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

Did you get this right?

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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Options

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

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

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

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

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

JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry