# p-d sigma bond - Why not possible?

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?

• 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. – Nicolau Saker Neto Dec 7 '15 at 12:42
• Are you sure? Can you show me the overlapping diagram for such a compound? – Shodai Dec 7 '15 at 12:45

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 $\mathrm{d}_{z^2}$.
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.