JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqmedium

The correct order of spin only magnetic moments for the following complex ions is

Official previous-year question

Held on 8 Apr 2023 · Verified 6 Jul 2026.

Options

A
${[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$
B
${[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}<{[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$
C
${[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}$
D
${[{\mathrm{MnBr}}_{4}]}^{2-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$

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Solution

${[{\mathrm{MnBr}}_{4}]}^{2-}$ is a tetrahedral complex. The electronic configuration ${\mathrm{Mn}}^{2+}$ is ${e}^{2}{t}_{2}^{3}$ and it has $5$ unpaired electrons.

$\therefore \mu =\sqrt{5(5+2)}=\sqrt{35}\text{ B. }M.$

${[{\mathrm{CoF}}_{6}]}^{3-}$ is an octahedral complex and it is high spin complex. The electronic configuration of ${\mathrm{Co}}^{3+}$ is ${t}_{2g}^{4}{{e}_{g}}^{2}$ , it has $4$ unpaired electrons

$\therefore \mu =\sqrt{4(4+2)}=\sqrt{24}\text{ B.M. }$

${[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$ is an octahedral complex and it is low spin complex. The electronic configuration ${\mathrm{Mn}}^{3+}$ is ${t}_{2g}^{4}{e}_{g}^{0}$ it has two unpaired electrons

$\therefore \mu =\sqrt{2(2+2)}=\sqrt{8}\text{ B.M }$

$[\mathrm{Fe}(\mathrm{CN}{)}_{6}{]}^{3-}$ is an octahedral complex and it is low spin complex. The electronic configuration of ${\mathrm{Fe}}^{3+}$ is ${t}_{2g}^{5}{\mathrm{eg}}^{0}$ and it has one unpaired electron.

$\therefore \mu =\sqrt{1(1+2)}=2\sqrt{3}\text{ B.M }$

Hence the correct answer is option $3$

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Options

${[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$

${[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}<{[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$

${[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[{\mathrm{MnBr}}_{4}]}^{2-}$

${[{\mathrm{MnBr}}_{4}]}^{2-}<{[{\mathrm{CoF}}_{6}]}^{3-}<{[\mathrm{Fe}(\mathrm{CN}{)}_{6}]}^{3-}<{[\mathrm{Mn}(\mathrm{CN}{)}_{6}]}^{3-}$

Solution

${[{\mathrm{MnBr}}_{4}]}^{2-}$ is a tetrahedral complex. The electronic configuration ${\mathrm{Mn}}^{2+}$ is ${e}^{2}{t}_{2}^{3}$ and it has $5$ unpaired electrons.

$\therefore \mu =\sqrt{5(5+2)}=\sqrt{35}\text{ B. }M.$

${[{\mathrm{CoF}}_{6}]}^{3-}$ is an octahedral complex and it is high spin complex. The electronic configuration of ${\mathrm{Co}}^{3+}$ is ${t}_{2g}^{4}{{e}_{g}}^{2}$ , it has $4$ unpaired electrons

$\therefore \mu =\sqrt{4(4+2)}=\sqrt{24}\text{ B.M. }$

$\therefore \mu =\sqrt{2(2+2)}=\sqrt{8}\text{ B.M }$

$\therefore \mu =\sqrt{1(1+2)}=2\sqrt{3}\text{ B.M }$

Hence the correct answer is option $3$

JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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