CBSE Class 10 Science 2019 · Paper 31/3/3 · Question 4

Sodium ($2,8,1$) has one outer electron to lose and oxygen ($2,6$) needs two more, so two sodium atoms each transfer one electron to one oxygen atom: $2\text{Na}⟶2{\text{Na}}^{+}+2e^{-}$ (each $2,8,1\to 2,8$) $\text{O}+2e^{-}⟶{\text{O}}^{2-}$ ($2,6\to 2,8$) $2{\text{Na}}^{+}+{\text{O}}^{2-}⟶{\text{Na}}_2\text{O}$ In the electron-dot diagram (step 4) the crosses of the two sodium atoms move to the oxygen atom, which becomes an oxide ion with eight outer electrons and charge $2-$, between two ${\text{Na}}^{+}$ ions; the ions are held by ionic bonds.