Write the electron-dot structures of (i) sodium, and (ii) oxygen. Using these structures, show the formation of sodium oxide. Mark the anion and cation present in this compound.
(At. No. – Sodium = 11 and Oxygen = 8)
Generated by claude-sonnet-4-6 · 2026-06-15 06:47 · grounding rag
Model Answer
(i) Electron-dot structures:
- Sodium (At. No. 11; configuration 2, 8, 1): one dot around the symbol Na.
Na·
- Oxygen (At. No. 8; configuration 2, 6): six dots around the symbol O.
:Ö: (six electrons shown around O)
(ii) Formation of sodium oxide (Na₂O):
Oxygen needs 2 electrons to complete its octet. Each sodium atom donates 1 electron. So 2 sodium atoms each transfer 1 electron to one oxygen atom:
$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$
The compound formed is Na₂O.
(iii) Ions present:
- Na⁺ → Cation (sodium ion)
- O²⁻ → Anion (oxide ion)
Source: Chapter 3, Section 3.3 — How do Metals and Non-metals React?
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Explanation
- Draw electron-dot (Lewis) structures by placing dots equal to valence electrons around the symbol: Na has 1 valence electron; O has 6.
- For Na₂O, show two Na atoms each losing one electron (since O needs 2 electrons to achieve octet).
- Examiners specifically expect you to label the cation (Na⁺) and anion (O²⁻) — missing this costs marks.
- The transfer of electrons makes this an ionic compound; no need to explain bonding in detail for 3 marks.