CBSE Class 10 Science HOTs Metals and Non Metals Set 07

This is Set 07 of higher-order thinking (HOTS) questions for CBSE Class 10 Science Chapter 3, Metals and Non-metals, with step-wise answers and a Teacher's Note for each of the 18 questions. The questions test identifying unknown metals from clues, displacement and electrolysis set-ups, rust protection, and reasoning about extraction and ionic compounds.

Short Answer HOTS Questions

Question 1. A copper plate was dipped into a solution of AgNO₃. After sometime, a black layer was deposited on the copper plate. State the reason for it. Write the chemical equation of the reaction involved.
Answer:
Reason: Copper is more reactive than silver, so copper displaces silver from the silver nitrate solution. The silver that forms is deposited on the copper plate as a thin dark layer, and the solution slowly turns blue because copper(II) nitrate is formed.
Equation: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
In simple words: Copper is more active than silver, so it pushes silver out of the solution and the silver sticks on the plate.

Teacher's Note:
1. In a displacement reaction the more reactive metal goes into the solution.
2. Mention the blue colour of copper(II) nitrate to complete the observation.

Question 2. Give reasons for the following : (i) Silicon is considered as a metalloid. (ii) Carbon is not used for extraction of aluminium from aluminium oxide. (iii) For producing hydrogen gas by reaction with hydrochloric acid, granulated zinc is preferred over block of zinc.
Answer:
(i) Silicon shows some properties of metals (it is shiny and conducts electricity slightly) and some of non-metals (it is brittle and forms an acidic oxide, SiO₂). An element with such mixed properties is called a metalloid.
(ii) Aluminium is high in the reactivity series and has a greater affinity for oxygen than carbon has. So carbon cannot reduce Al₂O₃; aluminium is obtained by electrolysis of molten alumina.
(iii) Granulated zinc has a much larger surface area in contact with the acid than a block, so the reaction is faster and hydrogen is produced steadily: Zn + 2HCl → ZnCl₂ + H₂.
In simple words: Silicon is part metal and part non-metal; carbon is too weak to pull oxygen away from aluminium; small pieces of zinc react faster.

Teacher's Note:
1. Always link a reducing-agent question to the position of the metal in the reactivity series.
2. Surface area decides the speed of a reaction between a solid and an acid.

Question 3. Give reasons for the following : (i) Na, K and Ca metals form hydrides by combination with hydrogen gas, but most other metals do not. (ii) Zinc oxide is called an amphoteric oxide.
Answer:
(i) Na, K and Ca are highly electropositive, so they can give electrons to hydrogen atoms and form ionic hydrides: 2Na + H₂ → 2NaH and Ca + H₂ → CaH₂. Most other metals do not lose electrons so easily to hydrogen.
(ii) Zinc oxide reacts with both acids and bases to form salt and water, so it shows both basic and acidic behaviour.
ZnO + 2HCl → ZnCl₂ + H₂O
ZnO + 2NaOH → Na₂ZnO₂ + H₂O (sodium zincate)
In simple words: Very reactive metals hand electrons to hydrogen; zinc oxide reacts with an acid and also with an alkali.

Teacher's Note:
1. Support every amphoteric oxide with one equation each for the acid and the base.
2. Only the most electropositive metals form hydrides.

Question 4. Metals X and Y can be recovered from the anode mud left behind after the electrolytic refining of copper metal. The coins made of metal X look new even after several years of use but the coins made of metal Y lose their shine gradually and get blackened soon. When metal X is alloyed with a small amount of metal Y, it becomes hard and hence suitable for making ornaments. What are metals X and Y ? Also state the colour of metal X.
Answer:
Step 1: Metals that are less reactive than copper do not dissolve at the anode; they collect as anode mud. These are silver, gold and platinum.
Step 2: A coin that stays new for years is made of a metal that does not react with air or moisture, so X is gold. Its colour is yellow.
Step 3: A coin that gradually turns black is attacked by sulphur compounds in air, forming Ag₂S, so Y is silver. Pure gold is soft; a little silver (or copper) makes it hard enough for ornaments.
In simple words: X is yellow gold, which never tarnishes; Y is silver, which turns black.

Teacher's Note:
1. The anode mud holds the unreactive metals, so think of Ag, Au and Pt.
2. Link tarnishing (Ag₂S) to the sulphur compounds present in air.

Question 5. A black metal oxide XO₂ is used as a catalyst in the preparation of oxygen gas from potassium chlorate. The oxide XO₂ is also used in ordinary dry cells. The metal oxide XO₂ cannot be reduced satisfactorily with carbon to form metal X. (i) Name the metal X. (ii) Name the metal oxide XO₂ (iii) Which reducing agent can be used to reduce XO₂ to obtain metal X? (iv) Name another metal which can also be extracted by the reduction of its oxide with the above reducing agent.
Answer:
(i) X is manganese (Mn).
(ii) XO₂ is manganese dioxide, MnO₂ (black; catalyst for KClO₃, and used in dry cells).
(iii) Aluminium powder is the reducing agent: 3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat.
(iv) Chromium can also be obtained by reducing its oxide with aluminium: Cr₂O₃ + 2Al → 2Cr + Al₂O₃ (iron from Fe₂O₃ is another example).
In simple words: MnO₂ is black, speeds up oxygen formation and is used in dry cells; aluminium powder is used to get manganese out of it.

Teacher's Note:
1. Remember the thermite-type reductions: Mn, Cr and Fe oxides are reduced by aluminium.
2. Write the oxide and metal names with their formulae.

Question 6. In an electrolytic tank, aluminium metal is being extracted by the electrolysis of molten aluminium oxide using carbon electrodes. It is observed that one of the carbon electrodes is gradually burnt away and has to be replaced. (i) Which carbon electrode (cathode or anode) is burnt away? (ii) Why is this carbon electrode burnt away?
Answer:
(i) The carbon anode (positive electrode) is burnt away.
(ii) Oxide ions move to the anode and give up electrons to form oxygen gas. This oxygen reacts with the hot carbon of the anode and forms carbon dioxide: C + O₂ → CO₂. So the anode is used up slowly and has to be replaced.
At the cathode: Al³⁺ + 3e⁻ → Al (aluminium collects at the bottom).
In simple words: The oxygen released at the carbon anode burns the anode away as carbon dioxide.

Teacher's Note:
1. Oxidation always happens at the anode, so non-metal ions give electrons there.
2. Metal is always deposited at the cathode.

Question 7. An element E combines with O₂ to form an oxide E₂O, which is a good conductor of electricity. Write the formula of the compound formed when it combines with chlorine.
Answer:
The formula E₂O shows that one atom of oxygen (valency 2) combines with two atoms of E, so E has valency 1. A metal oxide that conducts electricity (in the molten state) is ionic, so E is a metal that forms E⁺.
With chlorine (valency 1), one E⁺ ion combines with one Cl⁻ ion. The formula is ECl.
In simple words: E gives one electron, so it pairs one-to-one with chlorine to give ECl.

Teacher's Note:
1. Find the valency from the oxide formula first.
2. Use the criss-cross method with the valency of chlorine, 1.

Question 8. A metallic element M, has the following properties: (I) Floats on water (II) Can be cut with a knife (III) Occurs naturally as its chloride, of formula MCl (IV) Its oxide dissolves in water to form the hydroxide. Based on the information provided above, answer the following questions: (A) State the method of obtaining the metal M. (B) Name the major byproduct obtained in the process.
Answer:
Identifying M: It is soft, floats on water, occurs as MCl (rock salt) and forms a soluble hydroxide MOH. So M is sodium.
(A) Sodium is highly reactive, so it is obtained by the electrolysis of molten sodium chloride (electrolytic reduction). At cathode: Na⁺ + e⁻ → Na. At anode: 2Cl⁻ → Cl₂ + 2e⁻.
(B) The major by-product is chlorine gas, liberated at the anode.
In simple words: M is sodium; it is obtained by passing electricity through molten NaCl and chlorine gas comes out as the by-product.

Teacher's Note:
1. Soft, floating and cut by a knife are clues for an alkali metal.
2. Use molten, not aqueous, NaCl, otherwise hydrogen is formed at the cathode.

Long Answer HOTS Questions

Question 9. (a) What is corrosion of metals ? Name one metal which does not corrode and one which corrodes on being kept open. (b) How will you show that the rusting of iron needs oxygen and moisture at the same time ?
Answer:
(a) Corrosion is the slow eating away of a metal when its surface is attacked by air, moisture or acids and changes into oxide, carbonate or sulphide. Gold (or platinum) does not corrode; iron corrodes when kept open.
(b) Activity: Take three test tubes, each with a clean iron nail. In A put ordinary water and cork it (air + water). In B put boiled (air-free) distilled water, add a little oil on top and cork it (water, no air). In C put anhydrous calcium chloride and cork it (dry air, no moisture). Leave for a few days.
Observation: The nail rusts only in A. Conclusion: Both oxygen (air) and moisture must be present together for rusting.
In simple words: Rusting happens only when the iron gets both air and water; remove either one and it stays bright.

Teacher's Note:
1. Always state the three set-ups and what is excluded in each.
2. Boiled water has no dissolved air, and oil stops fresh air from entering.

Question 10. Four metals P, Q, R and S are all obtained by the reduction of their oxides with carbon. Metal P is used to form a thin layer over the sheets of metal S to prevent its corrosion. Metal Q is used for electroplating tin boxes made of metal S whereas metal R is used in making car batteries. Metals Q and R form an alloy called solder. What are metals P, Q, R and S ? How have you arrived at this conclusion?
Answer:
P = zinc, Q = tin, R = lead, S = iron.
Step 1: Solder is an alloy of lead and tin, so Q and R are tin and lead. Lead is used in car (lead-acid) batteries, so R is lead and Q is tin.
Step 2: Tin plating (on cans) and a protective thin layer on sheets both protect iron, so S is iron. Coating iron with zinc is galvanisation, so P is zinc.
Step 3: Zn, Sn, Pb and Fe all lie in the middle of the reactivity series, so their oxides can be reduced by carbon, which fits the clue.
In simple words: Solder gives tin and lead, the battery gives lead, the protective coating on iron gives zinc.

Teacher's Note:
1. Start with the clue that names an alloy; it fixes two metals at once.
2. Check your answer against the carbon-reduction clue.

Question 11. A teacher asks her students to identify a metal M, she gives them the following clues to help them. (I) Its oxide reacts with both HCl and NaOH. (II) It does not react with hot or cold water but reacts with steam. (III) It can be extracted by electrolysis of its ore. Based on the information provided above, answer the following questions: (A) Identify the metal. (B) Write the chemical equations for the reaction of the metal with HCl and NaOH respectively. (C) What would happen if the metal is reacted with iron oxide?
Answer:
(A) The oxide is amphoteric, the metal reacts only with steam, and it is obtained by electrolysis of its ore. So M is aluminium (zinc is also amphoteric, but it is extracted by carbon reduction, not electrolysis).
(B) 2Al + 6HCl → 2AlCl₃ + 3H₂
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂
(C) Aluminium displaces iron from iron(III) oxide in a highly exothermic thermite reaction. The iron is obtained in the molten state: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe + heat.
In simple words: The clues point to aluminium, which gives hydrogen with acids and alkalis and melts iron out of its oxide.

Teacher's Note:
1. Check clue III before you pick zinc over aluminium.
2. The thermite reaction is used to weld railway tracks.

Question 12. Three pieces of a rust free iron rod are completely coated with the following: (I) Plastic (II) Oil paint (III) Zinc. An identical scratch is made on each piece, thus, exposing the iron. The pieces of iron are kept exposed to moist air for 10 days and then checked for rust formation. (A) State if rusting will be observed at the point of the scratch on the three iron pieces. (B) Give reasons for your answer in each case. (C) Name the process of applying a protective zinc coating to steel or iron.
Answer:
(A) Rust will be seen at the scratch on the plastic-coated and the oil-paint-coated pieces. No rust will form on the zinc-coated piece.
(B) Plastic and paint only act as a barrier; once scratched, iron touches air and moisture and rusts. Zinc is more reactive than iron, so it is oxidised in preference to iron and goes on protecting the iron even where the coating is broken.
(C) The process is galvanisation.
In simple words: Plastic and paint protect only while they are unbroken; zinc keeps protecting even when scratched.

Teacher's Note:
1. Compare a barrier coating with a sacrificial coating.
2. Zinc corrodes first because it is more reactive than iron.

Question 13. No chemical reaction takes place when granules of a solid A are mixed with the powder of another solid B. However, when the mixture is heated, a reaction takes place between its components. One of the products, C, is a metal and settles down in its molten state, while the other product, D, floats over it. It was observed that the reaction is highly exothermic. (A) Based on the given information, make an assumption about A and B and write a chemical equation for the chemical reaction indicating the conditions of the reaction, physical state of reactions and products and thermal status of reaction. (B) Mention any two types of reaction under which the above chemical reaction can be classified.
Answer:
(A) This is the thermite reaction. Assume A is iron(III) oxide (granules) and B is aluminium powder. On heating, aluminium displaces iron: Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) + heat. C is molten iron and D is aluminium oxide, which floats on it.
(B) The reaction is (1) a displacement reaction (aluminium displaces iron), (2) a redox reaction (Fe₂O₃ is reduced, Al is oxidised) and (3) exothermic.
In simple words: Aluminium powder reacts with iron oxide to give molten iron plus aluminium oxide and a lot of heat.

Teacher's Note:
1. Write the physical state of every substance in the equation.
2. The heat produced is enough to melt the iron, which is used to join rails.

Question 14. Sheetal takes the colourless solution of a sulphate salt of metal ‘M’ in a beaker. She adds metal granules ‘A’, ‘B’ and ‘C’ one after the other to the beaker. The following colour changes are observed by her.
M₂SO₄ (colourless solution) —Metal ‘A’→ Blue solution —Metal ‘B’→ Light green solution —Metal ‘C’→ Colourless solution(A) After identifying all the metals M, A, B and C arrange them in increasing order of their reactivity. What would be the reason for colour changes? Why some metals are more reactive while others are less reactive? (B) Write all the chemical reactions involved.
Answer:
Identifying the metals: Colourless M₂SO₄ turns blue when A is added, so A displaced M and formed CuSO₄: A is copper and M is less reactive than copper (silver; M₂SO₄ is Ag₂SO₄). The blue solution turns light green with B, so B displaced copper and formed FeSO₄: B is iron. The light green solution turns colourless with C, so C displaced iron and formed ZnSO₄: C is zinc.
(A) Increasing order of reactivity: Ag (M) < Cu (A) < Fe (B) < Zn (C). The colours change because a more reactive metal displaces a less reactive one from its salt solution and a new coloured salt forms. A metal is more reactive when its atoms lose electrons more easily to form positive ions.
(B) Cu + Ag₂SO₄ → CuSO₄ + 2Ag
Fe + CuSO₄ → FeSO₄ + Cu
Zn + FeSO₄ → ZnSO₄ + Fe
In simple words: Each metal pushes out the one before it, so the metals get more reactive in the order silver, copper, iron, zinc.

Teacher's Note:
1. Use the colour of each sulphate (blue, pale green, colourless) to name the metal.
2. Write one displacement equation for every colour change.

Case Based HOTS Questions

Question 15. Kunal connected a copper plate and an iron plate to the positive and negative terminals of a battery respectively along with a switch. He immersed the plates into a beaker containing acidified copper sulphate solution.
(A) (i) After a few minutes, even before he turned the switch on, he noticed that copper was deposited on the iron plate. What would be the reason for this? (ii) What would happen when the current is supplied? (B) (i) Kunal now replaces the iron plate with a silver plate. He sees that there is no deposition of copper on the silver plate before starting the current. What could be the reason? (ii) What is likely to happen to the concentration of copper sulphate in the solution on passing electric current through the solution in the setup with the silver plate? (C) (i) What will happen to the weight of the silver and copper plates after passing the current for some time? (ii) Write balanced chemical equation between silver and copper sulphate.
Answer:
(A)(i) Iron is more reactive than copper, so the iron plate displaces copper from the copper sulphate solution on its own, even with the switch off. Reddish-brown copper is deposited on the iron plate: Fe + CuSO₄ → FeSO₄ + Cu.
(A)(ii) When current passes, the copper plate (anode, +) dissolves: Cu → Cu²⁺ + 2e⁻. Cu²⁺ ions are deposited on the iron plate (cathode, −): Cu²⁺ + 2e⁻ → Cu. So the layer of copper on the iron plate keeps growing (electroplating).
(B)(i) Silver is less reactive than copper, so it cannot displace copper from copper sulphate solution.
(B)(ii) The concentration of copper sulphate stays the same, because as much copper dissolves at the anode as is deposited at the cathode.
(C)(i) The mass of the silver plate increases (copper is deposited on it) and the mass of the copper plate decreases.
(C)(ii) There is no reaction: Ag + CuSO₄ → no reaction, because silver is less reactive than copper.
In simple words: Iron pushes copper out by itself; with a battery, copper moves from the + plate to the − plate.

Teacher's Note:
1. Separate the chemical (displacement) change from the electrical (electroplating) change.
2. Never write a displacement equation for a less reactive metal.

Teacher's Note:
1. Use both the boiling point and the conduction column to decide the type of bonding.
2. Diamond is a covalent solid, not an ionic one.

Teacher's Note:
1. Build the order first: II is above I; I is above copper; III and IV are at the bottom.
2. Read each row separately before comparing.

Teacher's Note:
1. Rank metals by how many of the three tests they show.
2. Only Mg and Mn evolve hydrogen with very dilute HNO₃.