Atomic Foundations of Matter — Complete Q&A
Think It Over · Pause and Ponder · Revise, Reflect, Refine
Think It Over
Q1. Water can be obtained from various sources. Are all these samples of water chemically identical?
Answer:
Yes, all pure samples of water are chemically identical. Water from any source — river, borewell, or ocean — when purified, always contains hydrogen and oxygen in the fixed mass ratio of 1:8. This is explained by the Law of Constant Proportions.
Q2. Oxygen is sometimes represented as O and sometimes as O₂. What is the difference between these symbols?
Answer:
O represents a single oxygen atom, while O₂ represents a molecule of oxygen gas consisting of two oxygen atoms bonded together. Oxygen atoms are unstable alone and exist as diatomic molecules (O₂) in nature.
Q3. Why does dissolved salt in water conduct electricity, but sugar does not?
Answer:
Salt (NaCl) dissociates into Na⁺ and Cl⁻ ions when dissolved in water, which are free to move and conduct electricity. Sugar dissolves in water but does not dissociate into ions — it remains as neutral molecules, which cannot conduct electricity.
Pause and Ponder
Q1. A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.
Answer:
No, the Law of Conservation of Mass is not violated. When ethanol burns, it reacts with oxygen to form carbon dioxide and water vapour, which escape into the air. If the escaping gases were collected and weighed along with the products, the total mass would equal the initial mass of ethanol and oxygen. In an open beaker, the mass appears to decrease because gases escape.
Q2. If atoms could combine in any ratio and not in a fixed ratio, how would this affect the substances around us?
Answer:
If atoms could combine in any ratio, substances would not have fixed properties. The composition of a compound would vary from sample to sample. For example, water could sometimes be 1 part hydrogen and 8 parts oxygen, sometimes 2 parts hydrogen and 8 parts oxygen — leading to different properties. This would make chemistry unpredictable and substances inconsistent.
Q3. A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?
Answer:
If a compound contains 40% sulfur and 60% oxygen by mass, the mass ratio of sulfur to oxygen is 40:60 = 2:3.
For 20 g of sulfur:
Mass of oxygen = (20 g × 60) ÷ 40 = 30 g
Therefore, 30 g of oxygen must be present.
Q4. Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?
Answer:
Mass ratio of carbon to oxygen in CO = 3:4
For 9 g of carbon:
Mass of oxygen = (9 g × 4) ÷ 3 = 12 g
Therefore, 12 g of oxygen will combine with 9 g of carbon to form CO.
Q5. The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.
Answer:
A compound is formed when elements chemically combine in a fixed ratio by mass. This ratio is constant for that compound regardless of its source. A mixture, on the other hand, is formed by physically combining substances in any proportion. The components of a mixture retain their individual properties and can be present in varying ratios. Hence, the Law of Definite Proportions applies only to compounds, not mixtures.
Q6. Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4:1 and 8:2, respectively. Do their results justify the Law of Constant Proportions? Explain.
Answer:
Yes, the results justify the Law of Constant Proportions. The ratio 4:1 is equivalent to 8:2 (both simplify to 4:1). Since both students obtained the same ratio of copper to oxygen, they produced the same compound — copper oxide. This shows that a compound always contains elements in a fixed ratio by mass.
Q7. Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. Reason (R): According to Dalton's Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds. 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:
Option (ii) Both A and R are true, but R is not the correct explanation of A.
While it is true that atoms combine in simple whole number ratios, Dalton's Atomic Theory states that atoms combine in simple whole number ratios by number, not by mass. The mass ratio comes from the atomic masses of the elements.
Q8. Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).
Answer:
Each nitrogen atom has 5 valence electrons. It needs 3 more electrons to complete its octet. Two nitrogen atoms share 3 pairs of electrons forming a triple bond (N≡N).
Q9. The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).
Answer:
Fluorine (atomic number 9) has the electronic configuration 2,7. Each fluorine atom has 7 valence electrons and needs 1 more electron to complete its octet. Two fluorine atoms share one pair of electrons, forming a single covalent bond (F-F).
Q10. Show the formation of the following molecules: (i) Carbon dioxide (CO₂) (ii) Hydrogen sulfide (H₂S) (iii) Ammonia (NH₃)
Answer:
(i) Carbon dioxide (CO₂): Carbon (2,4) shares its 4 valence electrons with 2 oxygen atoms (each 2,6). Each oxygen shares 2 electrons with carbon, forming two double bonds (O=C=O).
(ii) Hydrogen sulfide (H₂S): Sulfur (2,8,6) needs 2 electrons. Two hydrogen atoms each share one electron with sulfur, forming two single bonds (H-S-H).
(iii) Ammonia (NH₃): Nitrogen (2,5) needs 3 electrons. Three hydrogen atoms each share one electron with nitrogen, forming three single bonds.
Q11. Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Answer:
Neon has the electronic configuration 2,8. Its valence shell is completely filled with 8 electrons (octet). Since it already has a stable electronic configuration, it neither gains, loses, nor shares electrons. Neon exists as a monoatomic gas and is chemically inert.
Q12. What kind of ion will oxygen (O) form?
Answer:
Oxygen has the electronic configuration 2,6. It has 6 valence electrons and needs 2 more electrons to complete its octet. Therefore, oxygen will gain 2 electrons and form an oxide ion with a 2- charge (O²⁻).
Q13. Fill in the blanks. Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become ______. Now, ______ ion of magnesium and ______ ions of chlorine combine to give magnesium chloride.
Answer:
Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become Cl⁻. Now, one ion of magnesium and two ions of chlorine combine to give magnesium chloride (MgCl₂).
Q14. Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.
Answer:
Potassium chloride (KCl): K (2,8,8,1) loses 1 electron to form K⁺. Cl (2,8,7) gains 1 electron to form Cl⁻. They combine to form KCl.
Calcium chloride (CaCl₂): Ca (2,8,8,2) loses 2 electrons to form Ca²⁺. Two Cl atoms each gain 1 electron to form 2Cl⁻. They combine to form CaCl₂.
Q15. Illustrate how sodium sulfide (Na₂S) is formed.
Answer:
Na (2,8,1) loses 1 electron to form Na⁺. S (2,8,6) gains 2 electrons to form S²⁻. Two Na atoms transfer one electron each to sulfur. The ions Na⁺ and S²⁻ combine in the ratio 2:1 to form Na₂S.
Q16. Name the following: (i) CO₂ (ii) NO₂ (iii) SF₆ (iv) PCl₃
Answer:
(i) CO₂ — Carbon dioxide
(ii) NO₂ — Nitrogen dioxide
(iii) SF₆ — Sulfur hexafluoride
(iv) PCl₃ — Phosphorus trichloride
Q17. Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide
Answer:
(i) Sodium hydrogencarbonate — NaHCO₃
(ii) Sulfur dioxide — SO₂
(iii) Ferric chloride — FeCl₃
(iv) Cuprous oxide — Cu₂O
Q18. Write the formulae for the compounds formed from the following pairs of ions: (i) Fe³⁺ and OH⁻ (ii) K⁺ and CO₃²⁻
Answer:
(i) Fe³⁺ and OH⁻ → Fe(OH)₃ (Iron(III) hydroxide)
(ii) K⁺ and CO₃²⁻ → K₂CO₃ (Potassium carbonate)
Q19. What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?
Answer:
Such a compound has an ionic bond. In the solid state, ions are held in fixed positions and cannot move. When dissolved in water, the ions become free to move and conduct electricity.
Q20. Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its: (i) formula (ii) type of bond (iii) electrical conductivity of its aqueous solution.
Answer:
(i) Metal M has 2 valence electrons, so its valency is 2. Oxygen has valency 2. The formula of the compound is MO.
(ii) The bond is ionic because M will lose 2 electrons to form M²⁺ and O will gain 2 electrons to form O²⁻.
(iii) In aqueous solution, the compound will dissociate into M²⁺ and O²⁻ ions, which are free to move. Therefore, it will conduct electricity in its aqueous solution.
Q21. Find the molecular mass of nitric acid (HNO₃). Atomic mass — H = 1 u; N = 14 u; O = 16 u.
Answer:
Molecular mass of HNO₃ = (1 × 1) + (1 × 14) + (3 × 16)
= 1 + 14 + 48
= 63 u
Q22. Find the molecular mass of methane (CH₄). Atomic mass — C = 12 u; H = 1 u.
Answer:
Molecular mass of CH₄ = (1 × 12) + (4 × 1)
= 12 + 4
= 16 u
Q23. Find the formula unit mass of potassium chloride (KCl). Atomic mass — K = 39 u; Cl = 35.5 u.
Answer:
Formula unit mass of KCl = (1 × 39) + (1 × 35.5)
= 74.5 u
Q24. Find the formula unit mass of magnesium hydroxide, Mg(OH)₂. Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.
Answer:
Formula unit mass of Mg(OH)₂ = (1 × 24) + 2 × (1 × 16 + 1 × 1)
= 24 + 2 × (16 + 1)
= 24 + 2 × 17
= 24 + 34
= 58 u
Revise, Reflect, Refine
Q1. 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?
Answer:
(i) A has 1 electron in its third shell (valence shell). It will lose 1 electron to become stable.
(ii) It will form a cation with a +1 charge (A⁺).
(iii) B has 6 electrons in its second shell (valence shell). It needs 2 more electrons to complete its octet. It will gain 2 electrons.
(iv) It will form an anion with a -2 charge (B²⁻).
(v) A and B will form an ionic bond through transfer of electrons.
(vi) Two A atoms (each losing 1 electron) will combine with one B atom (gaining 2 electrons). Formula: A₂B.
Q2. 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.
Answer:
(i) X has 6 valence electrons and needs 2 more to complete its octet. Two X atoms share 2 pairs of electrons each, forming a diatomic molecule.
(ii) It forms a covalent bond (double bond) by sharing electrons.
(iii) X is oxygen (atomic number 8) — O₂ with double bond (O=O).
(iv) Y has 2 electrons in its second shell (2,8,2) — it is magnesium (Mg). X (O) needs 2 electrons, Y (Mg) needs to lose 2 electrons. Mg loses 2 electrons to become Mg²⁺, and O gains 2 electrons to become O²⁻. They form MgO (ionic bond).
Q3. 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) 2Al³⁺ and 3Cl⁻ (ii) 3Mg²⁺ and 1PO₄³⁻ (iii) 2Fe³⁺ and 3O²⁻ (iv) 3Ca²⁺ and 2SO₄²⁻
Answer:
The correct combination is (iii) 2Fe³⁺ and 3O²⁻.
Total positive charge = 2 × (+3) = +6
Total negative charge = 3 × (-2) = -6
The charges balance, so the compound is Fe₂O₃.
Q4. 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.
Answer:
(i) False. Elements are made up of atoms, and compounds are made up of molecules (or ions in ionic compounds).
(ii) False. A molecule of a compound is made up of two or more atoms of different elements (e.g., H₂O, CO₂).
(iii) False. One molecule of nitrogen gas (N₂) contains two nitrogen atoms.
(iv) True. Water (H₂O) is made of two hydrogen atoms covalently bonded with one oxygen atom.
Q5. Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide
Answer:
(i) Aluminium nitrate — Al(NO₃)₃
(ii) Calcium oxide — CaO
(iii) Ferric oxide — Fe₂O₃
Q6. 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⁻
Answer:
(i) Ca²⁺ and Br⁻ → CaBr₂
(ii) Al³⁺ and CO₃²⁻ → Al₂(CO₃)₃
(iii) K⁺ and SO₄²⁻ → K₂SO₄
(iv) NH₄⁺ and Cl⁻ → NH₄Cl
Q7. Which of the following, in Fig. 9.18, correctly represents Cl⁻ ion (Atomic number of chlorine = 17).
Answer:
Chlorine atom (atomic number 17) has electronic configuration 2,8,7. It has 7 valence electrons. When it gains 1 electron, it becomes Cl⁻ with 8 valence electrons. The correct representation will show 8 electrons in the valence shell.
Q8. Determine the formula unit mass of the following substances. (i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser (ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents (iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity
Answer:
(i) NH₄NO₃: N=14, H=1, O=16
= (1×14) + (4×1) + (1×14) + (3×16)
= 14 + 4 + 14 + 48
= 80 u
(ii) H₃PO₄: H=1, P=31, O=16
= (3×1) + (1×31) + (4×16)
= 3 + 31 + 64
= 98 u
(iii) NaHCO₃: Na=23, H=1, C=12, O=16
= 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 u
Q9. 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
Answer:
(i) Magnesium + Nitrogen → Mg₃N₂ (Magnesium nitride)
(ii) Lithium + Nitrogen → Li₃N (Lithium nitride)
(iii) Sodium + Sulfur → Na₂S (Sodium sulfide)
(iv) Aluminium + Oxygen → Al₂O₃ (Aluminium oxide)
Q10. Complete 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:
| 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 |
Q11. 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:
Mass of reactants:
Sodium carbonate = 5.3 g
Acetic acid = 6.0 g
Total reactants = 5.3 + 6.0 = 11.3 g
Mass of products:
Carbon dioxide = 2.2 g
Water = 0.9 g
Sodium acetate = 8.2 g
Total products = 2.2 + 0.9 + 8.2 = 11.3 g
Mass of reactants = Mass of products (11.3 g = 11.3 g)
Hence, the Law of Conservation of Mass is valid.
Q12. 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.
Answer:
(i) Atomic number = Number of protons = 11
Mass number = Protons + Neutrons = 11 + 12 = 23
(ii) It has 11 protons (+11) and 10 electrons (-10). Net charge = +1. Therefore, it is a cation (Na⁺).
(iii) Electronic configuration: 2, 8, 1 (for Na atom). Na⁺ has lost one electron: 2, 8.
(iv) The species is sodium ion (Na⁺).
Q13. Two elements, A and B, have the following configurations — A: 2, 8, 5 and 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.
Answer:
(i) B is more reactive because it has 7 valence electrons and needs only 1 electron to complete its octet.
(ii) A has 5 valence electrons (needs 3). B has 7 valence electrons (needs 1). A and B are both non-metals, so they will form a covalent bond by sharing electrons. A will share 3 electrons with 3 B atoms.
(iii) One A atom shares 3 electrons with 3 B atoms. Formula: AB₃.
Q14. 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.
Answer:
Option (iii) A is true, but R is false.
Copper sulfate conducts electricity in the molten state but not in the solid state. This is because in the solid state, ions are fixed in the lattice and cannot move. In the molten state, the lattice breaks down and ions become free to move. The Reason states the opposite, so it is false.
Q15. The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Answer:
²⁷Al:
Protons = 13, Atomic number = 13
Neutrons = 27 - 13 = 14
Electrons = 13 (neutral atom)
⁸⁰Br⁻:
Protons = 35, Atomic number = 35
Neutrons = 80 - 35 = 45
Electrons = 35 + 1 = 36 (gained 1 electron)
²⁰¹Hg²⁺:
Protons = 80, Atomic number = 80
Neutrons = 201 - 80 = 121
Electrons = 80 - 2 = 78 (lost 2 electrons)