Question

What is the number of valence electrons in (i) Sodium ion (${\mathrm{Na}}^{+}$) (ii) Oxide ion (${\mathrm{O}}^{2¯}$)

Answer 1

(i) Sodium ion (${\mathrm{Na}}^{+}$):

Number of electrons = 11-1 = 10

Electronic configuration = 2,8

(ii) Oxide ion (${\mathrm{O}}^{2¯}$)

Number of electrons = 8+2 = 10

Electronic configuration= 2,8

Conclusion: Therefore 8 valence electrons in ${\mathrm{Na}}^{+}$ and ${\mathrm{O}}^{2-}$.