NCERT Exemplar Problems and their Solutions

Grade X - Chemistry

Reviewed By:
Krishna Kant Majee
M.Sc., B.Ed.

Chapter 3-Metals and Non-metals- Short Answers

Q.37. 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 be identify the produced gas. Write chemical equations for both the reactions.

The divalent element must be some metal which releases hydrogen gas with both acid and bases. Bring a burning splinter near the gas, it will burn with a pop sound, hence it must be hydrogen gas. The divalent element could be zinc. Chemical equations for both the reactions are;
$\ce{Zn(s)+ 2NaOH(aq) ->[{Heat}]2NaOH(aq)+H2\uparrow}$
$\ce{Zn(s) + 2HCl(aq)-> ZnCl2(aq)+ H2\uparrow}$

Q.38. During extraction of metals, electrolytic refining is used to obtain pure metals.
(a) Which material will be used as anode and cathode for refining of silver metal by this process?
(b) Suggest a suitable electrolyte also.
(c) In this electrolytic cell, where do we get pure silver after passing electric current?

(a) Thin strip of pure silver is made the cathode and impure silver block is made the anode.
(b) Any salt solution of silver like $AgNO_{3}$ can be used as electrolyte.
(c) On passing electric current, from the electrolyte metal ions discharge on the cathode whereas metal ions from impure metal dissolves and goes into the electrolyte solution.

Q.39. Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them?

Usually metal sulphides and carbonates are converted to metal oxides in the process of extraction of metal. This is because metal oxides can easily reduce to metal with the help of some common reducing agent like coke.
Ores of metal can converted to metal oxides with the help of calcination or roasting which depends on the nature of ore. For example, metal carbonate can convert to metal oxide by calcinations whereas roasting is used for metal sulphide.
Q.40. Generally when metals are treated with mineral, acids. Hydrogen gas is liberated but when metals (except Mn and Mg) are treated with $HNO_{3}$, hydrogen is not liberated, why?
Unlike other acids, reaction of nitric acid is little different because it is a good oxidizing agent.
As hydrogen gas is formed during the reaction of nitric acid with metals, acid reacts with hydrogen gas and converts it to water. And nitric acid gets reduced to $NO_{2} or NO or N_{2}O.$
Q.41. Compound X and aluminium are used to join railway tracks.
(a) Identify the compound X.
(b) Name the reaction.
(c) Write down its reaction.
(a) Iron (III) oxide $(Fe_{2}O_{3})$ is used with aluminium (Al) to join railway tracks.
(b) The reaction is known as Thermite reaction or Alumino thermy process.
$(c) Fe_{2}O_{3}(s) + 2 Al(s)\rightarrow 2 Fe (l) + Al_{2}O_{3} (s) + Heat$
Q.42. 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.
Given, molecular formula of Y = XOH
And molecular mass OF Y = 40.
So, atomic mass of metal X would be;
M + 16 + 1 = 40 ⟹ M = 40 − 17=23
Metal with atomic mass 23 is sodium and reaction of it with cold water will form sodium hydroxide (NaOH).
$2Na + 2H_{2}O (Cold)\rightarrow 2NaOH + H_{2}\uparrow$
Hence; X = Sodium (Na), Y = Sodium hydroxide (NaOH), Z = Hydrogen gas $(H_{2})$
Q.43. A non-metal X exists in two different forms Y and Z. Y is hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.
Different forms of any element are called as allotropes. Since Y is the hardest substance so it must be diamond which is an allotrope of carbon. Hence non-metal X is carbon. Another allotrope of carbon is graphite which is very good conductor of electricity due to presence of free mobile electrons.
Hence; X = Carbon, Y= Diamond and Z = Graphite
Q.44. The following reaction takes place when aluminium powder is heated with $MnO_{2}$
$\ce{3MnO2 (s) + 4Al (s) ->[{Heat}] 3Mn (l) + 2Al2O3 (l) }$
(a) Is aluminium getting reduced?
(b) Is $MnO_{2}$ getting oxidized?
(a) As Al converts to ${Al_{2}O_{3}$, i.e., oxygen is added to Al so it is getting oxidized.
(b) Now $MnO_{2}$ gets converted to Mn, i.e. oxygen is being removed from $MnO_{2}$, so it is getting reduced.
Q.45. What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?
Solder is a fusible metal alloy of lead (Pb) and tin (Sn). It contains 40% lead and 60% of Sn metal.
Solder has a low melting point around 90°C to 450 °C as compared to the metals used in forming electrical wires. So, it acts as bridge between two metal pieces, hence used for welding electrical wires.
Q.46. 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.
Ans. Aluminium (Al) is used in thermite process so metal A is Aluminium. The reaction of Al with oxygen forms aluminium oxide $Al_{2}O_{3}$ which is amphoteric in nature as it can exhibit acidic as well as basic nature.
$4Al (s) + 3O_{2} (g)\rightarrow 2Al_{2}O_{3} (s)$
Hence; A = Aluminium and B = Aluminium oxide
Reactions of aluminium oxide with HCl and NaOH are as given below;
$Al_{2}O{3} (s) + 6 HCl (aq)\rightarrow 2AlCl_{3} (aq) + 3 H_{2}O(l)$
$Al_{2}O_{3} (s) + 2NaOH (aq)\rightarrow 2NaAlO_{2} (aq) + H_{2}O (l)$
Q.47. 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. Give the reactions involved.
Mercury is the only metal which exists in liquid state at room temperature. It is usually extracted from its sulphide ore, mercury (II) sulphide or cinnabar (HgS) by the process of roasting. Hence;
Metal = Mercury and Ore = Cinnabar
Roasting of ore occurs in the presence of air and form metal oxide which further reduce to metal.
$\ce{2HgS(s) +3O2(g)-> [{Roasting}] 2HgO(s)+ 2SO2(g)}$
Q.48. Give the formulae of the stable binary compounds that would be formed by the combination of following pairs of elements.
(a) Mg and $N_{2}$
(b) Li and $O_{2}$
(c) Al and $Cl_{2}$
(d) K and $O_{2}$
(a) $Mg and N_{2} – Mg_{3}N{2}$ (Magnesium nitride)
(b) $Li and O_{2}- Li_{2}O $ (Lithium oxide)
(c) $Al and Cl_{2}- AlCI_{3}$ (Aluminium chloride)
(d) $K and O_{2}- K_{2}O$ (Potassium oxide)
Q.49. What happens when
(a) $ZnCO_{3}$ is heated in the absence of oxygen?
(b) A mixture of $Cu_{2}O and Cu_{2}S$ is heated?
(a) Heating of carbonate ores in the absence of oxygen is called as calcinations. Zinc carbonate $(ZnCO_{3})$ also forms zinc oxide (ZnO) and carbon dioxide when it is heated in the absence of oxygen.
$\ce{ZnCO3(s)->[{Calcination}] ZnO(s)+ CO2(g) }$
$\ce{ZnO(s) + C(s) ->[{Heat}]Zn(s)+ CO(g)}$

This process is used to convert carbonate ores to metal oxides and also helps to remove moisture and volatile impurities.
(b) Reaction of copper sulphide $(Cu_{2}S)$ and copper oxide $(Cu_{2}O)$at high temperature forms copper metal and sulphur dioxide gas is expelled out.

$\ce{2Cu2O(s)->[{Heat}][{Reduction}] 6Cu(s)+ SO2(g)}$
Q.50. 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.
(a) Identify A, B and C.
(b) To which group of periodic table does A belong?
(a) Non-metal A must be carbon as it forms two oxides; carbon monoxide (CO) and carbon dioxide. Out of these two oxides; one is toxic that is CO and another is responsible for global warming so it must be $CO_{2}$ (carbon dioxide).
Hence; A = Carbon, B = Carbon monoxide and C = Carbon dioxide.
(b) The electronic configuration of carbon is 2, 4. So there are 4 valence electrons and group number should be 10 + number of valence electrons. So carbon belongs to 14th group of the periodic table.
Q.51. Give two examples each of the metals that are good conductors and comparatively poor conductors of heat respectively.
(a) Good conductors: Aluminium and copper are good conductors of heat.
(b) Poor conductors: Titanium and mercury are poor thermal conductors.
Q.52. 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 $(37^{°}C).$
(a) Metal that exists in liquid state at room temperature = Mercury
(b) Non-metal that exists in liquid state at room temperature = Bromine
(c) Metals having melting point less than 310 K = Caesium $(28.5^{o}C)$ and Gallium (29.76°C)
Q.53. 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.
The compound that is used in white wash is calcium hydroxide. So the element must be calcium that reacts with water to form calcium hydroxide. Heating of calcium hydroxide forms calcium oxide which again converts to hydroxide with the reaction of water.
Hence; A = Ca, B = $Ca(OH)_{2}$, C = CaO
Reactions involved are:
$Ca + 2H_{2}O\rightarrowCa(OH)_{2}+ H_{2}$
$\ce{Ca(OH)2->[{Heat}]CaO + H2O}$
$CaO + H_{2}O-> Ca(OH)_{2}$

Q.54. 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 reactions involved.
Alkali metals form MOH type of metal hydroxides with water. The molecular mass of hydroxide, B is 40 therefore mass of metal can be calculated as;
MOH = x + 16 + 1 = 40 ⟹ x = 23
Thus the atomic mass of alkali metal is 23. So it must be sodium that reacts with water to form sodium hydroxide (NaOH).
$2Na + 2H_{2}O\rightarrow 2NaOH (aq) + H_{2}2 (g)\uparrow$
Reaction of sodium hydroxide with aluminium oxide $(Al_{2}O_{3})$ gives soluble sodium aluminate $(NaAlO_{2})$.
$Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$
Hence; A = Sodium, B = Sodium hydroxide and C = Sodium aluminate
Q.55. Give the reaction involved during extraction of zinc from its ore by
(a) Roasting of zinc ore
(b) Calcination of zinc ore
(a) Roasting of zinc ore:
• Conversion of sulphide ore to metal oxide can be done with the help of roasting which is a heating process in the presence of air below melting point of metal. Roasting of zinc ore like zinc sulphide forms zinc oxide and sulphur dioxide.
• This metal oxide formed above can further reduce to metallic zinc in the presence of coke.

$\ce{2ZnS(s) + 3O2->[{Roasting}] 2ZnO(s)+ 2SO2(g)}$
$ce{ZnO(s) +\underset{\text{Carbon(reducing agents)}} C(s)->[{Heat}] Zn(s) + CO(g)}$

(b) Calcination of zinc ore:
• Calcination is a heating process for carbonate ore in the absence of air. Zinc carbonate $(ZnCO_{3})$ or calamine converts to zinc oxide and carbon dioxide.
• Then, like roasting, ZnO again reduces to metallic zinc in the presence of coke at high temperature.
$ce{ZnCO3(s)->[{Calcination}] ZnO(s)+ CO2(g)}$
$\ce{ZnO(s)+ \underset{\text{Carbon(reducing agents)}} C(s)->[{Delta}] Zn(s) + CO(g)}$

Q.56. 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.
Because of less reactivity, copper metal does not release hydrogen gas with acid.
$Cu(s) + HCl \rightarrow No reaction$
Hence metal M is copper.
Reaction of copper with oxygen on prolonged heating to form a black substance copper (II) oxide (CuO).
$2Cu(s) + O_{2} (g)\rightarrow 2CuO(s)$
Q.57. An element forms an oxide $ A_{2}O_{3}$ which is acidic in nature. Identify A as a metal or non-metal.
Metals usually form basic oxides like BaO, MgO, $Na_{2}O$ etc. Unlike metals, non-metals tend to form acidic oxides.
As we move from left to right in the periodic table, the acidity of oxides increases. So the element which forms acidic oxide must be non-metallic element.
The formula for oxide is $A_{2}O_{3}$ means the charge on element must be +3 or element should have 3 valence electrons that is boron and formula for oxide would be $B_{2}O_{3}$.
Q.58. A solution of $CuSO_{4}$ 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.
In reactivity series, iron is placed above copper. So we can say that iron is more reactive compare to copper and it can displace copper from its compounds.
When copper sulphate solution is placed in an iron pot, iron reacts with copper sulphate and form iron sulphate with copper.
So the blue colour of copper sulphate solution fade to light green due to formation of iron sulphate and holes are produced at places where iron metal has reacted.
$Fe(s) + CuSO_{4}(aq)\rightarrow FeSO_{4}(aq) + Cu(s)$