Question

The third ionization enthalpy is minimum for:

• A. $Mn$
• B. $Co$
• C. $Ni$
• D. $Fe$

Answer 1

The correct option is D $Fe$

Consider an Element E
$E^{2+}\to E^{3+}$ would be 3rd I.E of the element E.
Electronic configuration of Mn is $\left[Ar\right]4s^{2}3d^5,Co\text{is}\left[Ar\right]4s^{2}3d^7,Fe\text{is}\left[Ar\right]4s^{2}3d^6,Ni\text{is}\left[Ar\right]4s^{2}3d^8$
Electronic configuration of $M{n}^{2+}$ is $\left[Ar\right]3d^5,C{o}^{2+}\text{is}\left[Ar\right]3d^7,F{e}^{2+}\text{is}\left[Ar\right]3d^6,N{i}^{2+}\text{is}\left[Ar\right]3d^8$ .
As it is evident from the above configurations of the $E^{2+}$ for the given elements, $F{e}^{2+}$ would require the least amount of energy for removal of electron as it has the configuration $3d^{6}4s^0$. That means that its $E^{3+}$ form is the most stable among the four elements provided in their respective $E^{3+}$ states, i.e., when compared, the next electron removal will require least amount of energy.