Question

Write the electronic configuration of \[{\text{N}}{{\text{a}}^ + }\] ion.

Answer 1

Hint:The atomic number of sodium is 11. A positive charge is formed by loss of electrons. First of all s orbital is filled then p.

Complete step by step solution:
Sodium is a group number 1 element. It belongs to the third period. The atomic number of sodium is 11. This means it has 11 protons and 11 electrons. To write the electronic configuration we need to consider only electrons and not protons.
The electronic configuration of sodium involves s and p orbital. The electronic configuration can be written as:
\[{\text{Na : }}1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^1}\]
Now the presence of one positive charge means the loss of 1 electron. Hence the number of electrons in \[{\text{N}}{{\text{a}}^ + }\] ion will be 10; however the number of protons will be the same , that is 11. So we will just remove one electron from the electronic configuration of neutral sodium atom to get the electronic configuration of the ion:
\[{\text{N}}{{\text{a}}^ + }{\text{ : }}1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^0}\]

Additional information:
The removal of one electron from sodium makes it very stable due to attainment of the noble gas electronic configuration just like neon. Sodium is a metal and has a very strong tendency to release its one electron to get stability. Sodium is known as alkali metal because its oxide and hydroxide are basic in nature.

Note:
By looking at the electronic configuration we can predict the group number and period number of any element, for example the electronic configuration of sodium is \[{\text{Na : }}1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^1}\] . The group number is equal to the number of electrons present in the valence shell. Here valence shell is 3s and the number of electrons is 1 so the group number is 1. The period number is equal to the shell number. Here the maximum shell number used is 3 and hence the period number is 3.