To determine which transition metal has the highest third ionization enthalpy, we need to consider the electronic configurations and the stability of the resulting ions upon successive ionizations.
The electronic configuration for each element is as follows:
The third ionization enthalpy refers to the energy required to remove the third electron after removing the first two. When evaluating this, the stability of half-filled d-orbitals plays a significant role.
For manganese (Mn), removing the first two electrons results in a configuration of \(3d^5\), which is a stable half-filled configuration.
This stability makes it significantly more challenging to remove a third electron from manganese than from the other elements.
Therefore, the element with the highest third ionization enthalpy is: Mn (Manganese).
The 3rd ionisation enthalpy refers to the energy required to remove the third electron after the removal of the first and second electrons from a neutral atom.
Reasoning for Mn Having the Highest 3rd Ionisation Enthalpy
Manganese (Mn) has the electronic configuration [Ar] $3d^5 4s^2$.
The removal of the first and second electrons leads to a half-filled $3d^5$ configuration, which is highly stable due to symmetrical distribution and exchange energy.
Removing the third electron from this stable half-filled $3d^5$ configuration requires a significantly higher amount of energy compared to other transition elements such as Cr, V, and Fe.
Conclusion: The transition metal having the highest 3rd ionisation enthalpy is Mn.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: