Question

The values of $I{P}_1,I{P}_2,I{P}_3$ and $I{P}_4$ of an atom are $7.5eV,25.6eV,48.6eV$ and $170.6eV$ respectively. The electronic configuration of the atom will be:

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

Answer 1

The correct option is B $1s^2,2s^2,2p^6,3s^2,3p^1$

Whenever there is a sudden increase in the ionisation potential, it indicates that the species had a stable configuration.
It is given that $I{p}_4>>>I{p}_3$, which indicate that the species attained stability by loosing 3 electrons.
So, option (b) represents this high energy change due to the removal of the fourth electron from a fully filled $2p^6$ orbital.