JEE Main 2025 Jan 29 Shift 2, Chemistry Q1: Colour, Magnetism And Stability Of Co-ordination Compounds

JEE Main2025Jan 29, Shift 2Chemistry

The calculated spin-only magnetic moments of $K_{3} [F e (O H)_{6}]$ and $K_{4} [F e (O H)_{6}]$ respectively are :

Solution

Find the oxidation state of Fe and count unpaired electrons. $O H^{-}$ is a weak-field ligand, so both complexes are high-spin.

$K_{3} [F e (O H)_{6}]$ : Iron is in the $+ 3$ state, configuration $3 d^{5}$ . High-spin gives $5$ unpaired electrons.

$μ = n (n + 2) = 5 \times 7 = 35 \approx 5.92$ B.M.

$K_{4} [F e (O H)_{6}]$ : Iron is in the $+ 2$ state, configuration $3 d^{6}$ . High-spin gives $4$ unpaired electrons.

$μ = 4 \times 6 = 24 \approx 4.90$ B.M.

$∴$ The values are $5.92$ and $4.90$ B.M.

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