What is the electron configuration of Na${^+}$ ion?

When we want to write down an electron configuration for some element (for example 1s${^2}$ 2s${^2}$ 2p${^6}$) we have to mention that numbers 1 and 2 represent an energy level or a period in a periodic table of elements, letters s and p represent a shape of an orbital and the numbers in the superscript represent the number of electrons in that particular orbital.

• $\it{s}$ orbital can hold maximum 2e${^-}$
• $\it{p}$ orbital can hold maximum 6e${^-}$
• $\it{d}$ orbital can hold maximum 10e${^-}$
• $\it{f}$ orbital can hold maximum 14e${^-}$

The total number of electrons in a neutral sodium (Na) atom is 11, thus, the ground state electron configuration will be as following:

$$\mathrm{1s^2 2s^2 2p^6 3s^1}$$

In the case of Na${^+}$ ion, we have to detract one electron from the previous electron configuration (because of the positive charge), thus, the new electron configuration will be as following:

$$\mathrm{1s^2 2s^2 2p^6}$$