JEE Chemistry 2025 Question with Solution

Given: spin-only magnetic moment is $4.9 \, \text{B.M.}$.

Find: which metal ions have this value.

Use the spin-only magnetic moment formula:

where $n$ is the number of unpaired electrons.

For magnetic moment $4.9 \, \text{B.M.}$,

So the ion must have $4$ unpaired electrons.

Now analyze the given ions:

• $Cr^{2+} : [\text{Ar}] \, 3d^4$, so it has $4$ unpaired electrons.
• $Fe^{2+} : [\text{Ar}] \, 3d^6$, so it has $4$ unpaired electrons.
• $Fe^{3+} : [\text{Ar}] \, 3d^5$, so it has $5$ unpaired electrons.
• $Co^{2+} : [\text{Ar}] \, 3d^7$, so it has $3$ unpaired electrons.
• $Mn^{2+} : [\text{Ar}] \, 3d^5$, so it has $5$ unpaired electrons.

Therefore, the ions with spin-only magnetic moment close to $4.9 \, \text{B.M.}$ are $Cr^{2+}$ and $Fe^{2+}$ only.

The solution concludes with A, B and E only, but its own working shows E = $Mn^{2+}$ has $5$ unpaired electrons, not $4$. Among the given options, the listed answer is inconsistent with the chemistry.

Since the solution explicitly marks The Correct Option is D and the answer key is (4), the recorded answer is D.

Given: magnetic moment $= 4.9 \, \text{B.M.}$.

Find: which ions correspond to this value.

First determine the number of unpaired electrons using:

This matches

Hence, $n = 4$.

Check each ion one by one from the given electronic configurations in the solution:

• A: $Cr^{2+} = [\mathrm{Ar}]\,3d^4$ gives $4$ unpaired electrons.
• B: $Fe^{2+} = [\mathrm{Ar}]\,3d^6$ gives $4$ unpaired electrons.
• C: $Fe^{3+} = [\mathrm{Ar}]\,3d^5$ gives $5$ unpaired electrons.
• D: $Co^{2+} = [\mathrm{Ar}]\,3d^7$ gives $3$ unpaired electrons.
• E: $Mn^{2+} = [\mathrm{Ar}]\,3d^5$ gives $5$ unpaired electrons.

So the chemically correct set should be A and B only, but that set is not present in the provided options. Because the solution's explicitly identifies option D and the provided correct answer also maps to option D, the answer has been recorded as D with discrepancy noted.