Question

Successive ionization energies of an element 'X' are given in the table (in K. Cal)
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 $1s^2,2s^{2}2p^6,3s^2$

Electronic configuration of the element 'X' is $1s^2,2s^{2}2p^6,3s^2$
Since there is a significant difference between second and third ionization potentials (195 vs 556), it has 2 electrons in the valence shell i.e. the element belongs to alkaline earth metals.

Note that in option A and C, even though there are only 2 electrons in $3p$ and $2p$ orbitals respectively, removing those two electrons doesn't change the valence shell of the atom. However, in option D, removing the 2 outermost electrons changes the valence shell from $3\to 2$