NCERT Exemplar Solutions Class 10 Science Chapter 3

Match each word to a known metal property. I run down the four options and tick the ones that describe a metal. The leftover word is the answer.

Concept used. A typical metal is a good conductor, sonorous, malleable, ductile and lustrous. Lustre means a fresh-cut surface looks bright and reflective.

1. Electrical conduction. Metals carry current easily because of free electrons. A genuine metal property, so not the answer.
2. Sonorous. A metal plate or bell rings when struck. True of metals, so not the answer.
3. Ductility. Copper and aluminium are drawn into long wires. True of metals, so not the answer.
4. Dullness. Metals are shiny, not dull. This is the property metals do not show, so option (c) is correct.

Why this matters. Lustre is why gold and silver are used in jewellery and why a clean steel knife reflects light. Spotting "dull" as non-metallic links straight to how we tell metals from non-metals by eye.

Final Answer: Option (c): dullness is not a metallic property; metals are shiny.

Anchor each term to a picture. I keep one image per property in my head, then pick the term whose picture is "wire".

Concept used. Ductility describes wire-drawing; malleability describes sheet-beating. Both come from the way metal atoms slide over one another without breaking.

1. Picture a wire. A thin copper wire being pulled out of a block is ductility. That is the property asked for.
2. Picture a sheet. Aluminium foil hammered flat is malleability. Wrong picture for this question.
3. Eliminate the rest. Sonorousity is sound, conductivity is current. Neither describes shaping into wire.
4. So the term for drawing into wire is ductility, option (a).

Why this matters. Electrical cables exist because copper is ductile. Knowing the word lets students explain why some metals make good wiring while brittle materials cannot.

Final Answer: Option (a): ductility is the ability to be drawn into thin wire.

Ask what cooking actually needs. For a "why is it used for X" question, I list what task X demands and keep only the matching properties.

Concept used. Cooking needs heat to move from flame to food (thermal conduction) and needs the vessel to survive the heat (high melting point). Properties that do not serve cooking are distractors.

1. Need 1: move heat in. Aluminium conducts heat well, so food cooks quickly. Statement (i) qualifies.
2. Need 2: do not melt. Aluminium melts near 660 °C, well above stove temperatures, so the pan keeps its shape. Statement (iv) qualifies.
3. Reject the distractors. Electrical conductivity matters for wires, not pans. Ductility matters for drawing wire, not for a rigid utensil. Drop (ii) and (iii).
4. So the cooking-relevant pair is (i) and (iv), option (d).

Why this matters. Reading a "used for" question as "what does the job require" stops students from ticking every true property and instead keeps only the ones that fit the use.

Final Answer: Option (d): thermal conductivity and high melting point are the cooking-relevant properties.

Slot each metal on the water ladder. I rank the four metals by reactivity, then read off which one needs steam rather than water.

Concept used. Reactivity with water drops down the series. The top metals attack cold water; middle ones need hot water; iron and below need steam, so they look unreactive towards plain water.

1. Sodium. Top of the list, fizzes in cold water: 2Na + 2H₂O → 2NaOH + H₂. Reacts.
2. Calcium. Reacts with cold water more gently: Ca + 2H₂O → Ca(OH)₂ + H₂. Reacts.
3. Magnesium. Slow in cold water but reacts with hot water. Reacts.
4. Iron. Sits lower; only red-hot iron reacts with steam: 3Fe + 4H₂O → Fe₃O₄ + 4H₂. With cold or hot water it does not react. So iron is the answer, option (d).

Why this matters. This is exactly why iron tools survive contact with water for a while, whereas sodium must be stored under kerosene away from any moisture.

Final Answer: Option (d): iron reacts only with steam, so it is the metal that does not react with cold or hot water.

Write the steam reaction first. The product of iron and steam is a standard line; I recall the balanced equation and read the oxide off it.

Concept used. Hot iron reduces steam, taking its oxygen to form the mixed oxide Fe₃O₄ while hydrogen is set free. This is a high-temperature reaction, not slow rusting.

1. Recall the equation. 3Fe + 4H₂O → Fe₃O₄ + 4H₂. The oxide on the right is Fe₃O₄.
2. Check the balance. 3 Fe and 4 O on each side; this confirms Fe₃O₄ as the iron product.
3. Reject the others. FeO and Fe₂O₃ do not appear in this reaction, so (a), (b) and (d) are out.
4. Hence the oxide formed is Fe₃O₄, option (c).

Why this matters. Fe₃O₄ is magnetite, a black magnetic oxide and an important iron ore. This same reaction also produces hydrogen, once a way to make hydrogen gas in the lab.

Final Answer: Option (c): iron with steam yields the mixed oxide Fe₃O₄.

Compare calcium with sodium. I judge "violent or not" by ranking calcium against the most reactive metal sodium, then add the bubble observation.

Concept used. Calcium is below sodium in the reactivity series, so its reaction with cold water is gentler. The reaction still gives off hydrogen, whose bubbles adhere to the metal.

1. Does it react? Yes: Ca + 2H₂O → Ca(OH)₂ + H₂. So (i) "does not react" is false.
2. How violently? Less violently than sodium, since calcium is lower in reactivity. So (iii) is correct and (ii) is false.
3. What about the bubbles? Hydrogen forms and its bubbles cling to the calcium piece, making it float. So (iv) is correct.
4. Combining (iii) and (iv) gives option (d).

Why this matters. This observation, calcium floating on its own hydrogen bubbles, is a classic exam describe-the-experiment point and is the practical sign that the reaction is gentle, not explosive.

Final Answer: Option (d): statements (iii) and (iv) describe calcium with water correctly.

Spot the oxidising acid. Three acids are offered; I look for the one that behaves differently because it is a strong oxidiser.

Concept used. A metal plus a typical dilute acid gives salt and hydrogen. But HNO₃ re-oxidises the freed hydrogen to water, so the gas does not appear; instead oxides of nitrogen form.

1. Sulphuric acid. Zn + H₂SO₄ → ZnSO₄ + H₂. Hydrogen is released, so not the answer.
2. Hydrochloric acid. Zn + 2HCl → ZnCl₂ + H₂. Hydrogen is released, so not the answer.
3. Nitric acid. It oxidises the nascent hydrogen to water, so H₂ is not collected (except with Mn and Mg). This is the odd acid out.
4. Hence the acid that does not give hydrogen is HNO₃, option (c).

Why this matters. This single exception explains many later questions, including why aluminium can be stored in nitric acid and why gold dissolves only in aqua regia, not in any one acid alone.

Final Answer: Option (c): nitric acid does not give hydrogen gas with metals (except Mn and Mg).

Two checks settle it. First the strength of each acid, then the ratio. Both must be right for the option to be correct.

Concept used. Aqua regia gets its power from concentrated acids mixed 3 : 1 (HCl : HNO₃). The HNO₃ oxidises the gold and the HCl supplies chloride to lock it up as a soluble chloride complex.

1. Strength check. Both acids are concentrated. So eliminate every option that mentions "dilute": (a), (b) and (d) are out.
2. Ratio check. The standard mix is 3 volumes of conc. HCl to 1 volume of conc. HNO₃.
3. Match. Concentrated, concentrated, 3 : 1, this is exactly option (c).
4. So the composition is option (c).

Why this matters. Knowing aqua regia is concentrated and 3 : 1 explains Question 21 later, where gold dissolves only in tube C, the tube holding this very mixture.

Final Answer: Option (c): concentrated HCl and concentrated HNO₃ in a 3 : 1 ratio.

Read each bond as metal or non-metal. For every compound I check whether a metal is present. No metal usually means no ionic bond.

Concept used. Ionic bonding needs electron transfer, which happens between a metal (low electronegativity) and a non-metal. Two non-metals instead share electrons, forming covalent bonds.

1. KCl. K is a metal, Cl a non-metal. Electron transfer, ionic. Not an answer.
2. NaCl. Na is a metal, Cl a non-metal. Ionic. Not an answer.
3. HCl. Hydrogen and chlorine are both non-metals; they share electrons. Covalent, so (ii) is not ionic.
4. CCl₄. Carbon and chlorine are both non-metals; shared bonds. Covalent, so (iii) is not ionic. Hence option (b).

Why this matters. This distinction explains the next question: ionic compounds melt high and conduct when molten, while covalent ones like CCl₄ do not, because they have no free ions.

Final Answer: Option (b): HCl and CCl₄ are the non-ionic (covalent) compounds.

Tie conduction to ion movement. An ionic compound conducts only when its ions can move, so I check each property against that single idea.

Concept used. Conduction in ionic compounds is carried by mobile ions, not by free electrons. The ions move when the lattice is broken, that is, in the molten or dissolved state, but not in the solid.

1. Solubility (a). Many ionic solids dissolve in water, a normal property. Not the answer.
2. Solid conduction (b). In the solid, ions are clamped in the lattice and cannot move, so no current flows. This is the property they do not show.
3. High melting/boiling (c). Strong ionic attractions need a lot of energy to break, so high melting points are normal.
4. Molten conduction (d). Melting frees the ions, so molten ionic compounds do conduct. Normal. Hence the odd one is (b).

Why this matters. This is the basis of electrolysis: molten or aqueous ionic compounds conduct and can be split by current, which is exactly how reactive metals like sodium are extracted.

Final Answer: Option (b): solid-state conduction is the property ionic compounds do not generally show.

Pick the least reactive. Native means uncombined, so I look for the two metals lowest in the reactivity series.

Concept used. Reactivity decides occurrence. Highly reactive metals combine with oxygen, sulphur or carbonate and are found as ores; only the noble, least reactive metals stay free.

1. Rank the four. In the series, reactivity falls Zn > Cu > Ag > Au. Gold and silver are the two least reactive here.
2. Gold. So unreactive it never tarnishes; found as free metal nuggets. Native, so (ii).
3. Silver. Very low reactivity, often found native (though it can also occur combined). Native, so (iv).
4. Zinc. Reactive enough to be found only as ore, so (iii) is out. Hence the native pair is option (c).

Why this matters. This explains why coins and ornaments were historically made from gold and silver: they were the only metals available pure, straight from the ground.

Final Answer: Option (c): gold and silver are the metals that occur in the native state.

Separate refining from extraction. I split the four metals into "already a metal, just clean it" versus "so reactive it must be won by electrolysis".

Concept used. Electrolytic refining cleans a metal that is already extracted, depositing pure metal on the cathode. Reactive metals near the top of the series are instead extracted by electrolytic reduction of their molten compounds.

1. Copper. Blister copper is refined electrolytically to 99.9% purity. So (ii) qualifies for refining.
2. Gold. Gold is also purified by electrolytic refining (Wohlwill process). So (i) qualifies.
3. Sodium and potassium. These are won by electrolysis of molten NaCl/KCl, which is extraction, not refining of an impure lump. So (iii) and (iv) are out.
4. Hence the refined-by-electrolysis pair is (i) and (ii), option (a).

Why this matters. The same anode-to-cathode idea reappears in Question 36 and Question 62, where copper is refined by transferring metal from an impure anode to a pure cathode.

Final Answer: Option (a): gold and copper are refined by electrolytic refining.

Identify the gas that attacks silver. The black colour is the clue; I match it to the known reaction of silver with air-borne H₂S.

Concept used. Tarnishing of silver is corrosion by hydrogen sulphide in the air. The product, silver sulphide, is black and forms a thin film on the surface.

1. What attacks silver. Trace H₂S in air, not oxygen, is the main culprit for the black colour.
2. The reaction. 2Ag + H₂S → Ag₂S + H₂. The product is Ag₂S.
3. Reject the others. Ag₂O would be from oxygen and is not the black tarnish; Ag₃N is not formed here. So (a), (b), (d) are out.
4. Hence the black coating is Ag₂S, option (c).

Why this matters. Corrosion is not just rust on iron. Each metal corrodes differently: iron rusts (oxide), copper goes green (carbonate), and silver blackens (sulphide). Recognising the product is a common exam ask.

Final Answer: Option (c): silver blackens because of Ag₂S formation.

Recall the definition and the why. Galvanisation is a fixed term, so I state the metal used and back it up with the reason it works.

Concept used. A protective coating works either by blocking air and moisture or by being more reactive so it corrodes first. Zinc on iron does both, which is why galvanising is so effective.

1. State the coat. Galvanisation means coating with zinc, so the answer is (c).
2. Barrier action. The zinc layer physically keeps oxygen and water off the iron, stopping rust.
3. Sacrificial action. Zinc is above iron in the reactivity series, so even a scratched coat lets zinc corrode first, sparing the iron.
4. Hence the correct coating metal is zinc, option (c).

Why this matters. Galvanised iron sheets are everywhere, from roofing to buckets to fences, because this cheap zinc coat gives iron a long life outdoors.

Final Answer: Option (c): galvanisation is a zinc coating on iron.

Recall the alloy recipe. Stainless steel has a standard pair of added metals; I name them and explain what each contributes.

Concept used. An alloy mixes a base metal with others to improve its properties. In stainless steel, iron is the base, and chromium plus nickel give it strength and rust resistance.

1. Base metal. Iron forms the bulk of the alloy.
2. Chromium. It forms a thin protective oxide film that stops rust, the key to the "stainless" name.
3. Nickel. It adds strength and toughness and helps the rust-resistant finish.
4. So iron is mixed with Ni and Cr, option (a).

Why this matters. Knowing alloy compositions is a common exam ask. Pairing the base metal with the right additives (Fe + Ni + Cr for stainless steel) is a clean two-mark fact.

Final Answer: Option (a): stainless steel is iron with nickel and chromium.

Match the colour to the product. The green colour is the fingerprint; I link it to copper carbonate formed in moist, CO₂-rich air.

Concept used. Copper corrodes in damp air containing carbon dioxide. The product is basic copper carbonate, which is green, unlike copper's other compounds.

1. What is in the air. Moisture and carbon dioxide are the reactive species that attack copper over time.
2. The product. They form green basic copper carbonate, CuCO₃.Cu(OH)₂, written here as CuCO₃.
3. Reject by colour. CuSO₄ is blue and needs sulphuric acid; CuO is black; Cu(NO₃)₂ needs nitric acid. None is the green air-corrosion product.
4. Hence the green coating is CuCO₃, option (b).

Why this matters. Reading corrosion by colour is a fast exam skill: green means copper carbonate, brown means iron rust, black means silver sulphide.

Final Answer: Option (b): copper gains a green coating of CuCO₃ in moist air.

Recall the lone liquid metal. Only one metal breaks the "all metals are solid" rule, so I just name it and confirm the others are solids.

Concept used. Melting point decides physical state. Most metals have very high melting points and are solid, but mercury's is well below room temperature, so it stays liquid.

1. Sodium, iron, chromium. All have melting points far above room temperature, so they are solids. Rejected.
2. Mercury. Melting point about −39°C, below room temperature, so it flows as a silvery liquid.
3. Hence mercury is the liquid metal, option (d).

Why this matters. Mercury's liquid state is why it was used in old thermometers and barometers, where a liquid metal column rises and falls with temperature or pressure.

Final Answer: Option (d): mercury is the only metal liquid at room temperature.

Use the reactivity series as a sorter. Top metals need electrolysis; middle metals need carbon reduction. I place each metal and keep the top ones.

Concept used. Carbon cannot reduce the oxides of the most reactive metals because those metals hold their oxygen too strongly. Such metals are extracted by passing current through their molten salts.

1. Sodium. Very high in the series, so molten NaCl is electrolysed to get sodium. Statement (i) qualifies.
2. Calcium. Also high in the series; molten CaCl₂ is electrolysed. Statement (ii) qualifies.
3. Iron and copper. These sit in the middle/low part and are obtained by reducing their oxides with carbon. So (iii) and (iv) are out.
4. Hence sodium and calcium, option (d).

Why this matters. This links to Question 61(b): carbon cannot reduce Na or Mg oxides, so electrolysis is the only practical way to win these very reactive metals.

Final Answer: Option (d): sodium and calcium come from electrolysis of their molten chlorides.

Hunt for the exception. The question asks for the odd non-metal that breaks the "dull" rule, so I scan for the shiny one.

Concept used. Lustre is usually a metallic property, but iodine is a non-metal that still has a bright metallic shine on its crystal faces, making it a standard exception.

1. Sulphur. A dull, powdery yellow solid. No lustre.
2. Oxygen and nitrogen. Colourless gases; lustre does not apply. Rejected.
3. Iodine. Greyish-black crystals with a clear metallic shine. This is the lustrous non-metal.
4. Hence option (d).

Why this matters. Exceptions like iodine's lustre and graphite's conductivity remind students that "metal vs non-metal" rules are general trends, not absolute laws.

Final Answer: Option (d): iodine, though a non-metal, is lustrous.

Find the bottom of the ladder. The metal displaced by the other three must be the least reactive, so I rank the four and pick the lowest.

Concept used. A higher metal in the reactivity series displaces a lower one from its salt solution. So the most-displaced metal is the one at the very bottom of the given set.

1. Rank them. Mg > Zn > Cu > Ag, with magnesium most reactive and silver least.
2. Apply the rule. Each metal above silver can displace it, for example Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag.
3. Check the others. Magnesium is on top, so nothing here displaces it; zinc and copper are displaced by fewer metals than silver.
4. So silver is displaced by all three, option (b).

Why this matters. This logic powers electroplating and metal recovery: a cheaper, more reactive metal is used to push a valuable, less reactive metal like silver out of solution.

Final Answer: Option (b): silver is displaced from its salt solution by Mg, Zn and Cu.

Decode the three test tubes. I read what is in each tube, then ask which metal would behave exactly as described.

Concept used. Tube A is conc. HCl, tube B is conc. HNO₃, tube C is aqua regia. A metal untouched by single acids but dissolved by aqua regia is a noble metal, gold.

1. Tubes A and B. No reaction means the metal resists both concentrated HCl and concentrated HNO₃ on their own. Most common metals would react with at least one.
2. Tube C. The metal dissolves in aqua regia. Aqua regia is famous for dissolving gold.
3. Test the options. Aluminium and copper react with single acids, so they would change in A or B; gold fits "untouched by single acids, dissolved by aqua regia".
4. So the metal is gold, option (b).

Why this matters. This is a classic identify-the-metal experiment. It directly uses the aqua regia fact from Question 8 and shows why gold needs such a special reagent to dissolve.

Final Answer: Option (b): the metal is gold, dissolving only in the aqua regia of tube C.

Classify the substance. I sort the four choices into element, compound, or mixture, then pick the right kind of mixture.

Concept used. A homogeneous mixture has uniform composition throughout. Alloys are made by melting metals together so they blend evenly, giving a single uniform solid.

1. Not an element. An alloy has more than one kind of atom, so it is not an element. Drop (a).
2. Not a compound. Its components are not in a fixed chemical ratio bonded together, so it is not a compound. Drop (b).
3. Homogeneous vs heterogeneous. The metals are evenly blended, so the mixture is uniform throughout, that is homogeneous.
4. So an alloy is a homogeneous mixture, option (c).

Why this matters. Calling alloys homogeneous mixtures explains why their properties (like brass being harder than copper) can be tuned by changing the proportions, something only mixtures allow.

Final Answer: Option (c): an alloy is a homogeneous mixture.

Trace the battery terminals. The applied battery fixes the charges, so I connect each electrode to the matching terminal.

Concept used. An electrolytic cell uses an external battery. The electrode joined to the battery's positive terminal becomes the anode (positive); the one joined to the negative terminal becomes the cathode (negative).

1. Anode. Wired to the positive terminal, so the anode is positively charged. Statement (iii) is correct.
2. Cathode. Wired to the negative terminal, so the cathode is negatively charged. Statement (iv) is correct.
3. Reject the reversals. Statements (i) and (ii) swap the signs, so they are wrong.
4. Hence option (b).

Why this matters. Getting these signs right is essential for electrolytic refining (Questions 24, 36 and 62), where pure metal deposits on the negative cathode and impure metal dissolves from the positive anode.

Final Answer: Option (b): anode positive, cathode negative, statements (iii) and (iv).

Follow the metal ions. Pure metal always builds up where ions gain electrons, the cathode, so I trace the ion's journey.

Concept used. In refining, the impure anode loses metal as ions into the solution, and those ions are reduced to pure metal on the cathode. The cathode therefore grows while the anode shrinks.

1. At the anode. Impure zinc dissolves: Zn → Zn²⁺ + 2e^-. So nothing deposits on the anode.
2. In the solution. Zn²⁺ ions drift towards the negative cathode.
3. At the cathode. Zn²⁺ + 2e^- → Zn. Pure zinc plates out here.
4. So zinc is deposited on the cathode, option (a).

Why this matters. This cathode-deposition idea is the heart of all electrolytic refining, and it reappears for copper in Questions 36 and 62.

Final Answer: Option (a): refined zinc deposits on the cathode.

Collect the clues, then match. Each sentence is a property; I list them and find the one element that fits all three at once.

Concept used. Alkali metals such as sodium are soft, low-melting and extremely reactive. Their high reactivity means they react fast with both air and water and must be kept under kerosene.

1. Soft, cut with a knife. This rules out magnesium and calcium, which are harder, and rules out non-metal phosphorus.
2. Cannot be kept in air. Sodium oxidises instantly in air, so it is stored under kerosene. Strong sodium clue.
3. Reacts vigorously with water. 2Na + 2H₂O → 2NaOH + H₂, often catching fire. Classic sodium.
4. All three point to sodium, option (b).

Why this matters. Identify-the-element questions are common, and sodium's three-clue signature (soft, air-sensitive, vigorous with water) is one of the most tested in Class 10.

Final Answer: Option (b): the element is sodium.

List each alloy's parts. For every option I write its constituents and check whether any is a non-metal.

Concept used. Alloys can be metal--metal or metal--non-metal. Identifying the constituents tells us which alloy carries a non-metal.

1. Brass. Copper + zinc, two metals. No non-metal.
2. Bronze. Copper + tin, two metals. No non-metal.
3. Amalgam. A metal dissolved in mercury, both metals. No non-metal.
4. Steel. Iron + carbon. Carbon is a non-metal, so steel carries a non-metal. Option (d).

Why this matters. Knowing alloy compositions is a frequent exam fact, and steel's carbon content is what makes it harder and stronger than pure iron.

Final Answer: Option (d): steel contains the non-metal carbon.

Check each statement against magnesium's known reactions. I treat each line as a claim and test it for truth.

Concept used. Magnesium burns in oxygen to MgO, but with water it needs hot water or steam and gives the hydroxide Mg(OH)₂ plus hydrogen. It does not react with cold water.

1. (a) Burning. 2Mg + O₂ → 2MgO with a dazzling white flame. Correct.
2. (b) Cold water. Magnesium does not react with cold water at all, and the water product is the hydroxide, not the oxide. So (b) is false, the answer.
3. (c) Hot water. Mg + 2H₂O → Mg(OH)₂ + H₂. Correct.
4. (d) Steam. Magnesium reacts with steam to give Mg(OH)₂ (or MgO) and hydrogen. Treated as correct here. So the incorrect statement is (b).

Why this matters. Magnesium's position (reacts with hot water, not cold) is a fixed point on the reactivity ladder used in Questions 4 and 63, so getting it right pays off across the chapter.

Final Answer: Option (b): this statement is incorrect for magnesium.

Spot the keyword "amalgam". The term amalgam directly signals mercury, so I look for it among the options.

Concept used. An amalgam is, by definition, an alloy of a metal with mercury. So any alloy named "amalgam" must contain mercury.

1. Stainless steel. Iron, nickel, chromium. No mercury.
2. Alnico. Aluminium, nickel, cobalt. No mercury.
3. Solder. Lead and tin. No mercury.
4. Zinc amalgam. The word "amalgam" means mercury is present. So option (d).

Why this matters. Recognising amalgam = mercury alloy is a one-line fact that also explains real uses, from old dental fillings to the extraction of gold by amalgamation.

Final Answer: Option (d): zinc amalgam is the mercury-containing alloy.

Name the bond type first. Electron transfer means ionic, so I list ionic properties and pick the one missing from that list.

Concept used. Electron loss by X and gain by Y is ionic bonding. The resulting strong lattice gives high melting points, solid form and conduction only when molten or dissolved.

1. Identify Z. Electron transfer from X to Y makes Z an ionic compound.
2. List ionic properties. High melting point (a), molten conduction (c), solid at room temperature (d). All true of Z.
3. Find the misfit. Low melting point is the opposite of what ionic compounds show, so (b) is the property Z does not have.
4. Hence option (b).

Why this matters. This question ties bonding type to physical properties, the same chain of reasoning used in Questions 10 and 30 to tell ionic from covalent substances.

Final Answer: Option (b): a low melting point is not shown by the ionic compound Z.

Read the last number in each configuration. The valence electron count is the only thing I need to label each element.

Concept used. Metals have 1--3 valence electrons and tend to lose them; non-metals have 5--7 and tend to gain; a complete octet is a noble gas that does neither.

1. X (2, 8). Outer shell has 8 electrons, a complete octet, so X is a noble gas, not a metal. Rejects (a).
2. Y (2, 8, 7). Seven valence electrons; Y gains one to complete its octet, so Y is a non-metal. Rejects (b).
3. Z (2, 8, 2). Two valence electrons; Z loses them, so Z is a metal. This makes (c) "Z is a non-metal" wrong.
4. So "Y non-metal, Z metal" is correct, option (d).

Why this matters. This connects electronic structure to chemical behaviour, the foundation for why metals form positive ions and non-metals form negative ions, as seen throughout this chapter.

Final Answer: Option (d): Y is a non-metal and Z is a metal.

Recall the amphoteric oxides. Only a short list of metal oxides react with both acid and base; I match the option to that list.

Concept used. An amphoteric oxide shows both acidic and basic behaviour. Aluminium oxide is the standard Class 10 example, dissolving in both acids and alkalis.

1. Sodium, calcium oxides. Strongly basic only. Not amphoteric.
2. Copper oxide. Basic. Not amphoteric.
3. Aluminium oxide. Reacts with HCl: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O; and with NaOH: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. Both acid and base, so amphoteric.
4. Hence aluminium, option (c).

Why this matters. The amphoteric nature of Al₂O₃ is why aluminium is so versatile and why its oxide can be attacked by alkaline cleaners, a real-world consequence of this property.

Final Answer: Option (c): aluminium forms the amphoteric oxide Al₂O₃.

Look for the free electron. Conduction needs mobile charge, so I pick the carbon form that has spare electrons.

Concept used. In graphite, each carbon bonds to only three neighbours, leaving one electron free per atom. These delocalised electrons move between layers and conduct, unlike diamond where all four bonds are used.

1. Diamond. Every carbon uses all four electrons in rigid bonds, so none are free. Does not conduct.
2. Sulphur and fullerene. No free mobile electrons available for conduction. Reject both.
3. Graphite. One free electron per carbon glides between sheets, carrying current. So graphite conducts.
4. Hence option (b).

Why this matters. This is why graphite is used as electrodes in electrolysis and in batteries, and why the "lead" in a pencil leaves a conducting mark, all thanks to those free electrons.

Final Answer: Option (b): graphite is the good electrical conductor.

Pick the non-conductor. The coating must not carry current, so I reject anything that conducts and keep the insulator.

Concept used. Insulation works only with a material that has no free charges to move. Plastics like PVC are insulators; graphite, with its free electrons, is a conductor and would be dangerous.

1. Graphite. It conducts electricity (from Question 32), so it cannot insulate. Reject (b).
2. Sulphur. Not a practical coating material for wires. Reject (a).
3. PVC. A flexible, cheap plastic that does not conduct, ideal for wire coating. So (c).
4. Since only PVC works, "all can be used" (d) is wrong. Option (c).

Why this matters. This contrasts directly with Question 32: graphite conducts (used as electrodes), while PVC insulates (used as wire coating). Same topic, opposite role.

Final Answer: Option (c): PVC is the insulating material used to coat wires.

Recall the lone liquid non-metal. Just one non-metal is liquid at room temperature, so I name it and confirm the rest are solids.

Concept used. Physical state at room temperature depends on melting and boiling points. Bromine's are placed such that it is liquid at ordinary temperatures, unlike the other solid non-metals listed.

1. Carbon, phosphorus, sulphur. All solids at room temperature. Rejected.
2. Bromine. A dense, red-brown liquid at room temperature, the only liquid non-metal.
3. Hence option (b).

Why this matters. Remembering bromine (liquid non-metal) and mercury (liquid metal) together is the cleanest way to answer state-based questions, including the SA Question 52.

Final Answer: Option (b): bromine is the liquid non-metal.

Compare reactivities pair by pair. For each option I ask: is the free metal above the salt's metal? Only then does a reaction happen.

Concept used. A more reactive metal displaces a less reactive one from its salt solution. Reactivity order here: Mg > Zn > Fe > Pb > Cu.

1. (a) MgSO₄ + Fe. Iron is below magnesium, so it cannot displace Mg. No reaction.
2. (b) ZnSO₄ + Fe. Iron is below zinc, so it cannot displace Zn. No reaction.
3. (c) MgSO₄ + Pb. Lead is far below magnesium. No reaction.
4. (d) CuSO₄ + Fe. Iron is above copper, so Fe + CuSO₄ → FeSO₄ + Cu. A reaction occurs. Option (d).

Why this matters. This is the same chemistry that ruins an iron vessel holding copper sulphate (Question 58): iron, being more reactive, keeps displacing copper and slowly eats away.

Final Answer: Option (d): CuSO₄ + Fe is the only pair that reacts.

Audit the figure against the rule. I have one fixed mental diagram for refining and compare each option to it.

Concept used. During refining, the positive anode (impure metal) dissolves into the electrolyte and the negative cathode (pure metal) grows as ions are deposited on it. The electrolyte is a soluble salt of the metal.

1. Anode side. The thick impure block must be at the positive terminal, dissolving as M → Mⁿ⁺ + n e^-.
2. Cathode side. The thin pure strip must be at the negative terminal, gaining metal as Mⁿ⁺ + n e^- → M.
3. Electrolyte. A solution of the metal's salt carries the ions across.
4. Only figure (c) has impure anode at +, pure cathode at -, in a salt solution. So option (c).

Why this matters. This figure is the visual form of Questions 24 and 62. Knowing where each electrode goes turns an abstract process into a diagram you can draw and label in the exam.

Final Answer: Option (c): the figure with impure-metal anode and pure-metal cathode in salt solution.

Read the double reactivity as a clue. A metal fizzing with both acid and base is the textbook sign of an amphoteric metal releasing hydrogen, so I name the gas and write both equations.

Concept used. An amphoteric metal such as zinc reacts with acids and with alkalis, each time displacing hydrogen. Hydrogen is a light, flammable gas that pops when a flame is brought near.

1. What the bubbles are. The same gas appears with both NaOH and HCl, so it is hydrogen, released by the divalent metal M.
2. Confirm hydrogen. Hold a burning matchstick at the test tube mouth; the pop sound proves the gas is hydrogen.
3. With the alkali. M + 2NaOH → Na₂MO₂ + H₂, forming a sodium metallate and hydrogen.
4. With the acid. M + 2HCl → MCl₂ + H₂, forming the metal chloride and hydrogen.

Why this matters. Zinc is the everyday example of such a metal, and this double reaction is why galvanised (zinc-coated) iron protects so well, the zinc itself is reactive towards both kinds of attack.

Final Answer: Gas is hydrogen (pop sound test); M + 2NaOH → Na₂MO₂ + H₂ and M + 2HCl → MCl₂ + H₂.

Apply the universal refining rule to silver. Refining always uses impure metal as anode, pure as cathode, in the metal's own salt, so I just plug silver into that template.

Concept used. In refining, the anode dissolves and the cathode grows. The electrolyte must be a soluble salt of the metal so its ions can ferry the metal across.

1. Anode and cathode. Impure silver is the anode; pure silver is the cathode.
2. Electrolyte. A solution of a silver salt, AgNO₃, lets Ag+ ions move across.
3. At the anode. Ag → Ag+ + e^-, silver dissolves off the impure block.
4. At the cathode. Ag+ + e^- → Ag, so pure silver plates onto the cathode. That is where we collect it.

Why this matters. This is the same anode-to-cathode transfer used for zinc (Question 24) and copper (Questions 36, 62), showing one method refines many metals.

Final Answer: Anode: impure silver; Cathode: pure silver; Electrolyte: AgNO₃ solution; pure silver collects at the cathode.

Start from the easiest reduction. The extraction always tries to reach an oxide first because oxides are the easiest to turn into metal, so I explain why and how.

Concept used. Reduction of an oxide is thermodynamically and practically easier than reduction of a sulphide or carbonate. So the ore is first converted to oxide form before the reduction step.

1. The goal. Reduction (to free metal) works best on the oxide, so the oxide is the target intermediate.
2. Sulphide route. Roasting: heat the sulphide in air to get the oxide and release SO₂.
3. Carbonate route. Calcination: heat the carbonate in limited air to get the oxide and release CO₂.
4. Then reduce. The oxide is finally reduced (often by carbon) to the metal. This is why conversion to oxide comes first.

Why this matters. This single principle organises the whole of metallurgy in the chapter: every extraction in Questions 55, 60 and 65 passes through the metal oxide stage.

Final Answer: Oxides are far easier to reduce than sulphides or carbonates, so ores are first turned into oxides (roasting/calcination), then reduced.

Treat HNO₃ as the odd acid. The whole answer rests on nitric acid being an oxidiser, so I explain what it does to the freed hydrogen.

Concept used. Ordinary acids release hydrogen when a metal displaces it. But an oxidising acid like HNO₃ immediately oxidises that hydrogen to water, so no hydrogen gas is seen.

1. Normal acids. HCl and dilute H₂SO₄ give salt and H₂ gas with metals.
2. Nitric acid is different. It is a strong oxidiser, so it oxidises the just-formed hydrogen into water.
3. Result. Instead of H₂, the products are water and nitrogen oxides, so no hydrogen gas comes off.
4. Exception. Very dilute HNO₃ with Mg or Mn does give H₂, which is why these are named as exceptions.

Why this matters. This oxidising nature explains why aluminium can be stored in HNO₃ (Question 61a) and why gold needs aqua regia, where HNO₃ provides the oxidising punch.

Final Answer: Nitric acid oxidises the liberated hydrogen to water, so H₂ is not released (Mn and Mg with very dilute HNO₃ are the exceptions).

Recognise the railway-track clue. "Aluminium joins railway tracks" is the thermite reaction, so I name X, the reaction, and write the balanced equation.

Concept used. Aluminium is more reactive than iron, so it displaces iron from Fe₂O₃. The reaction is so exothermic that the iron comes out molten, ideal for welding.

1. Identify X. The oxide reduced by aluminium here is iron(III) oxide, Fe₂O₃.
2. Name it. This aluminothermic process is the thermite reaction.
3. Write it. Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + Heat, balanced with 2 Fe and 2 Al.
4. Use of the heat. The released heat melts the iron, which welds the rail joint.

Why this matters. This is the same reaction that identifies aluminium in Question 46 ("metal used in thermite process") and shows displacement put to practical use.

Final Answer: X = Fe₂O₃; thermite reaction; Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + Heat.

Use the mass to crack the identity. The molecular mass of XOH is a direct route to the metal, so I subtract the OH mass and match the result.

Concept used. The hydroxide XOH has mass 40. Subtracting OH (mass 17) gives the metal's atomic mass. A reactive metal that reacts with cold water giving a flammable gas is an alkali metal.

1. Find X. Atomic mass of X = 40 − 17 = 23. This is sodium.
2. Find Y. The basic salt (hydroxide) is NaOH.
3. Find Z. The flammable gas released is hydrogen, H₂.
4. Write the reaction. 2Na + 2H₂O → 2NaOH + H₂, with heat that can ignite the hydrogen.

Why this matters. This reaction explains why sodium is stored under kerosene (Question 25): its vigorous reaction with even cold water releases flammable hydrogen.

Final Answer: X = sodium, Y = NaOH, Z = hydrogen; 2Na + 2H₂O → 2NaOH + H₂.

Match each property to a carbon allotrope. The two clues, hardest and conducting, point straight to diamond and graphite, two forms of one non-metal.

Concept used. Allotropes are different physical forms of the same element. Carbon's allotropes include diamond (very hard, all four bonds used) and graphite (conducting, one free electron per atom).

1. The element. A non-metal in two forms with these properties is carbon, so X is carbon.
2. The hardest form. Diamond, where each carbon is bonded to four others in a rigid network. So Y is diamond.
3. The conducting form. Graphite, with free electrons moving between layers. So Z is graphite.

Why this matters. Allotropy explains why a single element can do two opposite jobs: diamond cuts glass while graphite lubricates and conducts, both from plain carbon.

Final Answer: X = carbon, Y = diamond, Z = graphite.

Track the oxygen for each species. Oxidation is gaining oxygen and reduction is losing it, so I follow where the oxygen goes.

Concept used. In this thermite-type reaction, aluminium pulls oxygen away from manganese dioxide. Whoever gains oxygen is oxidised; whoever loses it is reduced.

1. Aluminium's journey. Al Al₂O₃: it gains oxygen, so aluminium is oxidised. Hence "is aluminium reduced?" answer is No.
2. Manganese dioxide's journey. MnO₂ Mn: it loses oxygen, so MnO₂ is reduced. Hence "is MnO₂ oxidised?" answer is No.
3. Roles. Aluminium is the reducing agent; MnO₂ is the oxidising agent.

Why this matters. This is the thermite idea again (as in Question 41), used here to extract manganese, showing how a reactive metal reduces a less reactive metal's oxide.

Final Answer: (a) No, aluminium is oxidised. (b) No, MnO₂ is reduced.

Pair the alloy with the job it does. Solder's task is to melt and join wires, so its useful property must be a low melting point.

Concept used. Alloys are chosen for properties that suit a task. Solder, an alloy of lead and tin, melts at a low temperature, which is exactly what wire-joining needs.

1. Name the constituents. Solder = lead + tin.
2. Identify the task. It is used to weld (join) electrical wires.
3. Match the property. A low melting point lets solder melt with a small soldering iron, flow into the joint, and solidify to hold the wires.
4. So the suitable property is its low melting point.

Why this matters. This is why a cheap soldering iron can fix a wire joint: the lead-tin alloy melts at a temperature low enough not to harm the components around it.

Final Answer: Solder = lead + tin; its low melting point makes it ideal for welding electrical wires.

Start from the thermite clue. The metal in the thermite process is aluminium, so its oxide is Al₂O₃, which I then test against acid and base.

Concept used. An amphoteric oxide reacts with both acids and alkalis. Aluminium oxide does both, behaving as a base with HCl and as an acid with NaOH.

1. Metal A. Aluminium, the reducing metal in the thermite process.
2. Oxide B. 4Al + 3O₂ → 2Al₂O₃, so B is aluminium oxide, Al₂O₃.
3. With HCl. Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O, acting as a base.
4. With NaOH. Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O, acting as an acid. Both reactions confirm amphoteric character.

Why this matters. Amphoteric Al₂O₃ is why aluminium is so durable yet can be attacked by strong alkalis, a real concern when choosing cleaning agents for aluminium utensils.

Final Answer: A = Al, B = Al₂O₃; reacts with HCl (Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O) and NaOH (Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O).

Start from "liquid metal". The clue points to mercury, so I name its ore and write the two-step heating that frees the metal.

Concept used. Metals low in the reactivity series can be obtained by simply heating (roasting) their sulphide ores. Mercury's oxide is so unstable that heat alone decomposes it to the metal.

1. Identify the metal and ore. Liquid metal = mercury; sulphide ore = cinnabar, HgS.
2. Step 1, roasting. 2HgS + 3O₂ → 2HgO + 2SO₂ on heating in air.
3. Step 2, thermal decomposition. 2HgO → 2Hg + O₂ on further heating, releasing the free mercury.
4. So mercury is won from cinnabar by heating alone.

Why this matters. This is the classic example of extracting a low-reactivity metal: no electrolysis, no carbon, just heat, contrasting with the elaborate routes needed for reactive metals.

Final Answer: Mercury from cinnabar (HgS): 2HgS + 3O₂ → 2HgO + 2SO₂, then 2HgO → 2Hg + O₂, both on heating.

Balance the charges for each pair. I write the valency of each element and cross-multiply to get a neutral formula.

Concept used. In a stable ionic binary compound, total positive charge equals total negative charge. The criss-cross of valencies gives the smallest whole-number formula.

1. Mg and N. Mg is +2, N is −3; cross to Mg₃N₂.
2. Li and O. Li is +1, O is −2; cross to Li₂O.
3. Al and Cl. Al is +3, Cl is −1; cross to AlCl₃.
4. K and O. K is +1, O is −2; cross to K₂O.

Why this matters. Writing correct formulae is the backbone of balancing equations throughout this chapter, from oxides in extraction to chlorides in electrolysis.

Final Answer: (a) Mg₃N₂ (b) Li₂O (c) AlCl₃ (d) K₂O.

Name each process, then write the equation. Part (a) is calcination and part (b) is auto-reduction, so I label and balance both.

Concept used. Calcination decomposes a carbonate to its oxide and CO₂. Auto-reduction is when a metal's own sulphide reduces its oxide to the metal, releasing SO₂.

1. (a) Carbonate heated. ZnCO₃ → ZnO + CO₂. The carbonate loses CO₂ to leave zinc oxide.
2. (b) Oxide plus sulphide. 2Cu₂O + Cu₂S → 6Cu + SO₂. The sulphide's sulphur takes the oxide's oxygen as SO₂, freeing copper.
3. Why auto-reduction. No carbon or other reducer is added; the ore mix reduces itself.

Why this matters. Auto-reduction is the key final step in copper extraction (Question 62), showing how some metals are won without a separate reducing agent.

Final Answer: (a) ZnCO₃ → ZnO + CO₂; (b) 2Cu₂O + Cu₂S → 6Cu + SO₂, both on heating.

Match the clues to carbon. A food non-metal with one toxic and one warming oxide is carbon, so I name both oxides and place carbon in its group.

Concept used. Carbon forms two common oxides: carbon monoxide (toxic, from incomplete burning) and carbon dioxide (a greenhouse gas). It sits in Group 14 of the Periodic Table.

1. The non-metal. Carbon is essential in food (carbohydrates, fats, proteins), so A is carbon.
2. The toxic oxide. Carbon monoxide, CO, is poisonous, so B is CO.
3. The warming oxide. Carbon dioxide, CO₂, drives global warming, so C is CO₂.
4. The group. Carbon has 4 valence electrons and is in Group 14.

Why this matters. This links chemistry to current affairs: CO₂ from burning fossil fuels is the central greenhouse gas, and CO is the silent killer behind faulty heaters.

Final Answer: A = carbon, B = CO, C = CO₂; carbon is in Group 14.

Sort metals by thermal conductivity. The question wants two best and two worst, so I name the standard examples for each.

Concept used. Thermal conductivity varies across metals. Silver and copper top the list, which is why they are used where fast heat transfer matters; lead and mercury are relatively poor conductors.

1. Best conductors. Silver (Ag) is the best, closely followed by copper (Cu).
2. Poor conductors. Lead (Pb) and mercury (Hg) conduct heat much less efficiently.
3. So the two good examples are silver and copper, and the two poor examples are lead and mercury.

Why this matters. This is why cooking utensils often have copper or aluminium bases (good conductors) for even heating, a direct application of metal thermal conductivity.

Final Answer: Good: silver and copper; Poor: lead and mercury.

Collect the liquid-state facts. The question gathers several "unusual state" points, so I recall each one carefully.

Concept used. Physical state depends on melting point relative to room temperature. Mercury and bromine are liquid at room temperature, and caesium and gallium are low-melting metals that melt below body temperature.

1. Liquid metal. Mercury, melting at about −39°C, is liquid at room temperature.
2. Liquid non-metal. Bromine, a red-brown liquid at room temperature.
3. Low-melting metals. Caesium (melts near 302 K) and gallium (melts near 303 K), both below 310 K, so they melt with gentle warmth.

Why this matters. Gallium's hand-melting trick is a classic demonstration, and these state facts are reliable one-mark answers in board exams.

Final Answer: Mercury (liquid metal), bromine (liquid non-metal); caesium and gallium melt below 310 K.

Trace the lime cycle. White washing points to calcium hydroxide, so I run the calcium-lime cycle forwards and backwards.

Concept used. Calcium reacts with water to give slaked lime Ca(OH)₂. Heating it gives quick lime CaO, and adding water to quick lime returns slaked lime, a reversible lime cycle.

1. Identify the trio. A = calcium, B = Ca(OH)₂ (white washing), C = CaO.
2. Form B. Ca + 2H₂O → Ca(OH)₂ + H₂.
3. B to C on heating. Ca(OH)₂ → CaO + H₂O.
4. C back to B. CaO + H₂O → Ca(OH)₂, completing the cycle.

Why this matters. This lime cycle is everyday chemistry: from white washing walls to making mortar, calcium's oxide and hydroxide are among the most used compounds in construction.

Final Answer: A = Ca, B = Ca(OH)₂, C = CaO, linked by the lime cycle reactions above.

Use the mass, then the amphoteric clue. The molecular mass pins B as NaOH, and the reaction with Al₂O₃ reveals C as sodium aluminate.

Concept used. Sodium reacts with water to give NaOH. Because Al₂O₃ is amphoteric, it reacts with the base NaOH to form the soluble salt sodium aluminate.

1. Crack B. NaOH has mass 23 + 16 + 1 = 40, matching the clue. So B is NaOH and A is sodium.
2. Form B. 2Na + 2H₂O → 2NaOH + H₂.
3. React with Al₂O₃. Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O, giving soluble sodium aluminate.
4. So C is NaAlO₂.

Why this matters. The reaction of NaOH with Al₂O₃ is the same amphoteric behaviour from Questions 31 and 46, and is used industrially in the Bayer process to purify aluminium ore.

Final Answer: A = Na, B = NaOH, C = NaAlO₂; 2Na + 2H₂O → 2NaOH + H₂ and Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O.

Treat each ore separately. Zinc comes from a sulphide and a carbonate ore, so I roast the first and calcine the second.

Concept used. Roasting (sulphide + air) and calcination (carbonate, limited air) both aim to reach the oxide ZnO, the easiest form to reduce to zinc.

1. Roasting ZnS. 2ZnS + 3O₂ → 2ZnO + 2SO₂, releasing sulphur dioxide.
2. Calcination of ZnCO₃. ZnCO₃ → ZnO + CO₂, releasing carbon dioxide.
3. Common product. Both give ZnO, which is then reduced by carbon: ZnO + C → Zn + CO.

Why this matters. This is the practical application of Question 39: turn the ore into an oxide first, by roasting or calcination, then reduce it. Zinc shows both routes in one question.

Final Answer: (a) 2ZnS + 3O₂ → 2ZnO + 2SO₂; (b) ZnCO₃ → ZnO + CO₂, both on heating.

Use both clues together. "No hydrogen from acids" plus "black oxide" points to copper, so I confirm M and write its oxidation.

Concept used. Metals below hydrogen do not react with dilute acids to give hydrogen. Copper, on heating in air, is oxidised to black CuO.

1. No hydrogen clue. The metal sits below hydrogen in the series, so it cannot displace hydrogen from acids. Copper fits.
2. Black oxide clue. Heated copper forms black CuO, matching the description.
3. Reaction. 2Cu + O₂ → 2CuO; the copper gains oxygen, so it is oxidised.
4. Hence M is copper and the product is CuO.

Why this matters. This black CuO layer is what you see when a copper vessel is heated, and it links to Question 16, where copper's slow air-corrosion instead gives a green carbonate coat.

Final Answer: M = copper, black product = CuO; 2Cu + O₂ → 2CuO.

Read the oxide's acidity as the verdict. Acidic oxide means non-metal, so I state the rule and apply it directly.

Concept used. Non-metals form acidic oxides that turn blue litmus red and react with bases; metals form basic oxides. So the acidic nature of A₂O₃ settles the identity of A.

1. State the rule. Acidic oxides come from non-metals; basic oxides come from metals.
2. Apply it. A₂O₃ is acidic, so A is a non-metal.
3. Example. B₂O₃ (boron) and P₂O₃ (phosphorus) are acidic A₂O₃-type oxides of non-metals.

Why this matters. This is the reverse of Question 31 (aluminium's amphoteric oxide): the character of an oxide is a quick test to classify any element as metal or non-metal.

Final Answer: A is a non-metal, because an acidic oxide is formed by non-metals.

Compare iron and copper. The holes come from iron dissolving, so I check reactivity and write the displacement.

Concept used. A metal higher in the reactivity series displaces a lower one from its salt solution. Iron is above copper, so iron dissolves while copper deposits.

1. Reactivity check. Iron is more reactive than copper, so it can displace copper.
2. The reaction. Fe + CuSO₄ → FeSO₄ + Cu. Iron goes into solution as FeSO₄; copper settles out.
3. The holes. As iron from the pot continually dissolves, the pot thins and develops holes.

Why this matters. This is the practical face of Question 35, where CuSO₄ + Fe was the only reacting pair. Reactivity decides which metal survives contact with which solution.

Final Answer: Iron (more reactive) displaces copper from CuSO₄ and dissolves, making holes: Fe + CuSO₄ → FeSO₄ + Cu.

Follow the nitrogen trail. Each clue is a step in nitrogen chemistry, so I move from air to ammonia to oxide to acid.

Concept used. Nitrogen is the major gas in air. With hydrogen it forms ammonia (Haber); with oxygen it forms NO; and NO, oxidised and dissolved in water with air, gives nitric acid, a strong oxidiser.

1. A is nitrogen. It is the largest constituent of air, so A = N₂.
2. B is ammonia. N₂ + 3H₂ → 2NH₃ over an iron catalyst, the 1 : 3 ratio matching the equation.
3. C is nitric oxide. Heating nitrogen with oxygen: N₂ + O₂ → 2NO.
4. D is nitric acid. NO is further oxidised to NO₂, which with water and air gives HNO₃, a strong oxidising agent. Nitrogen is in Group 15.

Why this matters. This ties the chapter's "HNO₃ is a strong oxidiser" theme (Questions 7, 40) back to where nitric acid comes from, showing chemistry as a connected chain, not isolated facts.

Final Answer: A = N₂, B = NH₃, C = NO, D = HNO₃; nitrogen belongs to Group 15.

Lay out the standard three-step route. For sulphide ores I always roast first, then reduce, then refine, adjusting for low-reactivity metals.

Concept used. Sulphides are first roasted to oxides because oxides reduce far more easily. Medium-reactivity metals are then reduced with carbon; low-reactivity metals need only heat.

1. Roasting. 2ZnS + 3O₂ → 2ZnO + 2SO₂, turning the sulphide into the oxide.
2. Reduction with carbon. ZnO + C → Zn + CO for medium-reactivity metals like zinc, iron, lead.
3. Low-reactivity shortcut. Copper and mercury self-reduce or their oxides decompose on heating, needing no carbon.
4. Refining. Crude metal is purified (e.g. electrolytic refining) to high purity.

Why this matters. This single roadmap underlies every metallurgy question in the chapter (Questions 39, 55, 62, 65), so mastering the three steps pays off again and again.

Final Answer: Roast sulphide to oxide, reduce oxide to metal (carbon for medium-reactivity, heat alone for low-reactivity), then refine.

Answer each part with the one idea behind it. Five mini-questions, each resting on a single concept, so I name the concept and apply it.

Concept used. Passivation (a), reactivity limits of carbon (b), ionic conduction needing free ions (c), corrosion protection (d), and reactivity governing natural occurrence (e).

1. (a) HNO₃ forms a protective Al₂O₃ film on aluminium, sealing the surface and lowering reactivity.
2. (b) Carbon is less reactive than Na or Mg, so it cannot take oxygen from their oxides; electrolysis is needed instead.
3. (c) Conduction needs mobile ions. Solid NaCl has fixed ions (no conduction); molten or dissolved NaCl has free ions (conducts).
4. (d) and (e) Iron is galvanised with zinc to stop rust; and very reactive metals (Na, K, Ca, Mg) react so readily that they only ever exist as compounds.

Why this matters. Each part is a stand-alone board favourite, and together they show how one master idea, reactivity, explains corrosion, extraction and natural occurrence across the whole chapter.

Final Answer: (a) protective oxide layer; (b) Na/Mg more reactive than carbon; (c) free ions only when molten/aqueous; (d) zinc protects iron; (e) high reactivity means always combined.

Walk the three copper steps. Roast, auto-reduce, then refine, with a clear labelled diagram for the refining cell.

Concept used. Copper(I) sulphide is partly roasted to the oxide, which is then reduced by the leftover sulphide (auto-reduction). The blister copper is finally purified electrolytically.

1. Roasting. 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, making the oxide and releasing SO₂.
2. Auto-reduction. 2Cu₂O + Cu₂S → 6Cu + SO₂; the sulphide reduces the oxide to copper.
3. Refining cell. Impure copper anode (+) dissolves; pure copper cathode (-) grows, in CuSO₄ electrolyte (see the labelled diagram above).
4. Electrode reactions. Anode: Cu → Cu²⁺ + 2e^-; cathode: Cu²⁺ + 2e^- → Cu.

Why this matters. This is the full life of copper from ore to wire, combining roasting (Question 55), auto-reduction (Question 49) and refining (Questions 24, 36) into one complete extraction.

Final Answer: Roast (2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂), auto-reduce (2Cu₂O + Cu₂S → 6Cu + SO₂), then electrolytic refining with impure-copper anode and pure-copper cathode in CuSO₄.

Rank by water reactivity. Each metal's water behaviour fixes its place, so I order them from steam-only (least) to cold-water (most).

Concept used. The hotter the water a metal needs to react, the lower its reactivity. So cold-water metals top the series and steam-only metals sit lower.

1. X with cold water. Only the most reactive metals (sodium, potassium) react with cold water. So X is Na (or K).
2. Y with hot water. Magnesium and calcium need hot water. So Y is Mg (or Ca).
3. Z with steam only. Iron reacts only with steam. So Z is Fe.
4. Order. Reactivity increases Fe < Mg < Na.

Why this matters. This is the same water-reactivity ladder used in Questions 4 and 27, now applied to rank three unknown metals, a common board-exam reasoning task.

Final Answer: X = Na, Y = Mg, Z = Fe; increasing reactivity Fe < Mg < Na.

Read the flame and the atomic number. Golden flame means sodium; atomic number 17 means chlorine, so I build the salt and then electrolyse it.

Concept used. Sodium gives a characteristic golden flame and reacts with chlorine to form NaCl. Electrolysing aqueous NaCl (brine) yields NaOH at the cathode side with hydrogen, and chlorine at the anode.

1. A is sodium. The golden flame test identifies sodium.
2. B is chlorine. Atomic number 17 is chlorine.
3. C is NaCl. 2Na + Cl₂ → 2NaCl, the combination of the two.
4. Electrolyse C. 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂, so D is NaOH and hydrogen is set free.

Why this matters. This connects flame tests, ionic compound formation and electrolysis in one chain, and the chlor-alkali process is a real industry built entirely on this chemistry.

Final Answer: A = Na, B = Cl, C = NaCl, D = NaOH; 2Na + Cl₂ → 2NaCl and 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂.

Decode the gas, then apply the route. CO₂ flags a carbonate and SO₂ flags a sulphide, so I calcine one, roast the other, then reduce both.

Concept used. Carbonate ores release CO₂ on calcination; sulphide ores release SO₂ on roasting. Both give the metal oxide, which carbon then reduces to the metal.

1. Ore A is a carbonate. Calcine: MCO₃ → MO + CO₂.
2. Ore B is a sulphide. Roast: 2MS + 3O₂ → 2MO + 2SO₂.
3. Reduce both oxides. MO + C → M + CO gives the free metal in each case.
4. So both ores reach the metal through the oxide stage.

Why this matters. This is the capstone of the chapter's metallurgy: identify the ore from the gas, convert to oxide (calcination or roasting), then reduce, exactly the method used in Questions 39, 55 and 60.

Final Answer: Calcine carbonate A (MCO₃ → MO + CO₂) and roast sulphide B (2MS + 3O₂ → 2MO + 2SO₂), then reduce both oxides with carbon (MO + C → M + CO).