# Are there square planar complexes with sp2d hybridization?

We were taught (Under the section 'Valence Bond Theory') seven types of geometries a transition metal complex may assume and its corresponding hybridization states,

1. Linear - $\ce{sp}$
2. Trigonal planar - $\ce{sp^2}$
3. Tetrahedral - $\ce{sp^3}$
4. Square planar - $\ce{dsp^2}$ (Inner d-orbital involved)
5. Trigonal Bi-pyramidal - $\ce{dsp^3}$ (Inner d-orbital involved)
6. Square Pyramidal - $\ce{sp^3d}$ (Outer d-orbital involved)
7. Octahedral - $\ce{d^2sp^3}$ (Inner d-orbitals involved)

We were told that there are quite a few instances (for Octahedral complexes), where the outer, vacant d-orbital takes part in hybridization, so the hybridization state would thus become: $\ce{sp^3d^2}$

Now my question is:
Are there Square Planar complexes in which the outer, vacant d-orbitals take part in hybridization (i.e- are there square planar complexes with sp2d hybridization) ? If so, could someone provide a few examples.

• It has been proven again and again, that the contribution of d-orbitals in all these cases is minimal, almost negligible (usually <3%). The question you are asking is referring to an old, disproven model and therefore cannot really be answered in the way you would like to look at it. Have a look at this question to get an idea what I am talking about. – Martin - マーチン Aug 26 '16 at 8:36
• Except if OP is talking about transition metal coordination compounds, in which case d-orbital participation is very real and important but where I would not speak of hybridisation at all (CC @Mart). – Jan Aug 26 '16 at 23:30
• You used coordination-compounds. Are you talking about transition metal complexes or main-group non-metal molecules? – Jan Aug 26 '16 at 23:32
• @Jan Transition metal complexes, sorry. Since we only deal with those at school, I kinda forgot to specify it.... – paracetamol Aug 27 '16 at 6:05

1) ignoring 0.*, $\ce{PnX5}$ family where $\ce{Pn=P,As,Sb}$ and X is a halogen (typically $\ce{F}$ or $\ce{Cl}$) adopts trigonal-bipiramidal shape and was viewed as an example of $sp^3d$ hybridisation. Square planar compounds for p-elements are much rarer, but $\ce{XeF4}$ adopt such structure.
2) A rare anion $\ce{[Ni(CN)5]^{3-}}$ may adopt such structure, specifically in $\ce{ [Cr(NH3)6][Ni(CN)5]\cdot 2 H2O}$ Actually, it is often said that square planar complexes may coordinate weakly an additional ion to form a square pyramid and this is why they are typically much more reactive, than octahedral complexes.
• The point about $\ce{XeF4}$ still stands. It is square planar and was considered to have d-orbitals involved in hybridization before this model fell out of favor. – permeakra Aug 26 '16 at 8:00