Chapter 3 · Metals and Non-metals

NCERT Exemplar Class 10 Science Chapter 3 Short Answer Questions

All 22 questions of NCERT Exemplar Class 10 Science Chapter 3 Short Answer Questions (Metals and Non-metals), in the book's order, each with its answer and a step-by-step solution.

1. Q37
   Iqbal treated a lustrous, divalent ​element ​M with sodium hydroxide. He ​observed the formation of bubbles in reaction mixture. He made the ​same observations ​when this element was treated with hydrochloric acid. Suggest ​how can he identify the produced gas. Write chemical ​equations for both the reactions.
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   M is zinc and the gas is hydrogen, which burns with a pop sound near a burning splint.

   M is zinc. The gas is hydrogen; it is identified by bringing a burning splint to the mouth of the tube, where it burns with a characteristic pop sound. The two reactions are $\text{Zn}+2\text{NaOH}⟶{\text{Na}}_2{\text{ZnO}}_2+{\text{H}}_2$ and $\text{Zn}+2\text{HCl}⟶{\text{ZnCl}}_2+{\text{H}}_2$.

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2. Q38
   During extraction of metals, ​electolytic refining is used to obtain pure metals.

   1. (a)Which material ​will be used as anode and cathode for refining of silver metal by this process?
   2. (b)Suggest ​a suitable electrolyte also. ​
   3. (c)In this ​electrolytic cell, where do we get ​pure silver ​after passing electric ​current?
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   Impure silver anode, pure silver cathode and acidified silver nitrate solution; pure silver collects on the cathode.

   (a) The anode is a thick block of impure silver and the cathode is a thin strip of pure silver. (b) A suitable electrolyte is an aqueous solution of silver nitrate acidified with a little nitric acid. (c) Pure silver is deposited on the cathode; the insoluble impurities settle below the anode as anode mud.

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3. Q39
   Why ​should the metal ​sulphides ​and carbonates be converted to metal ​oxides in the process of ​extraction ​of ​metal from them?
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   Metals are obtained by reduction, which works far more easily on oxides, so sulphides are roasted and carbonates calcined to oxides first.

   Because the metal is finally obtained by reduction, and metal oxides are much easier to reduce than metal sulphides or carbonates. Carbon and carbon monoxide take oxygen away readily but cannot strip sulphur or a carbonate group as easily, so the ore is first brought to the oxide: a sulphide by roasting in excess air, $2\text{ZnS}+3{\text{O}}_2⟶2\text{ZnO}+2{\text{SO}}_2$, and a carbonate by calcination in limited air, ${\text{ZnCO}}_3⟶\text{ZnO}+{\text{CO}}_2$. The oxide is then reduced, $\text{ZnO}+\text{C}⟶\text{Zn}+\text{CO}$.

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4. Q40
   Generally, ​when metals ​are treated with ​mineral acids, hydrogen gas is ​liberated but ​when metals (except ​Mn and Mg), treated with ${\text{HNO}}_3$, ​hydrogen is not ​liberated, why?
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   Nitric acid is a strong oxidising agent: it oxidises the hydrogen to water and is itself reduced to oxides of nitrogen.

   Nitric acid is a strong oxidising agent. Any hydrogen the metal displaces is immediately oxidised to water, while the nitric acid is itself reduced to an oxide of nitrogen such as ${\text{N}}_2\text{O}$, $\text{NO}$ or ${\text{NO}}_2$. The metal therefore dissolves as its nitrate and an oxide of nitrogen is evolved instead of hydrogen. Manganese and magnesium are exceptions because with very dilute nitric acid they are reactive enough to release hydrogen before it can be oxidised.

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5. Q41
   Compound X ​and aluminium are ​used ​to join railway tracks.

   1. (a)​Identify the ​compound X ​
   2. (b)Name the reaction
   3. (c)Write down its reaction.
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   X is iron(III) oxide; it reacts with aluminium in the thermite reaction, giving molten iron that welds the rails.

   (a) X is iron(III) oxide, ${\text{Fe}}_2{\text{O}}_3$ (ferric oxide). (b) The reaction is the thermite reaction, a highly exothermic displacement reaction, used as thermite welding. (c) ${\text{Fe}}_2{\text{O}}_3(s)+2\text{Al}(s)⟶2\text{Fe}(l)+{\text{Al}}_2{\text{O}}_3(s)+\text{Heat}$.

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6. Q42
   When a metal X is ​treated with cold ​water, it gives a basic salt Y with ​molecular formula XOH (Molecular mass $=40$) and liberates a gas ​Z which easily ​catches ​fire. Identify ​X, Y and Z and also write the ​reaction involved.
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   X is sodium ($\text{Na}$, atomic mass 23), Y is $\text{NaOH}$ and Z is hydrogen: $2\text{Na}+2{\text{H}}_2\text{O}\to 2\text{NaOH}+{\text{H}}_2$.

   X is sodium ($\text{Na}$), Y is sodium hydroxide ($\text{NaOH}$) and Z is hydrogen (${\text{H}}_2$). The reaction is $2\text{Na}(s)+2{\text{H}}_2\text{O}(l)⟶2\text{NaOH}(aq)+{\text{H}}_2(g)$, and it gives out so much heat that the hydrogen catches fire.

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7. Q43
   A ​non-metal X exists in two different forms Y and Z. Y ​is the ​hardest natural ​substance, ​whereas Z is a good conductor of ​electricity. Identify X, ​Y and Z.
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   X is carbon; its allotropes are diamond (Y) and graphite (Z).

   X is carbon, Y is diamond and Z is graphite. Diamond and graphite are two allotropic forms of carbon: diamond is the hardest natural substance, and graphite is the one non-metal that conducts electricity well.

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8. Q44
   The following ​reaction takes ​place when aluminium powder is heated with ${\text{MnO}}_2$ $$3{\text{MnO}}_2(s)+4\text{Al}(s)⟶3\text{Mn}(l)+2{\text{Al}}_2{\text{O}}_3(l)+\text{Heat}$$

   1. (a)Is ​aluminium ​getting reduced?
   2. (b)​Is ${\text{MnO}}_2$ getting oxidised?
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   No: aluminium is oxidised because it gains oxygen, and ${\text{MnO}}_2$ is reduced because it loses oxygen.

   (a) No. Aluminium gains oxygen to become ${\text{Al}}_2{\text{O}}_3$, so it is oxidised, and it acts as the reducing agent. (b) No. Manganese dioxide loses its oxygen to become free manganese, so it is reduced, and it acts as the oxidising agent.

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9. Q45
   What are the ​constituents of ​solder alloy? ​Which property ​of ​solder makes it ​suitable for ​welding electrical wires?
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   Solder is an alloy of lead and tin; its very low melting point suits it to welding electrical wires.

   Solder is an alloy of lead and tin, taken in roughly equal proportions. Its very low melting point is the property that makes it suitable for welding electrical wires: it can be melted with a small soldering iron and run into the joint without heating the wires enough to damage them.

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10. Q46
    A metal A, ​which is ​used in thermite process, ​when heated with ​oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write ​down ​the reactions of oxide ​B ​with HCl and NaOH.
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    A is aluminium and B is amphoteric aluminium oxide, which reacts with $\text{HCl}$ as a base and with $\text{NaOH}$ as an acid.

    A is aluminium ($\text{Al}$) and B is aluminium oxide (${\text{Al}}_2{\text{O}}_3$), which is amphoteric. With hydrochloric acid it behaves as a base: ${\text{Al}}_2{\text{O}}_3+6\text{HCl}⟶2{\text{AlCl}}_3+3{\text{H}}_2\text{O}$. With sodium hydroxide it behaves as an acid: ${\text{Al}}_2{\text{O}}_3+2\text{NaOH}⟶2{\text{NaAlO}}_2+{\text{H}}_2\text{O}$.

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11. Q47
    A ​metal that ​exists as a liquid ​at room temperature is obtained by heating its sulphide in ​the ​presence of air. Identify the metal ​and ​its ore and give the reaction involved.
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    Mercury, obtained from its ore cinnabar ($\text{HgS}$) by roasting in air and then heating the oxide.

    The metal is mercury ($\text{Hg}$) and its ore is cinnabar, mercury(II) sulphide ($\text{HgS}$). Roasting the ore in air gives mercuric oxide, $2\text{HgS}+3{\text{O}}_2⟶2\text{HgO}+2{\text{SO}}_2$, and further heating decomposes that oxide directly to the metal, $2\text{HgO}⟶2\text{Hg}+{\text{O}}_2$.

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12. Q48
    Give the formulae ​of the ​stable binary ​compounds that would be formed ​by the combination of following ​pairs of elements.

    1. (a)​Mg and ${\text{N}}_2$
    2. (b)Li and ${\text{O}}_2$
    3. (c)Al and ${\text{Cl}}_2$
    4. (d)K ​and ${\text{O}}_2$
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    (a) ${\text{Mg}}_3{\text{N}}_2$, (b) ${\text{Li}}_2\text{O}$, (c) ${\text{AlCl}}_3$, (d) ${\text{K}}_2\text{O}$: the valencies are crossed over to get the formulae.

    (a) ${\text{Mg}}_3{\text{N}}_2$, magnesium nitride; (b) ${\text{Li}}_2\text{O}$, lithium oxide; (c) ${\text{AlCl}}_3$, aluminium chloride; (d) ${\text{K}}_2\text{O}$, potassium oxide.

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13. Q49
    What happens when

    1. (a)${\text{ZnCO}}_3$ is ​heated in the absence of ​oxygen?
    2. (b)​a ​mixture of ${\text{Cu}}_2\text{O}$ ​and ${\text{Cu}}_2\text{S}$ is ​heated?
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    Zinc carbonate decomposes to zinc oxide and carbon dioxide; copper(I) sulphide reduces copper(I) oxide to copper with sulphur dioxide.

    (a) Zinc carbonate is calcined: it decomposes on heating to zinc oxide and carbon dioxide, ${\text{ZnCO}}_3⟶\text{ZnO}+{\text{CO}}_2$. (b) Self-reduction takes place: copper(I) sulphide reduces copper(I) oxide, so copper metal is set free and sulphur dioxide escapes, $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}⟶6\text{Cu}+{\text{SO}}_2$.

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14. Q50
    A ​non-metal A ​is an important constituent of our food and forms two oxides B and C. Oxide B is toxic ​whereas C ​causes global warming

    1. (a)Identify ​A, B and C
    2. (b)To which Group of Periodic ​Table ​does A belong?
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    A is carbon, B is toxic carbon monoxide, C is carbon dioxide (global warming); carbon is in Group 14.

    (a) A is carbon ($\text{C}$), B is carbon monoxide ($\text{CO}$), which is toxic, and C is carbon dioxide (${\text{CO}}_2$), which causes global warming. (b) Carbon belongs to Group 14 of the periodic table (the carbon family, also written as Group IVA).

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15. Q51
    Give two examples each of the ​metals that are ​good conductors ​and poor ​conductors of heat respectively.
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    Good conductors of heat: silver and copper; poor conductors: lead and mercury.

    Good conductors of heat: silver ($\text{Ag}$) and copper ($\text{Cu}$). Poor conductors of heat: lead ($\text{Pb}$) and mercury ($\text{Hg}$). All four are metals, and poor here means poor in comparison with other metals.

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16. Q52
    Name one metal and one non-metal that ​exist in liquid state ​at room temperature. Also ​name ​two metals ​having melting point ​less than $310$ K ($37^{\circ }$C)
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    Mercury is the liquid metal and bromine the liquid non-metal; gallium and caesium melt below $310$ K.

    The metal that is liquid at room temperature is mercury ($\text{Hg}$) and the non-metal that is liquid at room temperature is bromine (${\text{Br}}_2$). The two metals whose melting points are below $310$ K are gallium ($\text{Ga}$) and caesium ($\text{Cs}$); a small piece of either will melt on the palm of the hand.

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17. Q53
    An element A reacts with water to form a compound B which is ​used in white washing. The compound B ​on heating forms ​an ​oxide C which on treatment with water ​gives back B. ​Identify A, ​B and C and give the reactions involved.
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    A is calcium, B is slaked lime ${\text{Ca(OH)}}_2$ and C is quick lime $\text{CaO}$; C with water gives B back.

    A is calcium ($\text{Ca}$), B is calcium hydroxide or slaked lime (${\text{Ca(OH)}}_2$) and C is calcium oxide or quick lime ($\text{CaO}$). The reactions are $\text{Ca}+2{\text{H}}_2\text{O}⟶{\text{Ca(OH)}}_2+{\text{H}}_2$; ${\text{Ca(OH)}}_2⟶\text{CaO}+{\text{H}}_2\text{O}$ on heating; and $\text{CaO}+{\text{H}}_2\text{O}⟶{\text{Ca(OH)}}_2$.

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18. Q54
    An ​alkali metal A gives a compound B (molecular mass $=40$) on ​reacting with ​water. The compound B gives a soluble compound C on ​treatment with aluminium oxide. Identify A, B ​and C and ​give the ​reaction involved.
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    A is sodium, B is sodium hydroxide ($\text{NaOH}$, mass $40$) and C is sodium aluminate (${\text{NaAlO}}_2$).

    A is sodium ($\text{Na}$), B is sodium hydroxide ($\text{NaOH}$, molecular mass $40$) and C is sodium aluminate (${\text{NaAlO}}_2$). The reactions are $2\text{Na}+2{\text{H}}_2\text{O}⟶2\text{NaOH}+{\text{H}}_2$ and ${\text{Al}}_2{\text{O}}_3+2\text{NaOH}⟶2{\text{NaAlO}}_2+{\text{H}}_2\text{O}$.

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19. Q55
    Give the ​reaction involved ​during extraction ​of zinc from its ore ​by

    1. (a)roasting of zinc ore
    2. (b)calcination ​of zinc ore
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    Roasting zinc blende gives $\text{ZnO}$ and ${\text{SO}}_2$; calcining calamine gives $\text{ZnO}$ and ${\text{CO}}_2$.

    (a) Roasting is carried out on zinc blende, the sulphide ore: $2\text{ZnS}(s)+3{\text{O}}_2(g)⟶2\text{ZnO}(s)+2{\text{SO}}_2(g)$ on strong heating in excess air. (b) Calcination is carried out on calamine, the carbonate ore: ${\text{ZnCO}}_3(s)⟶\text{ZnO}(s)+{\text{CO}}_2(g)$ on strong heating in limited air. Both give zinc oxide, which is then reduced with carbon to zinc.

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20. Q56
    A ​metal M does not liberate hydrogen from acids but reacts with oxygen to ​give ​a black colour product. Identify ​M and black coloured product and ​also explain the reaction of ​M with oxygen.
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    M is copper, and heating it in air forms black copper(II) oxide: $2\text{Cu}+{\text{O}}_2\to 2\text{CuO}$.

    M is copper ($\text{Cu}$) and the black product is copper(II) oxide ($\text{CuO}$). Copper lies below hydrogen in the reactivity series, so it cannot displace hydrogen from an acid; but when it is heated in air, oxygen combines with its surface and the shining red-brown metal becomes coated with black copper(II) oxide: $2\text{Cu}(s)+{\text{O}}_2(g)⟶2\text{CuO}(s)$.

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21. Q57
    An ​element forms an ​oxide ${\text{A}}_2{\text{O}}_3$ ​which ​is acidic ​in nature. Identify ​A ​as a metal or non-metal.
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    A is a non-metal, because non-metals form acidic oxides while metals form basic oxides.

    A is a non-metal. Metals form basic oxides while non-metals form acidic oxides, and the oxide ${\text{A}}_2{\text{O}}_3$ is stated to be acidic. Examples with this formula are ${\text{B}}_2{\text{O}}_3$, ${\text{N}}_2{\text{O}}_3$ and ${\text{P}}_2{\text{O}}_3$, all oxides of non-metals.

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22. Q58
    A ​solution of ${\text{CuSO}}_4$ was kept in an ​iron pot. ​After few days ​the iron pot ​was found to have a number of holes in it. Explain ​the reason in ​terms of reactivity. Write the equation ​of the reaction involved.
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    Iron is more reactive than copper, so it displaces copper from ${\text{CuSO}}_4$ and dissolves, leaving holes in the pot.

    Iron is more reactive than copper, so it displaces copper from copper sulphate solution. The iron of the pot goes into solution as soluble iron(II) sulphate while copper is deposited as the free metal, and the metal lost from the wall of the pot leaves holes in it. The reaction is $\text{Fe}(s)+{\text{CuSO}}_4(aq)⟶{\text{FeSO}}_4(aq)+\text{Cu}(s)$, and the blue solution slowly turns light green.

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