Atomic Foundations of Matter Class 9 Question Answer Science Exploration Chapter 9

Students can use Exploration Class 9 Science Solutions Chapter 9 Atomic Foundations of Matter Question Answer NCERT Solutions as a quick reference guide.

Class 9 Science Exploration Chapter 9 Question Answer

Class 9 Science Ch 9 Atomic Foundations of Matter Question Answer

Atomic Foundations of Matter Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine (NCERT Textbook Page No. 180 – 183)

Question 1.
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?
Solution:
Element A has 1 electron in its third shell → Electronic configuration: 2, 8, 1. This is sodium (Na). Element B has 6 electrons in its second shell
Electronic configuration: 2, 6. This is oxygen (O).
(i) Element A (Na) has 1 valence electron. It tends to give 1 electron to become stable (achieving the configuration of neon: 2, 8).
(ii) By losing 1 electron, A forms a cation with a charge of + 1. It would form Na⁺ (a positively charged ion / cation).
(iii) Element B (O) has 6 valence electrons. It needs 2 more to complete its octet. It tends to take 2 electrons to become stable.
(iv) By gaining 2 electrons, B forms an anion with a charge of 2^– . It would form O^2- (a negatively charged ion / anion).
(v) If A (Na) loses an electron and B (O) gains electrons, the bond formed is an ionic bond (electrostatic force of attraction between the oppositely charged ions Na⁺ and O^2- ).
(vi) A = Na₂⁺ (valency 1), B = O^2- (valency 2).
Formula: Na₂⁺ O (sodium oxide)

Question 2.
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.
Solution:
Element X has 6 valence electrons and needs 2 more to complete its octet. This is oxygen (O), atomic number 8.
(i) X has 6 valence electrons and needs 2 more to complete its octet, so two X atoms share electrons and form a diatomic molecule.
(ii) It forms a covalent bond by sharing electrons.
(iii) Structure of X₂ :

(iv) Element Y has 2 valence electrons, so it forms an ionic compound with X. Y donates 2 electrons and X gains 2 electrons, forming YX (like MgO) with ionic bonding.

Question 3.
You want to design a new ionic compound, where the total positive charge is 6⁺ and the total negative charge is 6^–. Which of the following combinations gives the correct number of ions?
(i) 2 Al³⁺ and 3 Cl^–
(ii) 3 Mg²⁺ and 1 PO^3-₄
(iii) 2 Fe³⁺ and 3 O^2-
(iv) 3 Ca²⁺ and 2 SO₄^2-
Solution:
Let’s check each option carefully.
The goal is total positive charge = + 6
and total negative charge = – 6.
(i) 2 Al³⁺ and 3 Cl^–
2 Al³⁺ → 2 × (+ 3) = + 6
3 Cl^– → 3 × (- 1) = – 3
Charges don’t balance (- 3 ≠ – 6).

(ii) 3 Mg²⁺ and 1 PO^3-₄
3 Mg²⁺ → 3 × (+ 2) = + 6
1 PO^3-₄ → – 3
Charges don’t balance (- 3 ≠ – 6).

(iii) 2 Fe³⁺ and 3 O^2-
2 Fe^3- → 2 × (+ 3) = + 6
3 O^2- → 3 × (- 2) = – 6
Charges balance perfectly (+ 6 and – 6).

(iv) 3 Ca²⁺ and 2 SO₄^2-
3 Ca²⁺ → 3 × (+ 2) = + 6
2 SO^2-₄ → 2 × (- 2) = – 4
Charges don’t balance (- 4 ≠ – 6).
Hence, option (iii) is correct.

Question 4.
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.
Solution:
False.
Correct statement: Elements are made up of atoms and compounds are made up of molecules (which contain atoms of two or more different elements). For example, iron is an element made up of iron atoms; water is a compound made up of molecules each containing 2 hydrogen atoms and 1 oxygen atom.

(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
Solution:
False.
Correct statement: The molecule of a compound is made up of two or more atoms of different kinds (different elements). For example, HCl is made of hydrogen and chlorine atoms (different kinds). It is the molecule of an element (like H₂ or O₂) that is made up of atoms of the same kind.

(iii) One molecule of nitrogen gas contains three nitrogen atoms.
Solution:
False.
Correct statement: One molecule of nitrogen gas (N₂) contains two nitrogen atoms, not three.

(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Solution:
True.
Correct statement: Water (H₂O) is indeed made of two hydrogen atoms covalently bonded with one oxygen atom.

Question 5.
Write the chemical formulae for the following compounds.
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
Solution:
(i) Aluminium nitrate:
Aluminium ion: Al³⁺ (Valency 3)
Nitrate ion: NO₃^– (Valency 1)
To balance charges: 1 Al³⁺ needs 3 NO₃^–
Formula: Al(NO₃)₃

(ii) Calcium oxide
Calcium ion: Ca²⁺(Valency 2)
Oxide ion: O^2- (Valency 2)
Charges balance 1 : 1
Formula: CaO

(iii) Ferric oxide (iron in + 3 oxidation state)
Iron ion: Fe³⁺ (Valency 3)
Oxide ion: O^2- (Valency 2)
To balance charges: 2 Fe³⁺ with 3 O^2-
Formula: Fe₂O₃

Question 6.
Write the formulae of the compounds formed from the following pairs of ions:
(i) Ca²⁺ and Br^–
(ii) Al³⁺ and CO₃^2-
(iii) K⁺ and SO₄^2-
(iv) NH₄⁺ and Cl^–
Solution:
(i) Ca²⁺ and Br^–
Calcium has a +2 charge.
Bromide has a -1 charge.
To balance: 1 Ca²⁺ needs 2 Br^–
Formula: CaBr₂ (Calcium Bromide)

(ii) Al³⁺ and CO₃^2-
Aluminium has a +3 charge.
Carbonate has a -2 charge.
Lowest common multiple of 3 and 2 = 6.
So, 2 Al³⁺ (+ 6) and 3 CO₃^2- (- 6).
Formula: Al₂(CO₃)₃ (Aluminium Carbonate)

(iii) K⁺ and SO₄^2-
Potassium has a +1 charge.
Sulfate has a -2 charge.
To balance: 2 K⁺+ (+ 2) and 1 SO₄^2- (- 2).
Formula: K₂SO₄ (Potassium Sulfate)

(iv) NH₄⁺ and Cl^–
Ammonium has a + 1 charge.
Chloride has a – 1 charge.
Charges balance 1 : 1.
Formula: NH₄Cl (Ammonium Chloride)

Question 7.
Which of the following, in Fig. (below), correctly represents Cl^– ion (Atomic number of Chlorine = 17)?

Answer:
Chlorine (atomic number = 17) has the electronic configuration 2, 8, 7. It has 7 electrons in its valence shell (third shell).
When chlorine gains one electron to form the chloride ion (Cl^–), it has 18 electrons total. The electronic configuration of Cl^– becomes 2, 8, 8.

So the correct diagram should show:
17 protons in the nucleus
18 electrons total

First shell: 2 electrons
Second shell: 8 electrons
Third shell: 8 electrons

The correct option is (ii) – which shows an atom with three shells having 2, 8, and 8 electrons (total 18 electrons), representing the Cl^– ion with a complete outermost shell.

Question 8.
Determine the formula unit mass of the following substances.
(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.
Answer:
(i) Ammonium nitrate (NH₄NO₃)
Atomic masses:
N = 14 u,
H = 1 u,
O = 16 u
Formula unit mass = (14) + (4 × 1) + (14) + (3 × 16)
= 14 + 4 + 14 + 48 = 80u

(ii) Phosphoric acid (H₃PO₄)
Atomic masses: H = 1 u, P = 31 u, O = 16 u
Formula unit mass = (3 × 1) + (1 × 3) + (4 × 16)
= 3 + 31 +64 = 98u

(iii) Sodium hydrogencarbonate (NaHCO₃)
Atomic masses:
Na = 23 u, H = 1 u, C = 12 u, O = 16 u
Formula unit mass = 23 + 1 + 12 + (3 × 16)
= 23 + 1 + 12 + 48 = 84 u

Question 9.
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
Solution:
(i) Magnesium and nitrogen:
Mg forms Mg²⁺ (valency 2);
N forms N^3- (valency 3).
Formula: Mg₃N₂ (magnesium nitride)

(ii) Lithium and nitrogen:
Li forms Li⁺ (valency 1);
N forms N^3- (valency 3).
Formula: Li₃N (eithium nitride)

(iii) Sodium and sulfur:
Na forms Na⁺ (valency 1);
S forms S^2- (valency 2).
Formula: Na₂S (sodium sulfide)

(iv) Aluminium and oxygen:
Al forms Al³⁺ (valency 3);
O forms O^2- (valency 2).
Formula: Al₂O₃, (aluminium oxide)

Question 10.
Complete the Table (given below) by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiN03 is given as an example.

Solution:
Here,
NH₄^– (valency 1) with NO₃^– (valency 1): NH₄NO₃
NH₄⁺ (valency 1) with SO₄^2- (valency 2): (NH₄)₂SO₄
NH₄⁺ (valency 1) with PO₄^3-(valency 3): (NH₄)₃PO₄
Li⁺ (valency 1) with NO₃^– (valency 1): LiNO₃ (given)

Li⁺ (valency 1) with SO₄^2- (valency 2): Li₂SO₄
Li⁺ (valency 1) with PO₄^3- (valency 3): Li₃PO₄
Al³⁺ (valency 3) with NO₃^– (valency 1): Al(NO₃)₃
Al³⁺ (valency 3) with SO₄^2- (valency 2): Al₂(SO₄)₃
Al³⁺ (valency 3) with PO₄^3- (valency 3): AlPO₄
(both valency 3, subscripts both 3, divide by 3 = 1 each, so AlPO₄)
Cu²⁺ (valency 2) with NO₃^– (valency 1): Cu(NO₃)₂
Cu²⁺ (valency 2) with SO₄^2- (valency 2): CuSO₄ (both valency 2, simplifies to 1 : 1)
Cu²⁺ (valency 2) with PO₄^3- (valency 3): Cu₃(PO₄)₂

Cation/anion	NO3–	SO4–	PO4–
NH4+	NH4NO3	(NH4)2SO4	(NH4)3PO4
Li+	LiNO3	Li2SO4	Li3PO4
Al3+	Al(NO3)3	Al2(SO4)3	AlPO4
Cu2+	Cu(NO3)2	CuSO4	Cu3(PO4)2

Question 11.
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:
Given
Reactants:
Sodium carbonate = 5.3 g
Acetic acid = 6.0 g
Total mass of reactants = 5.3 + 6.0 = 11.3 g

Products:
Carbon dioxide = 2.2 g
Water = 0.9 g
Sodium acetate = 8.2 g
Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g
Since, total mass of reactants = total mass of products (11.3 g), the law of conservation of mass is valid.

Question 12.
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.
Answer:
Given, species with 11 protons, 12 neutrons, and 10 electrons
(i) Here,
Atomic number = Number of protons = 11
Mass number = Number of protons + Number of neutrons
= 11 + 12 = 23

(ii) It is a cation. A neutral atom has equal numbers of protons and electrons. This species has 11 protons but only 10 electrons. Since the number of protons (positive charges) is greater than the number of electrons (negative charges), the species carries a net positive charge of +1. A positively charged species is called a cation. It is represented as Na⁺.

(iii) For neutral sodium (11 e^–): 2, 8, 1
For Na+ (10 e^–): 2, 8
Electronic configuration = 2, 8

(iv) Atomic number 11 corresponds to sodium.
With +1 charge → Sodium ion (Na⁺).

Question 13.
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.
Solution:
Let’s break this down clearly:

Element A: 2, 8, 5Total electrons = 15 → this is phosphorus (P).
Valence electrons = 5 → needs 3 more to complete octet.

Element B: 2, 8, 7
Total electrons = 17 → this is chlorine (Cl).
Valence electrons = 7 → needs 1 more to complete octet.

(i) Element B (Cl) is more reactive. Chlorine needs only 1 electron to complete its octet, whereas phosphorus needs 3 electrons. Elements that need fewer electrons to complete their octet tend to be more reactive non-metals (they have a higher tendency to gain electrons). Therefore, B (chlorine) is more reactive.

(ii) Both A (P) and B (Cl) are non-metals. A has 5 valence electrons and needs 3 more. B has 7valence electrons and needs 1 more.
When non-metals combine with each other, they generally form covalent bonds by sharing electrons. A (P) shares one electron each with three B (Cl) atoms. Each Cl atom completes its octet by sharing one electron from P, and P completes its octet by sharing one electron from each of the three Cl atoms. This is electron sharing, which forms covalent bonds.

(iii) Phosphorus (A) has a valency of 3 (needs 3 electrons). Chlorine (B) has a valency of 1 (needs 1 electron). Formula: PC13 (Phosphorus trichloride)

Question 14.
Assertion (A): Copper sulfate conducts electricity in the molten state hut not in the solid state.
Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
Assertion (A): Copper sulfate (an ionic compound) conducts electricity in the molten state but not in the solid state.
Reason (R): The reason stated in R is the opposite of what actually happens. In the solid state, copper and sulfate ions are fixed in the lattice (they cannot move freely), which is why the solid does not conduct electricity. In the molten state, the ions are free to move, which is why the molten state conducts electricity. The reason has incorrectly swapped the conditions for the solid and molten states.

Question 15.
The species ²⁷Al, ⁸⁰Br^– and 201 Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Answer:
We are given:
²⁷Al → Aluminium atom
⁸⁰Br^– → Bromine ion
²⁰¹Hg²⁺ → Mercury ion
(i) ²⁷Al
No. of protons = 13 (Given)
No. of electrons = No. of protons = 13 (neutral atom)
No. of neutrons = Mass number – No. of protons
= 27 – 13 = 14

(ii) ⁸⁰Br^–
No. of protons = 35 (Given)
No. of electrons = 35 + 1 = 36 (because of – 1 charge)
No. of neutrons = Mass number – No. of protons
= 80 – 35 = 45

(iii) ²⁰¹Hg²⁺
No. of. protons = 80 (Given)
No. of electrons = No. of protons
= 80 – 2 = 78 (because of + 2 charge)
No. of neutrons = Mass number – No. of protons
= 201 – 80 = 121

Class 9 Science Chapter 9 Atomic Foundations of Matter Question Answer (InText)

Think It Over (NCERT Textbook Page No. 162)

Question 1.
Water can be obtained from various sources. Are all these samples of water chemically identical?
Answer:
Yes. All pure water samples are chemically identical because they contain hydrogen and oxygen in a mass ratio of 1 : 8. However, natural water from rivers, lakes, wells, or taps often contains dissolved salts, minerals, gases, or impurities, which make their composition different even though the basic chemical identity remains the same.

Question 2.
Oxygen is sometimes represented as O and sometimes as O₂. What is the difference between these symbols?
Answer:
Difference between O and O₂

O (oxygen atom):

• It represents a single atom of oxygen.
• Atomic number = 8, meaning it has 8 protons.
• It is highly reactive and does not exist freely in nature for long, because lone oxygen atoms quickly combine with others.

O₂ (O₂ molecule):

• It represents a molecule made of two oxygen atoms bonded together.
• This is the stable form of oxygen found in the atmosphere.
• About 21 % of the air we breathe is O₂, and it is essential for respiration and combustion.

Question 3.
Why does dissolved salt in water conduct electricity, but sugar does not?
Answer:
Salt (NaCl) is an ionic compound.

• When dissolved in water, it dissociates into ions:
• NaCl → Na⁺ + Cl^–
• These free-moving ions act as charge carriers, allowing electric current to pass through the solution. Hence, salt water is a good conductor of electricity.

Sugar (C₁₂H₂₂O₁₁) is a covalent compound.

• When dissolved in water, sugar molecules simply disperse but do not break into ions.
• Since there are no free ions, the solution cannot conduct electricity.
• Sugar water is therefore a poor conductor (essentially an insulator).

Pause and Ponder (NCERT Textbook Page No. 166)

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 of Conservation of Mass is NOT violated. When ethanol burns, it reacts with oxygen from the air and produces carbon dioxide and water vapour, both of which escape into the atmosphere as gases. If we account for the mass of oxygen consumed from the air and the mass of carbon dioxide and water vapour released, the total mass of the products will be equal to the total mass of the reactants (ethanol + oxygen). The fact that no residue is left in the beaker does not mean mass is lost-the products have escaped as gases. The Law of Conservation of Mass is obeyed.

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:
According to the Law of Conservation of Mass, the total mass of products = the total mass of reactants.
∴ Total mass of reactants = 20 g (hydrogen) + 160 g (oxygen) = 180 g.
Therefore, mass of water formed = 180 g.

Pause and Ponder (NCERT Textbook Page No. 167)

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:
Given:
Compound composition: 40 % sulfur, 60 % oxygen by mass
The compound has sulfur and oxygen in the mass ratio of 40 : 60 = 2 : 3.
If 20 g of sulfur is present then 60 mass of O₂ = \(\frac {60}{40}\) × 20 = 30 g
In the sample containing 20 g of sulfur, 30 g of oxygen must be present. This shows the compound always has sulfur and oxygen in the fixed ratio of 40 : 60 by mass, satisfying the Law of Constant Proportions.

Question 4
Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3 : 4. How much oxygen wilt combine with 9 g of carbon to form carbon monoxide?
Solution:
Ratio of C : O = 3 : 4
If 3 g of carbon combines with 4 g of oxygen, then:
For 9 g of carbon, oxygen = \(\frac {4}{3}\) × 9 = 12 g
Hence, 12 g of oxygen will combine with 9 g of carbon to form carbon monoxide.

Question 5.
The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.
Solution:
In a compound, elements are chemically combined in a fixed ratio by mass, irrespective of the source or method of preparation. This fixed ratio is a characteristic property of the compound. For example, water always contains hydrogen and oxygen in the ratio 1 : 8 by mass.

In a mixture, however, the components are not chemically combined. They can be present in any ratio and the composition of a mixture can vary. For example, a mixture of sugar and water can be prepared in any proportion. Hence, the Law of Definite Proportions does not apply to mixtures.

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:
Here,
Student X: Ratio of Copper and Oxygen = 4 : 1
Student Y: Ratio of Copper and Oxygen = 8 : 2
Clearly, both ratios are the same.
Yes, their results justify the Law of Constant Proportions because both students obtained copper oxide with the same fixed ratio of copper to oxygen (4 : 1), showing that a compound always has elements in a definite proportion by mass.

Pause and Ponder (NCERT Textbook Page No. 168)

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.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
(ii) Assertion (A): This is true. The balanced chemical equation is:
2H₂ + O₂ → 2H₂O
Here, 2 g of hydrogen (1 mole of H₂) reacts with 16 g of oxygen (½ mole of O₂) to form 18 g of water.
This agrees with the Law of Conservation of Mass.

Reason (R):

According to Dalton’s Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds. This statement is also true. Dalton’s theory says atoms combine in fixed, simple ratios to form compounds.

Pause and Ponder (NCERT Textbook Page No. 170)

Question 8.
Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).
Solution:
Structure of the Nitrogen Molecule (N₂):

Valence Electrons:
Nitrogen (atomic number 7) has an electronic configuration of 2, 5, meaning each nitrogen atom has 5 valence electrons.

Bond Formation:
To achieve a stable octet (8 Valence electrons), each nitrogen atom needs 3 more electrons. Therefore, two nitrogen atoms share three pairs of electrons (6 electrons in total) between them.

Question 9.
The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).
Answer:
Formation of the Fluorine Molecule (F₂)
Atomic Structure:
Fluorine has an atomic number of 9. Meaning its electronic configuration is 2, 7.

Need for Bonding:
Fluorine has 7 valence electrons and needs 1 more electron to complete its octet (stable configuration of 8 electrons).

Formation Mechanism:

• Two fluorine atoms approach each other.
• Each atom has one unpaired electron in its outer shell.
• The two atoms share their unpaired electrons, resulting in a single shared pair of electrons between them.

Pause and Ponder (NCERT Textbook Page No. 171)

Question 10.
Show the formation of the following molecules:
(i) Carbon dioxide (CO₂)
(ii) Hydrogen sulfide (H₂S)
(iii) Ammonia (NH₃)
Solution:
(i) Carbon dioxide (CO₂): Carbon (atomic number 6) has 4 valence electrons and needs 4 more to complete its octet. Oxygen (atomic number 8) has 6 valence electrons and needs 2 more. Each oxygen atom shares 2 electrons with the carbon atom. Carbon shares 2 electrons with each of the two oxygen atoms. Thus, two double bonds are formed: O = C = O. Carbon dioxide has two double covalent bonds.

(ii) Hydrogen sulfide (H₂S): Sulfur (atomic number 16) has 6 valence electrons and needs
2 more electrons to complete its octet. Hydrogen needs 1 electron to complete its duplet. Sulfur shares one electron each with two hydrogen atoms. Each hydrogen atom shares one electron with sulfur atom.
Two single bonds are formed: H—S—H (each bond is a single covalent bond).

(iii) Ammonia (NH₃):
Nitrogen (atomic number 7) has 5 valence electrons and needs 3 more electrons to complete its octet. Hydrogen needs 1 electron to complete its duplet. Nitrogen shares one electron each with three hydrogen atoms. Each hydrogen atom shares one electron with nitrogen. Three single bonds are formed: H—N—H with one H below (or depicted as N at centre bonded to three H atoms). Ammonia has three single covalent bonds and one lone pair on nitrogen.

Question 11.
Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Solution:

• Neon has atomic number 10 → electronic configuration: 2, 8.
• This means its outer shell (valence shell) is completely filled with 8 electrons.
• Since neon already has a stable octet, it does not need to gain, lose, or share electrons.
• Therefore, neon is chemically inert and exists as single atoms, not molecules.
• This is why neon belongs to the noble gases group.

Pause and Ponder (NCERT Textbook Page No. 174)

Question 12.
What kind of ion will oxygen (O) form?
Solution:
Oxygen forms an anion known as an oxide ion (O₂).
Reasoning:
Oxygen has 6 valence electrons and needs to gain 2 more electrons to complete its valence shell (achieve a stable octet).

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 __________ and __________. Now, __________ ion of magnesium and __________ ions of chlorine combine to give magnesium chloride.
Solution:
Chloride ion (Cl^– ): 1 (one) and 2(two).

Question 14.
Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.
Solution:
Formation of Potassium cation (K⁺ ):
Potassium (atomic number 19) has electronic configuration 2, 8, 8, 1.
It has 1 electron in its valence shell.
To attain a stable configuration, it loses this 1 electron:
K → K⁺ + e^–.
Potassium becomes K⁺ (K⁺ has 19 protons and 18 electrons).

Formation of Potassium chloride (KCl):
K⁺ (valency 1) + Cl^– (valency 1) → KCl
One potassium ion combines with one chloride ion to form potassium chloride.

Formation of Calcium cation (Ca²⁺):

Calcium (atomic number 20) has electronic configuration 2, 8, 8, 2. It has 2 electrons in its valence shell.
To attain a stable configuration, it loses both electrons:
Ca → Ca²⁺ + 2e^– Calcium becomes Ca²⁺ (20 protons and 18 electrons).

Formation of Calcium chloride (CaCl₂):
Ca²⁺ (valency 2) + 2 Cl^– (valency 1 each) → CaCl₂
One calcium ion combines with two chloride ions to form calcium chloride.

Question 15.
Illustrate how sodium sulfide (Na₂S) is formed.
Solution:
Sodium (atomic number 11) has electronic configuration 2, 8, 1.
It has 1 valence electron, which it loses to form Na⁺: Na → Na⁺ + e^–

Sulfur (atomic number 16) has electronic configuration 2, 8, 6. It has 6 valence electrons and needs 2 more electrons to complete its octet.
It gains 2 electrons to form S^2- : S + 2e^– → S^2-

Two sodium atoms each lose one electron (total 2 electrons) which are gained by one sulfur atom:
2Na → 2Na⁺ + 2e^– and S + 2e^– → S^2-

The two Na⁺ ions and one S^2- ion are held together by electrostatic force of attraction (ionic bond) to form sodium sulfide:
2Na⁺ + S^2- → Na₂S

Pause and Ponder (NCERT Textbook Page No. 177)

Question 16.
Name the following:
(i) CO₂ __________
(ii) NO₂ __________
(iii) SF₆ __________
(iv) PCl₃ __________
Solution:
(i) CO₂ → Carbon dioxide
It contains one carbon atom and two oxygen atoms.

(ii) NO₂ → Nitrogen dioxide
It contains one nitrogen atom and two oxygen atoms.

(iii) SF₆ → Sulfur hexafluoride
It contains one sulfur atom and six fluorine atoms.

(iv) PCl₃ → Phosphorus trichloride
It contains one phosphorus atom and three chlorine atoms.

Question 17.
Write the formula for the following:
(i) Sodium hydrogencarbonate __________
(ii) Sulfur dioxide __________
(iii) Ferric chloride __________
(iv) Cuprous oxide __________
Solution:
(i) Sodium hydrogencarbonate → NaHCO₃
(ii) Sulfur dioxide → SO₂
(iii) Ferric chloride (iron in + 3 oxidation state) → FeCl₃
(iv) Cuprous oxide (copper in + 1 oxidation state) → Cu₂O

Question 18.
Write the formulae for the compounds formed from the following pairs of ions:
(i) Fe³⁺ and OH^–
(ii) K⁺ and CO₃^2-
Solution:
(i) Iron (Fe³⁺) has a charge of + 3.
Hydroxide (OH^–) has a charge of – 1.
To balance charges, 3 OH^– ions are needed for each Fe³⁺ ion.
Formula = Fe(OH)₃

(ii) K⁺ (charge of + 1) and CO₃^2- (Charge of – 2)
Two K⁺ ions are needed to balance the – 2 charge of a single carbonate ion.
2(+ 1) + (- 2) = 0
Formula = K₂CO₃

Pause and Ponder (NCERT Textbook Page No. 179)

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? (Page 179)
Solution:
The compound has an ionic bond. Ionic compounds do not conduct electricity in the solid state because their ions are held in fixed positions by strong electrostatic forces. However, when dissolved in water, the ions become free to move, which allows the compound to conduct electricity. This behaviour is characteristic of ionic compounds such as sodium chloride and copper sulfate.

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:
The metal M has two electrons in its valence shell (M shell, which is the third shell). This means M has electronic configuration 2, 8, 2. This is magnesium (Mg), which forms Mg²⁺ by losing 2 electrons.
(i) Formula: Mg²⁺ combines with O^2- (oxide ion, valency 2).
Formula: MgO (magnesium oxide)

(ii) Type of bond:
Metal loses electrons → forms M²⁺
Oxygen gains electrons → forms O^2-
Bond is ionic.

(iii) Electrical conductivity of aqueous solution:
MgO is slightly soluble in water and forms an aqueous solution. In this solution, Mg²⁺ and O^2- (or OH^–) ions are present and are free to move. Therefore, its aqueous solution will conduct electricity.

Question 21.
Find the molecular mass of nitric acid (HNO₃).
Atomic mass – H = 1u; N = 14 u; O = 16 u
Solution:
Given, atomic masses:
H = 1 u
N = 14 u
O = 16u
Molecular mass of HNO₃ = (1 × 1) + (1 × 14) + (3 × 16)
= 1 + 14 + 48 = 63u
∴ Molecular mass of nitric acid = 63 u

Question 22.
Find the molecular mass of methane (CH₄).
Atomic mass – C = 12 u; H = 1 u.
Solution:
Given, atomic masses:
C = 12 u
H = 1 u
Molecular mass of CH₄ = (1 × 12) + (4 × 1)
= 12 + 4= 16u
∴ Molecular mass of methane (CH₄) = 16 u

Pause and Ponder (NCERT Textbook Page No. 180)

Question 23.
Find the formula unit mass of potassium chloride (KCl).
Atomic mass – K = 39 u ; Cl = 35.5 u.
Solution:
Here,
Atomic mass of K = 39 u
Atomic mass of Cl = 35.5 u
Formula unit mass of KCl = 39 + 35.5 = 74.5 u
∴ Formula unit mass of potassium chloride = 74.5 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:
Atomic mass of Mg = 24 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Formula unit mass of Mg(OH)₂ = 24 + 2 × (16 + 1)
= 24 + 2 × 17
= 24 + 34 = 58u
Hence, formula unit mass of magnesium hydroxide is 58 u.