The third ionization enthalpy is minimum for

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(A) The electronic configurations of the given elements are:$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$$Co (Z=27): [Ar] 3d^7 4s^2$$Ni (Z=28

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$Fe$

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(A) The electronic configurations of the given elements are:$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$$Co (Z=27): [Ar] 3d^7 4s^2$$Ni (Z=28): [Ar] 3d^8 4s^2$After the removal of two electrons ($M^{2+}$ ions):$Mn^{2+}: [Ar] 3d^5$$Fe^{2+}: [Ar] 3d^6$$Co^{2+}: [Ar] 3d^7$$Ni^{2+}: [Ar] 3d^8$The third ionization enthalpy involves the removal of an electron from the $3d$ orbital. For $Fe^{2+}$,removing one electron results in the stable half-filled $3d^5$ configuration. Since the removal of an electron from $Fe^{2+}$ leads to a more stable state,the energy required (third ionization enthalpy) is the lowest among the given options.

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