MHTCET 18 Apr 2026 Shift 2 Paper

Concept:In transition metal ions, electrons are removed first from the $4s$ orbital and then from the $3d$ orbital.The number of unpaired electrons is decided by the $3d$ configuration using Hund's rule.Explanation:Electronic configuration of $\mathrm{Fe}$ is $[\mathrm{Ar}]\,3d^6 4s^2$.So, $\mathrm{Fe}^{2+}$ has $[\mathrm{Ar}]\,3d^6$.In $3d^6$, the distribution is $\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow$.Hence, unpaired electrons in $\mathrm{Fe}^{2+}$ = $4$.Electronic configuration of $\mathrm{Cr}$ is $[\mathrm{Ar}]\,3d^5 4s^1$.So, $\mathrm{Cr}^{2+}$ has $[\mathrm{Ar}]\,3d^4$.In $3d^4$, the distribution is $\uparrow \uparrow \uparrow \uparrow$ and one empty orbital.Hence, unpaired electrons in $\mathrm{Cr}^{2+}$ = $4$.Thus, $\mathrm{Fe}^{2+}$ and $\mathrm{Cr}^{2+}$ have the same number of unpaired electrons.Answer:Option C: $\mathrm{Fe}^{2+}$ and $\mathrm{Cr}^{2+}$.