Question

Successive ionisation energies of an element $X$ are given(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 $1s^2,2s^{2}2p^6,3s^2$

Because Ionization Energy rises, sleepy from $I{P}_2$ to $I{P}_3$, the element will have 2 electrons in its outermost shell. Only option (D) fulfills the above criteria.