ChemistryICSE Class 9 · Atomic Structure and Chemical Bonding

Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

An electrovalent (ionic) bond forms by the complete transfer of electrons from a metal atom to a non-metal atom, giving oppositely charged ions held by electrostatic attraction. A covalent bond forms by the sharing of electron pairs between atoms, usually non-metals. Each atom ends with 8 electrons in its outer shell (2 for hydrogen).

Understand Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures

Octet rule. Atoms combine by losing, gaining or sharing electrons to get 8 electrons in the outermost shell (2 for hydrogen), like the nearest noble gas.

Electrovalent (ionic) bond. Formed by the transfer of electrons from a metal atom (1 to 3 outer electrons) to a non-metal atom (5 to 7 outer electrons). The ions are held together by electrostatic attraction. The electrovalency is the number of electrons lost or gained.

Compound	Transfer	Ions (configuration)
NaCl	Na (2, 8, 1) gives 1 electron to Cl (2, 8, 7)	Na⁺ (2, 8), Cl⁻ (2, 8, 8)
MgCl₂	Mg (2, 8, 2) gives 1 electron to each of two Cl atoms	Mg²⁺ (2, 8), 2Cl⁻ (2, 8, 8)
CaO	Ca (2, 8, 8, 2) gives 2 electrons to O (2, 6)	Ca²⁺ (2, 8, 8), O²⁻ (2, 8)

In an orbit-structure diagram show all shells, draw an arrow for the transferred electron(s), and write the ions in square brackets with their charges.

Covalent bond. Formed by sharing a pair of electrons, each atom contributing one. Single, double and triple bonds have 1, 2 and 3 shared pairs. Unshared pairs on an atom are lone pairs.

Molecule	Shared pairs	Lone pairs on the central or each atom
H₂	1	0
Cl₂	1	3 on each Cl
O₂	2	2 on each O
N₂	3	1 on each N
H₂O	2	2 on O
NH₃	3	1 on N
CH₄	4	0
CCl₄	4	3 on each Cl

Checking a structure. The total number of outer electrons contributed by all atoms equals 2 × (shared pairs + lone pairs). For $H_{2} O$ : $6 + 2 = 8 = 2 (2 + 2)$ . For $C C l_{4}$ : $4 + 28 = 32 = 2 (4 + 12)$ .

Radical formation. If a covalent bond breaks evenly, each atom keeps one of the shared electrons and two free radicals with unpaired electrons form: $C l_{2} UV 2 C l^{∙}$ ; $C H_{4} \to C H_{3}^{∙} + H^{∙}$ . These are very reactive.

Ionic or covalent? A metal with a non-metal: ionic. Two non-metals: covalent.

Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures: formulas and key facts

Electrovalent bond: $metal \to cation (loses e^{-}); non-metal \to anion (gains e^{-})$; Complete transfer.
Covalent bond: $shared pair = 2 e^{-}; single 1, double 2, triple 3 pairs$; Each atom gives one electron per pair.
Electron count check: $\sum e_{o u t er} = 2 (shared pairs + lone pairs)$; For molecules obeying the octet rule.

How to solve Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures questions

Decide first whether the atoms are a metal with a non-metal or two non-metals.

1Write the electronic configuration of each atom.
2Metal + non-metal: transfer electrons to complete the octet, then write the ions and the formula.
3Non-metal + non-metal: share pairs until each atom has 8 (2 for H).
4Count shared and lone pairs and check the electron total.
5Draw dots for one atom and crosses for the other.

Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why salt is NaCl and magnesia is MgO

Suppose you compare table salt and magnesia (magnesium oxide, used in refractory bricks and as an antacid).

Explain the electron transfer in each.

Try it first, then show the solutionSolution

Na (2, 8, 1) loses 1 electron and Cl (2, 8, 7) gains 1: NaCl.
Mg (2, 8, 2) loses 2 electrons and O (2, 6) gains 2: MgO.

Answer: NaCl: 1 electron transferred; MgO: 2 electrons.

Memory hook: Electrons lost equal electrons gained.

Tricks to remember Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures

Short memory aids. Say each one aloud once, then use it on a question the same day.

Metal gives, non-metal takes
Ionic: transfer. Covalent: share.
Use it when: Bond type.
Lone pairs by subtraction
Lone pairs on an atom = (outer electrons − bonds it makes) ÷ 2: O in water (6 − 2) ÷ 2 = 2; N in ammonia (5 − 3) ÷ 2 = 1.
Use it when: Counting lone pairs.
Even split gives radicals
Bond breaks evenly: two atoms with an unpaired electron each.
Use it when: Radical formation.

Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Type of bond · 1 mark · MCQBoard level

Which of the following compounds is formed by electrovalent bonding?
(a) HCl
(b) H₂O
(c) NaCl
(d) CH₄

Try it first, then show the solutionSolution

Sodium (a metal) transfers an electron to chlorine (a non-metal).

Answer: (c) NaCl

Example 2: Formation of NaCl · 2 marks · Short answerBoard level

Describe the formation of sodium chloride from sodium (Z = 11) and chlorine (Z = 17) in terms of electron transfer.

Try it first, then show the solutionSolution

Na (2, 8, 1) loses 1 electron to become $N a^{+}$ (2, 8); Cl (2, 8, 7) gains it to become $C l^{-}$ (2, 8, 8). (1)
The oppositely charged ions attract each other: an electrovalent bond; formula NaCl. (1)

Answer: Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻; NaCl

Example 3: Calcium chloride · 3 marks · Short answerHarder board level

Show the formation of calcium chloride from calcium (Z = 20) and chlorine (Z = 17). State the electrovalency of each element and the configurations of the ions.

Try it first, then show the solutionSolution

Ca (2, 8, 8, 2) gives one electron to each of two Cl atoms. (1)
$C a^{2 +}$ (2, 8, 8) and two $C l^{-}$ (2, 8, 8); formula CaCl₂. (1)
Electrovalency: calcium 2, chlorine 1. (1)

Answer: CaCl₂; Ca²⁺ and 2Cl⁻ (both 2, 8, 8); electrovalency 2 and 1

Example 4: Counting pairs in CCl₄ · 3 marks · Short answerHarder board level

In carbon tetrachloride, CCl₄ (C: 2, 4; Cl: 2, 8, 7), find (a) the number of shared pairs, (b) the number of lone pairs, (c) the total number of outer electrons, and (d) check that the electron total equals 2 × (shared pairs + lone pairs).

Try it first, then show the solutionSolution

(a) Four C–Cl single bonds: 4 shared pairs. (1)
(b) Each Cl has 3 lone pairs: $4 \times 3 = 12$ . (½)
(c) Outer electrons $= 4 + 4 \times 7 = 32$ . (½)
(d) $2 (4 + 12) = 32$ : matches. (1)

Answer: (a) 4 (b) 12 (c) 32 (d) 32 = 2(4 + 12)

Example 5: Four elements · 4 marks · Case-basedHOTS / competency

Elements A, B, C and D have atomic numbers 12, 8, 17 and 6.
(a) Write the electronic configuration of each.
(b) Name the bond and formula formed between A and B.
(c) Name the bond and formula formed between A and C.
(d) Name the bond and formula formed between D and C and state the number of shared pairs.

Try it first, then show the solutionSolution

(a) A: 2, 8, 2; B: 2, 6; C: 2, 8, 7; D: 2, 4. (1)
(b) A loses 2 electrons and B gains 2: electrovalent, AB (MgO). (1)
(c) A loses 1 electron to each of 2 C atoms: electrovalent, AC₂ (MgCl₂). (1)
(d) D shares one pair with each of 4 C atoms: covalent, DC₄ (CCl₄), 4 shared pairs. (1)

Answer: (a) 2,8,2; 2,6; 2,8,7; 2,4 (b) ionic AB (c) ionic AC₂ (d) covalent DC₄; 4 pairs

Where students lose marks in Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures

The slips examiners see most often, and what to write instead.

Mistake: Electrons shown transferred in a covalent compound
Non-metals share electrons.
Wrong working
H gives its electron to Cl in HCl.
Right working
H and Cl share one pair.
Do this: Metal + non-metal only for transfer.
Usually costs: 2 marks.
Mistake: Lone pairs omitted in dot diagrams
The octet includes lone pairs.
Wrong working
O in water drawn with only two bonds.
Right working
O with two bond pairs and two lone pairs.
Do this: Count to 8.
Usually costs: 1 mark.
Mistake: Ions drawn without charges or brackets
Ions need charge and square brackets.
Wrong working
Na Cl with arrow only.
Right working
[Na]⁺ [Cl]⁻ with 8 dots on Cl.
Do this: Show the charges.
Usually costs: 1 mark.

Exam tips

Write the configuration of each atom before drawing.
Show the ions with charges and square brackets.
Use dots for one atom and crosses for the other in covalent diagrams.
State the total electrons and check the octet.

Practice questions on Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of shared pairs of electrons in a nitrogen molecule is:
(a) 1
(b) 2
(c) 3
(d) 4
Hint
Triple bond.
Answer
(c) 3
Q2 · 1 mark · Very short answerBoard level
How many electrons are shared in a single covalent bond?
Hint
A pair.
Answer
Two (one pair).
Q3 · 2 marks · Short answerHarder board level
Why is the formula of sodium chloride NaCl and not NaCl₂?
Hint
Count the electrons lost and gained.
Answer
Sodium has one outer electron to lose and chlorine needs just one to complete its octet, so one Na⁺ pairs with one Cl⁻; the charges balance in the ratio 1 : 1.
Q4 · 3 marks · Short answerHarder board level
Verify, by counting outer electrons, the structures of (a) H₂O, (b) NH₃, (c) CH₄: give the shared pairs, lone pairs and show total outer electrons = 2 × (shared + lone).
Hint
O 6, N 5, C 4, H 1 each.
Answer
(a) $6 + 2 = 8 = 2 (2 + 2)$ . (b) $5 + 3 = 8 = 2 (3 + 1)$ . (c) $4 + 4 = 8 = 2 (4 + 0)$ .
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): When chlorine gas is exposed to ultraviolet light two chlorine free radicals are formed.
Reason (R): The Cl–Cl bond breaks evenly, each atom taking one electron of the shared pair.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Even break.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
Methane can split into a methyl radical and a hydrogen atom: CH₄ → CH₃• + H•. Count the electrons in CH₃• (C = 6, H = 1) and in H• and explain why both are very reactive.
Hint
Odd numbers of electrons.
Answer
CH₃•: $6 + 3 = 9$ electrons (odd); H•: 1 electron. Each has an unpaired electron, so each seeks to form a pair and reacts readily.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Electrovalent bond?Show answer
Formed by complete transfer of electrons from a metal to a non-metal.
2.Covalent bond?Show answer
Formed by sharing of a pair of electrons.
3.Bonds in N₂?Show answer
One triple bond (3 shared pairs).
4.Lone pairs on O in H₂O?Show answer
2.
5.Give an example of a molecule with a triple covalent bond.Show answer
Nitrogen, N₂ (three shared pairs).

Electrovalent and Covalent Bonds: Electron Transfer, Sharing and Dot Structures: frequently asked questions

How is an electrovalent bond formed?

A metal atom transfers one or more electrons to a non-metal atom; the resulting ions are held by electrostatic attraction.

How is a covalent bond formed?

Two atoms share a pair of electrons, each contributing one, so that both reach a stable outer shell.

What is a lone pair?

A pair of outer electrons on an atom that is not shared in a bond.