Atomic Foundations of Matter - Chapter 9 Class 9 Exploration

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?
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A is 2, 8, 1 (like sodium): (i) gives
1
electron; (ii) forms a
cation A⁺
.
B is 2, 6 (like oxygen): (iii) takes
2
electrons; (iv) forms an
anion B²⁻
.
(v) A gives electrons and B takes them → an
ionic bond
.
(vi) Two A⁺ balance one B²⁻ → formula
A₂B
.
Ionic Compound from A
and B
A (2,8,1) gives 1 electron
B (2,6) takes 2 electrons
Formula A2B, an ionic bond
Back to: 9.4.2 Bonding by electron transfer — Ionic bond

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.
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(i) With 6 valence electrons X needs
2
more, so two X atoms share electrons.
(ii) It forms a
covalent (double) bond
.
(iii) X=X, sharing two pairs (like O=O).
(iv) Y has 2 electrons in the second shell (needs... actually Y is like beryllium/helium-type); X shares with Y to complete shells — draw X—Y sharing as appropriate.
Diatomic Molecule of X
X has 6 valence electrons
Needs 2 more, shares two
pairs
Forms a double covalent
bond
Back to: A. Molecules of elements

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 PO₄³⁻ (iii) 2 Fe³⁺ and 3 O²⁻ (iv) 3 Ca²⁺ and 2 SO₄²⁻
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(i) 2 Al³⁺ = 6+ but 3 Cl⁻ = 3− — not balanced.
(ii) 3 Mg²⁺ = 6+ and 1 PO₄³⁻ = 3− — not balanced.
(iii)
2 Fe³⁺ = 6+ and 3 O²⁻ = 6−
—
correct
(Fe₂O₃).
(iv) 3 Ca²⁺ = 6+ and 2 SO₄²⁻ = 4− — not balanced.
Answer: (iii)
Combination
Total charge → verdict
(i)
2 Al³⁺ and 3 Cl⁻
+6 and −3 —
Wrong
. Charges do not balance.
(ii)
3 Mg²⁺ and 1 PO₄³⁻
+6 and −3 —
Wrong
. Negative charge is only 3.
(iii)
2 Fe³⁺ and 3 O²⁻
+6 and −6 —
Correct
. This is the answer (Fe₂O₃).
(iv)
3 Ca²⁺ and 2 SO₄²⁻
+6 and −4 —
Wrong
. Charges do not balance.
Back to: 9.5.2 Writing chemical formulae of ionic compounds

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.
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(i)
False
— elements are made of atoms; compounds are made of atoms of different elements (often as molecules).
(ii)
False
— a molecule of a compound is made of atoms of
different
kinds.
(iii)
False
— a nitrogen molecule (N₂) contains
two
nitrogen atoms, not three.
(iv)
True
— water has two hydrogen atoms covalently bonded to one oxygen atom.
Statement
Verdict & corrected version
(i)
Elements are made up of molecules and compounds are made up of atoms.
False
— corrected: elements are made of
atoms
; compounds are made of
molecules
of two or more different elements.
(ii)
The molecule of a compound is always made up of two or more atoms of the same kind.
False
— corrected: a compound molecule has atoms of
different
kinds.
(iii)
One molecule of nitrogen gas contains three nitrogen atoms.
False
— corrected: nitrogen gas is N₂, so it contains
two
nitrogen atoms.
(iv)
Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Correct
— H₂O: two H atoms share electrons with one O atom.
Back to: 9.4 How Atoms Combine?

Question 5
Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide
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(i) Al³⁺ + NO₃⁻ →
Al(NO₃)₃
.
(ii) Ca²⁺ + O²⁻ →
CaO
.
(iii) Fe³⁺ + O²⁻ →
Fe₂O₃
.
Writing the Formulae
Aluminium nitrate is Al(NO3)3 Calcium oxide is CaO Ferric oxide is Fe2O3
Back to: 9.5.2 Writing chemical formulae of ionic compounds

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⁻
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(i) Ca²⁺ and Br⁻ →
CaBr₂
.
(ii) Al³⁺ and CO₃²⁻ →
Al₂(CO₃)₃
.
(iii) K⁺ and SO₄²⁻ →
K₂SO₄
.
(iv) NH₄⁺ and Cl⁻ →
NH₄Cl
.
Formulae from Ion Pairs
CaBr2 Al2(CO3)3 K2SO4 NH4Cl
Back to: 9.5.2 Writing chemical formulae of ionic compounds

Question 7
Which of the following, in Fig. 9.18, correctly represents Cl⁻ ion (Atomic number of chlorine = 17).
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Cl⁻ has gained one electron: 17 + 1 =
18 electrons
.
Configuration must be
2, 8, 8
(a complete octet).
The correct diagram is the one showing shells filled as 2, 8, 8.
Diagram (Fig. 9.18)
Electron count → verdict
(i)
Shows 17 electrons (2, 8, 7) —
Wrong
. That is a neutral Cl atom.
(ii)
Shows
18
electrons (2, 8, 8) —
Correct
. Cl⁻ gains one electron.
(iii)
Electron count does not give a complete octet —
Wrong
.
(iv)
Only 2, 8 shells shown —
Wrong
. Too few electrons for chlorine.
Back to: 9.4.2 Bonding by electron transfer — Ionic bond

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.
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(i) NH₄NO₃ = 14 + 4 + 14 + 48 =
80 u
.
(ii) H₃PO₄ = 3 + 31 + 64 =
98 u
.
(iii) NaHCO₃ = 23 + 1 + 12 + 48 =
84 u
.
Atomic masses: H=1, C=12, N=14, O=16, Na=23, P=31.
Formula Unit Masses
NH4NO3 equals 80 u H3PO4 equals 98 u NaHCO3 equals 84 u
Back to: 9.8 Formula Unit Mass of Ionic Compounds

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
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(i) Mg²⁺ and N³⁻ →
Mg₃N₂
.
(ii) Li⁺ and N³⁻ →
Li₃N
.
(iii) Na⁺ and S²⁻ →
Na₂S
.
(iv) Al³⁺ and O²⁻ →
Al₂O₃
.
Formulae from Reactions
Mg3N2 Li3N Na2S Al2O3
Back to: 9.5.2 Writing chemical formulae of ionic compounds

Question 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.
NO₃⁻
SO₄²⁻
PO₄³⁻
NH₄⁺
Li⁺
LiNO₃
Al³⁺
Cu²⁺
Table 9.3
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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
Completing Table 9.3
Cross the cation and anion charges Simplify to the smallest ratio Brackets for polyatomic ions
Back to: 9.5.2 Writing chemical formulae of ionic compounds

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.
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Reactants =
5.3 + 6.0 = 11.3
g
.
Products =
2.2 + 0.9 + 8.2 = 11.3
g
.
Reactants = products → the law is
valid
.
Verify Conservation of
Mass
Reactants equal 11.3 g
Products equal 11.3 g
So the law is valid
Back to: 9.1 Law of Conservation of Mass

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.
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(i) Atomic number = protons =
11
; mass number = 11 + 12 =
23
.
(ii) Protons (11) > electrons (10), so it is a
cation
(charge +1).
(iii) With 10 electrons:
2, 8
.
(iv) It is a
sodium ion, Na⁺
.
Identify the Species
Atomic number 11, mass
number 23
11 protons but 10 electrons
It is a sodium cation Na plus
Back to: 9.4.2 Bonding by electron transfer — Ionic bond

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.
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(i)
B
is more reactive — it needs only 1 electron to complete its octet.
(ii) A has 5 valence electrons and B has 7; both are non-metals, so they
share
electrons → a
covalent
bond.
(iii) A needs 3 electrons and B needs 1, so the formula is
AB₃
.
A (2,8,5) and B (2,8,7)
B is more reactive
Both non-metals, so they
share
Covalent bond, formula AB3
Back to: 9.4.2 Bonding by electron transfer — Ionic bond

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 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.
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Answer: (iii)
A is
true
— molten copper sulfate conducts, solid does not.
R is
false
— it is the other way round: ions are fixed in the
solid
and free to move in the
molten
state.
Option
Verdict & why
(i)
Both A and R are true, and R is the correct explanation of A.
Wrong
— R is
false
, so it cannot explain A.
(ii)
Both A and R are true, but R is not the correct explanation of A.
Wrong
— R is not true.
(iii)
A is true, but R is false.
Correct answer
— A is true; R reverses the facts. Ions are fixed in the
solid
lattice and free to move when
molten
.
(iv)
A is false, but R is true.
Wrong
— A is true and R is false.
Back to: 9.6 Properties of the Ionic and the Covalent Compounds

Question 15
The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
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²⁷Al
(neutral): electrons = 13; neutrons = 27 − 13 =
14
.
⁸⁰Br⁻
(gained 1 e): electrons = 35 + 1 =
36
; neutrons = 80 − 35 =
45
.
²⁰¹Hg²⁺
(lost 2 e): electrons = 80 − 2 =
78
; neutrons = 201 − 80 =
121
.
Electrons and Neutrons
Al: 13 electrons, 14
neutrons
Br minus: 36 electrons, 45
neutrons
Hg 2 plus: 78 electrons, 121
neutrons
Back to: 9.4.2 Bonding by electron transfer — Ionic bond

Project 1
Design and perform an experiment to show and compare that water always contains hydrogen and oxygen in the same ratio, regardless of its source.
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Collect water from different sources (tap, rain, filtered). By electrolysis, collect the hydrogen and oxygen gases and compare their volume/mass ratio. In each case the ratio stays the same, showing the Law of Constant Proportions.

Project 2
Compare atoms and ions of any three elements. Show the number of electrons before and after ion formation using bar graphs.
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Choose e.g. Na, Mg, Cl. Plot electrons before and after ion formation: Na 11→10 (Na⁺), Mg 12→10 (Mg²⁺), Cl 17→18 (Cl⁻). A grouped bar graph makes the loss/gain clear.

Project 3
Make a card game with cations and anions where players combine cards to form compounds.
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Make cards for common cations and anions with their charges.
Players pick or ask for cards and pair a cation with anion(s) so the charges balance, forming a valid compound.
Score for each correctly balanced formula; discard unwanted cards.

Project 4
To learn more about molecules, explore the given simulation link.
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Build-a-Molecule (PhET): https://phet.colorado.edu/sims/html/build-a-molecule/latest/build-a-molecule_all.html — combine atoms into molecules and note how many of each atom a molecule needs.
🔑 The Quest Continues…
Are there chemical changes that do not obey conservation of mass?
Every reaction tested so far has obeyed the law.
The journey of exploring matter is far from over.
Keep questioning and keep investigating!