Answer :

Formation of Sodium ion- The atomic number of sodium is 11. The arrangements of electrons in the shells of the sodium atom is in the combination of 2,8,1. The valence electron present in the outermost shell of sodium atom is 1. The sodium atom can lose one electron to form sodium ion(Na+). There will be reduction in number of electrons, E=10. Number of protons remain same as 11. Due to the decrease in one electron, there is positive charge on a sodium ion.

Formation of Chloride ion- The atomic number of chlorine is 17. The arrangements of electrons in the shells of the chlorine atom is in the combination of 2,8,7. The valence electron present in the outermost shell of chlorine atom is 1. The chlorine atom can lose one electron to form chlorine ion(Cl-). There will be increase in number of electrons, E=18. Number of protons remain same as 17. Due to the increase in one electron, there is negative charge on a chloride ion.


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