Question

What is the difference between electronic configuration of $Na$ and $N{a^ + }$?

Answer 1

Before solving this question we must know how cation and anion are formed. When a metal loses an electron, a positive charged species i.e. cation is formed and when a metal gains an electron, a negative charged species is formed i.e. anion. The species that have zero charge or no charge are neutral species.

Complete answer:
We must have to know that sodium is an element having atomic number $11$. It belongs to the group $1$ i.e. alkali metals. Sodium metal has one electron more to attain stable configuration.
Electronic configuration of $Na$: $1{s^2}2{s^2}2{p^6}3{s^1}$
Electronic configuration of $N{a^ + }$: $1{s^2}2{s^2}2{p^6}$
The difference between the electronic configuration of $Na$ and $N{a^ + }$ is that sodium metal has one electron more in order to get stable electronic configuration whereas sodium ion is formed when sodium metal loses one electron, therefore sodium ion has stable electronic configuration as its octet is completely filed.
$Na \to N{a^ + } + {e^ - }$
Sodium metal does not contain any charge it is a neutral moiety whereas sodium ion contains a positive charge, polyatomic species containing positive charge is known as cation. So $N{a^ + }$ is a cation and $Na$is a neutral species or metal. Sodium metal on losing an electron from outermost shell attains a noble gas configuration that has filled orbits.

Note:
Example of cation: $N{a^ + }$, example of anion: $C{l^ - }$and example of neutral species: $Na$. Sodium has molar mass which is approximately equal to $23g/mol$. Cation and anion are polyatomic species that contain charge. Noble gas has a stable electronic configuration that means it has completely filled orbits. Sodium is an alkali metal.