In [Ti(H2O)6]^3+ Ti has 1 electron in d orbital and this structure is octahedral. H2O is strong field ligand so pairing should occur in its case but here there is only 1 electron so what will be the hybridization? Will it be d2sp3? If yes then that means in spite of having 4 orbitals empty in d, H2O donates lone pair only to 2 d orbitals? How

[Ti(H2O)6]3+ is an octahedral complex.
Titanium exist here with +3 oxidation state. 
Ti​22 - 3d24s​2 configuration.
Ti3+ - 3d​1 

So the first 3d orbital is occupied with this electron. The remaining four 3d orbitals, and the 4s and 4p remain empty.

So the hybridisation is d2sp3 hybridisation, producing 6 hybrid orbitals which are occupied by electrons donated by the six H2O molecules which acts as ligands. Water is a weak field ligand. 

This arrangement makes the second and third 3d orbitals empty. 

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Tops compound is not possible
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