JEE Chemistry 2025 Question with Solution

Given: List-I contains $Ti^{3+}$, $V^{2+}$, $Ni^{2+}$ and $Sc^{3+}$. List-II contains the spin-only magnetic moment values $3.87$, $0.00$, $1.73$ and $2.84$.

Find: The correct matching between the ions and their magnetic moments.

Use the spin-only magnetic moment formula:

$$\mu_s = \sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

For $Ti^{3+}$: Titanium has configuration $[Ar]3d^24s^2$. On losing three electrons,

$$Ti^{3+} = [Ar]3d^1$$

So, $n = 1$.

$$\mu_s = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$$

Hence, $Ti^{3+} \rightarrow (III)\;1.73$.

For $V^{2+}$: Vanadium has configuration $[Ar]3d^34s^2$. On losing two electrons,

$$V^{2+} = [Ar]3d^3$$

So, $n = 3$.

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

Hence, $V^{2+} \rightarrow (I)\;3.87$.

For $Ni^{2+}$: Nickel has configuration $[Ar]3d^84s^2$. On losing two electrons,

$$Ni^{2+} = [Ar]3d^8$$

This ion has $2$ unpaired electrons, so $n = 2$.

$$\mu_s = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$$

Hence, $Ni^{2+} \rightarrow (IV)\;2.84$.

For $Sc^{3+}$: Scandium has configuration $[Ar]3d^14s^2$. On losing three electrons,

$$Sc^{3+} = [Ar]$$

So, $n = 0$.

$$\mu_s = \sqrt{0(0+2)} = 0.00$$

Hence, $Sc^{3+} \rightarrow (II)\;0.00$.

Therefore, the correct matching is

$$(A)-(III),\ (B)-(I),\ (C)-(IV),\ (D)-(II)$$

This corresponds to option C.

Discrepancy noted: the solution marks option D, but the actual spin-only magnetic moment calculation given by the formula leads to option C.