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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 $sp^3d^2$?

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

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    $\begingroup$ How are $sp^3d^2$ and $d^2sp^3$ different? $\endgroup$ – bon Feb 5 '16 at 16:06
  • $\begingroup$ In d2sp3, inner orbitals (in this case, 3d) will be used. In sp3d2, the outer 4s, 4p, 4d is used. $\endgroup$ – Aditya Dev Feb 6 '16 at 1:14
  • $\begingroup$ "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 $\endgroup$ – Nilay Ghosh Feb 6 '16 at 3:24
  • $\begingroup$ @NilayGhosh the t2g orbitals are already filled. The ligand electrons have to occupy eg orbitals. $\endgroup$ – Aditya Dev Feb 6 '16 at 3:55
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    $\begingroup$ @AdityaDev Read the second-last paragraph in my answer to the dupe question ;) $\endgroup$ – Jan Feb 6 '16 at 15:15
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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) .

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  • $\begingroup$ Can you elaborate your answer with a help of a hybridisation diagram? It looks like a comment rather than an answer right now. $\endgroup$ – Nilay Ghosh Feb 5 '16 at 17:55
  • $\begingroup$ With 40 reputation I cannot comment. $\endgroup$ – Akshay Pratap Singh Feb 5 '16 at 18:39
  • $\begingroup$ "Can you elaborate your answer with a help of a hybridisation diagram? " $\endgroup$ – Nilay Ghosh Feb 5 '16 at 19:18
  • $\begingroup$ I thought NH3 was a weak field ligand. $\endgroup$ – Aditya Dev Feb 6 '16 at 1:15
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    $\begingroup$ 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. $\endgroup$ – Jan Feb 6 '16 at 13:20

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