MHTCET 13 Apr 2026 Shift 2 Paper

Concept: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 value of $n$ will have the same spin only magnetic moment.Explanation:Find the number of unpaired electrons in each ion from its electronic configuration.For $\mathrm{Ti}^{3+}$: $\mathrm{Ti}$ is $[\mathrm{Ar}]3d^{2}4s^{2}$, so $\mathrm{Ti}^{3+}$ becomes $[\mathrm{Ar}]3d^{1}$. Thus $n=1$.For $\mathrm{Cr}^{3+}$: $\mathrm{Cr}$ is $[\mathrm{Ar}]3d^{5}4s^{1}$, so $\mathrm{Cr}^{3+}$ becomes $[\mathrm{Ar}]3d^{3}$. Thus $n=3$.For $\mathrm{Mn}^{2+}$: $\mathrm{Mn}$ is $[\mathrm{Ar}]3d^{5}4s^{2}$, so $\mathrm{Mn}^{2+}$ becomes $[\mathrm{Ar}]3d^{5}$. Thus $n=5$.For $\mathrm{Fe}^{3+}$: $\mathrm{Fe}$ is $[\mathrm{Ar}]3d^{6}4s^{2}$, so $\mathrm{Fe}^{3+}$ becomes $[\mathrm{Ar}]3d^{5}$. Thus $n=5$.For $\mathrm{Sc}^{3+}$: $\mathrm{Sc}$ is $[\mathrm{Ar}]3d^{1}4s^{2}$, so $\mathrm{Sc}^{3+}$ becomes $[\mathrm{Ar}]3d^{0}$. Thus $n=0$.Only $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{3+}$ have the same number of unpaired electrons, $n=5$.Therefore, both have the same spin only magnetic moment:$\mu=\sqrt{5(5+2)}=\sqrt{35}$ BM.Answer:Option C: (C) and (D) only.