# 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. 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.... Aug 27 '16 at 6:05

0.1) Hybridization is to be used with caution in inorganic chemistry above school level. It is proven not working for one-electron properties at the very least.

0.2) Depending on the details, you may or may not be taught about hypervalent compounds using d-orbitals of outer shell. While this concept fell out of favor, it still is taught here and there.

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. Aug 26 '16 at 8:00
• "... fell out of favour ..." is a bit lush I would say. It is proven to be wrong, at least the part about hypercoordination. I still think this pretty much sums up the predicament about the theory, hence thumbs up. Aug 26 '16 at 8:40

$$\ce{[Cu(NH3)4]^2+}$$ has $$\mathrm{sp^2d}$$ hybridisation.
A question might pop why not $$\mathrm{dsp^2}$$?
So basically if it were $$\mathrm{sp^2d}$$ then the paired electron which was promoted to $$\mathrm{4p}$$ orbital, would be less bound to nucleus and hence it would be easier to oxidise, but in contrary, $$\ce{[Cu(NH3)4]^3+}$$ doesn't exist. Hence finally Huggin suggested that the compound should be present in $$\mathrm{sp^2d}$$ hybridisation in order to support the experimental data.

Same hybridisation is observed in:

• $$\ce{[Cu(py)2]^2+}$$
• $$\ce{[Cu(en)2]^2+}$$
• $$\ce{[Cu(CN)4]^2-}$$
• Welcome to the chemistry part of stack exchange, sit down, put on your seat belt and try to enjoy the inflight movie. May 8 at 8:22