Question

What is the electronic configuration of Helium?

Answer 1

The electron distribution on the orbit of a particular atom is defined as the electronic configuration. Helium is the chemical element that is tasteless, colorless, odorless, non-toxic, monoatomic gas and it lies in the category of Nobel gas.

Complete answer:
Rule 1 (Aufbau Principle)- Electrons occupy the lowest energy orbitals possible, starting with 1s and continuing in the order dictated by quantum mechanics.
Rule 2 (Hund’s Rule)- Electrons occupy degenerate orbitals (i.e. same n and l quantum numbers).
$\bullet$ they must first occupy the empty orbitals before double occupying them. Furthermore, the most stable configuration results when the spins are parallel (i.e. all the same ${m_s}$ quantum number).
$\bullet$Rule $3$ (Pauli-exclusion Principle)- Each electron can be described with a unique set of four quantum numbers. Therefore, if two electrons occupy the same orbital, they have different spin magnetic quantum numbers (${m_s} = + \dfrac{1}{2}$ and ${m_s} = - \dfrac{1}{2}$).
We use numbers to indicate which shell an electron is in. The first shell, closest to the nucleus and with the lowest-energy electrons, is shell $1.$ This first shell has only one subshell, which is labeled as $1s$ and can hold a maximum of two electrons. We can combine the shell and subshell labels when referring to the organization of electrons about a nucleus and use a superscript to indicate how many electrons are in a subshell.
Helium has two electrons. Both electrons fit into the subshell because subshell can hold up to $2$ electrons. These two electrons which are present at its first orbit that is $1s$. So, the electronic configuration for helium ions is $1{s^2}$.
Now, the valence electrons. Valence electrons are the number of electrons present at the outermost shell. Here the helium has only two electrons which are present in its one shell. So, the number of valence electrons for helium is $2$.
The electronic configuration of this Noble gas element is $1{s^2}$ as $He$ consists of $2$ electrons.

Note:
Helium is a Noble gas and the electronic configuration of all the Noble gases is known as Noble gas configuration. The atom consists of the elemental symbol of the last Noble gas prior to that atom, followed by the configuration of the remaining electrons. So for lithium, we make a substitution of $[H]$ for the $1{s^2}$ part of the configuration. lithium’s Noble gas configuration becomes $[H]2{s^1}$.