How hybridisation of [Ti(H₂O)₆]³⁺ is d²sp³ and is paramagnetic?Please explain on the basis of VBT electronic configuration diagram.

Open in App

Solution

[Ti(H2O)6]3+ is an octahedral complex, and in it the metal is in +3 oxidation state.

In a free Ti(3+) ion, having the outer configuration 3d1, only the first 3d orbital is occupied with this single electron. The other four 3d orbitals, and also the 4s and 4p remain empty.

Now, the given octahedral complex ion, [Ti(H2O)6]3+, is formed with d2sp3 hybridisation, using the last two 3d, the 4s and the three 4p orbitals to accept the 12 electrons from the six H2O molecules (ligands). All the 6 hybridised orbitals get completely filled up with the 12 electrons, as it happens in bonding involving hybridisation.

This arrangement, therefore, leaves the second and third 3d orbitals empty.

Suggest Corrections

Similar questions

s p^{3} d^{2}

d^{2} s p^{3}

d^{2} s p^{3}

s p^{3} d^{2}

[C u (N H_{3})_{6}]^{2 +}

[T i (H_{2} O)_{6}]^{3 +}

Join BYJU'S Learning Program