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)

Q: 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)

**Electronic configurations** (from Table 3.3): Sodium (At. No. 11) = 2, 8, 1; Oxygen (At. No. 8) = 2, 6. **(i) Electron-dot structure of sodium:** Sodium has 1 valence electron, shown as a single dot around the symbol Na. $$\overset{\displaystyle\centerdot}{\text{Na}}$$ **(ii) Electron-dot structure of oxygen:** Oxygen has 6 valence electrons, shown as three lone pairs, needing 2 more electrons to complete its octet. $$:\overset{\displaystyle\centerdot\centerdot}{\underset{\displaystyle\centerdot\centerdot}{\text{O}}}:$$ (six dots placed around O, two positions left open) **Formation of sodium oxide ($\text{Na}_2\text{O}$):** Each sodium atom loses its single valence electron to attain the stable octet configuration of neon (2,8), forming a sodium cation: $$\text{Na} \rightarrow \text{Na}^+ + e^-$$ $$(2,8,1) \qquad (2,8)$$ Since oxygen needs 2 electrons to complete its octet (attaining the configuration of neon, 2,8), **two** sodium atoms are required to donate one electron each to a single oxygen atom: $$2\text{Na} \rightarrow 2\text{Na}^+ + 2e^-$$ $$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$ $$(2,6) \qquad\qquad (2,8)$$ Overall combination: $$2\text{Na} + \text{O} \rightarrow 2\text{Na}^+ + \text{O}^{2-} \rightarrow \text{Na}_2\text{O}$$ The two Na$^+$ ions and the O$^{2-}$ ion are held together by strong electrostatic forces of attraction to form the ionic (electrovalent) compound sodium oxide, $\text{Na}_2\text{O}$. **Cation and anion in Na$_{2}$O:** - **Cation:** Sodium ion, $\text{Na}^+$ (two per formula unit) - **Anion:** Oxide ion, $\text{O}^{2-}$ (one per formula unit)

Metals and Non-metals

Ionic Bond Formation — Electron-dot Structures

Official Answer

Electronic configurations (from Table 3.3): Sodium (At. No. 11) = 2, 8, 1; Oxygen (At. No. 8) = 2, 6.

(i) Electron-dot structure of sodium: Sodium has 1 valence electron, shown as a single dot around the symbol Na.

$\overset{\displaystyle\centerdot}{\text{Na}}$

(ii) Electron-dot structure of oxygen: Oxygen has 6 valence electrons, shown as three lone pairs, needing 2 more electrons to complete its octet.

$:\overset{\displaystyle\centerdot\centerdot}{\underset{\displaystyle\centerdot\centerdot}{\text{O}}}:$ (six dots placed around O, two positions left open)

Formation of sodium oxide ( $\text{Na}_2\text{O}$ ):

Each sodium atom loses its single valence electron to attain the stable octet configuration of neon (2,8), forming a sodium cation:

$\text{Na} \rightarrow \text{Na}^+ + e^-$

$(2,8,1) \qquad (2,8)$

Since oxygen needs 2 electrons to complete its octet (attaining the configuration of neon, 2,8), two sodium atoms are required to donate one electron each to a single oxygen atom:

$2\text{Na} \rightarrow 2\text{Na}^+ + 2e^-$

$\text{O} + 2e^- \rightarrow \text{O}^{2-}$

$(2,6) \qquad\qquad (2,8)$

Overall combination:

$2\text{Na} + \text{O} \rightarrow 2\text{Na}^+ + \text{O}^{2-} \rightarrow \text{Na}_2\text{O}$

The two Na $^+$ ions and the O $^{2-}$ ion are held together by strong electrostatic forces of attraction to form the ionic (electrovalent) compound sodium oxide, $\text{Na}_2\text{O}$ .

Cation and anion in Na $_{2}$ O:

Cation: Sodium ion, $\text{Na}^+$ (two per formula unit)
Anion: Oxide ion, $\text{O}^{2-}$ (one per formula unit)

Hint

Sodium (2,8,1) loses 1 electron; oxygen (2,6) needs 2 electrons to complete its octet. Two sodium atoms are needed per oxygen atom.

Analysis & Explanation

This question directly applies the method shown in Section 3.3 (Figures 3.5 and 3.6) for NaCl and MgCl $_{2}$ formation, extended to Na $_{2}$ O as posed in the in-text QUESTIONS after Section 3.3.1 (Q1: 'Show the formation of Na2O and MgO by the transfer of electrons'). The chapter establishes that metals lose electrons from their valence shell to attain the stable, completely-filled valence shell (octet) of the nearest noble gas, forming cations, while non-metals gain electrons to complete their octet, forming anions. Compounds formed by such electron transfer are called ionic or electrovalent compounds (Section 3.3), and as per Table 3.3, both Na (ending at 2,8 like Ne) and O (ending at 2,8 like Ne) attain noble gas configuration in Na $_{2}$ O. The key subtlety students must catch is the 2:1 stoichiometry — since O needs 2 electrons but each Na supplies only 1, two Na atoms combine with one O atom, unlike the 1:1 ratio seen in NaCl.

Common Mistakes

1Forgetting that two sodium atoms (not one) are required per oxygen atom, leading to an incorrectly balanced 'NaO' instead of Na2O.
2Drawing oxygen's electron-dot structure with the wrong number of dots (using 8 instead of 6 valence electrons for a neutral O atom).
3Mislabelling O2- as the cation and Na+ as the anion (reversing cation/anion identification).
4Showing crossed and dotted electrons for both original valence electrons of Na without indicating that Na has none left after losing its single electron.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2025?

This question carries 3 marks in the CBSE Class 10 Science 2025 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Metals and Non-metals" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Ionic Bond Formation — Electron-dot Structures" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2025 paper?

This is a Short Answer question from Section C in the CBSE Class 10 Science 2025 paper.