(i) Physical properties alone are not reliable for classifying elements as metals or non-metals, because there are many exceptions. Graphite (a non-metal) conducts electricity like a metal, and iodine (a non-metal) is lustrous. Such exceptions show that no single physical property is universal to all metals or non-metals.

(ii) Chemical properties provide a more reliable basis for classification.

Justification: Metals form basic oxides when burnt in air, while non-metals form acidic oxides. For example, magnesium burns to form MgO, which is basic, whereas sulphur burns to form SO₂, which dissolves in water to give an acidic solution. This chemical behaviour is consistent and does not show the exceptions seen in physical properties.

Source: Chapter 3, Section 3.1.2 and Activity 3.8

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• Part (i) expects you to state that physical properties have too many exceptions — the textbook itself lists graphite conducting electricity, iodine being lustrous, and alkali metals being very soft as key examples.
• Part (ii) is directly from the textbook conclusion after Activity 3.8: "Elements can be more clearly classified as metals and non-metals on the basis of their chemical properties." Always support with the metal oxide / non-metal oxide example (basic vs. acidic oxide) as the examiner expects it.
• Don't just say "chemical properties are better" — you must justify with an example as the question demands.

The two exceptions — iodine being lustrous and graphite conducting electricity — show that physical properties alone are not reliable for classifying elements as metals or non-metals.

Lustre (shiny surface) is generally considered a property of metals, yet iodine, a non-metal, is lustrous. Similarly, non-metals are generally bad conductors of electricity, yet graphite (an allotrope of carbon) conducts electricity. These exceptions prove that physical properties have limitations and can be misleading.

Therefore, chemical properties provide a more accurate and reliable basis for classifying elements as metals or non-metals.

Source: Chapter 3, Section 3.1.2 (Physical Properties — Non-metals)

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• Examiners expect you to name both exceptions clearly, explain what rule each one breaks, and then give the conclusion that chemical properties are more reliable.
• The textbook explicitly states: "we cannot group elements according to their physical properties alone, as there are many exceptions." Use this idea directly.
• Don't just list facts — connect them to what they tell us (the reliability question), since that is exactly what the question asks.

Flattening into thin sheets — Malleability: Copper is malleable, meaning it can be beaten or rolled into thin sheets without breaking. This property allows the metal to change shape under compressive force without cracking.

Stretching into wires — Ductility: Copper is ductile, meaning it can be drawn into long, thin wires. This property allows the metal to be stretched under tensile force without breaking.

Source: Chapter 3, Section 3.1.1

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• The examiner expects both property names (malleability, ductility) to be correctly matched to the right process — each named property with a brief explanation earns 1 mark each.
• A common error is confusing the two: malleable = beaten into sheets; ductile = drawn into wires. Remember: Ductile → Draw wires.
• You do not need to give examples of other metals; just define and link each property to the process described.

Metal core (copper/aluminium): Metals are good conductors of electricity, allowing electric current to flow easily through the wire with low resistance. Copper and aluminium are preferred because they are highly conductive, ductile, and relatively affordable.

Outer covering (PVC/rubber): The outer coating must be a non-metal because non-metals are bad conductors (insulators) of electricity. This insulating layer prevents electric shock to the user, avoids short-circuiting if two wires touch, and protects against current leakage.

Source: Chapter 12, Section 12.4 (Domestic Electric Circuits); Chapter 3 (Metals and Non-metals — What You Have Learnt)

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• The question tests two distinct concepts — conductivity of metals and insulating property of non-metals — so address both clearly for full marks (likely 1½ + 1½ or 1 + 2 split).
• Key phrase from Chapter 3: "Metals are good conductors of heat and electricity" and "Non-metals are bad conductors of heat and electricity."
• Examiners expect you to link the property (conductivity / non-conductivity) to the function (carrying current / preventing shock/short circuit). Don't just name the property — state why it matters for the wire.

(B) is incorrect. Alkali metals such as sodium and potassium are soft (can be cut with a knife) and have low densities and low melting points — not high densities and high melting points.

Source: Chapter 3, Section 3.1 Physical Properties

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The textbook explicitly states: "Alkali metals (lithium, sodium, potassium) are so soft that they can be cut with a knife. They have low densities and low melting points." All other options are factually correct as per the chapter — mercury is the only liquid metal at room temperature; silver and copper are the best conductors of heat; and one gram of gold can be drawn into ~2 km of wire. In MCQs, find the one statement that directly contradicts the textbook.

Two observations distinguishing sodium's reaction from calcium's:

1. Vigour/Speed: Sodium reacts very vigorously (violently) with cold water, often catching fire, whereas calcium's reaction is comparatively slower and less vigorous.

1. Movement on water: Sodium moves rapidly/erratically on the water surface (may melt into a ball) due to the large amount of heat produced, while calcium sinks and reacts steadily, producing bubbles.

Reasons: Sodium is higher in reactivity than calcium in the activity series — it loses its single valence electron more readily. Calcium has two valence electrons to lose, making it less reactive. Both form hydroxides and hydrogen gas:

$$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\uparrow$$
$$\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2\uparrow$$

Source: Chapter 3, Activity series / reactivity of metals with water

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• Examiners expect two distinct observations (vigour, fizzing/fire, movement) and a reason linked to reactivity/electronic configuration — losing 1 electron (Na) vs 2 electrons (Ca).
• Writing the balanced equations earns marks and shows understanding.
• "Sodium catches fire" is a valid observation for full credit.
• Do not confuse CaO + H₂O (combination reaction) with Ca metal + H₂O — the question is about the metal reacting with water.

Balanced chemical equation for iron reacting with steam:

$$3\text{Fe(s)} + 4\text{H}_2\text{O(g)} \rightarrow \text{Fe}_3\text{O}_4\text{(s)} + 4\text{H}_2\text{(g)}$$

Why iron cannot replace magnesium in a reaction with cold water:

Iron is much less reactive than magnesium. Magnesium reacts readily with hot water to produce hydrogen gas, whereas iron does not react with cold water or even hot water — it reacts only with steam at high temperatures. Therefore, iron cannot be used in the laboratory to produce hydrogen gas by reacting with cold water.

Source: Chapter 3, Section 3.2.2

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• The equation for iron + steam is a standard board exam question. Ensure the formula Fe₃O₄ (not Fe₂O₃) is written correctly and the equation is balanced: 3 Fe, 4 H₂O, 1 Fe₃O₄, 4 H₂.
• The second part tests your understanding of the reactivity series: Mg > Fe. Iron reacts only with steam (needs high temperature), not cold or hot water, so it cannot produce H₂ conveniently in the lab.
• Examiners award 1 mark for the balanced equation and 2 marks for the explanation comparing reactivity of iron and magnesium with water.

(B) Copper lies below hydrogen in the reactivity series and therefore cannot displace hydrogen from an acid.

The reactivity series determines which metals can displace hydrogen from dilute acids. Only metals placed above hydrogen in the reactivity series (like Zn, Fe, Mg) can displace it. Copper is below hydrogen, so it cannot reduce H⁺ ions to H₂ gas. This is the standard CBSE reason; options A, C, and D are factually incorrect.

(i) Balanced chemical equation:

$$\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$$

(ii) Type of reaction:
This is a displacement reaction, as zinc displaces copper from copper sulphate solution.

(iii) Relative reactivities:
Zinc is more reactive than copper. A more reactive metal can displace a less reactive metal from its salt solution. Since zinc displaces copper from copper sulphate, zinc is higher than copper in the reactivity series.

Source: Chapter 1, Section 1.2.3 Displacement Reaction; Chapter 3, Section 3.2.4

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• The equation (1.25) is directly given in the textbook — write it with state symbols for full marks.
• "Displacement reaction" is the exact NCERT term; do not write "single displacement" or "substitution."
• For part (iii), the key idea is: a more reactive metal displaces a less reactive one. The reactivity series (Table 3.2) places Zn above Cu, confirming this.
• The fading blue colour indicates CuSO₄ is being consumed; the reddish-brown deposit is metallic copper being formed — good details to mention if asked to describe observations.

Reaction with hydrochloric acid (HCl):

$$\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$$

Reaction with sodium hydroxide (NaOH):

$$\text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaAlO}_2 + \text{H}_2\text{O}$$

(Sodium aluminate)

Property demonstrated:
Since aluminium oxide reacts with both acids and bases to produce salt and water, it behaves as both a basic oxide (with HCl) and an acidic oxide (with NaOH). This property is called amphoteric nature. Aluminium oxide is therefore an amphoteric oxide.

Source: Chapter 2, Section 2.1.5 & 2.1.6

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• Examiners expect both balanced equations (1 mark each) and a clear explanation of the amphoteric property (1 mark).
• The key term to use is amphoteric — an oxide that reacts with both acids and bases. This is the exact terminology CBSE expects.
• Note that with NaOH, the product is sodium aluminate (NaAlO₂), not aluminium hydroxide — make sure to write the correct product.
• Don't forget: metallic oxides are basic oxides, but Al₂O₃ is an exception because it also behaves as an acidic oxide, hence amphoteric.

Property of HNO₃: Nitric acid is a strong oxidising agent. It oxidises the hydrogen gas produced to water and itself gets reduced to nitrogen oxides (N₂O, NO, NO₂). Therefore, H₂ is not evolved when metals react with HNO₃.

Exception with Mg and very dilute HNO₃: When the acid is extremely dilute, its oxidising ability is insufficient to oxidise all the hydrogen produced. Magnesium is also highly reactive (above H in activity series), reacting fast enough that H₂ escapes before it can be oxidised. This shows that the oxidising property of HNO₃ operates effectively only at sufficient concentration.

Source: Chapter 3, Section 3.2.3

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• The key term examiners look for is "strong oxidising agent" — this is the direct reason H₂ is not evolved.
• Mention that H₂ is oxidised to water and HNO₃ is reduced to nitrogen oxides.
• For the exception part, link it to very dilute conditions reducing the oxidising power, and high reactivity of Mg. Both parts together earn full marks.
• Don't write lengthy chemistry; one equation is not required here — the explanation in words suffices for 3 marks.

Since X displaces Y from Y's sulphate solution, X is more reactive than Y.
Since Y cannot displace Z from Z's sulphate solution, Y is less reactive than Z.
Since X can displace Z, X is more reactive than Z.

Decreasing order of reactivity: X > Z > Y

Source: Metals and Non-metals, Activity Series concept; Chapter 1, Displacement Reactions

• The key rule: a more reactive metal displaces a less reactive metal from its salt solution (from the activity series concept, Chapter 3).
• Work through each clue logically: X > Y (clue 1), Y < Z (clue 2), X > Z (clue 3).
• This fixes Z between X and Y: X > Z > Y.
• Examiners expect you to state the reasoning for each step, not just the final order — that's where the 2 marks lie.

The rate of reaction of a metal with oxygen depends not only on its reactivity but also on the surface area exposed. Iron filings have a very large surface area, allowing vigorous burning, whereas a solid piece of iron has a much smaller surface area, so it does not burn even on strong heating.

Source: Chapter 3, Section 3.2.1

The textbook explicitly states: "Iron does not burn on heating but iron filings burn vigorously when sprinkled in the flame." This points to surface area as the key additional factor. Examiners want you to name this factor clearly and link it to the observation given. Do not just repeat reactivity — the question asks what is beyond position in the reactivity series.

Electronic configurations:

• Na (11): 2, 8, 1 — has 1 electron in outermost shell
• Mg (12): 2, 8, 2 — has 2 electrons in outermost shell
• Cl (17): 2, 8, 7 — needs 1 electron to complete octet

NaCl: Sodium loses 1 electron to form Na⁺. One Cl atom gains that 1 electron to form Cl⁻. So one Na combines with one Cl → formula NaCl.

MgCl₂: Magnesium loses 2 electrons to form Mg²⁺. Each Cl atom can accept only 1 electron. So two Cl atoms are needed to accept both electrons, forming 2Cl⁻. Thus one Mg combines with two Cl → formula MgCl₂.

$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \quad;\quad 2\text{Cl} + 2e^- \rightarrow 2\text{Cl}^-$$

Source: Chapter 3, Section 3.3

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Examiners look for:

1. Correct electronic configurations of all three elements written explicitly.
2. Clear statement of how many electrons Na and Mg lose (valency).
3. Linking Mg's 2 electrons lost to needing 2 Cl atoms — this is the key reasoning for MgCl₂ vs NaCl.
4. Equations showing ion formation are a bonus that fetch full marks.

Remember: the formula of an ionic compound is determined by balancing the charges of the ions formed, which directly follows from the number of electrons lost/gained based on electronic configuration.

In the solid state, ions in an ionic compound are held in a fixed, rigid lattice structure. Although ions are present, they cannot move freely, so no electrical conduction occurs.

When the compound is melted (molten state), the lattice structure breaks down. The ions (cations and anions) become free to move. These freely moving ions act as charge carriers, allowing electricity to be conducted.

Thus, electrical conduction requires mobile charged particles — solid ionic compounds have ions but they are immobile, while molten ionic compounds have freely moving ions that carry current.

Source: Chapter 3 (ionic compounds), referenced in Chapter 4, Section 4.1

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• The key concept examiners look for: ions must be free to move to conduct electricity.
• In solid state → ions are in fixed positions (lattice) → no conduction.
• In molten state → lattice breaks → ions are mobile → conduction occurs.
• This question links Chapter 3 (ionic compounds) content; the passage in Chapter 4 confirms ionic compounds "conduct electricity in solution or in the molten state."
• Avoid confusing this with covalent compounds, which don't conduct even in molten state because they form no ions.

(B) A large amount of energy is needed to overcome the strong electrostatic forces between oppositely charged ions.

The textbook (Chapter 3, Section 3.3.1) states: "Ionic compounds have high melting and boiling points because a considerable amount of energy is required to break the strong inter-ionic attraction." Options A, C, and D are incorrect — ionic compounds do not consist of molecules, solubility is unrelated to melting point, and ionic bonds are not covalent bonds.

Error in the student's claim:
Aluminium has atomic number 13, so its electronic configuration is 2, 8, 3. It has 3 valence electrons and loses all three to form Al³⁺ (not Al⁺). The student incorrectly assumed aluminium forms a 1+ ion.

Electronic configurations:

• Al (2, 8, 3) → Al³⁺ (2, 8) + 3e⁻
• F (2, 7) + 1e⁻ → F⁻ (2, 8)

Correct formula:
Each Al³⁺ ion loses 3 electrons. Each F⁻ ion gains only 1 electron. Therefore, 3 fluoride ions are needed to accept the 3 electrons lost by one aluminium ion.

$$\text{Al}^{3+} + 3\text{F}^- \rightarrow \text{AlF}_3$$

The correct formula of the ionic compound formed is AlF₃ (aluminium fluoride).

Source: Chapter 3, Section 3.3 — How do Metals and Non-metals React?

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• Examiners award marks for: (1) identifying the specific error (Al forms Al³⁺, not Al⁺), (2) showing both electronic configurations with electron transfer, and (3) deriving the correct formula using charge balance.
• Always write electronic configurations in shell notation (2, 8, 3) and show the ion formed after electron loss/gain.
• The formula follows from charge neutrality: one Al³⁺ needs exactly three F⁻ ions → AlF₃. This mirrors the MgCl₂ example in the textbook.

The fundamental property is reactivity (position in the activity series).

Metals at the bottom of the activity series (like gold and silver) are least reactive, so they do not combine with other elements and are found in the free state. Metals at the top (like sodium and calcium) are highly reactive and readily combine with elements like oxygen, sulphur, etc., so they are never found in the free state in nature.

Source: Chapter 3, Section 3.4.1

• The examiner wants you to name the property (reactivity / position in activity series) and then link it to both examples given in the question.
• Key phrase: "least reactive → free state"; "highly reactive → combined state."
• Avoid vague answers like "chemical property" — be specific: reactivity or position in the activity series.

(i)

• ZnS (zinc sulphide): The process used is roasting — heating strongly in excess air.

$$2\text{ZnS(s)} + 3\text{O}_2\text{(g)} \xrightarrow{\text{Heat}} 2\text{ZnO(s)} + 2\text{SO}_2\text{(g)}$$

• ZnCO₃ (zinc carbonate): The process used is calcination — heating strongly in limited air.

$$\text{ZnCO}_3\text{(s)} \xrightarrow{\text{Heat}} \text{ZnO(s)} + \text{CO}_2\text{(g)}$$

(ii) It is easier to obtain a metal from its oxide than from its sulphide or carbonate, because carbon can readily reduce metal oxides. If sulphides or carbonates were reduced directly, the reaction would be less efficient and carbon cannot displace the metal as easily from these compounds.

Source: Chapter 3, Section 3.4.4

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• Examiners award 1 mark each for naming roasting and calcination correctly (with equations if space allows), and 1 mark for the reason.
• Key distinction: roasting = excess air; calcination = limited/no air. Don't mix these up.
• The reason must state that metal oxides are easier to reduce (with carbon) compared to sulphides/carbonates — this is the textbook's exact reasoning. Vague answers like "to purify the ore" will not score.

Carbon cannot reduce the oxides of sodium and aluminium because these metals have more affinity for oxygen than carbon. Since they are higher than carbon in the reactivity series, carbon cannot displace them from their oxides.

Extraction of Sodium: Sodium is extracted by electrolytic reduction of molten sodium chloride (NaCl).

• At cathode: Na⁺ + e⁻ → Na
• At anode: 2Cl⁻ → Cl₂ + 2e⁻

Extraction of Aluminium: Aluminium is extracted by electrolytic reduction of molten aluminium oxide (Al₂O₃).

Source: Chapter 3, Section 3.4.5

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• The key reason is relative affinity for oxygen: Na and Al bind oxygen more strongly than carbon does, so carbon cannot act as a reducing agent for them.
• The examiner expects you to name electrolysis/electrolytic reduction as the method and mention the molten compound used (NaCl for Na; Al₂O₃ for Al).
• Writing the electrode reactions for sodium earns extra credit and shows depth — include them if marks allow.
• Do not confuse electrolytic refining (purification) with electrolytic reduction (extraction). These are different processes.

Answer: B — Aluminium is more reactive than iron and has a greater affinity for oxygen than carbon does at the temperatures involved.

The thermit reaction is: Fe₂O₃ + 2Al → Al₂O₃ + 2Fe. Aluminium is the reducing agent because it is more reactive than iron and displaces it. At the high temperatures of this reaction, aluminium has a greater affinity for oxygen than carbon, so carbon cannot reduce Fe₂O₃ effectively here. The textbook (Chapter 1, Exercise Q2) classifies this as a displacement reaction. Examiners expect you to correctly name aluminium as the reducing agent and justify it by reactivity/oxygen affinity — not by cost or melting point.

At the Anode (impure copper): The impure copper dissolves into the electrolyte (acidified copper sulphate solution). Copper atoms lose electrons and go into solution as Cu²⁺ ions.

At the Cathode (pure copper strip): Cu²⁺ ions from the electrolyte gain electrons and get deposited as pure copper.

Why insoluble impurities don't deposit on cathode: The impurities present in impure copper (anode) that are insoluble do not dissolve into the electrolyte. They simply fall to the bottom of the anode and collect as anode mud. Since they never enter the solution as ions, they cannot be deposited at the cathode. Only copper ions migrate and deposit on the cathode.

Source: Chapter 3, Section 3.4.6 – Refining of Metals

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• Examiners expect you to clearly name both electrodes and state what happens at each — dissolution at anode, deposition at cathode.
• The key point for the second part is the word "insoluble" — impurities that don't dissolve can't form ions, so they can't migrate to the cathode. They settle as anode mud (this term is expected).
• Avoid vague statements like "they are heavy" — the correct reason is they don't dissolve into the electrolyte.

Observation:
Rusting will be observed only in Test Tube A (ordinary water open to air). No rusting will occur in Test Tubes B and C.

Reasons:

• Test Tube A: Both air (oxygen) and water are present → rusting occurs.
• Test Tube B: Boiled distilled water removes dissolved air; oil layer prevents air from entering → only water present, no rusting.
• Test Tube C: Desiccant (anhydrous CaCl₂) absorbs all moisture → only dry air present, no rusting.

Conditions necessary for rusting:
The presence of both oxygen (from air) and water simultaneously is essential for iron to rust.

Source: Chapter 3, Section 3.5

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• The examiner expects you to name the correct test tube (A) and give a reason for each of the three tubes — that's the core of the answer.
• Key phrase: "both air and water are necessary" — this must appear explicitly.
• Mentioning the role of boiled water (removes dissolved air) and oil layer shows deeper understanding and secures full marks.
• Anhydrous CaCl₂ as desiccant removes moisture — state this clearly for Test Tube C.

Zinc is more reactive than iron in the reactivity series. When the zinc coating is scratched, both zinc and iron are exposed to moisture and air. Since zinc is more reactive, it preferentially reacts with oxygen and moisture instead of iron. Zinc acts as a sacrificial metal, corroding itself while protecting the iron beneath. Thus, iron does not rust until all the zinc in the surrounding area is consumed.

Source: Chapter 3, Section 3.5.1 – Prevention of Corrosion

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• The key concept examiners want is reactivity series: zinc is higher (more reactive) than iron.
• The term sacrificial protection (or sacrificial metal) is expected — it often fetches a dedicated mark.
• Avoid just saying "zinc coats iron"; the question specifically asks about what happens after the coating breaks, so you must explain the electrochemical/reactivity reasoning.
• Three marks map to roughly three points: (1) zinc is more reactive than iron, (2) zinc reacts preferentially with oxygen/moisture, (3) iron is thus protected until zinc is fully consumed.

(B) Carbon and chromium — to increase hardness and resistance to corrosion

When iron is mixed with nickel and chromium, we get stainless steel, which is hard and does not rust.

The textbook (Section 3.5) explicitly states: "if it is mixed with a small amount of carbon (~0.05%), it becomes hard and strong. When iron is mixed with nickel and chromium, we get stainless steel, which is hard and does not rust." Option (B) is the closest correct match — carbon increases hardness and chromium provides corrosion resistance. Examiners expect you to link the alloying elements directly to the properties they impart.

The flaw in the student's reasoning is ignoring the protective oxide layer that forms on aluminium's surface.

Although aluminium is highly reactive, when exposed to air it reacts with oxygen to form a thin, tough layer of aluminium oxide (Al₂O₃):

$$4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$$

This oxide layer is stable and strongly adherent. It prevents further reaction of the underlying aluminium with air or moisture, thereby protecting it from corrosion. Because of this self-protective behaviour, aluminium does not corrode rapidly and is safe and durable for making cooking utensils.

Source: Chapter 3, Section 3.2.1

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• The key concept examiners look for is the protective oxide layer — name it, explain it forms immediately on exposure to air, and state it prevents further corrosion.
• Writing the balanced equation for aluminium oxide formation earns a mark and shows chemical understanding.
• Avoid just saying "aluminium doesn't corrode" — you must explain why (the oxide coat acts as a barrier). That reasoning is the heart of the answer.
• The word "amphoteric" is not required here; stay focused on the corrosion-resistance aspect.

Zinc is a metal of medium reactivity (middle of the activity series). Carbon can reduce zinc oxide because carbon has a greater affinity for oxygen than zinc at high temperatures:

$$\text{ZnO(s) + C(s)} \rightarrow \text{Zn(s) + CO(g)}$$

Sodium, however, is a highly reactive metal (top of the activity series). It has a much greater affinity for oxygen than carbon does, so carbon cannot reduce sodium oxide. Therefore, carbon reduction fails for sodium.

Instead, sodium is extracted by electrolytic reduction — electrolysis of molten sodium chloride, where sodium is deposited at the cathode:

$$\text{Na}^+ + e^- \rightarrow \text{Na}$$

The distinguishing property is reactivity (position in the activity series).

Source: Chapter 3, Sections 3.4.4 and 3.4.5

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Examiners look for three things: (1) identifying that zinc is mid-series and can be reduced by carbon, with the equation; (2) explaining why carbon fails for sodium — sodium's greater affinity for oxygen; (3) naming electrolysis as sodium's extraction method. Linking everything to "position in the activity series / reactivity" earns the reasoning mark. Don't just say "sodium is more reactive" — explain what that means for the reduction process.

(i) This observation illustrates that alloys generally have lower electrical (and thermal) conductivity than their pure parent metals. The irregular arrangement of different atoms in an alloy disrupts the free flow of electrons, thereby reducing conductivity.

(ii) One advantage of using brass (copper-zinc alloy) over pure copper for household items is that brass is harder and stronger than pure copper, making it more durable and resistant to wear and deformation in everyday use.

Source: Metals and Non-metals, Chapter 3 (alloy definition and properties)

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• Part (i) is worth ~1 mark: state the rule clearly — alloys have lower conductivity than pure constituent metals. Mention why briefly (disruption of electron flow) if you want to secure the mark fully.
• Part (ii) is worth ~2 marks: the key advantage examiners expect is greater hardness/strength/durability. Avoid repeating conductivity (that's the disadvantage given in the question). The textbook notes alloys are made to improve mechanical properties like hardness.
• Keep answers concise; one clear sentence per part is sufficient for the marks allocated.

Zinc is a moderately reactive metal found mainly as zinc sulphide (ZnS) or zinc carbonate (ZnCO₃) in nature.

Why conversion is needed: It is easier to reduce a metal oxide than a sulphide or carbonate. So the ore must first be converted to zinc oxide (ZnO).

Two conversion processes:

• Roasting – ZnS is heated strongly in excess air, converting it to ZnO:

$2\text{ZnS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{ZnO} + 2\text{SO}_2$

• Calcination – ZnCO₃ is heated strongly in limited air, converting it to ZnO:

$\text{ZnCO}_3 \xrightarrow{\Delta} \text{ZnO} + \text{CO}_2$

Reduction: ZnO is heated with carbon (coke), which acts as the reducing agent:
$\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}$

Purification: The impure zinc is refined by electrolytic refining. Impure zinc is made the anode, pure zinc the cathode, and zinc salt solution the electrolyte. On passing current, pure zinc deposits on the cathode; insoluble impurities collect as anode mud.

Source: Metals and Non-metals, Section 3.4.4 and 3.4.6

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• The question has four clearly labelled parts — examiners will award marks for each, so don't skip any.
• The key distinction between roasting (excess air) and calcination (limited air) is a favourite examiner point — get those details right.
• Writing the chemical equations earns marks and shows precision; keep them balanced.
• For electrolytic refining, remember the three essentials: impure metal = anode, pure metal = cathode, metal salt solution = electrolyte. Mentioning "anode mud" shows complete knowledge.
• Don't confuse calcination (for carbonates) with roasting (for sulphides) — a common error.

The fundamental tendency is that metals lose electrons easily to form positive ions (cations). This is driven by their need to attain a stable electronic configuration.

In ionic compound formation (e.g., NaCl): Sodium loses one electron to chlorine, forming Na⁺ and Cl⁻ ions. The oppositely charged ions attract each other, forming the ionic compound NaCl.

In electrolytic extraction of reactive metals: Highly reactive metals like sodium or magnesium have such strong affinity for oxygen that carbon cannot reduce their compounds. Electrolysis of molten chlorides is used; the metal cations (Na⁺) gain electrons at the cathode and are deposited as pure metal: Na⁺ + e⁻ → Na.

Thus, the same tendency — losing electrons to form cations — leads to compound formation in nature and enables isolation through forced electron gain during electrolysis.

Source: Chapter 3, Sections 3.3 and 3.4.5

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• The examiner wants you to name the tendency (metals lose electrons / form positive ions) and then link it explicitly to both processes.
• For NaCl: electron transfer → ionic bond. For electrolysis: metal ions already exist in molten ore; cathode supplies electrons to reduce them back to metal.
• The contrast is key: in NaCl formation the metal gives away electrons; in electrolysis, those same cations are given back electrons artificially — both driven by the metal's tendency to exist as cations.
• Always write the cathode half-reaction when discussing electrolytic extraction; it fetches a dedicated mark.

Metal oxides vs. Non-metal oxides in water:
Metals combine with oxygen to form basic oxides, which dissolve in water to give bases (alkalis).
Example: $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$ (calcium hydroxide — a base)

Non-metals form acidic oxides, which dissolve in water to give acids.
Example: $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ (sulphuric acid)

Give-and-take process (ionic bonding):
Metals have 1–3 electrons in their outermost shell, which they tend to lose, forming positive ions. Non-metals have 5–7 electrons and tend to gain electrons to complete their octet, forming negative ions.
For example, sodium (2,8,1) loses one electron to chlorine (2,8,7), forming Na⁺ and Cl⁻. The oppositely charged ions attract each other, producing an ionic (electrovalent) compound — sodium chloride (NaCl).

Source: Chapter 3 — Metals and Non-metals, Section 3.3

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• Examiners expect you to name the type of oxide (basic/acidic), give a chemical equation as the example, and name the product.
• For the give-and-take part: mention electron loss (metal → cation) and electron gain (non-metal → anion) with a specific example (Na + Cl is standard from the textbook), and name the resulting compound type as ionic/electrovalent.
• Do not forget to write the resulting compound name — that is usually the last mark point.