Question

The electronic configuration of an element is $\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^3$. What is the atomic number of the element which is just below the above element in the periodic table?

Answer 1

Electronic configuration:

• The distribution of electrons into orbitals of an atom is called its electronic configuration.

Atomic number of an unknown element:

• The electronic configuration of the given element is $\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^3$. So the element is Phosphorus$\left(\mathrm{P}\right)$.
• The unknown element is just below the phosphorus in the periodic table, so the valence shell electronic configuration of the unknown element is $4{\mathrm{s}}^{2}4{\mathrm{p}}^3$.
• So, the complete electronic configuration of an unknown element is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^3$.
• Hence, the atomic number of the element which is just below the above element in the periodic table is $\mathbf{33}$ and the element is Arsenic$\left(\mathrm{As}\right)$.