My textbook says that $\ce{[CrF6]^{3-}}$ is an example of $sp^3d^2$ hybridization. But I don't really see how. Chromium has an electronic configuration of:

$$3d: \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow}$$

Give three more electrons and it might look something like this:

$$3d: \boxed{\uparrow \downarrow}\boxed{\uparrow \downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow \downarrow}$$

This is the ground state of the electrons. I see no way for electrons to be promoted for the formation of six orbitals. And the electrons have to be promoted all the way up to the $4d$ orbital, which would require a significant amount of energy. What is the exact mechanism of bonding in this ion? Amazingly, I couldn't find anything on the Web about this.

EDIT: My textbook also says that $\ce{[Co(NH3)6]^{3+}}$ is an example of $d^2sp^3$ quite similar to $\ce{[CrF6]^{3-}}$. But when I try to understand the exact mechanism, I encounter a similar problem as a I did above.


In $\ce{[CrF6]^{3-}}$ the central atom is $\ce{Cr(III)}$ or $\ce{Cr^{3+}}$. Chromium is usually $d^6$, so now (take away 3 electrons) $d^3$.

When we start filling in the electron pairs from the ligands, we get the following: VB model for [CrF6]-

The electrons from the metal are marked in red.

The electrons from the ligands (blue) are filled into two $d$, one $s$ and three $p$ orbitals: $d^2sp^3$ hybridization as per solution!

Now try to do the same with $\ce{[Co(NH3)6]^{3+}}$!

  • 1
    $\begingroup$ This is a very tempting solution, but I'd like to issue the warning, that an $d^2sp^3$ hybrid orbital would mean a sixfold degeneracy, which is clearly impossible in an octahedral (ish) field. This application stretches the concept a little bit too far. $\endgroup$ Mar 19 '15 at 6:48

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.