Correct order of spin only magnetic… - JEE Main 2023 Chemistry PYQ | PrepForBharat

JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqmedium

Correct order of spin only magnetic moment of the following complex ions is:

(Given At. No. $\mathrm{Fe}:26,\mathrm{Co}:27$ )

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Options

A
${[{\mathrm{FeF}}_{6}]}^{3-}>{[{\mathrm{CoF}}_{6}]}^{3-}>{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}$
B
${[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}>{[{\mathrm{CoF}}_{6}]}^{3-}>{[{\mathrm{FeF}}_{6}]}^{3-}$
C
${[{\mathrm{FeF}}_{6}]}^{3-}>{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}>{[{\mathrm{CoF}}_{6}]}^{3-}$
D
${[{\mathrm{CoF}}_{6}]}^{3-}>{[{\mathrm{FeF}}_{6}]}^{3-}>{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}$

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Solution

(1) ${[{\mathrm{FeF}}_{6}]}^{3-}:{\mathrm{Fe}}^{3+}=3{d}^{5}({\Delta }_{O}<P)$

JEE Main 2023 Chemistry, Coordination Compounds — solution figure 1

Number of unpaired ${e}^{-}=5$

$\mu =\sqrt{n(n+2)}\mathrm{BM}$

$\therefore \mu =\sqrt{35}\mathrm{BM}$

(2) ${[{\mathrm{CoF}}_{6}]}^{3-}:{\mathrm{Co}}^{3+}=3{d}^{6}({\Delta }_{O}<P)$

JEE Main 2023 Chemistry, Coordination Compounds — solution figure 2

Number of unpaired ${e}^{-}=4$

$\mu =\sqrt{n(n+2)}\mathrm{BM}$

$\therefore \mu =\sqrt{24}\mathrm{BM}$

(3) ${[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}:{\mathrm{Co}}^{3+}=3{d}^{6}({\Delta }_{O}>P)$

JEE Main 2023 Chemistry, Coordination Compounds — solution figure 3

Number of unpaired ${e}^{-}=0$

$\mu =\sqrt{n(n+2)}\mathrm{BM}$

$\therefore \mu =0\mathrm{BM}$

So, the correct order is ${[{\mathrm{FeF}}_{6}]}^{3-}>{[{\mathrm{CoF}}_{6}]}^{3-}>{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}$

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