The calculated spin-only magnetic moments of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ and $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ respectively are :

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Coordination Compounds

The calculated spin-only magnetic moments of

K_{3} [Fe (OH)_{6}]

and

K_{4} [Fe (OH)_{6}]

respectively are :

(A)4.90 and 4.90 B.M.
(B)5.92 and 4.90 B.M.✓ Correct
(C)3.87 and 4.90 B.M.
(D)4.90 and 5.92 B.M.

Answer: (B)

Solution

K_{3} [F e (O H)_{6}]

, iron is present as

F e^{3 +}

(

3 d^{5}

);

O H^{-}

is a weak-field ligand, so the configuration remains high-spin with 5 unpaired electrons, giving

μ = 5 (5 + 2) = 35 = 5.92

B.M. In

K_{4} [F e (O H)_{6}]

, iron is

F e^{2 +}

(

3 d^{6}

);

O H^{-}

again keeps it high-spin, leaving 4 unpaired electrons, so

μ = 4 (4 + 2) = 24 = 4.90

B.M. The spin-only magnetic moments are 5.92 B.M. and 4.90 B.M. respectively.

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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