JEE Chemistry 2025 Question with Solution

Given: The ions are $Cu^+$, $Cu^{2+}$, $Cr^{2+}$ and $Cr^{3+}$.

Find: The correct decreasing order of their spin-only magnetic moments.

The spin-only magnetic moment depends on the number of unpaired electrons and is given by

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

where $n$ is the number of unpaired electrons.

For $Cu^+$:

• Electronic configuration of $Cu$ is $[Ar] \, 3d^{10}4s^1$.
• $Cu^+$ loses one electron, so configuration becomes $[Ar] \, 3d^{10}$.
• Number of unpaired electrons $n = 0$.

$$\mu = \sqrt{0(0+2)} = 0 \, \text{BM}$$

For $Cu^{2+}$:

• $Cu^{2+}$ has configuration $[Ar] \, 3d^9$.
• Number of unpaired electrons $n = 1$.

$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}$$

For $Cr^{2+}$:

• Electronic configuration of $Cr$ is $[Ar] \, 3d^54s^1$.
• $Cr^{2+}$ loses two electrons, so configuration becomes $[Ar] \, 3d^4$.
• Number of unpaired electrons $n = 4$.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}$$

For $Cr^{3+}$:

• $Cr^{3+}$ loses three electrons, so configuration becomes $[Ar] \, 3d^3$.
• Number of unpaired electrons $n = 3$.

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

Therefore, the decreasing order is $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^+$. The correct option is D.

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

Find: The decreasing order of magnetic moments.

Since

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

magnetic moment increases as the number of unpaired electrons increases.

So it is enough to compare unpaired electrons:

• $Cu^+ : 3d^{10} \Rightarrow 0$ unpaired electrons
• $Cu^{2+} : 3d^9 \Rightarrow 1$ unpaired electron
• $Cr^{3+} : 3d^3 \Rightarrow 3$ unpaired electrons
• $Cr^{2+} : 3d^4 \Rightarrow 4$ unpaired electrons

Hence the decreasing order is $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^+$. Therefore, the correct option is D.