JEE Main 2024 — Chemistry Inorganic Chemistry

2024mcqhard

Consider the following complex ions

$P=[{\mathrm{FeF}}_{6}{]}^{3–}$ , $Q=[V({H}_{2}O){}_{6}{]}^{2+}$ , $R=[\mathrm{Fe}({H}_{2}O{)}_{6}{]}^{2+}$

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

Options

A
$R<Q<P$
B
$R<P<Q$
C
$Q<R<P$
D
$Q<P<R$

Did you get this right?

Solution

$[{\mathrm{FeF}}_{6}{]}^{3–}:{\mathrm{Fe}}^{+3}:[\mathrm{Ar}]3{d}^{5}$

$F$ : Weak field Ligand

JEE Main 2024 Chemistry, Coordination Compounds — solution figure 1

No. of unpaired electrons $=5$

$\mu =\sqrt{(5(5+2))}$

$\mu =\sqrt{35}\mathrm{BM}$

${[V{({H}_{2}O)}_{6}]}^{+2}:{V}^{+2}:3{d}^{3}$ JEE Main 2024 Chemistry, Coordination Compounds — solution figure 2

No. of unpaired electrons $=3$

$\mu =\sqrt{3(3+2)}$

$\mu =\sqrt{15}\mathrm{BM}$

${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{+2}:{\mathrm{Fe}}^{+2}:3{d}^{6}$

${H}_{2}O$ : Weak field Ligand

JEE Main 2024 Chemistry, Coordination Compounds — solution figure 3

No. of unpaired electrons $=4$

$\mu =\sqrt{4(4+2)}$

$\mu =\sqrt{24}\mathrm{BM}$

Hence, the answer is option C.

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JEE Main 2024 — Chemistry Inorganic Chemistry

2024mcqhard

Consider the following complex ions

$P=[{\mathrm{FeF}}_{6}{]}^{3–}$ , $Q=[V({H}_{2}O){}_{6}{]}^{2+}$ , $R=[\mathrm{Fe}({H}_{2}O{)}_{6}{]}^{2+}$

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

Options

A
$R<Q<P$
B
$R<P<Q$
C
$Q<R<P$
D
$Q<P<R$

Did you get this right?

Solution

$[{\mathrm{FeF}}_{6}{]}^{3–}:{\mathrm{Fe}}^{+3}:[\mathrm{Ar}]3{d}^{5}$

$F$ : Weak field Ligand

JEE Main 2024 Chemistry, Coordination Compounds — solution figure 1

No. of unpaired electrons $=5$

$\mu =\sqrt{(5(5+2))}$

$\mu =\sqrt{35}\mathrm{BM}$

${[V{({H}_{2}O)}_{6}]}^{+2}:{V}^{+2}:3{d}^{3}$ JEE Main 2024 Chemistry, Coordination Compounds — solution figure 2

No. of unpaired electrons $=3$

$\mu =\sqrt{3(3+2)}$

$\mu =\sqrt{15}\mathrm{BM}$

${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{+2}:{\mathrm{Fe}}^{+2}:3{d}^{6}$

${H}_{2}O$ : Weak field Ligand

JEE Main 2024 Chemistry, Coordination Compounds — solution figure 3

No. of unpaired electrons $=4$

$\mu =\sqrt{4(4+2)}$

$\mu =\sqrt{24}\mathrm{BM}$

Hence, the answer is option C.

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 2024 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

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Physical Chemistry

Organic Chemistry

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JEE Main 2024 — 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