MHTCET 11 Apr 2026 Shift 1 Paper

Concept:An ion is paramagnetic when it contains one or more unpaired electrons.Explanation:For transition metal ions, electrons are removed first from the $4s$ orbital and then from the $3d$ orbital.Check each ion in Option C.For $\mathrm{Ti}^{2+}$:$\mathrm{Ti}: [\mathrm{Ar}]\,3d^{2}\,4s^{2}$, so $\mathrm{Ti}^{2+}: [\mathrm{Ar}]\,3d^{2}$.It has unpaired electrons, hence paramagnetic.For $\mathrm{Cu}^{2+}$:$\mathrm{Cu}: [\mathrm{Ar}]\,3d^{10}\,4s^{1}$, so $\mathrm{Cu}^{2+}: [\mathrm{Ar}]\,3d^{9}$.A $d^{9}$ configuration has one unpaired electron, hence paramagnetic.For $\mathrm{Mn}^{3+}$:$\mathrm{Mn}: [\mathrm{Ar}]\,3d^{5}\,4s^{2}$, so $\mathrm{Mn}^{3+}: [\mathrm{Ar}]\,3d^{4}$.A $d^{4}$ configuration has unpaired electrons, hence paramagnetic.Thus, all three ions in Option C are paramagnetic.Answer:Option C: $\mathrm{Ti}^{2+}, \mathrm{Cu}^{2+}, \mathrm{Mn}^{3+}$