JEE Main 2021 — Chemistry Inorganic Chemistry

2021mcqmedium

Arrange the following metal complex/ compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system.

(Atomic numbers $\mathrm{Ce}=58,\mathrm{Gd}=64$ and $\mathrm{Eu}=63$ .)

(a) ${({\mathrm{NH}}_{4})}_{2}[\mathrm{Ce}{({\mathrm{NO}}_{3})}_{6}]$

(b) $\mathrm{Gd}{({\mathrm{NO}}_{3})}_{3}$ and

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

Options

A
$(b)<(a)<(c)$
B
$(c)<(a)<(b)$
C
$(a)<(b)<(c)$
D
$(a)<(c)<(b)$

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Solution

(a) $\mathrm{Ce}58\rightarrow [\mathrm{Xe}]4{f}^{2}5{d}^{0}6{s}^{2}$

In complex ${\mathrm{Ce}}^{4+}\rightarrow [\mathrm{Xe}]4{f}^{0}5{d}^{0}6{s}^{0}$

there is no unpaired electron so ${\mu }_{m}=0$

(b) ${\mathrm{Gd}64}^{3+}\rightarrow [\mathrm{Xe}]4{f}^{7}5{d}^{0}6{s}^{0}$

contain seven unpaired electrons so, ${\mu }_{m}=\sqrt{7(7+2)}=\sqrt{63}B.M.$

contain six unpaired electron

so, ${\mu }_{m}=\sqrt{6(6+2)}=\sqrt{48}B.M.$

Hence, order of spin only magnetic movement

$b>c>a$

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Solution

(a) $\mathrm{Ce}58\rightarrow [\mathrm{Xe}]4{f}^{2}5{d}^{0}6{s}^{2}$

In complex ${\mathrm{Ce}}^{4+}\rightarrow [\mathrm{Xe}]4{f}^{0}5{d}^{0}6{s}^{0}$

there is no unpaired electron so ${\mu }_{m}=0$

(b) ${\mathrm{Gd}64}^{3+}\rightarrow [\mathrm{Xe}]4{f}^{7}5{d}^{0}6{s}^{0}$

contain seven unpaired electrons so, ${\mu }_{m}=\sqrt{7(7+2)}=\sqrt{63}B.M.$

contain six unpaired electron

so, ${\mu }_{m}=\sqrt{6(6+2)}=\sqrt{48}B.M.$

Hence, order of spin only magnetic movement

$b>c>a$

JEE Main 2021 — Chemistry Inorganic Chemistry

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JEE Main 2021 — Chemistry Inorganic Chemistry

Options

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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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