Question

Write the electronic configuration of the sodium.

Answer 1

Hint: We have to use the concept of electronic configuration of elements. We can write the general electronic configuration of elements as $1{s^2}2{s^2}2{p^x}3{s^y}.........$. For example, the element neon has an atomic number of 10. The electronic configuration of neon is $1{s^2}2{s^2}2{p^6}$. We can similarly write the electronic configuration of sodium.

Complete step by step answer:
We know that sodium is a chemical element that has the symbol $Na$ and its atomic number is $11$. Sodium is a highly reactive metal, appears as silver-white and is soft in nature. Sodium is an alkali metal, which belongs to group 1 of the periodic table. It has a single electron in its valence shell, which readily donates its electrons forming a positively charged sodium cation $\left( {N{a^ + }} \right)$.
We first need to know the atomic number and the number of electrons in sodium.
Sodium has atomic number 11 and the number of electrons in sodium is 11.
When we write the electronic configuration sodium, we can pull all 11 electrons in orbit around the nucleus of the sodium atom.
The first electrons will go into 1s orbital. Since 1s orbital can have only two electrons, the next two electrons go to 2s orbital. 2p orbitals will hold six electrons, so the next six electrons will go to 2p. The remaining one electron will go 3s orbital.
From this, we know the electron structure of sodium would be $1{s^2}2{s^2}2{p^6}3{s^1}$.
The atomic number of sodium is 11. Thus, the electronic configuration of sodium is $1{s^2}2{s^2}2{p^6}3{s^1}$ (or) $\left[ {Ne} \right]3{s^1}$.


Note:
We know the electrons move around the nucleus in imaginary paths known as shells (or) orbitals. The electrons are arranged in different orbits. The various orbits are K,L,M,N… etc. There is no difference between the energies of the various subshells within a shell, the 3s,3p and 3p orbitals, for an atom that contains only one electron.