Chapter 3 · Metals and Non-metals

NCERT Exemplar Class 10 Science Chapter 3 Long Answer Questions

All 7 questions of NCERT Exemplar Class 10 Science Chapter 3 Long Answer Questions (Metals and Non-metals), in the book's order, each with its answer and a step-by-step solution.

1. Q59
   A non-metal A which is ​the largest constituent ​of air, when heated with ${\text{H}}_2$ in $1:3$ ​ratio in the presence of catalyst (Fe) gives a gas B. On heating with ${\text{O}}_2$ it gives an ​oxide C. ​If this oxide is passed into water in the presence of ​air it gives an acid D which acts as a ​strong oxidising agent.

   1. (a)Identify A, B, C and D
   2. (b)​To which group of periodic table does this non-metal belong?
   Show answer

   A is nitrogen, B ammonia, C nitric oxide and D nitric acid; nitrogen belongs to Group 15.

   (a) A is nitrogen (${\text{N}}_2$), B is ammonia (${\text{NH}}_3$), C is nitric oxide ($\text{NO}$) and D is nitric acid (${\text{HNO}}_3$). The reactions are ${\text{N}}_2+3{\text{H}}_2⟶2{\text{NH}}_3$ over an iron catalyst, ${\text{N}}_2+{\text{O}}_2⟶2\text{NO}$ at a very high temperature (or $4{\text{NH}}_3+5{\text{O}}_2⟶4\text{NO}+6{\text{H}}_2\text{O}$ from B), then $2\text{NO}+{\text{O}}_2⟶2{\text{NO}}_2$ and $4{\text{NO}}_2+{\text{O}}_2+2{\text{H}}_2\text{O}⟶4{\text{HNO}}_3$, overall $4\text{NO}+3{\text{O}}_2+2{\text{H}}_2\text{O}⟶4{\text{HNO}}_3$. Naming C as nitrogen dioxide (${\text{NO}}_2$), the oxide that actually dissolves, with the same chain is also accepted. (b) Nitrogen belongs to Group 15 of the periodic table (the nitrogen family, also written as Group VA).

   Step-by-step solution →
2. Q60
   Give ​the steps involved in ​the extraction of ​metals ​of low and medium reactivity from ​their ​respective ​sulphide ores.
   Show answer

   Concentrate the ore, roast it to the oxide, reduce the oxide (heat alone for low, carbon for medium reactivity), then refine.

   For both classes the first steps are the same: the ore is crushed and concentrated, usually by froth flotation, and then roasted in excess air so that the sulphide becomes the oxide, for example $2\text{ZnS}+3{\text{O}}_2⟶2\text{ZnO}+2{\text{SO}}_2$. The routes then part at the reduction. For a metal of **low** reactivity such as mercury or copper, heat alone frees the metal: $2\text{HgO}⟶2\text{Hg}+{\text{O}}_2$, and for copper the unroasted sulphide reduces the oxide itself, $2{\text{Cu}}_2\text{O}+{\text{Cu}}_2\text{S}⟶6\text{Cu}+{\text{SO}}_2$. For a metal of **medium** reactivity such as zinc, iron or lead, the oxide is reduced with carbon: $\text{ZnO}+\text{C}⟶\text{Zn}+\text{CO}$. Finally the crude metal from either route is purified, usually by electrolytic refining.

   Step-by-step solution →
3. Q61
   Explain the ​following

   1. (a)Reactivity of Al decreases if it is dipped ​in ${\text{HNO}}_3$
   2. (b)Carbon cannot reduce the oxides of Na or Mg
   3. (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
   4. (d)Iron articles are ​galvanised.
   5. (e)Metals like ​Na, K, Ca ​and Mg are never found in their free state in ​nature.
   Show answer

   Aluminium forms a protective oxide film, carbon cannot reduce reactive metals, free ions carry current, zinc protects iron, and reactive metals occur combined.

   (a) Nitric acid oxidises the surface of the aluminium and forms a thin, hard, impervious film of aluminium oxide on it. The film seals the metal off, so no reagent can reach it and the metal is rendered passive. (b) Sodium and magnesium lie above carbon in the reactivity series and hold oxygen more strongly than carbon does, so carbon cannot take it from their oxides; these metals are obtained instead by electrolytic reduction of their molten chlorides or oxides. (c) Sodium chloride is ionic in every state, but in the solid the ions are locked in fixed positions in the crystal lattice and cannot move, so no current flows. On melting or dissolving, the electrostatic forces are overcome, the ions become free to move and the substance conducts. (d) Galvanising coats iron with a thin layer of zinc. The zinc excludes air and moisture, and being more reactive than iron it corrodes in preference to it, so the iron is protected even where the coating is scratched. (e) Sodium, potassium, calcium and magnesium are very reactive and combine at once with the oxygen, moisture and carbon dioxide of the air, so in nature they occur only in the combined state as their compounds.

   Step-by-step solution →
4. Q62
   (i) Given below are the steps for extraction of copper from ​its ore. Write ​the reaction involved.

   1. (a)Roasting of copper (I) sulphide
   2. (b)Reduction ​of copper (I) oxide with copper ​(I) sulphide.
   3. (c)Electrolytic refining (ii) Draw a neat and ​well labelled diagram for electrolytic refining of copper
   Show answer

   Roasting forms ${\text{Cu}}_2\text{O}$, self-reduction gives copper, and electrolytic refining deposits copper at the cathode; the refining cell is drawn and labelled.

   (i) (a) Roasting: $2{\text{Cu}}_2\text{S}(s)+3{\text{O}}_2(g)⟶2{\text{Cu}}_2\text{O}(s)+2{\text{SO}}_2(g)$ on strong heating in air. (b) Reduction of copper(I) oxide by copper(I) sulphide, a self-reduction: $2{\text{Cu}}_2\text{O}(s)+{\text{Cu}}_2\text{S}(s)⟶6\text{Cu}(s)+{\text{SO}}_2(g)$ on heating. (c) Electrolytic refining, in acidified copper sulphate solution: at the anode $\text{Cu}(s)⟶{\text{Cu}}^{2+}(aq)+2e^{-}$, and at the cathode ${\text{Cu}}^{2+}(aq)+2e^{-}⟶\text{Cu}(s)$. (ii) The labelled diagram is given above: a vessel of acidified copper sulphate solution, the thick block of impure copper as the anode joined to the positive terminal, the thin strip of pure copper as the cathode joined to the negative terminal, copper ions moving from anode to cathode, and the anode mud collecting below the anode.

   Step-by-step solution →
5. Q63
   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.
   Show answer

   X is sodium, Y is magnesium and Z is iron; reactivity increases in the order iron, magnesium, sodium.

   X is sodium ($\text{Na}$), Y is magnesium ($\text{Mg}$) and Z is iron ($\text{Fe}$). The order of increasing reactivity is $\text{Z}<\text{Y}<\text{X}$, that is iron $<$ magnesium $<$ sodium. The reactions are $2\text{Na}+2{\text{H}}_2\text{O}⟶2\text{NaOH}+{\text{H}}_2$ with cold water, $\text{Mg}+2{\text{H}}_2\text{O}⟶{\text{Mg(OH)}}_2+{\text{H}}_2$ with hot water, and $3\text{Fe}+4{\text{H}}_2\text{O}⟶{\text{Fe}}_3{\text{O}}_4+4{\text{H}}_2$ with steam. Other choices fit the same pattern and are equally acceptable: X could be potassium or calcium, and Z could be zinc or aluminium; the order of increasing reactivity is unchanged.

   Step-by-step solution →
6. Q64
   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.
   Show answer

   A is sodium, B chlorine, C sodium chloride and D sodium hydroxide, formed by electrolysis of brine.

   A is sodium ($\text{Na}$), B is chlorine ($\text{Cl}$, atomic number $17$), C is sodium chloride ($\text{NaCl}$) and D is sodium hydroxide ($\text{NaOH}$). The reactions are $4\text{Na}+{\text{O}}_2⟶2{\text{Na}}_2\text{O}$ when sodium burns in air with its golden flame, $2\text{Na}+{\text{Cl}}_2⟶2\text{NaCl}$, and $2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)⟶2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$ on electrolysis, which is the chlor-alkali process.

   Step-by-step solution →
7. Q65
   Two ores ​A and B were taken. On heating ore ​A gives ${\text{CO}}_2$ whereas, ​ore B gives ${\text{SO}}_2$. ​What steps will you take to convert ​them ​into ​metals?
   Show answer

   Crush and concentrate both ores; calcine the carbonate ore A and roast the sulphide ore B to oxides, reduce with carbon, then refine.

   Ore A gives carbon dioxide, so it is a carbonate ore; ore B gives sulphur dioxide, so it is a sulphide ore. Both are first crushed and concentrated. Ore A is then **calcined**, heated strongly in the absence or limited supply of air, ${\text{MCO}}_3⟶\text{MO}+{\text{CO}}_2$; ore B is **roasted**, heated strongly in excess air, $2\text{MS}+3{\text{O}}_2⟶2\text{MO}+2{\text{SO}}_2$. Both now give the metal oxide, which is reduced with carbon, $\text{MO}+\text{C}⟶\text{M}+\text{CO}$. Finally each crude metal is purified, usually by electrolytic refining. If the metal were a highly reactive one, the oxide could not be reduced with carbon and the metal would be obtained by electrolysis of its molten salt instead.

   Step-by-step solution →