JEE Main 2025 — D And F Block Elements Question with Solution

To determine which metal ions have a calculated spin-only magnetic moment of 4.9 Bohr Magnetons (B.M.), we can use the formula:

$\text{Magnetic Moment (M.M)}=\sqrt{n(n+2)}\text{B.M.}$

where $n$ is the number of unpaired electrons.

Given that the magnetic moment is 4.9 B.M., we can set up the equation:

$4.9=\sqrt{n(n+2)}$

Solving this equation, we find that $n=4$.

Let's analyze each metal ion:

(A) ${}_{24}{\mathrm{Cr}}^{2+}$: Electronic configuration: $[\mathrm{Ar}]3d^4$. This ion has 4 unpaired electrons.

(B) ${}_{26}{\mathrm{Fe}}^{2+}$: Electronic configuration: $[\mathrm{Ar}]3d^6$. This ion also has 4 unpaired electrons.

(C) ${}_{26}{\mathrm{Fe}}^{3+}$: Electronic configuration: $[\mathrm{Ar}]3d^5$. This ion has 5 unpaired electrons, which does not match the required $n$ value of 4.

(D) ${}_{27}{\mathrm{Co}}^{2+}$: Electronic configuration: $[\mathrm{Ar}]3d^7$. This ion has 3 unpaired electrons.

(E) ${}_{25}{\mathrm{Mn}}^{3+}$: Electronic configuration: $[\mathrm{Ar}]3d^4$. This ion has 4 unpaired electrons.

Based on the number of unpaired electrons, the metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are ${\mathrm{Cr}}^{2+}$, ${\mathrm{Fe}}^{2+}$, and ${\mathrm{Mn}}^{3+}$.