JEE Main 2025 Apr 3 Shift 1, Chemistry Q4: Colour, Magnetism And Stability Of Co-ordination Compounds

Spin only magnetic moment is $\mu =\sqrt{n(n+2)}\text{B.M.}$, where $n$ is the number of unpaired electrons.

Setting $\sqrt{n(n+2)}=4.9$ gives $n(n+2)=24$, so $n=4$.

Counting unpaired electrons in each $d$-configuration:

A. ${\text{Cr}}^{2+}$: $[\text{Ar}]3d^4$ → 4 unpaired ✓

B. ${\text{Fe}}^{2+}$: $[\text{Ar}]3d^6$ (high spin) → 4 unpaired ✓

C. ${\text{Fe}}^{3+}$: $[\text{Ar}]3d^5$ → 5 unpaired ✗

D. ${\text{Co}}^{2+}$: $[\text{Ar}]3d^7$ → 3 unpaired ✗

E. ${\text{Mn}}^{3+}$: $[\text{Ar}]3d^4$ → 4 unpaired ✓

$\therefore$ Ions A, B and E satisfy the condition.