P d F C l B r I 2 - has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilization energy C F S E of F e C N 6 n - 6 , respectively, are: [Note: Ignore the pairing energy]

Step-by-step solution

Number of Geometrical Isomers in square planar P d F C l B r 2 - are = 3 Hence, n = 3 Fe CN 6 3 - Fe 3 + = 3 d 5 , According to CFT, configuration is t 2 g 221 e g 00 ; it means number of unpaired electron ( n ) = 1 So spin-only magnetic moment μ = n n + 2 = 1 .73  BM CFSE = − 0 .4   Δ 0 × nt 2 g + 0 .6   Δ 0 × n eg = - 0.4 ∆ 0 × 5 = - 2.0 ∆ 0