JEE Main Chemistry Question (2026) — Solution

Step-by-step 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 ( _i H_1 : Mn > Cr). This makes statement C correct. After the first ionization, the electronic configurations become: Cr^+ : [Ar] 3d^5 Mn^+ : [Ar] 3d^5 4s^1 For the second ionization enthalpy, the electron is removed from the highly stable half-filled 3d^5 subshell in Cr^+, whereas in Mn^+, 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 ( _i H_2 : Cr > Mn). This makes statement B correct. Therefore, the correct statements are B and C. Answer: B and C only