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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 $sp^3d^2$?

Aditya Dev
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    How are $sp^3d^2$ and $d^2sp^3$ different? – bon Feb 05 '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 06 '16 at 01:14
  • "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 06 '16 at 03:24
  • @NilayGhosh the t2g orbitals are already filled. The ligand electrons have to occupy eg orbitals. – Aditya Dev Feb 06 '16 at 03:55
  • Reason for the above duplicate vote: The octahedral geometry is the same, the oxidation state of the metal is the same, in one case the ligands are the same and the question is equally wrong in its premises. – Jan Feb 06 '16 at 13:22
  • @Jan seems like you had an argument with a user. Even my teacher told that $\ce{NH_3}$ acts as a strong field ligand for certain metals. It's not specified in standard books like JD Lee or Atkins. But for the purpose of getting the correct answer, we are taught like that. This question comes up in my exam frequently and since I am in high school, the only way to get the correct answer is by learning it by heart. (I hate doing it). – Aditya Dev Feb 06 '16 at 13:45
  • @AdityaDev: Are you referring to $\ce{NH3}$ being strong field? Well … I can understand where you’re coming from; if you define water to be a neutral ligand and anything weaker being weak-field and anything stronger being strong-field then hmm … but at least it’s consequent. However, I was taught to distinguish strong-field ligands like $\ce{CO}$ or $\ce{CN}$ from medium-field ligands like most nitrogen-ligands $\ce{NH3, en}$, porphyrins and group in water as the weakest medium-field ligand. I agree that $\ce{NH3}$ is not weak-field though. – Jan Feb 06 '16 at 13:51
  • @Jan is it okay if I memorize that the complex is d2sp3? I am still at high school and there is only one month for my exam. (I have physics and maths to study as well). – Aditya Dev Feb 06 '16 at 13:58
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    @AdityaDev Read the second-last paragraph in my answer to the dupe question ;) – Jan Feb 06 '16 at 15:15

1 Answers1

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It has d2sp3 configuration since NH3 is a strong field ligand which causes the electron to pair up and use the inner d orbitals(3d) .

Akshay Pratap Singh
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