[Pd(F)(Cl)(Br)(I)] 2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)₆] n–6 , respectively, are:
[Note : Ignore the pairing energy]

Solution

Chemistry ma’am · 4 steps

First find n from the square planar isomers, then fill a low-spin Fe³⁺ into t_(2g) and count.

1. Count the isomers. Pd²⁺ is d⁸, and its complexes are square planar. A square planar [Mabcd] complex has 3 geometrical isomers, because each of the three ligands b, c, d can sit trans to a. So n = 3.
2. Identify the complex. The charge is n - 6 = -3, so the complex is [Fe(CN)₆]³⁻. Each CN⁻ carries -1, so iron is Fe³⁺, which is d⁵.
3. Low spin d5. CN⁻ is a strong-field ligand, so all five electrons go into t_(2g) as t_(2g)⁵ e_g⁰. That leaves 1 unpaired electron.

   μ = √n(n+2) = √(1 × 3) = 1.73 BM, CFSE = 5 × (-0.4 Δ₀) = -2.0 Δ₀

4. So. [Fe(CN)₆]³⁻ has μ = 1.73 BM and CFSE = -2.0 Δ₀, with pairing energy ignored. That is option (a).

The idea to keep. A square planar [Mabcd] complex always gives 3 geometrical isomers. With CN⁻, fill t_(2g) first: each t_(2g) electron adds -0.4 Δ₀ and each e_g electron adds +0.6 Δ₀.

Answer from the NTA final answer key. Solution by PrepWiser.