---
title: "NCERT Solutions Class 9 Science Ch 9 Atomic Foundations of Matter"
url: https://www.swavid.com/science/class/9/chapter/atomic-foundations-of-matter/ncert-solutions
dateModified: 2026-10-07T16:31:47+00:00
---

# NCERT Solutions Class 9 Science Ch 9 Atomic Foundations of Matter

This chapter contains questions focused on the fundamental laws of chemical combinations, including the Law of Conservation of Mass and the Law of Constant Proportions. It also covers atomic theory, the formation of covalent and ionic bonds, and the calculation of molecular and formula unit masses.

Free PDF (24 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-9/swavid-ncert-solutions-class-9-science-chapter-9-atomic-foundations-of-matter-b1b1f1cf8b.pdf

## Think It Over

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Water obtained from various sources such as rivers, borewells, or the ocean contains the same elements, hydrogen and oxygen.
2. According to the Law of Constant Proportions, these elements combine in a fixed mass ratio of 1:8 in any water sample.
3. Therefore, all these samples of water are chemically identical with the same molecular formula $$H_2O$$.

**Answer:** Yes, all water samples are chemically identical as they consist of hydrogen and oxygen combined in a fixed mass ratio of 1:8.

> Common mistake: Thinking that water from natural sources contains impurities and is therefore chemically different.

### Question 2

*3 marks · Short answer*

Oxygen is sometimes represented as O and sometimes as $O_2$. What is the difference between these symbols?

**Solution**

1. The symbol O represents a single oxygen atom, which is the basic unit of the element oxygen.
2. The symbol $O_2$ represents an oxygen molecule, which consists of two oxygen atoms chemically bonded together.
3. A single atom O is usually unstable and highly reactive, whereas the molecule $O_2$ is a stable, independent entity capable of existing freely.

**Answer:** O represents a single oxygen atom, while $O_2$ represents an oxygen molecule consisting of two oxygen atoms bonded together.

> Common mistake: Confusing atomic symbols with molecular formulae.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. Common salt (sodium chloride) is an ionic compound that dissociates into free mobile ions when dissolved in water.
2. These free ions carry electric current through the solution, allowing dissolved salt to conduct electricity.
3. Sugar is a covalent compound that dissolves in water to form molecules rather than ions, so it does not provide free ions to conduct electricity.

**Answer:** Dissolved salt conducts electricity because it dissociates into free ions in water, whereas sugar dissolves as neutral molecules without forming ions.

> Common mistake: Assuming that all substances dissolved in water can conduct electricity.

## Pause and Ponder

### Question 1

*3 marks · Short answer*

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**

1. No, the Law of Conservation of Mass is not violated.
2. Ethanol burns in an open beaker to produce carbon dioxide gas and water vapour, which escape into the atmosphere.
3. If the mass of the escaping gases is measured along with the products, the total mass before and after the reaction remains exactly the same.

**Answer:** The Law of Conservation of Mass is not violated; the mass is conserved, but the gaseous products escape into the surroundings.

> Common mistake: Assuming mass is lost simply because no residue is left in the open beaker, forgetting that gases escape.

### Question 2

*3 marks · Numerical*

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**

1. Given: Mass of hydrogen = $20\text{ g}$, Mass of oxygen = $160\text{ g}$
2. Formula: Total mass of reactants = Mass of hydrogen + Mass of oxygen
3. Substitution: Total mass of reactants = $20\text{ g} + 160\text{ g} = 180\text{ g}$
4. According to the Law of Conservation of Mass, the mass of products equals the mass of reactants.
5. Result: $180\text{ g}$ of water

**Answer:** $180\text{ g}$

> Common mistake: Adding the masses incorrectly or forgetting to state units.

## Pause and Ponder

### Question 3

*3 marks · Numerical*

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**

1. Given: Mass percentage of sulfur = $40\%$, Mass percentage of oxygen = $60\%$
2. Ratio of sulfur to oxygen by mass = $40 : 60 = 2 : 3$
3. Mass of sulfur in the given sample = $20\text{ g}$
4. Let the mass of oxygen be $x$
5. Formula: $\frac{\text{Mass of sulfur}}{\text{Mass of oxygen}} = \frac{2}{3}$
6. Substitution: $\frac{20}{x} = \frac{2}{3}$
7. Result: $x = \frac{20 \times 3}{2} = 30\text{ g}$ of oxygen

**Answer:** 30 g of oxygen

> Common mistake: Mixing up the ratio of elements or inverting the proportion fractions.

### Question 4

*3 marks · Numerical*

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**

1. Given: Mass ratio of carbon to oxygen in carbon monoxide (CO) = $3 : 4$
2. Mass of carbon given = $9\text{ g}$
3. Let the mass of oxygen required be $x$
4. Formula: $\frac{\text{Mass of carbon}}{\text{Mass of oxygen}} = \frac{3}{4}$
5. Substitution: $\frac{9}{x} = \frac{3}{4}$
6. Result: $x = \frac{9 \times 4}{3} = 12\text{ g}$ of oxygen

**Answer:** 12 g of oxygen

> Common mistake: Using the wrong ratio numbers or multiplying instead of dividing.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. Compounds are formed by elements combining in a fixed ratio by mass, irrespective of their source or method of preparation.
2. Mixtures consist of substances combined in any arbitrary ratio without any fixed chemical composition.
3. Therefore, the Law of Definite Proportions applies only to chemical compounds and not to mixtures.

**Answer:** Compounds have a fixed composition by mass, whereas mixtures do not have a fixed proportion of their components.

> Common mistake: Stating that mixtures have fixed ratios.

### Question 6

*3 marks · Short answer*

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**

1. Student X prepared the oxide of copper in the mass ratio of $4 : 1$, which simplifies to $4 : 1$.
2. Student Y prepared the oxide of copper in the mass ratio of $8 : 2$, which simplifies to $\frac{8}{2} = 4 : 1$.
3. Since both ratios simplify to the same mass ratio of $4 : 1$, their results fully justify the Law of Constant Proportions.

**Answer:** Yes, their results justify the Law of Constant Proportions because both mass ratios simplify to 4:1.

> Common mistake: Failing to simplify the ratio 8:2 to compare it with 4:1.

## Pause and Ponder

### Question 7

*1 mark · Assertion and reason*

Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. Reason (R): According to Dalton’s Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds. Choose the correct option:

- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. Assertion (A) is true because mass is conserved in a chemical reaction according to the Law of Conservation of Mass ($2\text{ g} + 16\text{ g} = 18\text{ g}$).
2. Reason (R) is true because Dalton's Atomic Theory states that atoms combine in a ratio of simple whole numbers to form compounds.
3. However, R is not the correct explanation of A because A illustrates the conservation of mass during a chemical reaction, whereas R explains the combining ratio of atoms.

**Answer:** (ii) Both A and R are true, but R is not the correct explanation of A.

> Common mistake: Confusing the law of conservation of mass with the postulate about simple whole number ratios of atoms.

## Pause and Ponder

### Question 8

*3 marks · Short answer*

Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule ($N_2$).

**Solution**

1. A nitrogen atom has five valence electrons and requires three more electrons to complete its octet.
2. Two nitrogen atoms share three pairs of electrons between them to attain a stable electronic configuration.
3. They are joined by a triple bond, represented as $\text{N}\equiv\text{N}$.

**Answer:** A nitrogen molecule ($\text{N}_2$) is formed by sharing three pairs of electrons, resulting in a triple bond ($\text{N}\equiv\text{N}$).

> Common mistake: Drawing a single or double bond instead of a triple bond for the nitrogen molecule.

### Question 9

*3 marks · Short answer*

The atomic number of fluorine is 9. Explain the formation of the fluorine molecule ($F_2$).

**Solution**

1. The atomic number of fluorine is 9, so its electronic configuration is 2, 7, having seven valence electrons.
2. Each fluorine atom requires one more electron to complete its octet and attain a stable configuration.
3. Two fluorine atoms share one electron each to form a fluorine molecule ($\text{F}_2$) joined by a single covalent bond ($\text{F}-\text{F}$).

**Answer:** Two fluorine atoms share one electron pair to form a fluorine molecule ($\text{F}_2$) with a single covalent bond.

> Common mistake: Confusing sharing of electrons with electron transfer in non-metal combination.

## Pause and Ponder

### Question 10

*3 marks · Short answer*

Show the formation of the following molecules: (i) Carbon dioxide ($CO_2$) (ii) Hydrogen sulfide ($H_2S$) (iii) Ammonia ($NH_3$)

**Part (i)**

1. Carbon has 4 valence electrons and needs 4 more, while each oxygen atom has 6 valence electrons and needs 2 more.
2. The carbon atom shares two pairs of electrons with each of the two oxygen atoms.
3. This forms a carbon dioxide ($CO_2$) molecule with double bonds between carbon and each oxygen atom.

Answer (i): Carbon shares two pairs of electrons with two oxygen atoms to form $CO_2$ with double bonds.

**Part (ii)**

1. Hydrogen has 1 valence electron and needs 1 more for its duplet, while sulfur has 6 valence electrons and needs 2 more for its octet.
2. Two hydrogen atoms share one electron each with one sulfur atom.
3. This forms a hydrogen sulfide ($H_2S$) molecule with single covalent bonds.

Answer (ii): Two hydrogen atoms share an electron each with a sulfur atom to form $H_2S$ with single bonds.

**Part (iii)**

1. Nitrogen has 5 valence electrons and needs 3 more, while each hydrogen atom needs 1 electron.
2. Three hydrogen atoms share one electron each with one nitrogen atom.
3. This forms an ammonia ($NH_3$) molecule with single covalent bonds.

Answer (iii): Three hydrogen atoms share one electron each with a nitrogen atom to form $NH_3$ with single bonds.

**Answer:** Carbon dioxide, hydrogen sulfide, and ammonia are formed by the sharing of valence electrons between atoms to complete their octet or duplet.

> Common mistake: Confusing the number of shared electron pairs with the number of atoms.

### Question 11

*3 marks · Short answer*

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

**Solution**

1. The atomic number of neon is 10, and its electronic configuration is 2, 8.
2. Neon has 8 electrons in its outermost valence shell, which represents a complete octet.
3. Atoms with an octet of electrons in their valence shell are already stable and do not need to lose, gain, or share electrons.

**Answer:** Neon has a stable electronic configuration with 8 valence electrons, so it does not need to transfer or share electrons.

> Common mistake: Stating neon has a duplet instead of an octet in its valence shell.

## Pause and Ponder

### Question 12

*2 marks · Very short answer*

What kind of ion will oxygen (O) form?

**Solution**

1. Oxygen has an atomic number of 8 with an electronic configuration of 2, 6.
2. It needs two more electrons to complete its octet and form an oxide anion represented as $O^{2-}$.
3. Therefore, oxygen forms a negatively charged anion.

**Answer:** Oxygen forms a negatively charged anion ($O^{2-}$).

> Common mistake: Confusing cations with anions.

### Question 13

*1 mark · Fill in the blank*

Fill in the blanks. Among magnesium and chlorine, magnesium atom can give two electrons to become $Mg^{2+}$. However, chlorine can take only one electron to become ____________. Now, __________ ion of magnesium and __________ ions of chlorine combine to give magnesium chloride.

**Solution**

1. Magnesium ion ($Mg^{2+}$) and two chloride ions ($Cl^-$) combine to form magnesium chloride ($MgCl_2$).

**Answer:** chloride ($Cl^-$), one, two

> Common mistake: Incorrect ratio of ions.

### Question 14

*3 marks · Short answer*

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

**Solution**

1. Potassium ($K$, atomic number 19) loses 1 valence electron to form a potassium cation ($K^+$): $K \rightarrow K^+ + e^-$.
2. Calcium ($Ca$, atomic number 20) loses 2 valence electrons to form a calcium cation ($Ca^{2+}$): $Ca \rightarrow Ca^{2+} + 2e^-$.
3. Potassium ion combines with one chloride ion to form $KCl$, and calcium ion combines with two chloride ions to form $CaCl_2$.

**Answer:** Potassium forms $K^+$ and $KCl$; calcium forms $Ca^{2+}$ and $CaCl_2$.

> Common mistake: Incorrect charge on the metal ions.

### Question 15

*3 marks · Short answer*

Illustrate how sodium sulfide ($Na_2S$) is formed.

**Solution**

1. A sodium atom ($Na$) loses 1 electron to form a sodium cation ($Na^+$).
2. A sulfur atom ($S$) gains 2 electrons to form a sulfide anion ($S^{2-}$).
3. Two sodium ions combine with one sulfide ion through an electrostatic force of attraction to form sodium sulfide ($Na_2S$).

**Answer:** Two $Na^+$ ions combine with one $S^{2-}$ ion to form $Na_2S$.

> Common mistake: Wrong stoichiometry in the formula.

## Pause and Ponder

### Question 16

*3 marks · Short answer*

Name the following: (i) $CO_2$ (ii) $NO_2$ (iii) $SF_6$ (iv) $PCl_3$

**Part (i)**

1. The first element C is carbon and the second element O with two atoms uses the prefix di-.
2. The name is carbon dioxide.

Answer (i): Carbon dioxide

**Part (ii)**

1. The first element N is nitrogen and the second element O with two atoms uses the prefix di-.
2. The name is nitrogen dioxide.

Answer (ii): Nitrogen dioxide

**Part (iii)**

1. The first element S is sulfur and the second element F with six atoms uses the prefix hexa-.
2. The name is sulfur hexafluoride.

Answer (iii): Sulfur hexafluoride

**Part (iv)**

1. The first element P is phosphorus and the second element Cl with three atoms uses the prefix tri-.
2. The name is phosphorus trichloride.

Answer (iv): Phosphorus trichloride

**Answer:** Named according to the number of atoms of each element using prefixes and the -ide suffix for the second element.

> Common mistake: Omitting the correct numerical prefix or failing to add the -ide suffix to the second element.

### Question 17

*3 marks · Short answer*

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

**Part (i)**

1. The ions are Na+ and HCO3-.
2. Criss-crossing the valencies gives NaHCO3.

Answer (i): $\text{NaHCO}_3$

**Part (ii)**

1. The constituent elements are sulfur and oxygen with valencies 2 and 2, simplified to 1 and 2.
2. Criss-crossing gives $\text{SO}_2$.

Answer (ii): $\text{SO}_2$

**Part (iii)**

1. The ions are Fe3+ and Cl-.
2. Criss-crossing the charges gives $\text{FeCl}_3$.

Answer (iii): $\text{FeCl}_3$

**Part (iv)**

1. The ions are Cu+ and O2-.
2. Criss-crossing the charges gives $\text{Cu}_2\text{O}$.

Answer (iv): $\text{Cu}_2\text{O}$

**Answer:** Derived using symbols, valencies or charges, and criss-crossing.

> Common mistake: Incorrectly criss-crossing charges or failing to reduce subscripts to the simplest ratio.

### Question 18

*3 marks · Short answer*

Write the formulae for the compounds formed from the following pairs of ions: (i) $Fe^{3+}$ and $OH^-$ (ii) $K^+$ and $CO_3^{2-}$

**Part (i)**

1. Write the symbols of the cation $\text{Fe}^{3+}$ and the anion $\text{OH}^-$.
2. Write their charges underneath as 3+ and 1-.
3. Criss-cross the numerical values to obtain $\text{Fe(OH)}_3$.

Answer (i): $\text{Fe(OH)}_3$

**Part (ii)**

1. Write the symbols of the cation $\text{K}^+$ and the anion $\text{CO}_3^{2-}$.
2. Write their charges underneath as 1+ and 2-.
3. Criss-cross the numerical values to obtain $\text{K}_2\text{CO}_3$.

Answer (ii): $\text{K}_2\text{CO}_3$

**Answer:** Derived by writing cation and anion symbols and criss-crossing their charges.

> Common mistake: Forgetting to enclose polyatomic ions in parentheses when a subscript greater than 1 is applied.

## Pause and Ponder

### Question 19

*3 marks · Short answer*

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**

1. Ionic compounds do not conduct electricity in the solid state because their ions are held in fixed positions by strong electrostatic forces.
2. When dissolved in water, the ions become free to move, allowing the solution to conduct electricity.
3. Therefore, the type of chemical bond present in the solid compound is an ionic bond.

**Answer:** Ionic bond

> Common mistake: Stating that covalent compounds conduct electricity in water.

### Question 20

*3 marks · Case-based*

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.

**Part (i)**

1. Metal M has two valence electrons, so its valency is 2 and it forms an $\text{M}^{2+}$ ion.
2. Oxygen has six valence electrons and requires two electrons, forming an $\text{O}^{2-}$ ion.
3. Criss-crossing the charges gives the formula $\text{MO}$.

Answer (i): $\text{MO}$

**Part (ii)**

1. Metal M donates its two valence electrons to oxygen to form ions.
2. The electrostatic force of attraction between the oppositely charged ions forms an ionic bond.

Answer (ii): Ionic bond

**Part (iii)**

1. In an aqueous solution, the ions of the ionic compound become free to move.
2. Therefore, its aqueous solution conducts electricity.

Answer (iii): Conducts electricity

**Answer:** (i) MO, (ii) Ionic bond, (iii) Conducts electricity

> Common mistake: Writing the formula as $\text{M}_{2}\text{O}_{2}$ instead of simplifying it to $\text{MO}$.

## Pause and Ponder

### Question 21

*3 marks · Numerical*

Find the molecular mass of nitric acid ($HNO_3$). Atomic mass — H = 1 u; N = 14 u; O = 16 u.

**Solution**

1. Given: Atomic masses of H = 1 u, N = 14 u, and O = 16 u
2. Formula: Molecular mass of $\text{HNO}_3 = (1 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of N}) + (3 \times \text{Atomic mass of O})$
3. Substitution: Molecular mass $= (1 \text{ u} \times 1) + (14 \text{ u} \times 1) + (16 \text{ u} \times 3)$
4. Result: $1\text{ u} + 14\text{ u} + 48\text{ u} = 63\text{ u}$

**Answer:** $63\text{ u}$

> Common mistake: Multiplying the atomic mass of nitrogen or hydrogen incorrectly by the number of atoms.

### Question 22

*3 marks · Numerical*

Find the molecular mass of methane ($CH_4$). Atomic mass — C = 12 u; H = 1 u.

**Solution**

1. Given: Atomic masses of C = 12 u and H = 1 u
2. Formula: Molecular mass of $\text{CH}_4 = (1 \times \text{Atomic mass of C}) + (4 \times \text{Atomic mass of H})$
3. Substitution: Molecular mass $= (12 \text{ u} \times 1) + (1 \text{ u} \times 4)$
4. Result: $12\text{ u} + 4\text{ u} = 16\text{ u}$

**Answer:** $16\text{ u}$

> Common mistake: Incorrectly adding the atomic masses without multiplying by the respective number of atoms.

## Pause and Ponder

### Question 23

*3 marks · Numerical*

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

**Solution**

1. Given: Atomic mass of $\text{K} = 39\text{ u}$, Atomic mass of $\text{Cl} = 35.5\text{ u}$
2. Formula: $\text{Formula unit mass of KCl} = (\text{Atomic mass of K} \times 1) + (\text{Atomic mass of Cl} \times 1)$
3. Substitution: $= (39\text{ u} \times 1) + (35.5\text{ u} \times 1)$
4. Result: $39\text{ u} + 35.5\text{ u} = 74.5\text{ u}$

**Answer:** 74.5 u

> Common mistake: Adding atomic masses incorrectly or forgetting units.

### Question 24

*3 marks · Numerical*

Find the formula unit mass of magnesium hydroxide, $Mg(OH)_2$. Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.

**Solution**

1. Given: Atomic mass of $\text{Mg} = 24\text{ u}$, $\text{O} = 16\text{ u}$, $\text{H} = 1\text{ u}$.
2. Formula: $\text{Formula unit mass of } \text{Mg(OH)}_2 = (1 \times \text{Atomic mass of Mg}) + 2 \times (\text{Atomic mass of O} + \text{Atomic mass of H})$
3. Substitution: $= (24\text{ u} \times 1) + 2 \times (16\text{ u} \times 1 + 1\text{ u} \times 1)$
4. Working: $= 24\text{ u} + 2 \times (17\text{ u}) = 24\text{ u} + 34\text{ u}$
5. Result: $58\text{ u}$

**Answer:** $58\text{ u}$

> Common mistake: Forgetting to multiply both oxygen and hydrogen by 2 due to the subscript outside the bracket.

## Revise, Reflect, Refine

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Element A has one electron in its valence shell, so it tends to give 1 electron to become stable.

Answer (i): 1 electron

**Part (ii)**

1. By losing one electron, it forms a positively charged ion called a cation (A+).

Answer (ii): Positive ion (cation)

**Part (iii)**

1. Element B has six electrons in its valence shell, so it tends to take 2 electrons to complete its octet.

Answer (iii): 2 electrons

**Part (iv)**

1. By gaining two electrons, it forms a negatively charged ion called an anion (B2-).

Answer (iv): Negative ion (anion)

**Part (v)**

1. Since A is a metal and B is a non-metal, they form an ionic bond by the transfer of electrons.

Answer (v): Ionic bond

**Part (vi)**

1. The valency of A is 1 and B is 2, so by criss-crossing, the formula is A2B.

Answer (vi): A2B

**Answer:** The compound formed is A2B.

> Common mistake: Confusing the number of electrons to gain/lose with the final charge.

### Question 2

*3 marks · Short answer*

An element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure of the molecule it would form. (iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.

**Part (i)**

1. Element X has six valence electrons and needs two more to complete its octet, so it shares two electrons with another X atom.

Answer (i): It shares two electrons to complete its octet.

**Part (ii)**

1. The sharing of two pairs of electrons results in a double covalent bond.

Answer (ii): Covalent bond (double bond)

**Part (iii)**

1. Diagram: Draw two X atoms with two shared pairs of electrons in the overlapping region.

Answer (iii): Structure with two shared pairs of electrons.

**Part (iv)**

1. Element Y has two valence electrons, so it transfers them to two X atoms to form an ionic compound YX2.

Answer (iv): Ionic structure Y2+ and 2X-.

**Answer:** X forms a double covalent bond with another X atom.

> Common mistake: Assuming all bonds are covalent.

### Question 3

*1 mark · MCQ*

You want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6–. Which of the following combinations gives the correct number of ions?

- 2 $Al^{3+}$ and 3 $Cl^-$
- 3 $Mg^{2+}$ and 1 $PO_4^{3-}$
- 2 $Fe^{3+}$ and 3 $O^{2-}$
- 3 $Ca^{2+}$ and 2 $SO_4^{2-}$

**Solution**

1. Total positive charge must equal total negative charge.
2. In option (iii), 2 Fe3+ gives 2 × 3+ = 6+ and 3 O2- gives 3 × 2- = 6-.

**Answer:** (iii) 2 Fe3+ and 3 O2-

> Common mistake: Miscalculating the total charge by ignoring the number of ions.

### Question 4

*3 marks · Short answer*

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

**Part (i)**

1. False. Elements are made up of atoms and compounds are made up of molecules or ions.

Answer (i): False

**Part (ii)**

1. False. The molecule of a compound is made up of two or more different kinds of atoms.

Answer (ii): False

**Part (iii)**

1. False. One molecule of nitrogen gas (N2) contains two nitrogen atoms.

Answer (iii): False

**Part (iv)**

1. True.

Answer (iv): True

**Answer:** Statement (iv) is correct.

> Common mistake: Confusing the definition of elements and compounds.

### Question 5

*3 marks · Short answer*

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

**Part (i)**

1. Al3+ and NO3- criss-cross to give Al(NO3)3.

Answer (i): Al(NO3)3

**Part (ii)**

1. Ca2+ and O2- have the same valency, so the formula is CaO.

Answer (ii): CaO

**Part (iii)**

1. Fe3+ and O2- criss-cross to give Fe2O3.

Answer (iii): Fe2O3

**Answer:** Formulae: (i) Al(NO3)3, (ii) CaO, (iii) Fe2O3.

> Common mistake: Forgetting to use brackets for polyatomic ions.

### Question 6

*3 marks · Short answer*

Write the formulae of the compounds formed from the following pairs of ions. (i) $Ca^{2+}$ and $Br^-$ (ii) $Al^{3+}$ and $CO_3^{2-}$ (iii) $K^+$ and $SO_4^{2-}$ (iv) $NH_4^+$ and $Cl^-$

**Part (i)**

1. Ca2+ and Br- criss-cross to give CaBr2.

Answer (i): CaBr2

**Part (ii)**

1. Al3+ and CO3 2- criss-cross to give Al2(CO3)3.

Answer (ii): Al2(CO3)3

**Part (iii)**

1. K+ and SO4 2- criss-cross to give K2SO4.

Answer (iii): K2SO4

**Part (iv)**

1. NH4+ and Cl- have the same valency, so the formula is NH4Cl.

Answer (iv): NH4Cl

**Answer:** Formulae: (i) CaBr2, (ii) Al2(CO3)3, (iii) K2SO4, (iv) NH4Cl.

> Common mistake: Incorrectly criss-crossing charges for polyatomic ions.

### Question 7

*1 mark · MCQ*

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

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

**Solution**

1. The atomic number of chlorine is 17, so its neutral atom has 17 electrons with the electronic configuration 2, 8, 7.
2. A chloride ion ($Cl^-$) is formed by gaining one electron, giving it 18 electrons with the electronic configuration 2, 8, 8.
3. Diagram (iii) in Fig. 9.18 correctly shows 2 electrons in the first shell, 8 in the second shell, and 8 in the outermost shell.

**Answer:** (iii)

> Common mistake: Confusing the chlorine neutral atom (2, 8, 7) with the chloride anion (2, 8, 8).

### Question 8

*3 marks · Numerical*

Determine the formula unit mass of the following substances. (i) Ammonium nitrate ($NH_4NO_3$), used as a nitrogen fertiliser, which is essential for plant growth. (ii) Phosphoric acid ($H_3PO_4$), used to make phosphate fertiliser and detergents. (iii) Sodium hydrogencarbonate ($NaHCO_3$), used to relieve acidity and helps in digestion.

**Part (i)**

1. Given: Atomic masses are N = 14 u, H = 1 u, O = 16 u.
2. Formula: Formula unit mass of $NH_4NO_3 = (14 \times 2) + (1 \times 4) + (16 \times 3)$
3. Substitution: $= 28 + 4 + 48$
4. Result: $80 \text{ u}$

Answer (i): 80 u

**Part (ii)**

1. Given: Atomic masses are H = 1 u, P = 31 u (standard atomic mass), O = 16 u.
2. Formula: Formula unit mass of $H_3PO_4 = (1 \times 3) + 31 + (16 \times 4)$
3. Substitution: $= 3 + 31 + 64$
4. Result: $98 \text{ u}$

Answer (ii): 98 u

**Part (iii)**

1. Given: Atomic masses are Na = 23 u, H = 1 u, C = 12 u, O = 16 u.
2. Formula: Formula unit mass of $NaHCO_3 = 23 + 1 + 12 + (16 \times 3)$
3. Substitution: $= 36 + 48$
4. Result: $84 \text{ u}$

Answer (iii): 84 u

**Answer:** Formula unit masses are (i) 80 u, (ii) 98 u, (iii) 84 u

> Common mistake: Forgetting to multiply the subscript of an element with its atomic mass inside parentheses.

### Question 9

*3 marks · Short answer*

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

**Part (i)**

1. Magnesium has a valency of 2 ($Mg^{2+}$) and nitrogen has a valency of 3 ($N^{3-}$).
2. Criss-crossing the valencies gives $Mg_3N_2$.

Answer (i): Mg3N2

**Part (ii)**

1. Lithium has a valency of 1 ($Li^+$) and nitrogen has a valency of 3 ($N^{3-}$).
2. Criss-crossing the valencies gives $Li_3N$.

Answer (ii): Li3N

**Part (iii)**

1. Sodium has a valency of 1 ($Na^+$) and sulfur has a valency of 2 ($S^{2-}$).
2. Criss-crossing the valencies gives $Na_2S$.

Answer (iii): Na2S

**Part (iv)**

1. Aluminium has a valency of 3 ($Al^{3+}$) and oxygen has a valency of 2 ($O^{2-}$).
2. Criss-crossing the valencies gives $Al_2O_3$.

Answer (iv): Al2O3

**Answer:** Formulae are (i) $Mg_3N_2$, (ii) $Li_3N$, (iii) $Na_2S$, (iv) $Al_2O_3$

> Common mistake: Writing incorrect valencies for elements from group numbers.

### Question 10

*Activity*

Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. $LiNO_3$ is given as an example.

**Solution**

1. For $NH_4^+$ and $NO_3^-$, the formula is $NH_4NO_3$.
2. For $NH_4^+$ and $SO_4^{2-}$, the formula is $(NH_4)_2SO_4$.
3. For $NH_4^+$ and $PO_4^{3-}$, the formula is $(NH_4)_3PO_4$.
4. For $Li^+$ and $SO_4^{2-}$, the formula is $Li_2SO_4$.
5. For $Li^+$ and $PO_4^{3-}$, the formula is $Li_3PO_4$.
6. For $Al^{3+}$ and $NO_3^-$, the formula is $Al(NO_3)_3$.
7. For $Al^{3+}$ and $SO_4^{2-}$, the formula is $Al_2(SO_4)_3$.
8. For $Al^{3+}$ and $PO_4^{3-}$, the formula is $AlPO_4$.
9. For $Cu^{2+}$ and $NO_3^-$, the formula is $Cu(NO_3)_2$.
10. For $Cu^{2+}$ and $SO_4^{2-}$, the formula is $CuSO_4$.
11. For $Cu^{2+}$ and $PO_4^{3-}$, the formula is $Cu_3(PO_4)_2$.

**Answer:** Table completed with correct chemical formulae.

> Common mistake: Omitting parentheses around polyatomic ions when their subscript is greater than 1.

### Question 11

*3 marks · Numerical*

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

**Solution**

1. Given: Mass of sodium carbonate = $5.3 \text{ g}$, Mass of acetic acid = $6.0 \text{ g}$.
2. Formula: Total mass of reactants = Mass of sodium carbonate + Mass of acetic acid
3. Substitution: Total mass of reactants = $5.3 \text{ g} + 6.0 \text{ g} = 11.3 \text{ g}$.
4. Given: Mass of carbon dioxide = $2.2 \text{ g}$, Mass of water = $0.9 \text{ g}$, Mass of sodium acetate = $8.2 \text{ g}$.
5. Formula: Total mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium acetate
6. Substitution: Total mass of products = $2.2 \text{ g} + 0.9 \text{ g} + 8.2 \text{ g} = 11.3 \text{ g}$.
7. Result: Since Mass of reactants ($11.3 \text{ g}$) = Mass of products ($11.3 \text{ g}$), the Law of Conservation of Mass is obeyed and valid.

**Answer:** Mass of reactants = Mass of products = 11.3 g. The Law of Conservation of Mass is valid.

> Common mistake: Adding reactant and product masses incorrectly.

### Question 12

*3 marks · Short answer*

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

**Part (i)**

1. Atomic number equals the number of protons, which is 11.
2. Mass number equals the sum of protons and neutrons = $11 + 12 = 23$.

Answer (i): Atomic number = 11, Mass number = 23

**Part (ii)**

1. The species has 11 protons (positive charge) and 10 electrons (negative charge).
2. Since protons exceed electrons, it is a positively charged ion called a cation.

Answer (ii): Cation

**Part (iii)**

1. The neutral atom has 11 electrons, with the electronic configuration 2, 8, 1.
2. The ion has 10 electrons, with the electronic configuration 2, 8.

Answer (iii): 2, 8, 1 (for neutral atom) or 2, 8 (for ion)

**Part (iv)**

1. An element with atomic number 11 is sodium, and its cation is the sodium ion ($Na^+$).

Answer (iv): Sodium ion

**Answer:** (i) Atomic number = 11, Mass number = 23. (ii) Cation, because it has more protons than electrons. (iii) Electronic configuration: 2, 8, 1. (iv) Sodium ion ($Na^+$).

> Common mistake: Confusing mass number with the number of neutrons, or confusing anions and cations.

### Question 13

*3 marks · Short answer*

Two elements, A and B, have the following configurations — A: 2, 8, 5 B: 2, 8, 7 (i) Which element is more reactive? (ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing. (iii) Predict the formula of the compound they would form.

**Part (i)**

1. Element A has electronic configuration $2, 8, 5$ with $5$ valence electrons, needing $3$ electrons to complete its octet.
2. Element B has electronic configuration $2, 8, 7$ with $7$ valence electrons, needing only $1$ electron to complete its octet.
3. Since B requires only one electron to attain stability, it is more reactive than A.

Answer (i): Element B is more reactive.

**Part (ii)**

1. Both elements A and B are non-metals as they have $5$ and $7$ valence electrons respectively.
2. Non-metals combine by sharing valence electrons rather than transferring them.
3. Therefore, A and B will form covalent bonds when they combine.

Answer (ii): They form covalent bonds by sharing electrons.

**Part (iii)**

1. Element A has a valency of $3$ (needs $3$ electrons) and Element B has a valency of $1$ (needs $1$ electron).
2. By criss-crossing the valencies, the formula of the compound formed is $\text{AB}_3$.

Answer (iii): $\text{AB}_3$

**Answer:** (i) Element B is more reactive. (ii) Covalent bonds by sharing electrons. (iii) $\text{AB}_3$ or $\text{AB}_5$.

> Common mistake: Assuming that non-metals can transfer electrons to form ionic bonds.

### Question 14

*1 mark · Assertion and reason*

Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state. Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely. Choose the correct option:

- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. Copper sulfate is an ionic compound which does not conduct electricity in the solid state because its ions are held in fixed positions in the crystal lattice.
2. In the molten state, the ions become free to move and can conduct electricity, making Assertion (A) true.
3. Reason (R) states that ions are fixed in the molten state and free in the solid state, which is exactly the opposite of the actual fact.
4. Therefore, Assertion (A) is true, but Reason (R) is false.

**Answer:** (iii) A is true, but R is false.

> Common mistake: Assuming ions are free to move in the solid state of ionic compounds.

### Question 15

*3 marks · Short answer*

The species $^{27}Al$, $^{80}Br^-$ and $^{201}Hg^{2+}$ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?

**Part (i)**

1. For $^{27}\text{Al}$ (mass number $A = 27$, atomic number $Z = 13$, protons = $13$):
2. Number of electrons = number of protons in a neutral atom = $13$.
3. Number of neutrons = $A - Z = 27 - 13 = 14$.

Answer (i): Electrons = $13$, Neutrons = $14$

**Part (ii)**

1. For $^{80}\text{Br}^-$ (mass number $A = 80$, protons = $35$, negative charge = $1$):
2. Number of electrons = protons + $1 = 35 + 1 = 36$.
3. Number of neutrons = $A - Z = 80 - 35 = 45$.

Answer (ii): Electrons = $36$, Neutrons = $45$

**Part (iii)**

1. For $^{201}\text{Hg}^{2+}$ (mass number $A = 201$, protons = $80$, positive charge = $2$):
2. Number of electrons = protons - $2 = 80 - 2 = 78$.
3. Number of neutrons = $A - Z = 201 - 80 = 121$.

Answer (iii): Electrons = $78$, Neutrons = $121$

**Answer:** For $^{27}\text{Al}$, electrons = $13$, neutrons = $14$; for $^{80}\text{Br}^-$, electrons = $36$, neutrons = $45$; for $^{201}\text{Hg}^{2+}$, electrons = $78$, neutrons = $121$.

> Common mistake: Subtracting or adding the charge incorrectly to find the number of electrons, or confusing mass number with the number of neutrons.

## Frequently asked questions

### How many questions and exercises are included in NCERT Solutions for Class 9 Science Chapter 9 Atomic Foundations of Matter?

This chapter features multiple interactive segments including 3 Think It Over questions and various Pause and Ponder sections spanning numericals, short answers, and fill-in-the-blanks. Additionally, the final Revise, Reflect, Refine exercise contains 15 questions covering activities, MCQs, assertion-reasons, and numericals. You can find complete step-by-step solutions for all these questions in SwaVid's free PDF available on this page.

### Which core topics and scientific laws do the questions in this chapter cover?

The questions thoroughly test fundamental concepts like the Law of Conservation of Mass, Law of Constant Proportions, and Dalton's Atomic Theory. Other major topics include atomic structure, mass number, ion formation, electronic configuration, and calculating molecular mass or formula unit mass. SwaVid's solutions on this page explain each of these topics clearly for the 2026-27 session following the new NCERT book.

### Which question types are considered the most challenging in this chapter and how should students approach them?

Numerical problems involving molecular mass calculation, formula unit mass of ionic compounds, and application-based assertion-reason questions are often found to be the hardest by students. To approach them, you should carefully memorize standard atomic masses and understand the underlying stoichiometry or bonding principles. SwaVid provides detailed, methodical breakdowns of these tough problems in the free PDF on this page.

### How can students write answers for full marks in Class 9 Science Chapter 9 examinations?

Scoring full marks requires precise definitions, clear chemical formulae writing using the criss-cross method, and proper unit inclusion in all numerical calculations. For descriptive topics like covalent bonding or ionic compound properties, structured point-wise answers work best. Referring to SwaVid's expert-crafted solutions on this page will guide you on how to format your answers correctly.

### Is a free PDF of these NCERT solutions available for download for the 2026-27 session?

Yes, comprehensive and accurate solutions aligned with the new NCERT book based on the NCF 2023 guidelines are fully accessible. You can easily download SwaVid's free PDF and view detailed step-by-step explanations directly on this page to aid your exam preparation.

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- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

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