JEE Main · 2025 · 04 Apr 2025 · Shift 2 · d and f Block Elements

Ionisation enthalpy is raised when the electron being lost comes from a stable half-filled $d^5$ shell, and it always increases as the positive charge on the ion grows. Removing an electron from $\mathrm{C}{\mathrm{r}}^{+}$ ($3d^5$) breaks that half-filled shell, so $\mathrm{C}{\mathrm{r}}^{+}$ needs more energy than $\mathrm{M}{\mathrm{n}}^{+}$, whose electron leaves to give the stable $3d^5$; the rise of ionisation enthalpy with charge likewise makes $\mathrm{M}{\mathrm{n}}^{+}$ lower than $\mathrm{M}{\mathrm{n}}^{2+}$ and $\mathrm{F}{\mathrm{e}}^{2+}$ lower than $\mathrm{F}{\mathrm{e}}^{3+}$. For the last pair, $\mathrm{M}{\mathrm{n}}^{2+}$ is the stable $3d^5$ while $\mathrm{F}{\mathrm{e}}^{2+}$ is $3d^6$; losing an electron from $\mathrm{M}{\mathrm{n}}^{2+}$ destroys the half-filled stability whereas from $\mathrm{F}{\mathrm{e}}^{2+}$ it produces the stable $d^5$, so $\mathrm{M}{\mathrm{n}}^{2+}$ has the higher ionisation enthalpy. The relationship written as $\mathrm{M}{\mathrm{n}}^{2+}<\mathrm{F}{\mathrm{e}}^{2+}$ is therefore the incorrect one.