The third ionisation enthalpy is the lowest for:

The correct option is D

$\mathrm{Fe}$

The explanation for the correct option:

D: $\mathrm{Fe}$

Consider an element E,

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

Explanation for the incorrect options:

• The third Ionization enthalpy for options A $\left(\mathrm{Mn}\right)$ , B $\left(\mathrm{Co}\right)$ and C $\left(\mathrm{Ni}\right)$ are higher than option D $\mathrm{Fe}$.

Therefore, option D is the correct option.