Given below are two statements : Statement I : The first ionization...

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 ($IE_1$)$of Cr$ ($653 \text{ kJ/mol}$) is lower than that of $Mn$ ($717 \text{ kJ/mol}$).
Statement I is true.

Statement II:
For the second ionization enthalpy ($IE_2$), $Cr^+$ has configuration $[Ar] 3d^5$ (stable half-filled)$and Mn^+$ has $[Ar] 3d^5 4s^1$.
Removing an electron from the stable $3d^5$ configuration of $Cr^+$ requires more energy than removing it from the $4s^1$ of $Mn^+$.
Thus, $IE_2$ of $Cr$ ($1592 \text{ kJ/mol}$) is higher than $Mn$ ($1509 \text{ kJ/mol}$).
For the third ionization enthalpy ($IE_3$), $Cr^{2+}$ is $[Ar] 3d^4$ and $Mn^{2+}$ is $[Ar] 3d^5$.
Removing an electron from the stable half-filled $3d^5$ configuration of $Mn^{2+}$ requires significantly more energy than from the $3d^4$ of $Cr^{2+}$.
Thus, $IE_3$ of $Mn$ ($3248 \text{ kJ/mol}$) is higher than $Cr$ ($2987 \text{ kJ/mol}$).
Since $IE_3$ of $Cr$ is lower than $Mn$, Statement II is false.