Question

Which of the following electronic configuration would exhibit the lowest ionization energy?

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

Answer 1

Answer: (D)

$1s^{2}2s^{2}2p^{6}3s^1$

The electronic configuration (D) will exhibit the lowest ionisation energy. As per the definition ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated neutral gaseous atom, molecule or ion. In the given configurations the electron in 3s subshell is far away from the nucleus and experience lowest attraction from the nucleus i.e its $Z_{eff}$ will be lowest among all hence (D) is the correct option.