A) Vanadium (Z = 23)
B) Chromium (Z = 24)
C) Iron (Z = 26)
D) Manganese (Z = 25)
Correct Answer: D
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 in third ionization enthalpy belongs to \[3{{d}^{3}}\]sub-shell. \[_{24}Cr=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{1}}\] Third electron which is removed in third ionisation potential belongs to \[3{{d}^{4}}\]sub-shell. \[_{26}Fe=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}},4{{s}^{2}}\] Third electron which is removed in third ionisation enthalpy belongs to \[3{{d}^{6}}\]sub-shell. \[_{25}Mn=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{5}},4{{s}^{2}}\] Third electron which is removed in third ionisation potential belongs to \[3{{d}^{5}}\] sub-shell. In all elements shell and sub-shells are same. Required amount of energy (enthalpy) is based upon the stability of d-subshell. The \[3{{d}^{5}}\]sub-shell has highest stability in all because it is half-filled sub-shell (while other are incomplete). So, Mn shows highest third ionisation potential.
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