Solved Examples

Comprehensive Solved Examples for Chapter 3

30+ board-focused examples — covering all topics.

Example 1: NCERT — Difference Between Metal and Non-metal

State 5 main physical differences between metal and non-metal.

Solution:

Property	Metal	Non-metal
Physical state	Mostly solid	All states
Lustre	Yes (metallic)	Mostly no
Malleability	Yes	No (brittle)
Ductility	Yes	No
Electrical/Thermal Conductivity	Good	Poor (some exceptions)
Sonority	Yes	No

Exceptions:

• Hg — liquid metal.
• Br — liquid non-metal.
• Iodine — lustrous non-metal.
• Graphite — electrically conducting non-metal.

[NCERT — fundamental question]

Example 2: NCERT — Combustion of Magnesium

What happens when Mg is burnt in air? Chemical reaction, nature of oxide, and precautions.

Solution:

Reaction: $$2Mg + O_2 \xrightarrow{\Delta} 2MgO + \text{heat} + \text{light}$$

Observations:

• Bright white flame.
• White MgO powder.
• Intense heat.

Nature of MgO:

• Basic.
• $MgO + H_2O \rightarrow Mg(OH)_2$.
• Forms alkali.

Precautions:

• Safety goggles (to protect eyes).
• Watch from a distance.
• Fire extinguisher ready.

Use: flash bulbs, fireworks, signal flares.

[NCERT — every year]

Example 3: NCERT — Reaction of Na with Water

Write the reaction of Na with water. Why is it dangerous? How is Na preserved?

Solution:

Reaction: $$2Na + 2H_2O \rightarrow 2NaOH + H_2\uparrow + \text{heat}$$

Observations:

1. Na floats (low density).
2. Vigorous bubbling (H₂).
3. Yellow flame (sometimes).
4. Possible explosion.

Why dangerous?

1. Highly exothermic.
2. H₂ catches fire immediately.
3. NaOH also corrosive.

Preservation: In kerosene. Kerosene — protects Na from O₂ and H₂O.

[NCERT — important]

Example 4: NCERT — Protective Layer on Al

Why do Al articles remain shiny for a long time?

Solution:

Principle: Protective layer of $Al_2O_3$.

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

Thin, dense $Al_2O_3$ layer on Al — protects inner Al from further corrosion.

Properties:

• Thin (a few micrometers).
• Dense, pore-free.
• Transparent (Al shine visible).
• Very hard.

Difference from Fe rust: Fe₂O₃ loose, porous → more Fe rust. Al₂O₃ dense → no further O₂ inside.

Anodising — process to make this layer thicker.

[Board: 3-mark]

Example 5: NCERT — Amphoteric Oxides

What are amphoteric oxides? Two examples and reactions.

Solution:

Definition: Metal oxides that react with both acids and bases.

Main Examples: $Al_2O_3$, $ZnO$

Al₂O₃:

With acid (acts as base): $$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$$

With base (acts as acid): $$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$ (Sodium aluminate)

ZnO:

With acid: $$ZnO + 2HCl \rightarrow ZnCl_2 + H_2O$$

With base: $$ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O$$

[Board: 5-mark]

Example 6: NCERT — Iron and Steam

Write reaction of $Fe + H_2O$ steam. Why not at normal temperature?

Solution:

With steam: $$3Fe + 4H_2O(\text{steam}) \xrightarrow{\Delta} Fe_3O_4 + 4H_2\uparrow$$

Products: Fe₃O₄ (black), H₂.

Why not at normal temperature?

• Fe is moderately reactive.
• Cold/hot water — not enough energy.
• Steam = higher temperature, more energy.
• Now reaction proceeds.

Other steam reactions:

• $Mg + H_2O(\text{steam}) \rightarrow MgO + H_2$
• $Zn + H_2O(\text{steam}) \rightarrow ZnO + H_2$
• $2Al + 3H_2O(\text{steam}) \rightarrow Al_2O_3 + 3H_2$

[Board: 3-mark]

Example 7: NCERT — Zn + Dilute Acid

Reaction of Zn + dilute $H_2SO_4$, observations, and confirmation of H₂.

Solution:

Reaction: $$Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\uparrow$$

Observations:

1. Vigorous bubbles (H₂).
2. Zn dissolves.
3. ZnSO₄ colourless.
4. Test tube warms (exothermic).

Confirmation of H₂: 'Pop' test:

• Collect gas in test tube.
• Bring burning matchstick close.
• 'Pop' sound → H₂.

Reaction: $2H_2 + O_2 \rightarrow 2H_2O + \text{heat}$.

[NCERT — every year]

Example 8: NCERT — Cu and Dilute $HCl$

Why doesn't Cu react with dilute $HCl$?

Solution:

Answer: Cu — below H (in reactivity series).

Series: $...Pb > **H** > Cu > Hg > Ag > Au$

Rule: Only metals above H release $H_2$.

Cu, below H → no reaction.

Exception — concentrated acid: $$Cu + 2H_2SO_4(\text{conc.}) \xrightarrow{\Delta} CuSO_4 + 2H_2O + SO_2$$

This is not displacement — redox.

Practical: Acidic foods can be stored in copper vessels — no direct reaction.

[NCERT — important]

Example 9: NCERT — Zn + CuSO₄ (Displacement)

Write $Zn + CuSO_4$ reaction. Observations and principle.

Solution:

Reaction: $$Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$$

Observations:

1. Blue CuSO₄ → colourless ZnSO₄.
2. Red-brown Cu coating on Zn strip.
3. Zn slowly dissolves.

Principle: Reactivity: Zn > Cu. Zn displaced Cu.

Reverse experiment: Cu + ZnSO₄ → no reaction. (Because Cu less reactive than Zn.)

Electrochemical:

• $Zn \rightarrow Zn^{2+} + 2e^-$ (oxidation)
• $Cu^{2+} + 2e^- \rightarrow Cu$ (reduction)

[Board — every year]

Example 10: NCERT — Reactivity Series

Write the complete reactivity series. 2 properties of each category.

Solution:

Complete series (high to low):

$$\text{K} > \text{Na} > \text{Ca} > \text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Pb} > \text{(H)} > \text{Cu} > \text{Hg} > \text{Ag} > \text{Au}$$

Three categories:

1. High (K to Mg/Al):

• React with cold water.
• Explosive with acids.
• Extracted by electrolysis.

2. Moderate (Zn to Pb):

• React with steam.
• React with dilute acids.
• Reduced by carbon.

3. Low (Cu to Au):

• No reaction with water/acid.
• Direct heating for extraction.
• Found in free form (Au, Pt).

Mnemonic: "Please Stop Calling Me A Zebra…"

[Board — every year]

Example 11: NCERT — Ionic Compounds

(a) Electron transfer in formation of NaCl, MgCl₂, CaO. (b) Behaviour of these compounds.

Solution:

NaCl:

• Na (2,8,1) → Na⁺ + e⁻
• Cl (2,8,7) + e⁻ → Cl⁻
• $2Na + Cl_2 \rightarrow 2NaCl$

MgCl₂:

• Mg (2,8,2) → Mg²⁺ + 2e⁻
• 2 × [Cl + e⁻ → Cl⁻]
• $Mg + Cl_2 \rightarrow MgCl_2$

CaO:

• Ca (2,8,8,2) → Ca²⁺ + 2e⁻
• O (2,6) + 2e⁻ → O²⁻
• $2Ca + O_2 \rightarrow 2CaO$

Behaviour

All three are ionic compounds:

• Solid, crystalline, high melting point.
• Soluble in water.
• Conduct electricity in molten/aqueous state.
• Brittle.

[NCERT — important]

Example 12: A Numerical — Mg + O₂

How much MgO is formed by burning 6 g of Mg? (Mg=24, O=16)

Solution:

Reaction: $2Mg + O_2 \rightarrow 2MgO$

Ratio: 48 g Mg → 80 g MgO (or 1:1.67).

From 6 g Mg: $$\text{MgO} = \frac{80}{48} \times 6 = 10 \text{ g}$$

Answer: 10 g MgO.

Verification (Mass conservation):

• O₂ = 10 - 6 = 4 g.
• mol Mg = 6/24 = 0.25.
• mol O₂ = 4/32 = 0.125.
• Ratio Mg:O₂ = 2:1 ✓

[Board: 2-3 mark]

Example 13: NCERT — Properties of Ionic Compounds

5 physical properties of ionic compounds and one chemical property.

Solution:

5 Physical Properties:

1. Solid state: hard, crystalline, brittle.
2. High melting/boiling point: NaCl - 801°C; MgO - 2852°C.
3. Soluble in water: most.
4. Insoluble in non-polar: kerosene, benzene.
5. Electrical conductivity: No in solid; Yes in molten/aqueous.

1 Chemical Property:

Metal oxides are basic: $$Na_2O + H_2O \rightarrow 2NaOH$$ $$CaO + H_2O \rightarrow Ca(OH)_2$$

Exceptions: $Al_2O_3$, $ZnO$ — amphoteric.

[Board — every year]

Example 14: NCERT — Mineral and Ore

Difference between mineral and ore. Names and formulas of 5 famous ores.

Solution:

Difference: Ore = mineral with metal in profitable amount. Every ore is a mineral, every mineral is not an ore.

5 Famous Ores:

Metal	Ore	Formula
Al	Bauxite	Al₂O₃·2H₂O
Fe	Hematite	Fe₂O₃
Fe	Magnetite	Fe₃O₄
Cu	Copper pyrites	CuFeS₂
Zn	Zinc blende	ZnS
Hg	Cinnabar	HgS
Pb	Galena	PbS

[NCERT — fundamental]

Example 15: NCERT — Froth Flotation Method

Description of froth flotation method. For which ores?

Solution:

For whom? For sulphide ores (ZnS, PbS, CuFeS₂).

Principle:

• Sulphide ores — wetted by oil (hydrophobic).
• Gangue — wetted by water (hydrophilic).

Method:

1. Powder of ore.
2. Tank: water + pine oil.
3. Air current.
4. What happens:

• Ore + oil → up in froth.
• Gangue → down in water.

1. Collect upper froth.
2. Dry → concentrated ore.

Chemicals:

• Collector: pine oil.
• Frother.
• Depressant: NaCN.

[NCERT — every year]

Example 16: NCERT — Fe Extraction (Blast Furnace)

Main reactions of Fe extraction in Blast Furnace.

Solution:

Setup: Tall, cylindrical furnace. Top: charge (ore + coal + limestone). Bottom: hot air.

In three zones:

1. Lower zone (~1500°C): $$C + O_2 \rightarrow CO_2$$ $$CO_2 + C \rightarrow 2CO$$

2. Middle zone (~1000°C) — main reaction: $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

3. Upper zone (~500°C): $$CaCO_3 \rightarrow CaO + CO_2$$ $$CaO + SiO_2 \rightarrow CaSiO_3 \text{ (slag)}$$

Final products:

• Bottom: molten Fe (pig iron, 4% C).
• Top: slag.

[Board — every year, 5-mark]

Example 17: NCERT — Al Extraction (Hall-Héroult)

Detailed extraction of Al.

Solution:

Ore: Bauxite (Al₂O₃·2H₂O)

Step 1: Concentration (Bayer Process):

• $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$
• $NaAlO_2 + 2H_2O \rightarrow NaOH + Al(OH)_3$
• $2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$

Step 2: Electrolysis (Hall-Héroult):

Dissolve Al₂O₃ in cryolite (Na₃AlF₆) — melting point 950°C.

Setup:

• Steel box + carbon lining = Cathode.
• Carbon rods = Anode.

Reactions:

• Cathode: $Al^{3+} + 3e^- \rightarrow Al$
• Anode: $2O^{2-} \rightarrow O_2 + 4e^-$

Overall: $2Al_2O_3 \xrightarrow{\text{electricity}} 4Al + 3O_2$

[Board: 5-mark]

Example 18: NCERT — Hg Extraction

How is Hg extracted? Why not carbon reduction?

Solution:

Ore: Cinnabar (HgS)

Step 1: Roasting $$2HgS + 3O_2 \xrightarrow{\Delta} 2HgO + 2SO_2$$

Step 2: Decomposition of HgO $$2HgO \xrightarrow{\Delta} 2Hg + O_2$$

That is, HgO decomposes itself — no external reducing agent needed.

Why not carbon needed? Hg less reactive — HgO unstable. Just heating — sufficient.

This — example of extraction of less reactive metals.

[Board: 3-mark]

Example 19: NCERT — Zn Extraction

Zn extraction from zinc blende (ZnS).

Solution:

Ore: ZnS (Zinc blende) or ZnCO₃ (Calamine)

Step 1: Concentration Froth flotation.

Step 2: Roasting (for ZnS) $$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$$

Or calcination (for ZnCO₃): $$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$$

Step 3: Reduction by carbon $$ZnO + C \xrightarrow{\Delta} Zn + CO\uparrow$$

Step 4: Refining By electrolytic method.

Key Insight: Zn is moderately reactive — carbon is enough.

[Board: 3-mark]

Example 20: NCERT — Electrolytic Refining

Electrolytic refining of Cu.

Solution:

Setup:

• Anode: impure Cu.
• Cathode: pure Cu.
• Electrolyte: CuSO₄ + H₂SO₄.

Reactions:

• Anode: $Cu \rightarrow Cu^{2+} + 2e^-$
• Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$

Impurities:

• Au, Ag, Pt → anode mud (valuable).
• Fe, Zn → in solution.

99.99% pure Cu at Cathode.

Importance of Anode mud: Bonus of Cu industry — gold, silver.

[Board — every year, 5-mark]

Example 21: NCERT — Rust

Elements required for rust. Three test tubes experiment.

Solution:

Required elements:

1. Water (moisture).
2. Oxygen (air).

Three test tubes experiment:

Tube	Conditions	Result
A	Water, no air	No rust
B	Air, no water	No rust
C	Water + air	Rust!

Reaction: $$4Fe + 3O_2 + xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O$$

Fe₂O₃·xH₂O = rust (brown-red).

Accelerators:

• Salt (NaCl).
• Acid rain.
• Moisture.

[Board — every year]

Example 22: NCERT — Galvanisation

What is galvanisation? Principle and benefits.

Solution:

Definition: Layer of Zn on Fe.

Principle — Cathodic Protection:

• Zn — more reactive than Fe.
• On scratch — Zn corrodes first.
• Fe protected.

Method: Dip Fe in molten Zn.

Benefits:

1. Protection even on scratches.
2. Long durability (20-50 years).
3. Economical.

Use:

• Taps, pipes, buckets.
• Roofs, fences.
• 'GI sheet'.

[Board: 3-5 mark]

Example 23: NCERT — Alloys

Names, compositions, uses of 5 famous alloys.

Solution:

Alloy	Composition	Use
Stainless Steel	Fe + Cr + Ni	Utensils, knives
Brass	Cu + Zn (70:30)	Decoration, utensils
Bronze	Cu + Sn (88:12)	Statues, medals
Solder	Pb + Sn (50:50)	Welding
Duralumin	Al + Cu + Mg + Mn	Aircraft
Amalgam	Hg + others	Dentistry

Benefits of alloys:

1. Hardness ↑.
2. Corrosion-resistance.
3. Specific properties.

Stainless steel — Cr₂O₃ layer from Cr — 'self-healing'.

[Board — every year]

Example 24: NCERT — Anodising

What is anodising? Method and uses.

Solution:

Definition: Electrolytic process of forming a thicker layer of $Al_2O_3$ on Al.

Method:

1. Make Al the Anode.
2. Dip in dilute $H_2SO_4$.
3. Pass electric current.
4. O₂ at anode — reacts with Al.

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

Thicker $Al_2O_3$ layer (tens of μm).

Benefits:

1. Better corrosion-resistance.
2. Colours can be added.
3. Scratch-resistant.
4. Durable.

Uses:

• Window frames.
• Kitchen utensils.
• Mobile cases.
• Decorative items.

[NCERT — important]

Example 25: A Numerical — Zn + HCl

How much H₂ from 13 g Zn? Volume too (NTP). (Zn=65)

Solution:

Reaction: $Zn + 2HCl \rightarrow ZnCl_2 + H_2$

Ratio: 65 g Zn → 2 g H₂.

From 13 g Zn: $$\text{H}_2 = \frac{2}{65} \times 13 = 0.4 \text{ g}$$

Moles H₂: $0.4/2 = 0.2$ mol

Volume (at NTP): $0.2 \times 22.4 = 4.48 \text{ L}$

Answer:

• Mass of H₂: 0.4 g
• Volume of H₂: 4.48 L (at NTP)

[Board: 3-mark]

Example 26: NCERT — Aqua Regia

What is aqua regia? Formula and uses.

Solution:

Definition: Aqua Regia = concentrated HNO₃ + concentrated HCl (in 1:3 ratio).

Meaning of name:

• 'Aqua' = water (Latin).
• 'Regia' = royal.
• 'Royal Water' — dissolves the 'king of metals' (gold).

Reaction with gold: $$Au + 3HCl + HNO_3 \rightarrow AuCl_3 + NO + 2H_2O$$

Why both together?

• Alone HCl or HNO₃ — don't dissolve Au.
• Together — Cl₂ released → reacts with Au.

Uses:

1. Refining of gold.
2. Dissolving platinum.
3. Laboratory testing.

[NCERT — interesting]

Example 27: NCERT — Thermite

Thermite reaction and uses.

Solution:

Reaction: $$2Al + Fe_2O_3 \xrightarrow{\text{heat}} Al_2O_3 + 2Fe + \text{lots of heat}$$

Properties:

• Highly exothermic.
• Temperature ~3000°C.
• Molten Fe drips down.

Principle: Al — more reactive than Fe. Al displaced Fe.

Uses:

1. Joining railway tracks (Welding):

• Thermite welding.
• Molten Fe in crack — solidifies on cooling.

1. Joining machine parts.
2. Emergency fuses.

Reduction of other metals: $$2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr$$

[Board: 5-mark]

Example 28: NCERT — A Comparative

Comparison of physical features of metals and non-metals.

Solution:

Property	Metal	Non-metal	Exceptions
State	Mostly solid	All	Hg liquid; Br liquid
Lustre	Metallic	Mostly no	I₂ lustrous
Malleability	Yes	No	Hg N/A
Ductility	Yes	No	—
Conductivity	Good	Poor	Graphite, Si, Ge
Sonority	Yes	No	—
Melting point	Mostly high	Low	C very high
Density	High	Low	Na, K light

Major Exceptions:

• Mercury (Hg): liquid metal.
• Iodine (I₂): lustrous non-metal.
• Graphite: conducting non-metal.
• Carbon (diamond): hard non-metal.

[Board — every year]

Example 29: NCERT — A Mixed Question

(a) Which is more reactive — Mg or Cu? How will we know? (b) Why is Na stored in kerosene? (c) Why is layer on Al protective, but not on Fe? (d) Why not direct reduction of ZnS?

Solution:

(a) Mg vs Cu

Series: $Mg > Cu$

How will we know: Experiment: Mg + CuSO₄ → MgSO₄ + Cu (reaction!) Mg displaced Cu → Mg more reactive.

(b) Na in kerosene

Na very reactive — reacts immediately with O₂ and H₂O in air. Kerosene — protects Na from both. Avoids fire and explosion risk.

(c) Al layer protective, Fe not

$Al_2O_3$ on Al — dense, pore-free. No further O₂ inside. Fe₂O₃·xH₂O on Fe — loose, porous. More Fe rust below.

(d) Why not direct reduction of ZnS

Direct reduction of sulphide by carbon is difficult. First roasting (ZnS → ZnO) — then reduction of ZnO by C.

[Board: 5-mark mixed]

Example 30: NCERT — Numerical — Al Extraction

How much Al from 100 g of Al₂O₃? (Al=27, O=16)

Solution:

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

Molecular mass:

• Al₂O₃ = 102 g/mol
• Al = 27 g/mol

Ratio: 102 g Al₂O₃ → 54 g Al (= 2 × 27)

From 100 g: $$\text{Al} = \frac{54}{102} \times 100 = 52.94 \text{ g}$$

Answer: ~53 g Al.

Additional: % of Al in Al₂O₃: $$\frac{54}{102} \times 100 = 53\%$$

Energy estimate: 53 g Al × 14 kWh/kg = ~750 kWh electricity.

That is, equivalent to one month's household consumption!

[Board: 3-mark numerical]

Example 31: NCERT — Which is More Reactive?

Which of the following is correct?

(a) Cu can displace Fe from CuSO₄. (b) Pb can displace Cu from CuSO₄. (c) Hg can displace Cu from CuSO₄. (d) Au can displace Cu from CuSO₄.

Solution:

Series: $Pb > H > Cu > Hg > Au$

(a) Cu + FeSO₄ → ?

Cu less reactive than Fe. Wrong. No reaction.

(b) Pb + CuSO₄ → ?

Pb more reactive than Cu. Correct! $$Pb + CuSO_4 \rightarrow PbSO_4 + Cu$$

(c) Hg + CuSO₄ → ?

Hg less reactive than Cu. Wrong. No reaction.

(d) Au + CuSO₄ → ?

Au less reactive than Cu. Wrong. No reaction.

Correct answer: (b).

Key Insight: For displacement — strip metal must be more reactive than salt metal.

[Board: 3-mark]

Example 32: NCERT — An Interesting

Why is Fe nail shiny? How does rust form?

Solution:

Reason for Shine

Fe — metallic lustre. Free electrons on surface. Reflection of light.

Properly polished Fe — stays shiny.

How Does Rust Form?

Fe + air + water:

Step 1: Micro-cells on Fe surface.

• Some areas Anode: $Fe \rightarrow Fe^{2+} + 2e^-$
• Some Cathode: $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$

Step 2: Fe²⁺ + OH⁻ → Fe(OH)₂.

Step 3: Fe(OH)₂ + O₂ → Fe(OH)₃.

Step 4: Fe(OH)₃ → Fe₂O₃·xH₂O (rust).

Overall: $$4Fe + 3O_2 + xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O$$

Rust = loose, porous. Fe inside also rusts.

Prevention: paint, galvanisation, alloys.

[Board: mixed question]

Example 33: NCERT — Formulas of Ionic Compounds

Find formulas for the following ionic compounds:

(a) Sodium sulphide (b) Calcium chloride (c) Aluminium oxide (d) Magnesium nitride (e) Potassium sulphate

Solution:

Formulas by cross-multiplication.

(a) Na⁺ + S²⁻

1 ↔ 2 → $Na_2S$

(b) Ca²⁺ + Cl⁻

2 ↔ 1 → $CaCl_2$

(c) Al³⁺ + O²⁻

3 ↔ 2 → $Al_2O_3$

(d) Mg²⁺ + N³⁻

2 ↔ 3 → $Mg_3N_2$

(e) K⁺ + SO₄²⁻

1 ↔ 2 → $K_2SO_4$

Summary

Compound	Formula
Na sulphide	Na₂S
Ca chloride	CaCl₂
Al oxide	Al₂O₃
Mg nitride	Mg₃N₂
K sulphate	K₂SO₄

Rule: For charge balance — cross-multiplication.

[NCERT — fundamental]

Example 34: NCERT — A Logical

Which of the following is correct? Give reasons:

(a) Au can be easily reduced. (b) Na is obtained from electrolysis of aqueous NaCl. (c) Froth flotation is for sulphide ores. (d) Mercury is solid at room temperature.

Solution:

(a) Au easily reduced — Wrong ✗

Au is very low reactivity. Found in free form in nature. No need for 'reduction' — directly obtained.

(b) Na from aqueous — Wrong ✗

In aqueous NaCl:

• H₂O less reactive than Na.
• H₂O reduced at Cathode (H₂).
• Na not obtained.

Correct method: electrolysis of molten NaCl.

(c) Froth flotation — Correct ✓

ZnS, PbS, CuFeS₂ — all sulphides. Sulphides get 'wet' by oil. Froth flotation — most suitable.

(d) Hg solid at room temperature — Wrong ✗

Hg — only liquid metal at room temperature. Melting point: -39°C (can solidify in cold regions). Boiling point: 357°C.

Use: thermometers, barometers.

Summary

Statement	T/F
(a)	✗
(b)	✗
(c)	✓
(d)	✗

[Board: 5-mark logical]

Example 35: NCERT — An Experimental

A student added different metals + dilute $HCl$ in 5 test tubes:

Tube A: Mg Tube B: Zn Tube C: Fe Tube D: Cu Tube E: Ag

What happens in each?

Solution:

Series: $Mg > Zn > Fe > H > Cu > Ag$

A: Mg + HCl → ?

Very vigorous reaction! $$Mg + 2HCl \rightarrow MgCl_2 + H_2\uparrow$$

Vigorous bubbles, lots of heat.

B: Zn + HCl → ?

Vigorous reaction! $$Zn + 2HCl \rightarrow ZnCl_2 + H_2\uparrow$$

Good bubbles.

C: Fe + HCl → ?

Slow reaction. $$Fe + 2HCl \rightarrow FeCl_2 + H_2\uparrow$$

Light green FeCl₂.

D: Cu + HCl → ?

No reaction. Cu, below H.

E: Ag + HCl → ?

No reaction. Ag, below H.

Order of intensity

$$Mg > Zn > Fe > Cu \approx Ag (\text{both no reaction})$$

Per the reactivity series.

Key Insight: Reactivity decides — intensity of reaction.

[Board: 5-mark]

Example 36: A Concluding Question — Mixed

(a) Difference between basic and amphoteric oxides. (b) Reactivity series. (c) Al extraction. (d) Rust and prevention. (e) 5 alloys.

Solution:

(a) Basic vs Amphoteric

Property	Basic	Amphoteric
Reacts with?	Only acid	Both acid and base
Examples	Na₂O, CaO	Al₂O₃, ZnO

(b) Reactivity Series

$$K > Na > Ca > Mg > Al > Zn > Fe > Pb > (H) > Cu > Hg > Ag > Au$$

(c) Al Extraction

1. Bayer: Al₂O₃ + NaOH → NaAlO₂
2. Hall-Héroult: Electrolysis (Al₂O₃ + cryolite).

• Cathode: Al³⁺ + 3e⁻ → Al
• Anode: 2O²⁻ → O₂ + 4e⁻

(d) Rust and Prevention

Rust: $4Fe + 3O_2 + xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O$

Required: water + air.

Prevention:

1. Paint.
2. Oil.
3. Galvanisation (Zn coating).
4. Cr plating.
5. Alloy (stainless steel).

(e) 5 Alloys

Name	Composition
Stainless	Fe + Cr + Ni
Brass	Cu + Zn
Bronze	Cu + Sn
Solder	Pb + Sn
Duralumin	Al + Cu + Mg + Mn

[Board — every year — summary of entire chapter]