_{{x}^{2} - {y}^{2}}$ level is at such a high level that it remains unoccupied in almost all ${\text{d}}^{8}$ complexes.

$\boldsymbol{{\text{AuCl}}_{4}^{-}}$

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

(b) The electron configuration of $\text{Au"^"3+}$ is ${\text{[Xe] 6s"^0 "4f"^14 "5d}}^{8}$. (c) As the 4 ${\text{Cl}}^{-}$ ions, which are weak-field ligands, approach the $\text{Au"^"3+}$ ion in preparation for bonding, the $\text{5d}$ electrons pair up since repulsions are minimal and the crystal-field splitting energy is relatively small.

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

These orbitals can accept a lone pair from each of the chloride ions and form $\text{Au-Cl}$ bonds.

The bonds point to the corners of a square, forming a square planar molecular geometry. (From www.chemtube3d.com)