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I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$.

In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$.

So there are $3$ more empty $\mathrm{d}$ orbitals.

Hence the hybridisation should be $\mathrm{d^3sp^2}$.

But it is $\mathrm{d^2sp^3}$ and paramagnetic .

I could not understand how .

Can you explain me?

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Basically, it isn't really wise to use hybridization on coordination complexes to begin with. However, the hybridization of $d^2sp^3$ makes sense if you think of it as just describing the shape of the molecule. This should be octahedral, due to its 6 equivalent groups around the metal center, and $d^2sp^3$ is the valence bond description of the hybridization of an octahedron.

Octahedral MO diagram

Using Ligand Field theory and MO diagrams is probably a better way to understand how the orbitals overlap and where electrons are in the final structure. This diagram is for $\ce{[Ni(H2O)6]^2+}$, but the diagram will look very similar for the $\ce{V}$ complex, just with fewer electrons. (Image source: dx.doi.org/10.1021/jp4100813 | J. Phys. Chem. B 2013, 117, 16512−16521)

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  • $\begingroup$ So actually it is not $d^2 sp^3$ . We write it as to denote the octahedral shape . $\endgroup$ – TrY iS CheM Mar 23 '17 at 2:11
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    $\begingroup$ @TrYiSCheM For the most part, I would avoid hybridization with complexes altogether, but yes the hybridizations are commonly used to represent the shape. We can see some of the common associated hybridizations/geometries here. $\endgroup$ – Tyberius Mar 23 '17 at 3:52
  • $\begingroup$ But actually in which sub shell the ligand donate their electrons $\endgroup$ – TrY iS CheM Mar 23 '17 at 4:00
  • $\begingroup$ In V what will be filled in empty orbitals. $\endgroup$ – TrY iS CheM Mar 24 '17 at 16:56
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    $\begingroup$ @TrY iS CheM I'm not sure I understand your question. Based on MO theory, we would say that the electrons from the ligands went into Molecular orbitals of the complex rather than the atomic orbitals of the original vanadium. $\endgroup$ – Tyberius Mar 24 '17 at 22:23

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