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

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.

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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 $\ce{d^2sp^3}$.

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  • $\begingroup$ So would [Co(NH3)Cl5] also have d2sp3 hybridisation? $\endgroup$ – Pewpaled Apr 21 '17 at 11:08
  • $\begingroup$ @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. $\endgroup$ – Berry Holmes Apr 21 '17 at 11:16
  • $\begingroup$ How do I find out the hybridisation for other transition metal ions then? $\endgroup$ – Pewpaled Apr 21 '17 at 11:21
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    $\begingroup$ @Pewpaled The ligands that lie above water are considered to be strong field ligands in the spectrochemical series. Here's the one I use: $\ce{I- < Br- < S^2- < SCN- < Cl- < NO3- < N3- < F- < OH- < ox^2- < H2O < NCS- < CH3NC < pyridine < NH3 < en < NO2- < PPh3 < CN- < CO < NO/NO+}$ $\endgroup$ – Berry Holmes Apr 21 '17 at 13:39
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    $\begingroup$ Here are the revelant messages from the chat room: As for the ligands if the complex has a CN=6 with 3SFL sand 3WFLs then consider all the ligands to be SFLs, if the complex has a CN=4 with 2SFL and 2WFL, take all the ligands to be WFL. If the complex is sp3 hybrid, it's tetrahedral and if it's dsp2 hybrid, it's square planar, the hybridization is solely dependent on the ability of the ligands to force the electrons to get paired up. I'm glad I could help you, if you get stuck, feel free to ask your queries, I'll be more than happy to help. :) $\endgroup$ – Berry Holmes Apr 21 '17 at 16:43

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