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 16 of 18

1. Q301 CBSE 2020
   (a) Explain the formation of ionic compound, ${\text{Al}}_2{\text{O}}_3$ with ​electron-dot structure ​: (Given : ​Atomic no. of Al and O are 13 and 8 respectively) (b) What happens ​when (Report only observations)

   1. (i)a ​reactive metal reacts ​with a dilute mineral acid ?
   2. (ii)an amphoteric oxide reacts with sodium hydroxide ​solution ?
   3. (iii)a metal of low reactivity is dropped in the salt solution of a metal of high ​reactivity ?
   4. (iv)a metal carbonate is treated with acid ?
   Show answer

   Al and O form ${\text{Al}}_2{\text{O}}_3$ by transferring 6 electrons; acid with a reactive metal gives hydrogen, with a carbonate gives carbon dioxide.

   (a) Electronic configurations: Al (13) is 2, 8, 3 and O (8) is 2, 6. Each Al atom loses its 3 valence electrons to form ${\text{Al}}^{3+}$ (2, 8), and each O atom gains 2 electrons to form ${\text{O}}^{2-}$ (2, 8). Since $2\times 3=3\times 2=6$, two Al atoms transfer 6 electrons to three O atoms, as in the electron-dot figure, giving $[{\text{Al}}^{3+}{]}_2[{\text{O}}^{2-}{]}_3$, that is ${\text{Al}}_2{\text{O}}_3$; the oppositely charged ions are held together by strong electrostatic attraction. (b) (i) Hydrogen gas is evolved: bubbles of a colourless gas are given off, and the gas burns with a pop sound. (ii) The oxide dissolves, forming a salt and water (with aluminium oxide, sodium aluminate and water). (iii) No reaction takes place; no change is seen. (iv) Brisk effervescence: carbon dioxide gas is given off, which turns lime water milky.

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2. Q302 CBSE 2020
   Carbon cannot reduce the oxides of sodium, ​magnesium and aluminium to their respective metals. Why ? Wherein the ​reactivity series these ​metals are placed? ​How ​are these ​metals ​obtained ​from their ores? Take an example to explain the process of extraction along with chemical equations.
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   These metals hold oxygen more strongly than carbon does, so they are obtained by electrolysis of molten compounds, for example sodium from molten $\text{NaCl}$.

   Carbon cannot reduce the oxides of sodium, magnesium and aluminium because these metals have more affinity for oxygen than carbon: they hold oxygen more strongly than carbon does, so carbon cannot remove the oxygen from their oxides. These metals are placed towards the top of the reactivity series ($\text{K}$, $\text{Na}$, $\text{Ca}$, $\text{Mg}$, $\text{Al}$, above zinc). They are obtained from their ores by electrolytic reduction, that is, by electrolysis of their molten compounds: sodium and magnesium from their molten chlorides, aluminium from its oxide. Example, sodium from sodium chloride: molten $\text{NaCl}$ is taken in an electrolytic cell and an electric current is passed; sodium is deposited at the cathode (negative electrode) and chlorine gas is liberated at the anode (positive electrode). At the cathode: ${\text{Na}}^{+}+e^{-}\to \text{Na}$; at the anode: $2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e^{-}$; overall: $2\text{NaCl}\to 2\text{Na}+{\text{Cl}}_2$. Any other correct example, such as magnesium from molten magnesium chloride, earns the same credit.

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3. Q303 CBSE 2019
   What are ​ionic compounds ? Why do ​ionic compounds not conduct electricity ​in ​the ​solid state ​?
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   Ionic compounds are formed by electron transfer; solid ones do not conduct because their ions are fixed and cannot move.

   Ionic compounds: Compounds formed by the transfer of electrons from a metal to a non-metal are called ionic (electrovalent) compounds. They are made of positive and negative ions held together by strong electrostatic forces of attraction, for example sodium chloride, ${\text{Na}}^{+}{\text{Cl}}^{-}$, and magnesium chloride, ${\text{MgCl}}_2$. Why the solid does not conduct: Electricity is conducted by the movement of charged particles. In the solid state the ions are held in fixed positions in a rigid crystal structure by strong electrostatic attraction, so they cannot move, and there are no free ions or free electrons. Hence solid ionic compounds do not conduct electricity. (When melted or dissolved in water, the ions become free to move and the compound conducts.)

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4. Q304 CBSE 2019
   Show ​the formation of ​magnesium ​oxide ​by the ​transfer ​of electrons.
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   Magnesium transfers its two valence electrons to oxygen, forming ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$, which attract to give $\text{MgO}$.

   Magnesium (atomic number $12$, configuration $2,8,2$) has two valence electrons, and oxygen (atomic number $8$, configuration $2,6$) needs two more electrons to complete its octet. The magnesium atom transfers its two valence electrons to the oxygen atom: $\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}$ ($2,8,2\to 2,8$) $\text{O}+2e^{-}\to {\text{O}}^{2-}$ ($2,6\to 2,8$) In electron-dot form, the two electrons (×) of Mg move to the O atom with its six electrons (•), giving ${\text{Mg}}^{2+}$ and the oxide ion with a complete octet, $[\text{O}{]}^{2-}$. The oppositely charged ions are held together by strong electrostatic attraction, the ionic bond: ${\text{Mg}}^{2+}+{\text{O}}^{2-}\to {\text{Mg}}^{2+}{\text{O}}^{2-}$, that is, magnesium oxide ($\text{MgO}$). Both ions have the stable configuration $2,8$ of neon.

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5. Q305 CBSE 2019
   Write ​the ​name, ​symbol and electronic configuration of ​an element X whose ​atomic ​number is ​11.
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   Element X is sodium, symbol Na, with electronic configuration $2,8,1$.

   Name: sodium. Symbol: Na. Electronic configuration: $2,8,1$ (K shell $2$, L shell $8$, M shell $1$).

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6. Q306 CBSE 2019
   Give reason ​:

   1. (a)Carbonate ​and sulphide ​ores are ​usually converted into oxides during the ​process ​of extraction.
   2. (b)Aluminium is a highly reactive metal; still it is widely used ​in making ​cooking utensils.
   Show answer

   Oxides are reduced by carbon directly, so ores are converted to oxides; aluminium's protective oxide layer stops further corrosion.

   (a) It is easier to obtain a metal from its oxide than from its carbonate or sulphide, because the oxide can be reduced to the metal directly by heating it with carbon. So a sulphide ore is roasted (heated strongly in excess air) and a carbonate ore is calcined (heated strongly in limited air) to change it into the oxide first: $2\text{ZnS}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2\text{ZnO}+2{\text{SO}}_2$, ${\text{ZnCO}}_3\stackrel{\text{heat}}{\to }\text{ZnO}+{\text{CO}}_2$; then $\text{ZnO}+\text{C}⟶\text{Zn}+\text{CO}$. (b) Aluminium reacts with the oxygen of the air to form a thin, tough, protective layer of aluminium oxide on its surface (made thicker by anodising). This layer does not let air and water reach the metal, so it does not react further or corrode. Being light and a good conductor of heat, aluminium is therefore widely used for cooking utensils.

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7. Q307 CBSE 2019
   Name ​a ​metal ​of ​medium reactivity ​and write three ​main steps ​in the extraction of this metal from its sulphide ore.
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   Zinc, a metal of medium reactivity, is extracted from zinc blende by concentration, roasting and reduction with carbon.

   Metal of medium reactivity: zinc (iron or lead are also accepted). Three main steps in extracting zinc from its sulphide ore, zinc blende ($\text{ZnS}$): (i) Concentration (enrichment) of the ore: the gangue is removed, for example by froth flotation. (ii) Roasting: the concentrated ore is heated strongly in excess air, which converts the sulphide into the oxide: $2\text{ZnS}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2\text{ZnO}+2{\text{SO}}_2$. (iii) Reduction: the oxide is heated with carbon (coke) to give the metal: $\text{ZnO}+\text{C}\stackrel{\text{heat}}{\to }\text{Zn}+\text{CO}$.

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8. Q308 CBSE 2019
   Show the formation of ${\text{Na}}_2\text{O}$ by ​the ​transfer of ​electrons.
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   Two sodium atoms transfer one electron each to an oxygen atom, forming ${\text{Na}}^{+}$ and ${\text{O}}^{2-}$ ions that make ${\text{Na}}_2\text{O}$.

   Sodium ($2,8,1$) has one outer electron to lose and oxygen ($2,6$) needs two more, so two sodium atoms each transfer one electron to one oxygen atom: $2\text{Na}⟶2{\text{Na}}^{+}+2e^{-}$ (each $2,8,1\to 2,8$) $\text{O}+2e^{-}⟶{\text{O}}^{2-}$ ($2,6\to 2,8$) $2{\text{Na}}^{+}+{\text{O}}^{2-}⟶{\text{Na}}_2\text{O}$ In the electron-dot diagram (step 4) the crosses of the two sodium atoms move to the oxygen atom, which becomes an oxide ion with eight outer electrons and charge $2-$, between two ${\text{Na}}^{+}$ ions; the ions are held by ionic bonds.

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9. Q309 CBSE 2019
   Give ​reasons ​:

   1. (a)​Platinum, gold and silver are ​used to make ​jewellery.
   2. (b)Metals like sodium and potassium are ​stored under oil.
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   Platinum, gold and silver are lustrous, malleable and very unreactive; sodium and potassium are highly reactive, so they are kept under oil.

   (a) Platinum, gold and silver are lustrous (shiny) and highly malleable and ductile, so they can be worked into fine ornaments, and they are among the least reactive metals, hardly attacked by air and water (silver only tarnishes slowly), so they stay bright for a long time. (b) Sodium and potassium are highly reactive: they react vigorously with the oxygen and moisture of the air and catch fire if kept in the open. They are kept immersed in kerosene (oil) so that air and moisture cannot reach them.

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10. Q310 CBSE 2019
    Silver ​articles become black when kept in open ​for some time, whereas ​copper ​vessels lose their shiny brown surfaces and gain a ​green coat when kept in open. Name ​the substances present ​in air with which these metals react and write the name of the ​products formed.
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    Silver reacts with sulphur in air to form black silver sulphide; copper reacts with moist carbon dioxide to form green basic copper carbonate.

    Silver reacts with sulphur present in the air (as hydrogen sulphide, ${\text{H}}_2\text{S}$) and forms a black coating of silver sulphide, ${\text{Ag}}_2\text{S}$. Copper reacts with moist carbon dioxide in the air (carbon dioxide and water vapour, with oxygen) and forms a green coating of basic copper carbonate, ${\text{CuCO}}_3\cdot {\text{Cu(OH)}}_2$.

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11. Q311 CBSE 2019
    (a) Silver articles generally turn black when kept in the open ​for a few days. The articles ​when rubbed with tooth-paste again start shining.

    1. (i)Why ​do ​silver ​articles turn black ​when kept in the open for a ​few days ? Name the phenomenon involved.
    2. (ii)​Name the black substance formed and give its chemical formula. (b) Write the formulae of (i) calcium phosphate and (ii) ammonium carbonate.
    Show answer

    Silver tarnishes (corrosion) to black silver sulphide, ${\text{Ag}}_2\text{S}$; calcium phosphate is ${\text{Ca}}_3({\text{PO}}_4{)}_2$ and ammonium carbonate $({\text{NH}}_4{)}_2{\text{CO}}_3$.

    (a)(i) Silver reacts with sulphur compounds, such as hydrogen sulphide, present in the air and forms a thin black coating of silver sulphide on its surface: $4\text{Ag}+2{\text{H}}_2\text{S}+{\text{O}}_2⟶2{\text{Ag}}_2\text{S}+2{\text{H}}_2\text{O}$. The phenomenon is corrosion (tarnishing of silver). (a)(ii) The black substance is silver sulphide, ${\text{Ag}}_2\text{S}$. (b)(i) Calcium phosphate: ${\text{Ca}}_3({\text{PO}}_4{)}_2$; (ii) Ammonium carbonate: $({\text{NH}}_4{)}_2{\text{CO}}_3$.

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12. Q312 CBSE 2019
    Out of three metals P, Q and R, ​P is less reactive ​than Q ​and ​R is more reactive ​than P and Q both. Suggest an activity ​to arrange P, Q and R in order of ​their decreasing reactivity.
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    A more reactive metal displaces a less reactive one from its salt solution; the tests give the order $\text{R}>\text{Q}>\text{P}$.

    Activity (displacement reactions): A more reactive metal displaces a less reactive metal from a solution of its salt. 1. Clean strips of P, Q and R with sandpaper. Take salt (for example, sulphate) solutions of P and Q in three test tubes. 2. Tube 1: put a strip of R in the salt solution of Q. Tube 2: put a strip of R in the salt solution of P. Tube 3: put a strip of P in the salt solution of Q. Leave them for some time. Observations and inferences: Tubes 1 and 2: a coating of the displaced metal forms on R and the colour of the solution changes (for coloured salts, such as those of iron or copper), so R displaces both Q and P; R is more reactive than both. Tube 3: no change, so P cannot displace Q; P is less reactive than Q. (For example, with R as zinc, Q as iron and P as copper, zinc turns both iron(II) sulphate and copper sulphate solutions colourless, while copper leaves iron(II) sulphate unchanged.) Order of decreasing reactivity: $\text{R}>\text{Q}>\text{P}$.

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13. Q313 CBSE 2019
    Name the ore of mercury. ​With ​the help of ​balanced chemical ​equations, explain the ​process of extraction ​of mercury from its ore.
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    Cinnabar (HgS) is roasted to HgO, which on further heating decomposes to mercury and oxygen.

    Ore of mercury: cinnabar, mercury(II) sulphide ($\text{HgS}$). Extraction: Mercury is low in the activity series, so it is obtained from its ore by heating alone. 1. Roasting: Cinnabar is heated strongly in the presence of excess air. It is converted into mercuric oxide, and sulphur dioxide is given off: $2\text{HgS}(\text{s})+3{\text{O}}_2(\text{g})\stackrel{\text{heat}}{\to }2\text{HgO}(\text{s})+2{\text{SO}}_2(\text{g})$ 2. Reduction by heating: The mercuric oxide is then heated further. It decomposes to give mercury (a liquid) and oxygen: $2\text{HgO}(\text{s})\stackrel{\text{heat}}{\to }2\text{Hg}(\text{l})+{\text{O}}_2(\text{g})$

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14. Q314 CBSE 2019
    What are amphoteric oxides ? Give ​an ​example. ​Write balanced ​chemical ​equations ​to justify your answer.
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    Amphoteric oxides react with both acids and bases to give salt and water; aluminium oxide is an example.

    Amphoteric oxides are metal oxides that react with both acids and bases to produce salt and water; that is, they show both basic and acidic behaviour. Example: aluminium oxide, ${\text{Al}}_2{\text{O}}_3$ (zinc oxide, $\text{ZnO}$, is another). With an acid (behaving as a base): ${\text{Al}}_2{\text{O}}_3+6\text{HCl}⟶2{\text{AlCl}}_3+3{\text{H}}_2\text{O}$ With a base (behaving as an acid): ${\text{Al}}_2{\text{O}}_3+2\text{NaOH}⟶2{\text{NaAlO}}_2+{\text{H}}_2\text{O}$ (sodium aluminate)

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15. Q315 CBSE 2019
    An ore on treatment with dil. $\text{HCl}$ ​gives the smell of rotten ​egg. ​Name ​the type ​of this ore. How can the metal be obtained ​from its concentrated ore ?
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    It is a sulphide ore; roast it in excess air to the oxide, reduce the oxide with carbon, then refine the metal.

    Type of ore: a sulphide ore. The smell of rotten eggs is hydrogen sulphide gas, which a metal sulphide gives with dilute hydrochloric acid, for example $\text{ZnS}+2\text{HCl}⟶{\text{ZnCl}}_2+{\text{H}}_2\text{S}\uparrow$. Obtaining the metal from the concentrated ore (for a metal of medium reactivity such as zinc): (1) Roasting: the ore is heated strongly in excess air, which converts the sulphide into the oxide: $2\text{ZnS(s)}+3{\text{O}}_2\text{(g)}\stackrel{\text{heat}}{\to }2\text{ZnO(s)}+2{\text{SO}}_2\text{(g)}$. (2) Reduction: the oxide is heated with carbon (coke), which reduces it to the metal: $\text{ZnO(s)}+\text{C(s)}\stackrel{\text{heat}}{\to }\text{Zn(s)}+\text{CO(g)}$. (3) Refining: the impure metal is purified, for example by electrolytic refining. For a metal low in reactivity, such as mercury, roasting gives the oxide, which on further heating decomposes to the metal ($2\text{HgO}\stackrel{\text{heat}}{\to }2\text{Hg}+{\text{O}}_2$); this route also earns credit.

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16. Q316 CBSE 2019
    During ​the reaction of some ​metals with dilute hydrochloric acid, the following observations were made by ​a student :

    1. (a)Silver does not show any ​change.
    2. (b)Some ​bubbles of a gas are seen ​when lead is reacted with the acid.
    3. (c)The ​reaction of sodium is ​found to be highly explosive.
    4. (d)The temperature of the reaction mixture rises when aluminium is added to the acid. Explain these observations giving appropriate reason.
    Show answer

    Silver lies below hydrogen so shows no change; lead displaces hydrogen slowly; sodium reacts explosively and aluminium exothermically.

    (a) Silver lies below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute hydrochloric acid; no reaction takes place, so no change is seen. (b) Lead lies just above hydrogen, so it displaces hydrogen from the acid, but slowly; the bubbles are of hydrogen gas: $\text{Pb}+2\text{HCl}\to {\text{PbCl}}_2+{\text{H}}_2$. (c) Sodium is a highly reactive metal: its reaction with the acid is very fast and highly exothermic, so the hydrogen gas evolved catches fire and the reaction is explosive: $2\text{Na}+2\text{HCl}\to 2\text{NaCl}+{\text{H}}_2$. (d) The reaction of aluminium with the acid is exothermic (it releases heat), so the temperature of the reaction mixture rises: $2\text{Al}+6\text{HCl}\to 2{\text{AlCl}}_3+3{\text{H}}_2$.

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17. Q317 CBSE 2019
    Given below are the steps ​for the extraction ​of ​copper from ​its ore. ​Write the chemical ​equation of the reactions involved in each case.

    1. (i)Roasting of copper (I) sulphide
    2. (ii)Reduction of ​copper (I) oxide from copper (I) sulphide
    3. (iii)Electrolytic refining
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    Roasting gives copper(I) oxide, the oxide and remaining sulphide give copper, and electrolytic refining deposits pure copper at the cathode.

    (i) Roasting: $2{\text{Cu}}_2\text{S}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2{\text{Cu}}_2\text{O}+2{\text{SO}}_2$ (ii) Reduction of copper(I) oxide by copper(I) sulphide: $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}\stackrel{\text{heat}}{\to }6\text{Cu}+{\text{SO}}_2$ (iii) Electrolytic refining (impure copper as the anode, a thin strip of pure copper as the cathode, acidified copper sulphate solution as the electrolyte): at the anode, $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$; at the cathode, ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$.

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18. Q318 CBSE 2019
    A ​metal X, which is used in thermit process, when heated with oxygen ​gives an ​oxide Y which is ​amphoteric in nature. ​Identify ​X and Y. Write balanced chemical equations of the reactions of oxide Y with ​hydrochloric acid and ​sodium hydroxide.
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    X is aluminium and Y is aluminium oxide, which reacts with both an acid and a base.

    X is aluminium (Al), the metal used in the thermit process: ${\text{Fe}}_2{\text{O}}_3+2\text{Al}⟶2\text{Fe}+{\text{Al}}_2{\text{O}}_3+\text{heat}$. Y is aluminium oxide (${\text{Al}}_2{\text{O}}_3$), formed when aluminium is heated in oxygen: $4\text{Al}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2{\text{Al}}_2{\text{O}}_3$. Reaction of Y with hydrochloric acid: ${\text{Al}}_2{\text{O}}_3+6\text{HCl}⟶2{\text{AlCl}}_3+3{\text{H}}_2\text{O}$ Reaction of Y with sodium hydroxide: ${\text{Al}}_2{\text{O}}_3+2\text{NaOH}⟶2{\text{NaAlO}}_2+{\text{H}}_2\text{O}$ (sodium aluminate) Y gives a salt and water with both an acid and a base, which is why it is amphoteric.

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19. Q319 CBSE 2019
    A solution of copper sulphate 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. Also write the chemical equation of ​the reaction ​involved.
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    Iron is more reactive than copper, so it displaces copper from copper sulphate solution and holes form in the pot.

    Iron is more reactive than copper: it lies above copper in the reactivity series. So the iron of the pot displaces copper from the copper sulphate solution. Iron from the wall of the pot goes into the solution as iron(II) sulphate and copper is deposited; as this goes on for days, the iron of the pot is used up wherever the solution touches it, and holes form. (The blue solution turns pale green and a reddish-brown deposit of copper appears.) Equation: $\text{Fe}(s)+{\text{CuSO}}_4(aq)⟶{\text{FeSO}}_4(aq)+\text{Cu}(s)$

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20. Q320 CBSE 2019
    Explain the following :

    1. (a)Sodium chloride is an ionic ​compound ​which ​does ​not ​conduct ​electricity in solid state ​where as it does conduct electricity in molten state as well as in aqueous solution.
    2. (b)Reactivity of aluminium decrease if it is dipped in nitric acid.
    3. (c)Metals like calcium and magnesium are never found in their free ​state in nature.
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    NaCl conducts only when its ions can move; nitric acid coats aluminium with oxide; calcium and magnesium are too reactive to occur free.

    (a) In solid sodium chloride the ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are held in fixed positions by strong electrostatic forces of attraction, so they cannot move and the solid does not conduct. On melting, heat overcomes these forces and the ions become free to move; in aqueous solution the ions separate and move freely in the water. These moving ions carry the current, so molten sodium chloride and its solution conduct electricity. (b) Nitric acid is a strong oxidising agent. It oxidises the surface of aluminium to a thin, tough layer of aluminium oxide, ${\text{Al}}_2{\text{O}}_3$, which covers the metal and protects it from further reaction, so its reactivity decreases. (c) Calcium and magnesium are highly reactive metals, near the top of the reactivity series. They react easily with other elements, such as oxygen, so in nature they occur only as compounds (for example carbonates, chlorides and sulphates) in their ores, never in the free state.

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