# Hybridisation of complex compound having octahedral geometry

$$\ce{K[Co(NH3)2Cl4]}$$

I have problem finding the hybridisation of this compound. The strong field ligand $\ce{NH3}$ forms the minority among the ligands, so do I consider the hybridisation to be $\ce{sp^3d^2}$ ? Next, in this compound, $\ce{[Co(NH3)Cl3]}$, the number of strong field ligands is equal to the number of weak field ones. Can someone please explain the basis on which I can find out the hybridisation using crystal field splitting.

If you calculate the oxidation state of the central metal atom in both of these cases, you'll find it to be $+3$. With $\ce{Co^3+}$, all ligands behave as strong field ligands except in the cases of $\ce{[CoF6]^3-}$ and $\ce{[Co(H2O)3F3]}$. Thus, in both of the cases you've mentioned, the ligands will cause pairing of electrons, the complex will be of a diamagnetic nature and the hybridization will be $\ce{d^2sp^3}$.
• @Pewpaled If you meant $\ce{[Co(NH3)Cl5]^2-}$, then my answer is yes, $\ce{Co^2+}$ also acts as a central metal atom so be sure to calculate the oxidation number of the complex you want to deal with. – Berry Holmes Apr 21 '17 at 11:16
• @Pewpaled The ligands that lie above water are considered to be strong field ligands in the spectrochemical series. Here's the one I use: $\ce{I- < Br- < S^2- < SCN- < Cl- < NO3- < N3- < F- < OH- < ox^2- < H2O < NCS- < CH3NC < pyridine < NH3 < en < NO2- < PPh3 < CN- < CO < NO/NO+}$ – Berry Holmes Apr 21 '17 at 13:39