The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is -

Coordination Compounds PYQ — JEE Main 2019 · 12 Jan · Shift 2 · Q1 with Solution

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Correct answer: C

Determine the electronic configuration of $\text{Mn}^{2+}$

Mn has atomic number $25$ .

\text{Mn}: [\text{Ar}]\,3d^5 4s^2

For $\text{Mn}^{2+}$ , remove two electrons from $4s$ first:

\text{Mn}^{2+}: [\text{Ar}]\,3d^5

So the metal ion is a $d^5$ system.

Use the given magnetic moment

The spin-only magnetic moment is:

\mu = \sqrt{n(n+2)}\ \text{BM}

Given:

\mu = 5.9\ \text{BM}

Now check for $n=5$ unpaired electrons:

\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}

This matches the given value very well.

So the complex has 5 unpaired electrons.

Interpretation for an octahedral $d^5$ complex

For an octahedral $d^5$ complex:

Weak-field ligand $\Rightarrow$ high-spin configuration
Strong-field ligand $\Rightarrow$ low-spin configuration

High-spin $d^5$ arrangement:

t_{2g}^3 e_g^2

This has 5 unpaired electrons.

Low-spin $d^5$ arrangement:

t_{2g}^5 e_g^0

This has only 1 unpaired electron.

Since the observed magnetic moment corresponds to 5 unpaired electrons, the ligand must be a weak-field ligand.

Check the options using spectrochemical series

A: $\text{CN}^-$ → strong-field ligand, gives low spin
B: ethylenediamine (en) → relatively strong-field ligand
C: $\text{NCS}^-$ → weaker-field ligand (especially N-bonded thiocyanato is weaker than CN $^-$ and CO)
D: CO → very strong-field ligand, gives low spin

Thus the suitable ligand is:

\boxed{\text{NCS}^-}

Final answer

The correct option is:

\boxed{\text{C}}

Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They agree.

More from Coordination Compounds

View written solutionFree

Correct answer: C

Determine the electronic configuration of $\text{Mn}^{2+}$

Mn has atomic number $25$ .

\text{Mn}: [\text{Ar}]\,3d^5 4s^2

For $\text{Mn}^{2+}$ , remove two electrons from $4s$ first:

\text{Mn}^{2+}: [\text{Ar}]\,3d^5

So the metal ion is a $d^5$ system.

Use the given magnetic moment

The spin-only magnetic moment is:

\mu = \sqrt{n(n+2)}\ \text{BM}

Given:

\mu = 5.9\ \text{BM}

Now check for $n=5$ unpaired electrons:

\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}

This matches the given value very well.

So the complex has 5 unpaired electrons.

Interpretation for an octahedral $d^5$ complex

For an octahedral $d^5$ complex:

Weak-field ligand $\Rightarrow$ high-spin configuration
Strong-field ligand $\Rightarrow$ low-spin configuration

High-spin $d^5$ arrangement:

t_{2g}^3 e_g^2

This has 5 unpaired electrons.

Low-spin $d^5$ arrangement:

t_{2g}^5 e_g^0

This has only 1 unpaired electron.

Since the observed magnetic moment corresponds to 5 unpaired electrons, the ligand must be a weak-field ligand.

Check the options using spectrochemical series

A: $\text{CN}^-$ → strong-field ligand, gives low spin
B: ethylenediamine (en) → relatively strong-field ligand
C: $\text{NCS}^-$ → weaker-field ligand (especially N-bonded thiocyanato is weaker than CN $^-$ and CO)
D: CO → very strong-field ligand, gives low spin

Thus the suitable ligand is:

\boxed{\text{NCS}^-}

Final answer

The correct option is:

\boxed{\text{C}}

Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They agree.

Coordination Compounds question

Coordination Compounds question

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Coordination Compounds question

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