Question 20.10: Describing the Bonding in Complexes in Terms of Valence Bond......

STRATEGY AND SOLUTION

The free $V^{3+}$ ion has the electron configuration [Ar] 3d² and the orbital diagram

Because $[V(NH_3)_6]^{3+}$ is octahedral, the $V^{3+}$ ion must use either d² sp³ or sp³ d² hybrid orbitals in accepting a share in six pairs of electrons from the six $NH_3$ ligands. The preferred hybrids are d² sp³ because several 3d orbitals are vacant and d² sp³ hybrids have lower energy than sp³ d² hybrids (because the 3d orbitals have lower energy than the 4d orbitals). Thus, $[V(NH_3)_6]^{3+}$ has the following orbital diagram: (the photo below)

The complex has two unpaired electrons and is therefore paramagnetic.