5
$\begingroup$

Why is a head on (sigma) bond between a p-orbital and a d-orbital not possible? Is it because extent of overlap becomes too little? Can you please illustrate?

$\endgroup$
  • 2
    $\begingroup$ I'm pretty sure your premise is wrong. I can't see why there wouldn't be tons of metallic compounds with p-d $\sigma$ bonds, such as $\ce{WCl6}$, which is undeniably a very covalent compound and whose only really available orbitals are the p orbitals in chlorine and d orbitals in tungsten. $\endgroup$ – Nicolau Saker Neto Dec 7 '15 at 12:42
  • $\begingroup$ Are you sure? Can you show me the overlapping diagram for such a compound? $\endgroup$ – Shodai Dec 7 '15 at 12:45
11
$\begingroup$

Your premise is incorrect. $\unicode[Times]{x3c3}_\mathrm{p-d}$ are very much possible. Consider any hexahalidometal complex such as $\ce{[FeF6]^3-}$. Since fluoride is essentially unhybridised as iron would be, too, we can separate the atomic orbitals into irreducible representations of the $O_\mathrm{h}$ point group. We find that iron’s d-orbitals transform as $\mathrm{t_{2g} + e_g}$ and the 18 fluoride p-orbitals as $\mathrm{a_{1g} + e_g + 2t_{1u} + t_{1g} + t_{2g} + t_{2u}}$. If we separate out those 12 p-orbitals that are perpendicular to the iron-fluoride axis, the ones that remain are $\mathrm{a_{1g} + e_g + t_{1u}}$. You may immediately realise that $\mathrm{e_g}$ is an irreducible representation of both and thus that the fluorides’ p-orbitals can and will mix with the iron’s $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$.

Professor Klüfers included the calculation results of said ion in his internet scriptum of his coordination chemistry course. The ones of relevance are:

You may be slightly surprised at the shape since you cannot recognise the typical $\mathrm{d}_{z^2}$ shape. But as Brandhorst, Grunenberg and Tamm explain, these two orbital shapes are mathematically equivalent descriptions for the $\mathrm{e_g}$ orbitals.[1]

The molecular orbitals shown are both of antibonding type (formally $\mathrm{e_g^*}$). The corresponding bonding orbitals are much lower in energy and centred on the fluoride ligands. Both of the depicted ones are occupied only by a single electron in α spin.


References

[1] K. Brandhorst, J. Grunenberg, M. Tamm J. Chem. Educ. 2008, 85, 1692. DOI: 10.1021/ed085p1692.

$\endgroup$

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.