Question

[PdFClBrI]^2– Number of Geometrical Isomers = n. For [Fe(CN)₆]^n–6 , Determine the spin only magnetic moment and CFSE (Ignore the pairing energy)

(1) 1.73 B.M., –2Δ₀ 

(2) 2.84 B.M., –1.6Δ₀ 

(3) 0, –1.6Δ₀ 

(4) 5.92 B.M., –2.4Δ₀ 

Answer 1

The answer is (3) 0, –1.6Δ₀ 

Number of Geometrical Isomers in square planar [PdFClBrI]^2– are = 3

Hence, n = 3

[Fe(CN)₆]^3– 

Fe³⁺ = 3d⁵, According to CFT configuration is t²²¹_2g e⁰⁰_g