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 7 of 18

1. Q121 CBSE 2026
   Show ​the formation of the ​following ionic compounds ​by transfer of electrons ​between the ​combining ​atoms ​:

   1. (i)Sodium oxide (${\text{Na}}_2\text{O}$)
   2. (ii)Magnesium chloride (${\text{MgCl}}_2$) (Given : Atomic number : $\text{Na}=11$, $\text{Mg}=12$, $\text{O}=8$, $\text{Cl}=17$)
   Show answer

   Two Na atoms give one electron each to O, forming ${\text{Na}}_2\text{O}$; Mg gives one electron to each of two Cl atoms, forming ${\text{MgCl}}_2$.

   (i) Sodium oxide: each of two sodium atoms $(2,8,1)$ transfers its one valence electron to an oxygen atom $(2,6)$, so $2\text{Na}⟶2{\text{Na}}^{+}+2e^{-}$ and $\text{O}+2e^{-}⟶{\text{O}}^{2-}$. The ions ${\text{Na}}^{+}$ $(2,8)$ and ${\text{O}}^{2-}$ $(2,8)$ attract one another as ${\text{Na}}_2\text{O}$. (ii) Magnesium chloride: a magnesium atom $(2,8,2)$ transfers one electron to each of two chlorine atoms $(2,8,7)$, so $\text{Mg}⟶{\text{Mg}}^{2+}+2e^{-}$ and $2\text{Cl}+2e^{-}⟶2{\text{Cl}}^{-}$. The ions ${\text{Mg}}^{2+}$ $(2,8)$ and ${\text{Cl}}^{-}$ $(2,8,8)$ attract one another as ${\text{MgCl}}_2$.

   Step-by-step solution →
2. Q122 CBSE 2026
   Comment on the following ​properties of ​ionic compounds ​and give reason in support ​of ​your ​answer :

   1. (i)Physical nature
   2. (ii)Melting and ​boiling points
   Show answer

   Ionic compounds are somewhat hard, brittle solids with high melting and boiling points, because of the strong attraction between their ions.

   (i) Physical nature: Ionic compounds are solids and are somewhat hard; they are brittle and break into pieces when pressure is applied. Reason: They are made of positive and negative ions held together by a strong force of attraction between the oppositely charged ions, so the ions stay in fixed places. (ii) Melting and boiling points: Ionic compounds have high melting and boiling points; sodium chloride melts at $1074$ K and boils at $1686$ K. Reason: A considerable amount of energy is required to break the strong force of attraction between the ions.

   Step-by-step solution →
3. Q123 CBSE 2026
   Show the formation ​of NaCl by ​transfer of electrons. Why does ​solid NaCl ​not conduct electricity ​? [Atomic number : $\text{Na}=11$, $\text{Cl}=17$]
   Show answer

   Sodium gives one electron to chlorine to form ${\text{Na}}^{+}{\text{Cl}}^{-}$; solid NaCl does not conduct because its ions are fixed and cannot move.

   Formation: sodium, $2,8,1$, loses its one valence electron to chlorine, $2,8,7$, so that $\text{Na}\to {\text{Na}}^{+}+e^{-}$ and $\text{Cl}+e^{-}\to {\text{Cl}}^{-}$. The ions ${\text{Na}}^{+}$ $(2,8)$ and ${\text{Cl}}^{-}$ $(2,8,8)$ both have complete outer shells and are held together by strong electrostatic attraction, giving ${\text{Na}}^{+}{\text{Cl}}^{-}$. Solid sodium chloride does not conduct electricity because its ions are held in fixed positions in the crystal by strong forces of attraction and are not free to move; it conducts only when molten or dissolved in water, when the ions can move to the electrodes.

   Step-by-step solution →
4. Q124 CBSE 2026
   Write chemical equations for the reactions taking ​place ​when :

   1. (i)Manganese dioxide is heated ​with ​aluminium powder.
   2. (ii)Steam is ​passed ​over ​red hot iron.
   Show answer

   Aluminium takes oxygen from manganese dioxide, setting manganese free with much heat; red hot iron and steam give magnetic iron oxide and hydrogen.

   (i) $3{\text{MnO}}_2+4\text{Al}⟶3\text{Mn}+2{\text{Al}}_2{\text{O}}_3+\text{Heat}$. Aluminium is more reactive than manganese, so it takes the oxygen and sets manganese free in a strongly exothermic displacement. (ii) $3\text{Fe}+4{\text{H}}_2\text{O}(\text{steam})⟶{\text{Fe}}_3{\text{O}}_4+4{\text{H}}_2$. Red hot iron reacts with steam to give the magnetic oxide of iron and hydrogen gas.

   Step-by-step solution →
5. Q125 CBSE 2026
   Write ​chemical equations for the reactions taking place ​when : (A) ${\text{Fe}}_2{\text{O}}_3$ is heated with aluminium ​powder. (B) Sodium ​metal ​reacts ​with water.
   Show answer

   Heated with aluminium, iron oxide gives aluminium oxide and molten iron; sodium and water give sodium hydroxide and hydrogen.

   (A) ${\text{Fe}}_2{\text{O}}_3(s)+2\text{Al}(s)\stackrel{\Delta }{\to }{\text{Al}}_2{\text{O}}_3(s)+2\text{Fe}(l)+\text{Heat}$ - the thermite reaction; so much heat is released that the iron is obtained molten. (B) $2\text{Na}(s)+2{\text{H}}_2\text{O}(l)⟶2\text{NaOH}(aq)+{\text{H}}_2(g)+\text{Heat}$ - the reaction is so exothermic that the hydrogen evolved catches fire.

   Step-by-step solution →
6. Q126 CBSE 2026
   Write ​balanced chemical equations ​for the reactions taking ​place when :

   1. (i)​Aluminium metal reacts ​with ​steam.
   2. (ii)Magnesium reacts with dilute $\text{HCl}$.
   Show answer

   Aluminium with steam gives ${\text{Al}}_2{\text{O}}_3$ and hydrogen; magnesium with dilute HCl gives ${\text{MgCl}}_2$ and hydrogen.

   (i) $2\text{Al}+3{\text{H}}_2\text{O}(\text{steam})⟶{\text{Al}}_2{\text{O}}_3+3{\text{H}}_2$. Aluminium takes the oxygen from steam to form aluminium oxide and hydrogen gas is set free. (ii) $\text{Mg}+2\text{HCl}(\text{dilute})⟶{\text{MgCl}}_2+{\text{H}}_2$. Magnesium displaces hydrogen from the dilute acid, giving magnesium chloride and hydrogen gas.

   Step-by-step solution →
7. Q127 CBSE 2026
   Define ​ionic compound. ​Write ​two ​properties ​of ​ionic ​compounds.
   Show answer

   Ionic compounds form by electron transfer from a metal to a non-metal, have high melting points, and conduct when molten or dissolved.

   Ionic compound: a compound formed by the transfer of electrons from a metal to a non-metal, so that it is made of positive and negative ions held together by strong electrostatic forces of attraction. It is also called an electrovalent compound. Example: sodium chloride, $\text{NaCl}$ (${\text{Na}}^{+}{\text{Cl}}^{-}$). Two properties: (1) Ionic compounds have high melting and boiling points, because much energy is needed to overcome the strong attraction between the ions. (2) They conduct electricity in the molten state or in aqueous solution, where the ions can move, but not in the solid state. Other correct properties: they are hard but brittle solids, and they are generally soluble in water but insoluble in kerosene and petrol.

   Step-by-step solution →
8. Q128 CBSE 2026
   Answer ​the following ​:

   1. (a)​Why can the oxides of highly reactive metals not be reduced by carbon ?
   2. (b)Why do ionic compounds have high ​melting and boiling points ?
   3. (c)​Why is a solder, an alloy of lead ​and ​tin, ​used for welding electrical wires together ?
   Show answer

   Carbon cannot reduce oxides of reactive metals, ionic compounds have strong inter-ionic attractions, and solder melts easily.

   (a) Highly reactive metals (potassium, sodium, calcium, magnesium, aluminium) lie above carbon in the reactivity series: they have a greater affinity for oxygen than carbon has, so carbon cannot remove oxygen from their oxides. These metals are obtained by electrolytic reduction instead. (b) In an ionic compound the oppositely charged ions are held in a crystal lattice by strong electrostatic forces of attraction. A large amount of energy is needed to overcome these strong inter-ionic attractions, so ionic compounds have high melting and boiling points. (c) Solder has a low melting point, lower than that of lead, tin or the copper wires. It melts easily with a soldering iron and flows over the joint without damaging the wires, and on cooling it sets into a firm joint that conducts electricity.

   Step-by-step solution →
9. Q129 CBSE 2026
   1. (i)Write one example each of one metal ​that is so soft ​that it can be ​cut with a knife, and one non-metal that is a ​good conductor of electricity.
   2. (ii)Using ​the electronic ​configuration, explain how Mg atom combines with ​oxygen ​to form magnesium oxide by transfer of electrons. [Atomic number : $\text{Mg}=12$, $\text{O}=8$]
   Show answer

   Sodium is soft enough to cut and graphite conducts; Mg gives two electrons to O, forming ionic $\text{MgO}$.

   (i) Sodium is so soft that it can be cut with a knife (potassium or lithium would also do); graphite, a form of carbon, is a non-metal that is a good conductor of electricity. (ii) The electronic configuration of Mg ($Z=12$) is $2,8,2$ and that of O ($Z=8$) is $2,6$. Magnesium loses its two valence electrons to form ${\text{Mg}}^{2+}$ ($2,8$), and oxygen gains these two electrons to form ${\text{O}}^{2-}$ ($2,8$), so both reach the stable configuration of neon: $\text{Mg}⟶{\text{Mg}}^{2+}+2e^{-}$ and $\text{O}+2e^{-}⟶{\text{O}}^{2-}$. The oppositely charged ions are held together by strong electrostatic attraction, an ionic bond, forming magnesium oxide, ${\text{Mg}}^{2+}{\text{O}}^{2-}$ ($\text{MgO}$).

   Step-by-step solution →
10. Q130 CBSE 2026
    Give reasons :

    1. (i)Metal sulphides and carbonates should be converted ​to their ​metal oxides ​in the ​process ​of extraction of metals.
    2. (ii)Aluminium oxide is considered as an amphoteric oxide.
    3. (iii)Metals like Na, K, Ca and Mg are never found in their free state ​in nature.
    Show answer

    Oxides are easier to reduce; aluminium oxide reacts with both acids and bases; Na, K, Ca and Mg are too reactive to occur free.

    (i) It is easier to obtain a metal from its oxide than from its sulphide or carbonate, because a reducing agent such as carbon works by removing oxygen. So sulphide ores are first roasted, $2\text{ZnS}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2\text{ZnO}+2{\text{SO}}_2$, and carbonate ores calcined, ${\text{ZnCO}}_3\stackrel{\text{heat}}{\to }\text{ZnO}+{\text{CO}}_2$, and the oxide is then reduced, $\text{ZnO}+\text{C}⟶\text{Zn}+\text{CO}$. (ii) Aluminium oxide reacts with both acids and bases to form a salt and water, so it shows both basic and acidic behaviour: ${\text{Al}}_2{\text{O}}_3+6\text{HCl}⟶2{\text{AlCl}}_3+3{\text{H}}_2\text{O}$ and ${\text{Al}}_2{\text{O}}_3+2\text{NaOH}⟶2{\text{NaAlO}}_2+{\text{H}}_2\text{O}$. (iii) Na, K, Ca and Mg are highly reactive metals at the top of the reactivity series. They react readily with oxygen, water and other substances in the environment, so they are always found combined, as compounds, and never in the free state.

    Step-by-step solution →
11. Q131 CBSE 2026
    (i) Which method ​will be used to convert each ​of them to their respective metal oxides ?

    1. (a)Carbonate ore
    2. (b)Sulphide ore (ii) Write a chemical reaction to illustrate the use of aluminium for joining cracked railway tracks. (iii) During extraction of metals, electrolytic ​refining is used to ​obtain pure metals. Which material will be used as anode and cathode for the ​refining ​of copper ​metal by this process ?
    Show answer

    Carbonate ore: calcination; sulphide ore: roasting; the thermite reaction welds rails; impure copper is the anode and pure copper the cathode.

    (i) (a) A carbonate ore is converted to its oxide by calcination, that is by heating it strongly in a limited supply of air: ${\text{ZnCO}}_3(s)\stackrel{\Delta }{\to }\text{ZnO}(s)+{\text{CO}}_2(g)$. (b) A sulphide ore is converted to its oxide by roasting, that is by heating it strongly in excess air: $2\text{ZnS}(s)+3{\text{O}}_2(g)\stackrel{\Delta }{\to }2\text{ZnO}(s)+2{\text{SO}}_2(g)$. (ii) The thermite reaction is used: ${\text{Fe}}_2{\text{O}}_3(s)+2\text{Al}(s)⟶2\text{Fe}(l)+{\text{Al}}_2{\text{O}}_3(s)+\text{heat}$. It is so strongly exothermic that the iron is produced molten, and it runs into the crack between the rails and solidifies there, welding them. (iii) Anode: a thick block of impure copper. Cathode: a thin strip or sheet of pure copper. The electrolyte is copper sulphate solution acidified with sulphuric acid, and the insoluble impurities collect below the anode as anode mud.

    Step-by-step solution →
12. Q132 CBSE 2026
    Define ​the ​following terms ​with an example ​:

    1. (i)Roasting
    2. (ii)​Alloy
    3. (iii)Corrosion
    Show answer

    Roasting heats a sulphide ore in excess air; an alloy is a homogeneous mixture of metals; corrosion is slow attack on a metal.

    (i) Roasting: The process of heating a sulphide ore strongly in the presence of excess air to convert it into the oxide of the metal. Example: $2\text{ZnS}(\text{s})+3{\text{O}}_2(\text{g})\stackrel{\text{heat}}{\to }2\text{ZnO}(\text{s})+2{\text{SO}}_2(\text{g})$. (ii) Alloy: A homogeneous mixture of two or more metals, or of a metal and a non-metal. Example: brass, an alloy of copper and zinc (bronze, solder and steel are others). (iii) Corrosion: The slow eating away of a metal when its surface is attacked by substances around it, such as air, moisture or acids. Example: rusting of iron - in moist air iron gets coated with reddish-brown rust, hydrated iron(III) oxide, ${\text{Fe}}_2{\text{O}}_3\cdot x{\text{H}}_2\text{O}$.

    Step-by-step solution →
13. Q133 CBSE 2026
    1. (i)Write all ​the ​reactions ​involved in the ​extraction ​of $\text{Zn}$ from $\text{ZnS}$.
    2. (ii)​Hydrogen gas is not evolved when most of the metals ​react with nitric ​acid. Give reason.
    Show answer

    Zinc sulphide is roasted to zinc oxide, then reduced with carbon; nitric acid oxidises the hydrogen formed to water.

    (i) Zinc is extracted from zinc blende in two reactions. Roasting, in which the concentrated ore is heated in excess air: $2\text{ZnS}(s)+3{\text{O}}_2(g)\stackrel{\Delta }{\to }2\text{ZnO}(s)+2{\text{SO}}_2(g)$. Reduction, in which the oxide is heated with coke: $\text{ZnO}(s)+\text{C}(s)\stackrel{\Delta }{\to }\text{Zn}(s)+\text{CO}(g)$. The zinc obtained is finally purified by electrolytic refining. (ii) Nitric acid is a strong oxidising agent. The hydrogen that the metal displaces from it is oxidised to water as soon as it is formed, while the nitric acid is itself reduced to an oxide of nitrogen (${\text{N}}_2\text{O}$, $\text{NO}$ or ${\text{NO}}_2$ according to the concentration of the acid and the metal used). No hydrogen therefore escapes as gas. Magnesium and manganese with very dilute nitric acid are the exceptions and do evolve hydrogen.

    Step-by-step solution →
14. Q134 CBSE 2026
    Give ​reasons for the following ​:

    1. (i)Alloying is a very good method of improving the properties of ​a metal.
    2. (ii)Carbon cannot be used as ​a reducing agent to obtain highly reactive metals from their oxides.
    3. (iii)Silver articles ​become ​black ​after some time when exposed to air ?
    Show answer

    Alloying improves properties; reactive metals have more affinity for oxygen than carbon; silver turns black as silver sulphide forms.

    (i) Mixing a metal with other metals, or with a non-metal, changes its properties, so they can be improved to suit a use. Pure iron is soft, but with a small amount of carbon it becomes hard and strong steel, and with nickel and chromium it becomes stainless steel, which does not rust. Pure gold is too soft, so it is alloyed with silver or copper for jewellery. (ii) Highly reactive metals such as sodium, calcium, magnesium and aluminium have more affinity for oxygen than carbon has, so carbon cannot remove oxygen from their oxides. These metals are obtained by electrolytic reduction instead. (iii) Silver reacts with sulphur compounds (such as hydrogen sulphide) in the air and forms a black coating of silver sulphide: $4\text{Ag}+2{\text{H}}_2\text{S}+{\text{O}}_2⟶2{\text{Ag}}_2\text{S}+2{\text{H}}_2\text{O}$. This is the corrosion of silver.

    Step-by-step solution →
15. Q135 CBSE 2026
    Read the following passage and answer the questions given below : Most of metals occur ​in combined state in form of ores. Carbonate ores are converted into oxides ​by calcination and ​sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like – Al, Mg are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. ​Alloying is a very good method of improving the properties of a metal. ​We can ​get desired properties by this ​method. The electrical conductivity and melting point of an alloy is less than that of pure ​metals.

    1. (a)Why carbonate or sulphide ores are converted to oxides before extraction of metal from it ?[1]
    2. (b)Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.[1]
    3. (c)(i)How is copper obtained from its ore (${\text{Cu}}_2\text{S}$) ? Give equations of the reactions.[2]
    4. (c)(ii)(I)OR : Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent ?[1]
    5. (c)(ii)(II)Why solder, an alloy of lead and tin, is used for welding electrical wires together ?[1]
    Show answer

    Oxides are easier to reduce; aluminium reduces iron oxide; copper sulphide is heated in air; carbon cannot reduce reactive metals; solder melts easily.

    (a) A metal is obtained much more easily from its oxide than from its carbonate or sulphide, because carbon readily reduces the oxide. So carbonate ores are calcined and sulphide ores roasted to the oxide first, e.g. ${\text{ZnCO}}_3\to \text{ZnO}+{\text{CO}}_2$ and $2\text{ZnS}+3{\text{O}}_2\to 2\text{ZnO}+2{\text{SO}}_2$, followed by $\text{ZnO}+\text{C}\to \text{Zn}+\text{CO}$. (b) Thermit reaction: ${\text{Fe}}_2{\text{O}}_3+2\text{Al}\to 2\text{Fe}+{\text{Al}}_2{\text{O}}_3+\text{heat}$. (c)(i) Copper is low in the reactivity series, so ${\text{Cu}}_2\text{S}$ is simply heated in air (self-reduction): $2{\text{Cu}}_2\text{S}+3{\text{O}}_2\to 2{\text{Cu}}_2\text{O}+2{\text{SO}}_2$, then $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}\to 6\text{Cu}+{\text{SO}}_2$. OR (c)(ii) (I) Highly reactive metals have a greater affinity for oxygen than carbon, so carbon cannot remove oxygen from their oxides; they are obtained by electrolytic reduction of their molten compounds. (II) Solder has a low melting point (lower than that of lead or of tin), so it melts easily and joins the wires without melting them, and the joint still conducts electricity.

    Step-by-step solution →
16. Q136 CBSE 2026
    Read the following passage and answer the questions that follow : An ​electrochemical process is used for ​refining ​metals ​by ​selectively dissolving and depositing ​them ​through electrolysis. Electrolytic refining is used for metals like Copper, Zinc, ​Tin, Nickel, Silver and Gold.

    1. (a)Define anode mud.[1]
    2. (b)In electrolytic refining of copper, name the anode and the cathode.[1]
    3. (c)(i)With the help of chemical equations, show the extraction of copper from its ore ${\text{Cu}}_2\text{S}$.[2]
    4. (c)(ii)OR : Draw a neat and labelled diagram showing electrolytic refining of copper.[2]
    Show answer

    Anode mud settles below the anode; the anode is impure copper and the cathode pure copper; heating copper sulphide in air gives copper.

    (a) Anode mud is the deposit of insoluble impurities that settles at the bottom of the tank, below the anode, during electrolytic refining; it contains less reactive metals such as silver and gold. (b) Anode: a thick block of impure copper (joined to the positive terminal). Cathode: a thin strip of pure copper (joined to the negative terminal). The electrolyte is acidified copper sulphate solution. (c)(i) ${\text{Cu}}_2\text{S}$ is heated in air: $2{\text{Cu}}_2\text{S}+3{\text{O}}_2\stackrel{\text{heat}}{\to }2{\text{Cu}}_2\text{O}+2{\text{SO}}_2$; the copper(I) oxide then reacts with the remaining sulphide (self-reduction): $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}\stackrel{\text{heat}}{\to }6\text{Cu}+{\text{SO}}_2$. OR (c)(ii) A tank of acidified copper sulphate solution; the impure copper anode joined to the positive terminal of a battery and the thin pure copper cathode joined to the negative terminal; ${\text{Cu}}^{2+}$ ions moving from the anode to the cathode; anode mud collecting below the anode (see the labelled figure).

    Step-by-step solution →
17. Q137 CBSE 2026
    The Statue of ​Liberty ​is one of the seven wonders. It is present on Liberty island in ​New ​York harbour in U.S.A. It is made up of a metal. Over a period of time, the metal’s colour has ​changed and statue gives greenish look. ​Answer the following questions based on ​the above ​information :

    1. (A)What is corrosion ?[1]
    2. (B)Why do silver articles turn black after some time ?[1]
    3. (C)Copper reacts with moist gas ‘A’ present in the air and slowly loses its shiny surface. It turns green. Identify ‘A’ and the substance of green coating formed on the surface of copper.[2]
    4. (C)OR : What is an alloy ? How is an alloy made ?[2]
    Show answer

    Corrosion eats away a metal's surface; silver turns black due to silver sulphide; gas A is carbon dioxide, the green coating basic copper carbonate.

    (A) Corrosion is the process in which the surface of a metal is slowly eaten away by the action of air, moisture, acids or other chemicals around it, the metal being converted into a surface compound such as an oxide, a carbonate or a sulphide. (B) Silver articles turn black because silver reacts with the sulphur present in the air, as hydrogen sulphide, to form a black coating of silver sulphide, ${\text{Ag}}_2\text{S}$: $4\text{Ag}+2{\text{H}}_2\text{S}+{\text{O}}_2⟶2{\text{Ag}}_2\text{S}+2{\text{H}}_2\text{O}$. (C) ‘A’ is carbon dioxide, ${\text{CO}}_2$, and the green coating formed on copper is basic copper carbonate, ${\text{CuCO}}_3\cdot {\text{Cu(OH)}}_2$: $2\text{Cu}+{\text{CO}}_2+{\text{H}}_2\text{O}+{\text{O}}_2⟶{\text{CuCO}}_3\cdot {\text{Cu(OH)}}_2$. (C) OR : An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal. It is made by first melting the main metal, then dissolving the other metals or non-metals in it in a fixed proportion, and then cooling the mixture to room temperature.

    Step-by-step solution →
18. Q138 CBSE 2026
    Read the following ​passage and answer the questions ​that follow : When metals are burnt in air, they form oxides. Metal ​oxides are ​generally basic in nature. But some metal oxides show both acidic as well as basic behaviour. Most metal oxides are insoluble in ​water, but some of these dissolve in water to form alkalis. Different metals show different reactivities ​towards oxygen. Metals such as sodium and potassium react ​vigorously whereas metals like silver and gold do not react with oxygen even at high temperature.

    1. (a)Why are metals like sodium and potassium kept immersed in kerosene oil ?[1]
    2. (b)Why does copper metal turn black on heating ?[1]
    3. (c)(i)What are the amphoteric oxides ? Give two examples.[2]
    4. (c)(ii)OR : (I) Name a water soluble metal oxide. Write the chemical equation for its reaction with water. (II) Why do silver and gold not react with oxygen even at high temperature ?[2]
    Show answer

    Kerosene keeps air and water away from reactive sodium and potassium; copper forms black copper oxide; amphoteric oxides react with both acids and bases.

    (a) Sodium and potassium are so reactive that they react with the oxygen and the moisture of air at room temperature, and the heat released can set the metal on fire. Kerosene oil does not react with them and keeps air and water vapour away, so the metals are stored immersed in it. (b) On heating, copper combines with the oxygen of air and a black layer of copper(II) oxide forms on its surface: $2\text{Cu}+{\text{O}}_2\stackrel{\text{heat}}{\to }2\text{CuO}$ (black). (c)(i) Amphoteric oxides are metal oxides that react with both acids and bases to give a salt and water, that is, they show basic as well as acidic behaviour. Examples: aluminium oxide ${\text{Al}}_2{\text{O}}_3$ and zinc oxide $\text{ZnO}$. OR : (c)(ii)(I) Sodium oxide, ${\text{Na}}_2\text{O}$, is soluble in water and gives an alkali: ${\text{Na}}_2\text{O}+{\text{H}}_2\text{O}⟶2\text{NaOH}$ (potassium oxide behaves the same way). (II) Silver and gold are the least reactive metals, at the bottom of the reactivity series; they hold their outer electrons so firmly that they do not give them up to oxygen, so no oxide forms even at high temperature and both are found in the free state.

    Step-by-step solution →
19. Q139 CBSE 2026
    1. (i)​Give reasons for the following : (I) Ionic compound have generally high ​melting points and boiling points. (II) Solder, an alloy of lead and tin, is ​used for welding electrical wires. (III) Carbon cannot reduce the oxides of ​Na or Mg.
    2. (ii)The reaction of compound ‘X’ with aluminium is used to join railway tracks : (I) Identify the ​compound ‘X’. (II) Name the reaction. (III) Write ​the ​balanced chemical equation of the reaction of compound ​‘X’ with aluminium.
    Show answer

    Strong ionic attractions give high melting points; solder melts easily; carbon cannot reduce Na or Mg oxides; X is ${\text{Fe}}_2{\text{O}}_3$ in the thermit reaction.

    (i)(I) In an ionic compound the oppositely charged ions are held together by strong electrostatic forces of attraction, and a considerable amount of energy is needed to overcome them; hence the high melting and boiling points. (II) Solder has a low melting point — lower than that of lead or of tin — so it melts easily and joins the wires without melting them, and the joint, being metallic, conducts electricity. (III) Sodium and magnesium lie above carbon in the reactivity series and have a greater affinity for oxygen than carbon has, so carbon cannot remove the oxygen from their oxides; such metals are obtained by electrolytic reduction. (ii)(I) ‘X’ is iron(III) oxide, ${\text{Fe}}_2{\text{O}}_3$. (II) The reaction is the thermit reaction. (III) ${\text{Fe}}_2{\text{O}}_3(\text{s})+2\text{Al}(\text{s})⟶2\text{Fe}(\text{l})+{\text{Al}}_2{\text{O}}_3(\text{s})+\text{heat}$

    Step-by-step solution →
20. Q140 CBSE 2026
    1. (i)Write the balanced chemical ​equation involved ​in the extraction of : (I) Hg from HgS (II) Zn from ${\text{ZnCO}}_3$ (III) ​Na from ​NaCl
    2. (ii)Why does sodium ​chloride : (I) not conduct electricity in solid ​state ? (II) have ​very high melting point ​?
    Show answer

    Hg is roasted and heated, Zn calcined and reduced, Na electrolysed; solid NaCl has fixed ions, and strong ionic forces give a high melting point.

    (i) (I) Hg from HgS — roasting, then thermal decomposition: $2\text{HgS}+3{\text{O}}_2\stackrel{\Delta }{\to }2\text{HgO}+2{\text{SO}}_2$ and $2\text{HgO}\stackrel{\Delta }{\to }2\text{Hg}+{\text{O}}_2$. (II) Zn from ${\text{ZnCO}}_3$ — calcination, then reduction with carbon: ${\text{ZnCO}}_3\stackrel{\Delta }{\to }\text{ZnO}+{\text{CO}}_2$ and $\text{ZnO}+\text{C}\stackrel{\Delta }{\to }\text{Zn}+\text{CO}$. (III) Na from NaCl — electrolytic reduction of the molten salt. At the cathode: ${\text{Na}}^{+}+e^{-}⟶\text{Na}$; at the anode: $2{\text{Cl}}^{-}⟶{\text{Cl}}_2+2e^{-}$; overall $2\text{NaCl}(\text{molten})\stackrel{\text{electrolysis}}{\to }2\text{Na}+{\text{Cl}}_2$. (ii) (I) Sodium chloride is ionic, and in the solid its ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are held in fixed positions in the rigid crystal lattice by strong electrostatic forces. There are no free ions and no free electrons to carry charge, so the solid does not conduct; molten or dissolved, the lattice breaks down, the ions become free to move, and it conducts. (II) Every ion in the crystal is held by strong electrostatic forces of attraction from the oppositely charged ions all around it, and these forces run through the whole lattice. A large amount of heat energy is needed to overcome them, so sodium chloride melts only at a very high temperature (about $1074$ K).

    Step-by-step solution →