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

The electronic configuration of Cr is $[Ar]3d^{5}4s^1$ and Mn is $[Ar]3d^{5}4s^2$.

For the first ionization enthalpy, the electron is removed from the $4s$ orbital. Mn has a fully filled $4s^2$ configuration and a higher effective nuclear charge compared to Cr. Thus, the first ionization enthalpy of Mn is greater than that of Cr (${\Delta }_i{H}_1:$ Mn $>$ Cr). This makes statement C correct.

After the first ionization, the electronic configurations become:
$C{r}^{+}:[Ar]3d^5$
$M{n}^{+}:[Ar]3d^{5}4s^1$

For the second ionization enthalpy, the electron is removed from the highly stable half-filled $3d^5$ subshell in $C{r}^{+}$, whereas in $M{n}^{+}$, it is removed from the $4s^1$ orbital. Removing an electron from a stable half-filled configuration requires significantly more energy. Thus, the second ionization enthalpy of Cr is greater than that of Mn (${\Delta }_i{H}_2:$ Cr $>$ Mn). This makes statement B correct.

Therefore, the correct statements are B and C.

Answer: B and C only