Which one of the following complexes will have $\Delta_{0}=0$ and $\mu=5.96$ B.M.?

(D) $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4-}$

(A) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$

(B) $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$

(C) $\left[\mathrm{FeF}_{6}\right]^{4-}$

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Coordination Compounds

Which one of the following complexes will have

Δ_{0} = 0

and

μ = 5.96

B.M.?

(A) $[Fe (CN)_{6}]^{4 -}$
(B) $[CO (NH_{3})_{6}]^{3 +}$
(C) $[FeF_{6}]^{4 -}$
(D) $[Mn (SCN)_{6}]^{4 -}$ ✓ Correct

Answer: (D)

Solution

A magnetic moment of

5.96 B.M.

fits

n (n + 2) = 5.96

, so

n = 5

unpaired electrons, while

Δ_{0}

being effectively zero demands a configuration with no net crystal-field stabilization.

Mn^{2 +}

is a

3 d^{5}

ion, and with the weak-field thiocyanate ligands it remains high-spin as

t_{2 g}^{3} e_{g}^{2}

, giving five unpaired electrons and

CFSE = (- 0.4 \times 3 + 0.6 \times 2) Δ_{0} = 0

. This matches

μ = 5 \times 7 = 5.92 \approx 5.96 B.M.

, so the complex meeting both conditions is

[Mn (SCN)_{6}]^{4 -}

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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