# Hybridization of Na in [Na(H2O)6]+

In the complex ion $\ce{[Na(H2O)6]+}$, the sodium cation forms 6 coordinate bonds with water ligands. Typically this octahedral form is associated with $\mathrm{sp^3d^2}$ hybridization as far as I know, but in the case of sodium the $d$ orbitals aren't readily available, so how can this be described as $\mathrm{sp^3d^2}$ hybridization?

I would guess that its hybridization is $\mathrm{s^2p^4}$?

On further thought I think the d orbitals probably are accessible and that it is indeed $\mathrm{sp^3d^2}$ but this has a high hybridisation energy and so we dont see $\ce{[Na(H2O)6]+}$ in large amounts instead we see the ion solvated by partial charges and electrostatic rather than dative covalent interactions

• Hybridisation is the wrong concept here and for most other coordination complexes. I am sorry I cannot answer right now, molecular orbital theory will give the answer though. I am confident it won't take long here. – Martin - マーチン May 16 '15 at 15:55
• $s^2p^4$ is definitely wrong. The 2s (or 4s) and 2p (or 4p) orbitals are waaay to far removed in energy to be involved in hybridisation. As Martin said though, hybridisation is a poor model for this type of bonding interaction. – bon Apr 20 '16 at 9:11

Also according to another letter, which reports research done on the stability of the different possible forms of the ion, there is not a consensus on only one form. This letter discusses three possible forms of the ion all in the form of $n_1 +n_2$, where $n_1$ is the number of molecules in the first shell of hydration and $n_2$ is the number of molecules in the second shell of hydration. The forms discussed are: