JEE Main 2022 — Chemistry Inorganic Chemistry

2022mcqhard

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $\mathrm{Mn}=25;\mathrm{Fe}=26$ )

A. ${[{\mathrm{FeF}}_{6}]}^{3-}$

B. ${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$

C. ${[{\mathrm{MnCl}}_{6}]}^{3-}$ (high spin)

D. ${[\mathrm{Mn}{(\mathrm{CN})}_{6}]}^{3-}$

Choose the correct answer from the options given below

Official previous-year question

Held on 27 Jun 2022 · Verified 6 Jul 2026.

Options

A
$A<B<D<C$
B
$B<D<C<A$
C
$A<C<D<B$
D
$B<D<A<C$

Did you get this right?

Solution

Magnetic moment is directly proportional to the number of unpaired in the central metal atom of a complex.

It is calculated by the formula, $\mu =\sqrt{n(n+2)}\mathrm{BM}$

(A) ${[{\mathrm{FeF}}_{6}]}^{3-}$

${\mathrm{Fe}}^{+3}\rightarrow 3{d}^{5}4{s}^{0}$

$n=5$

(B) ${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 1

${\mathrm{Fe}}^{+3}\rightarrow 3{d}^{5}4{s}^{0}$

$n=1$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 2

${\mathrm{Mn}}^{+3}\rightarrow 3{d}^{4}4{s}^{0}$

$n=4$

(D) ${[\mathrm{Mn}{(\mathrm{CN})}_{6}]}^{3-}$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 3

${\mathrm{Mn}}^{+3}\rightarrow 3{d}^{4}4{s}^{0}$

$n=2$

$\mathrm{Hence},\mathrm{correct}\mathrm{order}\mathrm{of}\mathrm{magnetic}\mathrm{moment}\mathrm{is}A>C>D>B$

Your note

JEE Main Inorganic Chemistry in other years

Inorganic Chemistry 2026 Inorganic Chemistry 2025 Inorganic Chemistry 2024 Inorganic Chemistry 2023 Inorganic Chemistry 2021 Inorganic Chemistry 2020 Inorganic Chemistry 2019 Inorganic Chemistry 2018 Inorganic Chemistry 2017 Inorganic Chemistry 2016 Inorganic Chemistry 2015 Inorganic Chemistry 2014

Keep practicing Inorganic Chemistry

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

Browse all Chemistry topics

Solution

Magnetic moment is directly proportional to the number of unpaired in the central metal atom of a complex.

It is calculated by the formula, $\mu =\sqrt{n(n+2)}\mathrm{BM}$

(A) ${[{\mathrm{FeF}}_{6}]}^{3-}$

${\mathrm{Fe}}^{+3}\rightarrow 3{d}^{5}4{s}^{0}$

$n=5$

(B) ${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 1

${\mathrm{Fe}}^{+3}\rightarrow 3{d}^{5}4{s}^{0}$

$n=1$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 2

${\mathrm{Mn}}^{+3}\rightarrow 3{d}^{4}4{s}^{0}$

$n=4$

(D) ${[\mathrm{Mn}{(\mathrm{CN})}_{6}]}^{3-}$

JEE Main 2022 Chemistry, Coordination Compounds — solution figure 3

${\mathrm{Mn}}^{+3}\rightarrow 3{d}^{4}4{s}^{0}$

$n=2$

$\mathrm{Hence},\mathrm{correct}\mathrm{order}\mathrm{of}\mathrm{magnetic}\mathrm{moment}\mathrm{is}A>C>D>B$

JEE Main 2022 — 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

JEE Main 2022 — 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