Class 9 Science NCERT Solutions Chapter 9: Atomic Foundations of Matter | Boundless Maths
📗 CBSE 2026-27 Unit IV · Matter and Its Composition ✨ Free — No Sign-up 59 Questions

Chapter 9: Atomic Foundations
of Matter

Complete NCERT Solutions for Chapter 9 of the new Class 9 Science Exploration textbook (CBSE 2026-27) — every Think It Over, Activity, Pause & Ponder, Think as a Scientist, Bridging Science and Society, Revise Reflect Refine, and Journey Beyond question on this one page, with full step-by-step working for every numerical.

Atomic Foundations of Matter puts the laws of chemical combination, atomic and molecular mass calculations, valency, and the difference between ionic and covalent bonding at the centre of the chapter. These NCERT solutions work through every numerical on formula unit mass and mole concepts step by step, alongside the ion-formation diagrams for common cations and anions — exactly the kind of content that shows up repeatedly in Class 9 Science chemistry important questions.

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Overview

What Chapter 9 Is Really About

Atomic Foundations of Matter moves from atoms to the substances they build. It establishes the Law of Conservation of Mass and the Law of Constant Proportions through hands-on activities, uses these to justify Dalton's Atomic Theory, and then explains how atoms combine — by sharing electrons in a covalent bond or transferring them in an ionic bond — to form molecules and ionic compounds. It closes with writing chemical formulae and calculating molecular mass and formula unit mass. Every question is solved here, section by section, exactly as the textbook presents them, with full working for every numerical.

⚖️

Conservation of Mass & Constant Proportions

Verifying with weighing-balance activities that mass is conserved in reactions, and that compounds always combine in fixed ratios.

🔗

Covalent & Ionic Bonding

How atoms share electrons to form molecules like H₂, O₂ and H₂O, or transfer electrons to form ions like Na⁺ and Cl⁻.

🧮

Formulae & Molecular Mass

Criss-cross method for writing chemical formulae, and calculating molecular mass and formula unit mass.

Quick Revision

Key Concepts & Formulae at a Glance

Laws of chemical combination

LawStatement
Law of Conservation of MassMass can neither be created nor destroyed in a chemical reaction — the total mass of reactants equals the total mass of products.
Law of Constant (Definite) ProportionsA pure chemical compound always contains the same elements combined together in the same fixed proportion by mass, no matter how or where it's prepared.

Key formulae

\[ \text{Molecular mass} = \text{Sum of atomic masses of all atoms in the molecule} \] \[ \text{Formula unit mass (for ionic compounds)} = \text{Sum of atomic masses of all ions in the formula unit} \]

Valency is the combining capacity of an element — the number of electrons it needs to lose, gain, or share to achieve a stable (usually 8-electron) outer shell.

Ions: cations vs. anions

PropertyCationAnion
ChargePositiveNegative
Formed byLosing electronsGaining electrons
Typical elementsMetals (e.g., Na⁺, Ca²⁺)Non-metals (e.g., Cl⁻, O²⁻)

An atom is the smallest particle of an element that can take part in a chemical reaction; a molecule is the smallest particle of an element or compound that can normally exist on its own.

Section A

Think It Over (Chapter Opener)

3 Questions
Q1Water can be obtained from various sources. Are all these samples of water chemically identical?

Answer: yes. According to the Law of Constant Proportions (also called the Law of Definite Proportions), a pure compound always contains the same elements in the same fixed ratio by mass, regardless of its source. Purified water from any source — rivers, borewells, oceans, or rainfall — always contains hydrogen and oxygen in the mass ratio 1:8. So all samples of pure water are chemically identical, even though the dissolved impurities may differ before purification.

Q2Oxygen is sometimes represented as O and sometimes as O₂. What is the difference between these symbols?

O represents a single atom of oxygen — the smallest particle of the element oxygen that retains its chemical identity. It is used when referring to the element in atomic terms (e.g., in formulae like MgO, CO₂, H₂O).

O₂ represents a molecule of oxygen — the form in which oxygen exists freely in nature. It consists of two oxygen atoms covalently bonded together by a double bond. O₂ is the actual smallest unit of oxygen gas that can exist independently and shows all the properties of oxygen gas.

Note

In short: O = one atom; O₂ = one molecule (two atoms bonded together). Similarly, H = one hydrogen atom; H₂ = hydrogen molecule.

Q3Why does dissolved salt in water conduct electricity, but sugar does not?

Answer: salt (sodium chloride, NaCl) is an ionic compound. When dissolved in water, it dissociates into free-moving ions — Na⁺ and Cl⁻. These mobile ions carry electric charge through the solution, allowing it to conduct electricity.

Sugar (sucrose, C₁₂H₂₂O₁₁) is a covalent compound. When dissolved in water, it does not break into ions — it dissolves as neutral molecules. Without free-moving charged particles (ions), no electric current can flow, so sugar solution does not conduct electricity.

Note

Electrical conductivity in a solution requires the presence of free-moving ions. Only ionic compounds (and certain covalent compounds that ionise, like acids) provide these in solution.

Section B

Activities 9.1 – 9.4

5 Questions
A9.1Common salt is added to water in a beaker on a digital balance. The mass is recorded before and after dissolving. What do you observe? What does this show?

Observation: the reading on the digital balance before dissolving (mass of water + undissolved salt) is the same as the reading after the salt has dissolved completely.

Conclusion: the mass of the solution (salt water) equals the sum of the masses of water and salt taken. There is no change in mass during the formation of a solution. This demonstrates that mass is conserved in physical changes.

Extension (paper tearing): if a piece of paper is weighed before and after tearing into pieces, the total mass remains the same, further confirming conservation of mass in all physical changes.

Note

Dissolving is a physical change — no new substance is formed. The salt merely disperses into water. Since no atoms are created or destroyed, mass is conserved.

A9.2 (Set-up 1)Baking soda from a balloon is poured into vinegar in an open conical flask. The flask and balloon are weighed before and after. Are the initial and final readings the same?

Observation: a brisk effervescence (fizzing) is observed as carbon dioxide gas is rapidly produced. The final reading on the balance is less than the initial reading.

Reason for the difference: the carbon dioxide gas produced during the reaction escapes into the surrounding air. Since CO₂ is no longer on the balance, the measured mass decreases. The mass appears to have been lost, but in reality it has not — it has simply left the system as a gas.

Conclusion: this set-up does not properly verify conservation of mass because the system is not closed — gas escapes.

A9.2 (Set-up 2)The balloon (containing baking soda) is fixed to the mouth of the conical flask (containing vinegar) before mixing. The entire closed system is weighed before and after the reaction. Are the readings the same?

Observation: again, brisk effervescence occurs, inflating the balloon as CO₂ is produced. This time, the final reading equals the initial reading.

Conclusion: when the system is closed (no gas can escape), the total mass before the reaction equals the total mass after the reaction. This demonstrates the Law of Conservation of Mass: matter is neither created nor destroyed in a chemical reaction, only rearranged.

Note

The balloon traps the CO₂ gas, keeping the entire mass — reactants and all products — on the balance. Only a closed system gives a true test of mass conservation.

A9.3Solutions of sodium sulfate and barium chloride are placed in separate conical flasks on a balance. Their combined mass is recorded. One solution is poured into the other. The reading is taken again. What do you observe?

Observation: when the two solutions are mixed, a white precipitate of barium sulfate (BaSO₄) is immediately formed, and the solution becomes cloudy. The reaction is:

Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2NaCl

The reading on the balance after mixing is the same as the reading before mixing. No gas is produced in this reaction, so no mass escapes. This is an open system, but since all products remain in the flask, mass is conserved.

Conclusion: the total mass of products (barium sulfate + sodium chloride in solution) equals the total mass of reactants (sodium sulfate solution + barium chloride solution). The Law of Conservation of Mass is verified.

Note

Both conical flasks are kept on the balance during transfer to avoid error from solution sticking to the flask walls.

A9.4Samples of camphor, sodium chloride, copper sulfate, sugar and naphthalene are tested for solubility in water, kerosene and petrol, and for electrical conductivity in solid state and aqueous solution. Complete Table 9.2 and group compounds with similar properties.
CompoundWaterKerosene/PetrolConducts (solid)Conducts (water)
Camphor (covalent)InsolubleSolubleNoNo
Sodium chloride (ionic)SolubleInsolubleNoYes
Copper sulfate (ionic)SolubleInsolubleNoYes
Sugar (covalent)SolubleInsolubleNoNo
Naphthalene (covalent)InsolubleSolubleNoNo

Grouping

  • Group 1 — Ionic compounds (NaCl, CuSO₄): soluble in water, insoluble in kerosene/petrol, non-conductor in solid state, conductor in aqueous solution.
  • Group 2 — Covalent compounds (camphor, naphthalene): insoluble in water, soluble in kerosene/petrol, non-conductor in all states.
  • Sugar is a special case: covalent but soluble in water, yet still non-conducting, because it does not produce ions in solution.

Why camphor and naphthalene do not conduct even dissolved: these covalent compounds dissolve without forming ions, so no charged particles exist to carry current.

Prediction for molten state: ionic compounds (like NaCl) will conduct electricity in the molten state because melting breaks the crystal lattice and releases free-moving ions. Covalent compounds will generally not conduct in the molten state since no ions are formed.

Section C

Pause and Ponder

24 Questions
P1A 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 from the air to form carbon dioxide (CO₂) and water vapour (H₂O), both of which are gases that escape into the atmosphere. The products are invisible and disperse into the air, so no residue is seen in the beaker. However, if the masses of all products (CO₂ and H₂O vapour) were collected and measured, their total mass would equal the mass of ethanol burnt plus the mass of oxygen used. The reaction takes place in an open system, so the products escape — but mass is still conserved.

P2When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?
Mass of water formed = mass of hydrogen + mass of oxygen = 20 g + 160 g = 180 g
P3A compound consists of 40% sulfur and 60% oxygen by mass. In a sample containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?

The ratio of sulfur to oxygen by mass = 40:60 = 2:3.

If sulfur = 20 g, then oxygen = 20 × (3/2) = 30 g
P4Carbon 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?
Mass ratio C:O = 3:4  ⟹  if carbon = 9 g, oxygen = 9 × (4/3) = 12 g
P5The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.

In a compound, elements combine chemically in a fixed ratio by mass (determined by their atomic masses and the formula), regardless of how the compound was prepared or from where it was obtained. For example, water always has H:O = 1:8 by mass.

In a mixture, components are simply mixed together physically, not chemically bonded. They can be present in any proportion — for example, air is a mixture of nitrogen, oxygen, etc., and their ratios vary, and salt water can have any amount of salt. There is no fixed ratio, so the Law of Definite Proportions does not apply to mixtures.

P6Students X and Y 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. Student X has ratio Cu:O = 4:1. Student Y has ratio Cu:O = 8:2 = 4:1 (dividing both by 2). Both ratios are identical. Both students produced the same copper oxide with the same fixed mass ratio, confirming the Law of Constant Proportions.

P7Assertion (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. Choose the correct option.

Answer: (ii) Both A and R are true, but R is not the correct explanation of A.

The Assertion (A) is true — 2 g H + 16 g O = 18 g H₂O (verified by the Law of Conservation of Mass and the Law of Definite Proportions). The Reason (R) is also true — Dalton stated that atoms combine in simple whole number ratios; in water, 2 atoms of H combine with 1 atom of O (2:1 atom ratio). However, R does not correctly explain A. The assertion is about mass ratios, while the reason talks about atom number ratios. The correct explanation for the mass ratio (2:16) comes from the atomic masses of hydrogen (1 u each) and oxygen (16 u) and the formula H₂O, not directly from the statement that atoms combine in whole number ratios.

P8Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).

Nitrogen atom (atomic number 7) has electronic configuration 2, 5. It has 5 valence electrons and needs 3 more electrons to complete its octet. Each nitrogen atom shares 3 electrons with the other nitrogen atom, forming 3 shared pairs (3 bonding pairs). This results in a triple bond.

Representation: N≡N (three lines = triple bond), with 1 lone pair on each N atom
N N

This is a very strong bond, which is why nitrogen gas (N₂) is very stable and unreactive under normal conditions.

P9The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).

Fluorine (atomic number 9) has electronic configuration 2, 7. It has 7 valence electrons and needs 1 more electron to complete its octet. Each fluorine atom shares 1 electron with the other fluorine atom, forming 1 shared pair — a single covalent bond.

Representation: F—F, with 3 lone pairs on each fluorine atom
F F

The fluorine molecule (F₂) is held together by a single covalent bond, with each F atom achieving a stable octet configuration.

P10Show the formation of the following molecules: (i) Carbon dioxide (CO₂), (ii) Hydrogen sulfide (H₂S), (iii) Ammonia (NH₃).

(i) Carbon dioxide (CO₂): carbon (atomic number 6) has 4 valence electrons and needs 4 more; oxygen (atomic number 8) has 6 valence electrons and needs 2 more. One carbon atom shares 2 electrons each with two oxygen atoms → two double bonds formed. Representation: O=C=O

(ii) Hydrogen sulfide (H₂S): sulfur (atomic number 16) has 6 valence electrons and needs 2 more; hydrogen needs 1 more. Two hydrogen atoms each share 1 electron with the sulfur atom → two single bonds. Representation: H—S—H (with 2 lone pairs on S).

(iii) Ammonia (NH₃): nitrogen has 5 valence electrons and needs 3 more; hydrogen needs 1 more. Three hydrogen atoms each share 1 electron with the nitrogen atom → three single bonds, with 1 lone pair on N. Formula: NH₃.

(i) Carbon dioxide — O=C=O O C O
(ii) Hydrogen sulfide — H—S—H H S H
(iii) Ammonia — NH₃ N H H H
P11Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.

Answer: neon has atomic number 10 and electronic configuration 2, 8. Its outermost shell (L shell) already contains 8 electrons — a complete octet. Neon is already in a stable electronic configuration and has no tendency to lose, gain, or share electrons, because doing so would disturb its stable arrangement. Therefore, neon does not form chemical bonds and exists as a monoatomic gas. It belongs to the noble gas group (Group 18), all of which are chemically inert for the same reason.

P12What kind of ion will oxygen (O) form?

Oxygen (atomic number 8) has electronic configuration 2, 6. It has 6 valence electrons and needs 2 more to complete its octet. Oxygen gains 2 electrons from another atom, acquiring 2 units of negative charge. It forms the oxide anion: O²⁻.

P13Fill in the blanks: Magnesium can give two electrons to become Mg²⁺. Chlorine can take only one electron to become ______. Now, ______ ion of magnesium and ______ ions of chlorine combine to give magnesium chloride.

Chlorine can take only one electron to become Cl⁻ (chloride anion). Now, 1 ion of magnesium (Mg²⁺) and 2 ions of chlorine (Cl⁻) combine to give magnesium chloride (MgCl₂) — two Cl⁻ ions are needed to balance the 2+ charge of one Mg²⁺ ion, making the compound neutral.

P14Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.

Potassium (K, atomic number 19): electronic configuration 2, 8, 8, 1. It has 1 valence electron. K loses 1 electron to form K⁺ (potassium cation). K⁺ + Cl⁻ → KCl (potassium chloride).

Calcium (Ca, atomic number 20): electronic configuration 2, 8, 8, 2. It has 2 valence electrons. Ca loses 2 electrons to form Ca²⁺ (calcium cation). Ca²⁺ + 2Cl⁻ → CaCl₂ (calcium chloride) — two chloride ions are needed to balance the 2+ charge.

K atom (2,8,8,1)
K
K⁺ ion (2,8,8)
K +
+
Cl⁻ ion
Cl
KCl
Ca atom (2,8,8,2)
Ca
Ca²⁺ ion (2,8,8)
Ca 2+
+ 2
Cl⁻ ions
Cl
CaCl₂
P15Illustrate how sodium sulfide (Na₂S) is formed.

Sodium (Na, atomic number 11): configuration 2, 8, 1. Has 1 valence electron, loses 1 to form Na⁺.

Sulfur (S, atomic number 16): configuration 2, 8, 6. Has 6 valence electrons, needs 2 to complete octet, gains 2 to form S²⁻.

Since each Na atom provides only 1 electron and S needs 2 electrons, 2 Na atoms each donate 1 electron to 1 S atom:

2Na → 2Na⁺ + 2e⁻;   S + 2e⁻ → S²⁻;   2Na⁺ + S²⁻ → Na₂S

The 2+ charge from 2 Na⁺ ions balances the 2− charge of S²⁻. Formula: Na₂S.

P16Name the following: (i) CO₂ (ii) NO₂ (iii) SF₆ (iv) PCl₃
  • (i) CO₂ → Carbon dioxide (one carbon + two oxygen atoms; "di" prefix for two O atoms).
  • (ii) NO₂ → Nitrogen dioxide (one nitrogen + two oxygen atoms).
  • (iii) SF₆ → Sulfur hexafluoride (one sulfur + six fluorine atoms; "hexa" prefix for six).
  • (iv) PCl₃ → Phosphorus trichloride (one phosphorus + three chlorine atoms; "tri" prefix for three).
P17Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide
  • (i) Sodium hydrogencarbonate: Na⁺ (valency 1) + HCO₃⁻ (valency 1) → NaHCO₃
  • (ii) Sulfur dioxide: "dioxide" means 2 O atoms → SO₂
  • (iii) Ferric chloride: Fe³⁺ (valency 3) + Cl⁻ (valency 1); criss-cross gives FeCl₃
  • (iv) Cuprous oxide: Cu⁺ (cuprous, valency 1) + O²⁻ (valency 2); criss-cross gives Cu₂O
P18Write the formulae for the compounds formed from the following pairs of ions: (i) Fe³⁺ and OH⁻ (ii) K⁺ and CO₃²⁻

(i) Fe³⁺ and OH⁻: valencies are 3 and 1. Criss-cross: Fe(OH)₃ (ferric hydroxide). Brackets needed as OH is polyatomic.

(ii) K⁺ and CO₃²⁻: valencies are 1 and 2. Criss-cross: K₂CO₃ (potassium carbonate). No brackets needed.

P19What 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: the compound has an ionic bond. In the solid state, the ions (cations and anions) are held in fixed positions in a crystal lattice by strong electrostatic forces, so they cannot move and the solid does not conduct electricity. When the ionic compound is dissolved in water, the lattice breaks down and the ions become free to move in solution — these mobile ions carry electric charge, allowing the solution to conduct electricity. Examples: sodium chloride (NaCl), copper sulfate (CuSO₄), potassium nitrate (KNO₃).

P20Metal M has two electrons in its valence shell (M shell). It reacts with oxygen to form a slightly soluble compound. Predict its: (i) formula, (ii) type of bond, (iii) electrical conductivity of its aqueous solution.

Metal M has 2 valence electrons in its M (third) shell, so its electronic configuration is 2, 8, 2 — this is magnesium (Mg, atomic number 12).

(i) Formula: Mg²⁺ (valency 2) + O²⁻ (valency 2); criss-cross gives MgO (magnesium oxide) — the two valencies are equal, so the formula simplifies to MgO.

(ii) Type of bond: ionic bond. Magnesium loses 2 electrons to form Mg²⁺; oxygen gains 2 electrons to form O²⁻. The electrostatic attraction between these oppositely charged ions forms an ionic bond.

(iii) Electrical conductivity of aqueous solution: MgO is slightly soluble in water (forms Mg(OH)₂, which partially ionises). The aqueous solution will show slight electrical conductivity due to the small number of Mg²⁺ and OH⁻ ions present.

P21Find the molecular mass of nitric acid (HNO₃). Atomic mass — H = 1 u; N = 14 u; O = 16 u.
Molecular mass = (1 × 1) + (1 × 14) + (3 × 16) = 1 + 14 + 48 = 63 u
P22Find the molecular mass of methane (CH₄). Atomic mass — C = 12 u; H = 1 u.
Molecular mass = (1 × 12) + (4 × 1) = 12 + 4 = 16 u
P23Find the formula unit mass of potassium chloride (KCl). Atomic mass — K = 39 u; Cl = 35.5 u.
Formula unit mass = (1 × 39) + (1 × 35.5) = 74.5 u
P24Find the formula unit mass of magnesium hydroxide, Mg(OH)₂. Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.
Formula unit mass = (1 × 24) + 2 × (16 + 1) = 24 + 34 = 58 u
Section D

Think as a Scientist

1 Question
TaS1Design an experiment to test the hypothesis that mass is conserved when zinc reacts with dilute hydrochloric acid. (Zn + HCl → ZnCl₂ + H₂)

Hypothesis: mass of reactants = mass of products in the reaction Zn + HCl(dil.) → ZnCl₂ + H₂↑

Challenge: H₂ gas is produced and could escape if the system is open.

Experimental Design (Closed System)

  • Take a conical flask fitted with a rubber stopper (gas-tight). Attach a deflated balloon to the stopper's outlet.
  • Place a measured mass of zinc pieces (e.g., 1.0 g) inside the flask.
  • Fill a small test tube with dilute HCl (enough to react with the zinc) and place it carefully inside the flask without mixing yet. Seal the flask.
  • Weigh the entire sealed apparatus (flask + zinc + HCl + balloon) on a digital balance. Record as initial mass.
  • Tilt the flask to mix the HCl with zinc. A vigorous reaction occurs; H₂ gas inflates the balloon.
  • After the reaction is complete, weigh the entire apparatus again (flask + products + inflated balloon). Record as final mass.

Expected result: initial mass = final mass. The H₂ gas is trapped in the balloon, so no mass escapes. This confirms the Law of Conservation of Mass.

Note

If using a fully sealed system (no balloon), the pressure builds up. The balloon design is safer and more practical for this experiment.

Section E

Worked Examples 9.1 – 9.7

7 Questions
Ex 9.1Students place 4.0 g of calcium carbonate with 2.92 g of hydrochloric acid in a closed container. After reaction: 1.76 g CO₂ + 0.72 g water + 4.44 g calcium chloride. Verify the Law of Conservation of Mass.
Total mass of reactants = 4.0 + 2.92 = 6.92 g
Total mass of products = 1.76 + 0.72 + 4.44 = 6.92 g

Mass of reactants = Mass of products = 6.92 g. Conclusion: the Law of Conservation of Mass is obeyed.

Ex 9.212 g of carbon combines with 32 g of oxygen to form 44 g of CO₂. If 2.4 g of carbon reacts completely, how much CO₂ is produced?

1 g of carbon produces 44/12 g of CO₂.

2.4 g of carbon produces = (44/12) × 2.4 = 8.8 g of CO₂
Ex 9.3NaCl contains Na and Cl in the mass ratio 23:35.5. If 46 g of sodium reacts completely, how much chlorine is needed?
Mass of chlorine = (35.5 ÷ 23) × 46 = 71 g
Ex 9.4Find the molecular mass of water (H₂O). (H = 1 u; O = 16 u)
Molecular mass = (2 × 1) + (1 × 16) = 2 + 16 = 18 u
Ex 9.5Find the molecular mass of carbon dioxide (CO₂). (C = 12 u; O = 16 u)
Molecular mass = (1 × 12) + (2 × 16) = 12 + 32 = 44 u
Ex 9.6Find the formula unit mass of sodium oxide (Na₂O). (Na = 23 u; O = 16 u)
Formula unit mass = (2 × 23) + (1 × 16) = 46 + 16 = 62 u
Ex 9.7Find the formula unit mass of calcium nitrate, Ca(NO₃)₂. (Ca = 40 u; N = 14 u; O = 16 u)
Formula unit mass = (1 × 40) + 2 × [(1 × 14) + (3 × 16)] = 40 + 2 × 62 = 40 + 124 = 164 u
Section F

Bridging Science and Society — Nuclear Energy

1 Question
BSS1Explain how nuclear energy is produced and its applications. Who is called the Father of the Indian Nuclear Programme?

Nuclear energy is released when the nuclei of atoms either split (nuclear fission) or combine (nuclear fusion) to form new elements. The mass of the products is slightly less than the mass of the reactants, and this mass difference (called mass defect) is converted into an enormous amount of energy, as described by Einstein's equation \(E = mc^2\).

In nuclear power plants, fission of uranium-235 nuclei releases heat energy. This heat produces steam, which drives turbines connected to generators, producing electricity. Nuclear power is a cleaner alternative to fossil fuels as it does not produce CO₂ or other greenhouse gases during operation.

Applications of Nuclear Energy

  • Electricity generation in nuclear power plants.
  • Medical applications: cancer treatment (radiation therapy), diagnostic imaging (PET scans, X-rays using radioisotopes).
  • Scientific research: nuclear reactors produce radioisotopes for research.
  • Space exploration: nuclear batteries (radioisotope thermoelectric generators) power spacecraft like Voyager and the Curiosity rover.

Raja Ramanna is often called the Father of the Indian Nuclear Programme. He made significant contributions in developing India's nuclear energy programme and promoting its peaceful use for national development.

Section G

Revise, Reflect, Refine (Back Exercise)

15 Questions
Q1Element A has one electron in its third shell. Element B has six electrons in its second shell. Answer (i)–(vi) on electrons given/taken, ions formed, bond type, and formula.

Element A: configuration 2, 8, 1 (third shell has 1 electron) — this is sodium (Na, atomic number 11).

(i) A tends to give 1 electron (easier to lose 1 than gain 7 to complete the octet). (ii) A forms a cation: A⁺ (Na⁺).

Element B: second shell has 6 electrons, so configuration is 2, 6 (oxygen, atomic number 8).

(iii) B tends to take 2 electrons (needs 2 more to complete its octet of 8). (iv) B forms an anion: B²⁻ (O²⁻).

(v) Bond type: ionic bond. A loses 1 electron; B gains 2 electrons. Since each A provides 1 electron and B needs 2, two A atoms combine with one B atom.

(vi) Formula: A⁺ valency 1; B²⁻ valency 2. Criss-cross: A₂B. If A = Na and B = O, the compound is Na₂O (sodium oxide).

Q2Element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why? (ii) Bond type? (iii) Structure of X₂? (iv) Element Y has two electrons in its second shell — structure of compound X forms with Y?

Element X has 6 valence electrons → needs 2 more for octet → this is oxygen (O, atomic number 8).

(i) Oxygen forms a diatomic molecule (O₂) because each oxygen atom needs 2 more electrons to achieve a stable octet; by sharing 2 electrons with another oxygen atom, both achieve stability.

(ii) Bond type: covalent bond (a double bond, since 2 electron pairs are shared between the two O atoms).

(iii) Structure of X₂ (O₂): O=O (two atoms joined by a double bond, with 2 lone pairs on each O atom).

(iv) Element Y, with configuration 2, 1 (one valence electron), is hydrogen. Two H atoms (Y) each share 1 electron with O (X), forming 2 single covalent bonds: H—O—H (water).

X₂ (O₂) — O=O O O
Compound with Y (H₂O) — H—O—H O H H
Q3Design a new ionic compound where total positive charge = 6+ and total negative charge = 6−. Which combination is correct? (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⁺: total positive = 6+; 3 Cl⁻: total negative = 3−. Not balanced. Incorrect.
  • (ii) 3 Mg²⁺: total positive = 6+; 1 PO₄³⁻: total negative = 3−. Not balanced. Incorrect.
  • (iii) 2 Fe³⁺: total positive = 6+; 3 O²⁻: total negative = 6−. Balanced! This is Fe₂O₃ (iron(III) oxide). Correct.
  • (iv) 3 Ca²⁺: total positive = 6+; 2 SO₄²⁻: total negative = 4−. Not balanced. Incorrect.

Answer: (iii) 2 Fe³⁺ and 3 O²⁻

Q4Choose the correct statement(s) and correct the false statement(s): (i)–(iv)
  • (i) FALSE. Correction: elements are made up of atoms (or molecules made of atoms of the same element). Compounds are made up of molecules (or formula units) containing atoms of two or more different elements bonded together.
  • (ii) FALSE. Correction: the molecule of a compound is always made up of two or more atoms of different kinds (different elements). For example, HCl has H and Cl atoms; H₂O has H and O atoms.
  • (iii) FALSE. Correction: one molecule of nitrogen gas (N₂) contains two nitrogen atoms (not three) — N₂ is a diatomic molecule.
  • (iv) TRUE. Water (H₂O) is indeed made of two hydrogen atoms covalently bonded with one oxygen atom, with each H—O bond being a single covalent bond.
Q5Write the chemical formulae for: (i) Aluminium nitrate, (ii) Calcium oxide, (iii) Ferric oxide.
  • (i) Aluminium nitrate: Al³⁺ (valency 3) + NO₃⁻ (valency 1). Criss-cross: Al(NO₃)₃
  • (ii) Calcium oxide: Ca²⁺ (valency 2) + O²⁻ (valency 2). Valencies equal; simplified to CaO
  • (iii) Ferric oxide: Fe³⁺ (ferric, valency 3) + O²⁻ (valency 2). Criss-cross: Fe₂O₃
Q6Write 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⁺ (valency 2) and Br⁻ (valency 1): CaBr₂ (calcium bromide)
  • (ii) Al³⁺ (valency 3) and CO₃²⁻ (valency 2): Al₂(CO₃)₃ (aluminium carbonate)
  • (iii) K⁺ (valency 1) and SO₄²⁻ (valency 2): K₂SO₄ (potassium sulfate)
  • (iv) NH₄⁺ (valency 1) and Cl⁻ (valency 1): NH₄Cl (ammonium chloride)
Q7Which of the diagrams in Fig. 9.18 correctly represents the Cl⁻ ion (atomic number of chlorine = 17)?

Cl atom (atomic number 17) has electronic configuration 2, 8, 7 (17 electrons). The Cl⁻ ion has gained 1 electron, so it has 18 electrons total: configuration 2, 8, 8 — all shells now complete.

Cl

The correct diagram must show 3 shells, with 2 electrons in the first (K), 8 electrons in the second (L), and 8 electrons in the third (M) shell, carrying a negative charge.

Note

Cl⁻ has the same electron configuration as argon (2, 8, 8) but with 17 protons (unlike Ar which has 18), so it carries a 1− charge.

Q8Determine the formula unit mass of: (i) Ammonium nitrate (NH₄NO₃), (ii) Phosphoric acid (H₃PO₄), (iii) Sodium hydrogencarbonate (NaHCO₃).
(i) NH₄NO₃: 2N + 4H + 3O = 28 + 4 + 48 = 80 u
(ii) H₃PO₄: 3H + 1P + 4O = 3 + 31 + 64 = 98 u
(iii) NaHCO₃: Na + H + C + 3O = 23 + 1 + 12 + 48 = 84 u
Q9Write the formulae for compounds formed by the reaction of: (i) Magnesium and nitrogen (ii) Lithium and nitrogen (iii) Sodium and sulfur (iv) Aluminium and oxygen
  • (i) Mg²⁺ and N³⁻: Mg₃N₂ (magnesium nitride)
  • (ii) Li⁺ and N³⁻: Li₃N (lithium nitride)
  • (iii) Na⁺ and S²⁻: Na₂S (sodium sulfide)
  • (iv) Al³⁺ and O²⁻: Al₂O₃ (aluminium oxide)
Q10Complete Table 9.3 by writing the formulae of the compounds formed by the cations (NH₄⁺, Li⁺, Al³⁺, Cu²⁺) with the anions (NO₃⁻, SO₄²⁻, PO₄³⁻).
NO₃⁻SO₄²⁻PO₄³⁻
NH₄⁺NH₄NO₃(NH₄)₂SO₄(NH₄)₃PO₄
Li⁺LiNO₃Li₂SO₄Li₃PO₄
Al³⁺Al(NO₃)₃Al₂(SO₄)₃AlPO₄
Cu²⁺Cu(NO₃)₂CuSO₄Cu₃(PO₄)₂
Q115.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of CO₂, 0.9 g of water, and 8.2 g of 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

Mass of reactants = Mass of products = 11.3 g. Conclusion: the Law of Conservation of Mass is valid for this reaction.

Q12A species has 11 protons, 12 neutrons and 10 electrons. (i) Atomic number and mass number? (ii) Neutral, cation or anion? (iii) Electronic configuration? (iv) Name the species.

(i) Atomic number = protons = 11. Mass number = protons + neutrons = 11 + 12 = 23.

(ii) It is a cation. Protons (11) > Electrons (10) → net positive charge of 1+ → it is Na⁺ (sodium ion).

(iii) Electronic configuration of Na⁺ (10 electrons): K = 2, L = 8. Configuration: 2, 8.

(iv) The species is the sodium ion (Na⁺) — the ion formed when a sodium atom loses 1 valence electron. Its electron configuration (2, 8) is the same as that of neon (a noble gas).

Q13Elements A (configuration 2,8,5) and B (configuration 2,8,7). (i) Which is more reactive? (ii) Ionic or covalent bond? (iii) Predict the formula.

Element A: configuration 2,8,5 → 5 valence electrons; atomic number = 15 → phosphorus (P). Needs 3 more electrons.

Element B: configuration 2,8,7 → 7 valence electrons; atomic number = 17 → chlorine (Cl). Needs 1 more electron.

(i) Reactivity: element B (chlorine) is more reactive — it needs only 1 electron to complete its octet, making it a highly reactive non-metal (halogen). Element A (phosphorus) needs 3 electrons, which is relatively harder.

(ii) Bond type: both A and B are non-metals, and non-metals generally form covalent bonds by sharing electrons. A will share 3 electrons with 3 B atoms, forming 3 single covalent bonds.

(iii) Formula: A needs 3 electrons; B provides 1 electron each. Three B atoms share one electron each with one A atom → AB₃ — in terms of P and Cl: PCl₃ (phosphorus trichloride).

Q14Assertion (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.

Answer: (iii) A is true, but R is false.

Explanation of A: the Assertion is true. CuSO₄ (an ionic compound) does not conduct electricity in the solid state (ions are fixed in the crystal lattice) but does conduct in the molten state (ions become free to move when the lattice melts).

Explanation of R: the Reason is false — it has the explanation exactly backwards. In the solid state, ions are fixed (locked in the lattice) and cannot move — so solid does not conduct. In the molten (melted) state, the lattice breaks down and ions become free to move — so it conducts electricity. The reason incorrectly states the opposite of the truth.

Q15The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?

²⁷Al (aluminium atom, neutral): protons = 13, so electrons = 13. Neutrons = mass number − protons = 27 − 13 = 14.

⁸⁰Br⁻ (bromide ion, charge 1−): protons = 35; gained 1 electron, so electrons = 36. Neutrons = 80 − 35 = 45.

²⁰¹Hg²⁺ (mercury ion, charge 2+): protons = 80; lost 2 electrons, so electrons = 78. Neutrons = 201 − 80 = 121.

Section H

The Journey Beyond

3 Questions
JB1Design and perform an experiment to show that water always contains hydrogen and oxygen in the same ratio, regardless of its source.

Hypothesis: all samples of pure water contain hydrogen and oxygen in a fixed mass ratio of 1:8 (or volume ratio of 2:1 for H₂:O₂), regardless of source.

Method — Electrolysis of Water

  • Collect water from three sources: tap water, river water, and rainwater. Purify each sample by distillation to remove all dissolved impurities.
  • Set up an electrolysis apparatus (Hofmann voltameter or two test tubes over electrodes) for each sample. Use dilute H₂SO₄ or NaOH solution as electrolyte (the electrolyte doesn't affect the ratio).
  • Pass electric current through each sample. Collect gases at each electrode — hydrogen at the cathode (negative) and oxygen at the anode (positive).
  • Measure the volume of H₂ and O₂ produced from each water sample.

Expected result: for all three samples, the volume ratio of H₂:O₂ = 2:1 (or mass ratio H:O = 1:8). This confirms that water from all sources has the same fixed composition, verifying the Law of Constant Proportions.

JB2Compare atoms and ions of any three elements. Show the number of electrons before and after ion formation using bar graphs.

Example elements: Sodium (Na), Chlorine (Cl), Magnesium (Mg).

  • Sodium: Na atom → 11 electrons; Na⁺ ion → 10 electrons (lost 1).
  • Chlorine: Cl atom → 17 electrons; Cl⁻ ion → 18 electrons (gained 1).
  • Magnesium: Mg atom → 12 electrons; Mg²⁺ ion → 10 electrons (lost 2).

Bar graph description: X-axis shows species (atom/ion for each element); Y-axis shows number of electrons. Sodium bars: 11 and 10. Chlorine bars: 17 and 18. Magnesium bars: 12 and 10.

JB3 — The Quest ContinuesAre there any chemical changes that do not obey the Law of Conservation of Mass?

For all ordinary chemical reactions (the type studied in this chapter), the Law of Conservation of Mass holds exactly. No chemical change violates this law.

However, in nuclear reactions (not ordinary chemical changes), the situation is different. In nuclear fission and fusion, a tiny fraction of mass is converted directly into energy according to Einstein's famous equation \(E = mc^2\). The total mass of products is very slightly less than the total mass of reactants. This mass difference (called mass defect) appears as an enormous release of energy (nuclear energy).

This means that at the nuclear level, mass and energy are interconvertible. The more fundamental conservation law is the conservation of mass-energy (not mass alone). For all practical purposes in chemistry, however, the energy changes in chemical reactions are so small that the corresponding mass changes are immeasurably tiny and can be completely ignored. The Law of Conservation of Mass holds to extremely high precision for all chemical reactions.

Note

In everyday chemistry, mass is always conserved. Only at the nuclear scale (involving the strong nuclear force and atomic nuclei) does the conversion of mass to energy become significant.

💡 Chapter 9's core idea, in one line

Mass is conserved and elements combine in fixed ratios in every chemical reaction — and these two experimentally-verified laws are exactly why Dalton's picture of atoms simply rearranging, never being created or destroyed, works, and why atoms combine by sharing electrons (covalent bonds) or transferring them (ionic bonds) in the same fixed whole-number ratios every single time.

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Common Questions

Frequently Asked Questions

Look at the number of electrons in the outermost shell. If there are 1, 2, or 3 electrons, the element tends to lose them, and valency equals that number. If there are 5, 6, or 7 electrons, the element tends to gain electrons to complete the octet, and valency equals 8 minus the number of outer electrons. Elements with a complete outer shell (8, or 2 for the first shell) have zero valency and don't readily form bonds.
An ionic bond forms when one atom completely transfers one or more electrons to another, creating oppositely charged ions that attract each other — this typically happens between a metal and a non-metal. A covalent bond forms when two atoms, usually both non-metals, share one or more pairs of electrons instead of transferring them, so that both atoms achieve a stable outer shell together.
The Law of Conservation of Mass (Lavoisier, 1789) states that mass is neither created nor destroyed in a chemical reaction — the total mass of reactants equals the total mass of products. The Law of Constant Proportions (Proust) states that a given compound always contains its constituent elements in the same fixed ratio by mass, regardless of its source or how it was prepared. Together, these two laws form the experimental basis of Dalton's Atomic Theory.
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