Question

Show the formation of Na₂O and MgO by the transfer of electrons.

Answer 1

Formation of Na₂O : 
The atomic number of sodium is 11 and it has only one valence electron. 

Hence, electronic configuration of ₁₁Na is 2, 8, 1.

The atomic number of oxygen is 8 and it has 6 electrons in its valence shell. 

Hence, electronic configuration of ₈O is 2, 6. 

Sodium has a tendency to lose the valence electron and oxygen has a tendency to gain the electron lost by sodium. Since sodium can lose only one electron of the valence shell and oxygen atom needs two electrons to complete its octet in the valence shell, two atoms of sodium combine with one atom of oxygen. By losing valence electron, sodium is changed into Na+ and by gaining two electrons lost by two sodium atoms, oxygen atom is changed into an oxide anion, O^2-. In this process, both the atoms, sodium and oxygen, obtain the stable electronic configuration of the noble gas neon.

The oppositely charged sodium ion, Na⁺ and oxide ion, O^2- are now held together by electrostatic forces of attraction or by ionic or electrovalent bond. Na₂O is, therefore, an ionic or electrovalent compound.

Formation of MgO: 

It has two electrons in its outermost shell. So, the magnesium atom donates its 2 valence electrons and forms a stable magnesium ion, Mg²⁺ to attain the electronic arrangement of neon gas.

It has six electrons in its valence shell. Therefore, it requires two more electrons to attain the stable electronic arrangement of neon gas. Thus, oxygen accepts two electrons donated by magnesium atom and forms a stable oxide ion, O^2-.

The oppositely changed magnesium ions, Mg²⁺ and oxide ions,O^2-, are held together by a  strong force of electrostatic attraction to form Magnesium oxide compound Mg²⁺O^2- or MgO.

MgO is an ionic compound.