Answer :

Solution :

F^⊝ with Fe⁺³ behaves as WFL, Hence pairing does not take place, so it forms high spin complex.

[FeF₆]³– ⇒ sp³d² hybridisation

Number of unpaired electron = 5 [Fe(CN)₆]^3- ⇒ d²sp³ hybridisation

Fe⁺³ = 3d⁵ Number of unpaired electron = 1

[Fe(H₂O)₆]²⁺ ⇒ sp³d²

Fe⁺² = 3d⁶

Number of unpaired electron = 4

[V(H₂O)₆]⁺² ⇒ d²sp³ hybridisation

V⁺² = 3d³

Number of unpaired electron = 3

Spin angular momentum = $\sqrt{S(S+1)}\frac{h}{2\mathrm{\pi <!--\; \pi \; -->}}$

S = total spin quantum no.

More the number of unpaired electron more will be spin angular momentum. [FeF₆]^3– has 5 unpaired electron hence maximum spin angular momentum value.