JEE Main 2026 — Chemistry Inorganic Chemistry

The electronic configuration of Cr is $[Ar] 3d^5 4s^1$[Ar]3d54s1 and Mn is $[Ar] 3d^5 4s^2$[Ar]3d54s2.

For the first ionization enthalpy, the electron is removed from the $4s$4s orbital. Mn has a fully filled $4s^2$4s2 configuration and a higher effective nuclear charge compared to Cr. Thus, the first ionization enthalpy of Mn is greater than that of Cr ($\Delta_i H_1 :$Δi​H1​: Mn $>$> Cr). This makes statement C correct.

After the first ionization, the electronic configurations become:

$$Cr^+ : [Ar] 3d^5$$Cr+:[Ar]3d5

$$Mn^+ : [Ar] 3d^5 4s^1$$Mn+:[Ar]3d54s1

For the second ionization enthalpy, the electron is removed from the highly stable half-filled $3d^5$3d5 subshell in $Cr^+$Cr+, whereas in $Mn^+$Mn+, it is removed from the $4s^1$4s1 orbital. Removing an electron from a stable half-filled configuration requires significantly more energy. Thus, the second ionization enthalpy of Cr is greater than that of Mn ($\Delta_i H_2 :$Δi​H2​: Cr $>$> Mn). This makes statement B correct.

Therefore, the correct statements are B and C.

Answer: B and C only