(i) Write the electron-dot structures for sodium, oxygen and magnesium. (ii) Show the formation of Na₂O and MgO by the transfer of electrons. (iii) What are the ions present in these compounds?
Hint. Count valence electrons first — sodium and magnesium are in groups 1 and 2, oxygen is in group 16.
Step 1 — (i) Electron-dot structures (valence electrons only). Sodium (Na): one dot around the symbol (1 valence electron). Magnesium (Mg): two dots (2 valence electrons). Oxygen (O): six dots (6 valence electrons, needing 2 more for a stable octet).
Step 2 — (ii) Formation of Na₂O. Each of two sodium atoms loses its one valence electron (Na → Na⁺ + e⁻), and one oxygen atom gains both of those electrons to complete its octet (O + 2e⁻ → O²⁻). The resulting ions combine as 2Na⁺ + O²⁻ → Na₂O.
Step 3 — Formation of MgO. One magnesium atom loses both its valence electrons (Mg → Mg²⁺ + 2e⁻), and one oxygen atom gains them (O + 2e⁻ → O²⁻), giving Mg²⁺ + O²⁻ → MgO.
Step 4 — (iii) Ions present. Na₂O contains Na⁺ and O²⁻ ions, since that's what the electron transfer in Step 2 produced. MgO contains Mg²⁺ and O²⁻ ions.
✦ Answer: Na has 1 valence electron, Mg has 2, O has 6. Na₂O forms from 2Na⁺ + O²⁻; MgO forms from Mg²⁺ + O²⁻. Ions present: Na⁺ and O²⁻ in Na₂O; Mg²⁺ and O²⁻ in MgO.
Where students slip. Giving oxygen only one dot pattern that doesn't add up to 6 valence electrons — oxygen is in group 16, so it starts with 6 valence electrons and needs exactly 2 more to reach a stable octet, which is why it always takes 2 electrons per atom in these reactions.
