1
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Martin - マーチン Aug 26 '16 at 8:36
  • 1
    $\begingroup$ 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). $\endgroup$ – Jan Aug 26 '16 at 23:30
  • 1
    $\begingroup$ You used coordination-compounds. Are you talking about transition metal complexes or main-group non-metal molecules? $\endgroup$ – Jan Aug 26 '16 at 23:32
  • 1
    $\begingroup$ @Jan Transition metal complexes, sorry. Since we only deal with those at school, I kinda forgot to specify it.... $\endgroup$ – paracetamol Aug 27 '16 at 6:05
6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – permeakra Aug 26 '16 at 8:00
  • 3
    $\begingroup$ "... 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. $\endgroup$ – Martin - マーチン Aug 26 '16 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.