JEE Chemistry 2023 Question with Solution

Given: The complex is $[\text{Mn(H}_2\text{O)}_6]^{2+}$ and atomic number of Mn is $25$.

Find: The spin-only magnetic moment in B.M. to the nearest integer.

The electronic configuration of Mn is $[\text{Ar}]\,3d^5\,4s^2$.

In $[\text{Mn(H}_2\text{O)}_6]^{2+}$, Mn is in the $+2$ oxidation state. Water is a weak field ligand.

Electronic configuration of $\text{Mn}^{2+}$ is $[\text{Ar}]\,3d^5$.

Since $\text{H}_2\text{O}$ is a weak field ligand, there will be no pairing of electrons in the $d$ orbitals.

Number of unpaired electrons:

$$n = 5$$

Spin-only magnetic moment:

$$\mu_{\text{spin}} = \sqrt{n(n+2)} \, \text{B.M.}$$

Substituting $n = 5$:

$$\mu_{\text{spin}} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.}$$

Nearest integer:

$$6$$

Therefore, the spin-only magnetic moment is nearest to $6$, so the answer is $6$.

Given: Mn has atomic number $25$ and the complex is $[\text{Mn(H}_2\text{O)}_6]^{2+}$.

Find: The nearest integer value of the spin-only magnetic moment.

First identify the metal ion state. Since water is neutral, the charge on manganese in the complex is $+2$.

Neutral Mn:

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

Removing two electrons for $\text{Mn}^{2+}$ gives:

$$[\text{Ar}]\,3d^5$$

Now use the ligand property stated in the solution: $\text{H}_2\text{O}$ is a weak field ligand, so the complex remains high spin and the five $d$ electrons stay unpaired.

Thus:

$$n = 5$$

Apply the spin-only formula:

$$\mu_{\text{spin}} = \sqrt{n(n+2)}$$ $$\mu_{\text{spin}} = \sqrt{5(7)} = \sqrt{35} \approx 5.92$$

So the nearest integer value is $6$.

Therefore, the numerical answer is $6$.