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

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)