Question & Answer Class 9 Science Exploration Chapter 9 Atomic Foundation of Matter

Revise, Reflect, Refine

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?

Answer:

• Element A: Has 1 electron in its third shell → Electronic configuration: 2, 8, 1 (like Sodium/Na)
• Element B: Has 6 electrons in its second shell → Electronic configuration: 2, 6 (like Oxygen/O)

(i) Element A tends to give away 1 electron from its third shell to achieve a stable octet (like the noble gas Neon with configuration 2, 8).

(ii) By losing 1 electron, A becomes positively charged. It forms a cation with a charge of +1, represented as A⁺.

(iii) Element B has 6 electrons in its outermost shell and needs 2 more electrons to complete its octet (to reach 8 electrons). So it tends to gain 2 electrons.

(iv) By gaining 2 electrons, B becomes negatively charged. It forms an anion with a charge of −2, represented as B²⁻.

(v) Since A loses electrons and B gains electrons, the bond formed is an Ionic Bond (formed by transfer of electrons).

(vi) Using the criss-cross method:

Criss-crossing the charges → A₂B

The formula of the compound is A₂B.

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.

Answer:

• Element X → 6 valence electrons (like Oxygen, configuration: 2, 6)
• Element Y → 2 electrons in second shell (configuration: 2, 2)

(i) Element X has 6 electrons in its outermost shell and needs 2 more electrons to complete its octet. To achieve stability, two atoms of X share 2 electrons each with one another. This sharing completes the octet of both atoms, making the molecule stable. Hence, X forms a diatomic molecule (X₂).

(ii) Since both atoms of X share electrons with each other, they form a Covalent Bond. Since each atom shares 2 electrons, a total of 2 pairs are shared, forming a Double Covalent Bond, represented as X=X.

(iii)

(iv) Element Y has 2 valence electrons and needs to share both to become stable. Since each X atom needs 2 electrons to complete its octet, and Y can only give 1 electron to each, two X atoms each form a single bond with one Y atom. This results in the molecule X₂Y — similar to water (H₂O) — with two single covalent bonds arranged as X–Y–X.

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 PO₄³⁻
(iii) 2 Fe³⁺ and 3 O²⁻
(iv) 3 Ca²⁺ and 2 SO₄²⁻

Answer: Rule: In a valid ionic compound, total positive charge must equal total negative charge, and both must equal 6+ and 6− as given.

Let us check each option:

(i) 2 Al³⁺ and 3 Cl⁻

• Total positive charge = 2 × 3 = 6+ ✓
• Total negative charge = 3 × 1 = 3− ✗
• 6+ ≠ 3− → Not correct

(ii) 3 Mg²⁺ and 1 PO₄³⁻

• Total positive charge = 3 × 2 = 6+ ✓
• Total negative charge = 1 × 3 = 3− ✗
• 6+ ≠ 3− → Not correct

(iii) 2 Fe³⁺ and 3 O²⁻

• Total positive charge = 2 × 3 = 6+ ✓
• Total negative charge = 3 × 2 = 6− ✓
• 6+ = 6− → Correct ✓

(iv) 3 Ca²⁺ and 2 SO₄²⁻

• Total positive charge = 3 × 2 = 6+ ✓
• Total negative charge = 2 × 2 = 4− ✗
• 6+ ≠ 4− → Not correct

Correct Option (iii) — 2 Fe³⁺ and 3 O²⁻

This is the only combination where both total positive charge (6+) and total negative charge (6−) are equal, satisfying the condition for a valid ionic compound. The compound formed is Fe₂O₃ (Ferric oxide).

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.

Answer: False
Correct statement:
Elements are made up of atoms or molecules, and compounds are made up of molecules formed by atoms of different elements.

(ii) The molecule of a compound is always made up of two or more atoms of the same kind.

Answer: False
Correct statement:
The molecule of a compound is made up of two or more atoms of different kinds.

(iii) One molecule of nitrogen gas contains three nitrogen atoms.

Answer: False
Correct statement:
One molecule of nitrogen gas contains two nitrogen atoms (N₂)

(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.

Answer: (iv) True
Water H₂O is made of two hydrogen atoms covalently bonded with one oxygen atom.

5. Write the chemical formulae for the following compounds.

(i) Aluminium nitrate

• Cation: Aluminium → Al³⁺ (valency = 3)
• Anion: Nitrate → NO₃⁻ (valency = 1)

Criss-cross the charges:

Al³⁺ and NO₃⁻ → subscript of Al = 1, subscript of NO₃ = 3
Since there are multiple nitrate ions, use brackets:
Formula = Al(NO₃)₃

(ii) Calcium oxide

• Cation: Calcium → Ca²⁺ (valency = 2)
• Anion: Oxide → O²⁻ (valency = 2)

Criss-cross the charges:

Ca²⁺ and O²⁻ → subscript of Ca = 2, subscript of O = 2
Since both valencies are equal, divide by common factor 2:
2 and 2 → becomes 1 and 1
Formula = CaO

(iii) Ferric oxide

• Cation: Ferric (Iron) → Fe³⁺ (valency = 3)
• Anion: Oxide → O²⁻ (valency = 2)

Criss-cross the charges:

Fe³⁺ and O²⁻ → subscript of Fe = 2, subscript of O = 3
Formula = Fe₂O₃

6. Write the formulae of the compounds formed from the following pairs of ions.

(i) Ca²⁺ and Br⁻
(ii) Al³⁺ and CO₃²⁻
(iii) K⁺ and SO₄²⁻
(iv) NH₄⁺ and Cl⁻

(i) Ca²⁺ and Br⁻

• Cation: Ca²⁺ (valency = 2)
• Anion: Br⁻ (valency = 1)

Criss-cross:

Ca²⁺ and Br⁻ → subscript of Ca = 1, subscript of Br = 2
Formula = CaBr₂
(Calcium bromide)

(ii) Al³⁺ and CO₃²⁻

• Cation: Al³⁺ (valency = 3)
• Anion: CO₃²⁻ (valency = 2)

Criss-cross:

Al³⁺ and CO₃²⁻ → subscript of Al = 2, subscript of CO₃ = 3
Since CO₃ is a polyatomic ion and appears more than once, use brackets:
Formula = Al₂(CO₃)₃
(Aluminium carbonate)

(iii) K⁺ and SO₄²⁻

• Cation: K⁺ (valency = 1)
• Anion: SO₄²⁻ (valency = 2)

Criss-cross:

K⁺ and SO₄²⁻ → subscript of K = 2, subscript of SO₄ = 1
Since SO₄ appears only once, no brackets needed:
Formula = K₂SO₄
(Potassium sulfate)

(iv) NH₄⁺ and Cl⁻

Cation: NH₄⁺ (valency = 1)
Anion: Cl⁻ (valency = 1)

Criss-cross:

NH₄⁺ and Cl⁻ → subscript of NH₄ = 1, subscript of Cl = 1
Since both valencies are equal and = 1, no subscripts are written:
Formula = NH₄Cl
(Ammonium chloride)

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

Answer:

Option (i)
Option (i) correctly represents the Cl⁻ ion because it shows:

• 3 electron shells
• 2 electrons in K shell
• 8 electrons in L shell
• 8 electrons in M shell (outermost)
• Total = 18 electrons

This matches the electronic configuration 2, 8, 8 of the chloride ion (Cl⁻), which is formed when a chlorine atom gains one electron to complete its octet.

8. Determine the formula unit mass of the following substances.

(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, whichis essential for plant growth.
(ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser anddetergents.
(iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.

Atomic masses used:

• H = 1 u
• N = 14 u
• O = 16 u
• P = 31 u
• Na = 23 u
• C = 12 u

(i) Ammonium nitrate — NH₄NO₃

Breaking the formula:

• N = 2 atoms
• H = 4 atoms
• O = 3 atoms

Formula unit mass = (14 × 2) + (1 × 4) + (16 × 3)
= 28 + 4 + 48
= 80 u

(ii) Phosphoric acid — H₃PO₄

Breaking the formula:

• H = 3 atoms
• P = 1 atom
• O = 4 atoms

Formula unit mass = (1 × 3) + (31 × 1) + (16 × 4)
= 3 + 31 + 64
= 98 u

(iii) Sodium hydrogencarbonate — NaHCO₃

Breaking the formula:

• Na = 1 atom
• H = 1 atom
• C = 1 atom
• O = 3 atoms

Formula unit mass = (23 × 1) + (1 × 1) + (12 × 1) + (16 × 3)
= 23 + 1 + 12 + 48
= 84 u

9. Write the formulae for the compounds formed by the reaction of:

(i) Magnesium and nitrogen

• Magnesium forms: Mg²⁺ (valency = 2)
• Nitrogen forms: N³⁻ (valency = 3)

Criss-cross:

Mg²⁺ and N³⁻ → subscript of Mg = 3, subscript of N = 2

Formula = Mg₃N₂

(Magnesium nitride)

(ii) Lithium and nitrogen

• Lithium forms: Li⁺ (valency = 1)
• Nitrogen forms: N³⁻ (valency = 3)

Criss-cross:

Li⁺ and N³⁻ → subscript of Li = 3, subscript of N = 1

Formula = Li₃N

(Lithium nitride)

(iii) Sodium and sulfur

• Sodium forms: Na⁺ (valency = 1)
• Sulfur forms: S²⁻ (valency = 2)

Criss-cross:

Na⁺ and S²⁻ → subscript of Na = 2, subscript of S = 1

Formula = Na₂S

(Sodium sulfide)

(iv) Aluminium and oxygen

• Aluminium forms: Al³⁺ (valency = 3)
• Oxygen forms: O²⁻ (valency = 2)

Criss-cross:

Al³⁺ and O²⁻ → subscript of Al = 2, subscript of O = 3

Formula = Al₂O₃

(Aluminium oxide)

10. 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. LiNO3 is given as an example.

Answer:

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.

Answer:

Given:

Reactants:

• Sodium carbonate = 5.3 g
• Acetic acid = 6.0 g

Products:

• Carbon dioxide = 2.2 g
• Water = 0.9 g
• Sodium acetate = 8.2 g

Step 1 — Total mass of reactants:

= 5.3 + 6.0

= 11.3 g

Step 2 — Total mass of products:

= 2.2 + 0.9 + 8.2

= 11.3 g

Step 3 — Comparison:

Total mass of reactants = Total mass of products

11.3 g = 11.3 g

Conclusion:

Since mass of reactants equals mass of products, the Law of Conservation of Mass is valid for this reaction. Matter is neither created nor destroyed during the chemical reaction.

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:

• Protons = 11
• Neutrons = 12
• Electrons = 10

(i) Atomic number and Mass number:

Atomic number = Number of protons = 11

Mass number = Number of protons + Number of neutrons

= 11 + 12

= 23

(ii) Is it neutral, a cation or an anion?

It is a CATION.

Reason: In a neutral atom, number of protons = number of electrons. But here:

• Protons = 11
• Electrons = 10

Since protons > electrons, the species has lost 1 electron and carries a positive charge of +1.

Therefore it is a cation, represented as Na⁺

(iii) Electronic configuration:

Total electrons = 10

Distribution:

• K shell = 2
• L shell = 8

Electronic configuration = 2, 8

(Note: The outermost shell has 8 electrons, which is a stable octet)

(iv) Name of the species:

• Atomic number = 11
• Element with atomic number 11 = Sodium (Na)
• Since it has lost 1 electron and carries +1 charge

The species is Sodium ion — Na⁺

(Also called Sodium cation)

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.

Answer:

Given:

• Element A: 2, 8, 5 → valence electrons = 5
• Element B: 2, 8, 7 → valence electrons = 7

(i) Which element is more reactive?

Element B is more reactive.

Reason:

• Element A has 5 valence electrons and needs 3 more to complete octet
• Element B has 7 valence electrons and needs only 1 more to complete octet
• The fewer electrons needed to complete the octet, the more reactive the non-metal
• Since B needs only 1 electron, it achieves stability much more easily

Therefore B is more reactive than A.

(ii) Ionic or Covalent bond?

A and B will form a Covalent Bond.

Reason:

• A has 5 valence electrons → needs 3 more → will share electrons
• B has 7 valence electrons → needs 1 more → will share electrons
• Both A and B are non-metals (having more than 4 valence electrons or close to 8)
• When two non-metals combine, they share electrons rather than transferring them
• Hence a covalent bond is formed between A and B

(iii) Formula of the compound:

• A needs 3 electrons → valency = 3
• B needs 1 electron → valency = 1

Using criss-cross method:

A (valency 3) and B (valency 1) → subscript of A = 1, subscript of B = 3

Formula = AB₃

(Similar to NCl₃ — Nitrogen trichloride, where N has config 2,5 and Cl has config 2,8,7)

14. Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.

Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.

Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Answer:

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.

Analysis:

Assertion (A) — TRUE ✓

Copper sulfate is an ionic compound. In solid state, ions are held tightly in fixed positions in the crystal lattice and cannot move, so it does NOT conduct electricity. In molten state, ions become free to move and hence it DOES conduct electricity. So assertion is correct.

Reason (R) — FALSE ✗

The reason is completely reversed. The correct explanation is:

• In solid state → ions are fixed in lattice → cannot conduct
• In molten state → ions are free to move → can conduct

The reason states the opposite, so it is incorrect.

Option (iii) — A is true, but R is false.

15. The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively.

How many electrons and neutrons do they have?

Answer:

Given species: ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺

Formula used:

• Electrons in neutral atom = Protons
• For cation: Electrons = Protons − charge
• For anion: Electrons = Protons + charge
• Neutrons = Mass number − Protons

(i) ²⁷Al

• Protons = 13
• Mass number = 27
• Charge = 0 (neutral atom)

Electrons = Protons = 13

Neutrons = Mass number − Protons = 27 − 13 = 14

(ii) ⁸⁰Br⁻

• Protons = 35
• Mass number = 80
• Charge = −1 (anion, gained 1 electron)

Electrons = Protons + 1 = 35 + 1 = 36

Neutrons = Mass number − Protons = 80 − 35 = 45

(iii) ²⁰¹Hg²⁺

• Protons = 80
• Mass number = 201
• Charge = +2 (cation, lost 2 electrons)

Electrons = Protons − 2 = 80 − 2 = 78

Neutrons = Mass number − Protons = 201 − 80 = 121