SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Science Metals and Non-metals

65 questions · 65 still being checked

Long Answer Questions 59–65 (part 7 of 7)

  1. Exercise 59

    A non-metal A which is the largest constituent of air, when heated with H2\displaystyle \mathrm{H}_2 in 1\displaystyle 1:3\displaystyle 3 ratio in the presence of catalyst (Fe) gives a gas B. On heating with O2\displaystyle \mathrm{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.
    (a)
    Identify A, B, C and D
    (b)
    To which group of periodic table does this non-metal belong?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    A — \(\displaystyle \mathrm{N}_2\); B — \(\displaystyle \mathrm{NH}_3\); C — NO; D — \(\displaystyle \mathrm{HNO}_3\)
    (b)
    Element A belongs to Group -$\displaystyle 15$ of the Periodic Table
    \(\displaystyle \mathrm{N_2}\to\mathrm{NH_3}\to\mathrm{NO}\to\mathrm{HNO_3}\):
    \[\mathrm{N_2} + 3\mathrm{H_2} \xrightarrow[\mathrm{Fe}]{\Delta,\,P} 2\mathrm{NH_3} \]
    \[4\mathrm{NH_3} + 5\mathrm{O_2} \xrightarrow{\mathrm{Pt}} 4\mathrm{NO} + 6\mathrm{H_2O} \]
    \[4\mathrm{NO} + 3\mathrm{O_2} + 2\mathrm{H_2O} \rightarrow 4\mathrm{HNO_3} \]
    (b)
    N has $\displaystyle 5$ valence electrons \(\displaystyle \Rightarrow\) Group 15.
    Answer: \(\displaystyle A=\mathrm{N_2}\), \(\displaystyle B=\mathrm{NH_3}\), \(\displaystyle C=\mathrm{NO}\), \(\displaystyle D=\mathrm{HNO_3}\); N is in Group 15.

  2. Exercise 60

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

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Sulphide ore of low reactivity metal \(\displaystyle \xrightarrow{\text{Roasting}}\) Metal \(\displaystyle \xrightarrow{\text{Refining}}\) Pure metal Sulphide ore of medium reactivity metal \(\displaystyle \xrightarrow{\text{Roasting}}\) Oxide of metal \(\displaystyle \xrightarrow{\text{Reduction}}\) Metal \(\displaystyle \xrightarrow{\text{Refining}}\) Pure metal
    Concentration (froth flotation) removes gangue; roasting converts sulphide to oxide; reduction gives the metal; electrolytic refining purifies it.Medium reactivity, e.g. Zn (reduced by carbon): \[2\mathrm{ZnS} + 3\mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{ZnO} + 2\mathrm{SO_2} \] \[\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO} \]Low reactivity, e.g. Cu (roasting alone self-reduces the oxide, no added reducing agent): \[2\mathrm{Cu_2S} + 3\mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{Cu_2O} + 2\mathrm{SO_2} \] \[2\mathrm{Cu_2O} + \mathrm{Cu_2S} \rightarrow 6\mathrm{Cu} + \mathrm{SO_2} \]Answer: Concentrate \(\displaystyle \to\) roast (sulphide \(\displaystyle \to\) oxide) \(\displaystyle \to\) reduce (carbon for medium reactivity; self-reduction for low reactivity) \(\displaystyle \to\) electrolytically refine.

  3. Exercise 61

    Explain the following
    (a)
    Reactivity of Al decreases if it is dipped in HNO3\displaystyle \mathrm{HNO}_3
    (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\displaystyle \mathrm{Na}, \mathrm{K}, \mathrm{Ca} and Mg are never found in their free state in nature.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Hint— (a) Due to the formation of a layer of oxide i.e., \(\displaystyle \mathrm{Al}_2 \mathrm{O}_3\)
    (b)
    Na or Mg are more reactive metals as compared to carbon
    (c)
    In solid NaCl, the movement of ions is not possible due to its rigid structure but in aqueous solution or molten state, the ions can move freely.
    (d)
    To protect from corrosion
    (e)
    They are highly reactive
    (a)
    Conc. \(\displaystyle \mathrm{HNO_3}\) forms a thin \(\displaystyle \mathrm{Al_2O_3}\) layer on Al \(\displaystyle \Rightarrow\) passivation stops further reaction.
    (b)
    Na, Mg are more reactive than C \(\displaystyle \Rightarrow\) C cannot pull oxygen from \(\displaystyle \mathrm{Na_2O}\) or \(\displaystyle \mathrm{MgO}\).
    (c)
    Solid NaCl's ions are fixed in the lattice; melting or dissolving frees them, so only the molten/aqueous forms conduct.
    (d)
    Zn (more reactive than Fe) corrodes first, sacrificially protecting the iron even if the coating is scratched.
    (e)
    Na, K, Ca, Mg react readily with air/moisture \(\displaystyle \Rightarrow\) they occur only as compounds, never free.
    Answer: (a) passivation (b) Na/Mg more reactive than C (c) ions mobile only in melt/solution (d) sacrificial Zn coating (e) high reactivity \(\displaystyle \Rightarrow\) always combined.

  4. Exercise 62

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

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)(a)
    Roasting of sulphide ore
    (i)
    \(\displaystyle 2 \mathrm{Cu}_2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_2(\mathrm{~s}) \xrightarrow{\text { Heat }} 2 \mathrm{Cu}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{SO}_2(\mathrm{~g})\)
    (b)
    \(\displaystyle 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \xrightarrow{\text { Heat }} 6 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_2(\mathrm{~g})\)
    This reaction is known as auto-reduction
    (c)
    Reaction for electrolytic refining
    At cathode: \(\displaystyle \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\)
    At anode: \(\displaystyle \mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\)
    (ii)
    Diagram for electroytic refining of copper
    NCERT_Solution_Class10_Science_Exemplar_Ch3_Q62_ncert
    (i)(a)
    Roasting:
    \[2\mathrm{Cu_2S} + 3\mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{Cu_2O} + 2\mathrm{SO_2} \]
    (b)
    Self-reduction (the sulphide reduces the oxide, no added reducing agent):
    \[2\mathrm{Cu_2O} + \mathrm{Cu_2S} \rightarrow 6\mathrm{Cu} + \mathrm{SO_2} \]
    (c)
    Electrolytic refining: impure Cu = anode, pure Cu = cathode, acidified \(\displaystyle \mathrm{CuSO_4}\) = electrolyte.
    \[\mathrm{Cu(anode)} \rightarrow \mathrm{Cu^{2+}} + 2e^- \]
    \[\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu(cathode)} \]
    Join impure Cu to the \(\displaystyle +\) terminal and pure Cu to the \(\displaystyle -\) terminal; if reversed, the pure copper dissolves instead.
    (ii)
    Answer: roast \(\displaystyle \to\) self-reduce \(\displaystyle \to\) electro-refine; cell shown above.

  5. Exercise 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.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    X, Y, Z sit at three reactivity tiers, read off by which form of water attacks each: \[2\mathrm{Na(X)} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2}\uparrow \quad (\text{cold water}) \] \[\mathrm{Mg(Y)} + 2\mathrm{H_2O} \xrightarrow{\text{hot}} \mathrm{Mg(OH)_2} + \mathrm{H_2}\uparrow \quad (\text{hot water}) \] \[3\mathrm{Fe(Z)} + 4\mathrm{H_2O} \xrightarrow{\text{steam}} \mathrm{Fe_3O_4} + 4\mathrm{H_2}\uparrow \quad (\text{steam}) \]A metal that needs steam to react is the least reactive of the three, and one that reacts with cold water is the most.Answer: \(\displaystyle X=\mathrm{Na}\), \(\displaystyle Y=\mathrm{Mg}\), \(\displaystyle Z=\mathrm{Fe}\); increasing reactivity \(\displaystyle Z<Y<X\), i.e. \(\displaystyle \mathrm{Fe}<\mathrm{Mg}<\mathrm{Na}\).

  6. Exercise 64

    An element A burns with golden flame in air. It reacts with another element B, atomic number 17\displaystyle 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.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    A = Na; B = \(\displaystyle \mathrm{Cl}_2\); C = NaCl; D = NaOH \(\displaystyle 2 \mathrm{Na}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{NaCl}\) \(\displaystyle 2 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{g})+\mathrm{H}_2(\mathrm{g})\)
    A burns with a golden flame \(\displaystyle \Rightarrow\) sodium; B, atomic number $\displaystyle 17$, is chlorine. \[2\mathrm{Na} + \mathrm{Cl_2} \rightarrow 2\mathrm{NaCl} \] Electrolysing the aqueous solution of C (brine) liberates \(\displaystyle \mathrm{H_2}\) and gives D: \[2\mathrm{NaCl} + 2\mathrm{H_2O} \xrightarrow{\text{electrolysis}} 2\mathrm{NaOH} + \mathrm{H_2}\uparrow + \mathrm{Cl_2}\uparrow \]Answer: \(\displaystyle A=\mathrm{Na}\), \(\displaystyle B=\mathrm{Cl_2}\), \(\displaystyle C=\mathrm{NaCl}\), \(\displaystyle D=\mathrm{NaOH}\).

  7. Exercise 65

    Two ores A and B were taken. On heating ore A gives CO2\displaystyle \mathrm{CO}_2 whereas, ore B gives SO2\displaystyle \mathrm{SO}_2. What steps will you take to convert them into metals?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Since ore A gives \(\displaystyle \mathrm{CO}_2\) and ore B gives \(\displaystyle \mathrm{SO}_2\). Therefore, ores are \(\displaystyle \mathrm{MCO}_3\) and MS. A can be obtained \(\displaystyle \mathrm{MCO}_3 \xrightarrow{\text{Calcination}} \mathrm{MO}+\mathrm{CO}_2\) \(\displaystyle \mathrm{MO}+\mathrm{C} \xrightarrow{\text{Reduction}} \mathrm{M}+\mathrm{CO}\) B can be obtained \(\displaystyle 2 \mathrm{MS}+3 \mathrm{O}_2 \xrightarrow{\text{Roasting}} 2 \mathrm{MO}+2 \mathrm{SO}_2\) \(\displaystyle \mathrm{MO}+\mathrm{C} \rightarrow \mathrm{M}+\mathrm{CO}\)
    A releasing \(\displaystyle \mathrm{CO_2}\) on heating is a carbonate ore; B releasing \(\displaystyle \mathrm{SO_2}\) is a sulphide ore.A: calcination, strong heating in limited air (thermal decomposition, no \(\displaystyle \mathrm{O_2}\) needed): \[\mathrm{MCO_3} \xrightarrow{\Delta} \mathrm{MO} + \mathrm{CO_2} \] B: roasting, strong heating in excess air (\(\displaystyle \mathrm{O_2}\) is a reactant): \[2\mathrm{MS} + 3\mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{MO} + 2\mathrm{SO_2} \] Both oxides are then reduced with carbon: \[\mathrm{MO} + \mathrm{C} \rightarrow \mathrm{M} + \mathrm{CO} \]Answer: Calcine A, roast B, then reduce both oxides with carbon to the metal.