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d- and f-Block Elements: Chemistry | JEE Main

Q: Is there a quick trick to find n from μ?

Yes, for , the integer part of in B.M. is always equal to (here, ).

The magnetic moment of a transition metal compound has been calculated to be

3.87 \mathrm{~B.M.}

The metal ion is

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What feels right?

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\mathrm{Cr}^{2+}

\mathrm{Ti}^{2+}

\mathrm{V}^{2+}

\mathrm{Mn}^{2+}

Step-by-step solutionView

Correct answer

A spin-only magnetic moment of $3.87 \text{ B.M.}$ corresponds to $\sqrt{n(n+2)} = \sqrt{15}$ , which indicates $n = 3$ unpaired electrons, matching $\mathrm{V}^{2+}$ ( $3d^3$ ).

Option analysis

Why each option works or fails

A ·

\mathrm{Cr}^{2+}

Confusing the $d$ -electron count of $\mathrm{Cr}^{2+}$ ( $3d^4$ , having 4 unpaired electrons with $\mu = 4.90 \text{ B.M.}$ ) with 3 unpaired electrons. Write the outer electronic configuration of chromium as $[ ext{Ar}]3d^5 4s^1$ , so $\mathrm{Cr}^{2+}$ is $3d^4$ with $n = 4$ unpaired electrons.

B ·

\mathrm{Ti}^{2+}

Assuming that titanium(II) retains 3 unpaired electrons instead of calculating its configuration properly. Titanium has $Z = 22$ with configuration $[ ext{Ar}]3d^2 4s^2$ ; loss of two electrons yields $\mathrm{Ti}^{2+}$ as $3d^2$ , giving $n = 2$ and $\mu = 2.83 \text{ B.M.}$

C ·

\mathrm{V}^{2+}

None. This option is correct. Vanadium has $Z = 23$ with configuration $[ ext{Ar}]3d^3 4s^2$ . The $\mathrm{V}^{2+}$ ion has a $3d^3$ configuration, giving $n = 3$ unpaired electrons and $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}$

D ·

\mathrm{Mn}^{2+}

Associating $\mathrm{Mn}^{2+}$ incorrectly with 3 unpaired electrons rather than recognizing its half-filled $d^5$ shell. Manganese has $Z = 25$ ( $[\text{Ar}]3d^5 4s^2$ ). The $\mathrm{Mn}^{2+}$ ion is $3d^5$ with $n = 5$ unpaired electrons, resulting in $\mu = \sqrt{5(7)} = 5.92 \text{ B.M.}$

Reviewed route

Solution

StepWorking

01given

Spin-only magnetic moment $\mu = 3.87\text{ B.M.}$

02strategise

Use spin-only formula $\mu = \sqrt{n(n+2)}\text{ B.M.}$ to determine the number of unpaired electrons $n$ , then match with the electronic configurations of the given divalent ions.

03execute

Solve $\sqrt{n(n+2)} = 3.87$ . Since $3.87 \approx \sqrt{15} = \sqrt{3(3+2)}$ , $n = 3$ .

✓verify

Writing $3d$ configurations of ions: $\mathrm{Cr^{2+}} = 3d^4$ (4 unpaired), $\mathrm{Ti^{2+}} = 3d^2$ (2 unpaired), $\mathrm{V^{2+}} = 3d^3$ (3 unpaired), $\mathrm{Mn^{2+}} = 3d^5$ (5 unpaired). Ion with $n=3$ is $\mathrm{V^{2+}}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Is there a quick trick to find n from μ?

Yes, for $\mu = \sqrt{n(n+2)}$ , the integer part of $\mu$ in B.M. is always equal to $n$ (here, $3.87 \implies n = 3$ ).

Answer

A spin-only magnetic moment of $3.87 \text{ B.M.}$ corresponds to $\sqrt{n(n+2)} = \sqrt{15}$ , which indicates $n = 3$ unpaired electrons, matching $\mathrm{V}^{2+}$ ( $3d^3$ ).

Why each option works or fails

A: $\mathrm{Cr}^{2+}$ - Confusing the $d$ -electron count of $\mathrm{Cr}^{2+}$ ( $3d^4$ , having 4 unpaired electrons with $\mu = 4.90 \text{ B.M.}$ ) with 3 unpaired electrons. Write the outer electronic configuration of chromium as $[ ext{Ar}]3d^5 4s^1$ , so $\mathrm{Cr}^{2+}$ is $3d^4$ with $n = 4$ unpaired electrons.
B: $\mathrm{Ti}^{2+}$ - Assuming that titanium(II) retains 3 unpaired electrons instead of calculating its configuration properly. Titanium has $Z = 22$ with configuration $[ ext{Ar}]3d^2 4s^2$ ; loss of two electrons yields $\mathrm{Ti}^{2+}$ as $3d^2$ , giving $n = 2$ and $\mu = 2.83 \text{ B.M.}$
C · correct: $\mathrm{V}^{2+}$ - None. This option is correct. Vanadium has $Z = 23$ with configuration $[ ext{Ar}]3d^3 4s^2$ . The $\mathrm{V}^{2+}$ ion has a $3d^3$ configuration, giving $n = 3$ unpaired electrons and $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}$
D: $\mathrm{Mn}^{2+}$ - Associating $\mathrm{Mn}^{2+}$ incorrectly with 3 unpaired electrons rather than recognizing its half-filled $d^5$ shell. Manganese has $Z = 25$ ( $[\text{Ar}]3d^5 4s^2$ ). The $\mathrm{Mn}^{2+}$ ion is $3d^5$ with $n = 5$ unpaired electrons, resulting in $\mu = \sqrt{5(7)} = 5.92 \text{ B.M.}$

Step-by-step solution

given: Spin-only magnetic moment $\mu = 3.87\text{ B.M.}$
strategise: Use spin-only formula $\mu = \sqrt{n(n+2)}\text{ B.M.}$ to determine the number of unpaired electrons $n$ , then match with the electronic configurations of the given divalent ions.
execute: Solve $\sqrt{n(n+2)} = 3.87$ . Since $3.87 \approx \sqrt{15} = \sqrt{3(3+2)}$ , $n = 5$ .
verify: Writing $3d$ configurations of ions: $\mathrm{Cr^{2+}} = 3d^4$ (4 unpaired), $\mathrm{Ti^{2+}} = 3d^2$ (2 unpaired), $\mathrm{V^{2+}} = 3d^3$ (3 unpaired), $\mathrm{Mn^{2+}} = 3d^5$ (5 unpaired). Ion with $n=3$ is $\mathrm{V^{2+}}$ .

Shortcut:

When to use it: Direct inspection of integer part of spin-only magnetic moment in B.M.

quick_kill: The integer before the decimal in $\mu = 3.87\text{ B.M.}$ directly gives the number of unpaired electrons $n = 3$ , which corresponds to $3d^3$ ( $\mathrm{V^{2+}}$ ).

d- and f-Block Elements: Chemistry | JEE Main

What feels right?

A spin-only magnetic moment of $3.87 \text{ B.M.}$ corresponds to $\sqrt{n(n+2)} = \sqrt{15}$ , which indicates $n = 3$ unpaired electrons, matching $\mathrm{V}^{2+}$ ( $3d^3$ ).

Why each option works or fails

Solution

01given

02strategise

03execute

✓verify

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Strong reducing and oxidizing agents among the following, respectively, are

Students also ask

d- and f-Block Elements: Chemistry | JEE Main

A spin-only magnetic moment of 3.87 B.M.3.87 \text{ B.M.}3.87 B.M. corresponds to n(n+2)=15\sqrt{n(n+2)} = \sqrt{15}n(n+2)​=15​, which indicates n=3n = 3n=3 unpaired electrons, matching V2+\mathrm{V}^{2+}V2+ (3d33d^33d3).

01given

02strategise

03execute

✓verify

Strong reducing and oxidizing agents among the following, respectively, are

Students also ask

A spin-only magnetic moment of $3.87 \text{ B.M.}$ corresponds to $\sqrt{n(n+2)} = \sqrt{15}$ , which indicates $n = 3$ unpaired electrons, matching $\mathrm{V}^{2+}$ ( $3d^3$ ).