Explain the electronic configuration of periods in the periodic table.

(N/A) The period number of an element indicates the value of the principal quantum number $(n)$.
As the period number increases,the energy level increases $(n=1, n=2, \ldots \text{ etc. })$.
The number of elements in any period is double the number of atomic orbitals available in that shell.

Principal $(n)$ Energy level	Obtained orbitals	Number of elements
$n=1$ $(K \text{ shell})$	One $(1s)$	$2$ elements: Hydrogen,Helium
$n=2$ $(L \text{ shell})$ Second period	One $2s +$ three $2p$,total four orbitals	$8$ elements: ${}_{3}Li$ to ${}_{10}Ne$
$n=3$ $(M \text{ shell})$ Third period	One $3s +$ three $3p$,total four orbitals	$8$ elements: ${}_{11}Na$ to ${}_{18}Ar$
$n=4$ $(N \text{ shell})$ Fourth period	One $4s +$ five $3d +$ three $4p$,total nine orbitals	$18$ elements: ${}_{19}K$ to ${}_{36}Kr$
$n=5$ $(O \text{ shell})$ Fifth period	One $5s +$ five $4d +$ three $5p$,total nine orbitals	$18$ elements: ${}_{37}Rb$ to ${}_{54}Xe$
$n=6$ $(P \text{ shell})$ Sixth period	One $6s +$ seven $4f +$ five $5d +$ three $6p$,total sixteen orbitals	$32$ elements: ${}_{55}Cs$ to ${}_{86}Rn$ (includes Lanthanide series)
$n=7$ $(Q \text{ shell})$ Seventh period	One $7s +$ seven $5f +$ five $6d +$ three $7p$,total sixteen orbitals	$32$ elements: ${}_{87}Fr$ to ${}_{118}Og$ (includes Actinide series)