Question

Spin only magnetic moment of an octahedral complex of Fe²⁺ in the presence of a strong field ligand in BM is :

(1) 4.89

(2) 2.82

(3) 0

(4) 3.46

Answer 1

Answer is (3) 0

In presence of SFL △₀ > p means pairing occurs therefore

For Fe²⁺ → 3d⁶

∴ No of unpaired e^-(s) = 0

∴ μ = \(\sqrt{n(n+2)}BM = 0\)

[n = No of unpaired e^– (s)] 

In NiCl₂ Ni⁺² is having configuration 3d⁸

∴ Number of unpaired electron = 2 

After formation of oxidised product 

[Ni(CN)₆] ^–2 Ni⁺⁴ is obtained 

Ni⁺⁴ ⇒ 3d⁶ and CN^– is strong field ligand  

∴ number of unpaired electrons = 0

∴ The charge is 2 – 0 = 2