NCERT Solutions for Class 9 Science Chapter 9: Atomic Foundations of Matter – Free PDF Download

Class 9 Science Chapter 9, “Atomic Foundations of Matter,” from the new NCERT “Exploration” textbook (2026-27 session) builds on the atomic structure basics from Chapter 8 and explains how atoms combine to form molecules and ionic compounds through covalent and ionic bonding, along with the mole concept, chemical formula writing, and molecular/formula unit mass calculations. These solutions were prepared by working through every question first-hand and then cross-checking the exercise and in-text answers against LearnCBSE and TiwariAcademy’s Exploration-edition coverage of this chapter; a few OCR-style typos and an ambiguous option-letter in the source material were corrected using the underlying chemistry rather than copied as-is (details noted below the relevant answers).

Last Updated: September 23, 2026

NCERT Solutions for Class 9 Science Chapter 9: Atomic Foundations of Matter

Revise, Reflect, Refine (NCERT Textbook, Page No. 182)

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) Element A has 1 electron in its third (outermost) shell. Losing this single electron is far easier than gaining 7 more, so A tends to give away 1 electron to attain a stable configuration.
(ii) Since A loses 1 electron, it forms a positively charged ion — a cation, written as A⁺.
(iii) Element B has 6 electrons in its second (outermost) shell, which can hold a maximum of 8. So B needs 2 more electrons to become stable.
(iv) Since B gains 2 electrons, it forms a negatively charged ion — an anion, written as B^2-.
(v) A gives away 1 electron and B needs 2 electrons, so 2 atoms of A are needed to satisfy 1 atom of B. The electron transfer results in an ionic (electrovalent) bond.
(vi) Using the criss-cross method (valency of A = 1, valency of B = 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) X has 6 valence electrons and needs 2 more to complete its octet. Being a non-metal, it attains stability by sharing 2 electrons with another atom of the same element (like oxygen forming O₂), giving a diatomic molecule X₂.
(ii) Each X atom shares 2 electrons with the other X atom, so a double covalent bond is formed (X=X).
(iii) Structure: each X atom starts with 6 dots (electrons) around it; two pairs of electrons are shown shared between the two X atoms, so both atoms end up with a complete octet of 8 electrons around them, joined by a double bond (X=X).
(iv) Element Y has 2 electrons in its outermost (second) shell, so its valency is 2 — it loses both electrons to become Y²⁺. Element X needs 2 electrons to complete its octet, so it gains both electrons released by Y to become X^2-. The two oppositely charged ions attract each other electrostatically, forming an ionic bond. The compound formed has the formula YX.

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 PO₄^3-
(iii) 2 Fe³⁺ and 3 O^2-
(iv) 3 Ca²⁺ and 2 SO₄^2-
Checking each option (total positive charge must equal total negative charge = 6):
(i) 2 Al³⁺ and 3 Cl^– → Positive = 2 × 3 = 6+; Negative = 3 × 1 = 3−. Incorrect.
(ii) 3 Mg²⁺ and 1 PO₄^3- → Positive = 3 × 2 = 6+; Negative = 1 × 3 = 3−. Incorrect.
(iii) 2 Fe³⁺ and 3 O^2- → Positive = 2 × 3 = 6+; Negative = 3 × 2 = 6−. Correct.
(iv) 3 Ca²⁺ and 2 SO₄^2- → Positive = 3 × 2 = 6+; Negative = 2 × 2 = 4−. Incorrect.
∴ The correct option is (iii).

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) False. Correction: Elements are made up of atoms (or, in molecular elements, of molecules containing atoms of the same kind), and compounds are made up of molecules.
(ii) False. Correction: The molecule of a compound is always made up of two or more atoms of different kinds.
(iii) False. Correction: One molecule of nitrogen gas (N₂) contains two nitrogen atoms, not three.
(iv) True. Water (H₂O) consists of two hydrogen atoms covalently bonded to one oxygen atom.

5. Write the chemical formulae for the following compounds.
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
(i) Aluminium nitrate: Al(NO₃)₃ (Al³⁺ combines with 3 NO₃^– ions)
(ii) Calcium oxide: CaO (Ca²⁺ and O^2- combine in a 1:1 ratio)
(iii) Ferric oxide: Fe₂O₃ (Fe³⁺ and O^2- criss-cross to give subscripts 2 and 3)

6. Write the formulae of the compounds formed from the following pairs of ions.
(i) Ca²⁺ and Br^–
(ii) Al³⁺ and CO₃^2-
(iii) K⁺ and SO₄^2-
(iv) NH₄⁺ and Cl^–
(i) CaBr₂ (calcium bromide)
(ii) Al₂(CO₃)₃ (aluminium carbonate)
(iii) K₂SO₄ (potassium sulfate)
(iv) NH₄Cl (ammonium chloride)

7. Which of the following, in Fig. 9.18, correctly represents Cl^– ion (Atomic number of chlorine = 17)?
A neutral chlorine atom (atomic number 17) has the electronic configuration 2, 8, 7. When it gains 1 electron to form the chloride ion (Cl^–), its outer shell becomes complete, giving the configuration 2, 8, 8 (18 electrons in total, arranged as 2 in the first shell, 8 in the second, and 8 in the third). The correct diagram is the one showing this 2, 8, 8 arrangement with an overall single negative charge — this is commonly labelled option (ii) in the printed answer keys, though you should match it to whichever lettered diagram in your copy of Fig. 9.18 actually shows 2, 8, 8 with 18 electrons, since some reprints of the figure re-order the options.

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.
Atomic masses used: H = 1 u, N = 14 u, O = 16 u, P = 31 u, Na = 23 u, C = 12 u.
(i) NH₄NO₃ = (2 × 14) + (4 × 1) + (3 × 16) = 28 + 4 + 48 = 80 u
(ii) H₃PO₄ = (3 × 1) + (1 × 31) + (4 × 16) = 3 + 31 + 64 = 98 u
(iii) NaHCO₃ = (1 × 23) + (1 × 1) + (1 × 12) + (3 × 16) = 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 (Mg²⁺) and nitrogen (N^3-): criss-crossing the charges gives Mg₃N₂ (magnesium nitride).
(ii) Lithium (Li⁺) and nitrogen (N^3-): gives Li₃N (lithium nitride).
(iii) Sodium (Na⁺) and sulfur (S^2-): gives Na₂S (sodium sulfide).
(iv) Aluminium (Al³⁺) and oxygen (O^2-): gives Al₂O₃ (aluminium oxide).

10. 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.
Combining each cation with each anion by balancing charges gives:
NH₄⁺ → NH₄NO₃ (with NO₃^–), (NH₄)₂SO₄ (with SO₄^2-), (NH₄)₃PO₄ (with PO₄^3-)
Li⁺ → LiNO₃ (given), Li₂SO₄, Li₃PO₄
Al³⁺ → Al(NO₃)₃, Al₂(SO₄)₃, AlPO₄ (3+ and 3− cancel exactly, so a 1:1 ratio)
Cu²⁺ → Cu(NO₃)₂, CuSO₄, Cu₃(PO₄)₂

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.
Total mass of reactants = 5.3 g + 6.0 g = 11.3 g
Total mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g
Since mass of reactants = mass of products (11.3 g = 11.3 g), the Law of Conservation of Mass is verified — matter is neither created nor destroyed in this 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) Atomic number (Z) = number of protons = 11. Mass number (A) = protons + neutrons = 11 + 12 = 23.
(ii) Protons (11) are greater than electrons (10), so the species carries a net positive charge of +1. It is a cation.
(iii) With 10 electrons, the electronic configuration is 2, 8.
(iv) The element with atomic number 11 is sodium (Na); with a +1 charge, this species is the 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) Element B is more reactive. A has 5 valence electrons and needs 3 more to complete its octet, while B has 7 valence electrons and needs only 1 more — the fewer electrons an atom needs, the stronger its tendency to attract them, so B (like chlorine) is more reactive.
(ii) Both A and B are non-metals with more than 4 valence electrons, so neither readily loses electrons; instead they share electrons to complete their octets, forming a covalent bond. Atom A needs 3 electrons, so it shares one electron each with three separate atoms of B; each B atom needs only 1 electron and completes its octet with that single shared pair.
(iii) Since one A atom bonds with three B atoms, the formula of the compound is AB₃ (analogous to PCl₃, phosphorus 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.
(iii) A is true, but R is false. The Assertion correctly describes how ionic compounds conduct electricity, but the Reason has the situation backwards — in the solid state, ions are fixed in the crystal lattice and cannot move, while in the molten state, the lattice breaks down and ions are free to move and carry current.

15. The species ²⁷Al, ⁸⁰Br^– and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
²⁷Al (neutral atom): protons = 13, electrons = 13 (same as protons, since neutral), neutrons = 27 − 13 = 14.
⁸⁰Br^– (gained 1 electron): protons = 35, electrons = 35 + 1 = 36, neutrons = 80 − 35 = 45.
²⁰¹Hg²⁺ (lost 2 electrons): protons = 80, electrons = 80 − 2 = 78, neutrons = 201 − 80 = 121.

In-Text Questions (Think It Over, Think as a Scientist, Pause and Ponder, What if…, and Activities)

Think It Over (Page No. 162) — Water can be obtained from various sources. Are all these samples of water chemically identical?
Yes, all samples of pure water are chemically identical because water always consists of hydrogen and oxygen combined in the same fixed ratio (2 atoms of hydrogen to 1 atom of oxygen). Water collected from different natural sources may carry different dissolved impurities, but once purified, every sample is chemically the same substance, H₂O.

Think It Over (Page No. 162) — Oxygen is sometimes represented as O and sometimes as O₂. What is the difference between these symbols?
O represents a single, free oxygen atom, while O₂ represents a molecule made of two oxygen atoms chemically bonded together — the form in which oxygen actually exists and is capable of independent existence in nature.

Think It Over (Page No. 162) — Why does dissolved salt in water conduct electricity, but sugar does not?
Salt (an ionic compound) dissociates in water into free-moving, charged Na⁺ and Cl^– ions, which carry electric current through the solution. Sugar is a covalent compound; it dissolves as neutral molecules without breaking into ions, so there are no charge carriers present and the solution cannot conduct electricity.

Think as a Scientist (Page No. 166) — Design and perform an experiment to test the hypothesis that mass is conserved when zinc reacts with dilute hydrochloric acid to form zinc chloride and hydrogen gas.
Aim: To verify that mass is conserved during the reaction Zinc + dilute HCl → Zinc chloride + Hydrogen.
Materials: Conical flask, tight-fitting rubber cork with a delivery tube (or a balloon fitted over the mouth of the flask), a sensitive weighing balance, zinc granules, dilute hydrochloric acid.
Procedure: Place the flask with dilute HCl on the balance and note its mass. Add a known mass of zinc granules and immediately seal the flask (e.g., with a balloon over the mouth) so that no hydrogen gas can escape. Let the reaction go to completion, then weigh the sealed flask again.
Observation: Effervescence occurs as hydrogen gas is released and inflates the balloon; the balance reading before and after the reaction stays the same.
Conclusion: Since the total mass of the sealed system does not change, the total mass of reactants equals the total mass of products, verifying the Law of Conservation of Mass.

Pause and Ponder (Page No. 166) — 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?
No. When ethanol burns, it reacts with oxygen from the air to form carbon dioxide and water vapour, both of which are gases that escape into the surrounding air in an open beaker. Since these gaseous products are not weighed, it can look as if mass has “disappeared,” but the mass is fully accounted for once the escaped gases are included — the law is not violated, only the open setup fails to capture all the products.

Pause and Ponder (Page No. 166) — When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?
By the Law of Conservation of Mass, total mass of reactants = total mass of products.
Mass of water formed = 20 g + 160 g = 180 g.

Pause and Ponder (Page No. 167) — A 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?
Mass ratio of sulfur : oxygen = 40 : 60 = 2 : 3.
If sulfur = 20 g, then oxygen = (3/2) × 20 = 30 g.

Pause and Ponder (Page No. 167) — 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?
3 g of carbon combines with 4 g of oxygen, so for 9 g of carbon: oxygen = (4/3) × 9 = 12 g.

Pause and Ponder (Page No. 167) — The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.
In a compound, elements are chemically combined in a fixed mass ratio (e.g., water is always 1:8 hydrogen to oxygen by mass), because this ratio is set by the atoms’ valencies and how they bond. In a mixture, substances are only physically combined and can be present in any proportion (like salt and sand, or a more or less concentrated sugar solution), so no fixed ratio applies.

Pause and Ponder (Page No. 167) — 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?
Yes. Simplifying Student Y’s ratio, 8:2 reduces to 4:1, which is identical to Student X’s ratio. Since both students obtained the same copper-to-oxygen mass ratio despite different absolute amounts, this confirms the Law of Constant Proportions.

Pause and Ponder (Page No. 168) — 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.
Both A and R are true, but R is not the correct explanation of A. The Assertion is a direct illustration of the Law of Conservation of Mass (2 g + 16 g = 18 g), while the Reason describes a separate idea from Dalton’s Atomic Theory about whole-number combining ratios — a true statement, but not what explains why the masses add up in the Assertion.

Pause and Ponder (Page No. 170) — Nitrogen has five valence electrons. Explain the structure of the nitrogen molecule (N₂).
Each nitrogen atom has 5 valence electrons and needs 3 more to complete its octet. Two nitrogen atoms share three electrons each, forming three shared electron pairs — a triple covalent bond (N≡N). Each atom is also left with one lone (unshared) pair of electrons.

Pause and Ponder (Page No. 170) — The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).
Fluorine (atomic number 9) has the electronic configuration 2, 7 — 7 valence electrons, needing just 1 more to complete its octet. Two fluorine atoms each contribute 1 electron to a single shared pair, forming a single covalent bond (F–F); both atoms then have a complete octet of 8 electrons.

Pause and Ponder (Page No. 171) — Show the formation of (i) carbon dioxide (CO₂), (ii) hydrogen sulfide (H₂S), (iii) ammonia (NH₃).
(i) Carbon has 4 valence electrons and needs 4 more; each oxygen atom has 6 valence electrons and needs 2 more. Carbon shares two electron pairs with each oxygen atom, forming two double covalent bonds: O=C=O.
(ii) Sulfur has 6 valence electrons and needs 2 more; each hydrogen atom needs 1 more electron. Sulfur shares one electron pair with each of two hydrogen atoms, forming two single covalent bonds: H–S–H.
(iii) Nitrogen has 5 valence electrons and needs 3 more; each hydrogen atom needs 1 more electron. Nitrogen shares one electron pair with each of three hydrogen atoms, forming three single covalent bonds and leaving one lone pair on nitrogen (pyramidal NH₃).

Pause and Ponder (Page No. 171) — Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Neon’s electronic configuration is 2, 8 — its outermost shell is already completely filled with 8 electrons, giving it a stable octet. Since it has no need to lose, gain, or share electrons, neon does not form chemical bonds and remains chemically inert (unreactive), like the other noble gases.

Pause and Ponder (Page No. 174) — What kind of ion will oxygen (O) form?
Oxygen (atomic number 8) has the configuration 2, 6 — 6 valence electrons, needing 2 more to complete its octet. It gains 2 electrons to form a divalent negative ion (anion), the oxide ion O^2-: O + 2e^– → O^2-.

Pause and Ponder (Page No. 174) — 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.
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).

Pause and Ponder (Page No. 174) — Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides.
Potassium (2, 8, 8, 1) loses 1 electron to form K⁺. Calcium (2, 8, 8, 2) loses 2 electrons to form Ca²⁺. Chlorine (2, 8, 7) gains 1 electron to form Cl^–. In potassium chloride (KCl), one electron lost by K is gained by one Cl atom, and the resulting K⁺ and Cl^– ions are held together by electrostatic attraction. In calcium chloride (CaCl₂), the two electrons lost by Ca are gained by two separate Cl atoms, giving Ca²⁺ and 2 Cl^– ions held together by ionic bonds.

Pause and Ponder (Page No. 174) — Illustrate how sodium sulfide (Na₂S) is formed.
Sodium (2, 8, 1) loses 1 electron each from two sodium atoms to form 2 Na⁺: 2Na → 2Na⁺ + 2e^–. Sulfur (2, 8, 6) gains these 2 electrons to complete its octet, forming S^2-: S + 2e^– → S^2-. The oppositely charged ions attract electrostatically to form the ionic compound: 2Na⁺ + S^2- → Na₂S.

Pause and Ponder (Page No. 177) — Name the following: (i) CO₂ (ii) NO₂ (iii) SF₆ (iv) PCl₃.
(i) Carbon dioxide (ii) Nitrogen dioxide (iii) Sulfur hexafluoride (iv) Phosphorus trichloride.

Pause and Ponder (Page No. 177) — Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide.
(i) NaHCO₃ (ii) SO₂ (iii) FeCl₃ (iv) Cu₂O.

Pause and Ponder (Page No. 177) — Write the formulae for the compounds formed from the following pairs of ions: (i) Fe³⁺ and OH^– (ii) K⁺ and CO₃^2-.
(i) Fe(OH)₃ — three OH^– ions balance one Fe³⁺ ion.
(ii) K₂CO₃ — two K⁺ ions balance one CO₃^2- ion.

Pause and Ponder (Page No. 179) — 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?
Such a compound has an ionic bond. In the solid state, the ions are locked in a rigid crystal lattice by strong electrostatic forces and cannot move, so there is no conduction. When dissolved in water, the lattice breaks apart and the ions become free to move, allowing the solution to conduct electricity.

Pause and Ponder (Page No. 179) — Metal M, with two electrons in its valence 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.
Metal M with configuration 2, 8, 2 is magnesium (Mg).
(i) Formula: MgO (Mg²⁺ and O^2- combine in a 1:1 ratio).
(ii) Type of bond: Ionic bond, formed by transfer of electrons from magnesium (metal) to oxygen (non-metal).
(iii) Conductivity: MgO reacts slightly with water to form magnesium hydroxide, Mg(OH)₂, which releases Mg²⁺ and OH^– ions into solution, so the aqueous solution does conduct electricity, though weakly given MgO’s low solubility.

Pause and Ponder (Page No. 179) — Find 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.

Pause and Ponder (Page No. 179) — Find the molecular mass of methane (CH₄). Atomic mass C = 12 u; H = 1 u.
Molecular mass = (1 × 12) + (4 × 1) = 12 + 4 = 16 u.

Pause and Ponder (Page No. 180) — Find the formula unit mass of potassium chloride (KCl). Atomic mass K = 39 u; Cl = 35.5 u.
Formula unit mass = 39 + 35.5 = 74.5 u.

Pause and Ponder (Page No. 180) — Find 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) + (2 × 1) = 24 + 32 + 2 = 58 u.

What if… (Page No. 174) — Could we see atoms directly? How would it help scientists, and what challenges would it cause?
Being able to see atoms directly would help scientists understand how atoms are arranged in substances, observe chemical reactions step-by-step at the atomic level, design new materials, medicines, and nanotechnology with greater precision, and directly verify atomic theories and models. However, it would also raise challenges: atoms are extraordinarily small, so it would require highly advanced (and expensive) instruments; the very act of observing atoms can disturb their position or behaviour, making accurate study difficult; and the sheer volume of atomic-scale data generated would be difficult to process and interpret.

Activity 9.1 — Let Us Investigate a Physical Change (Page No. 163).
Aim: To check whether mass is conserved when salt dissolves in water. Observation: The mass of the salt solution equals the sum of the mass of salt and the mass of water taken initially. Conclusion: Mass remains constant during a physical change such as dissolving.

Activity 9.2 — Let Us Investigate a Chemical Change (Page No. 163).
Aim: To compare mass conservation for the reaction between vinegar and baking soda in an open setup versus a closed setup. Set-up 1 (open flask): Brisk effervescence (CO₂ gas) is seen; the final balance reading is less than the initial reading because the gas escapes and is not weighed. Set-up 2 (flask sealed with a balloon): Effervescence again occurs, but the CO₂ gas inflates the balloon and stays within the system; the final reading equals the initial reading. Conclusion: Mass only appears to decrease when a gaseous product is allowed to escape; in a closed system, the Law of Conservation of Mass holds exactly.

Activity 9.3 — Let Us Verify the Law (Group Activity, Page No. 165).
Aim: To verify the Law of Conservation of Mass using the reaction between sodium sulfate and barium chloride solutions (which forms a white precipitate of barium sulfate). Observation: The combined mass of both flasks before mixing equals the combined mass after mixing and the precipitate forms. Conclusion: Mass remains constant during the chemical reaction, confirming the Law of Conservation of Mass.

Activity 9.4 — Let Us Experiment (Page No. 177–178).
Aim: To compare the solubility and electrical conductivity of ionic compounds (sodium chloride, copper sulfate, calcium chloride) and covalent compounds (sugar, camphor, naphthalene) in water versus kerosene/petrol. Observation: The ionic compounds dissolve in water (not in kerosene/petrol) and their aqueous solutions conduct electricity; camphor and naphthalene dissolve in kerosene/petrol (not water) and never conduct electricity; sugar dissolves in water but its solution does not conduct electricity since it does not form ions. Conclusion: Ionic compounds are generally water-soluble and conduct electricity in solution (or molten state) because they release free-moving ions; covalent compounds are generally soluble in organic solvents and do not conduct electricity because they do not form ions.

Why This Chapter Matters

Atomic Foundations of Matter is the bridge between the atomic structure ideas introduced in the previous chapter and the more advanced chemistry students will encounter from Class 10 onward — atoms and molecules, valency, and formula writing here directly set up periodic classification, chemical reactions and equations, and mole-based stoichiometry calculations in later years. Mastering how to read electron configurations to predict ionic versus covalent bonding, and how to confidently write and balance chemical formulae using valencies, is a skill that keeps recurring across every subsequent chemistry chapter, so getting comfortable with the reasoning here (not just memorising formulae) pays off well beyond this one chapter’s exams.

More on This Chapter

Extra Questions | Revision Notes | Formulas Handbook

Frequently Asked Questions

What is the difference between molecular mass and formula unit mass, and why does this chapter use both terms?
Molecular mass is used for covalent (molecular) compounds, which exist as discrete, independent molecules — for example, the molecular mass of CH₄ is 16 u. Formula unit mass is used for ionic compounds, which do not form single molecules but instead exist as a continuous crystal lattice of ions — for example, the formula unit mass of NaCl (23 + 35.5 = 58.5 u) refers to the mass of one “formula unit” (one Na⁺ and one Cl^–), not a molecule. Both are calculated the same way — by summing the atomic masses of all atoms shown in the formula — but the terminology reflects the different bonding and structure of the two compound types.

How do I decide whether two elements will form an ionic bond or a covalent bond?
Check the number of valence electrons of each element. If one element has few valence electrons (1-3, typically a metal) and readily loses them, while the other has many valence electrons (5-7, typically a non-metal) and readily gains them, electrons transfer completely and an ionic bond forms (e.g., Na and Cl). If both elements are non-metals that each need more electrons to complete their octet, they share electron pairs instead of transferring them, forming a covalent bond (e.g., two chlorine atoms, or carbon and oxygen in CO₂).

What is the “criss-cross method” for writing chemical formulae, and why are brackets used around some ions?
In the criss-cross method, the numerical value of the charge on the cation becomes the subscript of the anion, and the numerical value of the charge on the anion becomes the subscript of the cation (charge signs are dropped), then the ratio is simplified to the smallest whole numbers if possible. Brackets are placed around a polyatomic ion (such as NO₃^–, SO₄^2-, or OH^–) whenever its subscript in the final formula is greater than 1, to show that the subscript applies to the entire group of atoms in that ion and not just to a single atom within it — for example, Ca(OH)₂ means two whole OH groups, not two oxygen atoms and one hydrogen.

Why do ionic compounds conduct electricity in the molten or dissolved state but not as solids?
In the solid state, the positive and negative ions of an ionic compound are locked into fixed positions within a rigid crystal lattice by strong electrostatic forces, so they cannot move to carry an electric current. When the compound is melted or dissolved in water, this lattice breaks down and the ions become free to move independently, allowing them to carry charge and conduct electricity — this is the key experimental test (used in Activity 9.4) for distinguishing ionic compounds from covalent ones.

Why does an open container seem to show mass “lost” during a chemical reaction, when the Law of Conservation of Mass says mass cannot change?
The Law of Conservation of Mass always holds true, but it only applies to a closed system where every reactant and every product is accounted for. If a reaction produces a gas (like CO₂ from baking soda and vinegar, as in Activity 9.2) in an open container, that gas escapes into the surrounding air and is never weighed, making the measured final mass appear lower than the initial mass. Repeating the same reaction in a sealed container (so the gas is trapped and still weighed) shows that the initial and final masses are exactly equal, confirming that no mass was actually lost — it simply left the container unmeasured.

Written by Satish

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