Read the following passage and answer the next five questions based on it:

In any transition series, as we move from left to right the \(d-\)orbitals are progressively filled and their properties vary accordingly

The above are the two series of f-block elements in which the chemical properties won't change much. The 5f-series elements are radioactive in nature and mostly are artificially synthesized in laboratories and thus much is not known about their chemical properties.

Which of the following is the correct order of second ionization enthalpy?

V < Cr > Mn

The correct answer is option 3. V < Cr > Mn.

The second ionization enthalpy refers to the energy required to remove the second electron from an already singly ionized atom. Here's a breakdown of why the order is V < Cr > Mn:

Electron Configuration:

Vanadium (V): [Ar]3d³4s² (3 electrons in the 3d subshell)

Chromium (Cr): [Ar]3d⁵4s¹ (5 electrons in the 3d subshell) - Exceptional Case

Manganese (Mn): [Ar]3d⁵4s² (5 electrons in the 3d subshell)

Stability of Electron Configuration:

Generally, elements with completely filled or half-filled subshells (especially d orbitals) have extra stability due to symmetrical electron arrangements. This stability translates to higher ionization energy.

In this case, Chromium (Cr) has a half-filled 3d subshell (3d⁵) after losing its first electron. This configuration is more stable than the expected [Ar]3d⁴4s² for Cr⁺ because of the symmetrical arrangement of electrons in the 3d subshell.

Impact on Second Ionization Enthalpy:

Removing the second electron from Vanadium (V) or Manganese (Mn) involves taking away an electron from a 4s subshell. This requires a relatively lower amount of energy.

However, removing the second electron from Chromium (Cr) disrupts the stability of its half-filled 3d⁵ configuration. Overcoming this stability requires a significantly higher amount of energy compared to V and Mn.

Therefore, the order of increasing second ionization enthalpy is: 3. V < Cr > Mn