Where Mindful Learning Begins
Where Mindful Learning Begins
NCERT Class 9 Science Chapter 9 question answers for Atomic Foundations of Matter (CBSE 2026-27), with every Think It Over, Activity, Pause & Ponder, Think as a Scientist, What If, Revise Reflect Refine and Journey Beyond question solved step by step. Read your Class 9 Science chapter 9 notes first, then check your answers and numericals here.
Atomic Foundations of Matter explains how atoms combine, and these Class 9 Science Chapter 9 NCERT solutions cover the law of conservation of mass and law of constant proportions numericals, Dalton's atomic theory, valency, ionic and covalent bonds, writing chemical formulae by the criss-cross method, and molecular mass and formula unit mass calculations, with every numerical worked step by step.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
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 them 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, as complete Class 9 Science Chapter 9 NCERT solutions with answers.
Verifying with weighing-balance activities that mass is conserved in reactions, and that compounds always combine in fixed ratios.
How atoms share electrons to form molecules like H₂, O₂ and H₂O, or transfer electrons to form ions like Na⁺ and Cl⁻.
Criss-cross method for writing chemical formulae, and calculating molecular mass and formula unit mass.
Water from different sources (rivers, borewells, the ocean) contains different dissolved impurities.
When the water is purified and analysed, it is always found to contain hydrogen and oxygen in the mass ratio 1:8.
For example, 9 g of purified water from any source gives 1 g of hydrogen and 8 g of oxygen.
This is the Law of Constant Proportions (Law of Definite Proportions) proposed by Proust.
O represents one atom of oxygen.
O₂ represents a molecule of oxygen, which is made of two oxygen atoms.
Each oxygen atom has 6 valence electrons and needs 2 more to complete its octet.
So two oxygen atoms share two electrons each, and are joined by a double bond (O=O) to form the O₂ molecule.
Common salt (NaCl) is an ionic compound. It is made of Na⁺ and Cl⁻ ions.
When salt dissolves in water, the ions become free to move, and the moving ions carry the electric current.
Sugar is a covalent compound. It dissolves in water as neutral molecules and does not form ions.
So there are no charged particles to carry the current.
Observation (sample readings):
Mass of water (50 mL) ≈ 50 g, and mass of salt ≈ 2 g.
Reading before the salt dissolves (Fig. 9.1a) ≈ 52 g.
On swirling, the salt dissolves and a clear solution is formed.
Reading after the salt dissolves (Fig. 9.1b) ≈ 52 g.
Torn paper: the mass of the paper pieces is equal to the mass of the whole paper.
Mass of the solution = mass of water + mass of salt, so there is practically no change in mass.
Step 8: A brisk effervescence (bubbles of carbon dioxide gas) is seen when baking soda mixes with vinegar.
Step 9: No. The final reading is slightly less than the initial reading.
Reason: The flask is open, so the carbon dioxide gas produced escapes into the air.
The mass of the escaped gas is not weighed, so the reading falls.
The mass is not destroyed. Only the gas has left the system.
Step 8: A brisk effervescence occurs and the balloon gets inflated.
Step 11: Yes. The final reading is the same as the initial reading.
Reason: The flask and balloon form a closed system, so the carbon dioxide gas cannot escape. It is collected in the balloon.
Total mass of vinegar and baking soda before the reaction = total mass of carbon dioxide and other substances after the reaction.
Step 7: A white precipitate is formed on mixing the two solutions.
Step 9: No. The reading after mixing is the same as the reading before mixing.
No gas is formed in this reaction, so nothing escapes from the open flasks.
Mass of reactants (before mixing) = mass of products (after mixing).
Table 9.2 (observations):
| Compound | Water | Kerosene | Petrol | Solid state conducts? | In water conducts? |
|---|---|---|---|---|---|
| Camphor | Insoluble | Soluble | Soluble | No | No |
| Sodium chloride | Soluble | Insoluble | Insoluble | No | Yes |
| Copper sulfate | Soluble | Insoluble | Insoluble | No | Yes |
| Sugar | Soluble | Insoluble | Insoluble | No | No |
| Naphthalene | Insoluble | Soluble | Soluble | No | No |
Step 9: Grouping the compounds
Group 1 (sodium chloride, copper sulfate): soluble in water, insoluble in kerosene and petrol, conduct electricity in water. These are ionic compounds.
Group 2 (camphor, naphthalene): insoluble in water, soluble in kerosene and petrol, do not conduct electricity. These are covalent compounds.
Sugar: it is a covalent compound, soluble in water, but it does not form ions, so its solution does not conduct electricity.
No, the law is not violated.
Ethanol burns in the oxygen of the air to form carbon dioxide and water vapour.
Both products are gases, so they escape into the air from the open beaker and no residue is left.
Mass of ethanol + mass of oxygen used = mass of carbon dioxide + mass of water vapour formed.
The mass of the products is not lost. It is just not collected in the beaker.
Given: mass of hydrogen = 20 g, mass of oxygen = 160 g
To find: mass of water formed
According to the Law of Conservation of Mass, the mass of the product = the total mass of the reactants.
Given: sulfur : oxygen = 40 : 60 by mass; mass of sulfur in the sample = 20 g
To find: mass of oxygen in the sample
The ratio of the masses of the elements in a compound is fixed.
Sulfur : oxygen = 40 : 60 = 2 : 3
Check: the sample has 20 g + 30 g = 50 g of the compound. Sulfur = 20/50 = 40% and oxygen = 30/50 = 60%.
Given: carbon : oxygen = 3 : 4; mass of carbon = 9 g
To find: mass of oxygen
3 g of carbon combines with 4 g of oxygen.
So 1 g of carbon combines with 4/3 g of oxygen.
In a compound, the elements are combined chemically in a fixed ratio by mass. For example, water always has hydrogen and oxygen in the mass ratio 1:8.
In a mixture, the components are only mixed physically and are not chemically combined.
They can be mixed in any proportion. For example, sugar and salt can be mixed in any ratio, and the composition of air varies from place to place.
Ratio of copper to oxygen for student X = 4 : 1
Ratio of copper to oxygen for student Y = 8 : 2
Both students obtained the same ratio, 4 : 1.
Assertion (A): It is true. The mass of water = 2 g + 16 g = 18 g, which obeys the Law of Conservation of Mass. Also H : O = 2 : 16 = 1 : 8.
Reason (R): It is false. Dalton's postulate says that atoms combine in the ratio of simple whole numbers, i.e., in terms of the number of atoms and not their masses.
The ratio by mass is generally not a simple whole number ratio. For example, in NaCl the mass ratio of Na to Cl is 23 : 35.5.
Nitrogen has 5 valence electrons (electronic configuration 2, 5).
Each nitrogen atom needs 3 more electrons to complete its octet.
So two nitrogen atoms share three pairs of electrons (each atom gives 3 electrons).
The two atoms are joined by a triple bond, N≡N. Each atom also has one lone pair.
Electronic configuration of fluorine (Z = 9) = 2, 7. It has 7 valence electrons.
Each fluorine atom needs one more electron to complete its octet.
So two fluorine atoms share one pair of electrons (each gives 1 electron).
They are joined by a single covalent bond, F—F. Each atom has three lone pairs.
Carbon (2, 4) has 4 valence electrons and needs 4 more.
Oxygen (2, 6) has 6 valence electrons and needs 2 more.
Carbon shares two electrons with each oxygen atom, so there are two double bonds: O=C=O.
Sulfur (2, 8, 6) needs 2 more electrons. Hydrogen (1) needs 1 more electron.
Sulfur shares one electron with each of two hydrogen atoms: H—S—H.
Nitrogen (2, 5) needs 3 more electrons. Hydrogen needs 1 more electron.
Nitrogen shares one electron with each of three hydrogen atoms (three single bonds), and has one lone pair.
Electronic configuration of neon (Z = 10) = 2, 8.
Its valence shell (L shell) already has 8 electrons, which is a complete octet.
An atom with a complete octet is stable, so it has no tendency to lose, gain or share electrons.
Oxygen (Z = 8) has the electronic configuration 2, 6.
It needs 2 more electrons to complete its octet, so it gains 2 electrons.
Number of protons = 8 and number of electrons = 8 + 2 = 10.
Chlorine (2, 8, 7) takes one electron and becomes Cl⁻ (2, 8, 8).
One Mg²⁺ ion gives 2 positive charges, and each Cl⁻ ion carries 1 negative charge.
So 2 Cl⁻ ions are needed to balance the charge of one Mg²⁺ ion.
Chlorine takes one electron.
One Mg²⁺ ion has a charge of 2+.
Two Cl⁻ ions have a total charge of 2–.
Potassium (Z = 19): configuration 2, 8, 8, 1. It loses 1 electron and becomes K⁺ (2, 8, 8).
Calcium (Z = 20): configuration 2, 8, 8, 2. It loses 2 electrons and becomes Ca²⁺ (2, 8, 8).
Chlorine (Z = 17): configuration 2, 8, 7. It gains 1 electron and becomes Cl⁻ (2, 8, 8).
Potassium chloride: K⁺ + Cl⁻ → KCl (one K⁺ for one Cl⁻).
Calcium chloride: Ca²⁺ + 2 Cl⁻ → CaCl₂ (one Ca²⁺ for two Cl⁻).
Sodium (2, 8, 1) has 1 valence electron. It loses it and becomes Na⁺.
Sulfur (2, 8, 6) has 6 valence electrons. It needs 2 electrons, so it gains 2 electrons and becomes S²⁻.
One S²⁻ ion needs two Na⁺ ions, each giving 1 electron.
Prefixes show the number of atoms: mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6).
1 carbon atom and 2 oxygen atoms. Mono is omitted for the first element.
1 nitrogen atom and 2 oxygen atoms.
1 sulfur atom and 6 fluorine atoms.
1 phosphorus atom and 3 chlorine atoms.
Na⁺ (charge 1+) and HCO₃⁻ (charge 1–). The charges are equal, so one of each combines.
It is a covalent compound. The prefix 'di' means 2 oxygen atoms and one sulfur atom.
Ferric is Fe³⁺ and chloride is Cl⁻.
Criss-cross the charges: Fe₁Cl₃, so the formula is FeCl₃.
Cuprous is Cu⁺ and oxide is O²⁻.
Criss-cross the charges: Cu₂O₁, so the formula is Cu₂O.
Symbols: Fe and OH. Charges: 3+ and 1–.
Criss-cross the charges (only the numbers): Fe has 1 and OH has 3.
Brackets are used for the polyatomic ion OH when it occurs more than once.
Symbols: K and CO₃. Charges: 1+ and 2–.
Criss-cross the charges: K has 2 and CO₃ has 1.
Brackets are not required when only one polyatomic ion is present.
In the solid state, the ions of an ionic compound are held in fixed positions, so the compound does not conduct electricity.
When it is dissolved in water, the ions become free to move, so the solution conducts electricity.
Covalent compounds do not give ions in water.
The valence shell is the M shell (third shell), so the configuration of M is 2, 8, 2 (atomic number 12, magnesium).
M loses 2 electrons to form M²⁺. Oxygen gains 2 electrons to form O²⁻.
Charges: M is 2+ and O is 2–.
Criss-cross: M₂O₂, which is simplified to MO.
A metal transfers electrons to a non-metal.
The dissolved part gives free M²⁺ and O²⁻-derived ions in the solution.
Free ions can carry current.
Given: H = 1 u, N = 14 u, O = 16 u
HNO₃ has 1 hydrogen atom, 1 nitrogen atom and 3 oxygen atoms.
Molecular mass of HNO₃:
Given: C = 12 u, H = 1 u
CH₄ has 1 carbon atom and 4 hydrogen atoms.
Given: K = 39 u, Cl = 35.5 u
KCl has 1 potassium and 1 chlorine in its formula unit.
Formula unit mass of KCl:
Given: Mg = 24 u, O = 16 u, H = 1 u
Mg(OH)₂ has 1 magnesium atom and 2 OH groups (2 oxygen and 2 hydrogen atoms).
Hypothesis: The mass of the reactants (zinc and dilute hydrochloric acid) = the mass of the products (zinc chloride and hydrogen).
Problem: Hydrogen gas is formed and can escape from an open flask. So a closed set-up is needed.
Procedure:
1. Take a clean, dry conical flask and pour about 20 mL of dilute hydrochloric acid into it.
2. Put a few small pieces of zinc granules in a small balloon, and fix the balloon to the mouth of the flask with a thread, so that the zinc does not fall into the acid.
3. Place the whole set-up on a digital weighing balance and record the initial reading.
4. Lift the balloon upwards so that the zinc falls into the acid.
5. Hydrogen gas is formed with brisk bubbling and it inflates the balloon.
6. When the reaction is over, place the whole set-up on the balance again and record the final reading.
Expected observation: The final reading is the same as the initial reading.
The hydrogen gas is collected in the balloon and cannot escape, so the whole mass is accounted for.
Precaution: Keep flames away, as hydrogen is combustible. Do not close the flask with a rigid stopper, as the gas pressure may build up.
A compound would not have a fixed composition, so the Law of Constant Proportions would not hold.
For example, water from different sources could contain different amounts of hydrogen and oxygen.
A substance with the same name could then have different properties each time it is made.
A compound like sodium chloride could differ from sample to sample, and chemical formulae would lose their meaning.
It would be difficult to make reliable medicines, foods, fertilisers and other materials.
How it would help:
Scientists could see how atoms are arranged in molecules and crystals, and how bonds form and break during a reaction.
They could test atomic models directly, and design better medicines, batteries and materials.
Challenges:
An atom is only about 10⁻¹⁰ m in size, so it would need very powerful instruments (like electron microscopes).
Even a tiny sample has a huge number of atoms, so the picture would be very complicated to understand.
Atoms are always in motion, and electrons do not move in well-defined paths, so a clear picture would be hard to get.
From Table 8.4, elements with fewer than four valence electrons are:
| Element | Configuration | Valence electrons |
|---|---|---|
| Lithium (Li) | 2, 1 | 1 |
| Beryllium (Be) | 2, 2 | 2 |
| Sodium (Na) | 2, 8, 1 | 1 |
| Magnesium (Mg) | 2, 8, 2 | 2 |
| Aluminium (Al) | 2, 8, 3 | 3 |
A sodium atom has 11 protons and 11 electrons, so it is neutral.
After losing one electron, it has 11 protons and 10 electrons.
Positive charge = +11 and negative charge = –10, so the net charge = +1.
Ionic compounds: in the molten state, the ions are free to move, so they conduct electricity.
Covalent compounds: they are made of neutral molecules, which have no free ions or electrons, so they do not conduct electricity even in the molten state.
Camphor and naphthalene are covalent compounds made of neutral molecules.
They are insoluble in water, so no ions are formed in water.
They do not have free ions or free electrons to carry the electric charge.
Given: calcium carbonate = 4.0 g; hydrochloric acid = 2.92 g; carbon dioxide = 1.76 g; water = 0.72 g; calcium chloride = 4.44 g
Total mass of the reactants:
Total mass of the products:
Mass of reactants = mass of products.
Given: 12 g of carbon gives 44 g of carbon dioxide
To find: carbon dioxide formed from 2.4 g of carbon
12 g of carbon gives 44 g of carbon dioxide.
So 1 g of carbon gives 44/12 g of carbon dioxide.
So 2.4 g of carbon gives
Given: Na : Cl = 23 : 35.5; mass of sodium = 46 g
To find: mass of chlorine needed
23 g of sodium combines with 35.5 g of chlorine.
So 1 g of sodium combines with 35.5/23 g of chlorine.
H₂O has 2 hydrogen atoms and 1 oxygen atom.
CO₂ has 1 carbon atom and 2 oxygen atoms.
Na₂O has 2 sodium atoms and 1 oxygen atom in its formula unit.
Formula unit mass of Na₂O:
Ca(NO₃)₂ has 1 calcium atom and 2 nitrate (NO₃) groups.
Mass of one NO₃ group = 14 + (16 × 3) = 62 u.
Element A has 1 electron in the third shell, so its configuration is 2, 8, 1 (atomic number 11, sodium).
Element B has 6 electrons in the second shell, so its configuration is 2, 6 (atomic number 8, oxygen).
A has 1 valence electron. It gives away 1 electron to get the stable configuration 2, 8.
After losing 1 electron, it has 11 protons and 10 electrons.
B has 6 valence electrons. It takes 2 electrons to complete its octet.
After gaining 2 electrons, it has 8 protons and 10 electrons.
A transfers electrons and B accepts them.
Charges: A is 1+ and B is 2–.
Criss-cross: A₂B₁, so the formula is A₂B.
X has 6 valence electrons, so it needs 2 more electrons to complete its octet. X is oxygen if it has two shells (configuration 2, 6).
Each atom of X needs 2 electrons for an octet.
Two atoms of X share two electrons each, and both get 8 electrons in the outer shell.
The atoms share two pairs of electrons.
Two shared pairs and two lone pairs on each atom (with X = O):
Y has two electrons in its second shell, so its configuration is 2, 2 (atomic number 4, beryllium).
Y has 2 valence electrons. It loses them, and X (needing 2 electrons) accepts them.
So X and Y do not share electrons. Y²⁺ and X²⁻ ions are formed and held by an ionic bond.
The formula is YX (BeO). It is an ionic compound and exists as a crystal, not as a separate molecule.
Total charge = number of ions × charge on each ion.
Positive charge = 2 × 3 = 6+
Negative charge = 3 × 1 = 3–
Positive charge = 3 × 2 = 6+
Negative charge = 1 × 3 = 3–
Positive charge = 2 × 3 = 6+
Negative charge = 3 × 2 = 6–
Positive charge = 3 × 2 = 6+
Negative charge = 2 × 2 = 4–
Elements are made up of only one kind of atom (some exist as molecules of the same atoms). Compounds are made up of atoms of different elements, combined as molecules or ions.
A molecule of a compound is made from atoms of different elements (for example, H₂O). A molecule with atoms of the same kind is a molecule of an element.
Nitrogen is N₂, with two nitrogen atoms joined by a triple bond.
The formula of water is H₂O. Hydrogen and oxygen share electrons, so the bonds are covalent.
Ions: Al³⁺ and NO₃⁻.
Criss-cross: Al has 1 and NO₃ has 3.
Brackets are needed for NO₃.
Ions: Ca²⁺ and O²⁻.
Criss-cross: Ca₂O₂, which is simplified to CaO.
Ions: Fe³⁺ and O²⁻.
Criss-cross: Fe₂O₃.
Charges: 2+ and 1–.
Criss-cross: Ca₁Br₂.
Charges: 3+ and 2–.
Criss-cross: Al₂(CO₃)₃.
Charges: 1+ and 2–.
Criss-cross: K₂(SO₄)₁. Brackets are not needed for one polyatomic ion.
Charges: 1+ and 1–. One ion of each combines.
A chlorine atom has 17 electrons (2, 8, 7).
The Cl⁻ ion has gained one electron, so it has 17 + 1 = 18 electrons, with the configuration 2, 8, 8.
Counting the electrons in each diagram of Fig. 9.18:
| Diagram | K | L | M | Total electrons |
|---|---|---|---|---|
| (i) | 2 | 7 | 8 | 17 |
| (ii) | 2 | 8 | 8 | 18 |
| (iii) | 2 | 8 | 9 | 19 |
| (iv) | 2 | 8 | 7 | 17 |
(i) has 17 electrons with a wrong arrangement (2, 7, 8). (iii) has 19 electrons. (iv) is a neutral chlorine atom.
Atomic masses used: H = 1 u, C = 12 u, N = 14 u, O = 16 u, Na = 23 u, P = 31 u.
It has 2 N, 4 H and 3 O atoms.
It has 3 H, 1 P and 4 O atoms.
It has 1 Na, 1 H, 1 C and 3 O atoms.
Mg gives Mg²⁺ and nitrogen gives N³⁻.
Criss-cross: Mg₃N₂.
Li gives Li⁺ and nitrogen gives N³⁻.
Criss-cross: Li₃N.
Na gives Na⁺ and sulfur gives S²⁻.
Criss-cross: Na₂S.
Al gives Al³⁺ and oxygen gives O²⁻.
Criss-cross: Al₂O₃.
| NO₃⁻ | SO₄²⁻ | PO₄³⁻ | |
|---|---|---|---|
| NH₄⁺ | |||
| Li⁺ | LiNO₃ | ||
| Al³⁺ | |||
| Cu²⁺ |
Write the cation first and then the anion, and criss-cross the charges. Use brackets for a polyatomic ion that occurs more than once.
NH₄⁺: NH₄NO₃, (NH₄)₂SO₄, (NH₄)₃PO₄
Li⁺: LiNO₃ (given), Li₂SO₄, Li₃PO₄
Al³⁺: Al(NO₃)₃, Al₂(SO₄)₃. For Al³⁺ and PO₄³⁻, the charges are equal, so the formula is AlPO₄.
Cu²⁺: Cu(NO₃)₂. For Cu²⁺ and SO₄²⁻, the charges are equal, so the formula is CuSO₄. For PO₄³⁻, Cu₃(PO₄)₂.
| 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 |
Given: sodium carbonate = 5.3 g; acetic acid = 6.0 g; carbon dioxide = 2.2 g; water = 0.9 g; sodium acetate = 8.2 g
Total mass of the reactants:
Total mass of the products:
Given: protons = 11, neutrons = 12, electrons = 10
Atomic number = number of protons = 11.
Mass number = protons + neutrons = 11 + 12 = 23.
Positive charge = +11 and negative charge = –10.
Net charge = +1, because protons are more than electrons.
It has 10 electrons: K = 2, L = 8.
Atomic number 11 is sodium, and the charge is +1.
A has 15 electrons (phosphorus). A needs 3 more electrons for an octet, so its valency is 3.
B has 17 electrons (chlorine). B needs 1 more electron for an octet, so its valency is 1.
B needs only 1 electron to complete its octet, so it gains the electron very easily.
A needs 3 electrons, which is harder.
Both A and B are non-metals and both need electrons.
Neither can lose electrons easily, so they share electrons.
A shares 3 electrons, one with each of three B atoms.
Valency of A = 3 and valency of B = 1.
Criss-cross: A₁B₃, so the formula is AB₃.
Assertion (A): It is true. Copper sulfate is an ionic compound. In the molten state its ions are free to move, but in the solid state they are not.
Reason (R): It is false. It states the opposite. In the solid state, the ions are fixed in the lattice, and in the molten state they can move freely.
Number of neutrons = mass number − number of protons.
Number of electrons = number of protons − charge on the species (a positive charge means electrons are lost, and a negative charge means electrons are gained).
| Species | Protons | Electrons | Neutrons |
|---|---|---|---|
| \({}^{27}_{13}\text{Al}\) | 13 | 13 (neutral atom) | 27 − 13 = 14 |
| \({}^{80}_{35}\text{Br}\)⁻ | 35 | 35 + 1 = 36 | 80 − 35 = 45 |
| \({}^{201}_{80}\text{Hg}\)²⁺ | 80 | 80 − 2 = 78 | 201 − 80 = 121 |
Aim: To show that water from different sources has hydrogen and oxygen in the same ratio.
Procedure (electrolysis of water):
1. Collect water samples from different sources, such as tap water, rainwater and river water.
2. Purify each sample by distillation.
3. Take a Hofmann voltameter (or two inverted test tubes over electrodes) and fill it with the water sample, adding a few drops of dilute sulfuric acid to make it conduct.
4. Pass an electric current through the water.
5. Collect hydrogen at the negative electrode (cathode) and oxygen at the positive electrode (anode).
6. Note the volumes of the two gases, and repeat with each water sample.
Expected observation: The volume of hydrogen is twice the volume of oxygen for every sample (2 : 1).
The mass ratio is then (2 × 2) : (1 × 32) = 4 : 32 = 1 : 8 for every sample.
Take sodium, magnesium and chlorine.
The number of electrons is counted before (atom) and after (ion) the formation of the ion.
| Element | Ion formed | Electrons in atom | Electrons in ion |
|---|---|---|---|
| Sodium (Z = 11) | Na⁺ | 11 | 11 − 1 = 10 |
| Magnesium (Z = 12) | Mg²⁺ | 12 | 12 − 2 = 10 |
| Chlorine (Z = 17) | Cl⁻ | 17 | 17 + 1 = 18 |
Sodium and magnesium lose electrons (cations), so the bar for the ion is shorter.
Chlorine gains an electron (anion), so the bar for the ion is taller.
Making the cards: Make one card for each ion, showing its symbol and charge. Make 3 or 4 copies of each card.
| Cation cards | Anion cards |
|---|---|
| Na⁺, K⁺, Li⁺, Ag⁺, NH₄⁺ | Cl⁻, Br⁻, OH⁻, NO₃⁻ |
| Mg²⁺, Ca²⁺, Zn²⁺, Cu²⁺ | O²⁻, S²⁻, SO₄²⁻, CO₃²⁻ |
| Al³⁺, Fe³⁺ | PO₄³⁻ |
Rules:
1. Deal 5 cards to each player and keep the rest as the pile.
2. On your turn, pick a card from the pile or from the open cards, or ask another player for a card.
3. Match cations and anions so that the total positive charge = the total negative charge, and write the formula of the compound.
4. Discard one unwanted card at the end of your turn.
5. The player who forms the most correct compounds (with correct formulae) wins.
Examples of correct matches:
Mg²⁺ + 2 Cl⁻ → MgCl₂
Al³⁺ + 3 OH⁻ → Al(OH)₃
2 Na⁺ + SO₄²⁻ → Na₂SO₄
2 Al³⁺ + 3 O²⁻ → Al₂O₃
No chemical change violates the Law of Conservation of Mass. It applies to every chemical reaction.
Sometimes it looks as if mass is lost, as in Activity 9.2 (set-up 1), but this happens because a gas escapes from an open system.
In a closed system, the mass of the reactants is always equal to the mass of the products.
In nuclear reactions (nuclear energy), a very small amount of mass changes into energy. But these are nuclear changes and not chemical changes.
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.
Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.
Given: mass of carbon = 3 g; mass of carbon dioxide = 11 g
By the Law of Conservation of Mass, mass of reactants = mass of products.
3 g of carbon gives 11 g of carbon dioxide.
So 6 g of carbon (twice as much) gives
Ratio of hydrogen to oxygen in sample A = 2 : 16 = 1 : 8
Ratio of hydrogen to oxygen in sample B = 3 : 24 = 1 : 8
Both samples have hydrogen and oxygen in the same mass ratio, 1 : 8.
Given: atomic number \(Z = 13\), mass number \(A = 27\), charge = 3+
Number of protons = atomic number = 13.
Number of neutrons = mass number − atomic number = 27 − 13 = 14.
A neutral atom has 13 electrons. The ion has lost 3 electrons, so the number of electrons = 13 − 3 = 10.
Given: Ca = 40 u, Cl = 35.5 u
CaCl₂ has 1 calcium atom and 2 chlorine atoms.
Charges: Ca is 2+ and PO₄ is 3–.
Criss-cross: Ca has 3 and PO₄ has 2.
Brackets are needed for PO₄ because there are two of them.
Ca₃(PO₄)₂ has 3 Ca atoms and 2 PO₄ groups.
Mass of one PO₄ group = 31 + (16 × 4) = 95 u.
1 g of hydrogen needs 8 g of oxygen.
So 40 g of oxygen needs
Hydrogen available = 10 g, so hydrogen left over = 10 − 5 = 5 g.
The water formed uses all the oxygen (40 g) and 5 g of hydrogen.
P (Z = 12): K = 2, L = 8, M = 2.
Q (Z = 17): K = 2, L = 8, M = 7.
P has 2 valence electrons and loses them to form P²⁺.
Q needs 1 electron and forms Q⁻.
Electrons are transferred, so the bond is ionic.
Criss-cross the charges: P₁Q₂.
PQ₂ has 1 P and 2 Q.
In the solid state, the ions are fixed in the lattice and cannot move, so X does not conduct.
In water, the ions become free to move, so the solution conducts electricity.
Revisit Chapter 8 on the Journey Inside the Atom, browse the full Class 9 Science hub, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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