JEE Main 2016 — Chemistry Inorganic Chemistry

2016mcqmedium

The pair having the same magnetic moment is:

$[At. No. :Cr=24, Mn=25, Fe=26, Co=27]$

Official previous-year question

Held on 3 Apr 2016 · Verified 6 Jul 2026.

Options

A
${[Mn {({H}_{2}O)}_{6}]}^{2+} and {[Cr{({H}_{2}O)}_{6}]}^{2+}$
B
${[CoC{l}_{4}]}^{2-} and {[Fe{({H}_{2}O)}_{6}]}^{2+}$
C
${[Cr{({H}_{2}O)}_{6}]}^{2+} and {[CoC{l}_{4}]}^{2-}$
D
${[Cr{({H}_{2}O)}_{6}]}^{2+} and {[Fe{({H}_{2}O)}_{6}]}^{2+}$

Did you get this right?

Solution

The compounds having the same number of unpaired electrons will have same magnetic moment.

Magnetic moment $(\mu )=\sqrt{n(n+2)}$

where $n$ is the number of unpaired electrons

${ [ \text{C}\text{r}{ ( {\text{H}}_{2} \text{O} ) }_{6} ] }^{ +2 }$

$\text{C}{\text{r}}^{2+} .(3{\text{d}}^{4} )$ with weak field ligand so unpaired electrons are 4 JEE Main 2016 Chemistry, Coordination Compounds — solution figure 1

So, $\mu =\sqrt{4(4+2)}=\sqrt{24}$

$\text{F}{\text{e}}^{ +2 } ( 3{\text{d}}^{6} )$ Weak field ligand so unpaired electrons are 4 JEE Main 2016 Chemistry, Coordination Compounds — solution figure 2

$\mu =\sqrt{4(4+2)}=\sqrt{24}$

Your note

JEE Main Inorganic Chemistry in other years

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

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JEE Main 2016 — Chemistry Inorganic Chemistry

2016mcqmedium

The pair having the same magnetic moment is:

$[At. No. :Cr=24, Mn=25, Fe=26, Co=27]$

Official previous-year question

Held on 3 Apr 2016 · Verified 6 Jul 2026.

Options

A
${[Mn {({H}_{2}O)}_{6}]}^{2+} and {[Cr{({H}_{2}O)}_{6}]}^{2+}$
B
${[CoC{l}_{4}]}^{2-} and {[Fe{({H}_{2}O)}_{6}]}^{2+}$
C
${[Cr{({H}_{2}O)}_{6}]}^{2+} and {[CoC{l}_{4}]}^{2-}$
D
${[Cr{({H}_{2}O)}_{6}]}^{2+} and {[Fe{({H}_{2}O)}_{6}]}^{2+}$

Did you get this right?

Solution

The compounds having the same number of unpaired electrons will have same magnetic moment.

Magnetic moment $(\mu )=\sqrt{n(n+2)}$

where $n$ is the number of unpaired electrons

${ [ \text{C}\text{r}{ ( {\text{H}}_{2} \text{O} ) }_{6} ] }^{ +2 }$

$\text{C}{\text{r}}^{2+} .(3{\text{d}}^{4} )$ with weak field ligand so unpaired electrons are 4 JEE Main 2016 Chemistry, Coordination Compounds — solution figure 1

So, $\mu =\sqrt{4(4+2)}=\sqrt{24}$

$\text{F}{\text{e}}^{ +2 } ( 3{\text{d}}^{6} )$ Weak field ligand so unpaired electrons are 4 JEE Main 2016 Chemistry, Coordination Compounds — solution figure 2

$\mu =\sqrt{4(4+2)}=\sqrt{24}$

Your note

JEE Main Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

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

Browse all Chemistry topics

JEE Main 2016 — 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 2016 — 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