Metal-Non-metal Reactions — Ionic Compounds

Why Do Metals and Non-Metals Combine?

Nature's first discovery: Why are metals and non-metals attracted to each other?

Basic Principle — The Quest for Stability

Every atom seeks to attain the stable electronic configuration of its nearest noble gas — a complete outer shell.

What is a complete outer shell?

• 2 electrons (like helium), or
• 8 electrons (like neon/argon) — the octet rule

Behaviour of Metals

Metals have 1, 2, or 3 electrons in their outer shell — few.

To attain a complete shell — it is easier to lose electrons.

Examples:

• Na (2,8,1) → Na⁺ (2,8) [like neon]
• Mg (2,8,2) → Mg²⁺ (2,8) [like neon]
• Al (2,8,3) → Al³⁺ (2,8) [like neon]

Behaviour of Non-Metals

Non-metals have 5, 6, or 7 electrons in their outer shell — many.

To attain a complete shell — it is easier to gain electrons.

Examples:

• Cl (2,8,7) → Cl⁻ (2,8,8) [like argon]
• O (2,6) → O²⁻ (2,8) [like neon]
• N (2,5) → N³⁻ (2,8) [like neon]

This is How Ionic Bond Forms

Metal → loses electrons (becomes a cation) Non-metal → gains electrons (becomes an anion)

The two ions are held together by electrostatic attraction between opposite charges.

This bond — ionic bond or electrovalent bond.

Electronic Configuration and Ion Formation

Configuration of Important Elements

Major Metals:

Element	Atomic Number	Configuration	Valence Electrons	Ion
Li	3	2, 1	1	Li⁺
Na	11	2, 8, 1	1	Na⁺
K	19	2, 8, 8, 1	1	K⁺
Mg	12	2, 8, 2	2	Mg²⁺
Ca	20	2, 8, 8, 2	2	Ca²⁺
Al	13	2, 8, 3	3	Al³⁺

Major Non-Metals:

Element	Atomic Number	Configuration	Valence Electrons	Ion
H	1	1	1	H⁻ or H⁺
F	9	2, 7	7	F⁻
Cl	17	2, 8, 7	7	Cl⁻
O	8	2, 6	6	O²⁻
S	16	2, 8, 6	6	S²⁻
N	7	2, 5	5	N³⁻

Logic of Charges on Metals

The number of electrons a metal loses = its positive charge.

• Na (loses 1 electron) → Na⁺ (+1)
• Mg (loses 2 electrons) → Mg²⁺ (+2)
• Al (loses 3 electrons) → Al³⁺ (+3)

Logic of Charges on Non-Metals

The number of electrons a non-metal gains = its negative charge.

• Cl (gains 1 electron) → Cl⁻ (-1)
• O (gains 2 electrons) → O²⁻ (-2)
• N (gains 3 electrons) → N³⁻ (-3)

An Interesting Fact

A metal combines with a non-metal that can accept exactly the number of electrons it gives.

Therefore:

• $NaCl$: 1 Na gives 1, 1 Cl takes 1.
• $MgCl_2$: 1 Mg gives 2; 2 Cl take 1 each.
• $CaO$: 1 Ca gives 2; 1 O takes 2.
• $AlCl_3$: 1 Al gives 3; 3 Cl take 1 each.

Formation of Ionic Compounds — Detailed

1. Sodium Chloride ($NaCl$)

Basic Reaction:

$$2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$$

Electron Transfer:

Na atom: 2, 8, 1 — one valence electron. Cl atom: 2, 8, 7 — seven valence electrons.

Step by Step:

1. Na loses its 1 valence electron: $$Na \rightarrow Na^+ + e^-$$ Now Na⁺ has configuration 2, 8 (like neon — stable!)

2. Cl gains that 1 electron: $$Cl + e^- \rightarrow Cl^-$$ Now Cl⁻ has configuration 2, 8, 8 (like argon — stable!)

3. Na⁺ and Cl⁻ — opposite charges attract each other: $$Na^+ + Cl^- \rightarrow Na^+Cl^-$$

Appearance:

• Na: shiny, soft, reactive metal.
• Cl₂: yellow-green, toxic gas.
• $NaCl$: white, stable solid — common salt!

This is amazing: From two dangerous elements — life-sustaining everyday salt!

2. Magnesium Chloride ($MgCl_2$)

Reaction:

$$Mg(s) + Cl_2(g) \rightarrow MgCl_2(s)$$

Description:

• Mg (2,8,2) → Mg²⁺ (2,8) + 2e⁻
• Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8) — needs two Cl atoms.

1 Mg gave 1-1 electron to two Cl atoms.

Ionic Structure: Mg²⁺ + 2 Cl⁻ → $MgCl_2$

3. Calcium Oxide ($CaO$) — Quicklime

Reaction:

$$2Ca(s) + O_2(g) \rightarrow 2CaO(s)$$

Description:

• Ca (2,8,8,2) → Ca²⁺ (2,8,8) + 2e⁻
• O (2,6) + 2e⁻ → O²⁻ (2,8)

Ca's 2 electrons — directly to O.

Ionic Structure: Ca²⁺ + O²⁻ → $CaO$

This is 'quicklime' — used in cement, sugar refining, water treatment.

4. Aluminium Oxide ($Al_2O_3$)

Reaction:

$$4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3(s)$$

Description:

• 2 Al — make two Al³⁺ (release total 6e⁻).
• 3 O — make three O²⁻ (gain total 6e⁻).

Ionic Structure: 2 Al³⁺ + 3 O²⁻ → $Al_2O_3$

Coefficients adjusted to balance.

Nature of the Ionic Bond

What is an Ionic Bond?

A bond formed between a metal and a non-metal — by the electrostatic attraction between oppositely charged ions.

Electrovalency

The number of electrons a metal loses = its electropositive valency (Electrovalency). The number of electrons a non-metal gains = its electronegative valency.

Properties of Ionic Bond

1. Complete Transfer of Electron:

• Completely leaves one side, completely attaches to the other.
• No 'in-between' state.

2. Attraction between Opposite Charges:

• Coulomb's Law: $F = \frac{kq_1q_2}{r^2}$
• More charge = more attraction.

3. Crystalline Lattice Formation:

• Ions surround each other in three-dimensional structure.
• Called the 'crystal lattice'.

Crystal Structure of NaCl

Each Na⁺ surrounded by 6 Cl⁻ — and each Cl⁻ surrounded by 6 Na⁺. This is 'face-centered cubic' (FCC) structure.

Formulas of Ionic Compounds

The charge of metal and the charge of non-metal — give the formula by cross-multiplication.

Rule: Cross-down rule.

Examples:

• Al³⁺ + O²⁻ → formula: Al₂O₃ (3 from top to 2 below, 2 from top to 3 below)
• Ca²⁺ + N³⁻ → formula: Ca₃N₂
• Mg²⁺ + Br⁻ → formula: MgBr₂
• Na⁺ + S²⁻ → formula: Na₂S

Comparison with Covalent Bond

Non-metal — non-metal → covalent bond (electron sharing). Metal — metal → metallic bond (electron sea). Metal — non-metal → ionic bond (electron transfer).

Oxidation and Reduction — At a Glance

Principle

Formation of an ionic compound = a kind of redox reaction.

Oxidation: Loss of electrons — done by metal. Reduction: Gain of electrons — done by non-metal.

In NaCl Formation:

Na lost an electron → Oxidation of Na $$Na \rightarrow Na^+ + e^-$$

Cl gained an electron → Reduction of Cl $$Cl + e^- \rightarrow Cl^-$$

Both happen together — redox reaction.

In MgO:

Mg lost 2 electrons → oxidation. O gained 2 electrons → reduction.

Mg is not the oxidiser — it is the reducing agent (it reduces the other). O is the oxidising agent (oxidises the other).

A Fundamental Role of Metals

Metals — strong reducing agents. Non-metals — strong oxidising agents.

A Little Math

Charge balance in ionic compound:

• Total positive charge = Total negative charge.

Examples:

• $Al_2O_3$: 2 × (+3) = +6, and 3 × (-2) = -6 ✓
• $Mg_3N_2$: 3 × (+2) = +6, and 2 × (-3) = -6 ✓
• $Na_2S$: 2 × (+1) = +2, and 1 × (-2) = -2 ✓

Summary Table

Metal Ion	Non-Metal Ion	Formula
Na⁺	Cl⁻	NaCl
Mg²⁺	Cl⁻	MgCl₂
Ca²⁺	O²⁻	CaO
Al³⁺	O²⁻	Al₂O₃
K⁺	F⁻	KF
Mg²⁺	N³⁻	Mg₃N₂
Al³⁺	S²⁻	Al₂S₃
Na⁺	O²⁻	Na₂O

[Board Important] Formula formation and electron transfer — essential!

🧠 Memory Capsule

A quick glance just before the board exam.

1. Basic Principle

The quest for stability — complete outer shell (octet).

Metal → loses electrons (cation, positive ion). Non-metal → gains electrons (anion, negative ion). Opposite charges attract → ionic bond.

2. Important Ions

Metal	Configuration	Ion	Charge
Na	2,8,1	Na⁺	+1
Mg	2,8,2	Mg²⁺	+2
Al	2,8,3	Al³⁺	+3
Ca	2,8,8,2	Ca²⁺	+2

Non-Metal	Configuration	Ion	Charge
Cl	2,8,7	Cl⁻	-1
O	2,6	O²⁻	-2
N	2,5	N³⁻	-3

3. Main Examples

NaCl:

• Na (2,8,1) → Na⁺ (2,8) + e⁻
• Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8)

MgCl₂:

• Mg → Mg²⁺ + 2e⁻
• 2Cl + 2e⁻ → 2Cl⁻

CaO:

• Ca → Ca²⁺ + 2e⁻
• O + 2e⁻ → O²⁻

4. Cross-Multiplication — to Find Formula

Example: Al³⁺ + O²⁻

• 3 from top to bottom of O; 2 from top to bottom of Al
• Formula: $Al_2O_3$

5. Role of Redox

Metal = reducing agent (electron donor) Non-metal = oxidising agent (electron acceptor)

6. Board's 'Golden' Questions

1. NaCl formation — electron transfer.
2. MgCl₂ formation — detailed.
3. What is an ionic bond?
4. Formula of Al³⁺ and O²⁻.
5. Reaction of Ca + Cl₂.

Final Formula: Metal — donor; Non-metal — receiver; Opposite attraction — ionic compound.

Solved Examples

Example 1: NCERT — Formation of NaCl

Explain the formation of sodium chloride (NaCl) by electron transfer.

Solution:

1. Initial Configuration:

• Na: 2, 8, 1 (1 electron in outer shell)
• Cl: 2, 8, 7 (7 electrons in outer shell)

2. Quest for Stability:

• Na: 1 extra electron — hard to keep. Easy to lose.
• Cl: 1 short of 8. Easy to gain.

3. Electron Transfer:

Na donates its 1 outer electron to Cl.

$$Na (2,8,1) \rightarrow Na^+ (2,8) + e^-$$

Now Na⁺ configuration like neon — stable!

$$Cl (2,8,7) + e^- \rightarrow Cl^- (2,8,8)$$

Now Cl⁻ configuration like argon — stable!

4. Ionic Bond:

Na⁺ and Cl⁻ attract each other due to opposite charges.

$$Na^+ + Cl^- \rightarrow Na^+Cl^-$$

5. Overall Reaction:

$$2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$$

Special:

• Silvery-metallic Na + yellow-green Cl₂ gas.
• → White NaCl solid — everyday salt.
• Heat is released (exothermic).

[NCERT textbook — every year in board]

Example 2: NCERT — Formation of MgCl₂

Explain the formation of magnesium chloride (MgCl₂).

Solution:

1. Configuration:

• Mg: 2, 8, 2 (2 electrons in outer shell)
• Cl: 2, 8, 7 (7 electrons in outer shell)

2. Requirements:

• Mg: must give 2 electrons.
• Cl: must take 1 electron.

3. Ratio: 1 Mg → 2 Cl (Mg's 2 electrons — given 1-1 to two Cl atoms)

4. Electron Transfer:

$$Mg (2,8,2) \rightarrow Mg^{2+} (2,8) + 2e^-$$

Mg²⁺ — like neon.

$$2 \times [Cl (2,8,7) + e^- \rightarrow Cl^- (2,8,8)]$$

Each Cl⁻ — like argon.

5. Ionic Bond:

Mg²⁺ and 2 Cl⁻ → MgCl₂

6. Overall Reaction:

$$Mg(s) + Cl_2(g) \rightarrow MgCl_2(s)$$

Special:

• MgCl₂ — white crystalline solid.
• Highly soluble in water.
• Found in seawater.
• Compound near to Epsom salt.
• Used in liquid batteries.

Electrovalency

Mg's positive valency = 2 Cl's negative valency = 1

[NCERT — important]

Example 3: NCERT — Formation of CaO

Explain the formation of calcium oxide (CaO) by electron transfer.

Solution:

1. Configuration:

• Ca: 2, 8, 8, 2
• O: 2, 6

2. Requirements:

• Ca: must give 2 electrons (for octet).
• O: must take 2 electrons (for octet).

3. Ratio: 1 Ca ↔ 1 O (Ca's 2 → directly to O)

4. Electron Transfer:

$$Ca (2,8,8,2) \rightarrow Ca^{2+} (2,8,8) + 2e^-$$

Ca²⁺ — like argon.

$$O (2,6) + 2e^- \rightarrow O^{2-} (2,8)$$

O²⁻ — like neon.

5. Ionic Bond:

Ca²⁺ and O²⁻ → CaO

6. Reaction:

$$2Ca(s) + O_2(g) \rightarrow 2CaO(s)$$

Special:

• CaO ≠ slaked lime! That is $Ca(OH)_2$.
• CaO = quicklime.
• White, solid.
• Vigorous reaction with water (heat): $$CaO + H_2O \rightarrow Ca(OH)_2 + \text{heat}$$

Practical Uses

• Cement manufacturing.
• Sugar refining.
• Removing water hardness.
• Increasing soil fertility (in acidic soil).

[NCERT textbook]

Example 4: NCERT — Formation of Al₂O₃

Explain the formation and formula of aluminium oxide.

Solution:

1. Configuration:

• Al: 2, 8, 3
• O: 2, 6

2. Requirements:

• Al: must give 3 electrons.
• O: must take 2 electrons.

3. To find ratio — LCM:

LCM of 3 and 2 = 6.

That is, total of 6 electrons exchanged.

2 Al × 3e⁻ = will give 6e⁻. 3 O × 2e⁻ = will take 6e⁻.

4. Electron Transfer:

$$2 \times [Al \rightarrow Al^{3+} + 3e^-]$$

That is 2 Al → 2 Al³⁺ + 6e⁻

$$3 \times [O + 2e^- \rightarrow O^{2-}]$$

That is 3 O + 6e⁻ → 3 O²⁻

5. Ionic Compound:

2 Al³⁺ + 3 O²⁻ → Al₂O₃

6. Cross-Multiplication (Shortcut):

$$\text{Al}^{3+} \xrightarrow{\text{cross}} \text{O}^{2-} \Rightarrow Al_2O_3$$

(O below 3, Al below 2)

7. Overall Reaction:

$$4Al + 3O_2 \rightarrow 2Al_2O_3$$

Special:

• $Al_2O_3$ = alumina
• Transparent solid.
• High melting point (~2050°C).
• Found in gemstones: ruby (Cr-doped), sapphire (Fe/Ti-doped).

[Board: 3-5 marks]

Example 5: An Interesting — Formation of Na₂O

Explain the formation of Na₂O from sodium and oxygen.

Solution:

1. Configuration:

• Na: 2, 8, 1
• O: 2, 6

2. Requirements:

• Na: will give 1 electron.
• O: will take 2 electrons.

3. Ratio: 2 Na : 1 O (2 Na give 2 electrons, 1 O takes 2)

4. Electron Transfer:

2 Na (2,8,1) → 2 Na⁺ (2,8) + 2e⁻

O (2,6) + 2e⁻ → O²⁻ (2,8)

5. Compound:

$$2Na^+ + O^{2-} \rightarrow Na_2O$$

6. Overall Reaction:

$$4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$$

Special:

• Na₂O — white solid.
• Vigorous reaction with water — forms NaOH: $$Na_2O + H_2O \rightarrow 2NaOH$$
• Strongly basic.
• When Na is exposed to air — first Na₂O forms, then NaOH (from moisture).

Answer to 'Why is Na stored in kerosene?':

• Because Na with O₂ and moisture in air forms Na₂O and NaOH.
• Kerosene protects Na from both.

[Board: 3-mark]

Example 6: Formation of Mg₃N₂

Explain the formation of magnesium nitride.

Solution:

1. Configuration:

• Mg: 2, 8, 2
• N: 2, 5

2. Requirements:

• Mg: will give 2 electrons.
• N: will take 3 electrons.

3. Ratio:

LCM of 2 and 3 = 6.

3 Mg × 2 = will give 6e⁻. 2 N × 3 = will take 6e⁻.

Ratio: 3 Mg : 2 N

4. Electron Transfer:

3 Mg → 3 Mg²⁺ + 6e⁻

2 N + 6e⁻ → 2 N³⁻

5. Compound:

$$3Mg^{2+} + 2N^{3-} \rightarrow Mg_3N_2$$

6. By Cross-Multiplication:

$$\text{Mg}^{2+} \xrightarrow{\text{cross}} \text{N}^{3-} \Rightarrow Mg_3N_2$$

7. Overall Reaction:

$$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$$

Special:

• Mg₃N₂ — green-yellow solid.
• When Mg burns in air — some Mg reacts with N₂ to give Mg₃N₂.
• Reacts with water to give Mg(OH)₂ + NH₃: $$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3\uparrow$$

(NH₃ — smell of ammonia.)

Special Note: When Mg burns in air — mainly MgO (white), but some Mg₃N₂ (yellow). If residue appears yellow-green — N₂ has reacted.

[Board: 3-mark]

Example 7: NCERT — Find the Formula

Find the formula for the following (charges on ions are given):

(a) Sodium sulphide (Na⁺, S²⁻) (b) Potassium oxide (K⁺, O²⁻) (c) Calcium chloride (Ca²⁺, Cl⁻) (d) Aluminium sulphide (Al³⁺, S²⁻) (e) Iron(III) chloride (Fe³⁺, Cl⁻)

Solution:

Principle: Cross-multiplication — make the charge of one ion the subscript of the opposite ion.

(a) Na⁺ and S²⁻

1 above, 2 above → cross

$$Na_2S$$

Check: 2 × (+1) + 1 × (-2) = 0 ✓

(b) K⁺ and O²⁻

$$K_2O$$

Check: 2 × (+1) + 1 × (-2) = 0 ✓

(c) Ca²⁺ and Cl⁻

$$CaCl_2$$

Check: 1 × (+2) + 2 × (-1) = 0 ✓

(d) Al³⁺ and S²⁻

3 and 2 — direct cross.

$$Al_2S_3$$

Check: 2 × (+3) + 3 × (-2) = 0 ✓

(e) Fe³⁺ and Cl⁻

$$FeCl_3$$

Check: 1 × (+3) + 3 × (-1) = 0 ✓

Summary Table

Compound	Ions	Formula
Na sulphide	Na⁺, S²⁻	$Na_2S$
K oxide	K⁺, O²⁻	$K_2O$
Ca chloride	Ca²⁺, Cl⁻	$CaCl_2$
Al sulphide	Al³⁺, S²⁻	$Al_2S_3$
Fe(III) chloride	Fe³⁺, Cl⁻	$FeCl_3$

Key Insight: Without finding LCM — just cross-multiplication gives the formula.

[NCERT textbook — fundamental]

Example 8: NCERT — Identifying Redox

In the following reactions, identify oxidation and reduction:

(a) $2Na + Cl_2 \rightarrow 2NaCl$ (b) $2Mg + O_2 \rightarrow 2MgO$ (c) $4Al + 3O_2 \rightarrow 2Al_2O_3$

Solution:

Basic Rule:

• Loss of electron = oxidation.
• Gain of electron = reduction.

(a) 2Na + Cl₂ → 2NaCl

Change in Na: Na → Na⁺ + e⁻ Oxidation number: 0 → +1 Na is oxidised. Na — reducing agent (reduced the other).

Change in Cl: Cl + e⁻ → Cl⁻ Oxidation number: 0 → -1 Cl is reduced. Cl — oxidising agent.

(b) 2Mg + O₂ → 2MgO

Mg → Mg²⁺ + 2e⁻ → oxidation. Mg — reducing agent.

O + 2e⁻ → O²⁻ → reduction. O₂ — oxidising agent.

(c) 4Al + 3O₂ → 2Al₂O₃

Al → Al³⁺ + 3e⁻ → oxidation. Al — reducing agent.

O + 2e⁻ → O²⁻ → reduction. O₂ — oxidising agent.

General Pattern

Ionic Compound Formation = Redox Reaction

Always:

• Metal → oxidation (loses electrons)
• Non-metal → reduction (gains electrons)
• Metal = reducing agent
• Non-metal = oxidising agent

Remember: OIL RIG

• O = Oxidation
• I = Is
• L = Loss (of electrons)
• R = Reduction
• I = Is
• G = Gain (of electrons)

[Board: 5-mark]

Example 9: NCERT — Identifying Ionic Bond

Identify the ionic bonds in the following:

(a) $H_2O$ (b) $NaCl$ (c) $CO_2$ (d) $MgO$ (e) $CH_4$ (f) $K_2S$

Solution:

Rule: Ionic bond = metal + non-metal. Covalent bond = non-metal + non-metal.

Analysis

(a) $H_2O$ — H + O Both non-metals → covalent.

(b) $NaCl$ — Na (metal) + Cl (non-metal) ✓ Ionic!

(c) $CO_2$ — C + O Both non-metals → covalent.

(d) $MgO$ — Mg (metal) + O (non-metal) ✓ Ionic!

(e) $CH_4$ — C + H Both non-metals → covalent.

(f) $K_2S$ — K (metal) + S (non-metal) ✓ Ionic!

Summary

Compound	Type
$H_2O$	Covalent
NaCl	Ionic
$CO_2$	Covalent
MgO	Ionic
$CH_4$	Covalent
K₂S	Ionic

Insight

In the periodic table:

• Left side elements (metals): Na, K, Mg, Ca, Al etc.
• Right side (non-metals): O, N, F, Cl, S etc.
• When left + right → ionic.
• When right + right → covalent.

[NCERT textbook]

Example 10: An Interesting — Mg and S

Formation of magnesium sulphide. What will be the formula? How does it form?

Solution:

1. Configuration:

• Mg: 2, 8, 2
• S: 2, 8, 6

2. Requirements:

• Mg: will give 2 electrons (for octet).
• S: will take 2 electrons (for octet).

3. Ratio: 1 Mg : 1 S (Mg's 2 → directly to S)

4. Electron Transfer:

$$Mg (2,8,2) \rightarrow Mg^{2+} (2,8) + 2e^-$$

$$S (2,8,6) + 2e^- \rightarrow S^{2-} (2,8,8)$$

5. Ionic Compound:

$$Mg^{2+} + S^{2-} \rightarrow MgS$$

Both have same charge (2) — so 1:1.

6. Overall Reaction:

$$Mg(s) + S(s) \xrightarrow{\Delta} MgS(s)$$

Special:

• $MgS$ — pink-brown solid.
• Reacts with water to give Mg(OH)₂ + H₂S gas: $$MgS + 2H_2O \rightarrow Mg(OH)_2 + H_2S\uparrow$$

(H₂S — smell of rotten eggs.)

Comparison

Mg + O → MgO — charges 2:2 → 1:1 → MgO Mg + S → MgS — charges 2:2 → 1:1 → MgS Mg + Cl → MgCl₂ — charges 2:1 → 1:2 → MgCl₂ Mg + N → Mg₃N₂ — charges 2:3 → 3:2 → Mg₃N₂

Key Insight

Charge LCM determines the ratio.

[Board: 3-mark]

Example 11: A Numerical — How Many Ions?

How many Na⁺ and Cl⁻ ions are present in 5.85 g of $NaCl$? (Na=23, Cl=35.5)

Solution:

Molecular mass of $NaCl$ = 23 + 35.5 = 58.5 g/mol

Number of moles: $$n = \frac{5.85}{58.5} = 0.1 \text{ mol}$$

Number of ions:

1 NaCl contains: 1 Na⁺ + 1 Cl⁻

1 mol NaCl contains: $6.022 \times 10^{23}$ Na⁺ + $6.022 \times 10^{23}$ Cl⁻

0.1 mol contains:

Number of Na⁺ ions: $$0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$$

Number of Cl⁻ ions: $$0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$$

Total ions: $$2 \times 6.022 \times 10^{22} = 1.2 \times 10^{23}$$

Answer:

• Na⁺ ions: $6.022 \times 10^{22}$
• Cl⁻ ions: $6.022 \times 10^{22}$

Key Insight: Ionic compounds — not single molecules — but lattice of ions. But mole calculations from formula unit.

Additional — In Solution

If NaCl is dissolved in water:

This will be electrically conducting — because it is an ionic compound.

[Board: 3-mark numerical]

Example 12: A Comparison — Ionic vs Covalent

$NaCl$ and $CCl_4$ — both contain Cl. But one is ionic, the other covalent. Why?

Solution:

NaCl — Ionic

Elements: Na (metal) + Cl (non-metal)

Electronegativity:

• Na: 0.93 (low)
• Cl: 3.16 (high)
• Difference: 2.23 (very high!)

Difference > 1.7 → Ionic.

Behaviour of Electrons:

• Cl takes electron from Na completely.
• Complete transfer.
• Na⁺ Cl⁻ formed.

CCl₄ — Covalent

Elements: C (non-metal) + Cl (non-metal)

Electronegativity:

• C: 2.55
• Cl: 3.16
• Difference: 0.61 (low!)

Difference 0.5-1.7 → polar covalent, here mildly covalent.

Behaviour of Electrons:

• C and Cl share electrons.
• No complete transfer.
• No ions formed.

Comparative Table

Property	NaCl (Ionic)	CCl₄ (Covalent)
Bond type	Ionic	Covalent
Melting point	High (801°C)	Low (-23°C)
Boiling point	1465°C	77°C
Solubility in water	Soluble	Insoluble
Electrical conductivity (molten/in solution)	Yes	No
Structure	Crystal lattice	Separate molecules

Fundamental Difference

Ionic: Complete transfer → ions. Covalent: Sharing → molecules.

[Board: 5-mark comparative]

Example 13: NCERT — A Mixed Question

Answer the following questions:

(a) Between Na and Mg, which loses electrons more easily? (b) Between F and Cl, which gains electrons more easily? (c) Between NaCl and MgO, which is more stable?

Solution:

(a) Na and Mg, which?

Configuration:

• Na: 2, 8, 1 (1 electron)
• Mg: 2, 8, 2 (2 electrons)

Reasoning:

• Na must lose 1 electron → easy.
• Mg must lose 2 electrons → harder (the second one needs more effort).

Answer: Na loses electrons more easily.

That is, Na is more reactive than Mg.

(b) F and Cl, which?

Configuration:

• F: 2, 7 (1 vacancy in outer shell)
• Cl: 2, 8, 7 (1 vacancy in outer shell)

Reasoning:

• F's outer shell is very close to nucleus.
• Cl's outer shell is far (third shell).
• F's nucleus — attracts the incoming electron more strongly.

Answer: F gains electrons more easily.

That is, F is more reactive non-metal than Cl.

(c) NaCl and MgO, which is more stable?

Comparison:

• NaCl: Na⁺ (+1) and Cl⁻ (-1) — relatively weak attraction.
• MgO: Mg²⁺ (+2) and O²⁻ (-2) — attraction 4 times stronger.

Coulomb's Law: $F = \frac{kq_1q_2}{r^2}$

• More charge product → more attraction.

Answer: MgO is more stable.

Evidence:

• Melting point of NaCl: 801°C
• Melting point of MgO: 2852°C Much higher melting point of MgO — more stable bond.

[NCERT — combined question]

Example 14: A Challenge — Charges and Formulas

Find the formulas of the following ionic compounds:

(a) Sodium nitride (Na⁺, N³⁻) (b) Calcium phosphate (Ca²⁺, PO₄³⁻) (c) Ammonium sulphate (NH₄⁺, SO₄²⁻) (d) Aluminium carbonate (Al³⁺, CO₃²⁻)

Solution:

(a) Na⁺ and N³⁻

Cross: 1 ↔ 3

$$Na_3N$$

Check: 3 × (+1) + 1 × (-3) = 0 ✓

(b) Ca²⁺ and PO₄³⁻

Cross: 2 ↔ 3

Note: PO₄³⁻ — a 'polyatomic' ion. If multiple, put in brackets.

$$Ca_3(PO_4)_2$$

Check: 3 × (+2) + 2 × (-3) = 0 ✓

(c) NH₄⁺ and SO₄²⁻

Cross: 1 ↔ 2

Both polyatomic — both in brackets.

$$(NH_4)_2 SO_4$$

Check: 2 × (+1) + 1 × (-2) = 0 ✓

(d) Al³⁺ and CO₃²⁻

Cross: 3 ↔ 2

CO₃²⁻ multiple — in brackets.

$$Al_2(CO_3)_3$$

Check: 2 × (+3) + 3 × (-2) = 0 ✓

Key Insights

Rules:

1. Cross-multiplication for formula.
2. If polyatomic ion appears more than once — use brackets.
3. Subscript of one ion = charge of opposite ion.

Other Important Polyatomic Ions

Ion	Charge
OH⁻ (hydroxide)	-1
NO₃⁻ (nitrate)	-1
HCO₃⁻ (bicarbonate)	-1
CO₃²⁻ (carbonate)	-2
SO₄²⁻ (sulphate)	-2
PO₄³⁻ (phosphate)	-3
NH₄⁺ (ammonium)	+1

[Board: 5-mark]

Example 15: An Interesting — Cl₂ and H₂ Example

Do $Cl_2$ non-metal and $H_2$ non-metal — together form an ionic compound?

Solution:

Reaction:

$$H_2 + Cl_2 \rightarrow 2HCl$$

Answer: No! HCl is covalent — not ionic.

Why?

Rule: Ionic bond = metal + non-metal. Here — both non-metals → covalent.

Behaviour of Electrons

In HCl:

• H and Cl share electrons.
• No complete transfer.
• Hence polar covalent bond.

Although — Cl is more electronegative than H — so the shared electrons are pulled towards Cl. HCl is a polar molecule.

An Interesting Fact

HCl in gaseous state — covalent. HCl when added to water — gets ionised:

$$HCl + H_2O \rightarrow H_3O^+ + Cl^-$$

That is, in water, H⁺ (H₃O⁺) and Cl⁻ are formed. Then it shows ionic behaviour.

This is why — gaseous HCl is not electrically conducting, but aqueous HCl (dilute acid) conducts electricity.

Comparison

Property	NaCl (Ionic)	HCl (Covalent)
Bond	Ionic	Covalent (polar)
State	Solid	Gas
Melting point	801°C	-114°C
In water	Ionised	Ionised (forms acid)
Conductivity in pure form	No	No
Conductivity in water	Yes	Yes (when ionised)

Key Insight: Two non-metals → always covalent. But can be ionic in water.

[Board: 5-mark]

Example 16: A Concluding Question

(a) Difference between ionic bond and covalent bond. (b) NaCl, MgO, Al₂O₃ — electron transfer. (c) Why does $Mg^{2+}$ have charge +2, not +1? (d) Polyatomic ions — 3 examples.

Solution:

(a) Ionic vs Covalent

Property	Ionic	Covalent
Electrons	Complete transfer	Shared
Elements	Metal + Non-metal	Non-metal + Non-metal
Examples	NaCl, MgO	H₂O, CO₂
Melting point	High	Low
Electrical conductivity	Yes (in water/molten)	No

(b) Formation of all three

NaCl:

• Na (2,8,1) → Na⁺ (2,8) + e⁻
• Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8)

MgO:

• Mg (2,8,2) → Mg²⁺ (2,8) + 2e⁻
• O (2,6) + 2e⁻ → O²⁻ (2,8)

Al₂O₃:

• 2 Al (2,8,3) → 2 Al³⁺ (2,8) + 6e⁻
• 3 O (2,6) + 6e⁻ → 3 O²⁻ (2,8)

(c) Why Mg²⁺?

Configuration: Mg = 2, 8, 2

If Mg lost 1 electron → Mg⁺ (2,8,1) — unstable. If Mg lost 2 electrons → Mg²⁺ (2,8) — stable like neon.

Nature seeks 'complete' octet. Mg²⁺ is more stable — that's why it forms.

(d) Polyatomic Ions

1. Hydroxide: $OH^-$

• 1 O + 1 H, total charge -1
• In NaOH, KOH.

2. Nitrate: $NO_3^-$

• 1 N + 3 O, total charge -1
• In KNO₃, AgNO₃.

3. Carbonate: $CO_3^{2-}$

• 1 C + 3 O, total charge -2
• In CaCO₃, Na₂CO₃.

Others:

• $SO_4^{2-}$ (sulphate)
• $PO_4^{3-}$ (phosphate)
• $NH_4^+$ (ammonium — a positive ion)

[Board: 5-mark mixed question]