Question

The electronic configuration of an element is $1s^2$ $2s^2$ $2p^6$ $3s^2$ $3p^4$. The atomic number and the group number of element 'X' (which is just below the said element in the periodic table) respectively are:

• A. 24 & 6
• B. 34 & 6
• C. 24 & 16
• D. 34 & 16

Answer 1

The correct option is D 34 & 16

Electronic configuration of an element: $1s^2$ $2s^2$ $2p^6$ $3s^2$ $3p^4$. As last electron is in p-subshell, the element is from p-block.

In p-block:
Group number = Number of valence electrons + 10
= (2+4) + 10
= 6+10
= 16
Atomic number = Number of electrons in electronic configuration
= 2 + 2 + 6 + 2 +4
= 16 ( i.e. sulphur)
But an element 'X' is first below the above element. So, group number will be the same, i.e., ${16}^{\text{th}}$ and atomic number = 16 + 18
= 34
So, X is Selenium (Se) with atomic number = 34
and group number = 16