Class 10 Science · Chapter 3

Metals and Non-metals

344 Class 10 Science questions on Metals and Non-metals from the NCERT textbook, NCERT Exemplar and CBSE board papers 2017–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 12 of 18

1. Q221 CBSE 2024
   Name the chemical reaction ​that occurs between iron (III) ​oxide and aluminium metal. Write its ​balanced chemical ​equation. Why ​is ​this reaction called displacement reaction ? Give one use of this reaction.
   Show answer

   The thermit reaction: aluminium displaces iron from iron(III) oxide, forming molten iron that is used to join railway tracks.

   The reaction is the thermit reaction: ${\text{Fe}}_2{\text{O}}_3(\text{s})+2\text{Al}(\text{s})\to {\text{Al}}_2{\text{O}}_3(\text{s})+2\text{Fe}(\text{l})+\text{heat}$. It is a displacement reaction because aluminium, being more reactive than iron, displaces iron from iron(III) oxide. Use: joining railway tracks or cracked machine parts with the molten iron it produces.

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2. Q222 CBSE 2024
   A student is working in a laboratory with ​metal ​‘E’ which is stored ​under kerosene oil. Some how a small piece of this ​metal falls in a beaker containing ​water and starts burning.

   1. (i)Name the metal ‘E’.
   2. (ii)Write chemical ​equation for the reaction when metal ‘E’ reacts with water. State the nature (acidic/basic/neutral) of the product obtained.
   3. (iii)Name the process by ​which this metal ‘E’ is ​obtained from its molten chloride.
   Show answer

   E is sodium; it forms basic sodium hydroxide with water and is obtained by electrolysis of molten sodium chloride.

   (i) Metal E is sodium, Na. Potassium, which is also kept under kerosene and catches fire in water, fits the clues equally. (ii) $2\text{Na}(\text{s})+2{\text{H}}_2\text{O}(\text{l})\to 2\text{NaOH}(\text{aq})+{\text{H}}_2(\text{g})+\text{heat}$; the product, sodium hydroxide, is basic (an alkali). (iii) Electrolytic reduction, that is, electrolysis of molten sodium chloride: sodium is deposited at the cathode (${\text{Na}}^{+}+{\text{e}}^{-}\to \text{Na}$) and chlorine is given off at the anode.

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3. Q223 CBSE 2024
   Draw ​a labelled diagram to show electrolytic refining ​of ​copper. ​State what happens when electric current is ​passed through ​the electrolyte taken in ​this case.
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   The impure copper anode dissolves and pure copper deposits on the cathode; insoluble impurities settle as anode mud.

   Diagram: a tank containing acidified copper sulphate solution as the electrolyte; a thick block of impure copper as the anode, joined to the positive terminal of the battery; a thin strip of pure copper as the cathode, joined to the negative terminal; anode mud collecting at the bottom below the anode. When electric current is passed, copper from the impure anode dissolves into the electrolyte as copper ions ($\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$), and an equivalent amount of pure copper from the electrolyte is deposited on the cathode (${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$). The soluble impurities go into the solution, while the insoluble impurities settle down at the bottom of the anode as anode mud. The anode becomes thinner and the cathode becomes thicker with pure copper.

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4. Q224 CBSE 2024
   Write electronic configuration of Sodium (At. No.11) and ​Oxygen ​(At. No. 8) and show the formation of ​the ionic compound obtained when ​these ​two ​elements combine. Name anion and cation present ​in the ​compound.
   Show answer

   Two sodium atoms give one electron each to oxygen, forming ${\text{Na}}_2\text{O}$ with ${\text{Na}}^{+}$ cations and ${\text{O}}^{2-}$ anions.

   Electronic configurations: Na (11) = 2, 8, 1; O (8) = 2, 6. Formation: each sodium atom loses its one outer electron, $\text{Na}\to {\text{Na}}^{+}+e^{-}$, and the oxygen atom gains two electrons, $\text{O}+2e^{-}\to {\text{O}}^{2-}$; so two sodium atoms give one electron each to one oxygen atom, $2{\text{Na}}^{+}+{\text{O}}^{2-}\to {\text{Na}}_2\text{O}$ (sodium oxide). Both ions now have the configuration 2, 8. Cation: sodium ion, ${\text{Na}}^{+}$; anion: oxide ion, ${\text{O}}^{2-}$.

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5. Q225 CBSE 2024
   State reasons for the ​following ​:

   1. (a)​Zinc ​oxide is an amphoteric oxide.
   2. (b)Calcium starts floating when immersed in water.
   3. (c)In the reactions ​of nitric acid ​with metals, generally ​hydrogen gas is not evolved.
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   Zinc oxide reacts with both acids and bases; hydrogen bubbles make calcium float; nitric acid oxidises the hydrogen to water.

   (a) Zinc oxide reacts with both acids and bases to give salt and water: $\text{ZnO}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2\text{O}$ and $\text{ZnO}+2\text{NaOH}\to {\text{Na}}_2{\text{ZnO}}_2+{\text{H}}_2\text{O}$. It shows both basic and acidic behaviour, so it is amphoteric. (b) Calcium reacts with water to form calcium hydroxide and hydrogen; the heat is not enough to ignite the hydrogen, and the bubbles of hydrogen stick to the surface of the metal, so calcium starts floating. (c) Nitric acid is a strong oxidising agent: it oxidises the hydrogen produced to water and is itself reduced to an oxide of nitrogen (${\text{N}}_2\text{O}$, $\text{NO}$ or ${\text{NO}}_2$), so hydrogen gas is generally not evolved.

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6. Q226 CBSE 2024
   Name the ​ore of mercury and state the form in which it is found ​in nature. ​Write ​the chemical equations along ​with the condition required for the reactions involved in the ​extraction of mercury ​from its ore.
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   The ore is cinnabar, $\text{HgS}$, found as a sulphide; roasting gives $\text{HgO}$, which on heating gives mercury.

   Ore: cinnabar, $\text{HgS}$. Mercury is found in nature in the combined state, as its sulphide (mercury(II) sulphide), so cinnabar is a sulphide ore. Extraction: mercury is low in the activity series, so heating alone is enough. (1) Roasting, in which the ore is heated strongly in air: $2\text{HgS(s)}+3{\text{O}}_2\text{(g)}\stackrel{\text{Heat}}{\to }2\text{HgO(s)}+2{\text{SO}}_2\text{(g)}$. (2) On further heating the mercuric oxide is reduced to mercury: $2\text{HgO(s)}\stackrel{\text{Heat}}{\to }2\text{Hg(l)}+{\text{O}}_2\text{(g)}$.

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7. Q227 CBSE 2024
   Suggest an ​activity to differentiate between ​the chemical properties ​(acidic or basic character) ​of ​the ​product obtained on burning a metal (magnesium) and a non-metal (sulphur).
   Show answer

   The solution of burnt magnesium turns red litmus blue (basic), while that of burnt sulphur turns blue litmus red (acidic).

   Activity: Burn a cleaned magnesium ribbon held in tongs; it burns with a dazzling white flame and leaves white ash, magnesium oxide. Dissolve the ash in a little water and test it with litmus: red litmus turns blue, so the solution is basic. Burn sulphur powder in a deflagrating spoon and collect the gas, sulphur dioxide, in a test tube held upside down over it; add a little water, shake and test with litmus: blue litmus turns red, so the solution is acidic. Equations: $2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$, $\text{MgO}+{\text{H}}_2\text{O}\to {\text{Mg(OH)}}_2$; $\text{S}+{\text{O}}_2\to {\text{SO}}_2$, ${\text{SO}}_2+{\text{H}}_2\text{O}\to {\text{H}}_2{\text{SO}}_3$. Conclusion: the oxide of the metal (magnesium) is basic and the oxide of the non-metal (sulphur) is acidic.

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8. Q228 CBSE 2024
   State giving reason the reduction ​process to obtain the ​following ​metals from ​their compounds :

   1. (i)Mercury,
   2. (ii)Copper and
   3. (iii)Sodium
   Show answer

   Mercury and copper are obtained by heating alone as they are low in the activity series; sodium by electrolysis of molten sodium chloride.

   (i) Mercury: reduction by heating alone. Mercury is low in the activity series; its ore cinnabar is heated in air to give mercuric oxide, which on further heating is reduced to mercury: $2\text{HgS}+3{\text{O}}_2\to 2\text{HgO}+2{\text{SO}}_2$; $2\text{HgO}\to 2\text{Hg}+{\text{O}}_2$. (ii) Copper: reduction by heating alone. Copper is also low in the activity series; its ore ${\text{Cu}}_2\text{S}$ is heated in air: $2{\text{Cu}}_2\text{S}+3{\text{O}}_2\to 2{\text{Cu}}_2\text{O}+2{\text{SO}}_2$, then $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}\to 6\text{Cu}+{\text{SO}}_2$. (iii) Sodium: electrolytic reduction of molten sodium chloride. Sodium is very high in the activity series and has more affinity for oxygen than carbon, so carbon cannot reduce its compounds; sodium is deposited at the cathode (${\text{Na}}^{+}+{\text{e}}^{-}\to \text{Na}$) and chlorine is liberated at the anode.

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9. Q229 CBSE 2024
   State ​giving reason for the ​change in appearance ​observed ​when each of ​the following metal ​is exposed to atmospheric air for some time ​:

   1. (i)​Silver,
   2. (ii)Copper and
   3. (iii)Iron
   Show answer

   Silver turns black (silver sulphide), copper gets a green coating (basic copper carbonate) and iron gets reddish-brown rust; all are corrosion.

   (i) Silver: the shiny surface turns black, because silver reacts with sulphur (present in the air as compounds such as hydrogen sulphide) to form a black coating of silver sulphide, ${\text{Ag}}_2\text{S}$. (ii) Copper: it loses its shiny brown surface and gains a green coating, because copper reacts with the moist carbon dioxide in the air to form basic copper carbonate, ${\text{CuCO}}_3\cdot {\text{Cu(OH)}}_2$. (iii) Iron: it gets a coating of a reddish-brown flaky substance called rust, because iron reacts with the oxygen and moisture of the air to form hydrated iron(III) oxide, ${\text{Fe}}_2{\text{O}}_3\cdot x{\text{H}}_2\text{O}$. All three changes are examples of corrosion.

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10. Q230 CBSE 2024
    Name ​and explain ​the ​most widely used method ​for refining ​impure ​metals.
    Show answer

    Electrolytic refining: the impure metal is the anode and pure metal the cathode, so pure metal deposits on the cathode.

    The most widely used method is electrolytic refining. The impure metal is made the anode and a thin strip of the pure metal is made the cathode, and a solution of a salt of the metal is the electrolyte. For copper, the anode is impure copper, the cathode is a thin strip of pure copper and the electrolyte is acidified copper sulphate solution. On passing an electric current, copper from the anode dissolves into the electrolyte ($\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$) and an equivalent amount of pure copper from the electrolyte is deposited on the cathode (${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$). The soluble impurities go into the solution, whereas the insoluble impurities settle down at the bottom below the anode as anode mud. Copper, zinc, tin, nickel, silver and gold are refined in this way.

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11. Q231 CBSE 2024
    With ​the help ​of chemical equations differentiate between roasting and calcination. How is metal reduced from the product obtained ​after roasting/calcination of ​the ​ore ? Write the chemical equation ​for the reaction involved.
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    Roasting heats sulphide ore in excess air; calcination heats carbonate ore in limited air; the oxide is then reduced by carbon.

    Roasting: a sulphide ore is heated strongly in the presence of excess air and converted into the metal oxide, $2\text{ZnS}(\text{s})+3{\text{O}}_2(\text{g})\stackrel{\text{heat}}{\to }2\text{ZnO}(\text{s})+2{\text{SO}}_2(\text{g})$. Calcination: a carbonate ore is heated strongly in limited air and decomposes into the metal oxide, ${\text{ZnCO}}_3(\text{s})\stackrel{\text{heat}}{\to }\text{ZnO}(\text{s})+{\text{CO}}_2(\text{g})$. So roasting is used for sulphide ores, needs excess air because the ore reacts with oxygen, and gives off sulphur dioxide; calcination is used for carbonate ores, needs only limited air because the ore simply decomposes, and gives off carbon dioxide. The metal oxide obtained is then reduced to the metal by heating it with a reducing agent such as carbon: $\text{ZnO}(\text{s})+\text{C}(\text{s})\to \text{Zn}(\text{s})+\text{CO}(\text{g})$.

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12. Q232 CBSE 2024
    Write electronic configurations of two elements X (At. no. 12) and Y (At. no. 17). Show the formation of ionic compound when X and Y combine with each other. Write the name and formula ​of ​the compound formed. State what would happen if electric current is passed through ​the aqueous solution of this compound. Name ​the ion that will reach at ​

    1. (i)cathode, and
    2. (ii)anode.
    Show answer

    X is magnesium and Y is chlorine, forming ${\text{MgCl}}_2$; on electrolysis of its solution, ${\text{Mg}}^{2+}$ reaches the cathode and ${\text{Cl}}^{-}$ the anode.

    X (atomic number $12$) is magnesium with configuration $2,8,2$; Y (atomic number $17$) is chlorine with configuration $2,8,7$. Magnesium transfers one electron to each of two chlorine atoms — $\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}$ and $\text{Cl}+e^{-}\to {\text{Cl}}^{-}$ — and the oppositely charged ions attract to form the ionic compound magnesium chloride, ${\text{MgCl}}_2$. When a current is passed through its aqueous solution, the solution conducts electricity: its free ions move to the oppositely charged electrodes and electrolysis takes place. (i) The magnesium ion, ${\text{Mg}}^{2+}$, reaches the cathode. (ii) The chloride ion, ${\text{Cl}}^{-}$, reaches the anode.

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13. Q233 CBSE 2024
    1. (i)Write the name of two ​metals of moderate reactivity. Which one of the ​two is more reactive ? How can this fact ​be demonstrated experimentally in a ​school laboratory ?
    2. (ii)​Some metals of ​moderate reactivity are also found in the Earth’s crust in the ​form of their carbonates. Name the processes involved in the extraction ​of metal from them, giving chemical equations for the reactions that occur.
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    Zinc is more reactive than iron because it displaces iron from iron(II) sulphate but not conversely; carbonates are calcined, then reduced with carbon.

    (i) Zinc and iron are metals of moderate reactivity; zinc is the more reactive of the two. To show it, dip a clean zinc strip into iron(II) sulphate solution in one test tube and a clean iron nail into zinc sulphate solution in another. After some time the pale green colour of the first solution fades and a greyish-black deposit of iron forms on the zinc, $\text{Zn}+{\text{FeSO}}_4\to {\text{ZnSO}}_4+\text{Fe}$, while the second tube shows no change. Zinc displaces iron but iron cannot displace zinc, so zinc is more reactive than iron. (ii) The carbonate is first converted into the oxide by calcination (heating strongly in limited air): ${\text{ZnCO}}_3\stackrel{\text{heat}}{\to }\text{ZnO}+{\text{CO}}_2$. The oxide is then reduced to the metal by heating with carbon: $\text{ZnO}+\text{C}\to \text{Zn}+\text{CO}$.

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14. Q234 CBSE 2024
    The metals produced by ​various ​reduction processes are not very ​pure. ​They contain impurities, which must be removed to obtain pure metals. The most widely used method for ​refining impure metals is ​electrolytic refining.

    1. (i)What is the cathode and anode made of in the refining of copper by this process ?[1]
    2. (ii)Name the solution used in the above process and write its formula.[1]
    3. (iii)(a)How copper gets refined when electric current is passed in the electrolytic cell ?[2]
    4. (iii)(b)OR : You have two beakers ‘A’ and ‘B’ containing copper sulphate solution. What would you observe after about $2$ hours if you dip a strip of zinc in beaker ‘A’ and a strip of silver in beaker ‘B’? Give reason for your observations in each case.[2]
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    Impure copper anode and pure copper cathode in acidified copper sulphate; zinc displaces copper from its solution but silver does not.

    (i) Cathode: a thin strip of pure copper; anode: impure copper (a thick block or rod). (ii) Acidified copper sulphate solution; formula ${\text{CuSO}}_4$ (acidified with dilute ${\text{H}}_2{\text{SO}}_4$). (iii)(a) On passing current, copper from the impure anode dissolves into the solution as ${\text{Cu}}^{2+}$ ions ($\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$), and an equivalent amount of ${\text{Cu}}^{2+}$ ions from the solution is deposited on the cathode as pure copper (${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$). Soluble impurities go into the solution and insoluble impurities settle below the anode as anode mud, so pure copper collects on the cathode. (iii)(b) Beaker 'A': the blue colour of the solution fades and a reddish-brown coating of copper forms on the zinc strip, because zinc is more reactive than copper and displaces it ($\text{Zn}+{\text{CuSO}}_4\to {\text{ZnSO}}_4+\text{Cu}$). Beaker 'B': no change, because silver is less reactive than copper and cannot displace it from copper sulphate solution.

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15. Q235 CBSE 2024
    Three metal samples of magnesium, ​aluminium and iron were taken and rubbed with sand paper. These samples were then put separately in test tubes containing dilute hydrochloric acid. ​Thermometers were also suspended in each test tube so that their bulbs dipped in the acid. ​The rate ​of formation of bubbles ​was observed. The above ​activity was repeated with dilute nitric acid ​and the observations were recorded. Answer ​the following questions:

    1. (a)When activity was done with dilute hydrochloric acid, then in which one of the test tubes was the rate of formation of bubbles the fastest and the thermometer showed the highest temperature?[1]
    2. (b)Which metal did not react with dilute hydrochloric acid? Give reason.[1]
    3. (c)(i)Why is hydrogen gas not evolved when a metal reacts with dilute nitric acid? Name the ultimate products formed in the reaction.[2]
    4. (c)(ii)OR : Name the type of reaction on the basis of which reactivity of metals is decided. You have two metals X and Y. How would you decide which is more reactive than the other?[2]
    Show answer

    Magnesium reacts fastest with hydrochloric acid; none of the three metals fails to react; nitric acid oxidises hydrogen to water; displacement reactions decide reactivity.

    (a) The test tube with magnesium. Magnesium is the most reactive of the three, so it displaces hydrogen from dilute hydrochloric acid fastest and releases the most heat: $\text{Mg}+2\text{HCl}\to {\text{MgCl}}_2+{\text{H}}_2$. (b) None of the three metals fails to react. Magnesium, aluminium and iron all lie above hydrogen in the activity series, so each displaces hydrogen from dilute hydrochloric acid; iron reacts the most slowly. Only a metal below hydrogen, such as copper, does not react with dilute hydrochloric acid, because it cannot displace hydrogen from the acid. (c)(i) Nitric acid is a strong oxidising agent: it oxidises the hydrogen produced to water and is itself reduced to an oxide of nitrogen (${\text{N}}_2\text{O}$, $\text{NO}$ or ${\text{NO}}_2$). The ultimate products are the metal nitrate, water and an oxide of nitrogen. (c)(ii) Displacement reactions. Dip a strip of metal X in a solution of a salt of Y, and a strip of Y in a solution of a salt of X. If X displaces Y, so that Y is deposited on X, X is more reactive than Y; if instead Y displaces X, Y is more reactive. For example, iron displaces copper from copper sulphate solution, so iron is more reactive than copper.

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16. Q236 CBSE 2024
    Alloying ​is a very good method ​of improving the properties of metals. ​Most ​of the metallic objects (cooking ​utensils, ornaments, coins, guns, etc.) are made up of alloys. It is because ​pure metals are generally very soft and have low melting points. For example, iron is the most widely ​used metal, but it is never used in its pure form to make cooking pans because it is very ​soft in its purest form, it is corrosive and easily stretches when hot.

    1. (a)What is an alloy ?[1]
    2. (b)Write the constituents of bronze.[1]
    3. (c)(i)How is stainless steel made ? List two properties in which it differs from iron.[2]
    4. (c)(ii)OR : What is solder ? Write its main use stating the property of solder which makes it suitable for this specific use.[2]
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    An alloy is a homogeneous mixture; bronze is copper and tin; stainless steel is hard and does not rust; solder melts easily.

    (a) An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal. (b) Bronze is an alloy of copper (Cu) and tin (Sn). (c)(i) Stainless steel is made by mixing iron with nickel and chromium: the iron is melted, nickel and chromium are dissolved in it in definite proportions and the melt is cooled. It differs from iron in that (1) it is hard, whereas pure iron is very soft, and (2) it does not rust, whereas iron rusts easily. OR (c)(ii) Solder is an alloy of lead and tin (Pb and Sn). It is used for welding (joining) electrical wires together, because its low melting point lets it be melted to make the joint without melting the wires.

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17. Q237 CBSE 2023
    Metal oxides generally ​react ​with ​acids, but few oxides ​of metal also react with bases. ​Such metallic ​oxides ​are : I. ​MgO II. ZnO III. ${\text{Al}}_2{\text{O}}_3$ IV. CaO

    1. (a)I and II
    2. (b)II and III
    3. (c)III and IV
    4. (d)I and IV
    Show answer

    (b) II and III

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18. Q238 CBSE 2023
    Copper is used for ​making ​cooking utensils. Which of the ​following physical ​properties ​of copper is NOT responsible ​for the ​same ​?

    1. (a)Malleability
    2. (b)High melting point
    3. (c)Thermal conductivity
    4. (d)High reactivity
    Show answer

    (d) High reactivity

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19. Q239 CBSE 2023
    On placing a ​silver coin ​in a test ​tube containing blue solution of ${\text{CuSO}}_4$, a student would ​observe ​after some time, ​say 20 ​minutes, ​that the ${\text{CuSO}}_4$ solution

    1. (a)turns green and a grey substance is deposited on the coin.
    2. (b)turns colourless and a grey substance is deposited on the coin.
    3. (c)remains blue with no change in the silver coin.
    4. (d)turns colourless and reddish brown substance is deposited on the coin.
    Show answer

    (c) remains blue with no change in the silver coin.

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20. Q240 CBSE 2023
    Which ​of ​the following oxide(s) is/are soluble in ​water ​to ​form ​alkali/alkalies ?

    1. (i)${\text{SO}}_2$
    2. (ii)${\text{Na}}_2\text{O}$
    3. (iii)${\text{NO}}_2$
    4. (iv)${\text{K}}_2\text{O}$

    1. (a)(ii) and (iv)
    2. (b)Only (iii)
    3. (c)(i), (ii) and (iv)
    4. (d)Only (iv)
    Show answer

    (a) (ii) and (iv)

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