JEE Main 2021 — Chemistry Inorganic Chemistry

2021mcqhard

The calculated magnetic moments (spin only value) for species ${[{\mathrm{FeCl}}_{4}]}^{2-},{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}$ and ${\mathrm{MnO}}_{4}^{2-}$ respectively are:

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Options

A
$4.90,0$ and $1.73\mathrm{BM}$
B
$4.90,0$ and $2.83\mathrm{BM}$
C
$5.82,0$ and $0\mathrm{BM}$
D
$5.92,4.90$ and $0\mathrm{BM}$

Did you get this right?

Solution

JEE Main 2021 Chemistry, Coordination Compounds — solution figure 1

(i) ${[{\mathrm{FeCl}}_{4}]}^{2-}{\mathrm{Fe}}^{2+}3{d}^{6}\rightarrow 4$ unpaired electrons.

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

$=\sqrt{4(4+2)}\mathrm{BM}$

$=\sqrt{24}\mathrm{BM}\Rightarrow 4.90\mathrm{BM}$

(ii) ${[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{-3}$ is a strong field ligand & causes

pairing & hence no. unpaired electron

JEE Main 2021 Chemistry, Coordination Compounds — solution figure 2

$\mu =0$

(iii) ${[{\mathrm{MnO}}_{4}]}^{2-}{\mathrm{Mn}}^{+6}$ it has one unpaired electron.

${\mathrm{Mn}}^{+6}\Rightarrow [\mathrm{Ar}]3{d}^{1}\mu =\sqrt{n(n+2)}\mathrm{BM}$

$=\sqrt{1(1+2)}\mathrm{BM}$

$=\sqrt{3}\mathrm{BM}\Rightarrow 1.73\mathrm{BM}$

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JEE Main 2021 — Chemistry Inorganic Chemistry

2021mcqhard

The calculated magnetic moments (spin only value) for species ${[{\mathrm{FeCl}}_{4}]}^{2-},{[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{3-}$ and ${\mathrm{MnO}}_{4}^{2-}$ respectively are:

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Options

A
$4.90,0$ and $1.73\mathrm{BM}$
B
$4.90,0$ and $2.83\mathrm{BM}$
C
$5.82,0$ and $0\mathrm{BM}$
D
$5.92,4.90$ and $0\mathrm{BM}$

Did you get this right?

Solution

JEE Main 2021 Chemistry, Coordination Compounds — solution figure 1

(i) ${[{\mathrm{FeCl}}_{4}]}^{2-}{\mathrm{Fe}}^{2+}3{d}^{6}\rightarrow 4$ unpaired electrons.

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

$=\sqrt{4(4+2)}\mathrm{BM}$

$=\sqrt{24}\mathrm{BM}\Rightarrow 4.90\mathrm{BM}$

(ii) ${[\mathrm{Co}{({C}_{2}{O}_{4})}_{3}]}^{-3}$ is a strong field ligand & causes

pairing & hence no. unpaired electron

JEE Main 2021 Chemistry, Coordination Compounds — solution figure 2

$\mu =0$

(iii) ${[{\mathrm{MnO}}_{4}]}^{2-}{\mathrm{Mn}}^{+6}$ it has one unpaired electron.

${\mathrm{Mn}}^{+6}\Rightarrow [\mathrm{Ar}]3{d}^{1}\mu =\sqrt{n(n+2)}\mathrm{BM}$

$=\sqrt{1(1+2)}\mathrm{BM}$

$=\sqrt{3}\mathrm{BM}\Rightarrow 1.73\mathrm{BM}$

Your note

JEE Main Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

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JEE Main 2021 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

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JEE Main 2021 — 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

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