10Dq and B values for [V(H2O)6]+3

Calculate the 10Dq and B values for [V(H₂O)₆]⁺³ in terms of v₁, v₂ etc.

Water is a weak ligand hence, the given complex is a weak field octahedral complex.

V is in +3 state in the complex [V(H₂O)₆]⁺³

Spin multiplicity = n + 1 = 2 + 1 = 3
Total orbital angular momentum (L) = 3 = F

The ³F ground term has also the ³P is excited term. The term splitting is shown as-

Out of three expected d-d bonds, two are observed as-
v₁ = 17200 cm⁻¹, ε = 6
v₂ = 25600 cm⁻¹, ε = 8
v is due to ³T_2g ← ³T_1g at 8Dq.
If II bond is due to ³A_2g ← ³T_1g at 18 Dq then,
v₂ = (17200/8) × 18 = 36000 cm⁻¹
But, v₂ occurs at 25600 cm⁻¹
So, v₂ is not due to ³A_2g ← ³T_1g but actually it is due to ³T_1g(g) ← ³T_1g at 6Dq + 15B

ε values are low because, transition are symmetry forbidden.
Since, 8Dq = 17200 cm⁻¹
So, 10Dq = (17200/8) × 10 = 21500 cm⁻¹
And CFSE = −6Dq = − (21500/10) × 6 = −12900 cm⁻¹
Since, 6Dq = +15Dq = 25600cm⁻¹
SO, B = (25600 − 12900)/15 = 12700/15 = 846.6 cm⁻¹

The bonds are broad due to J.T.distortion in the excited ³T_2g terms and L-S coupling in the ground ³T_1g term. Bands are symmetric due to both J.T. and L-S coupling occuring simultaneously.