Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?

Solution :
In $${}_{23}V = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^3}$$  subshell.
$${}_{24}Cr = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^5}$$  subshell.
$${}_{26}Fe = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^6}$$  subshell.
$${}_{25}Mn = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^5}$$  subshell.
In all elements shell and subshells are same. Required amount of energy (enthalpy) is based upon the stability of $$d$$ - subshell.
The $$3{d^5}$$ -subshell has highest stability in all because it is half-filled subshell. So, $$Mn$$  shows highest third ionisation potential.