When sodium metal and chlorine gas are brought into contact, they react violently and we obtain sodium chloride. This reaction is shown below.
2 Na (s) + Cl₂ (g) ->2 NaCl (s)
The bonding in sodium chloride can be understood as follows:

Sodium (Na) has the atomic number 11 and we can write its electronic configuration as 2,8,1 i.e. it has one electron in its outermost (M) shell. If it loses this electron, it is left with 10 electrons. The resulting species is positively charged ion. Such a positively charged ion is called a cation. The cation in this case is called sodium cation Na⁺. This is shown below in Fig. 5.1.

Formation of sodium cation

Note that the sodium cation has 11 protons but 10 electrons only. It has 8 electrons in the outermost (L) shell. Thus, sodium atom has attained the noble gas configuration (that of Neon as shown in Table 5.1) by losing an electron present in its outermost shell. Thus, according to octet rule, sodium atom can acquire stability by changing to sodium cation.
The ionization of sodium atom to give sodium ion requires an energy of 496 kJ mol-1.
A chlorine atom having the atomic number 17, has the electronic configuration 2,8,7. It can complete its octet by gaining one electron. This is shown below in Fig.5.2.

Formation of chloride ion