Question

Show the formation of ${\mathrm{Na}}_2\mathrm{O}$ by transfer of electrons?

Answer 1

The ions present in the sodium oxide (${\mathrm{Na}}_2\mathrm{O}$ )are sodium ions ($2{\mathrm{Na}}^{+}$) and oxide ions (${\mathrm{O}}^{2-}$).

• The atomic number of Sodium is 11
• The electronic configuration of Sodium is:

$\mathrm{Na}(\mathrm{Z}=11)=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1$.

In order to obtain an octet configuration, it will lose one electron.

• The atomic number of oxygen is 8
• The electronic configuration of oxygen is:

O (Z = 8) = $\mathrm{O}(\mathrm{Z}=8)=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4$.

In order to obtain an octet configuration, it has to gain two electrons.

• Two sodium atoms transfer their 2 outermost electrons to an oxygen atom.
• By losing two electrons, the two sodium atoms form two sodium ions ($2{\mathrm{Na}}^{+}$).
• And by gaining two electrons, the oxygen atom forms an oxide ion (${\mathrm{O}}^{2-}$).

$\underset{\mathrm{Sodium}}{2\mathrm{Na}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Sodium}\mathrm{Oxide}}{{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)}$.