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Structure of Atom question

2023 · 24 Jan · Shift 1 · Q8

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Structure of Atom
2023 · 24 Jan · Shift 1 · Q8

Structure of Atom question

2023 · 24 Jan · Shift 1 · Q8

JEE MainChemistryStructure of AtomMCQ+4 / −1

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

A
Ti $^{2+}$
B
V $^{2+}$
C
Cr $^{2+}$
D
Mn $^{2+}$

View written solutionFree

Correct answer: B

Use the spin-only magnetic moment formula

For most first-row transition metal ions,

\mu = \sqrt{n(n+2)}\ \text{B.M.}

where $n$ is the number of unpaired electrons.

Given:

\mu = 3.87\ \text{B.M.}

Find the number of unpaired electrons

Check small integer values of $n$ :

If $n=1$ , $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$
If $n=2$ , $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$
If $n=3$ , $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$

So the ion has

n=3

unpaired electrons.

Find electronic configuration of each ion

For transition metal ions, electrons are removed first from $4s$ and then from $3d$ .

Ti: $Z=22$ Neutral: $[\mathrm{Ar}]\,3d^2 4s^2$
$\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$
Unpaired electrons $=2$
V: $Z=23$ Neutral: $[\mathrm{Ar}]\,3d^3 4s^2$
$\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$
Unpaired electrons $=3$
Cr: $Z=24$ Neutral: $[\mathrm{Ar}]\,3d^5 4s^1$
$\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$
Unpaired electrons $=4$
Mn: $Z=25$ Neutral: $[\mathrm{Ar}]\,3d^5 4s^2$
$\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$
Unpaired electrons $=5$

Match with $n=3$

Only $\mathrm{V}^{2+}$ has 3 unpaired electrons, so

\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\ \text{B.M.}

Evaluate options

A: $\mathrm{Ti}^{2+}$ $\rightarrow 2$ unpaired, incorrect
B: $\mathrm{V}^{2+}$ $\rightarrow 3$ unpaired, correct
C: $\mathrm{Cr}^{2+}$ $\rightarrow 4$ unpaired, incorrect
D: $\mathrm{Mn}^{2+}$ $\rightarrow 5$ unpaired, incorrect

Final Answer:

\boxed{\mathrm{V}^{2+}}

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