The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are - 0.8 \Delta 0 and 3.87 BM, respectively. Identify (Mz+) :

Coordination Compounds PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q5 with Solution

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Correct answer: D

Use the magnetic moment to find the number of unpaired electrons

For spin-only magnetic moment,

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

where $n$ is the number of unpaired electrons.

Given:

\mu = 3.87\ \text{BM}

Now,

\sqrt{3(3+2)} = \sqrt{15} \approx 3.87

So,

n=3

Thus, the complex has 3 unpaired electrons.

Given CFSE for octahedral complex

For an octahedral complex,

each electron in $t_{2g}$ contributes $-0.4\Delta_0$
each electron in $e_g$ contributes $+0.6\Delta_0$

Given:

\text{CFSE} = -0.8\Delta_0

We need an octahedral $d^n$ high-spin aqua complex with 3 unpaired electrons and CFSE $-0.8\Delta_0$ .

Since $H_2O$ is a weak field ligand, the aqua complex will be high spin.

Check each option

Option A: $V^{3+}$

Vanadium: $Z=23$

Neutral V:

[V] = [Ar]3d^34s^2

So,

V^{3+} = 3d^2

High-spin octahedral $d^2$ :

t_{2g}^2 e_g^0

Unpaired electrons $=2$

CFSE:

2(-0.4\Delta_0) = -0.8\Delta_0

CFSE matches, but magnetic moment does not.

So, A is incorrect.

Option B: $Cr^{3+}$

Chromium: $Z=24$

Neutral Cr:

[Cr]=[Ar]3d^54s^1

So,

Cr^{3+}=3d^3

High-spin octahedral $d^3$ :

t_{2g}^3 e_g^0

Unpaired electrons $=3$

Magnetic moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}

CFSE:

3(-0.4\Delta_0) = -1.2\Delta_0

Magnetic moment matches, but CFSE does not.

So, B is incorrect.

Option C: $Mn^{4+}$

Manganese: $Z=25$

Neutral Mn:

[Mn]=[Ar]3d^54s^2

So,

Mn^{4+}=3d^3

High-spin octahedral $d^3$ :

t_{2g}^3 e_g^0

Unpaired electrons $=3$

Magnetic moment:

\mu \approx 3.87\ \text{BM}

CFSE:

-1.2\Delta_0

Again, CFSE does not match.

So, C is incorrect.

Option D: $Co^{2+}$

Cobalt: $Z=27$

Neutral Co:

[Co]=[Ar]3d^74s^2

So,

Co^{2+}=3d^7

With $H_2O$ (weak field), octahedral complex is high spin:

t_{2g}^5 e_g^2

Unpaired electrons $=3$

Magnetic moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}

CFSE:

5(-0.4\Delta_0)+2(+0.6\Delta_0)

= -2.0\Delta_0 +1.2\Delta_0

= -0.8\Delta_0

Both conditions match.

So, D is correct.

Final Answer

The metal ion is:

\boxed{Co^{2+}}

Hence, the correct option is D.

More from Coordination Compounds

View written solutionFree

Correct answer: D

Use the magnetic moment to find the number of unpaired electrons

For spin-only magnetic moment,

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

where $n$ is the number of unpaired electrons.

Given:

\mu = 3.87\ \text{BM}

Now,

\sqrt{3(3+2)} = \sqrt{15} \approx 3.87

So,

n=3

Thus, the complex has 3 unpaired electrons.

Given CFSE for octahedral complex

For an octahedral complex,

each electron in $t_{2g}$ contributes $-0.4\Delta_0$
each electron in $e_g$ contributes $+0.6\Delta_0$

Given:

\text{CFSE} = -0.8\Delta_0

We need an octahedral $d^n$ high-spin aqua complex with 3 unpaired electrons and CFSE $-0.8\Delta_0$ .

Since $H_2O$ is a weak field ligand, the aqua complex will be high spin.

Check each option

Option A: $V^{3+}$

Vanadium: $Z=23$

Neutral V:

[V] = [Ar]3d^34s^2

So,

V^{3+} = 3d^2

High-spin octahedral $d^2$ :

t_{2g}^2 e_g^0

Unpaired electrons $=2$

CFSE:

2(-0.4\Delta_0) = -0.8\Delta_0

CFSE matches, but magnetic moment does not.

So, A is incorrect.

Option B: $Cr^{3+}$

Chromium: $Z=24$

Neutral Cr:

[Cr]=[Ar]3d^54s^1

So,

Cr^{3+}=3d^3

High-spin octahedral $d^3$ :

t_{2g}^3 e_g^0

Unpaired electrons $=3$

Magnetic moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}

CFSE:

3(-0.4\Delta_0) = -1.2\Delta_0

Magnetic moment matches, but CFSE does not.

So, B is incorrect.

Option C: $Mn^{4+}$

Manganese: $Z=25$

Neutral Mn:

[Mn]=[Ar]3d^54s^2

So,

Mn^{4+}=3d^3

High-spin octahedral $d^3$ :

t_{2g}^3 e_g^0

Unpaired electrons $=3$

Magnetic moment:

\mu \approx 3.87\ \text{BM}

CFSE:

-1.2\Delta_0

Again, CFSE does not match.

So, C is incorrect.

Option D: $Co^{2+}$

Cobalt: $Z=27$

Neutral Co:

[Co]=[Ar]3d^74s^2

So,

Co^{2+}=3d^7

With $H_2O$ (weak field), octahedral complex is high spin:

t_{2g}^5 e_g^2

Unpaired electrons $=3$

Magnetic moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}

CFSE:

5(-0.4\Delta_0)+2(+0.6\Delta_0)

= -2.0\Delta_0 +1.2\Delta_0

= -0.8\Delta_0

Both conditions match.

So, D is correct.

Final Answer

The metal ion is:

\boxed{Co^{2+}}

Hence, the correct option is D.

Coordination Compounds question

Coordination Compounds question

Option A: $V^{3+}$

Option B: $Cr^{3+}$

Option C: $Mn^{4+}$

Option D: $Co^{2+}$

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

Option A: $V^{3+}$

Option B: $Cr^{3+}$

Option C: $Mn^{4+}$

Option D: $Co^{2+}$

More from Coordination Compounds

Option A: V3+V^{3+}V3+

Option B: Cr3+Cr^{3+}Cr3+

Option C: Mn4+Mn^{4+}Mn4+

Option D: Co2+Co^{2+}Co2+

More from Coordination Compounds

Option A: V3+V^{3+}V3+

Option B: Cr3+Cr^{3+}Cr3+

Option C: Mn4+Mn^{4+}Mn4+

Option D: Co2+Co^{2+}Co2+

More from Coordination Compounds

Option A: $V^{3+}$

Option B: $Cr^{3+}$

Option C: $Mn^{4+}$

Option D: $Co^{2+}$

Option A: $V^{3+}$

Option B: $Cr^{3+}$

Option C: $Mn^{4+}$

Option D: $Co^{2+}$