JEE Main 2026 — d and f Block Elements Question with Solution

Statement I:
The electronic configuration of $Cr$ is $[Ar]3d^{5}4s^1$ and $Mn$ is $[Ar]3d^{5}4s^2$.
For $Cr$, the first electron is removed from $4s^1$, while for $Mn$, it is removed from a stable fully filled $4s^2$ subshell.
Additionally, $Mn$ has a higher nuclear charge.
Thus, the first ionization enthalpy ($I{E}_1$) of $Cr$ ($653\text{ kJ/mol}$) is lower than that of $Mn$ ($717\text{ kJ/mol}$).
Statement I is true.

Statement II:
For the second ionization enthalpy ($I{E}_2$), $C{r}^{+}$ has configuration $[Ar]3d^5$ (stable half-filled) and $M{n}^{+}$ has $[Ar]3d^{5}4s^1$.
Removing an electron from the stable $3d^5$ configuration of $C{r}^{+}$ requires more energy than removing it from the $4s^1$ of $M{n}^{+}$.
Thus, $I{E}_2$ of $Cr$ ($1592\text{ kJ/mol}$) is higher than $Mn$ ($1509\text{ kJ/mol}$).
For the third ionization enthalpy ($I{E}_3$), $C{r}^{2+}$ is $[Ar]3d^4$ and $M{n}^{2+}$ is $[Ar]3d^5$.
Removing an electron from the stable half-filled $3d^5$ configuration of $M{n}^{2+}$ requires significantly more energy than from the $3d^4$ of $C{r}^{2+}$.
Thus, $I{E}_3$ of $Mn$ ($3248\text{ kJ/mol}$) is higher than $Cr$ ($2987\text{ kJ/mol}$).
Since $I{E}_3$ of $Cr$ is lower than $Mn$, Statement II is false.