Question

Question: The atomic numbers of ${\mathbf{1}}^{\mathbf{st}}$three noble gases needs to be determined if there is no spin quantum number and each orbital can have a maximum of 1 electron only.

Short answer

According to the Aufbau rule, the electronic configuration of the second noble gas will be $1{\mathrm{s}}^{1}2{\mathrm{s}}^{1}2{\mathrm{p}}^3$ and therefore$Z$ is equal to 5. According to the Aufbau rule, the electronic configuration of the third noble gas will be$1{\mathrm{s}}^{1}2{\mathrm{s}}^{1}2{\mathrm{p}}^{3}3{\mathrm{s}}^{1}3{\mathrm{p}}^3$ and therefore$Z$ equal to 9.

Step-by-step solution

Stating Pauli’s Exclusion Principle

Electronic configuration of elements in the ground state is assigned to elements only when they follow certain rules such as Hund's rule, Pauli Exclusion Principle and Aufbau rule.If Z is the atomic number, then the Z number of electrons is filled in the orbitals arranged to increase energy. The two electrons in an atom can have the same group of four quantum numbers, which is called Pauli’s Exclusion Principle.

Stating Hund's Rule

While orbitals are filled, one electron enters each energy level with degenerate energy before pairing begins, which is called Hund's rule. If the spin quantum number doesn't exist, then only one electron can occupy an orbital of a many-electron atom. In that case, the electronic configuration of the first noble gas will be$1{\mathrm{s}}^1$ and, therefore,$Z$ equal to 1.