Chemistry Q61 – JEE-MAIN 2025

Home Chemistry Periodicity & ElementsQuestion #537

The incorrect relationship in the following pairs in relation to ionisation enthalpies is :

Clue 1 of 3: Conceptual Principle

Ionization enthalpy is the energy required to remove an electron. Stability of electronic configurations (half-filled or fully-filled orbitals) significantly affects ionization enthalpy.

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Ninja Strategy•Eliminate by Fundamental Rules

Quickly eliminate options that represent universally true principles of ionization enthalpy (like successive ionization energies always increasing) to narrow down the choices.

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Step 1: Understand Ionization Enthalpy Trends✦ Active

Ionization enthalpy (IE) is the energy required to remove an electron. Key factors influencing IE include effective nuclear charge, atomic size, and electronic configuration stability (half-filled or fully-filled subshells). Successive ionization enthalpies always increase, i.e., $I E_{1} < I E_{2} < I E_{3}$ .

Step 2: Determine Electronic Configurations of Ions○ Expand

Write the electronic configurations for the relevant neutral atoms and their ions:

C r : [A r] 3 d^{5} 4 s^{1} C r^{+} : [A r] 3 d^{5} M n : [A r] 3 d^{5} 4 s^{2} M n^{+} : [A r] 3 d^{5} 4 s^{1} M n^{2 +} : [A r] 3 d^{5} F e : [A r] 3 d^{6} 4 s^{2} F e^{2 +} : [A r] 3 d^{6} F e^{3 +} : [A r] 3 d^{5}

💡 Teacher's Secret Hint

Remember that electrons are removed first from the outermost s-orbital, then from the d-orbital for transition metals.

Step 3: Evaluate Each Option○ Expand

1. $M n^{+} < C r^{+}$ (i.e., $I E_{2} (M n) < I E_{2} (C r)$ ): Removing an electron from $M n^{+} ([A r] 3 d^{5} 4 s^{1})$ yields $M n^{2 +} ([A r] 3 d^{5})$ , which is stable. Removing an electron from $C r^{+} ([A r] 3 d^{5})$ yields $C r^{2 +} ([A r] 3 d^{4})$ , which destabilizes the half-filled $3 d^{5}$ configuration. Thus, $I E_{2} (M n)$ is less than $I E_{2} (C r)$ . This statement is **correct**.

2. $M n^{2 +} < F e^{2 +}$ (i.e., $I E_{3} (M n) < I E_{3} (F e)$ ): Removing an electron from $M n^{2 +} ([A r] 3 d^{5})$ yields $M n^{3 +} ([A r] 3 d^{4})$ , destabilizing the stable half-filled $3 d^{5}$ configuration. Removing an electron from $F e^{2 +} ([A r] 3 d^{6})$ yields $F e^{3 +} ([A r] 3 d^{5})$ , which achieves a stable half-filled $3 d^{5}$ configuration. Therefore, $I E_{3} (M n)$ is significantly higher than $I E_{3} (F e)$ . The statement $I E_{3} (M n) < I E_{3} (F e)$ is **incorrect**.

3. $F e^{2 +} < F e^{3 +}$ (i.e., $I E_{3} (F e) < I E_{4} (F e)$ ): This compares successive ionization enthalpies for Iron. Successive ionization enthalpies always increase. This statement is **correct**.

4. $M n^{+} < M n^{2 +}$ (i.e., $I E_{2} (M n) < I E_{3} (M n)$ ): This compares successive ionization enthalpies for Manganese. Successive ionization enthalpies always increase. This statement is **correct**.

The incorrect relationship is option 2.

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Chemistry Question 61 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

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