JEE Chemistry 2024 Question with Solution

Given: The options are $[Ar] \, 3d^7$, $[Ar] \, 3d^8$, $[Ar] \, 3d^3$ and $[Ar] \, 3d^6$.

Find: Which configuration has the highest magnetic moment.

The magnetic moment is determined from the number of unpaired electrons using

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

where $n$ is the number of unpaired electrons.

From the extracted solution working:

1. For $[Ar] \, 3d^7$, the number of unpaired electrons is $3$, so

$$\mu = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM}$$

2. For $[Ar] \, 3d^8$, the number of unpaired electrons is $2$, so

$$\mu = \sqrt{2(2+2)} = \sqrt{8} \, \text{BM}$$

3. For $[Ar] \, 3d^3$, the table in the solution shows $3$ unpaired electrons, so

$$\mu = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM}$$

4. For $[Ar] \, 3d^6$, the number of unpaired electrons is $4$, so

$$\mu = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM}$$

Since $\sqrt{24}$ is the largest among $\sqrt{15}$, $\sqrt{8}$, $\sqrt{15}$ and $\sqrt{24}$, the configuration $[Ar] \, 3d^6$ has the highest magnetic moment.

Therefore, the correct option is D.

Given: Spin-only magnetic moment depends on the number of unpaired electrons.

Find: The option with maximum magnetic moment.

Because

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

magnetic moment increases as the number of unpaired electrons $n$ increases.

So it is enough to compare the unpaired electrons in each option:

• $3d^7 \rightarrow 3$ unpaired electrons
• $3d^8 \rightarrow 2$ unpaired electrons
• $3d^3 \rightarrow 3$ unpaired electrons
• $3d^6 \rightarrow 4$ unpaired electrons

The largest value of $n$ is $4$ for $[Ar] \, 3d^6$. Hence it gives the maximum magnetic moment.

Therefore, the correct option is D.