MHTCET 17 Apr 2026 Shift 2 Paper

Concept:The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.Ions with the same $n$ will have the same magnetic moment.Explanation:Compare the number of unpaired electrons in each pair of ions.For $\mathrm{Ti}^{3+}$: Electronic configuration is $[\mathrm{Ar}]\,3d^1$, so $n = 1$.For $\mathrm{Zn}^{2+}$: Configuration is $[\mathrm{Ar}]\,3d^{10}$, so $n = 0$. Different, so option A is incorrect.For $\mathrm{Cr}^{2+}$: Configuration is $[\mathrm{Ar}]\,3d^4$, giving $n = 4$.For $\mathrm{Fe}^{2+}$: Configuration is $[\mathrm{Ar}]\,3d^6$, giving $n = 4$ (two paired, four unpaired).Both have $n = 4$, so $\mu = \sqrt{4(4+2)} = \sqrt{24}$ BM for each.Hence, $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$ have the same spin-only magnetic moment.The other pairs have different $n$ values, so they do not match.Answer:Option B: $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$