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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.

Berry Holmes
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Pewpaled
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1 Answers1

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

Berry Holmes
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  • So would [Co(NH3)Cl5] also have d2sp3 hybridisation? – Pewpaled Apr 21 '17 at 11:08
  • @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
  • How do I find out the hybridisation for other transition metal ions then? – Pewpaled Apr 21 '17 at 11:21
  • @Pewpaled Excluding the exceptional cases of cobalt (+3), metal ions with a $d^8$ configuration, $d^9$ configuration with a strong field ligand etc. the spectrochemical series of CFT will give a correct answer. You can even VBT if you you're aware of addtional properties, for instance the magnetic behavior of your complex. Afaik the configuration is experimentally determined, undertaking the task theoretically will be a mere speculation. – Berry Holmes Apr 21 '17 at 11:36
  • I also had a confusion regarding finding hybridisation of four coordination complex compounds. How do we know if a complex compound is square planar or tetrahedral given the compound. How does it depend upon the spectrochemical series. Uhhhh..The actual confusion is till which ligand in the spectrochemical series do I consider them to be strong and does it depend on the number of strong ligands. If yes, then how. I certainly need your help because I'm facing difficulties solving questions based on finding hybridisation of complex compounds – Pewpaled Apr 21 '17 at 11:58
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    @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
  • Let us continue this discussion in chat. – Berry Holmes Apr 21 '17 at 13:41
  • Thank you so much... that helped more than I can ever think. That is the ultimate answer to my doubts...The chat room was not working though...I'm logged in but it says to log in and then it fails logging me in.. Nevermind ..tysm – Pewpaled Apr 21 '17 at 15:32
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    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. :) – Berry Holmes Apr 21 '17 at 16:43
  • It read it but the problem was I couldn't chat back...Thank you again – Pewpaled Apr 21 '17 at 16:48
  • If in four coordination no. Complex compound if the no.of SFL are equal to 2 and WFL =2 for 3d elements should I consider it to be dsp2 or sp3? – Pewpaled Apr 21 '17 at 16:52
  • That depends, if there's a vacant d-orbital in the penultimate shell, the hybridization can be $dsp^2$, else it will be $sp^3$. – Berry Holmes Apr 21 '17 at 17:22
  • So what would be the hybridisation of [Ni(NH3)2Cl2] and K[Ni(NH3)Cl3] – Pewpaled Apr 21 '17 at 17:35
  • Would it be sp3 for both? – Pewpaled Apr 22 '17 at 03:59
  • Yes, you're correct. – Berry Holmes Apr 22 '17 at 04:51
  • And would it be dsp2 for [Ni(NH3)3Cl]+? – Pewpaled Apr 22 '17 at 06:16
  • Exactly, $dsp^2$, $NH_3$ is a strong field ligand. – Berry Holmes Apr 22 '17 at 07:25
  • Then would that mean four coordination complex compounds having SFL more than or equal to 3 would be square planar for 3d elements and always square planar for 4d and 5d elements irrespective of the number of strong ligands? – Pewpaled Apr 22 '17 at 07:29
  • https://chemistry.stackexchange.com/a/72729/43942 this might help in case of $d^8$ configuration. See, it depends on the configuration, you first write down the configuration of the CMA by using Hund's, Pauli's and Aufbau's rules. Then if it's a SFL, pair up the electrons in the $t_2g$ orbitals first and then proceed to fill $e_g$ orbitals in case of octahedral complexes. You might want to look that up if you're unsure what it means. – Berry Holmes Apr 22 '17 at 07:48
  • That helped a lot – Pewpaled Apr 22 '17 at 13:54
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    https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes – Mithoron Jun 24 '17 at 16:45