Step 1: Understanding the Concept:
The second ionisation enthalpy is the energy required to remove an electron from a +1 ion. To determine the order, we look at the electronic configurations of the +1 ions and the stability of the resulting +2 ions. Stable configurations (like half-filled or filled subshells) result in higher ionisation enthalpies.
Step 2: Detailed Explanation:
Write the electronic configuration for M\(^+\) ions:
- Ti\(^+\) (Z=22): [Ar] 3d\(^2\) 4s\(^1\)
- V\(^+\) (Z=23): [Ar] 3d\(^3\) 4s\(^1\)
- Cr\(^+\) (Z=24): [Ar] 3d\(^5\) (Note: 3d\(^5\) is half-filled and very stable)
- Mn\(^+\) (Z=25): [Ar] 3d\(^5\) 4s\(^1\)
Now consider the energy to remove the next electron (forming M\(^{2+}\)):
- Removing an electron from Cr\(^+\) ([Ar] 3d\(^5\)) to form Cr\(^{2+}\) ([Ar] 3d\(^4\)) is very difficult because Cr\(^+\) has a stable half-filled 3d subshell.
- Removing an electron from Mn\(^+\) ([Ar] 3d\(^5\) 4s\(^1\)) to form Mn\(^{2+}\) ([Ar] 3d\(^5\)) is relatively easier than Cr because the electron is being removed from the 4s orbital.
- However, comparing the trend, Cr and Mn have higher values due to the stable 3d configurations. Mn\(^{2+}\) is especially stable ([Ar] 3d\(^5\)).
Step 3: Comparing Values:
Following the general trend of second ionisation enthalpy in the 3d series:
- Mn (Z=25) has a high value as the resulting Mn\(^{2+}\) is [Ar] 3d\(^5\).
- Cr (Z=24) is also high because the removal of an electron disrupts the half-filled d-subshell of Cr\(^+\).
- V (Z=23) and Ti (Z=22) follow in decreasing order.
The experimental decreasing order of second ionisation enthalpy is Mn > Cr > V > Ti.
Step 4: Final Answer:
The correct decreasing order is (D), (C), (B), (A).