Formation of sodium chloride, NaCl ( ionic compound)

The formation of ionic bonds in sodium chloride, NaCl is explained as follows:

•    The electron arrangement of a sodium atom is 2.8.1.
•     Sodium atom releases one valence electron to form sodium ion, Na⁺

                             Na → Na⁺ + e

• The electron arrangement of sodium ion is 2.8.
• Sodium ion achieves a stable octet electron arrangement.
• The electron arrangement of a chlorine atom is 2.8.7.
• Chlorine atom receives one electron to form chloride ion, Cl^-

                             Cl + e → Cl^-

• The electron arrangement of chloride ion is 2.8.
•  Chloride ion achieves a stable octet electron arrangement.
•  Strong electrostatic force pulls the sodium ion and chloride ion together.
• An ionic bond is formed.
• Sodium chloride is an ionic compound.

Remark:
Students always confuse following terms.
chlorine is atom
choride is ion


When answer this question, please draw the diagram of formation of ionic bond in sodium chloride because some time the diagram will contribute marks for you.