# Hybridisation of hexaamminecobalt(III) complex [duplicate]

Does $\ce{[Cr(NH3)6]^3+}$ have $sp^3d^2$ or $d^2sp^3$ hybridisation?

Oxidation state of $\ce{Cr}$ is $+3$. The electronic configuration is $3d^3$ and the $d$ electrons occupy $t_{2g}$ orbitals. If the electrons from the ligands were to occupy the $e_g$ orbitals the stability of the complex decreases. So shouldn't the hybridisation be $sp^3d^2$?

## marked as duplicate by Jan, Community♦Feb 6 '16 at 13:45

• How are $sp^3d^2$ and $d^2sp^3$ different? – bon Feb 5 '16 at 16:06
• In d2sp3, inner orbitals (in this case, 3d) will be used. In sp3d2, the outer 4s, 4p, 4d is used. – Aditya Dev Feb 6 '16 at 1:14
• "If the electrons from the ligands were to occupy the $\ce{e_{g}}$ orbitals the stability of the complex decreases.". Why would the compound want to be unstable? The $\ce{d}$ electrons occupy $\ce{t_2 g}$ orbitals according to Hund's rule – Nilay Ghosh Feb 6 '16 at 3:24
• @NilayGhosh the t2g orbitals are already filled. The ligand electrons have to occupy eg orbitals. – Aditya Dev Feb 6 '16 at 3:55
• @AdityaDev Read the second-last paragraph in my answer to the dupe question ;) – Jan Feb 6 '16 at 15:15

• Calling $\ce{NH3}$ a strong-field ligand is really pushing it: It’s probably the weakest of all ligands stronger than water. Also, the answer is wrong because it is wrong to think of hybridisation here. – Jan Feb 6 '16 at 13:20