NCERT Solutions Class 9 Science Chapter 9 In-text Questions — Atomic Foundations of Matter

NCERT Class 9 Science — Atomic Foundations of Matter, the ten Pause and Ponder boxes on pages 166 to 180. All 24 questions solved.

Also here. Solutions to the end-of-chapter questions for this chapter and the activities are on their own pages.

The ten boxes run the length of the chapter, from the two laws of chemical combination through Dalton’s theory, covalent and ionic bonding, writing formulae, and finally molecular and formula unit mass. Four things carry all twenty-four questions:

• Mass is conserved in a chemical reaction, and compounds have a fixed composition by mass — the two laws that Questions 1 to 7 test.
• Atoms combine to complete their outermost shell: non-metals with non-metals share (covalent), a metal with a non-metal transfers (ionic).
• A formula balances the charges, with brackets whenever more than one polyatomic ion is needed.
• Molecular mass and formula unit mass are simply the atomic masses added up, counted with their subscripts.

Key insight. Whenever a question gives a ratio — 1 : 8 for water, 3 : 4 for carbon monoxide, 40% to 60% for the sulfur oxide — the calculation is one line of proportion and nothing more. The law of constant proportions is what guarantees the ratio holds for any sample, which is exactly why you are allowed to scale it. Questions 3, 4 and 6 are the same step three times.

Pause and Ponder, page 166. The box that follows Section 9.1, on the law of conservation of mass.

Question 1

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.

Solution. No, the law is not violated — the experiment simply does not account for everything.

Burning ethanol combines it with oxygen from the air, and the products are carbon dioxide and water vapour. Both are gases, and in an open beaker both escape into the air the moment they form. That is why nothing is left behind.

The mass has not disappeared; it has left the beaker. To test the law properly, the reaction must be carried out in a closed container, and the mass of the oxygen consumed must be counted along with the ethanol. Done that way, the total mass of the reactants — ethanol plus oxygen — equals the total mass of the products, carbon dioxide plus water.

This is the same point the chapter’s own note makes about Activity 9.2: the reaction there can be done in an open system precisely because no gas is formed.

Question 2

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?

Solution. The law says the total mass of the products equals the total mass of the reactants. Both reactants are used up completely and water is the only product, so all of that mass appears in the water:

$$\text{mass of water} = 20 + 160 = 180\ \text{g}$$

Pause and Ponder, page 167. The box that closes Section 9.2, on the law of constant proportions.

Question 3

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?

Solution. The law of constant proportions guarantees that every sample of this compound has the same composition by mass, whatever its size — so the ratio of oxygen to sulfur is fixed at

$$\frac{\text{oxygen}}{\text{sulfur}} = \frac{60}{40}$$

For 20 g of sulfur:

$$\text{oxygen} = \frac{60}{40} \times 20 = 30\ \text{g}$$

Check it: the sample would be $20 + 30 = 50$ g in all, of which sulfur is $20/50 = 40\%$ and oxygen $30/50 = 60\%$ — as required.

Question 4

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?

Solution. The ratio is fixed, so scale it. Three parts of carbon go with four parts of oxygen:

$$\frac{\text{oxygen}}{\text{carbon}} = \frac{4}{3} \quad\Longrightarrow\quad \text{oxygen} = \frac{4}{3} \times 9 = 12\ \text{g}$$

(9 g is three times 3 g, so the oxygen is three times 4 g — the same answer seen another way.)

Question 5

The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.

Solution. The difference is whether the components are chemically combined or merely mixed.

In a compound, atoms of the elements are joined by chemical bonds in a fixed whole-number ratio, and that ratio fixes the ratio by mass. Water is always 1 part hydrogen to 8 parts oxygen by mass, whether it comes from a river, a borewell or the ocean — because every water molecule is $\text{H}_2\text{O}$ and nothing else.

In a mixture, the components are simply put together, with no chemical bond between them, so they can be present in any proportion whatever. A salt solution can be made weak or strong, and air has no fixed composition either. Nothing fixes the ratio, so no definite proportion exists to be obeyed.

Question 6

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.

Solution. The two ratios look different, so simplify them before judging.

$$\text{X}: \quad 4 : 1 = 4$$

$$\text{Y}: \quad 8 : 2 = 4 : 1 = 4$$

They are the same ratio. Y simply used twice as much of everything as X — twice the copper and twice the oxygen — which leaves the proportion unchanged.

So both students made the same compound with the same composition by mass, and their results do justify the law of constant proportions. A ratio like 8 : 3 would not have done.

Pause and Ponder, page 168. A single assertion-and-reason question, after Dalton’s atomic theory.

Question 7

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.

Solution. A is true. Water contains hydrogen and oxygen in the mass ratio 1 : 8, so 2 g of hydrogen combines with 16 g of oxygen — and by conservation of mass the water formed is $2 + 16 = 18$ g.

R is false, and the error is in three words at the end. Dalton’s postulate is that atoms combine in the ratio of simple whole numbers — that is, by number of atoms, not by mass. In water two hydrogen atoms join one oxygen atom, a ratio of 2 : 1. Their ratio by mass is 1 : 8, which is not a simple whole-number ratio at all.

So R misstates the postulate. A is true and R is false: option (iii).

Pause and Ponder, page 170. The box that follows Section 9.4.1A, on the molecules of elements.

Question 8

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

Solution. Each nitrogen atom has five valence electrons and needs three more for an octet. Neither atom can take from the other, since both are identical — so they share.

Each atom contributes three electrons to be shared, making three shared pairs between the two atoms: a triple bond. Each nitrogen then counts its own two unshared electrons plus the six shared, which is eight.

$$\text{N} \equiv \text{N}$$

Each atom also keeps one lone pair, the two valence electrons it did not put into bonds.

The triple bond is very strong, which is why nitrogen gas is so unreactive that it makes up most of the air without taking part in much of anything.

Question 9

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

Solution. Atomic number 9 means nine electrons, arranged 2, 7 — so fluorine has seven valence electrons and needs just one more for an octet.

Two fluorine atoms are identical, so neither can pull an electron away from the other. Instead each contributes one electron to a single shared pair. That pair belongs to both, so each atom now counts its own seven electrons plus the one it gains by sharing — eight in all, a complete octet.

$$\text{F} – \text{F}$$

Each atom keeps its remaining six electrons as three lone pairs. The bond is a single covalent bond, since only one pair is shared.

Pause and Ponder, page 171. The box that follows Section 9.4.1B, on the molecules of compounds.

Question 10

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

Solution. In each case, write what every atom brings and what it needs, then share enough pairs to satisfy all of them.

(i) Carbon dioxide, $\text{CO}_2$. Carbon has 4 valence electrons and needs 4; each oxygen has 6 and needs 2. Carbon shares two pairs with each oxygen — a double bond on each side. Carbon then counts 8, and so does each oxygen.

$$\text{O} = \text{C} = \text{O}$$

(ii) Hydrogen sulfide, $\text{H}_2\text{S}$. Sulfur has 6 valence electrons and needs 2; each hydrogen has 1 and needs 1. Sulfur shares one pair with each of two hydrogen atoms. Sulfur reaches 8 and each hydrogen reaches 2, and sulfur keeps two lone pairs.

$$\text{H} – \text{S} – \text{H}$$

(iii) Ammonia, $\text{NH}_3$. Nitrogen has 5 valence electrons and needs 3; each hydrogen needs 1. Nitrogen shares one pair with each of three hydrogen atoms, reaching 8, while each hydrogen reaches 2. Nitrogen keeps one lone pair.

$$\text{H} – \text{N}(\text{H}) – \text{H}$$

Question 11

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

Solution. Atomic number 10 means ten electrons, arranged 2, 8. The K shell is full with 2, and the L shell — the outermost — holds eight, a complete octet.

Atoms transfer or share electrons only in order to complete their outermost shell. Neon’s is already complete, so it has nothing to gain: losing an electron would break a full shell, and gaining one would mean starting a new shell for a single electron. Either would make it less stable than it already is.

With no reason to bond, neon exists as single, unbonded atoms and is chemically inert — which is why it, and the other noble gases, are used where an unreactive atmosphere is needed.

Pause and Ponder, page 174. The box that follows Section 9.4.2, on ions and ionic bonding.

Question 12

What kind of ion will oxygen (O) form?

Solution. Oxygen has atomic number 8, so its configuration is 2, 6 — six valence electrons, two short of an octet.

Gaining two electrons is far easier than losing six, so oxygen gains two. That leaves it with 8 protons and 10 electrons, a net charge of $-2$.

A negatively charged ion is an anion, so oxygen forms the oxide ion, $\text{O}^{2-}$.

Question 13

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.

Solution. Chlorine is 2, 8, 7 — one electron short of an octet — so gaining one electron gives it a charge of $-1$: the chloride ion, $\text{Cl}^-$.

Now balance. Magnesium gives away two electrons, but each chlorine can accept only one. So two chlorine atoms are needed to take the two electrons that one magnesium gives up:

$$1 \times (+2) + 2 \times (-1) = 0 \quad\Longrightarrow\quad \text{MgCl}_2$$

So one ion of magnesium and two ions of chlorine combine to give magnesium chloride.

Question 14

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

Solution. Write each configuration and take off the valence electrons.

Potassium, atomic number 19, is 2, 8, 8, 1. One electron in the outermost shell, so it loses one to leave the full 2, 8, 8 beneath:

$$\text{K} \longrightarrow \text{K}^+ + e^-$$

Calcium, atomic number 20, is 2, 8, 8, 2. Two in the outermost shell, so it loses two:

$$\text{Ca} \longrightarrow \text{Ca}^{2+} + 2e^-$$

The chlorides. Each chlorine atom accepts exactly one electron, becoming $\text{Cl}^-$.

Potassium gives up one electron, so one chlorine suffices: $\text{K}^+$ and $\text{Cl}^-$ give KCl.

Calcium gives up two, so two chlorine atoms are needed: $\text{Ca}^{2+}$ and two $\text{Cl}^-$ give $\text{CaCl}_2$.

Question 15

Illustrate how sodium sulfide (Na₂S) is formed.

Solution. Write both configurations. Sodium is 2, 8, 1 — one valence electron, which it loses. Sulfur is 2, 8, 6 — six valence electrons, needing two more.

Each sodium can give only one electron, but sulfur needs two. So two sodium atoms are required for every one sulfur atom:

$$2\,\text{Na} \longrightarrow 2\,\text{Na}^+ + 2e^-$$

$$\text{S} + 2e^- \longrightarrow \text{S}^{2-}$$

Each sodium becomes $\text{Na}^+$ with the stable 2, 8 arrangement, and sulfur becomes the sulfide ion $\text{S}^{2-}$ with 2, 8, 8. The oppositely charged ions attract strongly, and the charges balance:

$$2 \times (+1) + 1 \times (-2) = 0 \quad\Longrightarrow\quad \mathbf{Na_2S}$$

The bond is ionic, formed by transfer rather than sharing, since sodium is a metal and sulfur a non-metal.

Pause and Ponder, page 177. The box that follows the section on naming and on writing formulae.

Question 16

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

Solution. These are covalent compounds, so they are named by prefixes that count the atoms of each element — mono-, di-, tri-, tetra-, penta-, hexa- — with the second element’s name ending in -ide. The prefix mono- is dropped on the first element.

(i) $\text{CO}_2$ — one carbon, two oxygen: carbon dioxide.

(ii) $\text{NO}_2$ — one nitrogen, two oxygen: nitrogen dioxide.

(iii) $\text{SF}_6$ — one sulfur, six fluorine: sulfur hexafluoride.

(iv) $\text{PCl}_3$ — one phosphorus, three chlorine: phosphorus trichloride.

Question 17

Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide

Solution. (i) Sodium hydrogencarbonate — $\text{Na}^+$ with the hydrogencarbonate ion $\text{HCO}_3^-$. Charges balance one to one: $\text{NaHCO}_3$.

(ii) Sulfur dioxide — a covalent compound, and the prefix di- gives the count directly: one sulfur, two oxygen, $\text{SO}_2$.

(iii) Ferric chloride — “ferric” is iron in its $+3$ state, so $\text{Fe}^{3+}$ with $\text{Cl}^-$. Three chlorides balance one iron: $\text{FeCl}_3$.

(iv) Cuprous oxide — “cuprous” is copper in its $+1$ state, so $\text{Cu}^+$ with $\text{O}^{2-}$. Two coppers balance one oxide: $\text{Cu}_2\text{O}$.

The old names are worth knowing: -ous is the lower charge and -ic the higher, so ferrous is $\text{Fe}^{2+}$ against ferric $\text{Fe}^{3+}$, and cuprous is $\text{Cu}^+$ against cupric $\text{Cu}^{2+}$.

Question 18

Write the formulae for the compounds formed from the following pairs of ions: (i) Fe³⁺ and OH⁻ (ii) K⁺ and CO₃²⁻

Solution. (i) $\text{Fe}^{3+}$ and $\text{OH}^-$. One iron carries $+3$, so three hydroxide ions are needed. Hydroxide is a group of two atoms, and there is more than one of it, so it must be bracketed:

$$\text{Fe(OH)}_3$$

Written $\text{FeOH}_3$ it would mean one oxygen and three hydrogens, which is a different thing entirely.

(ii) $\text{K}^+$ and $\text{CO}_3^{2-}$. Each potassium carries $+1$ and the carbonate $-2$, so two potassium ions are needed. Only one carbonate is present, so no bracket:

$$\text{K}_2\text{CO}_3$$

Pause and Ponder, page 179. Two boxes on this page. The first follows Section 9.6, on the properties of ionic and covalent compounds.

Question 19

What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?

Solution. Conduction needs charged particles free to move, so the description is really about where the charges are and whether they can travel.

The compound does not conduct as a solid, yet does once dissolved — so the charges are present all along and are merely immobile in the solid. That is exactly the behaviour of an ionic compound: its ions are locked in a rigid lattice in the solid state and cannot move, but dissolving in water separates them and lets them move freely, so the solution conducts.

A covalent compound would not fit: it contains no ions at all, so it conducts in neither state.

Question 20

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.

Solution. “Two electrons in the M shell” fixes the configuration as 2, 8, 2 — twelve electrons, so M is magnesium, and it loses two electrons to become $\text{M}^{2+}$.

(i) The formula. Oxygen gains two electrons to become $\text{O}^{2-}$. The charges are equal and opposite, so they combine one to one:

$$\mathbf{MO}$$

(With M as magnesium, that is MgO.)

(ii) The type of bond. M is a metal giving electrons away and oxygen a non-metal taking them, so the electrons are transferred, not shared. The bond is ionic.

(iii) Conductivity of the solution. The compound is stated to be slightly soluble, and the part that dissolves separates into free $\text{M}^{2+}$ and $\text{O}^{2-}$ ions. Free ions carry current, so the aqueous solution conducts electricity — as any ionic compound’s solution does.

Pause and Ponder, page 179 (second box). After Section 9.7, on the molecular mass of covalent compounds.

Question 21

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

Solution. Add the atomic masses, counting each atom with its subscript. There is one hydrogen, one nitrogen and three oxygens:

$$(1 \times 1) + (14 \times 1) + (16 \times 3) = 1 + 14 + 48 = 63\ \text{u}$$

Question 22

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

Solution. One carbon and four hydrogens:

$$(12 \times 1) + (1 \times 4) = 12 + 4 = 16\ \text{u}$$

Pause and Ponder, page 180. The last box, after Section 9.8 on the formula unit mass of ionic compounds.

Question 23

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

Solution. Potassium chloride is ionic, so it forms no molecules — its formula gives the simplest whole-number ratio of ions, and the mass of that formula unit is what is wanted. One potassium and one chloride:

$$39 + 35.5 = 74.5\ \text{u}$$

Question 24

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

Solution. The bracket matters here: the subscript 2 applies to the whole hydroxide group, so there are two oxygens and two hydrogens, not one of each.

$$24 + 2 \times (16 + 1) = 24 + 2 \times 17 = 24 + 34 = 58\ \text{u}$$

Counting the atoms individually gives the same thing: $24 + (16 \times 2) + (1 \times 2) = 24 + 32 + 2 = 58$ u.

Common mistakes

• Concluding the law of conservation of mass has failed in Question 1. The gases escaped from the open beaker; in a closed container, with the oxygen counted as a reactant, the masses balance exactly.
• Accepting R in Question 7 because it sounds like Dalton. His postulate is about the ratio of numbers of atoms, not of masses. Water is 2 : 1 by atoms and 1 : 8 by mass, and only the first is a simple whole-number ratio.
• Comparing 4 : 1 and 8 : 2 without simplifying in Question 6. They are the same ratio, so the law is obeyed; Y just used twice as much of both.
• Ignoring the bracket in $\text{Mg(OH)}_2$ in Question 24. The subscript applies to the whole group, so there are two oxygens and two hydrogens, giving 58 u rather than 41 u.
• Writing $\text{FeOH}_3$ instead of $\text{Fe(OH)}_3$ in Question 18. Without the bracket the formula says one oxygen and three hydrogens, which is a different compound altogether.
• Saying neon “shares its electrons weakly” in Question 11. It does neither. A complete octet gives it no reason to bond at all, which is why it stays as single atoms.

Practise next

• The end-of-chapter questions, Revise, Reflect, Refine, where the same formula-writing and mass calculations appear with harder cases — a whole grid of cations against anions, and the electron and neutron counts of charged species.
• Chapter 8, Journey Inside the Atom, whose electronic configurations are what every bonding answer here depends on.