Question

Successive ionization energies of an element ‘X’ are given in the table (in Kcal). Electronic configuration of the element ‘X’ is:

$I{{P}_{1}}$	$I{{P}_{2}}$	$I{{P}_{3}}$	$I{{P}_{4}}$
165	195	556	595

(a)- $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{2}}$
(b)- $1{{s}^{2}}2{{s}^{1}}$
(c)- $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$
(d)- $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$

Answer 1

The ionization energy is the energy required to extract the outermost electron from the valence shell. The higher the ionization energy the stronger the electron is attracted to the nucleus. The half-filled and fully-filled subshell has an exceptionally high value of ionization energy due to extra-stability.

Complete Solution :
- The ionization energy is the energy required to extract the outermost electron from the valence shell. The higher the ionization energy the stronger the electron is attracted to the nucleus. The half-filled and fully-filled subshell has an exceptionally high value of ionization energy due to extra-stability.
- So, from the table, we can see that the first and second ionization are small which indicates that the electrons are very far from the nucleus and both have the same outermost shell. The third ionization is very high as compared to the second enthalpy which will be either due to change in the shell or half-filled and fully-filled subshell. Therefore, by combining all these factors, $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$ configuration fulfills all the conditions above. The first and second enthalpies are small because the electrons are in the 3rd valence shell. When they removed the configuration becomes $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ which very stable and fully-filled, so its ionization is very high. Moreover, the valence shell is also changing from 3 to 2.
So, the correct answer is “Option D”.

Note: You may get confused that the option (a) and (c) also has two electrons in its last shell, but no one satisfies the condition of the third ionization enthalpy, so they are incorrect options.