Jahn-Teller distortion of [Fe(H2O)6]2+

Question

I am wondering whether Fe(II) hydrated to an octahedral complex would undergo a Jahn-Teller distortion.

As far as I am aware, this distortion comes about because of asymmetry in filling orbitals, in this case the t_2g and/or e_g orbitals.

I am also aware that the distortion is most obvious for asymmetry in the e_g orbitals because these interact most with the ligands.

However, I deduced the configuration t_2g⁴, e_g² for [Fe(H₂O)₆]²⁺, which has no asymmetry in e_g but does in t_2g.

My question is can it distort because of this asymmetry? Would it be possible that promotion to an excited state takes place for the distortion to then be possible, with an overall stabilising effect?

Answer 1

(1) It can and will distort. In general, any non-linear molecule with an orbitally degenerate ground state is susceptible to a first-order JT distortion. $\ce{[Fe(H2O)6]^2+}$ falls under this category.

However, the Jahn-Teller theorem has its roots in symmetry and group theory. These can only tell you whether an integral is zero or whether it is nonzero - i.e. whether there is no distortion, or whether there is a distortion. They cannot make any prediction as to the extent of the distortion, or the exact form of the distortion.

The ground-state distortion in $\ce{[Fe(H2O)6]^2+}$ is small, which may be rationalised as being due to an unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals, which point between ligands.

(2) No, the energy required to promote an electron is far larger than any stabilisation that can be derived from a JT distortion.