The number of unpaired electrons for the complex ion [VF₆]^3- is

Question

Question

This question was previously asked in

HPSC Assistant Professor 2018 Chemistry (Official Paper)

Answer 1

Crystal Field Theory (CFT):

Crystal Field Theory helps us understand the splitting of d-orbitals in transition metal complexes. Key points include:

Ligands cause the degeneration of d-orbital energies through electrostatic interactions.
In an octahedral field, the five degenerate d-orbitals split into two sets: t_2g (lower energy) and e_g (higher energy) orbitals.
In a tetrahedral field, the splitting pattern is reversed compared to the octahedral field, with e orbitals at lower energy and t₂ at higher energy.
The strength of ligand fields can vary, causing different degrees of splitting. Strong field ligands can enforce electron pairing, while weak field ligands do not.
The difference in energy between the split orbitals is known as the crystal field splitting energy (Δ_o for octahedral complexes and Δ_t for tetrahedral complexes).

For the complex ion [VF₆]^3-:

Vanadium (V) has an atomic number of 23, so its ground state configuration is [Ar] 3d³ 4s².
In the [VF₆]^3- complex, vanadium is in the +3 oxidation state.
The electronic configuration of V³⁺ is [Ar] 3d², as it loses two 4s electrons and one 3d electron.
F^- is a weak field ligand, so it does not cause pairing of electrons in the d-orbitals.
For V³⁺ in an octahedral field: 3d orbitals split into t_2g (lower energy) and e_g (higher energy) orbitals.
The two 3d electrons will occupy the lower energy t_2g orbitals, resulting in 2 unpaired electrons.

The number of unpaired electrons in the complex ion [VF₆]^3- is 2.