Ex-Allen Aakash Narayana Faculty | 99% Fail this Third Ionization Enthalpy Trap (JEE Main 2024)

99% Fail this Third Ionization Enthalpy Trap (JEE Main 2024) | Ex-Allen Aakash Narayana Faculty
Are you also one of those students who get confused by the half-filled d-subshell stability logic and pick the wrong option for the third ionization enthalpy? If yes, you are falling right into a classic d-and-f-block elements trap from JEE Main 2024!

In this quick video, Nishant Sir breaks down the electronic configurations of V, Cr, Mn, and Fe to show you exactly why Manganese has the highest third ionization enthalpy, and how to avoid this common mock-test mistake.

🔍 Search Keywords / Exact Question Text:
"The transition metal having highest third ionization enthalpy is:
(a) MN
(b) Fe
(c) Cr
(d) V"

📝 Key Concepts Covered:
1. Understanding Third Ionization Enthalpy (IE3):
IE3 corresponds to the energy required to remove the third electron, which means converting a dipositive ion (M2+) into a tripositive ion (M3+).
Equation: M2+ - M3+ + e-

2. Electronic Configurations of M2+ Ions:
- Mn2+ = 3d5 (Exactly half-filled d-subshell, exceptionally stable!)
- Fe2+ = 3d6 (Removing 1 electron leads to a stable 3d5 state, so it is favorable)
- Cr2+ = 3d4
- V2+ = 3d3

3. The Explanation:
Because Mn2+ has a highly stable, symmetrical, half-filled 3d5 configuration, removing an electron from it requires a massive amount of energy. Thus, Manganese (Mn) has the highest third ionization enthalpy among the options.
Hence, the correct option is A (MN).

💡 Nishant Sir's Pro-Tip:
Whenever you are asked about the "N-th" Ionization Enthalpy, always write the electronic configuration of the (N-1) ion. For Third Ionization Enthalpy, analyze the stability of the M2+ ion, not the neutral M atom!

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