Question

The electronic configuration of the atom having maximum difference between first and second ionization energies is :

• A. $1s^2,2s^2,2p^6,3s^1$
• B. $1s^2,2s^2,2p^6,3s^2$
• C. $1s^2,2s^2,2p^1$
• D. $1s^2,2s^2,2p^6,3s^{2}3p^3$

Answer 1

Answer: (A)

$1s^2,2s^2,2p^6,3s^1$

Consider the configurations of the options after the loss of an electron:-
A) $1s^2,2s^2,2p^6$
B) $1s^2,2s^2,2p^6,3s^1$
C) $1s^2,2s^2$
D) $1s^2,2s^2,2p^6,3s^2,3p^2$
The difference between $I{E}_1$ and $I{E}_2$ will be maximum for the case where there is a fully filled orbital. In this case, option A has attained noble gas configuration after the loss of an electron. Hence option A has the maximum difference between $I{E}_1$ and $I{E}_2$.