Atomic Foundations of Matter — Detailed Notes
This chapter explains the laws that govern how elements combine, Dalton’s atomic theory, how atoms join by chemical bonds to form molecules and ions, how to write chemical formulae, and how to calculate molecular / formula-unit mass.
9.1 Law of Conservation of Mass
Mass can neither be created nor destroyed in a chemical reaction — the total mass of reactants = total mass of products (Antoine Lavoisier, 1789). In an open container a gas can escape, making it seem violated (e.g. vinegar + baking soda in a balloon shows mass is conserved once the gas is trapped).
9.2 Law of Constant (Definite) Proportions
A compound always contains the same elements in a fixed ratio by mass, whatever its source (Joseph Proust). E.g. pure water always has hydrogen : oxygen = 1 : 8 by mass.
9.3 Dalton’s Atomic Theory (1808)
Postulates: all matter is made of tiny atoms; atoms are indivisible and cannot be created or destroyed in a reaction; atoms of a given element are identical in mass and chemical properties; atoms of different elements differ; atoms combine in simple whole-number ratios to form compounds. These explained the two laws above.
9.4 How atoms combine — chemical bonds
A molecule is an electrically neutral group of two or more atoms that can exist independently. Atoms combine to complete their octet (duplet for H/He) by either sharing or transferring electrons; the force holding them is a chemical bond.
Covalent bond — formed by sharing electron pairs. One shared pair = single bond (H–H, Cl–Cl, H–Cl); two shared pairs = double bond (O=O); in water each H shares one pair with O (H₂O). Covalent compounds are named with prefixes mono/di/tri/tetra… (CO = carbon monoxide, CO₂ = carbon dioxide, N₂O₅ = dinitrogen pentoxide); H₂S = hydrogen sulfide; NH₃ = ammonia.
Ionic bond — formed by transfer of electrons. A metal loses electrons to become a cation (Na → Na⁺), a non-metal gains electrons to become an anion (Cl → Cl⁻); the electrostatic attraction between opposite ions is the ionic bond (NaCl). Ionic compounds form 3-D crystal lattices, not molecules.
9.5 Writing chemical formulae
Write the symbols, put the valency/charge of each below, then criss-cross the valencies as subscripts (and simplify to the smallest ratio). Examples: CaCl₂, Al₂O₃, MgO, Mg(OH)₂, CaCO₃, Al₂(SO₄)₃. Use brackets for 2+ polyatomic ions. Common valencies: Na⁺, K⁺, Cl⁻, NO₃⁻ (1); Ca²⁺, Mg²⁺, O²⁻, SO₄²⁻, CO₃²⁻ (2); Al³⁺, PO₄³⁻ (3); NH₄⁺ (1).
9.6 Ionic vs covalent compounds
Ionic — usually soluble in water (insoluble in kerosene/petrol), conduct electricity when dissolved or molten (ions free to move) but not as solids, and have high melting/boiling points. Covalent — usually insoluble in water but soluble in kerosene/petrol, generally do not conduct electricity, and have low melting/boiling points.
9.7–9.8 Molecular mass & formula unit mass
Molecular mass (covalent) = sum of the atomic masses of all atoms in the molecule (H₂O = 2×1 + 16 = 18 u; CO₂ = 12 + 2×16 = 44 u). Formula unit mass (ionic, since they don’t form molecules) = sum of atomic masses in one formula unit (Na₂O = 2×23 + 16 = 62 u; Ca(NO₃)₂ = 40 + 2×(14 + 3×16) = 164 u).
Key Terms
law of conservation of mass, law of constant proportions, Dalton’s atomic theory, molecule, covalent bond (single/double), ionic bond, cation, anion, ion, polyatomic ion, valency, chemical formula (criss-cross), molecular mass, formula unit mass.
Revise, Reflect, Refine — All NCERT Exercise Answers
1Element A has 1 electron in its third shell; element B has 6 electrons in its second shell. (i)–(vi).
A is 2,8,1 (sodium-like); B is 2,6 (oxygen-like). (i) A tends to give 1 electron. (ii) It forms a cation (A⁺). (iii) B tends to take 2 electrons. (iv) It forms an anion (B²⁻). (v) A (metal) + B (non-metal) → ionic bond. (vi) Criss-crossing valencies 1 and 2 → formula A₂B (like Na₂O).
2Element X has six electrons in its outer shell and forms a diatomic molecule. (i)–(iv).
X is oxygen-like (2,6). (i) It has 6 valence electrons and needs 2 more, so two X atoms share two pairs of electrons to complete their octets — forming X₂. (ii) A covalent bond (a double bond). (iii) Structure: X=X (two shared pairs, e.g. O=O). (iv) Y has 2 valence electrons (valency 2); since X also has valency 2, they combine in a 1:1 ratio — the molecule is Y=X (formula YX, like a 1:1 oxide), the two atoms sharing two pairs of electrons.
3Design an ionic compound with total positive charge 6+ and total negative charge 6−. Which combination is correct?
(iii) 2 Fe³⁺ and 3 O²⁻. Positive = 2 × 3 = 6+, negative = 3 × 2 = 6−, so they balance (formula Fe₂O₃). The others don’t balance: (i) 2 Al³⁺ (6+) with 3 Cl⁻ (3−); (ii) 3 Mg²⁺ (6+) with 1 PO₄³⁻ (3−); (iv) 3 Ca²⁺ (6+) with 2 SO₄²⁻ (4−).
4Choose the correct statement(s) and correct the false ones.
(i) False — elements are made of atoms and compounds are made of molecules (not the other way round). (ii) False — a molecule of a compound is made of atoms of different kinds. (iii) False — a nitrogen gas molecule (N₂) contains two nitrogen atoms, not three. (iv) True — water is two hydrogen atoms covalently bonded to one oxygen atom (H₂O).
5Write the chemical formulae: (i) aluminium nitrate (ii) calcium oxide (iii) ferric oxide.
(i) Al³⁺ and NO₃⁻ → Al(NO₃)₃. (ii) Ca²⁺ and O²⁻ → CaO. (iii) Fe³⁺ and O²⁻ → Fe₂O₃.
6Write the formulae from these ion pairs: (i) Ca²⁺ & Br⁻ (ii) Al³⁺ & CO₃²⁻ (iii) K⁺ & SO₄²⁻ (iv) NH₄⁺ & Cl⁻.
(i) CaBr₂. (ii) Al₂(CO₃)₃. (iii) K₂SO₄. (iv) NH₄Cl.
7Which diagram correctly represents the Cl⁻ ion (atomic number of chlorine = 17)?
The Cl⁻ ion has 17 protons but has gained one electron, so it has 18 electrons arranged as 2, 8, 8 (a complete octet). The correct diagram is the one showing 2, 8, 8 electrons (a full outer shell), not 2,8,7.
8Determine the formula unit mass of (i) NH₄NO₃ (ii) H₃PO₄ (iii) NaHCO₃. (N=14, H=1, O=16, P=31, Na=23, C=12)
(i) NH₄NO₃ = (2×14) + (4×1) + (3×16) = 28 + 4 + 48 = 80 u. (ii) H₃PO₄ = (3×1) + 31 + (4×16) = 3 + 31 + 64 = 98 u. (iii) NaHCO₃ = 23 + 1 + 12 + (3×16) = 23 + 1 + 12 + 48 = 84 u.
9Write the formulae of compounds formed by: (i) Mg & N (ii) Li & N (iii) Na & S (iv) Al & O.
(i) Mg²⁺ and N³⁻ → Mg₃N₂ (magnesium nitride). (ii) Li⁺ and N³⁻ → Li₃N. (iii) Na⁺ and S²⁻ → Na₂S. (iv) Al³⁺ and O²⁻ → Al₂O₃.
10Complete the table of formulae (cations NH₄⁺, Li⁺, Al³⁺, Cu²⁺ with anions NO₃⁻, SO₄²⁻, PO₄³⁻).
NH₄⁺: NH₄NO₃, (NH₄)₂SO₄, (NH₄)₃PO₄. Li⁺: LiNO₃, Li₂SO₄, Li₃PO₄. Al³⁺: Al(NO₃)₃, Al₂(SO₄)₃, AlPO₄. Cu²⁺: Cu(NO₃)₂, CuSO₄, Cu₃(PO₄)₂.
115.3 g sodium carbonate + 6.0 g acetic acid → 2.2 g CO₂ + 0.9 g water + 8.2 g sodium acetate. Verify the law of conservation of mass.
Total mass of reactants = 5.3 + 6.0 = 11.3 g. Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g. Since reactant mass = product mass, the Law of Conservation of Mass is obeyed.
12A species has 11 protons, 12 neutrons, 10 electrons. (i) atomic & mass number (ii) ion type (iii) configuration (iv) name.
(i) Atomic number Z = protons = 11; mass number A = 11 + 12 = 23. (ii) Protons (11) > electrons (10), so it has a net +1 charge — it is a cation. (iii) With 10 electrons: 2, 8. (iv) It is the sodium ion, Na⁺.
13Elements A (2,8,5) and B (2,8,7). (i) more reactive (ii) ionic or covalent bond — explain (iii) formula of their compound.
(i) B (2,8,7) is more reactive — it needs only one electron to complete its octet, so it reacts readily. (ii) Both A (5 valence e⁻) and B (7 valence e⁻) are non-metals, so they complete their octets by sharing electrons → a covalent bond. (iii) A needs 3 electrons and B provides 1 each, so one A shares with three B atoms → AB₃ (like PCl₃).
14Assertion: copper sulfate conducts electricity in the molten state but not in the solid state. Reason: the ions are fixed in the lattice in the molten state, while in the solid state they move freely. Choose the option.
(iii) A is true, but R is false. It is true that copper sulfate conducts when molten but not when solid. The reason is reversed: in the solid state ions are fixed in the lattice (can’t conduct), while in the molten state the ions become free to move (so it conducts).
15²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons respectively. Find electrons and neutrons of each.
²⁷Al (neutral): electrons = 13; neutrons = 27 − 13 = 14. ⁸⁰Br⁻ (gained 1 e⁻): electrons = 35 + 1 = 36; neutrons = 80 − 35 = 45. ²⁰¹Hg²⁺ (lost 2 e⁻): electrons = 80 − 2 = 78; neutrons = 201 − 80 = 121.