Atomic Foundations of Matter: Class 9 Science · Chapter 9

Exploration · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

95 min
1 simulation
37 practice questions
30 MCQs

What you will learn in Atomic Foundations of Matter

Show by weighing that mass stays the same in physical and chemical changes, and explain why a gas must be trapped for this to show
Use the law of conservation of mass and the law of constant proportions to find unknown masses in reactions and compounds
State Dalton's postulates and use them to explain the two laws of chemical combination
Explain how atoms reach a stable octet or duplet by sharing electrons (covalent bonds) or transferring them (ionic bonds), and describe dot-and-cross diagrams of simple molecules and ions
Name simple covalent and ionic compounds and write their formulae from valencies, simplifying and using brackets for polyatomic ions
Compare ionic and covalent compounds by solubility, electrical conductivity and melting point, and explain the differences in terms of ions and molecules
Calculate molecular masses and formula unit masses from atomic masses

Original companion text aligned to NCERT Science Class 9, Chapter 9, pp. 162–183. Not affiliated with or endorsed by NCERT.

Does mass change when matter changes?

Chapter 8, Journey Inside the Atom, took you inside the atom: a nucleus of protons and neutrons, with electrons arranged in shells around it. It also showed that an atom is at its most stable when its outermost shell holds eight electrons, or two if that shell is the first (K) shell. This chapter picks up from there. It asks how atoms hold on to one another, how we write and name what they build, and how much the products of a reaction weigh.

Begin with a surprise from the kitchen shelf. Sodium is a soft, silvery metal that reacts violently with water, and chlorine is a choking, greenish-yellow gas. Joined together, they become sodium chloride: the harmless table salt in every namak-daani. A compound is a brand-new substance with properties of its own, not a mixture of its elements' properties. So here is a fair question: when one substance turns into another, does the total amount of matter change? A balance can answer that.

Activity 9.1NCERT Activity 9.1 · p. 163

Does dissolving change the weight?

You’ll need

a digital kitchen scale that reads to 0.1 g
a steel tumbler or glass
water
a small cube of jaggery (gur) or a spoon of sugar
a stick of chalk and a sheet of paper

Predict first

With the jaggery resting on the pan beside the water, the scale shows one reading. Once the jaggery has dissolved, the reading will be…

lower, because the jaggery has vanishedthe same as beforehigher, because the water now holds jaggery

Put the empty tumbler on the scale and press tare (or zero), so that the display reads 0.0 g.
Pour in water until the tumbler is about half full. Place the jaggery on a small square of paper beside the tumbler, still on the pan, and note the reading.
Drop the jaggery into the water, leaving the paper on the pan. Swirl the tumbler gently, without lifting it, until no lumps remain.
Note the reading again.
Next, weigh a stick of chalk lying on a sheet of paper. Crush it to powder on that same paper and weigh it once more.

Table 9.1

Change	Reading before (g)	Reading after (g)
Jaggery dissolving
Chalk crushed

Your entries are saved automatically.

A physical change keeps every particle on the pan, so the reading holds steady. A chemical change makes new substances, and some of them may be gases. Does the balance still agree when that happens?

Activity 9.2NCERT Activity 9.2 · p. 163

Fizz in a bottle: open or closed?

You’ll need

an empty 500 mL fizzy-drink bottle with its cap
vinegar or lemon juice
baking soda (sodium hydrogencarbonate)
a small steel bowl
tissue paper and thread
a digital scale

Predict first

In Set-up A the bottle stays open while the mixture fizzes. What will the final reading be?

Higher than the first readingExactly the same as the first readingLower than the first reading

Set-up A: bottle left open
- Pour vinegar into the bottle to a depth of about 3 cm. Put a quarter teaspoon of baking soda in the small bowl. Stand the bottle and the bowl on the scale together and note the reading.
- Tip the soda into the bottle and leave the empty bowl on the pan. Wait until the fizzing stops, then note the reading again.
Set-up B: cap screwed on
- Rinse the bottle and refill it with fresh vinegar to the same depth. Wrap the same amount of soda in a twist of tissue paper and tie it with a thread.
- Let the packet hang inside the neck, with the end of the thread trapped under the cap, and screw the cap on tightly. Weigh the closed bottle.
- Lift the bottle, turn it upside down once so the vinegar soaks the packet, and stand it back on the scale. Note the reading when the fizzing stops.

Table 9.2

Set-up	First reading (g)	Final reading (g)	Same or different?
A: open
B: capped

Your entries are saved automatically.

In Set-up A the reading falls by a few tenths of a gram. In Set-up B the trapped gas pushes on the cap and the bottle feels firm, yet the reading does not change. The fizz is carbon dioxide, a gas made by the reaction. In the open bottle it escaped, so the scale lost track of it; in the capped bottle it stayed on the pan. With every product kept on the pan, the total mass after the reaction equals the total mass before it. That is also why the empty bowl stays on the pan: a few grains of soda may cling to it.

Can’t do it now? Try the simulation · Fig. 9.2

4/4The gas stays inside the bottle: still 158.9 g.

Fig. 9.1Gas that escapes is not weighed; gas that is trapped is.NCERT Fig. 9.2 · p. 164

A plastic bottle of vinegar stands on a digital kitchen scale. In the first step the bottle is open and a small bowl of baking soda sits beside it; the display reads 152.6 g. In the second step the soda has been tipped in, bubbles fill the liquid, arrows show gas leaving the open neck and the display has dropped to 152.1 g, marked lighter. In the third step a fresh bottle has its cap screwed on, with a tissue packet of soda hanging inside the neck; the display reads 158.9 g. In the fourth step the packet has fallen into the vinegar, bubbles rise, a label says the gas stays inside, and the display still reads 158.9 g, marked same.

Quick check

1 of 2

Which change would make the reading on a balance go down if it happened in an open container?

Ice melting in a glassSugar dissolving in waterChalk being crushed to powderBaking soda fizzing in vinegar

Show answer

Answer (D) Baking soda fizzing in vinegar

The fizz is a gas, carbon dioxide, which leaves an open container. Melting, dissolving and crushing are physical changes in which nothing leaves the pan.

9.1 Law of Conservation of Mass

The capped bottle shows something that holds for every chemical reaction ever measured. The vinegar and soda turned into new substances, including a gas, yet the total mass on the pan stayed fixed. The open bottle only seemed to lose mass because one product walked out of the experiment. The French chemist Antoine Lavoisier reached this conclusion in 1789 after years of weighing reactions in sealed glass vessels.

LawLaw of conservation of mass

In a chemical reaction, matter is neither created nor destroyed: the total mass of the products equals the total mass of the reactants.

This is the law of conservation of mass (द्रव्यमान संरक्षण का नियम). It gives chemists a powerful bookkeeping rule. If you know the masses of all but one substance in a reaction, the law tells you the missing one.

You can watch the same idea at the level of particles. The simulation below shows a warm container of water. With the lid off, fast-moving particles escape as vapour, and the mass left in the container falls. With the lid on, every particle stays inside, just as the gas stayed inside the capped bottle.

SimulationOpen or closed: where do the particles go?

A container of warm water drawn as moving particles, with a thermometer, a temperature slider and a lid switch. With the lid off, fast particles leave the surface and a counter shows how many have escaped and how many are left. With the lid on, particles that leave the liquid bounce back from the lid, so the escaped count stops rising and every particle stays in the container.

Fig. 9.2Press Play. The counter shows how many particles have escaped from the container.

Try this

Leave the lid off and press Play. How many particles escape in about 20 seconds?
Switch the lid on. Does the escaped counter still rise? Which set-up in Activity 9.2 is this like?
Raise the temperature with the lid off. Why would an open pan of boiling dal weigh less every minute?

Activity 9.3NCERT Activity 9.3 · p. 165

Two clear solutions, one white solid

You’ll need

two test tubes and a small beaker
dilute sodium sulfate solution
dilute barium chloride solution
a balance that reads to 0.01 g, if your school has one

Predict first

Both solutions are colourless. When they are mixed, a white solid appears. Compared with the reading before mixing, the reading after mixing will be…

higher, because a solid has formedlower, because the liquids have been used upthe same, because nothing has left the pan

Label two clean test tubes P and Q and stand them in a small beaker on the balance. Press tare.
Pour about 5 mL of sodium sulfate solution into P and about 5 mL of barium chloride solution into Q. Note the total reading.
Pour the liquid in Q into P. Put tube Q back in the beaker, and stand P back in the beaker too.
Watch what forms in P, then note the reading again.

Table 9.3

What to record	Observation
Look of the two solutions before mixing
Look of the mixture
Reading before (g)
Reading after (g)

Your entries are saved automatically.

We can write the reaction of Activity 9.3 as a word equation and as a balanced chemical equation. Count the atoms on each side of the chemical equation: every atom on the left turns up on the right.

word equation

\begin{aligned} sodium sulfate + barium chloride \\ \to barium sulfate + sodium chloride \end{aligned}

(9.1)

N a_{2} S O_{4} + B a C l_{2} \overset{}{\to} B a S O_{4} + 2 N a C l

(9.2)

Example 9.1NCERT Example 9.1 · p. 166

Checking the law with measured masses

In a school laboratory, 1.20 g of magnesium ribbon is dropped into dilute hydrochloric acid containing 3.65 g of hydrogen chloride, inside a flask closed with a stopper. When the reaction is over, the flask contains 4.75 g of magnesium chloride and 0.10 g of hydrogen gas. Do these results obey the law of conservation of mass?

Try it yourself first. Then reveal the solution one step at a time.

Example 9.2NCERT Example 9.2 · p. 166

Scaling up from a known reaction

When magnesium burns in air, 24 g of magnesium combines with 16 g of oxygen to form 40 g of magnesium oxide. A student burns 6.0 g of magnesium completely. How much magnesium oxide forms, and how much oxygen is used?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A candle stands on a balance and burns for ten minutes. Why does the reading fall?

Mass is destroyed when wax burnsWax turns into products that are lighter than airThe flame pushes the candle upwardsGases made by the burning escape into the room

Show answer

Answer (D) Gases made by the burning escape into the room

Burning wax makes carbon dioxide and water vapour, which drift away. If they were trapped and weighed, the total mass would be at least as large as the wax burnt, since oxygen from the air joins in.

9.2 Law of Constant Proportions

The reactions so far all used compounds such as baking soda, magnesium oxide and sodium chloride. Here is a sharper question about them. Can hydrogen and oxygen join in any amounts to make water, a little more oxygen in one sample and a little less in another? Or is the recipe fixed?

In the 1790s the French chemist Joseph Proust answered this by analysing compounds collected from very different places and also made in the laboratory. Every pure sample of a compound had exactly the same make-up by mass. Try it with carbon dioxide. Collect some from three very different places: the air you breathe out, a glowing charcoal angithi and the fizz of a soda bottle. Purify each sample and analyse 11 g of it. Every one gives 3 g of carbon and 8 g of oxygen. Pure water is just as strict: one part hydrogen to eight parts oxygen by mass, every time.

3/3Carbon : oxygen is 3 : 8 by mass in every sample.

Fig. 9.3Pure carbon dioxide has the same composition by mass, whatever its source.NCERT Law of Constant Proportions · p. 167

Three rows are labelled breath, charcoal and soda. In the first step each row shows a grey bar labelled 11 g carbon dioxide. In the second step each bar splits into a shorter blue piece labelled 3 g carbon and a longer orange piece labelled 8 g oxygen. In the third step a line underneath reads always 3 : 8.

LawLaw of constant proportions

Every sample of a pure compound has the same make-up by mass: its elements are always in the same ratio, wherever the sample came from.

This is the law of constant proportions (स्थिर अनुपात का नियम). You will also see it called the law of definite proportions or Proust's law. It applies to compounds, not to mixtures: you can stir any amount of sugar into tea, but you cannot make water with a little extra oxygen.

Example 9.3NCERT Example 9.3 · p. 167

Using a fixed ratio, with a leftover

Iron sulfide ( $F e S$ ) always contains iron and sulfur in the mass ratio 7 : 4. A student heats 14 g of iron filings with 10 g of sulfur powder until no more reaction takes place. How much sulfur reacts, how much is left over, and how much iron sulfide forms?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

In pure carbon dioxide, the masses of carbon and oxygen are always in the ratio 3 : 8. How much oxygen is present in a sample that contains 6 g of carbon?

8 g11 g16 g22 g

Show answer

Answer (C) 16 g

Double the carbon means double the oxygen:

6 = 2 \times 3

, so oxygen

= 2 \times 8 = 16

g. 22 g is the mass of the whole sample, not of the oxygen.

Practice · 6 questions

Pause and ponder: questions with answers

Short answerEasyPause and Ponder Q1
During aarti, a small piece of camphor burns away completely in a diya and leaves nothing behind. Does this show that mass can be destroyed? Explain.
Show solution
AnswerNo. The camphor reacts with oxygen to form carbon dioxide and water vapour, which escape into the air.
Burning is a chemical reaction between camphor and oxygen from the air. Its products, carbon dioxide and water vapour, are gases that mix into the room, so the diya looks empty. If all the gases could be collected and weighed, their mass would equal the mass of camphor plus the mass of oxygen used. Mass is conserved; it has only moved out of sight.
NumericalEasyPause and Ponder Q2
A 4.8 g piece of magnesium ribbon burns completely, using up 3.2 g of oxygen from the air. What mass of white magnesium oxide forms?
g
Show answer
Answer 8 g
By conservation of mass, $4.8 + 3.2 = 8.0$ g of magnesium oxide. Check: $4.8 : 3.2 = 3 : 2$ , the same as $24 : 16$ , so neither reactant is left over.
NumericalMediumPause and Ponder Q3
Pure calcium carbonate, the substance of chalk and marble, is 40 % calcium by mass. How much calcium is there in a 25 g piece of pure calcium carbonate?
g
Show answer
Answer 10 g
$\frac{40}{100} \times 25 = 10$ g of calcium. Every pure sample has the same 40 %, whatever its size or source.
NumericalMediumPause and Ponder Q4
In ammonia, nitrogen and hydrogen are combined in the mass ratio 14 : 3. What mass of hydrogen combines with 28 g of nitrogen, and what mass of ammonia forms?
Show solution
Answer6 g of hydrogen; 34 g of ammonia.
$28 \div 14 = 2$ , so the hydrogen needed is $2 \times 3 = 6$ g. By conservation of mass, ammonia formed $= 28 + 6 = 34$ g.
Short answerMediumPause and Ponder Q5
Why does the law of constant proportions hold for compounds such as water but not for mixtures such as nimbu paani?
Show solution
AnswerA compound has a fixed recipe set by how its atoms join; a mixture's parts can be in any amounts.
In a compound, atoms of the elements are joined in a fixed number ratio, and each kind of atom has a fixed mass, so the mass ratio is fixed too. In nimbu paani the water, lemon juice, sugar and salt are simply mixed, not joined chemically. You can add more sugar or less lemon, so the proportions change from glass to glass.
Short answerMediumPause and Ponder Q6
Two samples of magnesium oxide are analysed. Sample 1, bought from a chemical supplier, contains 3.0 g of magnesium combined with 2.0 g of oxygen. Sample 2, made by burning ribbon in the school laboratory, contains 7.2 g of magnesium combined with 4.8 g of oxygen. Do the results support the law of constant proportions?
Yes: both ratios simplify to 3 : 2No: sample 2 contains more magnesiumNo: the masses of oxygen are different
Show answer
Answer (A) Yes: both ratios simplify to 3 : 2
$3.0 : 2.0 = 3 : 2$ and $7.2 : 4.8 = 3 : 2$ . The amounts differ, but the proportion is the same, which is exactly what the law predicts.

9.3 Dalton’s Atomic Theory

The two laws describe what happens; they do not say why. Why should the products always weigh exactly as much as the reactants? Why should water never be 1 : 7 or 1 : 9? In 1808 the English teacher and scientist John Dalton offered an answer built on one bold idea: matter is made of tiny, indestructible particles called atoms.

Dalton stated his theory as a set of postulates (अभिधारणा). A postulate is a basic idea that a theory takes as its starting point. It is not proved directly; instead, scientists test the predictions that follow from it, and the postulate survives as long as those predictions keep matching experiments. Here are his postulates, sorted into two groups.

About atoms
- Atoms are the tiny particles from which all matter is built, and they are what actually take part in a reaction.
- No chemical reaction can split, create or destroy an atom.
- Within one element every atom is alike in mass and behaviour, while different elements have atoms of different masses and behaviour.
About compounds
- To form a compound, atoms join in small whole-number ratios, such as 1 : 1, 1 : 2 or 2 : 3.
- Any particular compound always contains the same kinds of atoms, in the same relative numbers.

Now the two laws follow naturally. In a reaction, atoms only part company and join new partners, as Fig. 9.4 shows. No atom appears or vanishes, and each keeps its own mass, so the total mass cannot change. That is the law of conservation of mass. Then think of water. Every water molecule holds two hydrogen atoms (1 u each) and one oxygen atom (16 u). So every sample has 2 u of hydrogen for every 16 u of oxygen, which is 1 : 8. Fixed numbers of atoms with fixed masses give fixed mass ratios: the law of constant proportions.

4/4Four H and two O atoms before, four H and two O atoms after: the mass cannot change.

Fig. 9.4A reaction regroups atoms; it never makes or destroys them.

Atoms are drawn as balls: small blue balls marked H and larger orange balls marked O. In the first step, on the left, there are two hydrogen molecules, each a pair of H balls, and one oxygen molecule, a pair of O balls. In the second step each pair pulls slightly apart as the bonds break. In the third step the atoms move to the right and regroup as two water molecules, each an O ball with two H balls. In the fourth step labels underneath read before: 4 H, 2 O and after: 4 H, 2 O.

9.4 How Atoms Combine

Atoms seldom stay alone. Two oxygen atoms pair up to make an oxygen molecule, $O_{2}$ , the form of oxygen in the air we breathe. This settles one of the opening questions: O stands for a single oxygen atom, while $O_{2}$ stands for a molecule of two. Atoms of different elements join too. One carbon atom and two oxygen atoms make a molecule of carbon dioxide, $C O_{2}$ , a compound.

Definition

Molecule

The smallest particle of a substance that can exist by itself: two or more atoms joined together, with no overall charge, showing the properties of that substance.

Example $O_{2}$ is a molecule of an element; $C O_{2}$ is a molecule of a compound.

A few elements form no molecules at all. Helium, neon and argon already have full outer shells, so their atoms are stable on their own and the gases exist as separate atoms. That is why argon can fill an electric bulb without reacting with the hot filament.

Most other atoms have an outer shell that is not full: fewer than eight electrons, or fewer than two when the K shell is the only shell. Such an atom can reach a stable arrangement in one of two ways:

by sharing electrons with another atom, so that the shared electrons count for both atoms; or
by transferring electrons: one atom gives away one or more electrons and another atom takes them.

Either way, the joined atoms end up with less energy than they had apart. A system with lower energy is more stable, much as a marble resting at the bottom of a bowl is more settled than one balanced on its rim. Whatever keeps the atoms joined after this is called a chemical bond (रासायनिक आबंध).

Definition

Chemical bond

An attraction strong enough to keep two atoms (or ions) joined, so that they act as parts of one molecule or compound.

9.4.1 Bonding by sharing electrons: the covalent bond

Molecules of elements

Start with the smallest atom. Hydrogen (atomic number 1) has a single electron in its K shell, which has room for two. Two hydrogen atoms solve this together: each puts its electron into a shared pair that sits between the two nuclei. The pair counts for both atoms, so each now has two electrons around it, a full K shell. What holds the molecule together? The two positive nuclei are both pulled towards the negative pair between them, so neither atom can drift away.

Definition

Covalent bond

A chemical bond formed when two atoms share a pair of electrons, one electron usually coming from each atom.

ExampleThe bond in a hydrogen molecule, H–H.

A bond made of one shared pair is a single bond, drawn as one line: H–H. Chlorine (2, 8, 7) is one electron short of an octet, so two chlorine atoms also share one pair, Cl–Cl. Oxygen (2, 6) is two short. Two oxygen atoms therefore share two pairs, a double bond, drawn O=O. Nitrogen (2, 5) is three short, so the two atoms in $N_{2}$ share three pairs, a triple bond, N≡N. Step through Fig. 9.5 and count the electrons around each atom: in every molecule each atom ends up with a full shell.

4/5Oxygen atoms share two pairs: a double bond.

Fig. 9.5Shared pairs sit where the outer shells overlap. Dots and crosses only show where each electron came from; all electrons are identical.NCERT Figs 9.6–9.8 · p. 169

Each atom is drawn as a circle for its outer shell, with its symbol inside; electrons from one atom are dots and from the other atom are crosses. In the first step two hydrogen circles are apart, each with one electron. In the second step they overlap and the dot and cross sit together in the overlap, labelled H–H, one shared pair, single bond. In the third step two chlorine circles overlap with one shared pair and three unshared pairs on each atom, labelled Cl–Cl. In the fourth step two oxygen circles share two pairs, with two unshared pairs on each atom, labelled O=O, double bond. In the fifth step two nitrogen circles share three pairs, with one unshared pair on each atom, labelled N≡N, triple bond.

Molecules of compounds

Atoms of different elements share electrons in just the same way. Hydrogen needs one more electron and so does chlorine, so in hydrogen chloride they share one pair: H–Cl, a single bond. Oxygen needs two electrons, but a hydrogen atom can offer only one. So the oxygen atom shares one pair with each of two hydrogen atoms, making water, $H_{2} O$ , with two single bonds. A compound whose atoms are held together by shared electrons is called a covalent compound.

Notice the pattern in Fig. 9.6: each hydrogen atom makes up for one missing electron. Nitrogen is three short, so it bonds to three hydrogen atoms in ammonia, $N H_{3}$ . Carbon is four short, so it bonds to four in methane, $C H_{4}$ , the main gas in the biogas from a gobar gas plant.

2/4Oxygen is two short: two hydrogens, two unshared pairs.

Fig. 9.6The number of hydrogen atoms equals the number of electrons the central atom is short of an octet.NCERT Figs 9.9–9.10 · p. 171

A large green circle for the outer shell of a central atom sits in the middle; small blue circles for hydrogen atoms overlap it. Dots are the central atom's electrons and crosses are hydrogen electrons. In the first step the central atom is Cl with one hydrogen on the right and three unshared pairs; a side note says Cl needs 1, so 1 hydrogen. In the second step it is O with hydrogens on the right and below and two unshared pairs. In the third step it is N with three hydrogens and one unshared pair. In the fourth step it is C with four hydrogens on all sides and no unshared pairs.

In the real worldNCERT Bridging Science and Society · p. 171

Bridging science and society: energy from the nucleus

Chemical reactions only rearrange electrons; the nuclei are untouched. Nuclear reactions are different: when a heavy nucleus such as uranium splits, it releases millions of times more energy per atom than any burning fuel. India's first nuclear power station began working at Tarapur in Maharashtra in 1969, and plants such as Kudankulam in Tamil Nadu now generate electricity without burning coal. Homi J. Bhabha founded India's atomic energy programme. Raja Ramanna, a physicist who later headed the Bhabha Atomic Research Centre and the Atomic Energy Commission, shaped the country's nuclear research for decades. Substances made radioactive in reactors are also used to treat cancer and to sterilise medical equipment.

Naming covalent compounds

Many covalent compounds are made of just two non-metals. Their names tell you how many atoms of each element a molecule contains, using number prefixes taken from Greek.

Table 9.4Number prefixes

Number of atoms	Prefix
1	mono-
2	di-
3	tri-
4	tetra-
5	penta-
6	hexa-

Give each element a prefix for its number of atoms.
Exception: the first element never takes mono-. One carbon and one oxygen make carbon monoxide, not monocarbon monoxide.
Change the ending of the second element to -ide: oxygen to oxide, chlorine to chloride, sulfur to sulfide.
When a prefix ending in a or o runs into a vowel, its last letter drops out, so mono- + oxide is written monoxide. A prefix ending in i keeps it, as in trioxide.
A compound whose formula starts with hydrogen gets no prefix for the hydrogen at all.

$N O$ : nitrogen monoxide
$S O_{2}$ : sulfur dioxide, and $S O_{3}$ : sulfur trioxide
$C C l_{4}$ : carbon tetrachloride
$P C l_{5}$ : phosphorus pentachloride
$N_{2} O$ : dinitrogen monoxide, the 'laughing gas' once common in dentists' clinics
$H C l$ : hydrogen chloride, and $H_{2} S$ : hydrogen sulfide (hydrogen first, so no prefix)

A handful of compounds go by everyday names that you should always use: water, $H_{2} O$ ; ammonia, $N H_{3}$ ; and methane, $C H_{4}$ .

Quick check

1 of 3

How many electrons are shared between the two atoms of an oxygen molecule?

2468

Show answer

Answer (B) 4

A double bond is two shared pairs, which is four electrons. Eight is the number each oxygen atom ends up with around it, counting the shared ones.

Practice · 4 questions

Pause and ponder: questions with answers

DiagramMediumPause and Ponder Q8
Bromine (atomic number 35) has seven electrons in its outermost shell and exists as $B r_{2}$ molecules. Describe the dot-and-cross diagram of $B r_{2}$ and name the type of bond.
Show solution
AnswerOne shared pair (a single covalent bond, Br–Br) and three unshared pairs on each atom.
Each bromine atom has 7 outer electrons and needs 1 more. Each atom puts one electron into a shared pair, so each now counts 8 outer electrons: 2 in the shared pair and 6 in three unshared pairs. Draw two overlapping circles with one dot and one cross in the overlap and three pairs of dots (or crosses) on the outer side of each atom. One shared pair is a single covalent bond, Br–Br.
DiagramMediumPause and Ponder Q9
Fluorine has atomic number 9. Explain how a hydrogen atom and a fluorine atom reach stable shells in a molecule of hydrogen fluoride, $H F$ .
Show solution
AnswerThey share one pair: H gets a full K shell (2), F gets an octet (8). Single bond, H–F.
Fluorine is 2, 7, so it is one electron short of an octet; hydrogen is one short of a full K shell. Each gives one electron to a shared pair. Hydrogen now has 2 electrons around it, and fluorine has 8 (the shared pair plus three unshared pairs). The single shared pair is a single covalent bond, H–F.
10
DiagramHardPause and Ponder Q10
Show how electrons are shared in each molecule.
1. Carbon disulfide, $C S_{2}$ (carbon 2, 4; sulfur 2, 8, 6)
  Show solution
  AnswerS=C=S: two double bonds.
  Carbon needs 4 electrons and each sulfur needs 2. Carbon shares two pairs with each sulfur atom, giving two double bonds, S=C=S. Carbon then counts 8 electrons (four shared pairs) and each sulfur counts 8 (two shared pairs plus two unshared pairs). Carbon dioxide, O=C=O, is built the same way.
2. Phosphine, $P H_{3}$ (phosphorus has atomic number 15)
  Show solution
  AnswerThree P–H single bonds and one unshared pair on P.
  Phosphorus is 2, 8, 5, so it is three short of an octet, like nitrogen. It shares one pair with each of three hydrogen atoms. Phosphorus ends with three shared pairs and one unshared pair, 8 electrons; each hydrogen has 2.
Short answerEasyPause and Ponder Q11
A bulb is filled with argon (atomic number 18). Why does argon neither share nor transfer electrons with the metal filament?
Show solution
AnswerArgon is 2, 8, 8: its outer shell is already full, so it has nothing to gain by bonding.
Argon's electrons are arranged 2, 8, 8, so its outermost shell already holds a stable octet. Sharing, losing or gaining electrons would not make it more stable, so argon atoms stay single and unreactive. That is exactly why it is used to fill bulbs.

9.4.2 Bonding by electron transfer: the ionic bond

Sharing is not the only route to a full shell. An atom with only one, two or three electrons in its outer shell usually finds it easier to give them away than to collect five or more extra. An atom with five, six or seven outer electrons tends to gain the few it is missing, or to share. Lithium, sodium, magnesium and aluminium are typical givers among the first 18 elements. Which others can you find in Chapter 8's table of electron arrangements?

Take sodium (atomic number 11), arranged 2, 8, 1. If it loses its single outer electron, the full shell of eight beneath becomes its outer shell. But it now has 11 protons and only 10 electrons, so it carries one unit of positive charge. This charged particle is a sodium ion, written $N a^{+}$ . An ion with a positive charge is a cation (धनायन).

sodium loses an electron

N a \overset{}{\to} N a^{+} + e^{-}

(9.3)

Chlorine (2, 8, 7) does the opposite. It accepts one electron and completes its octet, giving 17 protons and 18 electrons: one unit of negative charge. This is a chloride ion, $C l^{-}$ , and an ion with a negative charge is an anion (ऋणायन). Cations and anions together are called ions (आयन).

chlorine gains an electron

C l + e^{-} \overset{}{\to} C l^{-}

(9.4)

3/4Now $N a^{+}$ (2, 8) and $C l^{-}$ (2, 8, 8): both have octets.

Fig. 9.7Transfer, then attraction: sodium gives one electron, chlorine takes it, and the oppositely charged ions hold together.NCERT Figs 9.11–9.13 · p. 172

Two atoms are drawn with their electron shells as rings and electrons as dots: sodium on the left with 2, 8, 1 electrons and chlorine on the right with 2, 8, 7. Sodium's single outer electron is drawn as an open circle. In the second step a dashed arrow labelled 1 electron carries it across to the gap in chlorine's outer ring. In the third step sodium's empty outer ring has gone and both atoms sit in square brackets as ions: sodium ion, 2, 8, with a plus charge, and chloride ion, 2, 8, 8, with a minus charge. In the fourth step the ions move closer with a double-headed arrow labelled attract, and the words ionic bond appear below.

Opposite charges attract. Once sodium has handed its electron to chlorine, the $N a^{+}$ and $C l^{-}$ ions pull on each other strongly, and sodium chloride is formed. This attraction is an ionic bond (आयनिक आबंध), and a compound held together this way is an ionic compound.

Definition

Ionic bond

The pull between ions of opposite charge, an electrostatic attraction, that holds them together in an ionic compound.

ExampleThe attraction between $N a^{+}$ and $C l^{-}$ ions in common salt.

2/3Ions repeat in a checkerboard: every neighbour has the opposite charge.

Fig. 9.8One layer of the crystal. Each sodium ion touches four chloride ions in its layer, plus one above and one below: six in all.NCERT Fig. 9.14 · p. 173

Small blue circles marked plus are sodium ions and larger green circles marked minus are chloride ions. In the first step only one sodium ion and one chloride ion are shown side by side. In the second step a whole layer appears, seven ions across and five down, alternating plus and minus like a checkerboard. In the third step the other ions fade, and a dashed square links the central sodium ion to its four chloride neighbours in the layer.

Some atoms move two electrons. Magnesium (2, 8, 2) gives away two and becomes $M g^{2 +}$ . Oxygen (2, 6) and sulfur (2, 8, 6) each take two and become the oxide ion, $O^{2 -}$ , and the sulfide ion, $S^{2 -}$ . But a chlorine atom can accept only one electron. So when magnesium meets chlorine, two chlorine atoms are needed to take magnesium's two electrons, and the compound is $M g C l_{2}$ : one $M g^{2 +}$ ion for every two $C l^{-}$ ions (Fig. 9.9).

3/3One $M g^{2 +}$ and two $C l^{-}$ : the formula is $M g C l_{2}$ .

Fig. 9.9One magnesium atom needs two chlorine atoms to take its two outer electrons.

Only outer shells are drawn. A magnesium circle in the middle has two dots; a chlorine circle on each side has seven crosses. In the second step dashed arrows carry one magnesium electron to each chlorine atom. In the third step square brackets enclose all three ions, the magnesium shell is faded, and the electron arrangements change to 2, 8 for magnesium and 2, 8, 8 for each chlorine; charge labels show one magnesium ion with charge 2 plus and two chloride ions with charge minus, and the formula MgCl2 appears below.

Naming ionic compounds

The name of an ionic compound reads like a pair: first the cation, then the anion. Because metals tend to lose electrons and non-metals tend to gain them, the first word is usually a metal and the second a non-metal whose ending has changed to -ide. Some toothpastes contain sodium fluoride, sunscreens often contain zinc oxide, and old black-and-white films used silver bromide. A cation simply keeps the name of its element.

Some ions are groups of atoms bonded together that move about as one unit and carry one overall charge. These are polyatomic ions, such as sulfate, $S O_{4}^{2 -}$ , and ammonium, $N H_{4}^{+}$ , the commonest positive one. Most polyatomic anions have names ending in -ate, not -ide; hydroxide, $O H^{-}$ , is the familiar exception.

The size of the charge on an ion is its valency (संयोजकता), the combining capacity of the atom or group. Iron and copper can each form two different ions, so their names carry a label: ferrous ( $F e^{2 +}$ ) and ferric ( $F e^{3 +}$ ), cuprous ( $C u^{+}$ ) and cupric ( $C u^{2 +}$ ). The two tables below, grouped by valency, list the ions used in this chapter.

Table 9.5Common cations, by valency

Valency	Cations
1	sodium $N a^{+}$ , potassium $K^{+}$ , lithium $L i^{+}$ , silver $A g^{+}$ , cuprous $C u^{+}$ , ammonium $N H_{4}^{+}$
2	magnesium $M g^{2 +}$ , calcium $C a^{2 +}$ , barium $B a^{2 +}$ , zinc $Z n^{2 +}$ , ferrous $F e^{2 +}$ , cupric $C u^{2 +}$
3	aluminium $A l^{3 +}$ , ferric $F e^{3 +}$

Table 9.6Common anions, by valency (the polyatomic ones are ammonium among the cations and hydroxide, nitrate, hydrogencarbonate, carbonate, sulfate and phosphate here)

Valency	Anions
1	fluoride $F^{-}$ , chloride $C l^{-}$ , bromide $B r^{-}$ , iodide $I^{-}$ , hydroxide $O H^{-}$ , nitrate $N O_{3}^{-}$ , hydrogencarbonate $H C O_{3}^{-}$
2	oxide $O^{2 -}$ , sulfide $S^{2 -}$ , carbonate $C O_{3}^{2 -}$ , sulfate $S O_{4}^{2 -}$
3	nitride $N^{3 -}$ , phosphate $P O_{4}^{3 -}$

Quick check

1 of 4

Aluminium is arranged 2, 8, 3. Which ion does an aluminium atom form?

A l^{+}

, by losing one electron

A l^{3 -}

, by gaining three electrons

A l^{5 -}

, by gaining five electrons

A l^{3 +}

, by losing three electrons

Show answer

Answer (D) $A l^{3 +}$ , by losing three electrons

Losing three electrons leaves the full 2, 8 arrangement and 13 protons against 10 electrons: a charge of 3+. Gaining five would be far harder, and losing an electron makes an ion positive, never negative.

Practice · 4 questions

Pause and ponder: questions with answers

Short answerEasyPause and Ponder Q12
Fluorine has atomic number 9. Which ion does a fluorine atom form, and how?
Show solution
AnswerThe fluoride ion, $F^{-}$ , by gaining one electron.
Fluorine is 2, 7, one electron short of an octet. It gains one electron to become 2, 8. With 9 protons and 10 electrons it carries one negative charge: the fluoride ion, $F^{-}$ .
Quick answerMediumPause and Ponder Q13
An aluminium atom gives away electrons to become $A l^{3 +}$ . An oxygen atom accepts electrons to become $O^{2 -}$ . So in aluminium oxide, aluminium ions combine with oxide ions.
Show answer
Answer (1) 3, (2) 2, (3) 2, (4) 3
Two Al atoms give away $2 \times 3 = 6$ electrons and three O atoms accept $3 \times 2 = 6$ , so the formula is $A l_{2} O_{3}$ .
14
DiagramMediumPause and Ponder Q14
Use electron arrangements to show how each metal forms its cation and then its fluoride.
1. Lithium (atomic number 3)
  Show solution
  Answer $L i$ (2, 1) → $L i^{+}$ (2); fluoride $L i F$ .
  Lithium (2, 1) loses its one outer electron and becomes $L i^{+}$ , arranged 2. A fluorine atom (2, 7) takes that electron and becomes $F^{-}$ (2, 8). One $L i^{+}$ balances one $F^{-}$ , so lithium fluoride is $L i F$ .
2. Magnesium (atomic number 12)
  Show solution
  Answer $M g$ (2, 8, 2) → $M g^{2 +}$ (2, 8); fluoride $M g F_{2}$ .
  Magnesium (2, 8, 2) loses two electrons and becomes $M g^{2 +}$ (2, 8). Each fluorine atom can take only one, so two fluorine atoms are needed, giving two $F^{-}$ ions. Magnesium fluoride is $M g F_{2}$ .
Short answerMediumPause and Ponder Q15
Explain how potassium oxide, $K_{2} O$ , forms from potassium atoms (2, 8, 8, 1) and an oxygen atom (2, 6).
Show solution
AnswerTwo K atoms each give one electron to one O atom: two $K^{+}$ ions and one $O^{2 -}$ ion.
Each potassium atom loses its single outer electron and becomes $K^{+}$ (2, 8, 8). The oxygen atom needs two electrons to complete its octet, so it takes one from each of two potassium atoms and becomes $O^{2 -}$ (2, 8). The two $K^{+}$ ions and one $O^{2 -}$ ion attract each other, and the charges balance: $2 \times (+ 1) + (- 2) = 0$ .

9.5 Writing Chemical Formulae

A chemical formula (रासायनिक सूत्र) shows which elements a compound contains and in what proportion, using symbols and small numbers called subscripts. In $H_{2} O$ , the subscript 2 belongs to hydrogen, and oxygen's unwritten subscript is 1. You could work out every formula by drawing electrons, as in Section 9.4, but valencies give a much quicker route.

9.5.1 Writing chemical formulae of covalent compounds

In a covalent compound, an element's valency is the number of electrons each of its atoms shares, which is also the number of bonds it forms. The common values are 1 for hydrogen and chlorine, 2 for oxygen and sulfur, 3 for nitrogen and 4 for carbon. Then follow four steps:

Write the symbols of the elements side by side.
Write each element's valency under its symbol.
Cross the valencies over, so that each number becomes the subscript of the other symbol.
Divide both subscripts by any common factor, and do not write a subscript of 1.

Table 9.7Formulae of covalent compounds from valencies

Elements	Valencies	After crossing	Formula
hydrogen, oxygen	1, 2	$H_{2} O_{1}$	$H_{2} O$
nitrogen, hydrogen	3, 1	$N_{1} H_{3}$	$N H_{3}$
carbon, oxygen	4, 2	$C_{2} O_{4}$	$C O_{2}$ (divide by 2)

Check each result against the bonds you counted in Fig. 9.5 and Fig. 9.6: water has two O–H bonds, ammonia three N–H bonds, and carbon dioxide two double bonds.

9.5.2 Writing chemical formulae of ionic compounds

For an ionic compound, the valency of each ion is the size of its charge, and the steps are almost the same:

Write the symbol of the cation first, then the anion.
Write each ion's charge number under its symbol.
Cross the numbers over to become subscripts.
Divide by any common factor, because the formula of an ionic compound shows the simplest whole-number ratio of its ions.
If a polyatomic ion is needed more than once, put it in brackets with the subscript outside.
Leave the charges out of the final formula.

4/4 $A l_{2} S_{3}$ : two ions of 3+ and three of 2− give $+ 6 - 6 = 0$ .

Fig. 9.10Each ion's charge number becomes the other ion's subscript.NCERT 9.5.2 Examples · p. 176

The symbols Al and S are written large, with the names aluminium and sulfide above them. In the second step their charges, 3 plus and 2 minus, appear underneath. In the third step a coloured 3 slides from under Al up to the lower right of S, and a coloured 2 slides from under S to the lower right of Al, along dashed crossing arrows. In the fourth step the formula Al2S3 appears in a box, with a check below: 2 times plus 3 and 3 times minus 2 give a total of 0.

Whatever the ions, the test is the same: the positive charges must cancel the negative charges, because a compound as a whole is neutral. The table shows the method on six pairs of ions, including the two situations that need extra care, simplifying and brackets.

Table 9.8Formulae of ionic compounds by criss-crossing

Ions	After crossing	Formula	Why
$B a^{2 +}$ , $C l^{-}$	$B a_{1} C l_{2}$	$B a C l_{2}$	a subscript 1 is not written
$Z n^{2 +}$ , $O^{2 -}$	$Z n_{2} O_{2}$	$Z n O$	divide by 2
$K^{+}$ , $S O_{4}^{2 -}$	$K_{2} (S O_{4})_{1}$	$K_{2} S O_{4}$	one sulfate ion: no brackets
$C a^{2 +}$ , $O H^{-}$	$C a_{1} (O H)_{2}$	$C a (O H)_{2}$	two hydroxide ions: brackets
$N H_{4}^{+}$ , $C O_{3}^{2 -}$	$(N H_{4})_{2} (C O_{3})_{1}$	$(N H_{4})_{2} C O_{3}$	two ammonium ions: brackets
$A l^{3 +}$ , $P O_{4}^{3 -}$	$A l_{3} (P O_{4})_{3}$	$A l P O_{4}$	divide by 3

Two slips are common. Calcium hydroxide is $C a (O H)_{2}$ , not $C a O H_{2}$ : without the brackets, the 2 would double only the hydrogen. And charges never appear in the finished formula. In barium chloride, for instance, you write just $B a C l_{2}$ . You can try any pair of ions yourself in the formula builder in Section 9.8.

Practice · 3 questions

Pause and ponder: questions with answers

16
Quick answerEasyPause and Ponder Q16
Name each covalent compound.
1. $S i O_{2}$ :
  Show answer
  Answer silicon dioxide
  One silicon (no mono-), two oxygens: silicon dioxide, the main substance in sand.
2. $N_{2} O_{3}$ :
  Show answer
  Answer dinitrogen trioxide
  Two nitrogens (di-) and three oxygens (tri-): dinitrogen trioxide.
3. $P B r_{3}$ :
  Show answer
  Answer phosphorus tribromide
  One phosphorus, three bromines: phosphorus tribromide.
4. $S F_{4}$ :
  Show answer
  Answer sulfur tetrafluoride
  One sulfur, four fluorines: sulfur tetrafluoride.
17
Short answerMediumPause and Ponder Q17
Write the formula of each compound.
1. calcium hydrogencarbonate
  Show solution
  Answer $C a (H C O_{3})_{2}$
  $C a^{2 +}$ and $H C O_{3}^{-}$ cross to 1 : 2. The polyatomic ion appears twice, so it goes in brackets: $C a (H C O_{3})_{2}$ .
2. nitrogen trichloride
  Show solution
  Answer $N C l_{3}$
  The name itself gives one nitrogen and three chlorines (valencies 3 and 1): $N C l_{3}$ .
3. ferrous sulfide
  Show solution
  Answer $F e S$
  Ferrous is $F e^{2 +}$ and sulfide is $S^{2 -}$ . Crossing gives $F e_{2} S_{2}$ , which simplifies to $F e S$ .
4. cupric nitrate
  Show solution
  Answer $C u (N O_{3})_{2}$
  Cupric is $C u^{2 +}$ and nitrate is $N O_{3}^{-}$ , so one copper ion needs two nitrate ions, in brackets: $C u (N O_{3})_{2}$ .
18
Short answerMediumPause and Ponder Q18
Write the formula of the compound formed by each pair of ions.
1. $B a^{2 +}$ and $O H^{-}$
  Show solution
  Answer $B a (O H)_{2}$
  Charges 2 and 1 cross to $B a_{1} (O H)_{2}$ ; two hydroxide ions need brackets: $B a (O H)_{2}$ , barium hydroxide.
2. $N H_{4}^{+}$ and $P O_{4}^{3 -}$
  Show solution
  Answer $(N H_{4})_{3} P O_{4}$
  Charges 1 and 3 cross to three ammonium ions and one phosphate ion. Ammonium is polyatomic and appears three times, so it is bracketed: $(N H_{4})_{3} P O_{4}$ , ammonium phosphate.

9.6 Properties of Ionic and Covalent Compounds

Ionic compounds are built from ions, and covalent compounds from molecules. Does that difference show up in how they behave? Activity 9.4 tests two everyday properties: whether a substance dissolves, and whether it lets an electric current through.

Activity 9.4NCERT Activity 9.4 · p. 177

Dissolve it, then test it with a bulb

You’ll need

small samples of common salt, washing soda, sugar, camphor and candle-wax shavings
water and kerosene
ten test tubes in a stand
a conductivity tester: a 9 V battery, a small torch bulb, wires, and two pencil leads pushed through a piece of card

Predict first

Which sample do you expect to light the bulb once it is in water?

SugarCommon saltCandle waxCamphor

A. Does it dissolve?
- Put a pinch of each sample into a test tube of water, and another pinch into a test tube of kerosene.
- Shake each tube gently for a minute and note whether the sample disappears.
B. Does it conduct?
- Press the two pencil leads of the tester into a dry heap of each solid sample. Does the bulb glow?
- Dip the leads into the water from part A for each sample, one at a time. Rinse and wipe the leads between samples.
- Group the samples that behaved alike.

Table 9.9

Sample	Dissolves in water?	Dissolves in kerosene?	Bulb glows: dry solid?	Bulb glows: in water?
Common salt
Washing soda
Sugar
Camphor
Candle wax

Your entries are saved automatically.

2/3Free ions drift to the oppositely charged rods: the bulb glows.

Fig. 9.11Only moving ions carry a current through the liquid.NCERT Fig. 9.15 · p. 178

A battery at the top left and a bulb at the top right are wired to two rods dipping into a beaker; the left rod is marked minus and the right rod plus. In the first step the beaker holds only dry salt crystals and the bulb is off. In the second step the beaker holds salt solution with plus ions and minus ions spread through it; the plus ions have shifted towards the minus rod and the minus ions towards the plus rod, and the bulb glows with rays around it. In the third step the beaker holds sugar solution with uncharged sugar molecules, labelled no charge, and the bulb is off.

Why the difference? An electric current is a flow of charge. In salt solution, the $N a^{+}$ and $C l^{-}$ ions are free to drift, cations towards the negative rod and anions towards the positive rod, and that movement carries the current. In the dry crystal, each ion is gripped by its neighbours in the lattice and can only jiggle on the spot, so no charge flows and the bulb stays dark. Sugar dissolves as whole molecules with no charge, so even its solution cannot carry a current. Camphor and wax give no ions either.

What if an ionic compound is melted instead of dissolved? Melting also frees the ions from the lattice, so molten sodium chloride conducts electricity. A melted covalent substance, such as liquid wax, still has no ions and does not.

Water molecules are attracted to ions and can tug them out of a crystal one by one, which is why many ionic compounds dissolve in water. Covalent substances like wax and camphor are chemically more like kerosene and mix with it instead. Breaking up a lattice of strongly attracting ions takes a great deal of energy, so ionic compounds melt only at high temperatures: sodium chloride melts at 801 °C. The molecules of most covalent compounds attract each other far more weakly, so naphthalene melts at about 80 °C and candle wax at about 60 °C.

Compare

Ionic and covalent compounds at a glance

Ionic compounds

Made of cations and anions in a crystal lattice
Usually dissolve in water but not in kerosene
Conduct when dissolved or melted, but not as solids
High melting points (sodium chloride: 801 °C)

Covalent compounds

Made of separate molecules
Often dissolve in kerosene; a few, like sugar, dissolve in water
Usually do not conduct, even when dissolved or melted
Low melting points (naphthalene: about 80 °C)

Practice · 2 questions

Pause and ponder: questions with answers

Short answerMediumPause and Ponder Q19
A student's notebook reads: 'Compound Z, white crystals. Dry crystals: bulb dark. Melted: bulb glows. Dissolved in water: bulb glows.' What kind of bonding does Z have? Explain.
Show solution
AnswerIonic bonding.
Z conducts when melted or dissolved but not as a solid. That is the pattern of an ionic compound: its ions are locked in the crystal, but free to move, and so carry a current, once the crystal melts or dissolves. A covalent compound would not conduct in any of the three states.
20
Short answerHardPause and Ponder Q20
A dull white coat forms on aluminium cookware as the metal reacts with oxygen in the air. Aluminium has atomic number 13.
1. Which ions do aluminium and oxygen form?
  Show solution
  Answer $A l^{3 +}$ and $O^{2 -}$
  Aluminium is 2, 8, 3, so it loses three electrons to form $A l^{3 +}$ . Oxygen is 2, 6, so it gains two to form $O^{2 -}$ .
2. What is the formula of the coat?
  Show solution
  Answer $A l_{2} O_{3}$
  Crossing the charge numbers 3 and 2 gives $A l_{2} O_{3}$ : $2 \times (+ 3) + 3 \times (- 2) = 0$ .
3. Would the oxide conduct electricity once melted?
  Yes, because its ions are free to moveNo, because it has no ions
  Show answer
  Answer (A) Yes, because its ions are free to move
  Aluminium oxide is ionic, and melting frees its ions. In fact, aluminium metal is extracted by passing electricity through aluminium oxide dissolved in a molten mineral.

9.7 Molecular Mass of Covalent Compounds

Atoms are far too light to weigh one at a time in grams, so chemists compare them on a scale of their own. The atomic mass unit, u, is one-twelfth of the mass of a carbon-12 atom, which is close to the mass of one proton or one neutron. On this scale a hydrogen atom has a mass of about 1 u and an oxygen atom 16 u. The table gives the rounded values used in school chemistry; chlorine's 35.5 u is an average, because natural chlorine is a mixture of two isotopes.

Table 9.10Rounded atomic masses

Atom	Mass (u)	Atom	Mass (u)	Atom	Mass (u)
H	1	Na	23	K	39
C	12	Mg	24	Ca	40
N	14	Al	27	Fe	56
O	16	P	31	Cu	63.5
F	19	S	32	Zn	65
Li	7	Cl	35.5	Ba	137

Finding a molecular mass is bookkeeping: count each kind of atom in the formula, multiply by its atomic mass, and total the results.

Definition

Molecular mass

The mass of one molecule in u, found by adding up the atomic masses of every atom it contains.

ExampleWater, $H_{2} O$ : $2 \times 1 + 16 = 18$ u.

3/3One $C a^{2 +}$ and two $C l^{-}$ : formula unit mass of $C a C l_{2}$ = 111 u.

Fig. 9.12Molecular mass and formula unit mass are both sums of atomic masses.NCERT 9.7–9.8 · p. 179

On the left, an ammonia molecule is drawn as one large N ball with three small H balls attached; tags read 14 u on nitrogen and 1 u on each hydrogen. In the second step the sum 14 + 1 + 1 + 1 = 17 u appears below. In the third step, on the right, a dashed box headed a formula unit holds one calcium ion and two chloride ions, tagged 40 u and 35.5 u each, with the sum 40 + 35.5 + 35.5 = 111 u below.

Example 9.4NCERT Example 9.4 · p. 179

Molecular mass of ammonia

Find the molecular mass of ammonia, $N H_{3}$ . (N = 14 u, H = 1 u)

Try it yourself first. Then reveal the solution one step at a time.

Example 9.5NCERT Example 9.5 · p. 179

Molecular mass of glucose

Glucose, the sugar in ORS and in your blood, has the formula $C_{6} H_{12} O_{6}$ . Find its molecular mass. (C = 12 u, H = 1 u, O = 16 u)

Try it yourself first. Then reveal the solution one step at a time.

Try the same sum for table salt and you hit a problem of words. A salt grain is one big lattice of ions (the salt crystal), with no separate NaCl molecules in it, so it has no molecular mass in the strict sense. Section 9.8 gives the word chemists use instead.

Quick check

1 of 3

Find the molecular mass of sulfuric acid,

H_{2} S O_{4}

. (H = 1 u, S = 32 u, O = 16 u)

Show answer

Answer 98 u

2 \times 1 + 32 + 4 \times 16 = 2 + 32 + 64 = 98

9.8 Formula Unit Mass of Ionic Compounds

The formula of an ionic compound gives the simplest whole-number ratio of its ions. One group of ions in that ratio, such as one $C a^{2 +}$ ion with two $C l^{-}$ ions in calcium chloride, is called a formula unit. Add the atomic masses in one formula unit and you get the formula unit mass (सूत्र इकाई द्रव्यमान). The arithmetic is the same as for molecular mass. Only the name changes, as a reminder that a formula unit is a way of counting ions, not a particle that exists on its own. Gaining or losing an electron changes an atom's mass far too little to matter here.

Definition

Formula unit mass

The mass in u of one formula unit of an ionic compound (its simplest whole-number group of ions), found by adding up the atomic masses of every atom in it.

ExampleSodium chloride, NaCl: $23 + 35.5 = 58.5$ u.

Example 9.6NCERT Example 9.6 · p. 179

Formula unit mass of calcium chloride

Calcium chloride, $C a C l_{2}$ , is sprinkled on roads in cold countries to melt ice. Find its formula unit mass. (Ca = 40 u, Cl = 35.5 u)

Try it yourself first. Then reveal the solution one step at a time.

Example 9.7NCERT Example 9.7 · p. 180

Formula unit mass with brackets

Calcium phosphate, $C a_{3} (P O_{4})_{2}$ , is sold as a calcium supplement and is added to some food powders to stop them caking. Find its formula unit mass. (Ca = 40 u, P = 31 u, O = 16 u)

Try it yourself first. Then reveal the solution one step at a time.

InteractiveFormula builder

An interactive formula builder. Two menus pick a cation from fourteen (from sodium to ferric) and an anion from thirteen (from chloride to phosphate), each shown with its charge. A panel shows the two ions with their charges and a crossing sign. Numbered steps explain the criss-cross: the charge numbers swap, a common factor is divided out when there is one, brackets are added when a polyatomic ion appears more than once, and a check shows the positive and negative charges adding to zero. A result box shows the formula, the compound's name and the formula unit mass worked out from rounded atomic masses. A checkbox hides the answer until a Show button is pressed, so learners can predict first, and buttons pick a random pair or reset.

Fig. 9.13Pick a cation and an anion. The builder crosses the charges, simplifies, adds brackets and works out the formula unit mass.

Quick check

1 of 3

Find the formula unit mass of magnesium chloride,

M g C l_{2}

. (Mg = 24 u, Cl = 35.5 u)

Show answer

Answer 95 u

24 + 2 \times 35.5 = 24 + 71 = 95

Recap

Atomic Foundations of Matter: key points and summary

In a chemical reaction the products weigh exactly as much as the reactants; mass only seems to change when a gas escapes or is taken in (Section 9.1).
A pure compound has a fixed recipe by mass, so samples from anywhere match: carbon dioxide is always 3 : 8 carbon to oxygen, and water 1 : 8 hydrogen to oxygen (Section 9.2).
Dalton explained both laws: atoms are not made or destroyed in reactions, they only regroup, and they join in fixed whole-number ratios (Section 9.3).
Atoms join into molecules such as $O_{2}$ and $C O_{2}$ , which can exist on their own; noble gases stay as single atoms (Section 9.4).
Atoms bond to reach a full outer shell and lower their energy, by sharing electrons or by transferring them; the attraction that results is a chemical bond (Section 9.4).
A covalent bond is a shared pair of electrons: one pair in H–H and Cl–Cl, two in O=O, three in N≡N (Section 9.4.1).
Two-element covalent compounds are named with number prefixes: $S O_{3}$ is sulfur trioxide and $N_{2} O$ is dinitrogen monoxide (Naming covalent compounds).
An ionic bond is the attraction between a cation (an atom that lost electrons) and an anion (an atom that gained them); ionic solids are 3-D crystal lattices (Section 9.4.2).
Ionic compounds are named cation first; polyatomic ions such as $S O_{4}^{2 -}$ and $N H_{4}^{+}$ act as single units (Naming ionic compounds).
To write a formula, cross over the valencies or charges, simplify, and bracket a polyatomic ion that appears more than once; the charges must add up to zero (Section 9.5).
Ionic compounds usually dissolve in water, conduct when dissolved or melted, and melt at high temperatures; covalent compounds usually do not conduct (Section 9.6).
Molecular mass (for molecules) and formula unit mass (for ionic compounds) are sums of atomic masses: $H_{2} S O_{4}$ is 98 u and $C a C l_{2}$ is 111 u (Section 9.7, Section 9.8).

Practice · 18 questions

Practice questions with answers (NCERT “Revise, reflect, refine”)

1
Short answerMediumNCERT Exercises Q1
A student's flash card reads: 'Element P: atomic number 12. Element Q: atomic number 9.'
1. Write the electron arrangement of each, and say which ion each forms.
  Show solution
  AnswerP is 2, 8, 2 and forms $P^{2 +}$ ; Q is 2, 7 and forms $Q^{-}$ .
  P (12 electrons) is 2, 8, 2. Losing its two outer electrons leaves a full shell, so it forms a cation with charge 2+. Q (9 electrons) is 2, 7; gaining one electron completes its octet, so it forms an anion with charge 1−.
2. What type of bond forms when P and Q combine?
  IonicCovalent
  Show answer
  Answer (A) Ionic
  P transfers electrons to Q, so oppositely charged ions form and attract: an ionic bond.
3. Write the formula of the compound.
  Show solution
  Answer $P Q_{2}$
  Charges 2+ and 1− cross to $P_{1} Q_{2}$ , written $P Q_{2}$ . (P is magnesium and Q is fluorine, so the compound is $M g F_{2}$ .)
2
DiagramHardNCERT Exercises Q2
A colourless gas that makes up most of the air is made of diatomic molecules. Its atoms have atomic number 7.
1. Write the electron arrangement of the atom and explain why two atoms join.
  Show solution
  Answer2, 5: each atom is three short of an octet; sharing completes both.
  Seven electrons fill the shells as 2, 5. On its own, each atom is three electrons short of an octet. If two atoms share three pairs, both can count eight electrons, so they join as $N_{2}$ .
2. What kind of bond holds the molecule together?
  A single covalent bondA double covalent bondA triple covalent bondAn ionic bond
  Show answer
  Answer (C) A triple covalent bond
  Three shared pairs make a triple bond, N≡N.
3. The same element combines with hydrogen. Give the formula of the compound and describe its bonds.
  Show solution
  Answer $N H_{3}$ : three single N–H bonds and one unshared pair on N.
  Nitrogen needs three electrons and each hydrogen atom can share one, so three hydrogen atoms bond to one nitrogen atom, each through a single covalent bond. Nitrogen keeps one unshared pair, so it counts 8 electrons; each hydrogen counts 2. The compound is ammonia, $N H_{3}$ .
Quick answerMediumNCERT Exercises Q3
A formula unit of an ionic compound contains three cations and two anions, and its charges balance. Which pair of ions could it be made of?
$A l^{3 +}$ and $O^{2 -}$ $N a^{+}$ and $S O_{4}^{2 -}$ $M g^{2 +}$ and $C l^{-}$ $C a^{2 +}$ and $P O_{4}^{3 -}$
Show answer
Answer (D) $C a^{2 +}$ and $P O_{4}^{3 -}$
Three $C a^{2 +}$ give +6 and two $P O_{4}^{3 -}$ give −6: $C a_{3} (P O_{4})_{2}$ . Aluminium oxide has two cations and three anions, sodium sulfate two and one, and magnesium chloride one and two.
4
Quick answerMediumNCERT Exercises Q4
Which of these statements are correct? Select all that apply.
A molecule of ozone, $O_{3}$ , contains three oxygen atoms.Every element exists as molecules of two or more atoms.A molecule of ammonia contains four atoms.In carbon dioxide, carbon is joined to each oxygen atom by a single bond.
Select all that apply
Show answer
Answer
(A) A molecule of ozone, $O_{3}$ , contains three oxygen atoms.
(C) A molecule of ammonia contains four atoms.
A is true (three O atoms) and C is true (one N and three H). B is false: noble gases such as helium and argon exist as single atoms. D is false: carbon shares two pairs with each oxygen atom, so both bonds are double bonds, O=C=O.
5
Short answerMediumNCERT Exercises Q5
Write the chemical formula of each compound.
1. zinc nitrate
  Show solution
  Answer $Z n (N O_{3})_{2}$
  $Z n^{2 +}$ and $N O_{3}^{-}$ : one zinc needs two nitrates, in brackets: $Z n (N O_{3})_{2}$ .
2. barium oxide
  Show solution
  Answer $B a O$
  $B a^{2 +}$ and $O^{2 -}$ cross to $B a_{2} O_{2}$ , which simplifies to $B a O$ .
3. ferric sulfate
  Show solution
  Answer $F e_{2} (S O_{4})_{3}$
  Ferric is $F e^{3 +}$ and sulfate is $S O_{4}^{2 -}$ . Crossing gives two iron ions and three sulfate ions: $F e_{2} (S O_{4})_{3}$ . Check: $2 \times 3 = 6$ and $3 \times 2 = 6$ .
4. cupric hydroxide
  Show solution
  Answer $C u (O H)_{2}$
  Cupric is $C u^{2 +}$ and hydroxide is $O H^{-}$ , so two hydroxide ions in brackets: $C u (O H)_{2}$ .
6
Short answerMediumNCERT Exercises Q6
Write the formula of the compound formed by each pair of ions.
1. $B a^{2 +}$ and $F^{-}$
  Show solution
  Answer $B a F_{2}$
  Charges 2 and 1 cross to $B a_{1} F_{2}$ : $B a F_{2}$ .
2. $N H_{4}^{+}$ and $S O_{4}^{2 -}$
  Show solution
  Answer $(N H_{4})_{2} S O_{4}$
  Two ammonium ions balance one sulfate ion. Ammonium appears twice, so it is bracketed: $(N H_{4})_{2} S O_{4}$ .
3. $L i^{+}$ and $C O_{3}^{2 -}$
  Show solution
  Answer $L i_{2} C O_{3}$
  Two lithium ions balance one carbonate ion. Carbonate appears once, so no brackets: $L i_{2} C O_{3}$ .
4. $F e^{3 +}$ and $N O_{3}^{-}$
  Show solution
  Answer $F e (N O_{3})_{3}$
  Three nitrate ions balance one ferric ion: $F e (N O_{3})_{3}$ .
Quick answerEasyNCERT Exercises Q7
Magnesium has atomic number 12. Which electron arrangement belongs to the magnesium ion, $M g^{2 +}$ ?
2, 8, 22, 82, 8, 42, 8, 8
Show answer
Answer (B) 2, 8
$M g^{2 +}$ has $12 - 2 = 10$ electrons, arranged 2, 8. The arrangement 2, 8, 2 is the neutral atom, and 2, 8, 4 would mean it had gained electrons.
8
NumericalMediumNCERT Exercises Q8
Find the molecular mass or formula unit mass of each substance. (H = 1, C = 12, N = 14, O = 16, S = 32, K = 39, Ca = 40, all in u)
1. ammonium sulfate, $(N H_{4})_{2} S O_{4}$ , a fertiliser
  u
  Show answer
  Answer 132 u
  $2 \times (14 + 4) + 32 + 4 \times 16 = 36 + 32 + 64 = 132$ u.
2. urea, $C O (N H_{2})_{2}$ , the most widely used nitrogen fertiliser (a covalent compound)
  u
  Show answer
  Answer 60 u
  Molecular mass $= 12 + 16 + 2 \times (14 + 2) = 28 + 32 = 60$ u.
3. potassium nitrate, $K N O_{3}$ , used in fertilisers and fireworks
  u
  Show answer
  Answer 101 u
  $39 + 14 + 3 \times 16 = 101$ u.
4. calcium carbonate, $C a C O_{3}$ , the substance of chalk and marble
  u
  Show answer
  Answer 100 u
  $40 + 12 + 3 \times 16 = 100$ u.
9
Short answerMediumNCERT Exercises Q9
Write the formula of the compound formed when each pair of elements reacts.
1. potassium and oxygen
  Show solution
  Answer $K_{2} O$
  $K^{+}$ and $O^{2 -}$ cross to $K_{2} O$ .
2. aluminium and sulfur
  Show solution
  Answer $A l_{2} S_{3}$
  $A l^{3 +}$ and $S^{2 -}$ cross to $A l_{2} S_{3}$ .
3. barium and nitrogen
  Show solution
  Answer $B a_{3} N_{2}$
  Barium forms $B a^{2 +}$ and nitrogen (2, 5) gains three electrons to form the nitride ion $N^{3 -}$ . Crossing gives $B a_{3} N_{2}$ .
4. lithium and fluorine
  Show solution
  Answer $L i F$
  $L i^{+}$ and $F^{-}$ combine 1 : 1: $L i F$ .
Short answerMediumNCERT Exercises Q10
Copy and complete the grid with the formula of the compound made by each cation (left) and anion (top). One cell is done for you.
Table 9.11
$O H^{-}$ $O^{2 -}$ $P O_{4}^{3 -}$
$K^{+}$ $K O H$ ? ?
$C a^{2 +}$ ? ? ?
$A l^{3 +}$ ? ? ?
Show solution
Table 9.12
$O H^{-}$ $O^{2 -}$ $P O_{4}^{3 -}$
$K^{+}$ $K O H$ $K_{2} O$ $K_{3} P O_{4}$
$C a^{2 +}$ $C a (O H)_{2}$ $C a O$ $C a_{3} (P O_{4})_{2}$
$A l^{3 +}$ $A l (O H)_{3}$ $A l_{2} O_{3}$ $A l P O_{4}$
Equal charges cancel one-for-one ( $C a O$ , $A l P O_{4}$ ). Brackets appear only where a polyatomic ion occurs more than once, as in $C a (O H)_{2}$ and $C a_{3} (P O_{4})_{2}$ .
11
NumericalMediumNCERT Exercises Q11
Zinc granules react with dilute sulfuric acid. In a closed flask, 6.5 g of zinc and 9.8 g of sulfuric acid give 16.1 g of zinc sulfate and 0.2 g of hydrogen gas.
1. Show that these masses obey the law of conservation of mass.
  Show solution
  AnswerReactants 16.3 g = products 16.3 g.
  Reactants: $6.5 + 9.8 = 16.3$ g. Products: $16.1 + 0.2 = 16.3$ g. The totals are equal, so the law holds.
2. The same reaction is done in an open beaker on a balance, and all the hydrogen escapes. What mass does the balance show for the contents at the end?
  g
  Show answer
  Answer 16.1 g
  Only the gas leaves: $16.3 - 0.2 = 16.1$ g. The law still holds; the balance simply no longer weighs the hydrogen.
12
Short answerMediumNCERT Exercises Q12
A particle has 16 protons, 16 neutrons and 18 electrons.
1. What are its atomic number and mass number?
  Show solution
  AnswerAtomic number 16; mass number 32.
  Atomic number = number of protons = 16. Mass number = protons + neutrons = $16 + 16 = 32$ .
2. Does it carry a charge? If so, is it a cation or an anion, and why?
  Show solution
  AnswerAn anion with charge 2−.
  It has two more electrons (18) than protons (16), so its charge is $16 - 18 = - 2$ : an anion.
3. Write its electron arrangement.
  Show solution
  Answer2, 8, 8
  18 electrons fill the shells as 2, 8, 8.
4. Name the particle.
  Show solution
  AnswerThe sulfide ion, $S^{2 -}$ .
  Atomic number 16 is sulfur, and a sulfur atom that has gained two electrons is the sulfide ion, $S^{2 -}$ .
13
Short answerHardNCERT Exercises Q13
Atoms of two non-metals, A and B, contain 6 and 8 electrons respectively.
1. Write each electron arrangement and say how many electrons each atom needs to complete its octet.
  Show solution
  AnswerA is 2, 4 and needs 4; B is 2, 6 and needs 2.
  A (6 electrons) is 2, 4, so it needs 4 more. B (8 electrons) is 2, 6, so it needs 2 more.
2. Will A and B form an ionic or a covalent compound? Explain.
  Show solution
  AnswerCovalent: they share electrons.
  Both are non-metals. A would have to lose or gain four electrons, which takes far too much energy, and B holds its own electrons too tightly to give any away. So neither atom hands electrons over; instead they share pairs, which makes covalent bonds.
3. Write the formula of the compound and name it.
  Show solution
  Answer $A B_{2}$ , which is $C O_{2}$ , carbon dioxide.
  Valencies 4 and 2 cross to $A_{2} B_{4}$ , which simplifies to $A B_{2}$ . A is carbon and B is oxygen, so the compound is carbon dioxide, $C O_{2}$ , with two double bonds, O=C=O.
Quick answerMediumNCERT Exercises Q14
Assertion (A): Naphthalene dissolves in kerosene but not in water. Reason (R): Naphthalene is an ionic compound.
A and R are both true, and R explains AA and R are both true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (C) A is true, but R is false
A is true. R is false: naphthalene is a covalent compound made of molecules, which is exactly why it dissolves in kerosene and not in water, does not conduct, and melts at only about 80 °C.
15
Short answerMediumNCERT Exercises Q15
Find the numbers of electrons and neutrons in each particle.
1. $_{}^{}_{}^{24} M g^{2 +}$ (atomic number 12)
  Show solution
  Answer10 electrons, 12 neutrons
  Electrons: $12 - 2 = 10$ . Neutrons: $24 - 12 = 12$ .
2. $_{}^{}_{}^{19} F^{-}$ (atomic number 9)
  Show solution
  Answer10 electrons, 10 neutrons
  Electrons: $9 + 1 = 10$ . Neutrons: $19 - 9 = 10$ .
3. $_{}^{}_{}^{56} F e^{3 +}$ (atomic number 26)
  Show solution
  Answer23 electrons, 30 neutrons
  Electrons: $26 - 3 = 23$ . Neutrons: $56 - 26 = 30$ .
16
Quick answerEasy
Put the steps for writing the formula of an ionic compound in order.
1. Divide by a common factor
2. Cross the numbers over
3. Write the charges below
4. Check the charges cancel
5. Write cation, then anion
Show answer
Answer
Write cation, then anion
Write the charges below
Cross the numbers over
Divide by a common factor
Check the charges cancel
Symbols first, then charges, cross, simplify, and finally check that the total charge is zero.
17
Quick answerEasy
Match each substance with the particles it is made of.
Tap an item on the left, then its match on the right.
Show answer
Answer
sodium chloride : ions in a lattice
argon : single atoms
oxygen gas : molecules of an element
water : molecules of a compound
Salt is ionic, argon is a noble gas, oxygen exists as $O_{2}$ molecules and water as $H_{2} O$ molecules.
Quick answerMedium
A particle has 13 protons and 10 electrons. Its charge is .
Show answer
Answer 3+
$13 - 10 = 3$ more protons than electrons, so the charge is 3+. It is the aluminium ion, $A l^{3 +}$ .

	$O H^{-}$	$O^{2 -}$	$P O_{4}^{3 -}$
$K^{+}$	$K O H$	?	?
$C a^{2 +}$	?	?	?
$A l^{3 +}$	?	?	?

	$O H^{-}$	$O^{2 -}$	$P O_{4}^{3 -}$
$K^{+}$	$K O H$	$K_{2} O$	$K_{3} P O_{4}$
$C a^{2 +}$	$C a (O H)_{2}$	$C a O$	$C a_{3} (P O_{4})_{2}$
$A l^{3 +}$	$A l (O H)_{3}$	$A l_{2} O_{3}$	$A l P O_{4}$

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Quick check: MCQs on Atomic Foundations of Matter

Does mass change when matter changes?2 questions
9.1 Law of Conservation of Mass3 questions
9.2 Law of Constant Proportions3 questions
9.3 Dalton’s Atomic Theory3 questions
Naming covalent compounds3 questions
Naming ionic compounds4 questions
9.5.2 Writing chemical formulae of ionic compounds3 questions
9.6 Properties of Ionic and Covalent Compounds3 questions
9.7 Molecular Mass of Covalent Compounds3 questions
9.8 Formula Unit Mass of Ionic Compounds3 questions

Glossary · 15

Key terms (English–Hindi)

anionऋणायनrinayan: A negatively charged ion, formed when an atom gains electrons, such as $C l^{-}$ .First seen in the chapter →

cationधनायनdhanayan: A positively charged ion, formed when an atom loses electrons, such as $N a^{+}$ .First seen in the chapter →
chemical bondरासायनिक आबंधrasayanik aabandh: The force of attraction that holds atoms or ions together.First seen in the chapter →
chemical formulaरासायनिक सूत्रrasayanik sutra: Symbols and subscripts showing the elements in a substance and their proportions, such as $H_{2} O$ .First seen in the chapter →
covalent bondसहसंयोजी आबंधsahsanyoji aabandh: A bond formed when two atoms share a pair of electrons.

formula unit massसूत्र इकाई द्रव्यमानsutra ikai dravyaman: The mass in u of one formula unit (the simplest ion ratio) of an ionic compound, found by adding atomic masses.First seen in the chapter →

ionआयनaayan: An atom or group of atoms that carries an electric charge because it has lost or gained electrons.First seen in the chapter →
ionic bondआयनिक आबंधaayanik aabandh: The electrostatic attraction between oppositely charged ions.First seen in the chapter →

law of conservation of massद्रव्यमान संरक्षण का नियमdravyaman sanrakshan ka niyam: In a chemical reaction the total mass of the products equals the total mass of the reactants; matter is neither created nor destroyed.First seen in the chapter →
law of constant proportionsस्थिर अनुपात का नियमsthir anupat ka niyam: Every pure sample of a compound has its elements in the same ratio by mass, wherever it comes from.First seen in the chapter →

molecular massआण्विक द्रव्यमानaanvik dravyaman: The mass of one molecule in u, found by adding up the atomic masses of every atom in it.
moleculeअणुanu: The smallest particle of a substance that can exist by itself: two or more joined atoms with no overall charge.

polyatomic ionबहुपरमाणुक आयनbahuparmanuk aayan: A charged group of bonded atoms that acts as one unit, such as $S O_{4}^{2 -}$ or $N H_{4}^{+}$ .First seen in the chapter →
postulateअभिधारणाabhidharana: A basic idea taken as the starting point of a theory; it is tested through the predictions that follow from it.First seen in the chapter →

valencyसंयोजकताsanyojakta: The combining capacity of an atom or ion: the number of electrons it shares, or the size of its charge.First seen in the chapter →

Does mass change when matter changes?

9.1 Law of Conservation of Mass

9.2 Law of Constant Proportions

9.3 Dalton’s Atomic Theory

9.4 How Atoms Combine

9.4.1 Bonding by sharing electrons: the covalent bond

Molecules of elements

Molecules of compounds

Naming covalent compounds

9.4.2 Bonding by electron transfer: the ionic bond

Naming ionic compounds

9.5 Writing Chemical Formulae

9.5.1 Writing chemical formulae of covalent compounds

9.5.2 Writing chemical formulae of ionic compounds

9.6 Properties of Ionic and Covalent Compounds

9.7 Molecular Mass of Covalent Compounds

9.8 Formula Unit Mass of Ionic Compounds

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