NCERT Solutions for Class 9 Science Exploration Chapter 9: Atomic Foundations of Matter
Revise, Reflect, Refine
1. A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.
(i) How many electrons does A tend to give or take to become stable?
(ii) What kind of ion would it form?
(iii) How many electrons does B tend to give or take to become stable?
(iv) What kind of ion would it form?
(v) If A and B were to combine, what kind of bond would be formed?
(vi) What would be the formula for the compound thus formed?
(i) How many electrons does A tend to give or take to become stable?
(ii) What kind of ion would it form?
(iii) How many electrons does B tend to give or take to become stable?
(iv) What kind of ion would it form?
(v) If A and B were to combine, what kind of bond would be formed?
(vi) What would be the formula for the compound thus formed?
Answer:
(i) Element A has 1 electron in its third shell. To become stable, it tends to give away 1 electron (since losing 1 is easier than gaining 7 more).
(ii) Since A loses 1 electron, it forms a cation (positive ion) with charge +1, written as A⁺.
(iii) Element B has 6 electrons in its second shell. The second shell can hold 8 electrons, so B needs 2 more electrons to become stable.
(iv) Since B gains 2 electrons, it forms an anion (negative ion) with charge −2, written as B²⁻.
(v) A gives 1 electron and B needs 2 electrons. So 2 atoms of A combine with 1 atom of B. The bond formed is an Ionic Bond (electron transfer from A to B).
(vi) Using criss-cross method: A has valency 1, B has valency 2. Formula = A₂B.
(i) Element A has 1 electron in its third shell. To become stable, it tends to give away 1 electron (since losing 1 is easier than gaining 7 more).
(ii) Since A loses 1 electron, it forms a cation (positive ion) with charge +1, written as A⁺.
(iii) Element B has 6 electrons in its second shell. The second shell can hold 8 electrons, so B needs 2 more electrons to become stable.
(iv) Since B gains 2 electrons, it forms an anion (negative ion) with charge −2, written as B²⁻.
(v) A gives 1 electron and B needs 2 electrons. So 2 atoms of A combine with 1 atom of B. The bond formed is an Ionic Bond (electron transfer from A to B).
(vi) Using criss-cross method: A has valency 1, B has valency 2. Formula = A₂B.
2. An element X has six electrons in its outer shell and forms a diatomic molecule.
(i) Why would that be so?
(ii) What kind of bond would it form?
(iii) Draw the structure of the molecule it would form.
(iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
(i) Why would that be so?
(ii) What kind of bond would it form?
(iii) Draw the structure of the molecule it would form.
(iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Answer:
(i) X has 6 valence electrons and needs 2 more to complete its octet. Since it’s a non-metal, it achieves stability by sharing electrons with another atom of the same element, forming a diatomic molecule (X₂).
(ii) Each atom shares 2 electrons with the other, so they form a Double Covalent Bond (like in O₂).
(iii) Structure of X₂ molecule showing double bond (like oxygen molecule O=O with two shared electron pairs).
(iv) Element Y has 2 electrons in its second shell, meaning its valency is 2 (it needs 6 more to complete the octet — but sharing 2 is simpler). X and Y would form a molecule where Y shares 2 electrons with X — just like in water (H₂O). Formula: YX (e.g., if Y=Mg type metal — ionic; if Y=Be type — covalent). If Y is like oxygen-type element itself: XY with double bond.
(i) X has 6 valence electrons and needs 2 more to complete its octet. Since it’s a non-metal, it achieves stability by sharing electrons with another atom of the same element, forming a diatomic molecule (X₂).
(ii) Each atom shares 2 electrons with the other, so they form a Double Covalent Bond (like in O₂).
(iii) Structure of X₂ molecule showing double bond (like oxygen molecule O=O with two shared electron pairs).
(iv) Element Y has 2 electrons in its second shell, meaning its valency is 2 (it needs 6 more to complete the octet — but sharing 2 is simpler). X and Y would form a molecule where Y shares 2 electrons with X — just like in water (H₂O). Formula: YX (e.g., if Y=Mg type metal — ionic; if Y=Be type — covalent). If Y is like oxygen-type element itself: XY with double bond.
3. You want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6–. Which of the following combinations gives the correct number of ions?
(i) 2 Al³⁺ and 3 Cl⁻
(ii) 3 Mg²⁺ and 1 PO₄³⁻
(iii) 2 Fe³⁺ and 3 O²⁻
(iv) 3 Ca²⁺ and 2 SO₄²⁻
(i) 2 Al³⁺ and 3 Cl⁻
(ii) 3 Mg²⁺ and 1 PO₄³⁻
(iii) 2 Fe³⁺ and 3 O²⁻
(iv) 3 Ca²⁺ and 2 SO₄²⁻
Answer: Checking each option to see if charges balance (total + = total −):
(i) 2 Al³⁺ and 3 Cl⁻ → Positive: 2×3 = 6+, Negative: 3×1 = 3− Incorrect
(ii) 3 Mg²⁺ and 1 PO₄³⁻ → Positive: 3×2 = 6+, Negative: 1×3 = 3− Incorrect
(iii) 2 Fe³⁺ and 3 O²⁻ → Positive: 2×3 = 6+, Negative: 3×2 = 6− Correct
(iv) 3 Ca²⁺ and 2 SO₄²⁻ → Positive: 3×2 = 6+, Negative: 2×2 = 4− Incorrect
(i) 2 Al³⁺ and 3 Cl⁻ → Positive: 2×3 = 6+, Negative: 3×1 = 3− Incorrect
(ii) 3 Mg²⁺ and 1 PO₄³⁻ → Positive: 3×2 = 6+, Negative: 1×3 = 3− Incorrect
(iii) 2 Fe³⁺ and 3 O²⁻ → Positive: 2×3 = 6+, Negative: 3×2 = 6− Correct
(iv) 3 Ca²⁺ and 2 SO₄²⁻ → Positive: 3×2 = 6+, Negative: 2×2 = 4− Incorrect
4. Choose the correct statement(s) and correct the false statement(s).
(i) Elements are made up of molecules and compounds are made up of atoms.
(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
(iii) One molecule of nitrogen gas contains three nitrogen atoms.
(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
(i) Elements are made up of molecules and compounds are made up of atoms.
(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
(iii) One molecule of nitrogen gas contains three nitrogen atoms.
(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Answer:
(i) False. Correction: Elements are made up of atoms and compounds are made up of molecules (which contain atoms of different elements).
(ii) False. Correction: A molecule of a compound is always made up of two or more atoms of different kinds of elements, not the same kind.
(iii) False. Correction: One molecule of nitrogen gas (N₂) contains two nitrogen atoms, not three.
(iv) True. Water (H₂O) is made of two hydrogen atoms covalently bonded with one oxygen atom. This statement is correct.
(i) False. Correction: Elements are made up of atoms and compounds are made up of molecules (which contain atoms of different elements).
(ii) False. Correction: A molecule of a compound is always made up of two or more atoms of different kinds of elements, not the same kind.
(iii) False. Correction: One molecule of nitrogen gas (N₂) contains two nitrogen atoms, not three.
(iv) True. Water (H₂O) is made of two hydrogen atoms covalently bonded with one oxygen atom. This statement is correct.
5. Write the chemical formulae for the following compounds.
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
Answer:
(i) Aluminium nitrate – Al(NO₃)₃
(ii) Calcium oxide – CaO
(iii) Ferric oxide – Fe₂O₃
(i) Aluminium nitrate – Al(NO₃)₃
(ii) Calcium oxide – CaO
(iii) Ferric oxide – Fe₂O₃
6. Write the formulae of the compounds formed from the following pairs of ions.
(i) Ca²⁺ and Br⁻
(ii) Al³⁺ and CO₃²⁻
(iii) K⁺ and SO₄²⁻
(iv) NH₄⁺ and Cl⁻
(i) Ca²⁺ and Br⁻
(ii) Al³⁺ and CO₃²⁻
(iii) K⁺ and SO₄²⁻
(iv) NH₄⁺ and Cl⁻
Answer:
(i) Ca²⁺ and Br⁻ : CaBr₂ (Calcium bromide)
(ii) Al³⁺ and CO₃²⁻ : Al₂(CO₃)₃ (Aluminium carbonate)
(iii) K⁺ and SO₄²⁻ : K₂SO₄ (Potassium sulfate)
(iv) NH₄⁺ and Cl⁻ : NH₄Cl (Ammonium chloride).
(i) Ca²⁺ and Br⁻ : CaBr₂ (Calcium bromide)
(ii) Al³⁺ and CO₃²⁻ : Al₂(CO₃)₃ (Aluminium carbonate)
(iii) K⁺ and SO₄²⁻ : K₂SO₄ (Potassium sulfate)
(iv) NH₄⁺ and Cl⁻ : NH₄Cl (Ammonium chloride).
7. Which of the following, in Fig. 9.18, correctly represents Cl⁻ ion (Atomic number of chlorine = 17).
Answer: Chlorine has atomic number 17. Its electronic configuration is 2, 8, 7 (shells K, L, M). When chlorine gains 1 electron to form Cl⁻ ion, the M shell becomes complete with 8 electrons. Configuration of Cl⁻ = 2, 8, 8 (total 18 electrons). So diagram (iii) correctly represents the Cl⁻ ion.
8. Determine the formula unit mass of the following substances.
(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.
(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.
Answer:
(i) Ammonium nitrate — NH₄NO₃: N = 14 u, H = 1 u, O = 16 u. Mass = (14×1) + (1×4) + (14×1) + (16×3) = 14 + 4 + 14 + 48 = 80 u.
(ii) Phosphoric acid — H₃PO₄: H = 1 u, P = 31 u, O = 16 u. Mass = (1×3) + (31×1) + (16×4) = 3 + 31 + 64 = 98 u.
(iii) Sodium hydrogencarbonate — NaHCO₃: Na = 23 u, H = 1 u, C = 12 u, O = 16 u. Mass = 23 + 1 + 12 + (16×3) = 23 + 1 + 12 + 48 = 84 u.
(i) Ammonium nitrate — NH₄NO₃: N = 14 u, H = 1 u, O = 16 u. Mass = (14×1) + (1×4) + (14×1) + (16×3) = 14 + 4 + 14 + 48 = 80 u.
(ii) Phosphoric acid — H₃PO₄: H = 1 u, P = 31 u, O = 16 u. Mass = (1×3) + (31×1) + (16×4) = 3 + 31 + 64 = 98 u.
(iii) Sodium hydrogencarbonate — NaHCO₃: Na = 23 u, H = 1 u, C = 12 u, O = 16 u. Mass = 23 + 1 + 12 + (16×3) = 23 + 1 + 12 + 48 = 84 u.
9. Write the formulae for the compounds formed by the reaction of:
(i) Magnesium and nitrogen
(ii) Lithium and nitrogen
(iii) Sodium and sulfur
(iv) Aluminium and oxygen
(i) Magnesium and nitrogen
(ii) Lithium and nitrogen
(iii) Sodium and sulfur
(iv) Aluminium and oxygen
Answer:
(i) Magnesium (Mg²⁺) and nitrogen (N³⁻): Criss-cross → Mg gets 3, N gets 2. Formula: Mg₃N₂ (Magnesium nitride).
(ii) Lithium (Li⁺) and nitrogen (N³⁻): Criss-cross → Li gets 3, N gets 1. Formula: Li₃N (Lithium nitride).
(iii) Sodium (Na⁺) and sulfur (S²⁻): Criss-cross → Na gets 2, S gets 1. Formula: Na₂S (Sodium sulfide).
(iv) Aluminium (Al³⁺) and oxygen (O²⁻): Criss-cross → Al gets 2, O gets 3. Formula: Al₂O₃ (Aluminium oxide).
(i) Magnesium (Mg²⁺) and nitrogen (N³⁻): Criss-cross → Mg gets 3, N gets 2. Formula: Mg₃N₂ (Magnesium nitride).
(ii) Lithium (Li⁺) and nitrogen (N³⁻): Criss-cross → Li gets 3, N gets 1. Formula: Li₃N (Lithium nitride).
(iii) Sodium (Na⁺) and sulfur (S²⁻): Criss-cross → Na gets 2, S gets 1. Formula: Na₂S (Sodium sulfide).
(iv) Aluminium (Al³⁺) and oxygen (O²⁻): Criss-cross → Al gets 2, O gets 3. Formula: Al₂O₃ (Aluminium oxide).
10. Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO₃ is given as an example.
Answer: (Table completion expected based on criss-cross rules for the given ions).
11. 5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.
Answer:
Total mass of reactants = 5.3 + 6.0 = 11.3 g.
Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g.
Mass of reactants = Mass of products = 11.3 g.
Therefore, the Law of Conservation of Mass is verified. Matter is neither created nor destroyed in this chemical reaction.
Total mass of reactants = 5.3 + 6.0 = 11.3 g.
Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g.
Mass of reactants = Mass of products = 11.3 g.
Therefore, the Law of Conservation of Mass is verified. Matter is neither created nor destroyed in this chemical reaction.
12. If a species has 11 protons, 12 neutrons and 10 electrons then
(i) what is its atomic number and mass number?
(ii) is it neutral, a cation or an anion? Explain.
(iii) write its electronic configuration.
(iv) name the species.
(i) what is its atomic number and mass number?
(ii) is it neutral, a cation or an anion? Explain.
(iii) write its electronic configuration.
(iv) name the species.
Answer:
(i) Atomic number = number of protons = 11. Mass number = protons + neutrons = 11 + 12 = 23.
(ii) Protons = 11, Electrons = 10. Since protons > electrons, the species has a net positive charge. Charge = 11 − 10 = +1. So it is a Cation with charge +1.
(iii) Total electrons = 10. Electronic configuration: 2, 8 (K shell = 2, L shell = 8).
(iv) Atomic number 11 = Sodium (Na). This species with charge +1 is Sodium ion — Na⁺.
(i) Atomic number = number of protons = 11. Mass number = protons + neutrons = 11 + 12 = 23.
(ii) Protons = 11, Electrons = 10. Since protons > electrons, the species has a net positive charge. Charge = 11 − 10 = +1. So it is a Cation with charge +1.
(iii) Total electrons = 10. Electronic configuration: 2, 8 (K shell = 2, L shell = 8).
(iv) Atomic number 11 = Sodium (Na). This species with charge +1 is Sodium ion — Na⁺.
13. Two elements, A and B, have the following configurations – A: 2, 8, 5 B: 2, 8, 7
(i) Which element is more reactive?
(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.
(iii) Predict the formula of the compound they would form.
(i) Which element is more reactive?
(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.
(iii) Predict the formula of the compound they would form.
Answer:
(i) A has 5 valence electrons (needs 3 more). B has 7 valence electrons (needs only 1 more). B needs fewer electrons and has a stronger tendency to attract electrons. B is more reactive (like chlorine, which is highly reactive).
(ii) A has 5 valence electrons and B has 7. Neither has a very low number of valence electrons (less than 4) to simply donate. However, B needs only 1 electron and A can share electrons. If A shares electrons with B: A needs 3 electrons and B needs 1. So A can share one electron with B. But this creates an imbalance — B with 7+1 = 8 (stable), but A still needs 2 more. Therefore A would share electrons with 3 atoms of B. Type of bond formed: Covalent Bond (sharing of electrons between non-metals).
(iii) A needs 3 electrons (valency = 3), B needs 1 electron (valency = 1). Using criss-cross rule: AB₃ (e.g., similar to NCl₃ — Nitrogen trichloride).
(i) A has 5 valence electrons (needs 3 more). B has 7 valence electrons (needs only 1 more). B needs fewer electrons and has a stronger tendency to attract electrons. B is more reactive (like chlorine, which is highly reactive).
(ii) A has 5 valence electrons and B has 7. Neither has a very low number of valence electrons (less than 4) to simply donate. However, B needs only 1 electron and A can share electrons. If A shares electrons with B: A needs 3 electrons and B needs 1. So A can share one electron with B. But this creates an imbalance — B with 7+1 = 8 (stable), but A still needs 2 more. Therefore A would share electrons with 3 atoms of B. Type of bond formed: Covalent Bond (sharing of electrons between non-metals).
(iii) A needs 3 electrons (valency = 3), B needs 1 electron (valency = 1). Using criss-cross rule: AB₃ (e.g., similar to NCl₃ — Nitrogen trichloride).
14. Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state. Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer: (iii) A is true, but R is false.
15. The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Answer:
²⁷Al (neutral) : Mass No. = 27, Protons = 13, Electrons = 13, Neutrons = 14.
⁸⁰Br⁻ (gained 1e⁻) : Mass No. = 80, Protons = 35, Electrons = 36, Neutrons = 45.
²⁰¹Hg²⁺ (lost 2e⁻) : Mass No. = 201, Protons = 80, Electrons = 78, Neutrons = 121.
²⁷Al (neutral) : Mass No. = 27, Protons = 13, Electrons = 13, Neutrons = 14.
⁸⁰Br⁻ (gained 1e⁻) : Mass No. = 80, Protons = 35, Electrons = 36, Neutrons = 45.
²⁰¹Hg²⁺ (lost 2e⁻) : Mass No. = 201, Protons = 80, Electrons = 78, Neutrons = 121.
Very Short Answer Type Questions
1. State the Law of Conservation of Mass. Who proposed it and when?
Answer: Matter can neither be created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products. Proposed by Antoine Lavoisier in 1789, known as the Father of Modern Chemistry.
2. State the Law of Constant Proportions.
Answer: In any compound formed by two or more elements, the elements always combine in a fixed ratio by mass, irrespective of the source of the compound or the method of its preparation. Also called Proust’s Law.
3. Give two postulates of Dalton’s Atomic Theory.
Answer: ► All matter is made up of tiny particles called atoms, which participate in chemical reactions. ► Atoms are indivisible and cannot be created or destroyed in a chemical reaction. ► Dalton proposed six postulates in 1808 that explained the two fundamental laws of chemistry.
4. Define a molecule.
Answer: A molecule is an electrically neutral entity consisting of more than one atom that is capable of independent existence and shows all the properties of that substance. Example: H₂ (hydrogen molecule), H₂O (water molecule).
5. What is a covalent bond? Give one example.
Answer: A covalent bond is formed by the sharing of one or more pairs of electrons between atoms to achieve stable electronic configurations. Example: Two hydrogen atoms share one electron each to form H₂ with a single covalent bond H—H.
6. What is an ionic bond? Give one example.
Answer: An ionic bond is the electrostatic force of attraction between oppositely charged ions (cation and anion) formed by the transfer of electrons. Example: Na loses one electron to Cl, forming Na⁺ and Cl⁻, held together by ionic bond in NaCl.
7. Distinguish between a cation and an anion.
Answer: A cation is a positively charged ion formed when an atom loses one or more electrons. An anion is a negatively charged ion formed when an atom gains one or more electrons. Example: Na⁺ is a cation; Cl⁻ is an anion. Together they are called ions.
8. Why does ionic compound sodium chloride not conduct electricity in the solid state?
Answer: In solid NaCl, the Na⁺ and Cl⁻ ions are held rigidly in fixed positions within the crystal lattice by strong electrostatic forces. Since ions cannot move freely in solid state, the compound cannot conduct electricity. Conductivity occurs only when dissolved in water.
9. What is the difference between molecular mass and formula unit mass?
Answer: Molecular mass is the sum of atomic masses of all atoms in one molecule — applies to covalent compounds. Formula unit mass is the sum of atomic masses of all atoms in a formula unit — applies to ionic compounds, which form crystal lattices and not discrete molecules.
10. Name the covalent compound with formula SF₆ using IUPAC prefix system.
Answer: SF₆ is named sulfur hexafluoride. ‘Hexa’ indicates 6 fluorine atoms. The second element (fluorine) ends in -ide, becoming fluoride, which with the prefix ‘hexa’ gives hexafluoride. The first element (sulfur) retains its name without a prefix.
11. Write the chemical formula of calcium carbonate using the criss-cross method.
Answer: Calcium ion = Ca²⁺ (charge 2); Carbonate ion = CO₃²⁻ (charge 2). Criss-crossing equal charges (2 and 2) gives Ca₂(CO₃)₂. Dividing by common factor 2 gives the simplest formula: CaCO₃. The charge on each ion is not shown in the formula.
12. Why does sugar dissolve in water but not conduct electricity?
Answer: Sugar is a covalent compound. When dissolved in water, it does not break into ions — it remains as sugar molecules in solution. Since electrical conductivity requires free-moving charged particles (ions), and no ions are produced, sugar solution does not conduct electricity.
13. What type of bond is present in an oxygen molecule (O₂)? How many pairs of electrons are shared?
Answer: Oxygen molecule (O₂) has a double covalent bond. Oxygen has 6 valence electrons and needs 2 more. Two oxygen atoms each share 2 electrons, resulting in two shared pairs of electrons. It is represented as O=O (double bond with two lines).
14. Write the formula of magnesium hydroxide. Why are brackets used?
Answer: Magnesium hydroxide formula is Mg(OH)₂. Criss-crossing Mg²⁺ (charge 2) and OH⁻ (charge 1) gives Mg(OH)₂ — two hydroxide ions per magnesium ion. Brackets are used around the polyatomic ion (OH) to show that the subscript 2 applies to the entire OH group, not just O.
15. Calculate the molecular mass of water (H₂O).
Answer: Atomic masses: H = 1 u; O = 16 u. Molecular mass of H₂O = (1 u × 2) + (16 u × 1) = 2 + 16 = 18 u. Water has 2 hydrogen atoms and 1 oxygen atom. The molecular mass (18 u) applies to covalent compounds whose atoms form discrete molecules.
Short Answer Type Questions
1. Explain why the apparent mass decreases in Experimental Set-up 1 of Activity 9.2 but not in Set-up 2. Does this violate the Law of Conservation of Mass?
Answer: In Set-up 1, the balloon is separated from the flask — when baking soda reacts with vinegar, CO₂ gas is produced and escapes into the surrounding air. Since the escaped gas is not weighed, the final reading appears less than the initial reading. In Set-up 2, the balloon is attached to the flask’s mouth. When baking soda falls into the vinegar, the CO₂ produced inflates the balloon and is trapped within the system. The total mass (flask + balloon + all products) remains equal to the initial mass. This does NOT violate the Law of Conservation of Mass. In both cases, the actual total mass is conserved. Set-up 1 simply fails to account for the escaped gas. The law always holds in a closed system where all products are accounted for.
2. Why does the Law of Constant Proportions apply to compounds but not to mixtures?
Answer: In a compound, atoms of different elements are chemically bonded in a specific, fixed ratio determined by their valencies and atomic structure. This ratio is constant regardless of the source or method of preparation. For example, water always contains H and O in a 1:8 mass ratio. In a mixture, substances are physically combined without chemical bonding. They can be mixed in any proportion — air can contain varying percentages of nitrogen and oxygen, and salt can be dissolved in water at different concentrations. No fixed ratio is required for physical mixing. Therefore, the Law of Constant Proportions is a property of chemical combination (compounds) and not of physical mixing (mixtures).
3. Explain the formation of a water molecule (H₂O) with the help of electron sharing. What type of bond is formed?
Answer: Oxygen (Z=8) has electronic configuration 2, 6 — it has 6 valence electrons and needs 2 more to complete its octet. Hydrogen (Z=1) has 1 electron in K-shell and needs 1 more to complete its duplet. To satisfy both needs, two hydrogen atoms each share one electron with the oxygen atom. Oxygen shares one electron each with the two hydrogen atoms. This results in two single covalent bonds forming the water molecule H₂O. The structure is H—O—H, where each dash represents one shared pair of electrons (single covalent bond). The molecule is electrically neutral and stable. Since atoms achieve stability through sharing, water is a covalent compound.
4. Explain the formation of sodium chloride (NaCl) through ionic bonding. Why is NaCl electrically neutral overall?
Answer: Sodium (Z=11) has electronic configuration 2, 8, 1 — one valence electron that it readily loses to achieve a stable octet in the L-shell. When sodium loses this electron, it becomes Na⁺ cation (11 protons, 10 electrons, net charge +1). Chlorine (Z=17) has configuration 2, 8, 7 — seven valence electrons, needing one more to complete its octet. Chlorine gains the electron released by sodium, becoming Cl⁻ anion (17 protons, 18 electrons, net charge −1). Na⁺ and Cl⁻ are attracted to each other by strong electrostatic force — this is the ionic bond. They form NaCl (sodium chloride). NaCl is electrically neutral overall because the +1 charge of Na⁺ exactly cancels the −1 charge of Cl⁻.
5. Compare the properties of ionic and covalent compounds under three headings: solubility, electrical conductivity, and melting/boiling points.
Answer: Solubility: Ionic compounds (NaCl, CuSO₄) are generally soluble in polar solvents like water but insoluble in non-polar organic solvents like kerosene and petrol. Covalent compounds (camphor, naphthalene) are generally insoluble in water but dissolve in non-polar organic solvents. Electrical Conductivity: Ionic compounds do not conduct electricity in solid state (ions fixed in lattice) but conduct when dissolved in water (ions free to move). Covalent compounds do not conduct electricity even in solution because they do not produce ions. Exception: sugar dissolves in water but still does not conduct. Melting and Boiling Points: Ionic compounds have high melting and boiling points due to strong electrostatic forces of attraction between oppositely charged ions. Covalent compounds generally have low melting and boiling points because the intermolecular forces are much weaker than ionic bonds.
6. Write the chemical formulae of the following ionic compounds using criss-cross method: (a) Aluminium oxide, (b) Calcium carbonate, (c) Magnesium hydroxide.
Answer: (a) Aluminium oxide: Al³⁺ (charge 3) and O²⁻ (charge 2). Criss-cross 3 and 2 → Al₂O₃. (b) Calcium carbonate: Ca²⁺ (charge 2) and CO₃²⁻ (charge 2). Same charges criss-cross → Ca₂(CO₃)₂ → divide by 2 → CaCO₃. (c) Magnesium hydroxide: Mg²⁺ (charge 2) and OH⁻ (charge 1). Criss-cross → Mg(OH)₂. Brackets are essential because subscript 2 applies to the entire polyatomic OH group. Formula: Mg(OH)₂. Note: In all formulae, charges on ions are not written in the final formula.
7. How does Dalton’s Atomic Theory explain both the Law of Conservation of Mass and the Law of Constant Proportions?
Answer: Dalton proposed that atoms are indivisible particles that cannot be created or destroyed in a chemical reaction — they merely rearrange. This directly explains the Law of Conservation of Mass: since no atoms are gained or lost, the total mass of all atoms before the reaction equals the total mass after the reaction. For the Law of Constant Proportions, Dalton proposed that atoms of a given element are identical in mass and that compounds are formed when atoms combine in fixed whole number ratios. Since atoms always combine in the same fixed ratio, the mass ratio of elements in a compound is always constant, regardless of the source or method of preparation.
8. Name the following covalent compounds using IUPAC prefix rules: (a) CO, (b) CO₂, (c) N₂O₄, (d) PCl₃, (e) SF₆.
Answer: (a) CO: Carbon monoxide. ‘Mono’ indicates 1 oxygen atom. Mono- is used for the second element even when omitted for the first. (b) CO₂: Carbon dioxide. ‘Di’ indicates 2 oxygen atoms. (c) N₂O₄: Dinitrogen tetroxide. ‘Di’ for 2 nitrogen; ‘tetra’ for 4 oxygen, but the final ‘a’ of ‘tetra’ is dropped before ‘oxide’ → tetroxide not tetraoxide. (d) PCl₃: Phosphorus trichloride. ‘Tri’ indicates 3 chlorine atoms. (e) SF₆: Sulfur hexafluoride. ‘Hexa’ indicates 6 fluorine atoms.
9. Calculate the molecular mass of nitric acid (HNO₃) and the formula unit mass of calcium nitrate Ca(NO₃)₂.
Answer: Atomic masses: H = 1 u, N = 14 u, O = 16 u, Ca = 40 u. Molecular mass of nitric acid (HNO₃): = (1×1) + (14×1) + (16×3) = 1 + 14 + 48 = 63 u. Formula unit mass of calcium nitrate Ca(NO₃)₂: = (40×1) + {(14×1) + (16×3)} × 2 = 40 + (14 + 48) × 2 = 40 + 62 × 2 = 40 + 124 = 164 u. Note: Molecular mass is used for covalent compounds (discrete molecules). Formula unit mass is used for ionic compounds (crystal lattice — no discrete molecules exist).
10. 4.0 g of calcium carbonate reacts with 2.92 g of hydrochloric acid in a closed container to produce 1.76 g of CO₂, 0.72 g of water, and 4.44 g of calcium chloride. Verify the Law of Conservation of Mass.
Answer: Total mass of reactants: = Mass of CaCO₃ + Mass of HCl = 4.0 g + 2.92 g = 6.92 g. Total mass of products: = Mass of CO₂ + Mass of H₂O + Mass of CaCl₂ = 1.76 g + 0.72 g + 4.44 g = 6.92 g. Since mass of reactants (6.92 g) = mass of products (6.92 g), the Law of Conservation of Mass is verified and obeyed in this chemical reaction. This demonstration is valid because the reaction was carried out in a closed container — all gaseous products (CO₂) were retained and weighed.
Long Answer Type Questions
1. State the Law of Conservation of Mass. Describe Activity 9.2 to demonstrate it. Why is a closed system essential?
Answer: Law of Conservation of Mass: Matter can neither be created nor destroyed in a chemical reaction. Mass of reactants = Mass of products. Activity 9.2 uses vinegar and baking soda. In Set-up 1 (open system), baking soda is poured directly into vinegar in an open flask. CO₂ gas escapes into the air. The final mass is less than initial — apparently violating the law. In Set-up 2 (closed system), the balloon filled with baking soda is attached to the flask. When baking soda mixes with vinegar, CO₂ inflates the balloon but stays within the system. Final mass exactly equals initial mass. A closed system is essential because if any product (gas) escapes, it is not weighed — giving an incorrect impression that mass was lost. The law holds universally only when all reactants and products are accounted for.
2. State Dalton’s Atomic Theory. How does it explain the Law of Conservation of Mass and the Law of Constant Proportions? What are its limitations?
Answer: Dalton’s Atomic Theory (1808) postulates: (i) All matter is made of tiny atoms; (ii) atoms are indivisible and cannot be created or destroyed; (iii) atoms of a given element are identical; (iv) atoms of different elements differ in mass; (v) atoms combine in whole number ratios to form compounds; (vi) the relative number and kinds of atoms in a compound are constant. Law of Conservation of Mass is explained because atoms only rearrange — no atom is created or destroyed, so total mass is unchanged. Law of Constant Proportions is explained because atoms always combine in the same fixed whole number ratios — giving a constant mass ratio in any compound. Limitations: Atoms are actually divisible (contain electrons, protons, neutrons). Atoms of the same element can have different masses (isotopes). These facts were discovered after Dalton’s time, requiring modifications to his theory.
3. Explain the formation of a hydrogen chloride (HCl) molecule. Compare single and double covalent bonds with examples. How are covalent compounds named?
Answer: Hydrogen (Z=1) has 1 electron in K-shell and needs 1 more for a stable duplet. Chlorine (Z=17) has configuration 2, 8, 7 — with 7 valence electrons needing 1 more for a stable octet. Since both atoms need exactly 1 electron, each shares 1 electron with the other. One shared pair of electrons forms the molecule H—Cl (single covalent bond). HCl is a covalent compound. Single bond: One shared pair of electrons. Examples: H₂ (H—H), Cl₂ (Cl—Cl), HCl (H—Cl). Double bond: Two shared pairs of electrons. Example: O₂ (O=O) — oxygen has 6 valence electrons and needs 2 more, so two O atoms share 2 electrons each. Naming covalent compounds: Use IUPAC prefix system — mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6). First element retains its name; second element ends in -ide. Mono is omitted for the first element. Example: CO = carbon monoxide; CO₂ = carbon dioxide; PCl₃ = phosphorus trichloride.
4. Explain the formation of an ionic bond using the example of NaCl. Describe the crystal structure of NaCl. Why does NaCl conduct electricity in solution but not in solid state?
Answer: Sodium (Z=11, configuration 2, 8, 1) has 1 valence electron. It readily loses this electron to become Na⁺ (11 protons, 10 electrons, charge +1), achieving the stable neon configuration 2, 8. Chlorine (Z=17, configuration 2, 8, 7) has 7 valence electrons. It accepts sodium’s electron to become Cl⁻ (17 protons, 18 electrons, charge −1), achieving the stable argon configuration 2, 8, 8. Na⁺ and Cl⁻ are held together by electrostatic attraction — this is the ionic bond. NaCl is electrically neutral (total charge = 0). Crystal structure: Ionic compounds don’t form single molecules. In solid NaCl, each Na⁺ is surrounded by 6 Cl⁻ ions, and each Cl⁻ is surrounded by 6 Na⁺ ions, forming a regular 3-D crystal lattice pattern. Conductivity: In solid state, all ions are rigidly fixed in the lattice by strong electrostatic forces — they cannot move, so no conductivity. When dissolved in water, the crystal lattice breaks apart and Na⁺ and Cl⁻ ions become free to move independently through the solution. These freely moving charged ions carry electric current, enabling conductivity.
5. Explain the criss-cross method for writing chemical formulae. Write the formulae for (a) magnesium hydroxide, (b) aluminium sulfate, (c) ferric chloride. Calculate the formula unit mass of magnesium hydroxide.
Answer: Criss-cross method for ionic compounds: ► Write cation symbol first, then anion ► Write charges (numbers only) below each symbol ► Swap the charge numbers as subscripts ► Reduce subscripts by common factor if needed ► Use brackets around polyatomic ions when subscript > 1.
(a) Magnesium hydroxide Mg(OH)₂: Mg²⁺ and OH⁻ → criss-cross charges (2 and 1) → Mg(OH)₂. Brackets are essential because subscript 2 applies to the entire OH group.
(b) Aluminium sulfate Al₂(SO₄)₃: Al³⁺ and SO₄²⁻ → criss-cross (3 and 2) → Al₂(SO₄)₃.
(c) Ferric chloride FeCl₃: Fe³⁺ and Cl⁻ → criss-cross (3 and 1) → FeCl₃.
Formula Unit Mass of Mg(OH)₂: Atomic masses: Mg = 24 u; O = 16 u; H = 1 u = (24 × 1) + {(16 × 1) + (1 × 1)} × 2 = 24 + (17 × 2) = 24 + 34 = 58 u.
(a) Magnesium hydroxide Mg(OH)₂: Mg²⁺ and OH⁻ → criss-cross charges (2 and 1) → Mg(OH)₂. Brackets are essential because subscript 2 applies to the entire OH group.
(b) Aluminium sulfate Al₂(SO₄)₃: Al³⁺ and SO₄²⁻ → criss-cross (3 and 2) → Al₂(SO₄)₃.
(c) Ferric chloride FeCl₃: Fe³⁺ and Cl⁻ → criss-cross (3 and 1) → FeCl₃.
Formula Unit Mass of Mg(OH)₂: Atomic masses: Mg = 24 u; O = 16 u; H = 1 u = (24 × 1) + {(16 × 1) + (1 × 1)} × 2 = 24 + (17 × 2) = 24 + 34 = 58 u.