The electronic configuration of helium is $1{s^2}$ and of oxygen is:
A. $1{s^2},2{s^2},2{p^5}$
B. $1{s^2},2{s^2},2{p^4}$
C. $1{s^2},2{s^2},2{p^3}$
D. $1{s^2},2{s^2},2{p^6}$

Hint: The way in which electrons are distributed among the orbitals of an atom is known as electronic configuration. It is also known as electronic structure. There are three rules that govern the electronic configuration of an element.

Complete step by step answer:
In case of oxygen the atomic number is 8 and the atomic number is equal to the number of protons which is equal to the number of electrons. Hence the number of electrons in oxygen is 8.
According to aufbau principle, the electrons are filled in various orbitals in order of their increasing energies. Thus an orbital with lowest energy is filled first. Also, the energy content of two subshells can be compared by means of (n+1) rule. Where n is the principal quantum number which indicates the main energy level. In case of oxygen $1s$ is filled first then $2s{\text{ and }}2p$.
According to the Pauli Exclusion Principle, an orbital cannot accommodate more than two electrons and the two electrons in an orbital have two spins.
According to Hund’s rule of maximum multiplicity, no pairing takes place in the orbitals of equivalent energy until each orbital in the given subshell contains one electron and the spin of any unpaired electrons are parallel. In this case all 3 p orbitals contain one electron then pairing starts.
By considering the above the above laws, we can say the electronic configuration of oxygen is $1{s^2},2{s^2},2{p^4}$

Hence the correct answer is option B .

Note:
Based on the rule of magnetic effect of electron spin, when the different orbitals of same energy are half or less than half filled, the electrons should have parallel spin because parallel arrangement of spin is more stable than antiparallel arrangement.