JEE Main 2020 — Coordination Compounds Question with Solution

Complex [Pd(F)(Cl)(Br)(I)]^2– (square planar geometry)-has 3 geometrical isomers.

$$\therefore$$ n = 3

[Fe(CN)₆]^n–6 becomes [Fe(CN)₆]^3-

Here Oxidation state of Fe = +3

$$\therefore$$ Fe⁺³ = [Ar]3d⁵4s⁰

CN^- is strong field ligand so it pairing of electrons happens.

E.C. according to CFT = ($$t_{2g}^{5}e{g}^0$$)

$$\therefore$$ No. of unpaired e^- = 1

Then Magnetic moment = $$\sqrt{n\left(n+2\right)}$$ = $$\sqrt{3}$$ = 1.73 B.M

CFSE = [(–0.4 × 5) + (0.6 × 0)] $$\Delta$$₀ = –2.0 $$\Delta$$₀