Class 9 Science NCERT Solution chapter 9 Atomic Foundations of Matter

Revise, Reflect, Refine — Complete Solutions

🟦 QUESTION 1

A particular element A has one electron in its third shell. Another element B has 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 combine, what kind of bond would be formed?
(vi) What would be the formula of the compound formed?

🟩 ANSWER

Element A has the electronic configuration: $2,8,1$ 2,8,1

Therefore, element A is sodium (Na).

Element B has the electronic configuration: $2,6$ 2,6

Therefore, element B is oxygen (O).

(i) A tends to lose one electron to obtain a stable octet. $\text{Na}\rightarrow \text{Na}^{+}+e^{-}$ Na→Na++e−

(ii) It forms a positively charged ion called a cation, Na⁺.

(iii) B tends to gain two electrons to complete its octet. $\text{O}+2e^{-}\rightarrow \text{O}^{2-}$ O+2e−→O2−

(iv) It forms a negatively charged ion called an anion, O²⁻.

(v) A transfers electrons to B. Therefore, an ionic bond is formed.

Two sodium atoms are required to provide the two electrons needed by one oxygen atom.

(vi) Formula of the compound: $\boxed{\text{Na}_2\text{O}}$ Na2O

The compound formed is 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) Another element Y has two electrons in its second shell. Draw the structure of the compound that X would form with Y.

🟩 ANSWER

Element X has six electrons in its valence shell. It requires two more electrons to complete its octet.

Therefore, X represents oxygen, whose electronic configuration is: $2,6$ 2,6

(i) Why does X form a diatomic molecule?

Each oxygen atom requires two electrons to complete its octet. Two oxygen atoms share two pairs of electrons with each other.

As a result, a stable oxygen molecule, O₂, is formed.

(ii) Type of bond

Two pairs of electrons are shared. Therefore, a double covalent bond is formed.

(iii) Structure of the molecule

$\boxed{\text{O}=\text{O}}$ O=O

Each oxygen atom has two lone pairs of electrons in addition to the two shared pairs.

(iv) Compound formed between X and Y

Element Y has the electronic configuration: $2,2$ 2,2

Therefore, Y is beryllium (Be).

Beryllium loses two electrons: $\text{Be}\rightarrow \text{Be}^{2+}+2e^{-}$ Be→Be2++2e−

Oxygen accepts two electrons: $\text{O}+2e^{-}\rightarrow \text{O}^{2-}$ O+2e−→O2−

The oppositely charged ions attract each other: $\text{Be}^{2+}+\text{O}^{2-}\rightarrow \text{BeO}$ Be2++O2−→BeO $\boxed{\text{Formula}=\text{BeO}}$ Formula=BeO

An ionic bond is formed between Be²⁺ and O²⁻.

🟦 QUESTION 3

You want to design a new ionic compound in which the total positive charge is 6+6+6+ and the total negative charge is 6−6-6−. 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

Correct option: (iii) 2 Fe³⁺ and 3 O²⁻

For 2 Fe³⁺ ions: $2\times(+3)=+6$ 2×(+3)=+6

For 3 O²⁻ ions: $3\times(-2)=-6$ 3×(−2)=−6

Therefore, the total positive and negative charges balance each other. $2\text{Fe}^{3+}+3\text{O}^{2-}\rightarrow \text{Fe}_2\text{O}_3$ 2Fe3++3O2−→Fe2O3

The compound formed is ferric oxide or iron(III) oxide.

Why the other options are incorrect

(i) $2\text{Al}^{3+}=+6$ 2Al3+=+6, but $3\text{Cl}^{-}=-3$ 3Cl−=−3
(ii) $3\text{Mg}^{2+}=+6$ 3Mg2+=+6, but $1\text{PO}_4^{3-}=-3$ 1PO43−=−3
(iv) $3\text{Ca}^{2+}=+6$ 3Ca2+=+6, but $2\text{SO}_4^{2-}=-4$ 2SO42−=−4

🟦 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.
(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.

🟩 ANSWER

(i) False

Correct statement: Elements are made up of atoms of the same kind. Compounds are made up of atoms of two or more different elements chemically combined in a fixed ratio.

Some elements, such as oxygen and nitrogen, may exist as molecules.

(ii) False

Correct statement: A molecule of a compound contains atoms of two or more different elements chemically bonded together.

For example, H₂O contains hydrogen and oxygen atoms.

(iii) False

One molecule of nitrogen gas contains two nitrogen atoms. $\boxed{\text{N}_2}$ N2

(iv) True

A water molecule contains two hydrogen atoms covalently bonded to one oxygen atom. $\boxed{\text{H}_2\text{O}}$ H2O

🟦 QUESTION 5

Write the chemical formulae for the following compounds:

(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide

🟩 ANSWER

(i) Aluminium nitrate

Ions: $\text{Al}^{3+},\qquad \text{NO}_3^{-}$ Al3+,NO3−

Criss-crossing the charges: $\boxed{\text{Al(NO}_3\text{)}_3}$ Al(NO3)3

(ii) Calcium oxide

Ions: $\text{Ca}^{2+},\qquad \text{O}^{2-}$ Ca2+,O2−

The charges reduce to the simplest ratio of $1:1$ 1:1. $\boxed{\text{CaO}}$ CaO

(iii) Ferric oxide

Ferric ion: $\text{Fe}^{3+}$ Fe3+

Oxide ion: $\text{O}^{2-}$ O2−

Criss-crossing the charges: $\boxed{\text{Fe}_2\text{O}_3}$ Fe2O3

🟦 QUESTION 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⁻

🟩 ANSWER

Ions	Balancing of charges	Formula
Ca²⁺ and Br⁻	One Ca²⁺ requires two Br⁻ ions	CaBr₂
Al³⁺ and CO₃²⁻	Two Al³⁺ require three CO₃²⁻ ions	Al₂(CO₃)₃
K⁺ and SO₄²⁻	Two K⁺ ions require one SO₄²⁻ ion	K₂SO₄
NH₄⁺ and Cl⁻	Charges combine in a $1:1$ 1:1 ratio	NH₄Cl

🟦 QUESTION 7

Which of the structures shown in Fig. 9.18 correctly represents the chloride ion, Cl⁻? The atomic number of chlorine is 17.

🟩 ANSWER

Correct option: (ii)

A neutral chlorine atom has 17 electrons with the electronic configuration: $2,8,7$ 2,8,7

To form a chloride ion, it gains one electron: $\text{Cl}+e^{-}\rightarrow \text{Cl}^{-}$ Cl+e−→Cl−

The chloride ion therefore has 18 electrons: $\boxed{2,8,8}$ 2,8,8

In Fig. 9.18 on textbook page 182, diagram (ii) shows:

2 electrons in the K shell
8 electrons in the L shell
8 electrons in the M shell

Therefore, diagram (ii) correctly represents Cl⁻.

🟦 QUESTION 8

Determine the formula unit mass of the following substances:

(i) Ammonium nitrate, NH₄NO₃
(ii) Phosphoric acid, H₃PO₄
(iii) Sodium hydrogencarbonate, NaHCO₃

Use: $\text{H}=1,\ \text{C}=12,\ \text{N}=14,\ \text{O}=16,\ \text{Na}=23,\ \text{P}=31$ H=1, C=12, N=14, O=16, Na=23, P=31

🟩 ANSWER

(i) Ammonium nitrate, NH₄NO₃

NH₄NO₃ contains:

2 nitrogen atoms
4 hydrogen atoms
3 oxygen atoms

$=(2\times14)+(4\times1)+(3\times16)$ =(2×14)+(4×1)+(3×16) $=28+4+48$ =28+4+48 $\boxed{80\text{ u}}$ 80 u

(ii) Phosphoric acid, H₃PO₄

$=(3\times1)+(1\times31)+(4\times16)$ =(3×1)+(1×31)+(4×16) $=3+31+64$ =3+31+64 $\boxed{98\text{ u}}$ 98 u

(iii) Sodium hydrogencarbonate, NaHCO₃

$=(1\times23)+(1\times1)+(1\times12)+(3\times16)$ =(1×23)+(1×1)+(1×12)+(3×16) $=23+1+12+48$ =23+1+12+48 $\boxed{84\text{ u}}$ 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

🟩 ANSWER

(i) Magnesium and nitrogen

$\text{Mg}^{2+},\qquad \text{N}^{3-}$ Mg2+,N3− $\boxed{\text{Mg}_3\text{N}_2}$ Mg3N2

The compound is magnesium nitride.

(ii) Lithium and nitrogen

$\text{Li}^{+},\qquad \text{N}^{3-}$ Li+,N3− $\boxed{\text{Li}_3\text{N}}$ Li3N

The compound is lithium nitride.

(iii) Sodium and sulfur

$\text{Na}^{+},\qquad \text{S}^{2-}$ Na+,S2− $\boxed{\text{Na}_2\text{S}}$ Na2S

The compound is sodium sulfide.

(iv) Aluminium and oxygen

$\text{Al}^{3+},\qquad \text{O}^{2-}$ Al3+,O2− $\boxed{\text{Al}_2\text{O}_3}$ Al2O3

The compound is aluminium oxide.

🟦 QUESTION 10

Complete Table 9.3 by writing the formulae of compounds formed by the cations on the left and the anions at the top. LiNO₃ is given as an example.

🟩 ANSWER

Cation / Anion	NO₃⁻	SO₄²⁻	PO₄³⁻
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

Brackets are used when more than one polyatomic ion is required in the formula.

🟦 QUESTION 11

A mass of 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

Total mass of reactants

$5.3+6.0=11.3\text{ g}$ 5.3+6.0=11.3 g

Total mass of products

$2.2+0.9+8.2=11.3\text{ g}$ 2.2+0.9+8.2=11.3 g

Therefore: $\text{Mass of reactants}=\text{Mass of products}$ Mass of reactants=Mass of products $11.3\text{ g}=11.3\text{ g}$ 11.3 g=11.3 g

Final Answer:

The total mass remains unchanged. Therefore, the Law of Conservation of Mass is valid.

🟦 QUESTION 12

If a species has 11 protons, 12 neutrons and 10 electrons, determine:

(i) Its atomic number and mass number
(ii) Whether it is neutral, a cation or an anion
(iii) Its electronic configuration
(iv) The name of the species

🟩 ANSWER

(i) Atomic number and mass number

Atomic number: $Z=\text{Number of protons}=11$ Z=Number of protons=11

Mass number: $A=\text{Protons}+\text{Neutrons}$ A=Protons+Neutrons $A=11+12=23$ A=11+12=23

(ii) Nature of the species

The species has:

11 positively charged protons
10 negatively charged electrons

It has one more proton than electrons. Therefore, its charge is $+1$ +1. $\boxed{\text{It is a cation.}}$ It is a cation.

(iii) Electronic configuration

There are 10 electrons: $\boxed{2,8}$ 2,8

(iv) Name of the species

Atomic number 11 belongs to sodium.

Since it has lost one electron, the species is a sodium ion: $\boxed{\text{Na}^{+}}$ Na+

🟦 QUESTION 13

Two elements A and B have the following electronic configurations: $A: 2,8,5$ A:2,8,5 $B: 2,8,7$ 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 sharing.
(iii) Predict the formula of the compound they would form.

🟩 ANSWER

Element A has five valence electrons, while element B has seven valence electrons.

(i) More reactive element

Element B requires only one electron to complete its octet, whereas A requires three electrons.

Therefore: $\boxed{\text{Element B is more reactive.}}$ Element B is more reactive.

A represents phosphorus and B represents chlorine.

(ii) Type of bond

Both A and B are non-metals. They do not readily transfer electrons completely.

They achieve stable electronic configurations by sharing electrons. Therefore, they form covalent bonds.

One phosphorus atom requires three electrons. It shares one electron each with three chlorine atoms.

(iii) Formula of the compound

Valency of phosphorus = 3
Valency of chlorine = 1 $\boxed{\text{PCl}_3}$ PCl3

The compound is phosphorus trichloride.

🟦 QUESTION 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 the molten state, while in the 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

Correct option: (iii) Assertion is true, but Reason is false.

The Assertion is correct because copper sulfate is an ionic compound.

In the solid state, its ions are held in fixed positions in the crystal lattice and cannot move freely.
In the molten state, the ions become free to move and carry electric charge.

The Reason is false because it reverses the actual conditions.

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

🟦 QUESTION 15

The species 27Al{}^{27}\text{Al}27Al, 80Br−{}^{80}\text{Br}^{-}80Br− and 201Hg2+{}^{201}\text{Hg}^{2+}201Hg2+ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?

🟩 ANSWER

The number of neutrons is: $n=A-Z$ n=A−Z

The number of electrons depends on the charge:

A negative ion has gained electrons.
A positive ion has lost electrons.

Species	Mass number	Protons	Electrons	Neutrons
${}^{27}\text{Al}$ 27Al	27	13	13	$27-13=14$ 27−13=14
${}^{80}\text{Br}^{-}$ 80Br−	80	35	$35+1=36$ 35+1=36	$80-35=45$ 80−35=45
${}^{201}\text{Hg}^{2+}$ 201Hg2+	201	80	$80-2=78$ 80−2=78	$201-80=121$ 201−80=121

Final Answer:

Aluminium atom: 13 electrons and 14 neutrons
Bromide ion: 36 electrons and 45 neutrons
Mercury ion: 78 electrons and 121 neutrons