NCERT Solutions Class 9 Science Chapter 9 Atomic Foundations of Matter

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 9 · SCIENCE

NCERT Solutions

Chapter 9: Atomic Foundations
of Matter

NCERT Textbook — Exploration

BOOK PAGES SECTIONS QUESTIONS MEDIUM

162 – 183 23 63 English

Solutions, notes, sample papers & more at 69 pages

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

CLASS 9 · SCIENCE · EXPLORATION

NCERT Solutions — Chapter 9: Atomic Foundations of
Matter
Two experiments with a digital balance — one with salt and water, one with vinegar and baking soda — settle
a question chemistry had argued over for centuries: does matter get made or destroyed when substances
react? From those weighings this chapter builds the two laws (conservation of mass, constant proportions),
Dalton's atomic theory, the covalent and ionic bond, chemical formulae, and the arithmetic of molecular
mass and formula unit mass.

TEXTBOOK BOOK PAGES

Exploration (Class 9) 162 – 183

SECTIONS QUESTIONS

23 63

MEDIUM

English

Think It Over — Page 162
Chapter opener

THINK IT OVER

Q1 Water can be obtained from various sources. Are all these samples of water
chemically identical?

The water itself is identical; the samples are not. Once each sample is purified, every one of
them is the same compound H₂O, with hydrogen and oxygen combined in the fixed mass ratio 1
: 8 — that is exactly what the Law of Constant Proportions says.
What differs is everything the water is carrying with it:

River water — suspended silt, dissolved oxygen, traces of organic matter.
Borewell water — dissolved calcium and magnesium salts (this is what makes water “hard”),
sometimes iron or fluoride.
Sea water — about 35 g of dissolved salts per kilogram, mostly sodium chloride.
Rain water — dissolved carbon dioxide and other gases from the air.

So these are mixtures, and a mixture may have its components in any proportion. Distil any of
them and what comes over is pure water — 9 g of it will always decompose into 1 g of hydrogen
and 8 g of oxygen.

Page 1 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: the composition of a compound is fixed by how its atoms are
bonded — two H atoms to one O atom, always. Nothing about the source can
change that. The composition of a mixture is not fixed by any bond, so it can vary
freely.

Q2 Oxygen is sometimes represented as O and sometimes as O2. What is the difference
between these symbols?

O stands for one atom of oxygen; O₂ stands for one molecule of oxygen gas, which
contains two oxygen atoms joined by a double bond.

SYMBOL WHAT IT NUMBER OF MASS DOES IT EXIST ON ITS OWN?
NAMES ATOMS

O one oxygen atom 1 16 u No — a lone O atom has only 6
valence electrons, so it is unstable

O₂ one molecule of 2 32 u Yes — both atoms have a complete
oxygen gas octet

The symbol O is used when you are counting atoms inside a formula — in H₂O, in CO₂, in CaO.
The formula O₂ is used when you mean the gas you breathe.

Why it happens: an oxygen atom is 2, 6 — it is two electrons short of an octet. Two
oxygen atoms solve the problem together by sharing two pairs of electrons, so each
one now “sees” eight valence electrons. That is the double bond O=O, and it is why
free oxygen in air is O₂ and not O.

Did you know? Oxygen atoms can also join in threes to give ozone, O₃. Same
element, different molecule, completely different behaviour — O₂ keeps us alive at
ground level, O₃ shields us from ultraviolet rays high in the atmosphere.

Q3 Why does dissolved salt in water conduct electricity, but sugar does not?

Because salt breaks into free ions in water and sugar does not. An electric current in a
solution is carried by charged particles that are free to move; sugar solution has none.

Page 2 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Common salt, NaCl, is an ionic compound. The solid is already made of Na⁺ and Cl⁻ ions
locked in a lattice. Water pulls that lattice apart, and the ions become free to move. Switch on
the battery and Na⁺ drifts to the negative electrode, Cl⁻ to the positive one — that drift is the
current, and the bulb glows.
Sugar is a covalent compound. It dissolves well, but what floats away into the water are
whole, electrically neutral sugar molecules. No charges are released, nothing drifts to either
electrode, and the bulb stays dark.

Why it happens: dissolving and ionising are two different things. Both salt and
sugar dissolve; only salt ionises. That is the whole difference between an ionic and a
covalent compound in water.

Check it yourself: solid salt does not conduct either. In the dry crystal the ions are
held in fixed positions, so although the charges are there, they cannot move. Free
charge is not enough — it must also be mobile.

Activity 9.1 — Page 163
Let us investigate a physical change

ACTIVITY 9.1

Q1 What do you observe? (after swirling until the added salt dissolves)

The reading on the balance does not change. Before swirling, the pan carries water plus a
heap of undissolved salt; after swirling it carries a clear salt solution — and the number on the
display is the same.

Mass of water taken (50 mL) ≈ 50.0 g

Mass of salt added (one spatula) ≈ 2.0 g

Expected mass of solution = 50.0 g + 2.0 g = 52.0 g

Mass actually read after dissolving = 52.0 g (within ±0.1 g)

So mass of solution = mass of water + mass of salt. Dissolving a solid in water is a physical
change, and no mass is gained or lost in it.

Page 3 of 69

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Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the salt has not gone anywhere. Its Na⁺ and Cl⁻ ions have simply

m l a se
spread out among the water molecules instead of sitting stacked in a crystal. Every
o
.c was on the pan before is still on the pan. Nothing has
a gbeen added,
m
particle that

l a se has escaped — so the balance cannot read anything different.
ag
nothing

o m
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Try This: repeat with a sheet of paper. Weigh it, tear it into small pieces on the pan,
.
m the shape and size of the pieces, not
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weigh again. Same reading. Tearing changes
a
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the amount of matter.

co m
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coma chemical change
Activity 9.2 — Pages 163 – 165
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Let us investigate

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ACTIVITY 9.2

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Q1 (Experimental set-up 1, step 8) What do you observe?

se m
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ANSWER a
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A brisk effervescence — the mixture fizzes strongly as soon as the baking soda meets the

. co
vinegar, and the balance reading starts falling and then settles at a value lower than the initial
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com is as
one.

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The reaction

a s
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Vinegar + Baking soda (sodium hydrogencarbonate) → Carbon dioxide + Other substances

a se
. com a g l
e m
s and leaves the balance pan altogether.
The fizzing is carbon dioxide gas being produced. In this set-up the flask is open, so the gas

g l aflask
a
bubbles out of the mouth of the

co
Why it happens: the gas is a product, so its mass belongs in the product total. But a
m
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balance can only weigh what is standing on it. Once CO₂ has drifted away into the
m l as
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room, it is no longer being weighed — and the display drops by exactly the mass of

em a
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gas that left.
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.co


a g l Page 4 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Q2 (Experimental set-up 1, step 9) Are the initial and the final readings same?

No. The final reading is less than the initial reading.
Estimate how much less. About 2 g of baking soda is used, and vinegar is in excess, so
essentially all of it reacts.

Formula unit mass of NaHCO₃ = 23 u + 1 u + 12 u + (16 u × 3) = 84 u

Molecular mass of CO₂ = 12 u + (16 u × 2) = 44 u

Each 84 u of NaHCO₃ gives 44 u of CO₂

Mass of CO₂ from 2.0 g = 2.0 g × 44 ÷ 84 = ≈ 1.0 g

So you should expect a drop of roughly 1 g — easily visible on a digital balance, and far bigger
than the ±1 in the last digit that any measurement carries.

Tip: a fall of about a gram is the signature of a gas escaping. If your reading had
dropped by only 0.01 g you would be looking at experimental error, not at a real loss.

Q3 A brisk effervescence is observed. The final reading does not match the initial
reading. What can be the reason for this?

Because the carbon dioxide produced escapes from the open flask, and mass that has left
the pan cannot be weighed.
Mass is conserved in the reaction. Write the full account:

Mass of vinegar + Mass of baking soda = Mass of CO₂ + Mass of other substances

But the balance now reads only: Mass of other substances (+ flask + balloon)

So, Final reading = Initial reading − Mass of CO₂ that escaped

The Law of Conservation of Mass has not been broken. The experiment was faulty, not the law
— it was carried out in an open system, where a product is free to leave.

Page 5 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

CO₂ escapes

final < initial final = initial

Open system (set-up 1) Closed system (set-up 2)
gas leaves the balance gas stays on the balance

Left: in the open flask the CO₂ walks off the balance and the reading falls. Right: with the balloon tied
on, the same gas is trapped, stays on the pan, and the reading holds.

Why it happens: a chemical balance measures the matter inside its boundary. To test
a conservation law you must first make sure nothing can cross that boundary. That
is exactly the correction made in set-up 2, where the balloon seals the flask.

Q4 (Experimental set-up 2, step 8) What do you observe?

The same brisk effervescence — and this time the balloon inflates. As soon as the baking
soda is tipped into the vinegar the mixture fizzes, and the carbon dioxide, having nowhere else
to go, blows the balloon up.
The balance reading, meanwhile, stays where it was.

Why it happens: the gas is still being made in exactly the same amount as in set-up
1 — about 1 g of CO₂ from 2 g of baking soda. The difference is only that the balloon
keeps it inside the system. Its mass is still resting on the pan, so the balance has no
reason to change.

Page 6 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Check it yourself: the inflated balloon does push a little air aside and so feels a
slight upthrust, but the effect is a few milligrams and is swallowed by the ±1 in the
last digit of the balance.

Q5 (Experimental set-up 2, step 11) Are the initial and the final readings same in this
case?

Yes. The final reading matches the initial reading (allowing for the usual uncertainty of ±1 in the
last digit).

Mass of vinegar + Mass of baking soda (before) = Mass of CO₂ + Mass of other

substances (after)

Total mass before reaction = Total mass after reaction

This is the experimental proof of the Law of Conservation of Mass: in a chemical reaction,
matter can neither be created nor destroyed.

Why it happens: the atoms present at the start — the sodium, carbon, hydrogen
and oxygen atoms of baking soda and vinegar — are all still present at the end. They
have only been rearranged into new substances. Since no atom was destroyed and
none appeared from nowhere, the total mass cannot change.

Tip: set-ups 1 and 2 differ in exactly one thing — whether the system is sealed.
Changing one variable at a time is how you make an experiment prove something.

Activity 9.3 — Page 165

Page 7 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Let us verify the law — Group activity

ACTIVITY 9.3

Q1 (Step 7) What do you observe?

A milky white precipitate appears at once when the barium chloride solution is poured into
the sodium sulfate solution. The clear liquid turns cloudy white.

Sodium sulfate + Barium chloride → Barium sulfate + Sodium chloride

Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2 NaCl

The white solid is barium sulfate, which is insoluble in water and so settles out. The sodium
chloride formed alongside it stays dissolved and is invisible.

Why it happens: both starting solutions contain free ions — Na⁺ and SO₄²⁻ in one
flask, Ba²⁺ and Cl⁻ in the other. When they meet, Ba²⁺ and SO₄²⁻ attract each other so
strongly that they lock together into a solid lattice instead of staying in solution. Na⁺
and Cl⁻ have no such tendency, so they remain free. A new insoluble solid formed in
this way is called a precipitate, and its appearance is proof that a chemical change
has occurred.

Q2 (Step 9) Do you observe any change in the reading after mixing the solutions?

No — the reading is unchanged. The total mass of both conical flasks together is the same
after mixing as it was before.

Mass of Na₂SO₄ solution + Mass of BaCl₂ solution

= Mass of BaSO₄ precipitate + Mass of NaCl solution
Total mass before = Total mass after

This verifies the Law of Conservation of Mass for a reaction in which a solid is formed.

Page 8 of 69

ase
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: unlike Activity 9.2, this reaction produces no gas. Nothing can

m l a se
leave the flasks, so an ordinary open beaker is a closed system as far as mass is
o g same.
.c Every atom stays on the pan; the balance must readathe
m
concerned.

l a se
ag Tip: the book asks you to keep both flasks on the pan throughout, even the empty

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one. A little solution always clings to the walls of flask B during the transfer; if you
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take flask B off, that clinging film leaves
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loss of mass.

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com of Mass
Think as a Scientist — Page 166
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9.1 Law of Conservation

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THINK AS A SCIENTIST

om a s
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Q1 You are given a chemical reaction in which zinc reacts with dilute hydrochloric acid

s e m gas. Zinc + Hydrochloric acid (dilute) → Zinc
to form zinc chloride and hydrogen
a and perform an experiment to test the hypothesis that
agl
chloride + Hydrogen. Design
mass is conserved during the chemical reaction. You may use a set-up different
from the one shown in Activity 9.2.
co m
m .
m as e
.co a g l
a s em
Hypothesis: in the reaction Zn + 2 HCl → ZnCl₂ + H₂, the total mass after the reaction equals the

agl total mass before it, provided no hydrogen is allowed to escape.
The design problem. Hydrogen is the lightest of all gases and it comes off fast. So the whole
se m
com
experiment turns on one thing: the system must be sealed, and it must be sealed before the zinc
g l a
m . a
ase
touches the acid.

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Apparatus: a 250 mL conical flask, a small test tube that fits inside it, a balloon and thread (or a
rubber stopper carrying a small balloon), granulated zinc, dilute hydrochloric acid, and a digital
balance reading to 0.01 g.
co m
m .
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Method
m l as
.co
1. Put about 25 mL of dilute HCl into the conical flask.
a g
a s em2. Put about 1.3 g of granulated zinc into the small test tube and lower the test tube upright
agl into the flask, so that the zinc cannot yet fall into the acid.
.c
3. Stretch the balloon over the mouth of the flask and tie it on tightly with thread.
s e m
. c om
4. Place the whole sealed assembly on the balance and record the initial reading, m₁.
a g la
s e mpan, tilt it so that the test tube topples and the zinc falls into
glaat once and the balloon starts to inflate.
5. Without lifting the flask off the

a
the acid. Bubbles appear
6. Wait until the fizzing stops completely, then record the final reading, m₂.

co m
m .
m ase
.co


a g l Page 9 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

7. Repeat the whole run three times.

Why 1.3 g of zinc? Because you must produce enough hydrogen for the balance to notice.

Atomic mass of Zn = 65 u; molecular mass of H₂ = 2 u

Zn + 2 HCl → ZnCl₂ + H₂

65 u of Zn gives 2 u of H₂

Mass of H₂ from 1.3 g of Zn = 1.3 g × 2 ÷ 65 = 0.04 g

That is four times the 0.01 g resolution of the balance, so an open-flask control run would show
a clear, readable loss. Check the mass balance of the equation itself:

Reactants: 65 u (Zn) + 2 × 36.5 u (HCl) = 138 u

Products: 136 u (ZnCl₂) + 2 u (H₂) = 138 u ✓

Expected result

RUN INITIAL READING M₁ FINAL READING M₂ M₂ − M₁

1 (sealed) 176.42 g 176.42 g 0.00 g

2 (sealed) 178.05 g 178.04 g −0.01 g

3 (open control) 175.90 g 175.86 g −0.04 g

Conclusion: in the sealed runs m₂ = m₁ to within ±0.01 g — the uncertainty of the last digit — so
mass is conserved. The open control loses 0.04 g, exactly the calculated mass of hydrogen,
which shows that the “loss” in an open vessel is a loss of gas from the pan and not a loss of
matter from the universe.

Safety first: hydrogen is highly flammable. Do this away from any flame, do not use
a stopper so tight that pressure can build up, and let the balloon take the gas. Dilute
acid burns skin — wear eye protection.

Why it happens: the zinc atoms have not vanished; they are in solution as Zn²⁺. The
hydrogen that was attached to chlorine in HCl has simply been set free as H₂. Atoms
were rearranged, not created or destroyed — which is precisely Dalton's second
postulate, and the reason the Law of Conservation of Mass holds.

Page 10 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Pause and Ponder — Page 166
9.1 Law of Conservation of Mass

PAUSE AND PONDER

Q1 A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is
left in the beaker. Does this mean the Law of Conservation of Mass is violated?
Explain.

No, the law is not violated. Nothing is left in the beaker because every product of the
combustion is a gas, and in an open beaker the gases float away. The matter has moved, not
disappeared.
Ethanol burns in the oxygen of the air:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

Now do the mass accounting. Take H = 1 u, C = 12 u, O = 16 u.

Molecular mass of ethanol C₂H₅OH = (12 × 2) + (1 × 6) + 16 = 46 u

Oxygen used: 3 O₂ = 3 × 32 u = 96 u, so for 10 g of ethanol

Mass of O₂ = 10 g × 96 ÷ 46 = 20.87 g

Carbon dioxide formed: 2 CO₂ = 88 u → 10 g × 88 ÷ 46 = 19.13 g

Water formed: 3 H₂O = 54 u → 10 g × 54 ÷ 46 = 11.74 g

Total mass of reactants = 10 g + 20.87 g = 30.87 g

Total mass of products = 19.13 g + 11.74 g = 30.87 g

Mass of reactants = Mass of products ✓

So the balance would have shown a fall — but only because 30.87 g of gas has drifted out of the
beaker while 20.87 g of oxygen quietly came in from the air. Neither of those crossings was
weighed.

Page 11 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: a conservation law is a statement about a closed system. Burning
in an open beaker is about as open as a system gets — matter enters as oxygen and
leaves as carbon dioxide and steam. To test the law you must seal the system,
exactly as set-up 2 of Activity 9.2 does.

Check it yourself: burn a candle inside a closed jar standing on a balance. The
reading stays constant while the flame burns, even though wax is disappearing —
because the CO₂ and water vapour are trapped inside.

Q2 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?

180 g of water.

By the Law of Conservation of Mass,

Mass of water = Mass of hydrogen + Mass of oxygen

= 20 g + 160 g

= 180 g

Cross-check with the Law of Constant Proportions. Water always has hydrogen and oxygen
in the mass ratio 1 : 8.

20 g : 160 g = 1 : 8 ✓

The ratio is exactly right, which is why the question can say the hydrogen reacts completely —
neither gas is left over. If you had taken 20 g of hydrogen and only 100 g of oxygen, all 100 g of
oxygen would react with just 12.5 g of hydrogen, giving 112.5 g of water and leaving 7.5 g of
hydrogen unused.

Why it happens: the two laws work together. Conservation of mass fixes the total;
constant proportions fixes how much of each reactant can actually take part. Adding
more of one reactant beyond the fixed ratio does not make more product — the
extra simply stays behind.

Page 12 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Pause and Ponder — Page 167
9.2 Law of Constant Proportions

PAUSE AND PONDER

Q3 A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the
same compound containing 20 g of sulfur, what mass of oxygen must be present to
satisfy the Law of Constant Proportions?

30 g of oxygen.

In the compound, S : O = 40 : 60 = 2 : 3 by mass

So, mass of oxygen = mass of sulfur × 3 ÷ 2

= 20 g × 3 ÷ 2

= 30 g

Check: total mass of sample = 20 g + 30 g = 50 g. Percentage of sulfur = 20 ÷ 50 × 100 = 40 %,
and of oxygen = 30 ÷ 50 × 100 = 60 % ✓

Why it happens: the Law of Constant Proportions says a compound has a fixed
composition by mass no matter how big the sample or where it came from. The
percentages therefore stay 40 % and 60 % whether you have 5 g of the compound or
5 kg, so the S : O mass ratio 2 : 3 must hold in every sample.

Did you know? With S = 32 u and O = 16 u, the ratio 20 g : 30 g means 20/32 : 30/16
= 0.625 : 1.875 = 1 : 3 by number of atoms — so the compound is SO₃, sulfur trioxide.

Q4 Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How
much oxygen will combine with 9 g of carbon to form carbon monoxide?

12 g of oxygen.

Page 13 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
em.
C : O = 3 : 4 by mass
m l as
.co
Mass of oxygen = mass of carbon × 4 ÷ 3
m a g
l a se
g
=9g×4÷3
a= 12 g

. c om ag
s e m monoxide formed = 9 g + 12 g = 21 g, by the
Check: 9 g : 12 g = 3 : 4 ✓. Total mass of carbon
a
agl
Law of Conservation of Mass.

Why it happens: the ratio 3 : 4 is just the atomic masses in disguise. One CO
co m
molecule has one C atom (12 u) and one O atom (16 u), and 12 : 16 = 3 : 4. Because
em.
m l as
.co g
every molecule of the compound is built the same way, the mass ratio in any
m a
se
quantity of it is the same 3 : 4.

g l a
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Tip: if you were asked for carbon dioxide instead, the ratio would be 12 : 32 = 3 : 8,
m a
.co
and 9 g of carbon would need 24 g of oxygen. Same two elements, different
m agl
a se
compound, different fixed ratio — the law is about one compound at a time.
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The Law of Definite Proportions holds true for compounds but not for mixtures.

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Q5

m l
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Give reason.

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Because a compound is held together by chemical bonds in a fixed pattern of atoms, while

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a mixture is only a physical jumble whose parts can be present in any amount.
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COMPOUND ase
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MIXTURE

How the parts By chemical bonds — a fixed number of atoms of Not held at all; the components
each element per molecule or formula unit
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simply lie together
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are held

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Composition by Fixed — water is always 1 : 8 H : O
g l as Any value you like

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salt in water, in any ratio

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New properties, quite unlike those of the Each component keeps its own
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Properties
elements
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m ase
.co


a g l Page 14 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

In water, two hydrogen atoms are bonded to one oxygen atom, always. That count cannot be
changed without destroying the substance; change it and you no longer have water. So the
mass ratio is locked at 2 × 1 u : 1 × 16 u = 1 : 8.
A mixture has no such lock. You can stir 1 g of salt into a litre of water, or 30 g, and both are still
salt solution.

Why it happens: the fixed ratio comes from Dalton's postulate that atoms combine
in the ratio of simple whole numbers. A whole number cannot slide gradually —
there is no molecule of water with 2.3 hydrogen atoms. Mixing involves no atoms
combining at all, so nothing constrains the proportions.

Q6 Students X and Y, both prepared an oxide of copper by combining copper and
oxygen in the ratios of 4:1 and 8:2, respectively. Do their results justify the Law of
Constant Proportions? Explain.

Yes, their results justify the law — because 8 : 2 is the same ratio as 4 : 1.

Student X: Cu : O = 4 : 1
Student Y: Cu : O = 8 : 2 = (8 ÷ 2) : (2 ÷ 2) = 4 : 1

Both ratios are identical.

Compare them as percentages, which is the fairest test:

STUDENT CU : O AS TAKEN % COPPER % OXYGEN

X 4:1 4/5 × 100 = 80 % 1/5 × 100 = 20 %

Y 8:2 8/10 × 100 = 80 % 2/10 × 100 = 20 %

Same composition by mass, from two independent preparations — which is exactly what the
Law of Constant Proportions predicts. Student Y has simply made twice as much of the same
oxide.

Why it happens: a ratio measures proportion, not amount. Doubling both numbers
doubles the size of the sample but leaves the composition untouched, just as 2
chapatis for 1 bowl of dal is the same recipe as 4 chapatis for 2 bowls.

Page 15 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Did you know? With Cu = 63.5 u and O = 16 u, a 4 : 1 mass ratio means 4/63.5 : 1/16
≈ 1 : 1 by number of atoms — so both students have made CuO, black copper(II)
oxide.

What if … — Page 168
9.3 Dalton's Atomic Theory

WHAT IF …

Q1 What if … atoms could combine in any ratio and not in a fixed ratio? How would this
affect the substances around us?

Nothing around us would have a reliable identity. Fixed ratios are what make a substance a
substance; drop them and chemistry as a science collapses.

No substance could be named. “Water” would not mean one thing. One sample might be
H₂O, another H₃O, another H₁.₄O — each with a different boiling point, density and taste. The
word would describe a whole family of liquids, not a compound.
The Law of Constant Proportions and the Law of Multiple Proportions would both fail,
and with them Dalton's fifth postulate. There would be nothing for an atomic theory to
explain.
Chemical formulae, valency and molecular mass would be meaningless. You could not
write H₂O, could not calculate a formula unit mass, could not predict how much of one thing
reacts with how much of another.
Medicine would be impossible. A paracetamol tablet works because every molecule in it is
identical. If the ratio of atoms drifted from batch to batch, the same tablet could be useless
one day and poisonous the next.
Living things could not exist. Haemoglobin carries oxygen because it has an exact
structure; a protein assembled in a random ratio would not fold, would not work, and could
not be copied faithfully into the next generation.

Why it happens in the real world: the ratio is fixed because atoms bond to
complete their octet, and an octet is a whole number of electrons. Oxygen is short
of exactly two electrons, hydrogen of exactly one — so two hydrogens go with one
oxygen, never 2.3 of them. Whole numbers in the electron count are what force
whole numbers in the formula.

Page 16 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Pause and Ponder — Page 168
9.3 Dalton's Atomic Theory

PAUSE AND PONDER

Q7 Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.
Reason (R): According to Dalton’s Atomic Theory, atoms combine in a simple whole
number ratio by mass to form compounds. Choose the correct option: (i) Both A and
R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is
not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is
true.

Option (ii) — Both A and R are true, but R is not the correct explanation of A.
Is A true? Yes, on two counts.

Ratio taken, H : O = 2 g : 16 g = 1 : 8 — the fixed ratio in water ✓

Mass of water = 2 g + 16 g = 18 g — by conservation of mass ✓

(Check with the formula: H₂O = (1 u × 2) + 16 u = 18 u, and H : O = 2 : 16 = 1 : 8)

Is R true? It is a statement of Dalton's fifth postulate — atoms combine in the ratio of simple
whole numbers to form compounds — and it is a correct principle.
Does R explain A? No. A makes two claims, and Dalton's whole-number postulate delivers
neither of them:

That the mass ratio is 1 : 8 comes from Proust's Law of Constant Proportions, not from
Dalton's whole-number rule.
That 2 g + 16 g gives exactly 18 g comes from the Law of Conservation of Mass.

Dalton's postulate tells you how many atoms join (2 H to 1 O). It does not by itself tell you what
masses will be used up or produced. So R is a true statement placed beside a true statement —
but it is not the reason for it.

Tip: read R strictly and it has one more flaw — Dalton's postulate is about the ratio of
the number of atoms, not of masses. A student who marks (iii) on that reading has
spotted something real; but since 1 : 8 does happen to be a simple whole-number
mass ratio here, R is best treated as true and the deciding point is that it does not
explain A. Hence (ii).

Page 17 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Pause and Ponder — Page 170
9.4.1 A Molecules of elements

PAUSE AND PONDER

Q8 Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule
(N2).

Nitrogen (atomic number 7) has the configuration 2, 5. With five valence electrons it needs
three more to reach an octet. Neither atom can take three from the other, so the two atoms
share three pairs — a triple bond.

× ×
× N ×
×
N

N≡N
× electron of one nitrogen atom
electron of the other nitrogen atom

Each nitrogen atom contributes three electrons to the shared region and keeps one lone pair.
Counting round either atom: 6 shared + 2 lone = 8 electrons, a complete octet.

Valence electrons on each N = 5

Electrons needed to complete the octet = 8 − 5 = 3

So three pairs are shared → a triple bond, written N ≡ N

Page 18 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: a triple bond is the strongest of the common covalent bonds, and

m l a se
it is why nitrogen gas is so unreactive. Air is 78 % N₂ and yet it does not burn, rust or
o
.c g of energy,
— breaking three bonds at once needs a greatadeal
m
corrode anything

l a seis why industrial ammonia manufacture requires high temperature, high
g
which
apressure and a catalyst.

co m
e m . ag
g l as
a
Q9 The atomic number of fluorine is 9. Explain the formation of the fluorine molecule
(F2).

co m
em.
m l as
.co a g
Fluorine has atomic number 9, so its electronic configuration is 2, 7 — seven electrons in the

s em
valence (L) shell, one short of an octet.
a
gAlfluorine atom cannot gain that electron from another fluorine atom, because that atom needs
a one just as badly. The way out is for each atom to put one electron into a shared pair. The pair
m a s
.co agl
belongs to both nuclei at once, so both atoms now count eight valence electrons.

se m
g l a
a

co m
m .
m as e
.co ×× a g l
se m ×
g l a × F × F
a ××
se m
com g l a
m . a
ase
agl
F—F
× electron of one fluorine atom
co m
electron of the other fluorine atom
m .
o m l a se
m agcarries three lone pairs. Around
.c One shared pair between the two fluorine atoms; each atom also
l a se
g
either atom: 2 shared + 6 lone = 8 electrons.
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

F: 2, 7 → 7 valence electrons, needs 1 more

Electrons shared by each atom = 1 → one shared pair

Bond formed = single covalent bond, written F — F

The shared pair attracts both nuclei at the same time, and it is that double attraction that holds
the two atoms together as one molecule, F₂.

Why it happens: the molecule forms because the total energy of the shared
arrangement is lower than the energy of two separate fluorine atoms. Atoms
combine for exactly this reason — a lower-energy arrangement is a more stable one.

Pause and Ponder — Page 171
9.4.1 B Molecules of compounds

PAUSE AND PONDER

Q10 Show the formation of the following molecules: (i) Carbon dioxide (CO2) (ii)
Hydrogen sulfide (H2S) (iii) Ammonia (NH3)

In each case, first write the configuration, then count how many electrons each atom is short of,
then match the shortages.
(i) Carbon dioxide, CO₂

C (Z = 6): 2, 4 → 4 valence electrons, needs 4

O (Z = 8): 2, 6 → 6 valence electrons, needs 2

One carbon (4) is matched by two oxygens (2 + 2) → formula CO₂

Page 20 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

×× × ×
××
×
× O × C × O ××
×× ××

O=C=O
× electron of an oxygen atom
electron of the carbon atom

Carbon shares two electrons with each oxygen, forming two double bonds: O = C = O. Carbon now
has 8 valence electrons, and so does each oxygen.

(ii) Hydrogen sulfide, H₂S

S (Z = 16): 2, 8, 6 → needs 2 electrons

H (Z = 1): 1 → needs 1 electron to complete its duplet

Two hydrogen atoms are needed for one sulfur → formula H₂S

H
× ×
×
S ×
× ×

H
H—S—H
× electron of the sulfur atom
electron of a hydrogen atom

Page 21 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Two single covalent bonds, H — S — H. Sulfur keeps two lone pairs; each hydrogen has its duplet of 2
electrons.

(iii) Ammonia, NH₃

N (Z = 7): 2, 5 → needs 3 electrons

H: needs 1 each → three hydrogen atoms are needed

Formula NH₃

H
×× ×

H × N
×

H
NH₃
× electron of the nitrogen atom
electron of a hydrogen atom

Three single covalent bonds around nitrogen, with one lone pair left over on the nitrogen atom.
Nitrogen: 6 shared + 2 lone = 8 electrons.

Why it happens: the number of bonds an atom forms is simply the number of
electrons it is short of. Carbon is short of 4 and forms 4 bonds; nitrogen is short of 3
and forms 3; oxygen and sulfur are short of 2 and form 2; hydrogen is short of 1 and
forms 1. That single rule generates all three formulae above.

Tip: notice that NH₃ keeps a lone pair. That leftover pair is the reason ammonia
solution is basic and smells sharp — you will meet it again in higher classes.

Page 22 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Q11 Neon (atomic number 10) neither transfers nor shares its valence electrons.
Explain.

Because neon's valence shell is already full.

Neon, Z = 10 → electronic configuration 2, 8

Electrons in the valence (L) shell = 8 — a complete octet

Atoms lose, gain or share electrons for one reason only — to reach an octet. Neon has already
reached it, so it has nothing to gain from bonding:

It will not lose an electron — that would break up a complete octet and leave it less stable,
and pulling an electron out of a full shell needs a great deal of energy.
It will not gain an electron — there is no room in the L shell, so the extra electron would
have to start a new M shell far from the nucleus, which is not a stable arrangement.
It will not share — sharing only pays when it completes a shell, and neon's shell is
complete.

Neon therefore exists as single, separate atoms, not as Ne₂ molecules, and forms no ordinary
compounds. Helium behaves the same way with a complete duplet (2).

Why it happens: when atoms combine, the total energy of the system falls — that
energy drop is the reward that drives bonding. For a noble gas there is no reward:
any bond would raise the energy. This is why Group 18 elements are called inert or
noble gases and are used where chemical inactivity matters, such as in neon signs
and in argon-filled bulbs.

In-text Questions — Page 172

Page 23 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

9.4.2 Bonding by electron transfer — Ionic bond
co m
e m.
m l as
.co a
If the valence shell of an atom has less than four electrons, it would generally donate it
g
em to achieve a stable electronic configuration. Identify four such elements amo
Q1

a s
electrons

a gl elements by using their electronic configurations given in Table 8.4 of the Chapter 8, Jo
Atom.

com
m . ag
l a se
ag
NAME OF SYMBOL ATOMIC NUMBER NUMBER NUMBER DIS
THE NUMBER OF OF OF

com
ELEMENT PROTONS NEUTRONS ELECTRONS K

m .1
as e
com
Hydrogen H 1 1 - 1

e m. Helium agl
las
He 2 2 2 2 2

ag Lithium Li 3 3 4 3 2

m a s
.co agl
Beryllium Be 4 4 5 4 2

s5e m
Boron B
g l a 5 6 5 2
a
Carbon C 6 6 6 6 2

. com7
Nitrogen N 7 7 7
e m 2

m 8gl
as
.co Oxygen a
sem
O 8 8 8 2

a
agl Fluorine F 9 9 10 9 2

se m
com l a
Neon Ne 10 10 10 10 2

. ag
e m
as
Sodium Na 11 11 12 11 2

Magnesium Mg agl 12 12 12 12 2

co m 13
.
Aluminium Al 13 13 14 2

em
m l as
.co g
Silicon Si 14 14 14 14 2

m a
l a se
g
Phosphorus P 15 15 16 15 2
a c
m .
e
Sulfur S 16 16 16 16 2

a s
. co17m ag2 l
e m
Chlorine Cl 17 18 17

g l as
Argon
a Ar 18 18 22 18 2

com
m .
m ase
.co


a g l Page 24 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Table 8.4, Chapter 8 “Journey Inside the Atom”, page 152 — the first eighteen elements and th
their electrons in the K, L, M and N shells.

Lithium, beryllium, sodium, magnesium and aluminium — any four of these.

ELEMENT Z ELECTRONIC VALENCE ION FORMED ON
CONFIGURATION ELECTRONS DONATING THEM

Lithium (Li) 3 2, 1 1 Li⁺

Beryllium 4 2, 2 2 Be²⁺
(Be)

Sodium (Na) 11 2, 8, 1 1 Na⁺

Magnesium 12 2, 8, 2 2 Mg²⁺
(Mg)

Aluminium 13 2, 8, 3 3 Al³⁺
(Al)

Why it happens: giving away 1, 2 or 3 electrons is a short journey; gaining 5, 6 or 7
to complete an octet is a long one. When sodium (2, 8, 1) loses its single M-shell
electron, the shell underneath is already a full 2, 8 — the same stable arrangement
as neon. So one small loss buys a complete octet. That is why elements with fewer
than four valence electrons are metals and form cations.

Tip: boron (2, 3) also has fewer than four valence electrons, but with three electrons
to give and a small, tightly holding atom it prefers to share. The “less than four →
donate” rule is a good guide, not a law.

Q2 Will it still be neutral after losing one electron? If not, what charge would it carry
and why?

No — a sodium atom that has lost one electron is no longer neutral. It carries a charge of
+1 and is called the sodium cation, Na⁺.

Page 25 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Sodium atom, Z = 11 → 11 protons (+11) and 11 electrons (−11)

Net charge = (+11) + (−11) = 0 — neutral

After losing one electron: 11 protons (+11) and 10 electrons (−10)

Net charge = (+11) + (−10) = +1

The number of protons never changes in a chemical reaction — protons sit in the nucleus and
chemistry only ever moves the outer electrons. So removing one electron leaves one proton
whose positive charge is no longer balanced.

Why it happens: an atom is neutral only because protons and electrons happen to
be equal in number and equal and opposite in charge. Break that equality and a
charge appears. Lose electrons and you get a cation (Na⁺, Mg²⁺, Al³⁺); gain them and
you get an anion (Cl⁻, O²⁻, S²⁻).

Check it yourself: Na⁺ now has the configuration 2, 8 — the same as neon. It is
chemically nothing like a sodium atom: sodium metal reacts violently with water,
while Na⁺ in solution is what you eat every day in common salt.

What if … — Page 174
9.4.2 Bonding by electron transfer — Ionic bond

WHAT IF …

Q1 What if … we could see atoms directly? How would it help scientists and what
challenges would it cause?

It would turn chemistry from inference into observation — but “seeing” an atom is not
like seeing a marble, and that is where the difficulty starts.
How it would help

Structures could be read off, not deduced. Instead of arguing from valency and bond
counts that water is bent and CO₂ is straight, you would simply look.
Reactions could be watched happening. You could see a bond break, an electron transfer
from sodium to chlorine, or a catalyst holding two molecules in place while they join.

Page 26 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Materials and medicines could be designed atom by atom — a drug molecule shaped to
fit a particular site, or a battery electrode built to hold ions exactly where you want them.
Defects could be found instantly — a single misplaced atom in a semiconductor chip is
enough to ruin it.

What makes it hard

An atom is far smaller than the wavelength of light. An atom is about 10⁻¹⁰ m across;
visible light has a wavelength of roughly 5 × 10⁻⁷ m — some five hundred times larger. No
optical microscope, however good its lenses, can resolve something that much smaller than
the wave it is using.
Looking disturbs the thing you look at. To “see” an atom you must bounce something off
it, and anything small enough to do the job carries enough energy to knock the atom out of
place.
There is no sharp surface to see. Electrons are not tiny balls on tracks; they occupy a fuzzy
cloud of probability. A photograph would show a smudge, and a student who took it literally
would learn the wrong thing.
Data and cost. Mapping every atom in even a speck of matter would generate more data
than we could store, and the instruments run in vacuum, at very low temperature, and cost a
fortune.

Why it happens: we already do something close to this. A scanning tunnelling
microscope drags an atomically sharp needle over a surface and records how easily
electrons jump the gap, and an electron microscope uses electrons instead of light
because their wavelength is thousands of times shorter. Both give real images of
individual atoms — but they are maps of electron density, not photographs, which is
exactly the subtlety this question is pointing at.

Pause and Ponder — Page 174
9.4.2 Bonding by electron transfer — Ionic bond

PAUSE AND PONDER

Q12 What kind of ion will oxygen (O) form?

Oxygen forms the oxide anion, O²⁻ — a negatively charged ion carrying two units of
negative charge.

Page 27 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Oxygen, Z = 8 → configuration 2, 6

Valence electrons = 6; electrons needed for the octet = 8 − 6 = 2

O + 2 e⁻ → O²⁻ (configuration 2, 8)
Protons = 8 (+8), electrons = 10 (−10) → net charge = −2

So oxygen is a divalent anion, valency 2. It gains rather than loses because with six valence
electrons it is far closer to an octet by gaining two than by giving away six.

Why it happens: O²⁻ has the configuration 2, 8 — identical to neon. That is the
stable arrangement every atom is heading for, and gaining two electrons is the
cheapest route to it for oxygen.

Tip: this is why metal oxides come out the way they do. Na⁺ with O²⁻ gives Na₂O,
Mg²⁺ with O²⁻ gives MgO, and Al³⁺ with O²⁻ gives Al₂O₃ — in each case the charges
balance out to zero.

Q13 Fill in the blanks. Among magnesium and chlorine, magnesium atom can give two
electrons to become Mg2+. However, chlorine can take only one electron to
become ____________. Now, __________ ion of magnesium and __________ ions of chlorine
combine to give magnesium chloride.

Completed sentence: “However, chlorine can take only one electron to become the chloride
ion, Cl⁻. Now, one ion of magnesium and two ions of chlorine combine to give magnesium
chloride.”

Mg (2, 8, 2) → Mg²⁺ (2, 8) + 2 e⁻ — two electrons are released

Cl (2, 8, 7) + 1 e⁻ → Cl⁻ (2, 8, 8) — each chlorine can take only one

So two chlorine atoms are needed to absorb the two electrons:

Mg²⁺ + 2 Cl⁻ → MgCl₂

Charge check: (+2) + 2(−1) = 0 ✓

Page 28 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: electrons cannot be left over. Magnesium must dispose of exactly

m l a se
two, and each chlorine atom has room for exactly one, so the numbers must be 1 : 2.
o g — it is nothing
This is the.cwhole idea behind criss-crossing charges to get a formula
a
se m
butaelectron
a g l book-keeping.

o m
Show the formation of cations of m
e
. c ag
s
potassium (K) and calcium (Ca) atoms, and the
a chlorides using diagrams.
Q14

agl
formation of their corresponding

c o m
.
s em
Potassium (Z = 19): 2, 8, 8, 1. One electron in the outermost shell — losing it leaves the stable 2,
m a
. co
8, 8 arrangement of argon.
a gl
a s em
agl
m a s
m.co agl
l a se
a g − 1 e⁻

co m
m .
m l a se
m .co ag K⁺
l a se K 2, 8, 8

ag 2, 8, 8, 1

se m
com g l a
m . a
ase
K (2, 8, 8, 1) loses its single valence electron to become K⁺ (2, 8, 8), a monovalent cation.

agl
Calcium (Z = 20): 2, 8, 8, 2. Two electrons in the outermost shell — losing both leaves 2, 8, 8.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

− 2 e⁻

Ca²⁺
Ca 2, 8, 8
2, 8, 8, 2

Ca (2, 8, 8, 2) loses two valence electrons to become Ca²⁺ (2, 8, 8), a divalent cation.

Potassium chloride, KCl. The one electron potassium gives away is taken by one chlorine atom
(2, 8, 7 → 2, 8, 8).

K⁺ Cl⁻
2, 8, 8 2, 8, 8

KCl

K⁺ and Cl⁻, held together by electrostatic attraction. One cation to one anion, so the formula is KCl.

Calcium chloride, CaCl₂. Calcium must dispose of two electrons, and each chlorine can accept
only one — so two chlorine atoms are needed.

Page 30 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Ca²⁺ Cl⁻ Cl⁻
2, 8, 8 2, 8, 8 2, 8, 8

CaCl₂

One Ca²⁺ with two Cl⁻ ions. Charge check: (+2) + 2(−1) = 0, so the formula is CaCl₂.

K → K⁺ + e⁻; Cl + e⁻ → Cl⁻; K⁺ + Cl⁻ → KCl

Ca → Ca²⁺ + 2 e⁻; 2 Cl + 2 e⁻ → 2 Cl⁻; Ca²⁺ + 2 Cl⁻ → CaCl₂

Why it happens: both metals are in the same situation as sodium — a nearly empty
outermost shell sitting on top of a complete one. Emptying it is the quickest way to
an octet. The chlorine atoms are in the opposite situation, one electron short, so the
transfer suits both sides and the resulting opposite charges then hold the ions
together.

Q15 Illustrate how sodium sulfide (Na2S) is formed.

Sodium is 2, 8, 1 — one electron to give. Sulfur is 2, 8, 6 — two electrons short of an octet. Since
each sodium can supply only one electron, two sodium atoms are needed for one sulfur atom.

Page 31 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

+ +

Na S Na
2, 8, 1 2, 8, 6 2, 8, 1

transfer of 2 e⁻

Na⁺ Na⁺
2, 8 S²⁻ 2, 8
2, 8, 8

Na₂S

Two sodium atoms each transfer one electron to a single sulfur atom. Both Na⁺ ions end up with 2, 8
(like neon) and S²⁻ with 2, 8, 8 (like argon).

2 Na (2, 8, 1) → 2 Na⁺ (2, 8) + 2 e⁻

S (2, 8, 6) + 2 e⁻ → S²⁻ (2, 8, 8)

2 Na⁺ + S²⁻ → Na₂S

Charge check: 2(+1) + (−2) = 0 ✓

The two Na⁺ ions and the S²⁻ ion are then held together by the electrostatic force of attraction
between opposite charges — an ionic bond. As with all ionic compounds, Na₂S does not exist as
a separate molecule but as a 3-D crystal lattice; Na₂S is its formula unit, the simplest whole-
number ratio of its ions.

Page 32 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: the subscript 2 in Na₂S is not decoration — it is forced by the
electron count. Sulfur needs two electrons; sodium can spare one each; so the ratio
must be 2 : 1. The same reasoning gives Na₂O, K₂S and Li₂O.

Pause and Ponder — Page 177
9.5 Writing Chemical Formulae

PAUSE AND PONDER

Q16 Name the following: (i) CO2 (ii) NO2 (iii) SF6 (iv) PCl3

These are all covalent compounds, so they are named with the prefix system: the first element
keeps its ordinary name, the second ends in -ide, and prefixes (mono-, di-, tri-, tetra-, penta-,
hexa-) give the number of atoms. Mono- is dropped for the first element.

FORMULA ATOMS OF THE SECOND ELEMENT NAME

(i) CO₂ 2 oxygen → di- + oxide Carbon dioxide

(ii) NO₂ 2 oxygen → di- + oxide Nitrogen dioxide

(iii) SF₆ 6 fluorine → hexa- + fluoride Sulfur hexafluoride

(iv) PCl₃ 3 chlorine → tri- + chloride Phosphorus trichloride

Why it happens: two different elements can form more than one compound — CO
and CO₂, NO and NO₂ and N₂O₄. The prefix is what tells them apart, so it cannot be
dropped from the second element. Note also that CO is carbon monoxide, not
monooxide: when a prefix ending in ‘o’ or ‘a’ meets an element starting with a vowel,
the last vowel of the prefix is dropped.

Did you know? SF₆ is one of the most powerful greenhouse gases known and is
used as an insulating gas in high-voltage electrical switchgear. NO₂ is the reddish-
brown gas responsible for the haze over busy traffic junctions.

Page 33 of 69

ase
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur
e
Q17

m l as
.co
dioxide (iii) Ferric chloride (iv) Cuprous oxide

a g
se m
g l a
a

Use the criss-cross method with the charges from Table 9.1: write cation first, put the charge

m
numbers below the symbols, swap them as subscripts, then divide by any common factor.
. co ag
em
COMPOUND
l as
IONS / VALENCIES
g
CRISS-CROSS FORMULA

(i) Sodium
aNa⁺ (1), HCO₃⁻ (1) 1 and 1 → both NaHCO₃

com
hydrogencarbonate dropped

.
m SO₂
as e
com l
(ii) Sulfur dioxide S (valency 4), O (valency S₂O₄ ÷ 2

. a g
m
2)
e
as Ferric chloride
a g l(iii) Fe³⁺ (3), Cl⁻ (1) Fe₁Cl₃ FeCl₃

Cuprous oxide
m
Cu⁺ (1), O²⁻ (2) Cu₂O₁
a s
(iv) Cu₂O

m .co agl
l a se
a g
Why it happens: the criss-cross is a quick way of making the total positive charge
cancel the total negative charge. In FeCl₃, (+3) + 3(−1) = 0; in Cu₂O, 2(+1) + (−2) = 0.

m
Every ionic formula must come out electrically neutral.
. co
em
m l as
.co
Tip: “-ous” always means the lower charge and “-ic” the higher one — cuprous is Cu⁺
m cupric is Cu²⁺; ferrous is Fe²⁺ and ferric is Fe³⁺. Get that wrong and the whole a g
a s eand
agl formula changes: cupric oxide is CuO, not Cu₂O.
se m
com g l a
m . a
e
s compounds formed from the following pairs of ions: (i)
Write the formulae forathe
Q18
g l
a and CO3 2–
Fe3+ and OH‒ (ii) K+

co m
m .
m as e
.co
(i) Fe³⁺ and OH⁻ → Fe(OH)₃
a g l
se m
g l a
a Charges: Fe is 3+, OH is 1−
c
m .
Criss-cross: Fe₁(OH)₃ → Fe(OH)₃ — ferric hydroxide
m a s e
Charge check: (+3) + 3(−1) = 0 ✓
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 34 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Brackets are needed because there are three hydroxide ions. Fe(OH)₃ means one iron and three
OH groups; FeOH₃ would wrongly read as one O and three H.
(ii) K⁺ and CO₃²⁻ → K₂CO₃

Charges: K is 1+, CO₃ is 2−

Criss-cross: K₂(CO₃)₁ → K₂CO₃ — potassium carbonate

Charge check: 2(+1) + (−2) = 0 ✓

No brackets here, because only one carbonate ion is present. Brackets are used only when two
or more identical polyatomic ions appear.

Why it happens: a polyatomic ion such as OH⁻ or CO₃²⁻ behaves as a single unit —
its own atoms are covalently bonded to each other and the whole group carries one
charge. So in criss-crossing you treat it exactly as you would a single symbol, and the
bracket is what keeps that unit together.

Activity 9.4 — Pages 177 – 178

Page 35 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

9.6 Properties of the Ionic and the Covalent Compounds

ACTIVITY 9.4

Q1 (Step 9) Group the compounds showing similar properties listed in Table 9.2.

COMPOUND EXPERIMENTS

SOLUBILITY IN ELECTRICAL
CONDUCTIVITY OF
COMPOUNDS IN

WATER KEROSENE PETROL SOLID STATE WATER

Camphor

Sodium
chloride

Copper
sulfate

Sugar

Naphthalene

Any other

Table 9.2, page 178 — the observation table of Activity 9.4, listing the compounds that
were tested.

These are the observations you should record in Table 9.2.

Page 36 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

COMPOUND SOLUBILITY IN ELECTRICAL CONDUCTIVITY IN

WATER KEROSENE PETROL SOLID STATE WATER

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

The compounds fall into three groups.

Group 1 — ionic compounds: sodium chloride and copper sulfate. Soluble in water,
insoluble in kerosene and petrol; do not conduct as solids but do conduct in water. Both also
have high melting points.
Group 2 — covalent compounds that are insoluble in water: camphor and naphthalene.
They dissolve instead in kerosene and petrol, and never conduct. Both have low melting
points and a strong smell — they sublime easily.
Group 3 — a covalent compound that is soluble in water: sugar. It behaves like Group 1 in
the solubility column but like Group 2 in the conductivity column — it dissolves without
producing ions.

Why it happens: the rule behind every entry is “like dissolves like”. Water
molecules are polar — they have slightly charged ends — so they can surround and
pull apart the charged ions of an ionic lattice. Kerosene and petrol are non-polar, so
they have no such grip on ions, but they mix happily with non-polar molecules like
camphor and naphthalene. Sugar is a special case: it is covalent, but it carries several
—OH groups that water can hold on to, so it dissolves — as whole neutral molecules,
which is why the bulb stays dark.

Safety first: petrol and kerosene are highly flammable, so keep them away from
flames and work in a ventilated place. Do not touch the electrodes while the battery
is connected, and use a low-voltage battery only.

In-text Questions — Page 178

Page 37 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

9.6 Properties of the Ionic and the Covalent Compounds

Q1 Other covalent compounds, such as camphor and naphthalene, also do not conduct
electricity. Can you give a reason?

Because they contain no charged particles that are free to move — neither ions nor free
electrons.
A current is nothing but charge in motion. For a substance to conduct, it must supply mobile
charge carriers. Camphor and naphthalene supply none:

They are made of electrically neutral molecules. Every electron in them is either locked in a
shared pair inside a covalent bond or sitting as a lone pair on an atom — none of them can
wander off through the solid.
They are insoluble in water, so they never get the chance to release ions into solution the
way sodium chloride does.
Even when melted, they stay as molecules. Melting camphor loosens the molecules from one
another, but it does not break the covalent bonds inside them, so still no ions appear.

Why it happens: compare the three cases side by side — a metal conducts because
it has free electrons; molten or dissolved sodium chloride conducts because it has
free ions; camphor has neither, so the circuit stays broken and the bulb never glows.

Q2 Predict whether ionic and covalent compounds would conduct electricity in the
molten state (the melted state of a substance).

Ionic compounds conduct in the molten state; covalent compounds do not.

STATE IONIC COMPOUND (E.G. NACL) COVALENT COMPOUND (E.G.
NAPHTHALENE)

Solid Does not conduct — ions fixed in the Does not conduct — no ions at all
lattice

Molten Conducts — lattice breaks, ions move Does not conduct — still neutral molecules
freely

Aqueous Conducts — ions separated by water Does not conduct (if it dissolves at all)
solution

Page 38 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

Reasoning for the ionic case. In a solid ionic compound the Na⁺ and Cl⁻ ions are already
co m
se
charged, but they are locked in fixed positions by strong electrostatic forces. Heat the solid past
m.
o m l a
.cthat can move is a current, so the compound conducts. ag
its melting point and those positions break down: the ions are still charged, but now they can

m
se for the covalent case. Melting a covalent solid separates the molecules from one
move. Charge

l a
ag
Reasoning
another, but the covalent bonds inside each molecule survive. The liquid is a crowd of neutral
molecules, and neutral particles cannot carry a current.
co m
e m . ag
g l as
Why it happens: conduction needs two things at once — charge and mobility. Solid
a
NaCl has the charge but not the mobility; molten naphthalene has the mobility but

m
not the charge; molten NaCl has both, and only it conducts.
co
em.
m l as
.co
Did you know? Aluminium is extracted industrially by passing a current through
a g
a s em aluminium oxide. The process works only because the melt is full of free Al³⁺
molten

a gl and O²⁻ ions — exactly the prediction you have just made.
m a s
e m .co agl
las
Pause and Ponder — Page 179
ag
9.6 Properties of the Ionic and the Covalent Compounds

co m
.
PAUSE AND PONDER

se m
Q19 .c o m l a
g that does not conduct
type of chemical bond is present in a solid
a
em electricity in the solid state but conducts electricity when dissolved in water?
What compound

l a s
ag
se m
o m g l a
m
An ionic bond. That pair of observations
.c is the standard fingerprint of an ionic compound.
a
l a se ions are there, but they are clamped in fixed positions in the
ag
No conduction as a solid — the
crystal lattice by strong electrostatic forces, so no charge can move.

m
Conduction in water — water pulls the lattice apart and sets the ions free. Once mobile, the

. co
m
cations drift towards the negative electrode and the anions towards the positive one, and

o m
that drift is the current.
l a se
g as Activity 9.4 shows.
.c chloride and copper sulfate both behave exactly like athis,
se m
Sodium
a
agl c
Why it happens: the two clues must be read together. Conduction in solution alone
m .
m a s e
co agl
would not settle it — you would still have to rule out a covalent compound like sugar,

m .
e
which dissolves but does not conduct. It is the combination — ions present but

g l as
immobile in the solid, mobile once dissolved — that points uniquely to ionic
bonding. a

com
m .
m ase
.co


a g l Page 39 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Check it yourself: such compounds also melt at high temperatures, because pulling
apart a lattice of oppositely charged ions takes a lot of energy. Common salt melts at
about 800 °C; camphor, a covalent solid, melts below 180 °C.

Q20 Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to
form a compound that is slightly soluble in water. Predict its: (i) formula (ii) type of
bond (iii) electrical conductivity of its aqueous solution.

First identify the metal. Its valence shell is the M shell and it holds two electrons, so the
configuration is 2, 8, 2 — that is Z = 12, magnesium.
(i) Formula — MO (magnesium oxide, MgO)

M (2, 8, 2) → M²⁺ (2, 8) + 2 e⁻
O (2, 6) + 2 e⁻ → O²⁻ (2, 8)

Criss-cross the charges: M₂O₂, divide by 2 → MO

Charge check: (+2) + (−2) = 0 ✓

(ii) Type of bond — ionic. A metal with two valence electrons meets a non-metal that is two
electrons short. Electrons are transferred, not shared, giving M²⁺ and O²⁻, which are then held
together by electrostatic attraction.
(iii) Conductivity of its aqueous solution — it conducts, but only weakly. Whatever little of it
dissolves goes into solution as free M²⁺ and O²⁻ (in practice OH⁻) ions, and mobile ions carry a
current. But since the compound is only slightly soluble, the number of ions in the solution is
small, so the bulb glows dimly rather than brightly.

Why it happens: conductivity depends on how many mobile ions there are per unit
volume, not merely on whether the compound is ionic. The bond type decides
whether ions can appear; the solubility decides how many do.

Did you know? A suspension of this same slightly soluble compound in water is sold
as milk of magnesia and taken to relieve acidity — a use it has precisely because it
dissolves so little.

Pause and Ponder — Page 179

Page 40 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

9.7 Molecular Mass of Covalent Compounds

PAUSE AND PONDER

Q21 Find the molecular mass of nitric acid (HNO3). Atomic mass — H = 1 u; N = 14 u; O =
16 u.

63 u.

Molecular mass = sum of the atomic masses of all the atoms in the formula

HNO₃ contains: 1 H, 1 N, 3 O

= (1 u × 1) + (14 u × 1) + (16 u × 3)

= 1 u + 14 u + 48 u

= 63 u

Why it happens: nearly all the mass of an atom sits in its nucleus, and bonding only
rearranges the outer electrons — whose mass is negligible. So no mass is gained or
lost when atoms join, and the mass of the molecule is simply the sum of the masses
of its atoms. This is the Law of Conservation of Mass applied to a single molecule.

Tip: the subscript multiplies only the symbol immediately before it. In HNO₃ the 3
belongs to O alone, so it is 16 × 3, not (14 + 16) × 3.

Q22 Find the molecular mass of methane (CH4). Atomic mass — C = 12 u; H = 1 u.

16 u.

CH₄ contains: 1 C, 4 H

Molecular mass = (12 u × 1) + (1 u × 4)

= 12 u + 4 u

= 16 u

Page 41 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Methane is a covalent compound: carbon needs four electrons and each hydrogen needs one,
so carbon forms four single covalent bonds, one to each hydrogen.

Did you know? Methane is the main constituent of biogas and of CNG. It is also a
powerful greenhouse gas, which is why leaks from gas pipelines and from paddy
fields matter for the climate.

Pause and Ponder — Page 180
9.8 Formula Unit Mass of Ionic Compounds

PAUSE AND PONDER

Q23 Find the formula unit mass of potassium chloride (KCl). Atomic mass — K = 39 u; Cl
= 35.5 u.

74.5 u.

KCl contains: 1 K, 1 Cl

Formula unit mass = (39 u × 1) + (35.5 u × 1)

= 74.5 u

Why is it called formula unit mass and not molecular mass? Because potassium
chloride does not exist as molecules. Its solid is a 3-D lattice in which every K⁺ is
surrounded by Cl⁻ ions and every Cl⁻ by K⁺ ions — you cannot point to one K⁺ and
one Cl⁻ and say “that pair is a molecule”. KCl simply records the simplest whole-
number ratio of ions, 1 : 1, and that ratio is the formula unit.

Tip: 35.5 u is not a mistake. Chlorine occurs as two isotopes, of mass 35 u and 37 u,
in roughly a 3 : 1 ratio, and 35.5 u is the weighted average you met in Chapter 8.

Page 42 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Q24 Find the formula unit mass of magnesium hydroxide, Mg(OH)2. Atomic mass — Mg
= 24 u; O = 16 u; H = 1 u.

58 u.

Mg(OH)₂ contains: 1 Mg, and 2 hydroxide groups — so 2 O and 2 H

Formula unit mass = (24 u × 1) + {(16 u × 1) + (1 u × 1)} × 2

= 24 u + (17 u × 2)

= 24 u + 34 u

= 58 u

Why the bracket matters: the subscript 2 sits outside the bracket, so it multiplies
everything inside — both the oxygen and the hydrogen. Ignore the bracket and
you would write 24 + 16 + 2 = 42 u, which is wrong by 16 u. The bracket is there
because there are two complete hydroxide ions bound to one Mg²⁺ ion.

Check it yourself: apply the same rule to calcium nitrate in Example 9.7 — Ca(NO₃)₂
= 40 + {14 + (16 × 3)} × 2 = 40 + 124 = 164 u ✓

Revise, Reflect, Refine — Pages 181 – 183
End-of-chapter questions

REVISE, REFLECT, REFINE

Q1 A particular element (A) has one electron in its third shell. There is another element
(B) with six electrons in its second shell. (i) How many electrons does A tend to give
or take to become stable? (ii) What kind of ion would it form? (iii) How many
electrons does B tend to give or take to become stable? (iv) What kind of ion would
it form? (v) If A and B were to combine, what kind of bond would be formed? (vi)
What would be the formula for the compound thus formed?

First fix the configurations. A has one electron in the third shell, so the first two shells must be
full: A = 2, 8, 1 (Z = 11, sodium). B has six electrons in the second shell: B = 2, 6 (Z = 8, oxygen).

Page 43 of 69

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Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

(i) A gives away 1 electron. Removing its single M-shell electron leaves 2, 8 — a complete octet.
co m
Taking seven electrons instead would be far harder.
e m.
m l as
.co a g
(ii) A forms a monovalent cation, A⁺.

se m
g l a
aA: 11 protons (+11), 11 electrons (−11) → neutral

co m
A⁺: 11 protons (+11), 10 electrons (−10) → net charge +1

e m . ag
g l as
a
(iii) B takes 2 electrons. With six valence electrons it is two short of an octet; gaining two is
much easier than losing six.
(iv) B forms a divalent anion, B²⁻ (configuration 2, 8).
co m
(v) An ionic bond. One element hands electrons over and the other accepts them, so this is a
em.
c o m g l as
.
transfer, not a sharing. The oppositely charged ions that result are then held together by

e m attraction. a
s
electrostatic
la A₂B.
ag(vi)
m a s
.co agl
Each A supplies 1 electron; each B needs 2 → two A atoms per B atom

se m
g l a
2 A⁺ + B²⁻ → A₂B (in real terms, Na₂O — sodium oxide)

Charge check: 2(+1) + (−2) = 0 ✓ a

co m
m .
o m l a se
Why it happens: reading “one electron in the third shell” correctly is the whole

m .c A shell cannot begin filling until the one below itagis full, so the third shell
asehaving any electrons at all tells you the first two hold 2 and 8. That single deduction
question.

agl fixes the element, the ion, the bond and the formula.

se m
com g l a
m . a
gl ase
a
Q2 An element X has six electrons in its outer shell and forms a diatomic molecule. (i)
Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure

co
of the molecule it would form. (iv) A certain other element Y has two electrons in its
m
second shell. Draw the structure of the molecule that X would form with Y.
m .
as e
. com a g l
sem
a
agl X has six electrons in its outermost shell, so it is two electrons short of an octet. Taking the
c
smallest such element, X = oxygen (2, 6).
m .
m a s e
co agl
(i) Why a diatomic molecule? Because X cannot complete its octet alone, and the atom nearest

m .
as e
to hand is another X atom, which is short by exactly the same amount. Neither can take two

a g l
electrons from the other, since that would leave the donor four short. The only arrangement

com
m .
m ase
.co


a g l Page 44 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

that satisfies both is for each atom to contribute two electrons to a shared region — and once
both octets are complete there is no reason to add a third atom. So the molecule stops at two,
X₂.
(ii) A covalent bond — specifically a double bond, since two pairs of electrons are shared.
(iii) Structure of X₂

×× ×
×
× O × O
××

O=O
× electron of one oxygen atom
electron of the other oxygen atom

Two shared pairs between the atoms (a double bond, X = X) and two lone pairs on each atom. Around
either atom: 4 shared + 4 lone = 8 electrons.

(iv) The compound of X with Y. Y has two electrons in its second shell, so its configuration is 2,
2 (Z = 4, beryllium). Y has 2 valence electrons to dispose of and X needs exactly 2, so they
combine in a 1 : 1 ratio.

Page 45 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

+

Be O
2, 2 2, 6
transfer of 2 e⁻

BeO
Be²⁺
2 O²⁻
2, 8

Y (2, 2) transfers both valence electrons to X (2, 6). Y becomes Y²⁺ and X becomes X²⁻, and the
compound is YX — here BeO.

Y (2, 2) → Y²⁺ + 2 e⁻; X (2, 6) + 2 e⁻ → X²⁻

Y²⁺ + X²⁻ → YX (BeO)

Charge check: (+2) + (−2) = 0 ✓

Tip: at this level the expected answer is the 1 : 1 electron-transfer picture shown
above. In reality beryllium is such a small, tightly holding atom that the Be—O bond
has a great deal of covalent character, and BeO is written as a formula unit rather
than as a true molecule — a refinement you will meet in higher classes.

Why it happens: whether atoms share or transfer depends on how far apart their
pull on electrons is. Two identical X atoms pull equally, so neither can win — they
must share, and the bond is covalent. A metal like Y holds its outer electrons loosely,
so X can take them outright and the bond becomes ionic.

Page 46 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Q3 You want to design a new ionic compound, where the total positive charge is 6+ and
the total negative charge is 6 –. Which of the following combinations gives the
correct number of ions? (i) 2 Al3+ and 3 Cl– (ii) 3 Mg2+ and 1 PO4 3– (iii) 2 Fe3+ and 3
O2– (iv) 3 Ca2+ and 2 SO4 2–

Option (iii) — 2 Fe³⁺ and 3 O²⁻.

COMBINATION TOTAL POSITIVE TOTAL NEGATIVE BALANCED?

(i) 2 Al³⁺ and 3 Cl⁻ 2 × 3 = 6+ 3 × 1 = 3− No

(ii) 3 Mg²⁺ and 1 PO₄³⁻ 3 × 2 = 6+ 1 × 3 = 3− No

(iii) 2 Fe³⁺ and 3 O²⁻ 2 × 3 = 6+ 3 × 2 = 6− Yes ✓

(iv) 3 Ca²⁺ and 2 SO₄²⁻ 3 × 2 = 6+ 2 × 2 = 4− No

Only (iii) has both totals equal to 6, so only (iii) gives a neutral compound. The formula is Fe₂O₃
— ferric oxide, the red-brown compound in rust and in haematite ore.

Why it happens: every compound must be electrically neutral overall — if it were
not, the leftover charge would attract more ions until it was. So in any ionic formula,
(number of cations × cation charge) must equal (number of anions × anion charge).
That is exactly what the criss-cross method guarantees: Fe³⁺ and O²⁻ criss-cross to
Fe₂O₃.

Tip: the other three can be corrected easily — 1 Al³⁺ with 3 Cl⁻ gives AlCl₃, 3 Mg²⁺ with
2 PO₄³⁻ gives Mg₃(PO₄)₂, and 1 Ca²⁺ with 1 SO₄²⁻ gives CaSO₄.

Q4 Choose the correct statement(s) and correct the false statement(s). (i) Elements are
made up of molecules and compounds are made up of atoms. (ii) The molecule of a
compound is always made up of two or more atoms of the same kind. (iii) One
molecule of nitrogen gas contains three nitrogen atoms. (iv) Water is made of two
hydrogen atoms, covalently bonded with one oxygen atom.

Only statement (iv) is correct. The other three are false; corrected versions are given below.
(i) False. The two halves have been swapped and the words used loosely.

Page 47 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Corrected: Elements are made up of atoms of only one kind — those atoms may exist singly
(He, Ne) or joined as molecules of the element (H₂, O₂, N₂, Cl₂). Compounds are made up of
atoms of two or more different elements combined in a fixed ratio, forming either molecules
(H₂O, CO₂) or formula units (NaCl, CaCO₃).
(ii) False. Atoms of the same kind give a molecule of an element, not of a compound.
Corrected: The molecule of a compound is always made up of two or more atoms of
different kinds. For example, HCl has one hydrogen and one chlorine atom; H₂SO₄ has three
different elements.
(iii) False. Nitrogen has five valence electrons (2, 5), needs three more, and gets them by
sharing three pairs with one other nitrogen atom.
Corrected: One molecule of nitrogen gas contains two nitrogen atoms, joined by a triple
bond — N ≡ N, formula N₂.
(iv) True.

O (2, 6) needs 2 electrons; each H needs 1

Two H atoms each share one electron with the O atom

→ two single covalent bonds → H₂O, molecular mass 18 u

Why it happens: the trap in this question is the word “molecule”. A molecule is any
electrically neutral group of more than one bonded atom that can exist
independently. Same kind of atoms → molecule of an element; different kinds →
molecule of a compound. Keeping those two apart settles (i), (ii) and (iii) at once.

Q5 Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii)
Calcium oxide (iii) Ferric oxide

Write the cation first, put the charge numbers beneath, criss-cross them as subscripts, and
divide by any common factor.

COMPOUND IONS CRISS-CROSS FORMULA CHARGE CHECK

(i) Aluminium nitrate Al³⁺, NO₃⁻ Al₁(NO₃)₃ Al(NO₃)₃ (+3) + 3(−1) = 0

(ii) Calcium oxide Ca²⁺, O²⁻ Ca₂O₂ ÷ 2 CaO (+2) + (−2) = 0

(iii) Ferric oxide Fe³⁺, O²⁻ Fe₂O₃ Fe₂O₃ 2(+3) + 3(−2) = 0

Page 48 of 69

ase
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

Note the brackets in Al(NO₃)₃: there are three nitrate ions, and the bracket keeps each NO₃
co m
group intact. Without it, AlNO₃₃ would be meaningless.
e m.
m l as
.co a g
s em
Tip: in (ii) the criss-cross first gives Ca₂O₂, which must then be divided by the
a
gl
acommon factor 2. A chemical formula always shows the simplest whole-number
ratio, so CaO is the answer.

co m
m . ag
l a se
Q6 Write the formulae of the a g
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–

co m
e m.
m as

.co a g l
se mIONS
l a
CRISS-CROSS FORMULA NAME CHARGE

ag CHECK

a s
com agl
(i) Ca²⁺, Br⁻ Ca₁Br₂ CaBr₂ Calcium bromide (+2) + 2(−1) = 0

m .Al₂(CO₃)₃
(ii) Al³⁺, Al₂(CO₃)₃
as e Aluminium 2(+3) + 3(−2) = 0

agl
CO₃²⁻ carbonate

com
(iii) K⁺, SO₄²⁻ K₂(SO₄)₁ K₂SO₄ Potassium sulfate 2(+1) + (−2) = 0

m .
ase
(iv) NH₄⁺, Cl⁻ 1 and 1, both NH₄Cl Ammonium chloride (+1) + (−1) = 0
m
. co agl
dropped

m
se are brackets used? Only when two or more identical polyatomic ions are present. So
l a
ag
When
Al₂(CO₃)₃ needs them (three carbonate ions), K₂SO₄ does not (one sulfate ion), and NH₄Cl does

se m
com a
not (one ammonium ion).

. a g l
e m
g l as ion is a single charged unit — its atoms are covalently
Why it happens: a polyatomic
bonded to one anotheraand the whole group moves as one. Treat it in the criss-cross
exactly as you would a single symbol, and use a bracket whenever a subscript has to
co m
apply to the whole group.
m .
m as e
.co a g l
s e m Did you know? NH₄Cl is an unusual ionic compound — both its ions contain
agla covalent bonds inside them. The four N—H bonds in NH₄⁺ are covalent, while the
.c
bond between NH₄⁺ and Cl⁻ is ionic.
s e m
m a
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 49 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Q7 Which of the following, in Fig. 9.18, correctly represents Cl– ion (Atomic number of
chlorine = 17).

(i) (ii) (iii) (iv)

Fig. 9.18, page 182: four shell diagrams labelled (i) to (iv). The black dot at the centre is
the nucleus; each blue dot is an electron.

Option (ii).
A chloride ion is a chlorine atom that has gained one electron, so the diagram must show:

Chlorine atom, Z = 17 → 17 electrons → 2, 8, 7

Cl + 1 e⁻ → Cl⁻ → 18 electrons → 2, 8, 8

Count the dots on each of the four diagrams printed in the book:

DIAGRAM K L M TOTAL VERDICT
SHELL SHELL SHELL ELECTRONS

(i) 2 7 8 17 Wrong — the L shell holds only 7
while the M shell already has 8. A
shell must be filled before the
next one starts.

(ii) 2 8 8 18 Correct — this is Cl⁻ ✓

(iii) 2 8 9 19 Wrong — 19 electrons, two more
than chlorine's 17 protons, and 9
in the outermost shell, which can
never hold more than 8.

(iv) 2 8 7 17 Wrong — this is the neutral
chlorine atom, not the ion.

Page 50 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Cl⁻
2, 8, 8

The chloride ion: 17 protons in the nucleus but 18 electrons arranged 2, 8, 8 — hence a net charge of
−1.

Charge on Cl⁻ = (+17 from protons) + (−18 from electrons) = −1 ✓

Page 51 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: the trap in this question is that the diagram alone cannot tell you
the number of protons — the nucleus is drawn as a single dot. You are told Z = 17, so
the correct picture is the one with 18 electrons filled in the proper order 2, 8, 8.
Diagram (iv) has the right total for an atom but the wrong total for an ion; diagram
(i) has the right total but breaks the filling rule; diagram (iii) breaks both the total and
the maximum of 8 in the outermost shell.

Q8 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 (from Chapter 8): H = 1 u, C = 12 u, N = 14 u, O = 16 u, Na = 23 u, P = 31 u.
(i) Ammonium nitrate, NH₄NO₃ → 80 u

Atoms present: 2 N, 4 H, 3 O

= (14 u × 2) + (1 u × 4) + (16 u × 3)

= 28 u + 4 u + 48 u

= 80 u

(ii) Phosphoric acid, H₃PO₄ → 98 u

Atoms present: 3 H, 1 P, 4 O

= (1 u × 3) + (31 u × 1) + (16 u × 4)

= 3 u + 31 u + 64 u

= 98 u

(iii) Sodium hydrogencarbonate, NaHCO₃ → 84 u

Page 52 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Atoms present: 1 Na, 1 H, 1 C, 3 O

= (23 u × 1) + (1 u × 1) + (12 u × 1) + (16 u × 3)

= 23 u + 1 u + 12 u + 48 u

= 84 u

Tip: in NH₄NO₃ the nitrogen appears twice, once in the ammonium ion and once in
the nitrate ion. Add up all the atoms of each element in the formula before
multiplying — forgetting the second N costs you 14 u.

Why it happens: ammonium nitrate is a good fertiliser precisely because of this
arithmetic — 28 u of its 80 u is nitrogen, so it is 28/80 × 100 = 35 % nitrogen by
mass, and nitrogen is the element plants need most for leaf growth.

Q9 Write the formulae for the compounds formed by the reaction of: (i) Magnesium
and nitrogen (ii) Lithium and nitrogen (iii) Sodium and sulfur (iv) Aluminium and
oxygen

Work out the ion each element forms, then balance the charges.

ELEMENTS CONFIGURATIONS IONS FORMED FORMULA CHARGE CHECK

(i) Mg and N 2, 8, 2 and 2, 5 Mg²⁺, N³⁻ Mg₃N₂ 3(+2) + 2(−3) = 0

(ii) Li and N 2, 1 and 2, 5 Li⁺, N³⁻ Li₃N 3(+1) + (−3) = 0

(iii) Na and S 2, 8, 1 and 2, 8, 6 Na⁺, S²⁻ Na₂S 2(+1) + (−2) = 0

(iv) Al and O 2, 8, 3 and 2, 6 Al³⁺, O²⁻ Al₂O₃ 2(+3) + 3(−2) = 0

Names: (i) magnesium nitride, (ii) lithium nitride, (iii) sodium sulfide, (iv) aluminium oxide.

Why it happens: nitrogen has five valence electrons and is three short of an octet,
so it takes three electrons and forms N³⁻. Magnesium can supply two each and
lithium one each — hence three Mg for two N (six electrons moved either way) and
three Li for one N. The subscripts are always the smallest whole numbers that make
the electrons balance exactly.

Page 53 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
Did you know? Al₂O₃ is the tough, invisible film that forms on aluminium the

m l a se
moment it meets air. It is what stops aluminium vessels from corroding further, and
o g by traces of
.c substance as the gemstones ruby and sapphire, coloured
a
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it is the same

l a semetal ions.
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other

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Complete the Table 9.3 by writingm
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the formulae of the compounds formed by the
a at the top. LiNO3 is given as an example.
Q10

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cations on the left and the anions

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NO 3 − SO 4 2− PO 4 3−
m l
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NH4+

agl Li+ LiNO3

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Al3+
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Cu2+
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Table 9.3, page 182 — as printed, with LiNO3 filled in as the example.
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a s em
gl

a Criss-cross the charge numbers in every cell, then simplify. Brackets go round a polyatomic ion

se m
a
whenever its subscript is 2 or more.

. com a g l
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NO₃⁻ SO₄²⁻ PO₄³⁻

NH₄⁺ NH₄NO₃ (NH₄)₂SO₄ (NH₄)₃PO₄

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Li⁺ LiNO₃ Li₂SO₄
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Li₃PO₄

m ase AlPO₄
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Al³⁺ Al(NO₃)₃ Al₂(SO₄)₃
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Cu(NO₃)₂ CuSO₄ Cu₃(PO₄)₂

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Two cells deserve a second look:

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AlPO₄ — criss-crossing 3 and 3 gives Al₃(PO₄)₃, which must be divided by the common factor agl
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3 to reach the simplest ratio 1 : 1.as
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CuSO₄ — criss-crossing 2 and 2 gives Cu₂(SO₄)₂, which divides by 2 to give CuSO₄.

co m
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m as e
.co


a g l Page 54 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: a chemical formula records the simplest whole-number ratio of
ions, not the number of ions in a lump of the solid. So after criss-crossing you must
always cancel any common factor — the same reason MgO is not written Mg₂O₂ and
CaCO₃ is not written Ca₂(CO₃)₂.

Tip: (NH₄)₂SO₄ and (NH₄)₃PO₄ both need brackets, because more than one
ammonium ion is present. NH₄NO₃ needs none — there is only one of each ion.

Q11 5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon
dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of
conservation of mass is valid.

The law is obeyed — both sides come to 11.3 g.

Reactants
Mass of sodium carbonate = 5.3 g

Mass of acetic acid = 6.0 g

Total mass of reactants = 5.3 g + 6.0 g = 11.3 g

Products

Mass of carbon dioxide = 2.2 g

Mass of water = 0.9 g
Mass of sodium acetate = 8.2 g

Total mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g

Mass of reactants = Mass of products = 11.3 g ✓

Hence the Law of Conservation of Mass is valid for this reaction.

Page 55 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: one of the products here is carbon dioxide, a gas. The figures add
up only because the reaction was carried out in a closed container, so the 2.2 g of
CO₂ was still there to be weighed. Run the same reaction in an open beaker and the
balance would show 11.3 g before and only 9.1 g after — and a careless student
might conclude the law had failed.

Tip: whenever a question of this type gives you a gas among the products, check
first whether the vessel is closed. That single word decides whether the arithmetic
will balance.

Q12 If a species has 11 protons, 12 neutrons and 10 electrons then (i) what is its atomic
number and mass number? (ii) is it neutral, a cation or an anion? Explain. (iii) write
its electronic configuration. (iv) name the species.

(i) Atomic number Z = 11, mass number A = 23.

Z = number of protons = 11

A = protons + neutrons = 11 + 12 = 23

Note that electrons play no part in either — their mass is negligible, and it is the proton count
that fixes the identity of the element.
(ii) It is a cation, carrying a charge of +1.

Charge from protons = +11
Charge from electrons = −10

Net charge = (+11) + (−10) = +1

There is one electron fewer than the number of protons, so one positive charge is left
unbalanced. An atom that has lost electrons is a cation.
(iii) Electronic configuration: 2, 8. The species has only 10 electrons, so the K shell takes 2 and
the L shell the remaining 8 — the same arrangement as neon. (The neutral atom would have
been 2, 8, 1.)

(iv) It is the sodium ion, Na⁺ — more precisely 2311Na⁺.

Page 56 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: identify the element from the protons alone. Z = 11 is sodium,
whatever the electron count may be — changing electrons changes the charge, not
the element. This is exactly what happens when sodium metal reacts: it loses its
single valence electron and becomes the stable Na⁺ ion of common salt.

Q13 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.

A is 2, 8, 5 — five valence electrons, three short of an octet (Z = 15, phosphorus). B is 2, 8, 7 —
seven valence electrons, one short (Z = 17, chlorine).
(i) B is more reactive.

A needs 8 − 5 = 3 electrons to complete its octet

B needs 8 − 7 = 1 electron to complete its octet

B is only one electron away from stability, so it reacts readily and with almost anything. A must
gather three electrons, which is a much harder demand, so it reacts less eagerly.
(ii) They form covalent bonds. Both A and B have more than four valence electrons, so neither
will hand electrons over — giving away five or seven electrons would need an enormous
amount of energy. And neither can simply take electrons from the other, because both are
trying to gain. The only workable arrangement is sharing.
So A shares one electron with each of three B atoms. Each shared pair completes one gap: A
gains three shared pairs and reaches eight, while each B gains one shared pair and reaches
eight.
(iii) Formula: AB₃ (in real terms PCl₃, phosphorus trichloride).

Electrons A must acquire = 3, and each B can supply a share of only 1

→ three B atoms per A atom → AB₃

Valency check: A has valency 3, B has valency 1; criss-cross → A₁B₃

Page 57 of 69

Class 9 Science Chapter 9 Atomic Foundations of Matter AglaSem · NCERT Solutions

Why it happens: a rough rule from section 9.4.2 — fewer than four valence
electrons and the atom donates (metal, ionic bonding); more than four and it gains
or shares (non-metal). Two non-metals together therefore always share, giving a
covalent compound. An ionic bond needs one partner willing to give, and here there
is none.

Q14 Assertion (A): Copper sulfate conducts electricity in the molten state but not in the
solid state. Reason (R): Copper and sulfate ions are fixed in the lattice in molten
state, while in solid state they can move freely. Choose the correct option: (i) Both
A and R are true, and R is the correct explanation of A. (ii) Both A and R are true,
but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false,
but R is true.

Option (iii) — A is true, but R is false.
A is true. Copper sulfate is an ionic compound. In the solid its Cu²⁺ and SO₄²⁻ ions are locked in
fixed lattice positions, so no charge can move and it does not conduct. Melt it and the lattice
breaks down; the ions become mobile and the melt conducts.
R is false — it has the two states exactly the wrong way round.

STATE WHAT R CLAIMS WHAT IS ACTUALLY TRUE

Solid ions can move freely ions are fixed in the lattice

Molten ions are fixed in the lattice ions move freely

Since R is a false statement, options (i) and (ii) are ruled out; and since A is true, (iv) is ruled out.
The answer is (iii).

Why it happens: melting is precisely the process of overcoming the forces that hold
particles in fixed positions. For an ionic solid those forces are the strong electrostatic
attractions between opposite charges, which is why the melting point is high — and
once they are overcome, the ions are free. A statement that has ions fixed in a liquid
and free in a solid is describing melting backwards.

Tip: in assertion-reason questions, test A and R separately first, and only then ask
whether R explains A. Here R fails at the first step, so the second step never arises.

Page 58 of 69

as e
Class 9 Science Chapter 9 Atomic Foundations of Matter
a g l AglaSem · NCERT Solutions

co m
m.
The species 27Al, 80Br– and 201Hg2+ have 13, 35 and 80 protons, respectively. How
e
Q15

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many electrons and neutrons do they have?

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a

Use two rules: neutrons = mass number − protons, and electrons = protons − charge (a

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positive charge means electrons are missing, a negative charge means extra electrons are
. co ag
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present).

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SPECIES MASS NUMBER A g
aPROTONS CHARGE ELECTRONS NEUTRONS

27Al 27 13 0 (neutral atom) 13
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27 − 13 = 14

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80Br⁻ 80 35 −1 (gained 1 e⁻) 35 + 1 = 36 80 − 35 = 45

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201Hg²⁺ 201 80 +2 (lost 2 e⁻) 80 − 2 = 78 201 − 80 = 121

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27Al → 13 electrons, 14 neutrons

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80Br⁻ → 36 electrons, 45 neutrons
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201Hg²⁺ → 78 electrons, 121 neutrons
a

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Why it happens: forming an ion touches only the electrons. Protons and neutrons
a g l
a s em sit in the nucleus and are untouched by chemistry, so the mass number of Br⁻ is the

agl same 80 as that of a bromine atom, and Hg²⁺ still has 121 neutrons. Only the
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electron count — and therefore the charge — has changed.

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Check it yourself: Br⁻ with 36 electrons has the configuration 2, 8, 18, 8 — the same

agl
as krypton. Gaining that one electron is exactly what a halogen wants, which is why
bromine is so reactive.

com
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.co a g l
a s em
The Journey Beyond — Page 183

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