Home › MCQs › NEET Chemistry › Question #7400
Back to Questions
NEET Chemistry QUESTION #7400
Question 1
The correct order for decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24), and Mn(25) is:
  • \(\text{Mn} > \text{Cr} > \text{Ti} > \text{V}\)
  • \(\text{Ti} > \text{V} > \text{Cr} > \text{Mn}\)
  • \(\text{Cr} > \text{Mn} > \text{V} > \text{Ti}\)✔️
  • \(\text{V} > \text{Mn} > \text{Cr} > \text{Ti}\)
Correct Answer Explanation
Second IE involves removing the 2nd electron (from M⁺ ion):
After first ionisation (losing 4s electron):
  • Ti⁺: [Ar]3d³ → lose 3d electron
  • V⁺: [Ar]3d⁴
  • Cr⁺: [Ar]3d⁵ (half-filled, very stable) → IE₂ very high
  • Mn⁺: [Ar]3d⁶ → less stable than half-filled
Cr⁺ (3d⁵) has maximum stability, so IE₂ of Cr is highest. Order: Cr > Mn > V > Ti