If the successive ionisation energies of an element $A$ are 165, 190, 550 and $595 \mathrm{kcal}$,…
If the successive ionisation energies of an element $A$ are 165, 190, 550 and $595 \mathrm{kcal}$, respectively, then the ground state electronic configuration of element $A$ is
$[\mathrm{Ne}] 3 s^2 3 p^2$
$[\mathrm{He}] 2 \mathrm{~s}^1$
$[\mathrm{He}] 2 s^2 2 p^2$
$[\mathrm{Ne}] 3 \mathrm{~s}^2$
Solution
$
\begin{aligned}
& \mathrm{IE}_1=165 \mathrm{kcal}, \mathrm{IE}_2=190 \mathrm{kcal} . \\
& \mathrm{IE}_3=550 \mathrm{kcal}, \mathrm{IE}_4=595 \mathrm{kcal}
\end{aligned}
$
As ionisation energy value (first and second) of element $A$ is lower, than third and fourth ionisation energy.
So, it indicates that valence shell posess 2 electrons.
$\therefore$ Configuration of element $A$ is $[\mathrm{Ne}] 3 \mathrm{~s}^2$