Chemistry Question 70 – JEE-MAIN 2026

We need to compare the energy required to remove the next electron from ${\text{Fe}}^{2+}$, ${\text{Fe}}^{3+}$, and ${\text{Cr}}^{2+}$.

- For ${\text{Fe}}^{2+}$ ($[\text{Ar}]3d^6$): The next ionization is ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ ($[\text{Ar}]3d^5$). This corresponds to the 3rd ionization enthalpy of Iron (${\text{IE}}_3(\text{Fe})$).

- For ${\text{Fe}}^{3+}$ ($[\text{Ar}]3d^5$): The next ionization is ${\text{Fe}}^{3+}\to {\text{Fe}}^{4+}$ ($[\text{Ar}]3d^4$). This corresponds to the 4th ionization enthalpy of Iron (${\text{IE}}_4(\text{Fe})$). Removing an electron from a stable half-filled $3d^5$ configuration requires a very high amount of energy.

- For ${\text{Cr}}^{2+}$ ($[\text{Ar}]3d^4$): The next ionization is ${\text{Cr}}^{2+}\to {\text{Cr}}^{3+}$ ($[\text{Ar}]3d^3$). This corresponds to the 3rd ionization enthalpy of Chromium (${\text{IE}}_3(\text{Cr})$).

Using approximate standard ionization enthalpy values (in kJ/mol):

- ${\text{IE}}_3(\text{Fe})\approx 2957\text{ kJ/mol}$

- ${\text{IE}}_4(\text{Fe})\approx 5290\text{ kJ/mol}$ (Highest among the three)

- ${\text{IE}}_3(\text{Cr})\approx 2987\text{ kJ/mol}$

Comparing these values, ${\text{IE}}_4(\text{Fe})$ is the highest. Therefore, ${\text{Fe}}^{3+}$ is the ion that shows a positive borax bead test and has the highest ionization enthalpy (for the next electron removal) among the given options.