Question

Sucessive ionisation energies of an element 'X' are given below (in K. Cal)

$I{P}_1$	$I{P}_2$	$I{P}_3$	$I{P}_4$
165	195	556	595

Electronic configuration of the element 'X' is:

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

Answer 1

The correct option is D $1s^2,2s^{2}2p^6,3s^2$

$I{P}_1$ and $I{P}_2$ are less that indicates that there are only two valence electrons and they are in p orbital. $I{P}_3$ and $I{P}_4$ are large which means the next electrons are in lower orbitals which are closer to nucleus. From given options the only satisfying option is $1s^{2}2s^{2}2p^{6}3s^2$.