JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Classification of Elements and Periodicity

The configurations are $\mathrm{Cr}=[\mathrm{Ar}]3d^{5}4s^1$ and $\mathrm{Mn}=[\mathrm{Ar}]3d^{5}4s^2$. Removing the single $4s$ electron of chromium is easier than removing one of the paired $4s$ electrons of manganese, so $\mathrm{I}{\mathrm{E}}_1(\mathrm{Cr})<\mathrm{I}{\mathrm{E}}_1(\mathrm{Mn})$ and Statement-I is correct. For the second ionization $\mathrm{C}{\mathrm{r}}^{+}$ ($3d^5$) is an extra-stable half-filled ion and hard to ionise, whereas $\mathrm{M}{\mathrm{n}}^{+}$ ($3d^{5}4s^1$) loses its $4s$ electron easily, giving $\mathrm{I}{\mathrm{E}}_2(\mathrm{Cr})>\mathrm{I}{\mathrm{E}}_2(\mathrm{Mn})$; but for the third ionization $\mathrm{M}{\mathrm{n}}^{2+}$ ($3d^5$) is the stable half-filled ion while $\mathrm{C}{\mathrm{r}}^{2+}$ is $3d^4$, so $\mathrm{I}{\mathrm{E}}_3(\mathrm{Cr})<\mathrm{I}{\mathrm{E}}_3(\mathrm{Mn})$. As the third ionization enthalpy of chromium is not higher than that of manganese, Statement-II is false. Statement-I is true but Statement-II is false.