NCERT Exemplar Class 10 Science Solutions Chapter 3 Metals And Non Metals

Question 1. Which of the following properties is generally not shown by metals?
(a) Electrical conduction
(b) Sonorous in nature
(c) Dullness
(d) Ductility

Answer. (c)
Properties of metals:
1. High melting point
2. Good conductors of electricity
3. Good conductors of heat
4. High density
5. Malleable
6. Ductile
7. Sonorous nature
8. Shiny (lustrous) in nature only when it is fresh. Metals lose their shine on keeping in the air for a long time due to the formation of a thin layer of oxide, carbonate or sulphide on their surface.
Therefore option (C) is correct

Question 2. The ability of metals to be drawn into thin wire is known as
(a) ductility
(b) malleability
(c) sonorousity
(d) conductivity
Answer. (a)
(a)Ductility: The ability of metals to be drawn into thin wires.
(b)Malleability: Property which allows the metals to be hammered into thin sheets.
(c)Sonorousity: Property of metals due to which they make sound when hit with an object.
(d)Conductivity: Property of metals that allows heat to pass through them easily.
Therefore option (A) is correct

Question 3. Aluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?
(i) Good thermal conductivity
(ii) Good electrical conductivity
(iii) Ductility
(iv) High melting point
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer. (d)
Aluminium is used for making cooking utensils. This is so because aluminium has good thermal conductivity, it is malleable (can be hammered into thin sheets), it has a light weight and high melting point.
Hence (i) and (iv) are correct
Therefore option (D) is correct

Question 4. Which one of the following metals do not react with cold as well as hot water?
(a) Na
(b) Ca
(c) Mg
(d) Fe
Answer. (d)
The reactivity of metals with water is based on the reactivity series.
Metals like (Na, K and Ca) react with cold water.
Magnesium (Mg) reacts with hot water but Zinc (Zn) and Iron (Fe) react with steam.
Metals like lead, copper, silver and gold do not react with water (or even steam).
Red hot iron reacts with steam to form iron oxide and hydrogen.

Question 5. Which of the following oxide(s) of iron would be obtained on prolonged reaction of iron with steam?
(a) FeO
(b) ${\mathrm{Fe}}_2{\mathrm{O}}_3$
(c)${\mathrm{Fe}}_3{\mathrm{O}}_4$
(d)${\mathrm{Fe}}_2{\mathrm{O}}_3$ and ${\mathrm{Fe}}_3{\mathrm{O}}_4$
Answer. (c)
Iron does not react directly with cold or hot water. It forms a metal oxide when steam is passed over it. When red hot iron reacts with steam it forms iron(II, III) oxide and hydrogen gas is liberated. The reaction is reversible and is shown as follows:
3Fe(s)+4${\mathrm{H}}_2\mathrm{O}$→${\mathrm{Fe}}_3{\mathrm{O}}_4$(s)+4${\mathrm{H}}_2$(g)

Question 6. What happens when calcium is treated with water?
(i) It does not react with water
(ii) It reacts violently with water
(iii) It reacts less violently with water
(iv) Bubbles of hydrogen gas formed stick to the surface of calcium
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer. (d)
Calcium reacts less violently with water and the bubbles of hydrogen gas produced stick to the surface of calcium. Due to which it floats over the water surface.
Ca(s)+2${\mathrm{H}}_2\mathrm{O}$(l)→$\mathrm{Ca}{(\mathrm{OH})}_2$ (aq)+${\mathrm{H}}_2$(g)
Much less heat is produced in this reaction due to which hydrogen gas formed does not catch fire.

Question 7. Generally metals react with acids to give salt and hydrogen gas. Which of the following acids does not give hydrogen gas on reacting with metals (except Mn and Mg)?
(a) ${\mathrm{H}}_2{\mathrm{SO}}_4$
(b) HCl
(c) $\mathrm{H}{\mathrm{NO}}_3$
(d) All of these
Answer. (c)
Nitric acid $\mathrm{H}{\mathrm{NO}}_3$ does not give hydrogen gas on reacting with metals (except Manganese and Magnesium)
3Mn + 8${\mathrm{H}}_2{\mathrm{NO}}_3$ (dil.) → 3$\mathrm{Mn}{({\mathrm{NO}}_3)}_2$ + 2NO + 4${\mathrm{H}}_2\mathrm{O}$
Manganese reacts with nitric acid to produce manganese(II) nitrate, nitric oxide and water.
4Mg + 10$\mathrm{H}{\mathrm{NO}}_3$ (dil.) → 4$\mathrm{Mg}{({\mathrm{NO}}_3)}_2$ + ${\mathrm{N}}_2\mathrm{O}$ + 5${\mathrm{H}}_2\mathrm{O}$
Manganese reacts with nitric acid to produce manganese(II) nitrate, nitric oxide and water.
4Mg + 10$\mathrm{H}{\mathrm{NO}}_3$(dil.) → 4$\mathrm{Mg}{({\mathrm{NO}}_3)}_2$ + ${\mathrm{N}}_2\mathrm{O}$ + 5${\mathrm{H}}_2\mathrm{O}$
Magnesium reacts with nitric acid to produce nitrate magnesium, dinitrogen monoxide and water.
Nitric acid is a strong oxidizing agent so, as soon as hydrogen gas is formed in the reaction between metal and $\mathrm{H}{\mathrm{NO}}_3$, it oxidizes this hydrogen to water.
Therefore option (C) is correct

Question 8. The composition of aqua-regia is
(a) Dil. $\mathrm{H}\mathrm{Cl}$: Conc. $\mathrm{H}{\mathrm{NO}}_3$ 3 : 1
(b) Conc.$\mathrm{H}\mathrm{Cl}$ : Dil. $\mathrm{H}{\mathrm{NO}}_3$ 3 : 1
(c) Conc. $\mathrm{H}\mathrm{Cl}$ : Conc.$\mathrm{H}{\mathrm{NO}}_3$ 3 : 1
(d) Dil.$\mathrm{H}\mathrm{Cl}$ : Dil.$\mathrm{H}{\mathrm{NO}}_3$ 3 : 1
Answer. (c)
Aqua regia is a mixture of nitric acid and hydrochloric acid, in a molar ratio of 1:3.
So, Conc.$\mathrm{H}\mathrm{Cl}$ and conc. $\mathrm{H}{\mathrm{NO}}_3$ in ratio 3 : 1 forms aqua-regia.
Aqua-regia is a highly corrosive, yellow-orange fuming liquid. It can dissolve all metals (even gold and platinum).

Question 9. Which of the following are not ionic compounds?
(i) KCl
(ii) HCl
(iii) $\mathrm{C}{\mathrm{Cl}}_4$
(iv) NaCl
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iii)
Answer. (b)
Ionic compounds are neutral compounds made up of positively charged ions called cations and negatively charged ions called anions.
(i) KCl: formed by the transfer of electrons and hence ionic.
(ii) HCl: polar covalent compound (formed by sharing of electrons)
(iii) $\mathrm{C}{\mathrm{Cl}}_4$: is a non-polar covalent compound (formed by sharing of electrons)
(iv) NaCl: formed by the transfer of electrons and hence ionic.
Hence (ii) and (iii) are not ionic.
Therefore option (B) is correct

Question 10. Which one of the following properties is not generally exhibited by ionic compounds?
(a) Solubility in water
(b) Electrical conductivity in solid state
(c) High melting and boiling points
(d) Electrical conductivity in molten state

Answer. (b)
Properties of ionic compounds:

1. Can conduct electricity in molten or aqueous state only.

2. High melting and boiling points

3. Most ionic compounds are soluble in water and form aqueous solutions.

4. In solid states, these are non-conductors of electricity because of the absence of free ions.

5. The ions are held together in a fixed position by strong electrostatic force and cannot move freely.

Hence, electrical conductivity in the solid state is not generally exhibited by ionic compounds.
Therefore option (B) is correct

Question 11. Which of the following metals exist in their native state in nature?
(i) Cu
(ii) Au
(iii) Zn
(iv) Ag
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer. (c)
Gold and silver (Au and Ag) are noble metals and are very less reactive, so they are usually found in a free state (native state) in nature.

Question 12. Metals are refined by using different methods. Which of the following metals are refined by electrolytic refining?
(i) Au
(ii) Cu
(iii) Na
(iv) K
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)
Answer. (a)
The method to be used for refining an impure metal depends on the nature of the metal as well as on the nature of impurities present in it.
Electrolytic refining is a process of refining a metal by the process of electrolysis.
Electrolytic refining is used for metals like Ag, Au, Cu, Zn, etc
Electrolytic refining of Cu: A large piece of impure copper is used as the anode with a thin strip of pure copper as the cathode. Copper(II) sulphate solution is the electrolyte.
Electrolytic refining of Au: A thin sheet of gold is made the cathode and a gold alloy is made the anode. Hydrochloric acid is used as an electrolyte.
Therefore option (A) is correct

Question 13. Silver articles become black on prolonged exposure to air. This is due to the formation of
(a) ${\mathrm{Ag}}_3\mathrm{N}$
(b) ${\mathrm{Ag}}_2\mathrm{O}$
(c) ${\mathrm{Ag}}_2\mathrm{S}$
(d) ${\mathrm{Ag}}_2\mathrm{S}$ and ${\mathrm{Ag}}_3\mathrm{N}$
Answer. (c)
Silver articles become black because silver reacts with H2S gas present in the air to form a black coating of ${\mathrm{Ag}}_2\mathrm{S}$.
2 Ag + ${\mathrm{O}}_2$ → ${\mathrm{Ag}}_2\mathrm{O}$
4 Ag + 2 ${\mathrm{H}}_2\mathrm{S}$ + 2${\mathrm{O}}_2$ → 2${\mathrm{Ag}}_2\mathrm{S}$ + 2${\mathrm{H}}_2\mathrm{O}$
${\mathrm{Ag}}_2\mathrm{O}$ is brown in colour and ${\mathrm{Ag}}_2\mathrm{S}$ is black.
Therefore option (C) is correct

Question 14. Galvanisation is a method of protecting iron from rusting by coating with a thin layer of:
(a) Gallium
(b) Aluminium
(c) Zinc
(d) Silver

Answer. (c)
Galvanization is a method of protecting iron from rusting by coating it with a thin layer of zinc (Zn) metal.
Therefore option (C) is correct

Question 15. Stainless steel is very useful material for our life. In stainless steel, iron is mixed with
(a) Ni and Cr
(b) Cu and Cr
(c) Ni and Cu
(d) Cu and Au
Answer. (a)
Stainless steel: Iron (Fe) is mixed with Nickel (Ni), Chromium (Cr) and Carbon (C).
It contains approximately 17% chromium, 10% nickel and a very small amount of carbon.
Therefore option (A) is correct

Question 16. If copper is kept open in air, it slowly loses its shining brown surface and gains a green coating. It is due to the formation of
(a) $\mathrm{Cu}{\mathrm{SO}}_4$
(b) $\mathrm{Cu}{\mathrm{CO}}_3$

(c) $\mathrm{Cu}{({\mathrm{NO}}_3)}_2$
(d) $\mathrm{Cu}\mathrm{O}$

Answer. (b)
Copper reacts with oxygen present in the air and forms a green coating on its surface due to the formation of basic copper carbonate [$\mathrm{Cu}{\mathrm{CO}}_3$ and $\mathrm{Cu}{(\mathrm{OH})}_2$]
The reaction is given as:
2Cu + ${\mathrm{O}}_2$ (From air) + ${\mathrm{CO}}_2$ + ${\mathrm{H}}_2\mathrm{O}$ →$\mathrm{Cu}{\mathrm{CO}}_3$ + $\mathrm{Cu}{(\mathrm{OH})}_2$
Basic copper carbonate(green coating)
Therefore option (B) is correct

Question 17. Generally, metals are solid in nature. Which one of the following metals is found in liquid state at room temperature?
(a) Na
(b) Fe
(c) Cr
(d) Hg
Answer. (d)
Mercury (Hg) is the only metal which is found in liquid state at room temperature
Therefore option (D) is correct

Question 18. Which of the following metals are obtained by electrolysis of their chlorides in molten state?
(i) Na
(ii) Ca
(iii) Fe
(iv) Cu
(a) (i) and (iv)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (i) and (ii)
Answer. (d)
Sodium (Na) and calcium (Ca) being highly reactive are obtained by electrolysis of their chlorides in the molten state.
During electrolysis of NaCl
$\mathrm{NaCl}\to {\mathrm{Na}}^{+}(l)+{\mathrm{Cl}}^{-}(l)$
At cathode: reduction of$2{\mathrm{Na}}^{+}(aq)+e^{-}\to \mathrm{Na}(s)$
At anode: oxidation of $2{\mathrm{Cl}}^{-}(\mathrm{aq})\to {\mathrm{Cl}}_2(\mathrm{g})+2e^{-}$
Net Reaction: $2{\mathrm{Na}}^{+}(aq)+2{\mathrm{Cl}}^{-}(aq)\to 2\mathrm{Na}(s)+{\mathrm{Cl}}_2(g)$
During electrolysis of aqueous $\mathrm{Ca}{\mathrm{Cl}}_2$
We have,
At anode: $2{\mathrm{Cl}}^{-}\to {\mathrm{Cl}}_2+2e^{-}$
At cathod: ${\mathrm{Ca}}^{2+}+2e^{-}\to \mathrm{Ca}(s)$
Therefore option (D) is correct

Question 19. Generally, non-metals are not lustrous. Which of the following non metal is lustrous?
(a) Sulphur
(b) Oxygen
(c) Nitrogen
(d) Iodine

Answer. (d)
Shining metals are called lustrous.
Sulphur, oxygen, and nitrogen are non-lustrous.
Iodine is a non-metal having a lustrous appearance.
It has a shining surface (like that of metals).
Therefore option (D) is correct

Question 20. Which one of the following four metals would be displaced from the solution of its salts by other three metals?
(a) Mg
(b) Ag
(c) Zn
(d) Cu
Answer. (b)
We know that Ag is less reactive than Mg, Zn and Cu.
Hence Silver (Ag) metal would be displaced from the solution of its salts by the other three metals.
Mg + 2 $\mathrm{Ag}\mathrm{Cl}$ → $\mathrm{Mg}{\mathrm{Cl}}_2$+ 2 Ag
Zn + $\mathrm{Ag}\mathrm{Cl}$ → $\mathrm{Zn}{\mathrm{Cl}}_2$ + Ag
Cu + 2 $\mathrm{Ag}\mathrm{Cl}$ →$\mathrm{Cu}{\mathrm{Cl}}_2$ + 2 Ag
Therefore option (B) is correct

Question 21. 2 mL each of concentrated HCl, HNO3 and a mixture of concentrated HCl and concentrated HNO3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No change occurred in test tubes A and B but the metal got dissolved in test tube C respectively. The metal could be
(a) Al
(b) Au
(c) Cu
(d) Pt

Answer. (b)
A mixture of conc. HCl and conc. $\mathrm{H}{\mathrm{NO}}_3$ in the ratio of 3 : 1 is known as aqua-regia. Gold (Au) dissolve in aqua-regia.
Hence option (b) is correct.

Question 22. An alloy is
(a) an element
(b) a compound
(c) a homogeneous mixture
(d) a heterogeneous mixture
Answer. (c)
An alloy is a homogeneous mixture of different metals or a metal with one or more other element.
It is done in order to give greater strength or resistance to corrosion.
Examples: Brass, Steel, amalgam etc.
Therefore option (c) is correct

Question 23. An electrolytic cell consists of
(i) positively charged cathode
(ii) negatively charged anode
(iii) positively charged anode
(iv) negatively charged cathode
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) ad (iv)
Answer. (b)
An electrolytic cell consists of a positively charged anode and a negatively charged cathode.
Anode - positively charged - attracts negative ions or anions towards it
Cathode - negatively charged - attracts positive ions or cations towards it
Hence statements (iii) and (iv) are correct
Therefore option (b) is correct

Question 24. During electrolytic refining of zinc, it gets
(a) deposited on cathode
(b) deposited on anode
(c) deposited on the cathode as well as the anode
(d) remains in the solution
Answer. (a)
Electrolytic refining is a process of refining a metal by the process of electrolysis.
The impure metal is made the anode and a thin strip of pure metal is made the cathode.
The pure zinc metal deposited on the cathode is connected to the negative terminal of the battery.
Zn2+ ions from the solution move towards the negatively charged cathode.
Hence, Zn is deposited on the cathode.
Therefore option (a) is correct

Question 25. An element A is soft and can be cut with a knife. This is very reactive to air and cannot be kept open in air. It reacts vigorously with water. Identify the element from the following
(a) Mg
(b) Na
(c) P
(d) Ca
Answer. (b)
Given that A is soft and can be cut with a knife. This is very reactive to air and cannot be kept open in air. It reacts vigorously with water.
From the properties of the element described in the question, it can be said surely that the element is sodium (Na).
Therefore option (b) is correct

Question 26. Alloys are homogeneous mixtures of a metal with a metal or non metal. Which among the following alloys contain non-metal as one of its constituents?
(a) Brass
(b) Bronze
(c) Amalgam
(d) Steel
Answer. (d)
Alloys are homogeneous mixtures of a metal with a metal or non-metal.
Steel is an alloy which contains iron (a metal) and carbon (a non-metal).
Brass is an alloy which contains copper (a metal) and zinc (a metal).
Bronze is an alloy which contains copper (a metal) and tin (a metal).
Amalgam is formed by mixing mercury (a metal) with other metals.
Therefore option (d) is correct

Question 27. Which among the following statements is incorrect for magnesium metal?
(a) It burns in oxygen with a dazzling white flame
(b) It reacts with cold water to form magnesium oxide and evolves into hydrogen gas
(c) It reacts with hot water to form magnesium hydroxide and evolves into hydrogen gas
(d) It reacts with steam to form magnesium hydroxide and evolves into hydrogen gas
Answer. (b)
(a) When the magnesium metal burns it reacts with oxygen found in the air to form Magnesium Oxide (dazzling white flame).
$2\mathrm{Mg}(s)+{\mathrm{O}}_2(g)\to 2\mathrm{MgO}(s)+$ energy
(b) Magnesium metal does not react with cold water.
(c, d) Mg reacts with hot water and steam both. The reaction is shown below:
$\mathrm{Mg}(\mathrm{s})+{\mathrm{HO}}_2($ hot or steam $)\to \mathrm{Mg}(\mathrm{OH}{)}_2+{\mathrm{H}}_2(\mathrm{g})$
It gives magnesium hydroxide and hydrogen gas
Therefore option (b) is the correct answer

Question 28. Which among the following alloys contain mercury as one of its constituents?
(a) Stainless steel
(b) Alnico
(c) Solder
(d) Zinc amalgam
Answer. (d)
(a) Stainless steel: iron and carbon
(b) Alnico: Aluminium, Nickel, Cobalt and Iron plus varying levels of Copper, Titanium and Niobium.
(c) Solder: One is a soft solder, which is an alloy of tin and lead. The other is a hard solder, which is an alloy of copper and zinc.
(d) Zinc amalgam: When mercury reacts with a zinc metal, it forms the zinc amalgam
Therefore option (d) is correct

Question 29. Reaction between X and Y, forms compound Z. X loses electron and Y gains electron. Which of the following properties is not shown by Z?
(a) Has a high melting point
(b) Has a low melting point
(c) Conducts electricity in molten state
(d) Occurs as solid
Answer. (b)
During the reaction between X and Y, the compound Z is formed.
X loses electrons and Y gains electrons. Hence Z is an ionic compound.
Ionic compounds:
1 .Have a high melting point
2. Conducts electricity in molten state
3. Usually occur as solids.
Therefore option (b) is correct

Question 30. The electronic configurations of three elements X, Y and Z are X — 2, 8; Y — 2, 8, 7 and Z — 2, 8, 2. Which of the following is correct?
(a) X is a metal
(b) Y is a metal
(c) Z is a non-metal
(d) Y is a non-metal and Z is a metal
Answer. (d)
X — 2, 8: Has a complete octet, hence it is a noble gas.
Y — 2, 8, 7: It requires one electron to complete its octet.
Hence Y is a non-metal
Z — 2, 8, 2: It has a tendency to lose two electrons to complete its octet.
Hence Z is a metal
Therefore option (d) is correct

Question 31. Although metals form basic oxides, which of the following metals form an amphoteric oxide?
(a) Na
(b) Ca
(c) Al
(d) Cu
Answer. (c)
(a) Na - Sodium oxide (${\mathrm{Na}}_2\mathrm{O}$) is a strongly basic oxide.
It contains an oxide ion, ${\mathrm{O}}_2$-, which is a very strong base and has a high tendency to combine with hydrogen ions.
(b) Ca - Calcium oxide (CaO) is basic in nature. It dissolves in water to form calcium hydroxide
(c) Al - Aluminium oxide (${\mathrm{Al}}_2{\mathrm{O}}_3$) shows basic as well as acidic behaviour hence it is called an amphoteric oxide.
(d) Cu - Copper(II) oxide or cupric oxide (CuO)
Copper(I) oxide or cuprous oxide (${\mathrm{Cu}}_2\mathrm{O}$)
Both are basic in nature
Therefore option (c) is correct

Question 32. Generally, non-metals are not conductors of electricity. Which of the following is a good conductor of electricity?
(a) Diamond
(b) Graphite
(c) Sulphur
(d) Fullerene
Answer. (b)
Graphite is a non-metal which conducts electricity.
Diamond and graphite are allotropes of Carbon (a non-metal).
A graphite crystal comprises of layers of carbon molecules or sheets of carbon particles. Each carbon atom in a graphite layer is joined to three other carbon atoms by strong covalent bonds to form flat hexagonal rings. The fourth valence electron of every carbon atom is free to move. Due to the presence of free electrons in a graphite crystal, it conducts electricity.
Therefore option (b) is correct

Question 33. Electrical wires have a coating of an insulting material. The material, generally used is
(a) Sulphur
(b) Graphite
(c) PVC
(d) All can be used
Answer. (c)
(a) Sulphur is insulating in nature. But it has a very low ignition temperature. Sulphur is also highly ductile because it is a non-metal, which makes it difficult to give it the required shape. So it is not used as a coating material for an electric wire.
(b) Graphite: A graphite crystal comprises layers of carbon molecules or sheets of carbon particles. Each carbon atom in a graphite layer is joined to three other carbon atoms by strong covalent bonds to form flat hexagonal rings. The fourth valence electron of every carbon atom is free to move. Due to the presence of free electrons in a graphite crystal, it conducts electricity.
So it is not used as a coating material for an electric wire.
(c) PVC or Polyvinyl chloride is an insulating substance. It does not allow an electric current to pass through it. The electrical wires have a covering of PVC around them.
Therefore option (c) is correct

Question 34. Which of the following non-metals is a liquid?
(a) Carbon
(b) Bromine
(c) Phosphorus
(d) Sulphur
Answer. (b)
(a) Carbon: At room temperature, it is in a solid state.
(b) Bromine: is a liquid at room temperature, solidifying at -7.2ºC.
(c) Phosphorus: is a colourless, semitransparent, soft, waxy solid at room temperature.
(d) Sulphur: At room temperature,e it is a solid
Therefore option (b) is correct

Question 35. Which of the following can undergo a chemical reaction?
(a) $\mathrm{Mg}{\mathrm{SO}}_4$ + Fe
(b) $\mathrm{Zn}{\mathrm{SO}}_4$ + Fe
(c) $\mathrm{Mg}{\mathrm{SO}}_4$ + Pb
(d) $\mathrm{Cu}{\mathrm{SO}}_4$ + Fe
Answer. (d)
(a) and (b),
Fe is less reactive than Mg and Zn, so it is impossible to replace them.
No reaction takes place.
(c)
Pb is less reactive than Mg, so it is impossible to replace Mg.
No reaction takes place.
(d)
Fe is more reactive than Cu, so it replaces Cu from $\mathrm{Cu}{\mathrm{SO}}_4$
Hence reaction takes place.
Therefore option (d) is correct

Question 36. Which one of the following figures correctly describes the process of electrolytic refining?

Answer. (c)
Electrolytic refining is a process of refining a metal by the process of electrolysis.
Impure metal is made anode and is connected to the positive terminal of the battery. Pure metal is made cathode and is connected to the negative terminal of the battery.
Electrolytic refining of Cu: A large piece of impure copper is used as the anode with a thin strip of pure copper as the cathode. Copper(II) sulphate solution is the electrolyte.
${\mathrm{Cu}}^{2+}$ ions from the solution are deposited on the cathode while Cu from the impure anode dissolves into the solution and the impurities settle down below the anode as anode mud.
Therefore option (c) is correct

Question 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?
Answer. (a)
Electrolytic refining is a process of refining a metal by the process of electrolysis.
Impure metal is made anode and is connected to the positive terminal of the battery. Pure metal is made cathode and is connected to the negative terminal of the battery.
(a) Cathode: Thin strip of pure silver (It is connected to the negative terminal of the battery).
Anode: Impure, silver block (It is connected to the positive terminal of the battery).
(b) Electrolyte: any salt solution of silver like $\mathrm{Ag}{\mathrm{NO}}_3$ or $\mathrm{Ag}\mathrm{Cl}$
but sodium argentocyanide ($\mathrm{Na}{[\mathrm{Ag}(\mathrm{CN})}_2]$ ) is preferred.
(c) Pure metal from the electrolyte deposits on the cathode.

Question 39. Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them?
Answer.
It is easier to obtain metals from their oxides (by reduction) than from carbonate or sulphide ores. So, before reduction, the ore must be converted into metal oxide.
Calcination: method by which a carbonate ore is converted into oxide
Roasting: method by which a sulphide ore is converted into oxide
Calcination or roasting is done depending on the nature ofthe ore
Example for calcination:
$\mathrm{Ca}{\mathrm{CO}}_3$→$\mathrm{Ca}\mathrm{CO}$+${\mathrm{CO}}_2$
Example for roasting:
2ZnS+3${\mathrm{O}}_2$→2ZnO+${\mathrm{SO}}_2$

Question 40. Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), treated with HNO3, hydrogen is not liberated, why?
Answer.
Nitric acid (${\mathrm{HNO}}_3)$) is a strong oxidising agent. As soon as hydrogen gas is formed in the reaction between a metal and dilute nitric acid, the nitric acid oxidises this hydrogen to water and itself gets reduced to ${\mathrm{NO}}_2$ or NO or ${\mathrm{N}}_2\mathrm{O}$.
$\mathrm{Mg}+2{\mathrm{HNO}}_3\to \mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2$
$4\mathrm{Zn}+10{\mathrm{HNO}}_3\to 4\mathrm{Zn}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{N}}_2\mathrm{O}+5{\mathrm{H}}_2\mathrm{O}$
$4\mathrm{Sn}+10{\mathrm{HNO}}_3\to 4\mathrm{Sn}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{NH}}_4{\mathrm{NO}}_3+3{\mathrm{H}}_2\mathrm{O}$
$3\mathrm{Pb}+8{\mathrm{HNO}}_3\to 3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$
So, in the reaction of metals (except Mn and Mg), with dilute nitric acid, no hydrogen gas is evolved.

Question 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.
Answer.
(a) Compound X is iron oxide,${\mathrm{Fe}}_2{\mathrm{O}}_3$
(b) Thermite reaction
(c) The reaction that takes place is as follows

${\mathrm{Fe}}_2{\mathrm{O}}_3(s)+2\mathrm{Al}(s)\to 2\mathrm{Fe}(l)+{\mathrm{Al}}_2{\mathrm{O}}_3(s)+$ Heat

Question 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
Answer.
Sodium (Na) and potassium (K) react with cold water to form basic salt NaOH and KOH respectively.
Molecular formula = XOH
Molecular mass = 40
Let the atomic weight of metal X is x
Then, molecular mass of BOH = x + 16 + 1= 40 (Given)
x = 40 – 17 = 23
Thus, metal X is sodium (Na)
Sodium reacts with water as
$2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+{\mathrm{H}}_2$
Sodium (Na) liberates hydrogen gas (Z) in reaction with cold water.
So, Y is NaOH (sodium hydroxide)
Z Is H₂ (hydrogen gas).

Question 43. 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
Answer.
X exists in two different forms Y and Z
Y is the hardest natural substance, whereas Z is a good conductor of electricity
We know that diamond is the hardest natural substance, hence Y is diamond.
Now carbon (C) exists in two different forms called the allotropes of carbon. These allotropes are diamond and graphite. So X is carbon.
Z is graphite which is a good conductor of electricity.
X: Carbon
Y: Diamond
Z: Graphite

Question 44. The following reaction takes place when aluminium powder is heated with MnO₂
$3{\mathrm{MnO}}_2(s)+4\mathrm{Al}(s)\to 3\mathrm{Mn}(l)+2{\mathrm{Al}}_2{\mathrm{O}}_3(l)+$ Heat
(a) Is aluminium getting reduced?
(b) Is MnO₂ getting oxidised?
Ans.
Oxidation is the loss of electrons and reduction is the gain of electrons.
In the reaction:
$3{\mathrm{MnO}}_2(s)+4\mathrm{Al}(s)\to 3\mathrm{Mn}(l)+2{\mathrm{Al}}_2{\mathrm{O}}_3(l)+$ Heat
(a) Aluminium is getting oxidised because there is an addition of oxygen
(b) MnO₂ is getting reduced because there is removal of oxygen.

Question 45. What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?
Answer.
One is a soft solder, which is an alloy of tin and lead. The other is a hard solder, which is an alloy of copper and zinc.
The most commonly used solder is soft solder: an alloy of lead (Pb) and tin (Sn).
It contains 50% lead and 50% tin.
Solder is suitable for welding electrical wires because it has a low melting point. Hence, it is used for soldering (or welding) electrical wires together.

Question 46. 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
Answer.
Thermite Reaction:

$4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3(\mathrm{s})$

Metal A is aluminum (Al) which is used in the thermite process.
Al reacts with oxygen to form aluminium oxide Al­₂O₃ which is amphoteric in nature. So B is Al­₂O₃
${\mathrm{Al}}_2{\mathrm{O}}_3(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})\to 2{\mathrm{AlCl}}_3(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$(ii) ${\mathrm{Al}}_2{\mathrm{O}}_3(s)+2\mathrm{NaOH}(aq)\to 2{\mathrm{NaAlO}}_2(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Question 48. Give the formulae of the stable binary compounds that would be formed by the combination of following pairs of elements.
(a) Mg and N₂
(b) Li and O₂
(c) Al and Cl₂
(d) K and O₂
Answer.
(a)${\mathrm{Mg}}_3{\mathrm{N}}_2$2 Magnesium nitride
(b) ${\mathrm{Li}}_2\mathrm{O}$ Lithium oxide
(c) ${\mathrm{AlCl}}_3$ Aluminium chloride
(d) ${\mathrm{K}}_2\mathrm{O}$ Potassium oxide

Question 49. What happens when
(a) ZnCO₃ is heated in the absence of oxygen?
(b) a mixture of Cu₂O and Cu₂S is heated
Answer.
(a) Calcination takes place
${\mathrm{ZnCO}}_3(s)\to \mathrm{ZnO}(s)+{\mathrm{CO}}_2(g)$
When ZnCO₃ is heated in the absence of oxygen, Zinc oxide and carbon dioxide are produced.
(b) Roasting takes place
$2{\mathrm{Cu}}_2\mathrm{O}(s)+{\mathrm{Cu}}_2\mathrm{S}(s)\to 6\mathrm{Cu}(s)+{\mathrm{SO}}_2(g)$
When a mixture of Cu₂O and Cu₂S is heated, copper metal and sulphur dioxide gas are produced

Question 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?
Answer.
(a) Non-metal A is carbon.
Two oxides are carbon monoxide and carbon dioxide.
Oxide B is toxic whereas C causes global warming
So B is carbon monoxide (CO) and C is carbon dioxide (CO₂)
(b) Atomic weight of C = 6
The electronic configuration of C is 2, 4.
It is present in the 14th group (10 + valence electrons), i.e., IV A of the periodic table.

Question 51. Give two examples each of the metals that are good conductors and poor conductors of heat respectively.
Answer.
Good conductors: Silver, Bronze, Iron and copper. (All metals are good conductors of electricity)
Poor conductors: Lead and mercury.

Question 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)
Answer.
The only metal that exists as a liquid at room temperature is mercury (Hg).
Non-metal which exists in the liquid state at room temperature is bromine (Br).
Two metals having melting points less than 310 K (37°C) are caesium (Cs) with a melting point of 28.5°C and gallium (Ga) with a melting point of 30°C

Question 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

Answer.

Element A is calcium (Ca). When it reacts with water, it forms calcium hydroxide.
$\mathrm{Ca}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{H}}_2$
Thus, compound B is calcium hydroxide [Ca(OH)₂], which on heating gives CaO.
$\mathrm{Ca}(\mathrm{OH}{)}_2\to \mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}$
Thus, C is calcium oxide (CaO)
Hence the answer is:
A: Calcium
B: Ca(OH)₂
C: CaO

Question 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 reaction involved.
Answer:

Let, the atomic weight of alkali metal A is x. When it reacts with water it forms a compound B having molecular mass 40.
Let, the reaction is
$2\mathrm{A}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{AOH}+{\mathrm{H}}_2$
Molecular mass of B = x + 16 + 1 = 40
x = 40 - 17 = 23
Hence A is Sodium (Na)
The reaction involved is:
$2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+{\mathrm{H}}_2$
So, compound B is sodium hydroxide (NaOH)
Sodium hydroxide reacts with aluminium oxide (Al₂O₃) to give sodium aluminate (NaAIO₂)
Thus, C is sodium aluminate NaAIO₂
The reaction involved is:${\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{NaOH}\to 2{\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O}$

Question 55. Give the reaction involved during extraction of zinc from its ore by
(a) roasting of zinc ore
(b) calcination of zinc ore
Answer:

Roasting: method by which a sulphide ore is converted into oxide
Calcination: method by which a carbonate ore is converted into oxide
(a) Roasting of zinc ore: Sulphide ore of zinc is ZnS (zinc blende)
$2\mathrm{ZnS}(s)+3{\mathrm{O}}_2\to 2\mathrm{ZnO}(s)+2{\mathrm{SO}}_2$
ZnO is then reduced to zinc by heating with coke (carbon)
$ZnO(s)+C(s)\to Zn(s)+CO(g)$
(b) Calcination of zinc ore: Carbonate ore of Zinc is calamine ZnCO
${\mathrm{ZnCO}}_3(\mathrm{s})\to \mathrm{ZnO}(\mathrm{s})+{\mathrm{CO}}_2(\mathrm{g})$
ZnO is then reduced to zinc by heating with coke (carbon)

$\mathrm{ZnO}(s)+C(s)\to \mathrm{Zn}(s)+\mathrm{CO}(g)$

Question 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.
Answer:

Metal M is Cu as Cu is less reactive than hydrogen. So, it does not react with acids to release hydrogen.
Cu(s)+HCl→ No reaction
Copper metal does not burn in the air even on strong heating.
Copper reacts with the oxygen on prolonged heating to form a black substance, i.e., copper (Il) oxide.

$2\mathrm{Cu}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{CuO}(\mathrm{s})$

Question 57. An element forms an oxide A2O3 which is acidic in nature. Identify A as a metal or non-metal.
Answer:

Oxides of non-metals are acidic in nature
Oxides of metals are basic in nature
Given that the element A forms an acidic oxide, therefore, A is a non-metal.
Also, it forms an oxide A₂O₃ so it should have 3 valence electrons, i.e., +3 charge,
Therefore, an electronic configuration is 2, 3
Hence, A is boron and its oxide is B₂O₃.

Question 58. A solution of CuSO₄ 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.
Answer:

Iron displaces copper from copper sulphate solution as it is a more reactive metal than copper. The reaction is as follows:
$\mathrm{Fe}(s)+{\mathrm{CuSO}}_4\to {\mathrm{FeSO}}_4+\mathrm{Cu}$
A portion of the iron pot gets dissolved and causes holes in this process.
Iron taking part in this reaction, produces holes at places where iron metal has reacted to form iron (Il) sulphate.

Question 60. Give the steps involved in the extraction of metals of low and medium reactivity from their respective sulphide ores.
Answer:

Extraction of metals of low and medium reactivity:
1. Roasting: a respective sulphide ore is heated in air. So, metal oxide is formed.
2. Heating of metal oxide to obtain the metal
Mercury has low reactivity.
We can see the entire process for mercury (Hg) as follows:
(i) Roasting: Metal sulphide is converted into metal oxide by heating the ore strongly in the presence of excess of air.
$2\mathrm{HgS}+3{\mathrm{O}}_2\to 2\mathrm{HgO}+2{\mathrm{SO}}_2$ (Metal oxide is formed)
(ii) Reduction: Metal oxide is then reduced to metal by heating it.
$2\mathrm{HgO}\to 2\mathrm{Hg}+{\mathrm{O}}_2$
Hence mercury is extracted.
Similarly, we can see for zinc (medium reactivity).
Roasting: $2\mathrm{ZnS}+3{\mathrm{O}}_2\to 2\mathrm{ZnO}+2{\mathrm{SO}}_2$
Reduction: $\mathrm{ZnO}+\mathrm{C}\to \mathrm{Zn}+\mathrm{CO}$

Question 61. Explain the following
(a) Reactivity of Al decreases if it is dipped in HNO₃
(b) Carbon cannot reduce the oxides of Na or Mg
(c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as in molten state
(d) Iron articles are galvanised.
(e) Metals like Na, K, Ca and Mg are never found in their free state in nature.
Answer:

(a) Nitric acid is a strong oxidizing agent. When aluminium (Al) is dipped in HNO₃, an oxide layer of aluminium (Al₂O₃) is formed on the surface of the metal, which prevents it from further reaction. Thus, the reactivity of Al decreases.
(b) The affinity of Na, and Mg with oxygen is higher than that of carbon. This is because Na, and Mg metals are quite reactive and present towards the top of the reactivity series. Therefore, their oxides (Na₂O, MgO) are stable. A very high temperature is required to reduce them with carbon, and at that temperature, they will form their corresponding carbides. Hence, their oxides cannot be reduced by carbon
(c) NaCl is not a conductor of electricity in a solid state whereas it does conduct electricity in an aqueous solution as well as in a molten state. Ions of NaCl are free to move in an aqueous solution to carry the charge. But in a dry or solid state, ions of NaCl cannot move to carry the charge in as they are fixed. Hence, NaCl conducts electricity in the molten state.
(d) Galvanization is a method of protecting iron from rusting by coating it with a thin layer of zinc (Zn) metal.
A thin layer of zinc is formed over the iron articles by dipping them in molten zinc. As zinc is more reactive than iron, this prevents iron from corrosion
(e) Metals like Na, K, Ca and Mg are never found in their free state in nature because these metals are quite reactive so they cannot exist in the free state and hence, they are found in nature in the form of their compounds.

Question 62. (i) Given below are the steps for extraction of copper from its ore. Write the reaction involved.
(a) Roasting of copper (1) sulphide
(b) Reduction of copper (1) oxide with copper (1) sulphide.
(c) Electrolytic refining
(ii) Draw a neat and well labelled diagram for electrolytic refining of copper

Answer.
(i) $2{\mathrm{Cu}}_2\mathrm{S}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{Cu}}_2\mathrm{O}(\mathrm{s})+2{\mathrm{SO}}_2(\mathrm{g})$
(ii)$2{\mathrm{Cu}}_2\mathrm{O}(s)+{\mathrm{Cu}}_2\mathrm{S}(s)\to 6\mathrm{Cu}(s)+{\mathrm{SO}}_2(g)$
(iii) Electrolytic refining: Electrolytic refining is a process of refining a metal by the process of electrolysis.
A thick block of the impure metal is made anode (+ ve)
A thin strip of pure metal is made cathode (- ve)
CuSO4 solution is taken as an electrolyte.
At cathode:${\mathrm{Cu}}^{2+}(aq)+2e^{-}\to \mathrm{Cu}$
At anode $\mathrm{Cu}(\mathrm{s})\to {\mathrm{Cu}}^{2+}(\mathrm{aq})+2e^{-}$
(b)

Question 63. Of the three metals X, Y and Z. X reacts with cold water, Y with hot water and Z with steam only. Identify X, Y and Z and also arrange them in order of increasing reactivity.
Answer:

X is sodium or potassium (Na or K) as it reacts with cold water.
Na or K reacts violently with cold water.
$2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+{\mathrm{H}}_2$
$2\mathrm{K}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{KOH}+{\mathrm{H}}_2$
Y is magnesium (Mg) as it reacts with hot water.
$Ma+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ma}(\mathrm{OH}{)}_2+{\mathrm{H}}_2$
Z is iron (Fe)
$3\mathrm{Fe}+4{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2$
Order of reactivity
Z < Y< X
Or, Fe < Mg < Na (or) K

Question 64. An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

Answer:

An element "A" burns with a golden flame in the air. So, A is sodium (Na) because it burns with a golden flame in the air.
The atomic number of B is given as 17 so B is chlorine (Cl).
When Sodium reacts with Chlorine, sodium chloride (NaCl) is formed. Hence, C is sodium chloride (NaCl)
$2\mathrm{Na}(s)+{\mathrm{Cl}}_2(g)\to 2\mathrm{NaCl}(s)$
Now it is given that an aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen
Aqueous solution of NaCl, on electrolysis, gives sodium hydroxide.
Electrolysis Reaction:
The reaction at the cathode is:
${\mathrm{H}}_2\mathrm{O}(l)+2e^{-}\to {\mathrm{H}}_2(g)+2{\mathrm{OH}}^{-}$
The reaction at the anode is:
${\mathrm{Cl}}^{-}\to \frac{1}{2}{\mathrm{Cl}}_2(\mathrm{g})+1e^{-}$
The overall reaction is:
$2\mathrm{NaCl}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{NaOH}(\mathrm{aq})+{\mathrm{Cl}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$
Thus, D is sodium hydroxide (NaOH)

Question 65. Two ores A and B were taken. On heating ore A gives CO₂ whereas, ore B gives SO₂. What steps will you take to convert them into metals?
Answer:

We will use the following definitions:
1. Calcination: method by which a carbonate ore is converted into oxide
2 .Roasting: method by which a sulphide ore is converted into oxide
3 .Reduction (to convert metal oxide to metal
Given that ore A gives CO2 on heating, it must be a carbonate of moderately reactive metal like Zn.
Also given, B gives SO2, so B must be a sulphide ore, e.g., Cu2S, ZnS, HgS, etc.
A will be first subjected to calcination followed by reduction
${\mathrm{ZnCO}}_3\to \mathrm{ZnO}+{\mathrm{CO}}_2$(calcination)
To convert it into metal treat it with carbon
$\mathrm{ZnO}+\mathrm{C}\to \mathrm{Zn}+\mathrm{CO}$ (reduction)
B will be subjected to roasting followed by a reduction
$2\mathrm{ZnS}+3{\mathrm{O}}_2\to 2\mathrm{ZnO}+2{\mathrm{SO}}_2$ (roasting)
$2{\mathrm{Cu}}_2\mathrm{S}+3{\mathrm{O}}_2\to 2{\mathrm{Cu}}_2\mathrm{O}+2{\mathrm{SO}}_2$ (roasting)
$2\mathrm{HgS}+3{\mathrm{O}}_2\to 2\mathrm{HgO}+2{\mathrm{SO}}_2$ (roasting)
Metal is obtained from these oxides either by using a reducing agent like C or by auto-reduction
ZnO+C→Zn+CO (reduction)
${\mathrm{Cu}}_2\mathrm{S}+2{\mathrm{Cu}}_2\mathrm{O}\to 6\mathrm{Cu}+{\mathrm{SO}}_2$ (reduction)