The third ionization enthalpy is minimum for: 2+ 3+ 2 5 2 7 2 6 2 8 2+ 5 2+ 7 2+ 6 2+ 8 2+ 2+ 6 0 3+ — d and f Block Elements Chemistry Question

💡 Solution & Explanation

Consider an element E E E would be the 3rd I.E. of the element E. Electronic configuration of Mn is [Ar]4s 3d , Co is [Ar]4s 3d , Fe is [Ar]4s 3d , Ni is [Ar]4s 3d Electronic configuration of Mn is [Ar]3d , Co is [Ar]3d , Fe is [Ar]3d , Ni is [Ar]3d As it is evident from the above configurations of the E for the given elements, Fe would require the least amount of energy for removal of electron as it has the configuration 3d 4s . That means that its E form is the most stable among the four elements provided in their respective E states, i.e., when compared, the next electron removal will require least amount of energy.