(i) Element X has atomic number 11 (Sodium, Na).
Electronic configuration: 2, 8, 1
Valency = 1 (it has 1 electron in the outermost shell, which it loses to attain a stable octet).

(ii) Sodium forms the oxide Na₂O (sodium oxide).
Its nature is basic (metals combine with oxygen to form basic oxides).

---

• For (i), write the shell-wise distribution (K=2, L=8, M=1) and state valency as the number of electrons lost/gained to complete the octet.
• For (ii), since Na is monovalent (Na⁺) and oxygen is divalent (O²⁻), the formula is Na₂O. Examiners expect the word "basic" as the nature — remember: metal oxides are basic oxides (except amphoteric ones like Al₂O₃, ZnO).

(a) A metal that reacts vigorously with both air and water belongs to the high reactivity group. The process used is electrolysis of its molten ore.

(b) Aluminium is a highly reactive metal, more reactive than carbon. Therefore, carbon cannot displace aluminium from its oxide — carbon is not a strong enough reducing agent. Electrolysis is used instead.

(c) Cinnabar (HgS) is first roasted in air to convert it into mercuric oxide (HgO). On further heating, HgO decomposes to give mercury metal:

$$\text{2HgS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{HgO} + 2\text{SO}_2$$

$$2\text{HgO} \xrightarrow{\Delta} 2\text{Hg} + \text{O}_2$$

Source: Metals and Non-metals, Extraction of Metals section

---

• (a) Points directly to high-reactivity metals → electrolysis, as stated in the passage.
• (b) Standard NCERT reasoning: Al is higher than C in the reactivity series, so C cannot reduce Al₂O₃.
• (c) Examiners expect both equations clearly written. Cinnabar = HgS; the two-step process (roasting → thermal decomposition) is the key. Mercury is a low-reactivity metal, consistent with the passage.

(i) Gas 'G' is Hydrogen (H₂).

(ii) Bring a burning matchstick near the mouth of the test tube. If the gas burns with a 'pop' sound, it confirms the presence of hydrogen gas.

(iii) Metal 'M' is Zinc (Zn) (reacts with both dilute acid and base to liberate H₂).

(1) Reaction with acid (dilute H₂SO₄):

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$

(2) Reaction with base (NaOH):

$$\text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\uparrow$$

Source: Metals and Non-metals, Chapter 3, Section 3.2.3

---

• The key clue is that metal M reacts with both dilute acid and a base to give H₂ — this is a property of amphoteric metals like Zinc (or Aluminium). Zinc is the standard NCERT example.
• The pop test is the standard CBSE test for hydrogen gas — must be stated exactly this way.
• In the base reaction, the salt formed is sodium zincate (Na₂ZnO₂) — a negative ion composed of the metal and oxygen, as described in the chapter.
• Examiners award 1 mark each for (i) name of gas, (ii) test, and (iii) two equations. Write both equations balanced with state symbols (↑ for gas) for full credit.

(d) Copper and Tin

Bronze is an alloy of copper (Cu) and tin (Sn).

Source: Metals and Non-metals, Section 3.5 (Corrosion/Alloys)

---

The textbook explicitly states: "bronze, an alloy of copper and tin (Cu and Sn)." Option (a) describes brass (copper + zinc), so students must not confuse the two. Option (c) is a common distractor — bronze contains only Cu and Sn, not zinc. Remembering the pair Brass = Cu+Zn, Bronze = Cu+Sn is a frequent exam tip.

(b) II and III

ZnO and Al₂O₃ are amphoteric oxides — they react with both acids and bases. MgO and CaO are purely basic oxides that react only with acids.

The key term here is amphoteric oxide — an oxide that reacts with both acids and bases. The textbook explicitly names only zinc oxide (ZnO) and aluminium oxide (Al₂O₃) as examples. MgO and CaO are basic oxides (their metals are high in the activity series and form strongly basic compounds). Examiners expect you to recall these two specific examples from the chapter.

(a) When copper is heated in air, it reacts with oxygen to form copper(II) oxide (a black coating).

$$2Cu_{(s)} + O_{2(g)} \xrightarrow{\Delta} 2CuO_{(s)}$$

(b) Some metal oxides are amphoteric because they react with both acids and bases. Example: Aluminium oxide (Al₂O₃).

(c)
(i)
$$Na_2O_{(s)} + H_2O_{(l)} \rightarrow 2NaOH_{(aq)}$$

(ii)
$$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$
(Sodium aluminate + water)

Source: Metals and Non-metals, Reaction of Metals with Oxygen

---

• (a): The black coating on copper is CuO — mention this observation for full marks.
• (b): "Amphoteric" = reacts with both acid and base. Al₂O₃ is the standard CBSE example; ZnO is also acceptable.
• (c)(i): Na₂O is a basic oxide; it dissolves in water to give NaOH — balance correctly.
• (c)(ii): Al₂O₃ acting as an acid reacting with NaOH gives sodium aluminate (NaAlO₂). Product names earn marks.

(b) Al and Al₂O₃

Aluminium (Al) is used in the thermite process. When burnt in air, it forms aluminium oxide (Al₂O₃), which is an amphoteric oxide — it reacts with both acids and bases.

Key facts to remember: (1) Thermite process uses aluminium powder mixed with iron oxide. (2) Al₂O₃ is the classic example of an amphoteric oxide in CBSE Class 10. Option (d) is wrong because Al₃O₄ does not exist; the correct formula is Al₂O₃.

Source: Chapter 3, Section 3.2.1

(i) Silver turns black when exposed to air. This is because silver reacts with sulphur present in the air to form a black coating of silver sulphide (Ag₂S).

(ii) Copper loses its shiny brown surface and gains a green coating. This is because copper reacts with moist carbon dioxide in the air to form basic copper carbonate (green).

(iii) Iron develops a brown flaky coating called rust when exposed to moist air. Iron reacts with oxygen and moisture to form hydrated iron oxide (rust, Fe₂O₃·xH₂O).

Source: Chapter 3, Section 3.5 Corrosion; Chapter 1, Section 1.3.1

---

• Each metal = 1 mark: change in appearance + reason (chemical cause).
• Key terms examiners expect: silver sulphide (black), basic copper carbonate (green), rust/hydrated iron oxide (brown flaky).
• Don't just say "reacts with air" — name the specific substance (sulphur, CO₂ + moisture, O₂ + moisture) and the product formed. That's where the mark lies.

(i) Mercury – Mercury is low in the activity series. HgS (cinnabar) is first roasted in air to form HgO, which on further heating is reduced to mercury by thermal decomposition (heating alone):
$$2\text{HgO} \xrightarrow{\Delta} 2\text{Hg} + \text{O}_2$$

(ii) Copper – Copper is also low in the activity series. Cu₂S is heated in air; the Cu₂O formed reacts with remaining Cu₂S to give copper by thermal reduction (heating alone):
$$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \xrightarrow{\Delta} 6\text{Cu} + \text{SO}_2$$

(iii) Sodium – Sodium is at the top of the activity series and is highly reactive. Carbon cannot reduce its oxide. So sodium is obtained by electrolytic reduction of molten NaCl:
$$\text{Na}^+ + e^- \rightarrow \text{Na (at cathode)}$$

Source: Chapter 3, Sections 3.4.3 and 3.4.5

---

• The key to this question is linking reactivity position to extraction method: low reactivity → heating alone; high reactivity → electrolysis.
• Always state the reason (position in activity series) before the method — examiners look for it explicitly.
• Writing the chemical equation earns marks and shows precision; include it for each metal.
• Sodium cannot be reduced by carbon because it has greater affinity for oxygen than carbon does — this is the reason the examiner wants.

Nitric acid is a strong oxidising agent. When metals react with nitric acid, the hydrogen gas produced is immediately oxidised to water by the nitric acid itself. Instead of hydrogen, oxides of nitrogen (such as NO₂, NO) are evolved as by-products. Thus, hydrogen is generally not liberated in reactions of metals with nitric acid.

Source: Metals and Non-metals, Chapter 3

---

• The key reason is the oxidising nature of HNO₃ — examiners specifically look for this phrase.
• Mention that H₂ is oxidised to H₂O before it can escape, and that nitrogen oxides (NO₂ or NO) are released instead.
• This is a common 2-mark reason-based question; one clear explanation with the oxidising agent point scores full marks. Avoid writing a long paragraph — two to three crisp sentences are enough.

Sodium is a highly reactive metal. It reacts vigorously with oxygen in air and also reacts violently with moisture/water, producing heat and hydrogen gas which can catch fire:

$$2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)} + \text{heat}$$

Kerosene oil does not react with sodium. It covers the metal completely, preventing contact with atmospheric oxygen and moisture. This protects sodium from accidental fires and unwanted reactions.

Source: Chapter 3, Section 3.2.1 and 3.2.2

---

The examiner expects two key points: (1) sodium's high reactivity with both air (oxygen) and water, and (2) kerosene acts as an inert barrier preventing this contact. Writing the reaction with water earns full credit for a 2-mark question. Do not just say "it is reactive" — explain what it reacts with and why kerosene works (it is inert/does not react with sodium).

Zinc oxide is amphoteric because it shows the properties of both a basic oxide and an acidic oxide.

• It reacts with acids (behaving as a base):

ZnO + H₂SO₄ → ZnSO₄ + H₂O

• It reacts with bases (behaving as an acid):

ZnO + 2NaOH → Na₂ZnO₂ + H₂O

Since it neutralises both acids and bases, it is called an amphoteric oxide.

Source: Metals and Non-metals, Chapter 3

---

• The examiner wants both reactions (with acid and with base) to award full marks. Simply stating "reacts with both acids and bases" without equations may cost marks in a 2–3 mark question.
• The key term to use is amphoteric — defined as an oxide that behaves as both acidic and basic.
• Aluminium oxide (Al₂O₃) is the other standard example of an amphoteric oxide from the same chapter.

Option (C): (A) is true, but (R) is false.

Assertion is true — Zinc displaces copper from copper sulphate solution, forming a reddish-brown copper coating: $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$. But Reason is false — Zinc is more reactive than copper, not less.

Source: Chemical Reactions and Equations, Section 1.2.3; Metals and Non-metals, Section 3.2.5

---

• The Assertion describes a displacement reaction — Zinc (being more reactive) displaces copper from CuSO₄, depositing reddish-brown copper on the zinc piece. This is true.
• The Reason states copper is more reactive than zinc — this is false. The reactivity series clearly places Zn above Cu, meaning Zn is more reactive.
• Since A is true but R is false, the answer is (C).
• Examiners expect you to state both the correct option and a brief justification referencing the reactivity series.

Labelled Diagram — Electrolytic Refining of Copper:

```
Battery
+ –
| |
Anode Cathode
(Impure (Pure
Copper) Copper)
| |
___________________
| |
| CuSO₄ solution |
|___________________|
(Electrolyte)
```

(Diagram should show: battery connected to two copper electrodes dipped in copper sulphate (CuSO₄) solution, with anode labelled "impure copper" and cathode labelled "pure copper".)

What happens when electric current is passed:

When electric current is passed, the impure copper anode slowly dissolves into the CuSO₄ solution. Cu²⁺ ions from the solution get deposited as pure copper on the cathode. Insoluble impurities (like silver, gold) collect below the anode as anode mud.

• 1 mark for the correctly labelled diagram (anode = impure Cu, cathode = pure Cu, electrolyte = CuSO₄ solution, battery shown).
• 1 mark for stating anode dissolves / copper ions migrate to cathode.
• 1 mark for mentioning pure copper deposits at cathode AND impurities settle as anode mud.
• Examiners expect the term "anode mud" to be written explicitly — it fetches a mark.
• The diagram need not be elaborate; clear labels are what matter.

Calcium floats on water because the hydrogen gas ($H_2$) produced during the reaction lifts it to the surface. Also, calcium is less dense than water. The heat evolved is not enough to ignite the gas, so it just bubbles and causes calcium to float.

Reaction of calcium with water:

$$Ca(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + H_2(g)$$

The product formed is calcium hydroxide $[Ca(OH)_2]$. When its aqueous solution reacts with carbon dioxide, calcium carbonate is formed:

$$Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l)$$

Source: Chapter 1, Chemical Reactions and Equations

---

• Floating reason: Two points are expected — hydrogen gas evolution (buoyancy effect) and low density of calcium. Many students write only one; mention both for full marks.
• The first equation is not directly in the passage (Ca + H₂O is standard Class 10 chemistry); the passage gives the CaO + H₂O reaction, but the Ca(OH)₂ + CO₂ equation (1.14) is directly from the text — make sure to reproduce it exactly.
• Note the state symbols: $CaCO_3$ is a precipitate $(s)$, which is important.
• This is a 3-mark question: one mark for the explanation of floating, one for each balanced equation.

(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Both statements are independently correct facts about the reactivity series, but the position of a metal in the reactivity series explains extraction methods, not the difference in reactivity with water and acids.

• (A) is true: the reactivity series itself shows metals differ in reactivity with water and dilute acids.
• (R) is true: metals high in the series need electrolysis, middle ones need reduction, low ones are found free.
• However, (R) talks about extraction, which is a separate application of the reactivity series and does not explain why metals react differently with water/acids. Hence (B) is correct.

(D) (b) and (d)

In (b) Ca is more reactive than Cu, so Ca displaces Cu from CuSO₄. In (d) Mg is more reactive than Zn, so Mg displaces Zn from ZnSO₄. In (a) Al cannot displace Ca (Ca is more reactive), and in (c) Cu cannot displace Fe (Fe is more reactive than Cu).

This question tests the reactivity series (displacement reactions). A metal can displace another metal from its salt solution only if it is more reactive than the metal in the salt. The reactivity order is: Mg > Ca > Al > Zn > Fe > Cu. Students often confuse Ca and Al positions — remember Ca is above Al in the series, so Al cannot displace Ca. Always check the reactivity series before predicting if a displacement reaction occurs.

(B) Al₂O₃

Al₂O₃ is an amphoteric oxide — it reacts with HCl to form AlCl₃ + H₂O, and with KOH to form KAlO₂ + H₂O. CuO, Na₂O, and K₂O are basic oxides; they react with acids only.

The key concept is amphoteric oxides — oxides that react with both acids and bases. Al₂O₃ is the classic example at the Class 10 level. CuO, Na₂O, and K₂O are purely basic oxides, so they react with HCl but not with KOH. Examiners expect you to recall that amphoteric oxides show dual behaviour.

(i) In the electrolytic refining of copper:

• Anode — impure copper
• Cathode — pure (thin strip of) copper

(ii) The solution used is acidified copper sulphate solution.
Formula: CuSO₄

(iii) When electric current is passed:

• The impure copper anode slowly dissolves into the solution, releasing Cu²⁺ ions.
• These Cu²⁺ ions migrate to the cathode and deposit as pure copper.
• The insoluble impurities (such as gold, silver) from the anode settle at the bottom as anode mud.
• Thus, pure copper gets collected at the cathode.

Source: Chapter 3 – Metals and Non-metals, Electrolytic Refining

---

• (i) is a direct recall fact — anode = impure metal, cathode = pure metal. Don't reverse them.
• (ii) Examiners expect both the name AND formula; writing only one loses the mark.
• (iii) For 2 marks, cover two key steps: anode dissolving + cathode deposition. Mentioning anode mud is a bonus point that shows thorough knowledge.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Nitric acid is a strong oxidising agent; it oxidises the H₂ produced to water and itself gets reduced, so no hydrogen gas is evolved.

The key fact is that HNO₃ acts as an oxidising agent — it immediately oxidises any nascent hydrogen to water and is itself reduced (to oxides of nitrogen). This directly explains why H₂ is not evolved when Zn reacts with HNO₃, making R the correct and complete explanation of A. Choose option (a).

(c) amphoteric

Oxides of aluminium and zinc react with both acids and bases to produce salt and water. Such oxides are called amphoteric oxides.

Source: Chapter 3, Section 3.2.1

---

The key term here is amphoteric — meaning the oxide behaves as both acidic and basic. The textbook explicitly states: "aluminium oxide and zinc oxide show the properties of both basic as well as acidic oxides — these are known as amphoteric oxides." Students must not confuse amphoteric with basic (most metal oxides are basic, but Al₂O₃ and ZnO are exceptions). Remembering Al and Zn as the two standard examples of amphoteric oxides is essential for MCQs and short-answer questions.

(c) Mercury and Bromine

Mercury is a metal that exists as a liquid at room temperature. Bromine is a non-metal that exists as a liquid at room temperature.

The textbook explicitly states: "Metals are solids at room temperature, except mercury which is a liquid" and "The non-metals are either solids or gases except bromine which is a liquid." These two facts are standard exam points from Chapter 3. Remember: mercury = liquid metal; bromine = liquid non-metal. Options (a) and (d) are reversed, and (b) incorrectly lists iodine (a solid non-metal) instead of bromine.

(I) When a metal is converted to an alloy, its electrical conductivity decreases and its melting point also decreases compared to the pure metal.

(II) Solder is used for welding two wires together in an electric circuit.
Major constituents: Lead (Pb) and Tin (Sn) (roughly 1:2 ratio).

(III) (Note: Sub-parts (a) and (b) are not printed in the question. A student should answer whichever option is given in their exam paper. No option text is available here to answer.)

Source: Metals and Non-metals, Alloys section

---

• For (I), remember the two key changes in properties on alloy formation: electrical conductivity decreases (irregular atomic arrangement resists electron flow) and melting point decreases (e.g., solder melts easily — this is why it's useful for welding wires).
• For (II), solder is the classic CBSE example for electrical welding; always mention both constituents (Pb + Sn) to secure full marks.
• For (III), since the (a)/(b) options were not included in the question as typed, the student must refer to their actual paper and answer accordingly. Common related topics include: amalgams (Hg alloys), steel/stainless steel, or why iron articles are coated with zinc (galvanisation).

(a) Formation of Aluminium Nitride (AlN):

Electronic configuration:

• Al (At. no. 13): 2, 8, 3 → has 3 valence electrons
• N (At. no. 7): 2, 5 → needs 3 electrons to complete octet

Al loses 3 electrons to form Al³⁺:
$$\text{Al} \rightarrow \text{Al}^{3+} + 3e^-$$
$$2,8,3 \quad\quad 2,8$$

N gains 3 electrons to form N³⁻:
$$\text{N} + 3e^- \rightarrow \text{N}^{3-}$$
$$2,5 \quad\quad\quad 2,8$$

Al³⁺ and N³⁻ attract each other to form AlN.

---

(b) Reason for brittleness of ionic compounds:

Ionic compounds consist of a regular arrangement of oppositely charged ions held together by strong electrostatic forces. When pressure is applied, the layers of ions shift slightly. Ions of the same charge come opposite each other, causing strong repulsion between like charges. This repulsion breaks the crystal apart, making ionic compounds brittle.

Source: Chapter 3, Section 3.3.1

---

• (a) Always show electronic configuration, electron loss/gain equations, and name the ions formed — examiners award marks for each step. [2 marks]
• (b) The key concept is the shift of ionic layers causing like-charge repulsion, not just "strong forces." Simply saying "strong forces make it hard" is incomplete — explain why it breaks. [1 mark]
• Do not confuse hardness (due to strong attraction) with brittleness (due to repulsion when layers shift).

Tube A (Moist Air + Metal M): Metal M will corrode/rust. Both moisture and air (oxygen) are present, which are the two necessary conditions for corrosion.

Tube B (Moist Air + Oil + Metal M): Metal M will not corrode. The layer of oil prevents air (oxygen) from reaching the metal surface, so corrosion does not occur.

Tube C (Dry Air + Drying Agent + Metal M): Metal M will not corrode. The drying agent absorbs all moisture, so only dry air is present. Absence of moisture prevents corrosion.

Conclusion regarding galvanisation: Since metal M is protected by galvanisation, it is iron/steel. Galvanisation involves coating iron with a thin layer of zinc, which prevents contact with moisture and air, thus preventing corrosion even if the coating is slightly damaged.

Source: Chapter 3 – Metals and Non-metals, Section 3.5 Corrosion (Activity 3.14) and Section 3.5.1 Prevention of Corrosion

---

• The key concept is that both moisture and oxygen are required simultaneously for corrosion (rusting) to occur.
• In Tube B, oil blocks dissolved air; in Tube C, the drying agent removes moisture — either condition missing stops corrosion.
• The mention of galvanisation is a clue that metal M is iron, since galvanisation specifically protects iron/steel with a zinc coating.
• Examiners expect observations for all three tubes with reasons — never skip any. Award marks are split one per tube.

Option A — Impure copper (anode), pure copper (cathode), acidified copper sulphate solution (electrolyte).

Source: Metals and Non-metals, Section 3.4.6 Refining of Metals

The textbook clearly states: "the impure metal is made the anode and a thin strip of pure metal is made the cathode. A solution of the metal salt is used as an electrolyte." For copper, the electrolyte is acidified copper sulphate solution. Options B and C wrongly swap anode/cathode; Option D uses distilled water, which does not conduct electricity.

Option C: Zinc and Hydrogen

Zinc reacts with NaOH to give Na₂ZnO₂ + H₂, and with dilute H₂SO₄ to give ZnSO₄ + H₂. Thus X = Zinc, G = Hydrogen.

Zinc is an amphoteric metal — it reacts with both acids and bases to release hydrogen gas. Copper does not react with NaOH or dilute H₂SO₄, so options A and D are eliminated. Sulphur dioxide is not produced in these reactions, eliminating B. Examiners expect you to recall the reaction: 2NaOH + Zn → Na₂ZnO₂ + H₂↑, as given in Activity 2.4 of Chapter 2.

Option D — Zinc

Zinc is more reactive than iron but less reactive than aluminium in the activity series, so it displaces Fe from FeSO₄ but cannot displace Al from Al₂SO₄.

The activity series order is: Al > Zn > Fe > Cu. A metal displaces only those metals that are below it in the series. Zinc is below aluminium (cannot displace it) but above iron (can displace it). Magnesium is above both, so it would displace both. Copper and Lead are below iron, so they cannot displace iron either. Zinc is the only option that fits both conditions in the question.

Activity (Activity 3.5):

Materials required: Aluminium or copper wire, a stand, wax, a pin, spirit lamp/candle.

Procedure:

1. Clamp an aluminium or copper wire on a stand.
2. Fix a pin to the free end of the wire using wax.
3. Heat the wire with a spirit lamp near the clamped end.

Observation: After some time, the wax melts and the pin falls. However, the metal wire does not melt.

Conclusion: Heat travels from the heated end to the free end, showing metals are good conductors of heat. The wire remaining intact (not melting) shows metals have high melting points.

Source: Chapter 3, Activity 3.5, Section 3.1.1

---

• Examiners expect: materials, procedure (steps), observation, and conclusion — all four components, even in 3 marks.
• The key two conclusions must both appear: (i) good conductor of heat, (ii) high melting point.
• The pin/wax trick is the standard NCERT method — describe it exactly.
• Don't write extra theory; keep it crisp and activity-focused.

Zinc (as ZnS or ZnCO₃) is extracted by two processes — roasting/calcination (to convert ore to oxide) and reduction (to obtain metal).

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

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

Reduction (ZnO reduced by carbon):
$$\text{ZnO(s)} + \text{C(s)} \rightarrow \text{Zn(s)} + \text{CO(g)}$$

Source: Chapter 3, Section 3.4.4

---

• The question asks for processes named + equations, so name all three (roasting, calcination, reduction) and give all three equations.
• Zinc is a medium-reactivity metal, so carbon-reduction is used (not electrolysis).
• Examiners expect correct state symbols and balanced equations — these are easy marks, don't skip them.
• For 2 marks: naming the processes (½ + ½) + key equations (1 mark) is the likely split. Keep it concise but complete.

Mg has electronic configuration 2, 8, 2. It loses 2 electrons to form Mg²⁺.
Cl has electronic configuration 2, 8, 7. Each Cl atom gains 1 electron to form Cl⁻.

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

$$\text{Cl} + e^- \rightarrow \text{Cl}^-$$
$$(2,8,7) \quad (2,8,8)$$

Since Mg releases 2 electrons and each Cl needs 1, two Cl atoms are required → MgCl₂ is formed.

• Cation: Magnesium ion (Mg²⁺)
• Anion: Chloride ion (Cl⁻)

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

---

• Examiners expect the electron transfer equations with electronic configurations written below each species — this is the standard format shown in the textbook (Fig. 3.6).
• Clearly state that two Cl atoms are needed because Mg loses two electrons.
• Name both the cation and anion explicitly — easy 1 mark if not missed.
• Avoid drawing dot-cross diagrams unless specifically asked; the transfer equations are sufficient here.

(b) ductility

The ability of metals to be drawn into thin wires is called ductility.

Source: Chapter 3, Section 3.1.1

The question tests knowledge of physical properties of metals. "Malleability" refers to being beaten into thin sheets, not wires. "Ductility" is specifically defined in the textbook as the property of being drawn into thin wires. Examiners expect the exact term as stated in the NCERT text.

(a) An ore of Mercury is Cinnabar (HgS). Mercury is present in it in the form of mercuric sulphide (HgS).

(b) When zinc carbonate is heated strongly (calcination), it decomposes to form zinc oxide and carbon dioxide:
$$\text{ZnCO}_3 \xrightarrow{\Delta} \text{ZnO} + \text{CO}_2$$

(c)
(I) Metal A is Aluminium (Al). The reaction is called Thermite reaction.

(II)
$$\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} + \text{Heat}$$

This reaction is highly exothermic, producing molten iron which is used to join/weld railway tracks.

Source: Metals and Non-metals, Extraction of Metals

---

• (a) Cinnabar is the standard expected ore of mercury; students must name both ore and form of metal present.
• (b) This is calcination — heating a carbonate in limited/no air to get an oxide. Write the equation for full marks.
• (c) Thermite reaction is a classic CBSE board question. Examiners look for: correct identification of Al, the name "Thermite reaction," and a balanced equation. Don't forget to balance — 2Al and 2Fe on each side.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

In ZnO + C → Zn + CO, carbon removes oxygen from ZnO (reducing it to Zn), so ZnO is reduced and C acts as the reducing agent.

ZnO loses oxygen → it undergoes reduction. Carbon gains oxygen → it is oxidised, acting as a reducing agent. Both statements are correct and R directly explains why A is true, making option (a) the right choice. Examiners expect students to identify the role of each substance in terms of gain/loss of oxygen.

(c) It reacts with both acid as well as base to form salt and water.

An amphoteric oxide shows both acidic and basic behaviour, reacting with acids and bases to produce salt and water (not hydrogen gas).

Source: Chapter 2, Section 2.1.5 & 2.1.6

---

• The key term is amphoteric — it means the oxide can act as both an acid and a base.
• The products are always salt + water (like a neutralisation reaction), NOT hydrogen gas.
• Options (b) and (d) are wrong because hydrogen gas is not produced; option (a) is wrong because amphoteric oxides react with both acid AND base.
• Common example: ZnO, Al₂O₃ — these react with both HCl and NaOH to give salt and water.

(i) Electron transfer in formation of MgCl₂:

Mg has 2 valence electrons; each Cl needs 1 electron.

Mg → Mg²⁺ + 2e⁻
Cl + e⁻ → Cl⁻ (×2)

So Mg²⁺ + 2Cl⁻ → MgCl₂

(ii) Two properties of ionic compounds:

1. They are soluble in water (polar solvents).
2. They conduct electricity in molten state or in aqueous solution (due to free ions).

(iii) (The sub-question (A)/(B) options are not printed in the passage provided. Please provide the actual text of option A or B to receive a complete answer.)

Source: Chapter 3 – Metals and Non-metals, Chemical Bonding / Ionic Compounds section

---

• (i) Examiners expect you to show the dot-structure or ionic equation clearly indicating Mg loses 2e⁻ and each Cl gains 1e⁻. Even a simple equation with charges is accepted.
• (ii) Do NOT repeat high melting/boiling points — the question says "other than." Conductivity and solubility in water are the two standard NCERT properties to cite.
• (iii) Since the A/B options were not included in the extract given, the answer cannot be written. In your actual exam, you will see both options and must choose one — budget roughly 40–50 words for a 2-mark answer.

(b) X = Mg ; Y = MgO ; Type of reaction = Combination

Magnesium ribbon burns in oxygen with a dazzling white flame to form white ash magnesium oxide: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. This is a combination reaction.

The key identifiers are: metal ribbon (Mg is described as a ribbon), dazzling white flame, and white ash (MgO) — all matching the textbook description in Activity 1.1 and Section 3.2.1. The reaction is a combination (two substances combine to form one product), not decomposition or thermal decomposition. Option (d) is wrong because this reaction is exothermic, not endothermic.

Differences between Roasting and Calcination:

| | Roasting | Calcination |
|---|---|---|
| 1. | Ore is heated in excess of air/oxygen | Ore is heated in limited/absence of air |
| 2. | Applied to sulphide ores | Applied to carbonate/hydroxide ores |

Chemical equations:

Roasting: $2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2$

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

Reduction of metal from oxide:
The metal oxide obtained is reduced using a suitable reducing agent like carbon (coke):

$$2\text{ZnO} + \text{C} \rightarrow 2\text{Zn} + \text{CO}_2$$

Source: Metals and Non-metals, Section 3.3 (Extraction of Metals)

---

• Examiners expect both differences clearly stated — the key distinguishing points are (i) presence/absence of air and (ii) type of ore.
• One equation for each process is sufficient; ZnS/ZnCO₃ are standard textbook examples.
• For reduction, carbon reduction is the most common expected example. Some questions also accept thermite/aluminium reduction for highly reactive metal oxides, but carbon reduction suffices here.
• Do not mix up roasting (sulphides → oxides) with calcination (carbonates → oxides).

(b)(i) When copper powder is heated in air, it reacts with oxygen to form copper oxide (CuO), which is black in colour. This is an oxidation reaction:
$$2Cu + O_2 \xrightarrow{\Delta} 2CuO$$

(b)(ii) When hydrogen gas is passed over the hot black copper oxide (CuO), it gets reduced back to copper, regaining its original shiny brown colour:
$$CuO + H_2 \xrightarrow{\Delta} Cu + H_2O$$

Source: Chapter 1, Chemical Reactions and Equations

---

• (b)(i): Examiners expect students to identify this as oxidation (gain of oxygen) and name the product as copper(II) oxide / CuO. The equation is a bonus but helps secure the mark.
• (b)(ii): The reverse process is reduction (removal of oxygen using hydrogen). Students must state that hydrogen is passed over CuO and that copper is restored. This concept directly links to Q17 of the exercises ("shiny brown coloured element X") and the oxidation-reduction topic in Chapter 1.

(d) Assertion (A) is false, but Reason (R) is true.

Sodium oxide (Na₂O) is a basic oxide, not amphoteric. Amphoteric oxides (e.g., Al₂O₃, ZnO) react with both acids and bases — the definition in Reason (R) is correct.

Source: What you have learnt, Chapter 3

---

The key distinction: Na is an active metal, so Na₂O is purely basic. Only oxides like Al₂O₃ and ZnO (mentioned explicitly in the textbook) are amphoteric. The definition of amphoteric oxide given in Reason (R) is accurate, making (d) the correct option. Examiners expect students to recall specific examples of amphoteric oxides from Chapter 3.

(c) 5, 6 or 7

Non-metals gain electrons to complete their octet. From Table 3.3, non-metals like Nitrogen (5), Oxygen (6), Fluorine/Chlorine (7) have 5, 6, or 7 electrons in their outermost shell.

Source: Chapter 3, Section 3.3

• Non-metals need to gain electrons to attain a stable octet, so they already have close to 8 electrons in their valence shell (5, 6, or 7).
• Option (d) is a trap — 8 electrons in the outermost shell means the atom is already a noble gas (completely stable), not a non-metal.
• Option (a)/(b) describes metals or semi-metals (1–3 or 3 electrons → tendency to lose electrons).
• Always link the answer to examples from the table: N=5, O=6, F/Cl=7.

(a) Lithium

Alkali metals like lithium have low densities. Among the given options, lithium has the lowest density (~0.53 g/cm³).

Source: Physical Properties, Chapter 3, Section 3.1.1

The textbook states that alkali metals (lithium, sodium, potassium) have low densities. Among the options — Lead (very dense), Magnesium and Aluminium (moderate density) — lithium is the lightest metal overall. Examiners expect you to recall this directly from the exceptions listed in Section 3.1.1.

(b) Sodium zincate

Zinc reacts with sodium hydroxide to form sodium zincate (Na₂ZnO₂) and hydrogen gas, as zinc is amphoteric and reacts with bases.

Zinc is an amphoteric metal — it reacts with both acids and bases. With NaOH, it forms sodium zincate (Na₂ZnO₂), not zinc oxide or zinc hydroxide. The NCERT chapter on metals mentions zinc oxide as an amphoteric oxide, which is the key concept behind this reaction. Remember: amphoteric metals/oxides react with bases to form zincate/aluminate salts.

Ore of Mercury: Cinnabar (HgS). It is found in the combined state (as a sulphide compound) in nature.

Extraction of Mercury:

Step 1: Cinnabar is heated strongly in air, converting it to mercuric oxide:

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

Step 2: Mercuric oxide is further heated, and being a low-activity metal, it decomposes to give mercury:

$$2\text{HgO(s)} \xrightarrow{\text{Heat}} 2\text{Hg(l)} + \text{O}_2\text{(g)}$$

No reducing agent is needed; heat alone is sufficient because mercury is low in the activity series.

Source: Chapter 3, Section 3.4.3

---

• The question awards 1 mark for naming the ore and its form; 2 marks for the two balanced equations with conditions.
• Always write the condition (Heat) above the arrow and include state symbols — examiners check both.
• The key idea to state: since Hg is low in the activity series, simple heating suffices — no carbon reduction needed. Mentioning this shows conceptual understanding and can fetch full marks.

(B) Al, Al₂O₃

Aluminium is used in the thermite process (Fe₂O₃ + 2Al → 2Fe + Al₂O₃). When heated in air, it forms Al₂O₃, which is amphoteric — reacting with both acids and bases.

Source: Chapter 3, Sections 3.2.1 and 3.4.4

Two clues identify the metal: (1) use in the thermite reaction — the textbook explicitly states aluminium reduces Fe₂O₃ in the thermit reaction; (2) the oxide must be amphoteric — the textbook names Al₂O₃ and ZnO as the amphoteric oxides. Only option (B) satisfies both conditions. Fe₂O₃ is basic, not amphoteric, so (C) is wrong despite iron being involved in the thermite reaction.

(a) The test tube containing magnesium showed the fastest rate of bubble formation and the highest temperature rise, because magnesium is the most reactive among the three metals.

(b) Iron did not react with dilute nitric acid (when the activity was repeated with it). However, with dilute HCl, all three metals react. (Note: If the question refers to dilute HCl specifically, none of the three metals is completely unreactive with it; all three displace hydrogen.)

> (Since sub-question (c) is incomplete — options (i) and (ii) are not provided — it cannot be answered.)

Source: Metals and Non-metals, Activity on reactivity of metals with dilute acids.

---

• For (a): Reactivity order is Mg > Al > Fe. More reactive metals react faster and release more heat, so the thermometer reads highest in the Mg test tube.
• For (b): The passage says the activity was also repeated with dilute nitric acid. Iron (and other metals) generally do not react with concentrated HNO₃ due to passivation, but with dilute HCl all three do react. If the examiner intends dilute HNO₃, iron may be the expected answer due to passivation with nitric acid. Read the full question context carefully.
• For (c): The sub-options were missing from the question — always write what is given; do not attempt incomplete sub-questions.

Answer: C

Assertion (A) is true — sulphide ores must be roasted (converted to oxides) before reduction to extract the metal. Reason (R) is false — roasting converts sulphide ores into metal oxides, not directly into metals.

The key distinction: roasting ($2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2$) produces a metal oxide, which is then reduced (e.g., by carbon) to get the metal. Examiners expect you to know the two-step process. The Reason wrongly claims roasting gives the metal directly — making R false while A remains correct.

Answer: (C) Copper and silver

According to the textbook, gold, silver, platinum and copper are found in the free state, but copper and silver are also found in the combined state as sulphide or oxide ores.

The key distinction here is "both free state and combined state." Gold and platinum occur only in free (native) state. Copper and silver occur in both — hence option C is correct. This is directly stated in section 3.4.1.

(i) Extraction of metals:

(I) Mercury from cinnabar (HgS):
Cinnabar (HgS) is heated in air. The ore gets converted to mercuric oxide, which on further heating decomposes to give mercury.

$$\text{HgS} + \text{O}_2 \xrightarrow{\Delta} \text{HgO} + \text{SO}_2$$
$$2\text{HgO} \xrightarrow{\Delta} 2\text{Hg} + \text{O}_2$$

(II) Copper from copper sulphide (Cu₂S):
Copper sulphide ore is heated in air to obtain copper directly.

$$\text{Cu}_2\text{S} + \text{O}_2 \xrightarrow{\Delta} 2\text{Cu} + \text{SO}_2$$

---

(ii) Compounds formed as coating:

(I) Silver articles: A black coating of silver sulphide (Ag₂S) is formed when silver reacts with sulphur compounds present in air.

(II) Copper articles: A green coating of basic copper carbonate / copper carbonate [CuCO₃·Cu(OH)₂] is formed when copper reacts with CO₂ and moisture in air.

This process of metals being attacked by substances in their surroundings is called corrosion.

Source: Chapter 1, Chemical Reactions and Equations, Section on Corrosion

---

• For (i), the examiner expects balanced equations with state symbols or at least correct formulae. Two-step extraction of mercury is important. For copper, a single-step roasting equation is standard at this level.
• For (ii), the exact names of the compounds are the key scoring points: Ag₂S (silver sulphide) for silver, and basic copper carbonate / Cu(OH)₂·CuCO₃ (green coating) for copper. Mentioning the term "corrosion" adds value. Avoid writing "rust" — rust is specifically for iron.

(i)

I. Ag (Silver) — does not react with oxygen even at high temperature.
II. Al (Aluminium) — reacts at ordinary temperature and forms a protective Al₂O₃ layer.
III. K (Potassium) — catches fire when kept in the open.
IV. Fe (Iron) — does not burn but hot iron is coated with black Fe₃O₄ layer.

(ii) Amphoteric oxides are metal oxides that react with both acids and bases to form salt and water.

Aluminium oxide (Al₂O₃) is amphoteric:

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

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

(iii) Alkalis are bases that dissolve in water to give OH⁻ ions.
Example: Sodium hydroxide (NaOH).

---

• For (i), students often confuse Cu and Fe — Cu forms black CuO on heating (not Fe). K is so reactive it ignites in air; Ag is the least reactive metal listed.
• For (ii), the key is to state the definition AND show both reactions (with acid and with base) with balanced equations — examiners expect both.
• For (iii), one line with one example is sufficient for 1 mark. Don't write more.

(a) The reactivity series is a list of metals (and some non-metals) arranged in decreasing order of their reactivity (tendency to react). It is developed by comparing the ability of elements to displace each other from solutions or oxides and by observing reactions with water and acids.

Order (most to least reactive):
Calcium > Aluminium > Lead > Copper

(b) Balanced chemical equation for the reaction of Fe₂O₃ with aluminium (thermite reaction):

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

Source: Chapter 1 (Chemical Reactions and Equations), Exercises Q.2

---

• For (a), examiners expect a definition covering "decreasing order of reactivity" and at least one method of development (displacement reactions). The correct order — Ca > Al > Pb > Cu — must be stated explicitly; partial credit is lost if the sequence is wrong.
• For (b), this exact equation appears in the chapter exercises. Ensure coefficients (2Al, 2Fe) are correct; an unbalanced equation loses the mark. This reaction is also a displacement reaction (Al displaces Fe) — useful to note if asked for the type.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is correct — hydrogen gas is not evolved when metals react with nitric acid. However, Reason is false because nitric acid is a strong oxidising agent (not reducing agent), which oxidises the hydrogen produced to water.

The key error in the Reason is the wrong identification of nitric acid as a "reducing agent." Nitric acid is a strong oxidising agent — this is a common factual trap in Assertion-Reason questions. Examiners specifically test whether students can spot this reversal. Always remember: HNO₃ oxidises H₂ to H₂O, hence no hydrogen gas is released when metals react with it (except with very dilute HNO₃ and magnesium/manganese).

(A) Reduction with carbon

The most common method of extracting metals from their oxide ores is reduction with carbon (coke). Metals in the middle of the activity series (Zn, Fe, Pb, Cu) are commonly extracted this way, e.g., ZnO + C → Zn + CO.

Source: Metals and Non-metals, Section 3.4.4

---

The textbook (Fig. 3.9 and Section 3.4.4) clearly states that Zn, Fe, Pb, Cu — the largest group of commercially important metals — are extracted by reduction using carbon. Electrolytic reduction is used only for highly reactive metals (K, Na, Ca, Mg, Al), which is a special case, not the "most common" method. Reduction with Al (thermite) and H₂ are used in specific/limited cases only. Carbon (coke) is cheap and widely used, making option (A) correct.

(i) Electron-dot structures:

• Sodium (At. No. 11; configuration 2, 8, 1): one dot around the symbol Na.

Na·

• Oxygen (At. No. 8; configuration 2, 6): six dots around the symbol O.

:Ö: (six electrons shown around O)

(ii) Formation of sodium oxide (Na₂O):

Oxygen needs 2 electrons to complete its octet. Each sodium atom donates 1 electron. So 2 sodium atoms each transfer 1 electron to one oxygen atom:

$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$

The compound formed is Na₂O.

(iii) Ions present:

• Na⁺ → Cation (sodium ion)
• O²⁻ → Anion (oxide ion)

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

---

• Draw electron-dot (Lewis) structures by placing dots equal to valence electrons around the symbol: Na has 1 valence electron; O has 6.
• For Na₂O, show two Na atoms each losing one electron (since O needs 2 electrons to achieve octet).
• Examiners specifically expect you to label the cation (Na⁺) and anion (O²⁻) — missing this costs marks.
• The transfer of electrons makes this an ionic compound; no need to explain bonding in detail for 3 marks.

Silver is refined by electrolytic refining. In this method, impure silver is made the anode, pure silver is the cathode, and silver nitrate solution acts as the electrolyte.

The examiner expects you to name the method (electrolytic refining) and may award the mark for the name alone, but briefly stating the anode/cathode/electrolyte shows understanding and secures the mark confidently. This applies to refining of any metal, including silver.

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Metals high in the reactivity series are extracted by electrolytic reduction, not displacement.

The Assertion is true — highly reactive metals (Na, Mg, Al, Ca, K) cannot be reduced by carbon because they have more affinity for oxygen than carbon. The Reason is also true — displacement reactions (e.g., thermite reaction) are used to obtain metals. However, displacement reactions explain how mid-series metals are extracted, not why top-series metals cannot be reduced by carbon. So R does not explain A → option (B).

(C) Brass and Bronze

Zinc dissolved in molten copper gives brass, and tin dissolved in molten copper gives bronze.

The question tests knowledge of common copper-based alloys. Remember: Brass = Copper + Zinc; Bronze = Copper + Tin. The order in the question is zinc first, then tin — so the answer must be Brass first, then Bronze, making (C) correct. Solder is an alloy of lead and tin, not copper-based.

(B) When it is heated with iron (III) oxide, molten iron is obtained.

The reaction is: $\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$ (molten). This highly exothermic displacement reaction produces molten iron used to join railway tracks.

Source: Chapter 3, Section 3.2.4 (Reactivity Series / displacement reactions)

---

• Option (B) is correct because aluminium, being more reactive than iron, displaces iron from Fe₂O₃ in a highly exothermic thermite reaction, producing molten iron that fills the gap between railway tracks and solidifies to join them.
• Option (A) is partially true but incomplete — it doesn't explain what the reaction produces.
• Option (C) is wrong: molten Al₂O₃ is a by-product, not the substance that joins tracks.
• Option (D) is wrong: no iron-aluminium alloy is formed; pure molten iron is produced.
• Examiners expect you to pick the most complete and accurate statement — that is (B).

(i) Two metals that react violently with cold water: Sodium (Na) and Potassium (K).

Observations:

1. The metal melts and moves rapidly on the surface of water.
2. A hissing sound is produced and the metal may catch fire (potassium burns with a lilac flame).
3. The water turns pink/red when a few drops of phenolphthalein indicator are added, showing the solution formed is alkaline (metal hydroxide is produced).

(ii) Test for hydrogen gas: Bring a burning candle/matchstick near the mouth of the inverted test tube collecting the gas. If the gas burns with a 'pop' sound, it confirms the gas is hydrogen.

---

• For (i), the question asks for two metals + three observations — both marks depend on getting all five points. Common metals chosen are Na and K. Observations must be distinct: physical behaviour, sound/flame, and alkaline nature of solution.
• For (ii), the standard test for H₂ is the pop test — always mention the burning splint and the characteristic pop sound. This is a 1-mark point so one clear sentence is enough.
• These metals react as: $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\uparrow$ — knowing the equation helps justify observations.

(a) Metals in the middle of the reactivity series (e.g., iron, copper) can be extracted using displacement reactions where a more reactive metal displaces them from their salt solutions.

Example 1: $\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$
(Aluminium displaces iron from iron oxide — thermite reaction)

Example 2: $\text{CuSO}_4(aq) + \text{Fe}(s) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$
(Iron displaces copper from copper sulphate solution)

(b) Metals high up in the reactivity series (e.g., Na, K, Ca, Al) are very reactive and have a stronger affinity for oxygen than carbon does. Carbon cannot reduce their oxides because it is less reactive than these metals, so carbon reduction fails.

---

• For (a), the examiner expects the term "displacement reaction" used correctly, with two balanced equations clearly showing a more reactive element displacing a less reactive one.
• For (b), the key idea is relative reactivity: carbon must be more reactive than the metal to reduce its oxide. Since Na, K, Ca, Mg, Al are more reactive than carbon, carbon cannot displace them. One clear reason + one example metal is enough for 1 mark.

(B) Calcium and magnesium

Metals high in the reactivity series (K, Na, Ca, Mg, Al) are obtained by electrolytic reduction of their molten chlorides. Gold and silver occur in free state; iron is extracted by carbon reduction.

Source: Metals and Non-metals, Section 3.4.5

---

The key fact from Section 3.4.5: sodium, magnesium, and calcium are obtained by electrolysis of their molten chlorides. Note that aluminium is extracted by electrolysis of aluminium oxide (not chloride), so option (C) is wrong. Iron uses carbon reduction; gold/silver occur native. Only option (B) lists two metals (Ca and Mg) correctly extracted from molten chlorides.

(C) It has weak electrostatic forces of attraction between its oppositely charged ions.

Compound C is ionic (A loses electrons, B gains electrons). Ionic compounds have strong electrostatic forces, not weak ones.

When A loses electrons and B gains electrons, an ionic compound (C) is formed. Ionic compounds have high melting points (A is correct), are highly soluble in water (B is correct), and conduct electricity in molten state or aqueous solution (D is correct). The defining feature of ionic compounds is strong electrostatic forces of attraction — so option (C) is the incorrect property, making it the right answer.

(B) Al₂O₃ and MgO

When aluminium burns in air: $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$; when magnesium burns in air: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$.

Source: Chapter 3, Section 3.2.1

The correct formulae come from the valencies: Al has valency 3 and O has valency 2, giving Al₂O₃; Mg has valency 2 and O has valency 2, giving MgO. Options with Al₃O₄ or MgO₂ are chemically incorrect. The textbook explicitly states the reaction $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_�\text{O}_3$ and magnesium oxide (MgO) is standard knowledge from the activity series chapter.

(a) During electrolytic refining, the impurities (like gold, silver, platinum) that settle below the anode at the bottom of the electrolytic solution are called anode mud.

(b)

• Anode: Impure copper (thick block)
• Cathode: Pure copper (thin strip)

(c) Extraction of copper from Cu₂S:

Step 1 – Roasting:
$$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$

Step 2 – Reduction:
$$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$$

Copper is obtained as a molten mass.

Source: Metals and Non-Metals, Electrolytic Refining & Extraction of Metals

---

• Anode mud is a standard definition question — mention that it settles below the anode and contains precious metal impurities.
• For (b), examiners expect both "impure copper = anode" and "pure copper = cathode" clearly stated.
• For (c), both steps (roasting + auto-reduction) with balanced equations are required for full 2 marks. Missing either equation costs a mark.

(i) Formation of Sodium Oxide (Na₂O):

Na has 2,8,1 configuration. Each Na atom loses 1 electron to form Na⁺.
O has 2,6 configuration. O atom gains 2 electrons to form O²⁻.

$$\text{Na} \rightarrow \text{Na}^+ + e^- \quad \text{(×2)}$$
$$2,8,1 \quad 2,8$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$
$$2,6 \qquad 2,8$$

Two Na⁺ and one O²⁻ combine to form Na₂O.

---

(ii) Formation of Magnesium Chloride (MgCl₂):

Mg has 2,8,2 configuration. Mg loses 2 electrons to form Mg²⁺.
Cl has 2,8,7 configuration. Each Cl atom gains 1 electron to form Cl⁻.

$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$
$$2,8,2 \qquad 2,8$$
$$\text{Cl} + e^- \rightarrow \text{Cl}^- \quad \text{(×2)}$$
$$2,8,7 \quad 2,8,8$$

One Mg²⁺ and two Cl⁻ combine to form MgCl₂.

Source: Chapter 3, Section 3.3

---

• Always write the electronic configuration alongside each atom/ion to show the before-and-after electron arrangement — examiners specifically look for this.
• Show which atom loses electrons (metal → cation) and which gains (non-metal → anion).
• For Na₂O: two Na atoms are needed because O needs 2 electrons and each Na gives only 1. This is the key logic to state.
• Mention the final ionic formula to complete the answer.
• The (×2) notation neatly shows that the half-equation is repeated twice without rewriting it.

(C) Na₂ZnO₂

Zinc is an amphoteric metal; it reacts with NaOH to form sodium zincate (Na₂ZnO₂) and hydrogen gas.

Zinc reacts with NaOH because zinc oxide/zinc is amphoteric — it reacts with both acids and bases. The reaction is: Zn + 2NaOH → Na₂ZnO₂ + H₂. This is a standard fact linked to the amphoteric nature of zinc discussed in Chapter 3.

(C) Exothermic, but the heat evolved is not sufficient for the hydrogen to catch fire.

Calcium reacts with cold water more vigorously than magnesium but less violently than sodium/potassium. The reaction is exothermic and produces hydrogen gas, but the heat released is not enough to ignite the hydrogen. (Sodium/potassium reactions are so violent that hydrogen catches fire.) This is a standard reactivity series application tested in CBSE MCQs.

(i) Metal sulphides and carbonates cannot be directly reduced to metals. They must first be converted to metal oxides because oxides are more easily reduced (e.g., by carbon or CO) to obtain the free metal.

(ii) Aluminium oxide (Al₂O₃) reacts with both acids and bases. It reacts with acids to form salt and water, and with bases (like NaOH) to form aluminate. Since it shows properties of both acidic and basic oxides, it is called amphoteric.

(iii) Na, K, Ca, and Mg are highly reactive metals (high reactivity). They readily react with oxygen, water, and other elements in nature, so they always exist as compounds and are never found in free (native) state.

---

• (i) Key point: sulphides/carbonates → oxides (by roasting/calcination) → then reduction gives metal. Examiners want the reason: oxides are easier to reduce.
• (ii) "Amphoteric" means reacts with both acid and base — always state both reactions briefly.
• (iii) The reason is high chemical reactivity / high position in reactivity series. These metals instantly combine with environmental substances.

(i)

• Metal that can be cut with a knife: Sodium (Na)
• Non-metal that is a good conductor of electricity: Graphite (carbon)

(ii)
Electronic configuration:

• Mg (atomic number 12): 2, 8, 2
• O (atomic number 8): 2, 6

Mg has 2 electrons in its outermost shell. To attain stable configuration, it loses 2 electrons to become Mg²⁺ (2, 8).

Oxygen needs 2 electrons to complete its octet. It gains the 2 electrons from Mg to become O²⁻ (2, 8).

$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$

The oppositely charged ions (Mg²⁺ and O²⁻) attract each other to form magnesium oxide (MgO).

Source: Chapter 3, Sections 3.1 Physical Properties and 3.3 How do Metals and Non-metals React?

---

• For (i): Sodium is the textbook example of a soft metal (alkali metals); graphite is the standard exception among non-metals that conducts electricity. Both are directly stated in Section 3.1.
• For (ii): Show the electron configurations clearly, state that Mg loses 2e⁻ and O gains 2e⁻, write the ionic equations, and name the product. Examiners award marks for: correct configurations, direction of electron transfer, ion charges, and naming MgO as the product.

(i) Test tube B (containing CuSO₄) will show the most vigorous reaction, because iron is more reactive than copper and can displace it from its salt solution. The difference in reactivity between Fe and Cu is the greatest among the four options.

(ii) Balanced equation:

$$\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$$

---

• Iron cannot displace Zn or Al (both more reactive than Fe), and FeSO₄ already contains Fe, so no reaction occurs in A, C, D.
• Examiners expect you to justify the choice using the reactivity series — just naming the test tube without reason loses the mark.
• The balanced equation must show correct formulae and be properly balanced; state symbols are a bonus but not mandatory for full marks here.

(B) Mg reacts with water to produce H₂ gas which helps in floating.

Mg reacts with hot water to form magnesium hydroxide and hydrogen gas. The H₂ bubbles stick to the metal surface, making it float.

Source: Chapter 3, Section 3.2.2

---

The textbook explicitly states: "Magnesium does not react with cold water. It reacts with hot water to form magnesium hydroxide and hydrogen. It also starts floating due to the bubbles of hydrogen gas sticking to its surface." Options C and D are wrong gases; Option A is incorrect because Mg is denser than water — it floats only because of H₂ bubbles, not its own density.

(i)

(I) Heating a mixture of Cu₂O and Cu₂S:

$$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2\uparrow$$

(II) ZnS heated in the presence of oxygen:

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

---

(ii)

(I) PVC (polyvinyl chloride) is a poor conductor of electricity (insulator). It prevents electric shock to persons handling the wires.

(II) Copper does not react with hot water and does not corrode easily, making it suitable for hot water tanks. Steel (iron) reacts with hot water/steam and corrodes, contaminating water.

---

(iii) Formation of CaO:

Ca has 2 valence electrons; O needs 2 electrons to complete its octet.

$$\text{Ca} [\cdot\cdot] + [\cdot\cdot\underset{\cdot\cdot}{\text{O}}\cdot\cdot] \rightarrow \text{Ca}^{2+}[\underset{\cdot\cdot}{\ddot{\text{O}}}\underset{\cdot\cdot}{}]^{2-}$$

Ca donates 2 electrons to O. Ca becomes Ca²⁺ and O becomes O²⁻, forming the ionic compound CaO.

---

• Q(i): These are standard metallurgy reactions. The Cu₂O–Cu₂S reaction is the self-reduction used in copper extraction. ZnS roasting converts sulphide ore to oxide — a key step in metallurgy.
• Q(ii): For PVC coating, the keyword is "insulator/poor conductor → prevents electric shock." For copper tanks, focus on resistance to corrosion and non-reactivity with hot water. Avoid vague answers like "copper is better."
• Q(iii): Draw the electron dot (Lewis) structure clearly. Show electron transfer: Ca loses 2e⁻ → Ca²⁺; O gains 2e⁻ → O²⁻. Mention ionic bond formation. Even a rough diagram with dots/crosses earns the mark if the transfer is correctly shown.

(i)

(I) Ionic compounds are held together by strong electrostatic forces of attraction between oppositely charged ions. A large amount of energy is needed to overcome these forces, so they have high melting and boiling points.

(II) Solder is an alloy of lead and tin with a low melting point. This allows it to melt easily and join electrical wires without damaging them due to excessive heat.

(III) Carbon can act as a reducing agent only at very high temperatures. Na and Mg are more reactive than carbon; they are higher in the reactivity series. Hence carbon cannot displace Na or Mg from their oxides — in fact, Na or Mg can reduce carbon dioxide.

---

(ii)

(I) Compound 'X' is Iron(III) oxide (Fe₂O₃).

(II) The reaction is called the Thermite reaction.

(III) Balanced chemical equation:

$$\text{Fe}_2\text{O}_3 (s) + 2\text{Al} (s) \rightarrow \text{Al}_2\text{O}_3 (s) + 2\text{Fe} (l) + \text{Heat}$$

Source: Chemical Reactions and Equations, Chapter 1

---

• For (i)(I): Always mention "strong electrostatic forces" — that phrase earns the mark.
• For (i)(II): The key point is low melting point of solder (not just that it's an alloy).
• For (i)(III): The logic is reactivity series — Na and Mg are above carbon, so carbon cannot reduce their oxides.
• For (ii): The thermite reaction (Fe₂O₃ + Al) is a standard 2-mark question. Examiners expect the correct name "Thermite reaction," correct identification of Fe₂O₃, and a fully balanced equation. Note that Fe is produced in liquid/molten form due to the large heat released — writing (l) is a good addition.

(D) Al₂O₃

Al₂O₃ (aluminium oxide) is an amphoteric oxide because it reacts with both acids and bases to form salt and water.

• Na₂O and K₂O are basic oxides (formed from active metals); CO₂ is an acidic oxide (non-metallic).
• Amphoteric oxides show dual behaviour — they react with acids as well as bases. Al₂O₃ is the classic CBSE example of an amphoteric oxide.
• Remember: Zn, Al, Pb, Sn form amphoteric oxides — useful for MCQs.

(a) Carbonate and sulphide ores are converted to oxides because oxides are easier to reduce using reducing agents like carbon to obtain the free metal.

(b) Thermite reaction — Aluminium reduces iron(III) oxide:

$$Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe$$

(c) Cu₂S (copper glance) is roasted in the presence of air (oxygen) to obtain copper directly:

Step 1 — Partial roasting:
$$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$

Step 2 — Reduction of Cu₂O by remaining Cu₂S:
$$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$$

Thus, copper metal is obtained.

Source: Metals and Non-metals, Extraction of Metals

---

• (a) Key point: oxides are easier to reduce than carbonates/sulphides — one line is enough.
• (b) Any thermite-type reaction works; Fe₂O₃ + Al is the standard NCERT example. Write the balanced equation correctly.
• (c) This is a 2-mark question — examiners expect both steps with balanced equations. Cu₂S roasting is a special case where no external reducing agent is needed; the ore itself acts as the reducing agent in the second step. Award marks are split: 1 for each correct equation.

Amphoteric oxides are metal oxides that show the properties of both acidic and basic oxides, i.e., they react with both acids and bases. Example: Aluminium oxide (Al₂O₃) and zinc oxide (ZnO).

The answer is directly from Chapter 3 ("What you have learnt"): "Aluminium oxide and zinc oxide show the properties of both basic as well as acidic oxides. These oxides are known as amphoteric oxides." Examiners expect the definition plus at least one example for full marks.

Answer: B — Calcium

Calcium reacts with cold water and the hydrogen gas bubbles formed stick to its surface, causing it to float.

Source: Chapter 3, Section 3.2.2

The key detail from the passage is: "Calcium starts floating because the bubbles of hydrogen gas formed stick to the surface of the metal." Sodium and potassium also react with water but they react violently (hydrogen catches fire) — they don't simply float. Iron does not react with cold or hot water at all. Calcium is the correct answer because it reacts moderately and floats due to hydrogen bubbles.

Answer: A — Pb (Lead)

Lead (Pb) is a poor conductor of electricity compared to Cu (copper), Ag (silver), and Al (aluminium), which are all good conductors.

Cu, Ag, and Al are metals widely used as electrical conductors (Ag is the best conductor, Cu and Al are used in wiring). Lead (Pb) is a metal but has significantly higher resistivity, making it a poor conductor relative to the others. The textbook mentions Al, Cu as fuse/wire materials for good conductivity; Pb is notably absent from such applications.

(D) Steel

Steel is an alloy of iron (metal) and carbon (non-metal). Brass (copper + zinc) and Bronze (copper + tin) contain only metals, and amalgam contains mercury + another metal.

Examiners expect you to recall the composition of common alloys. Steel = Fe + C (carbon is a non-metal) — this is the key fact. Brass = Cu + Zn; Bronze = Cu + Sn; Amalgam = Hg + another metal — all metal-only alloys.

(D) Magnesium

Magnesium does not react with cold water. It reacts only with hot water to form magnesium hydroxide and hydrogen gas.

Source: Chapter 3, Section 3.2.2

---

The textbook clearly states: "Magnesium does not react with cold water. It reacts with hot water…" Sodium, potassium, and calcium all react with cold water (K and Na vigorously; Ca less violently). This is a common MCQ testing knowledge of the reactivity series — remember the order K > Na > Ca > Mg for reactions with water.