JEE Chemistry 2023 Question with Solution

Given: The metal ions are $V^{3+}$, $Cr^{3+}$, $Fe^{2+}$ and $Ni^{3+}$ in the gaseous state.

Find: How many of these ions have the same spin-only magnetic moment.

The spin-only magnetic moment is given by

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

where $n$ is the number of unpaired electrons.

For $V^{3+}$:

$$[Ar] \, 3d^2$$

So, $n = 2$ and

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

For $Cr^{3+}$:

$$[Ar] \, 3d^3$$

So, $n = 3$ and

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

For $Fe^{2+}$:

$$[Ar] \, 3d^6$$

So, $n = 4$ and

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

For $Ni^{3+}$:

$$[Ar] \, 3d^7$$

So, $n = 3$ and

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

Thus, $Cr^{3+}$ and $Ni^{3+}$ have the same spin-only magnetic moment.

Therefore, the number of metal ions having a similar value of spin-only magnetic moment is $2$.

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

Find: Which ions have equal values of $\mu$.

Using

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

ions with the same $n$ will have the same magnetic moment.

Now compare the unpaired electrons:

• $V^{3+} : 3d^2 \Rightarrow n = 2$
• $Cr^{3+} : 3d^3 \Rightarrow n = 3$
• $Fe^{2+} : 3d^6 \Rightarrow n = 4$
• $Ni^{3+} : 3d^7 \Rightarrow n = 3$

Since $Cr^{3+}$ and $Ni^{3+}$ both have $n = 3$, they have equal spin-only magnetic moments.

Hence, the required count is $2$.