d and f Block Elements Part 3

Concept:Ionization enthalpy depends on the stability of electronic configurations, especially half-filled and fully filled subshells.Explanation:The electronic configurations are:$\mathrm{Cr} = [\mathrm{Ar}]\,3d^5 4s^1$$\mathrm{Mn} = [\mathrm{Ar}]\,3d^5 4s^2$For first ionization enthalpy ($\mathrm{IE}_1$):Removing the $4s^1$ electron from Cr gives $\mathrm{Cr}^+$ with stable half‑filled $3d^5$ configuration.This removal is easier, so $\mathrm{IE}_1(\mathrm{Cr}) < \mathrm{IE}_1(\mathrm{Mn})$.Thus Statement I is true.For second ionization enthalpy ($\mathrm{IE}_2$):$\mathrm{Cr}^+$ already has $3d^5$ stability; removing another electron disturbs it, requiring high energy.In Mn, after losing one $4s$ electron, removing the next gives $\mathrm{Mn}^{2+}$ with stable $3d^5$, so easier.Hence $\mathrm{IE}_2(\mathrm{Cr}) > \mathrm{IE}_2(\mathrm{Mn})$. (Matches part of Statement II.)For third ionization enthalpy ($\mathrm{IE}_3$):$\mathrm{Mn}^{2+}$ is very stable due to $3d^5$; removing another electron is difficult.$\mathrm{Cr}^{2+}$( $3d^4$ ) lacks that extra stability, so removal is easier.Thus $\mathrm{IE}_3(\mathrm{Cr}) < \mathrm{IE}_3(\mathrm{Mn})$, which contradicts Statement II (which claims $\mathrm{IE}_3(\mathrm{Cr}) > \mathrm{IE}_3(\mathrm{Mn})$).So Statement II is false.Answer:Only Statement I is true, therefore the correct option is A.