In which of the following complexes the CFSE, $\Delta_{0}$ will be equal to zero?

(D) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{SCN})_{6}\right]$

(A) $\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Br}_{2}$

(B) $\left[\mathrm{Fe}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}$

(C) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Coordination Compounds

In which of the following complexes the CFSE,

Δ_{0}

will be equal to zero?

Answer: (D)

CFSE depends on how the d-electrons are distributed between the

t_{2 g}

and

e_{g}

sets in an octahedral field. In

K_{3} [F e (S C N)_{6}]

, iron is

F e^{3 +}

(

d^{5}

), and

S C N^{-}

is a weak-field ligand, so the complex is high spin with configuration

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

C F S E = (- 0.4 \times 3 + 0.6 \times 2) Δ_{0} = (- 1.2 + 1.2) Δ_{0} = 0

The half-filled high-spin

d^{5}

arrangement puts three electrons in

t_{2 g}

and two in

e_{g}

, exactly cancelling any net stabilization.

In

[F e (N H_{3})_{6}] B r_{2}

and

K_{4} [F e (C N)_{6}]

, iron is

F e^{2 +}

(

d^{6}

);

N H_{3}

gives a high-spin

t_{2 g}^{4} e_{g}^{2}

arrangement (CFSE

= - 0.4 Δ_{0}

) while the strong-field

C N^{-}

forces low spin,

t_{2 g}^{6}

(CFSE

= - 2.4 Δ_{0}

). In

[F e (e n)_{3}] C l_{3}

F e^{3 +}

with the strong-field chelate en is low spin,

t_{2 g}^{5}

(CFSE

= - 2.0 Δ_{0}

). Only the high-spin

d^{5}

thiocyanate complex gives zero net CFSE.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option