Question

The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is

• A. Cr > Mn > V > Ti
• B. V > Mn > Cr > Ti
• C. Mn > Cr > Ti > V
• D. Ti > V > Cr > Mn

Answer 1

The Correct Option is A

To solve the question of identifying the order of decreasing second ionization enthalpy for the transition metals Titanium (Ti), Vanadium (V), Chromium (Cr), and Manganese (Mn), we need to understand the concept of second ionization enthalpy.

The second ionization enthalpy is the energy required to remove an electron from a singly charged positive ion. The general rule is that the ionization enthalpy increases with the removal of electrons that bring an incomplete sub-shell closer to a stable electronic configuration.

Now, let's analyze the electronic configurations of each element after the first ionization:

• Titanium (Ti): Atomic number = 22
  Electronic configuration: [Ar] 3d² 4s² → After first ionization: [Ar] 3d² 4s¹
• Vanadium (V): Atomic number = 23
  Electronic configuration: [Ar] 3d³ 4s² → After first ionization: [Ar] 3d³ 4s¹
• Chromium (Cr): Atomic number = 24
  Electronic configuration: [Ar] 3d⁵ 4s¹ → After first ionization: [Ar] 3d⁵ (a half-filled, stable configuration)
• Manganese (Mn): Atomic number = 25
  Electronic configuration: [Ar] 3d⁵ 4s² → After first ionization: [Ar] 3d⁵ 4s¹

Cr⁺ has a half-filled 3d sub-shell ([Ar] 3d⁵), which is especially stable. Removing another electron from this stable configuration requires more energy than removing an electron from other ions.

Based on stability, Mn⁺ and Cr⁺ both have similarly stable half-filled d sub-shells, but Cr⁺ has only one electron removed initially, making the second ionization more energy-demanding for Chromium than for Manganese.

Overall, this leads us to the conclusion that the order of decreasing second ionization enthalpy is:

• Cr > Mn > V > Ti

The correct option is therefore Cr > Mn > V > Ti.