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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 ionization enthalpy :-
A
Vanadium (Z = 23)
B
Manganese (Z = 25)
C
Chromium (Z = 24)
D
Iron (Z = 26)
Detailed Solution
The third ionisation enthalpy is the energy needed to remove an electron from the $M^{2+}$ ion.
Configurations of the $M^{2+}$ ions: $V^{2+}$ $3d^3$, $Cr^{2+}$ $3d^4$, $Mn^{2+}$ $3d^5$, $Fe^{2+}$ $3d^6$.
$Mn^{2+}$ has the stable half-filled $3d^5$ configuration. Removing a third electron would disturb this extra stable arrangement, so a very large amount of energy is required.
$Fe^{2+}$ ($3d^6$) easily loses one electron to attain the stable $3d^5$ configuration, so its third ionisation enthalpy is comparatively low.
Approximate values (kJ/mol): V 2828, Cr 2987, Mn 3248, Fe 2957.
So manganese has the highest third ionisation enthalpy.
Configurations of the $M^{2+}$ ions: $V^{2+}$ $3d^3$, $Cr^{2+}$ $3d^4$, $Mn^{2+}$ $3d^5$, $Fe^{2+}$ $3d^6$.
$Mn^{2+}$ has the stable half-filled $3d^5$ configuration. Removing a third electron would disturb this extra stable arrangement, so a very large amount of energy is required.
$Fe^{2+}$ ($3d^6$) easily loses one electron to attain the stable $3d^5$ configuration, so its third ionisation enthalpy is comparatively low.
Approximate values (kJ/mol): V 2828, Cr 2987, Mn 3248, Fe 2957.
So manganese has the highest third ionisation enthalpy.