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NCERT Exemplar · Class 10 Science Metals and Non-metals

65 questions · 65 still being checked

Short Answer Questions 47–58 (part 6 of 7)

  1. Exercise 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 and give the reaction involved.

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    NCERT’s answer
    Metals low in activity series can be obtained by reducing their sulphides or oxides by heating. Mercury is the only metal that exists as liquid at room temperature. It can be obtained by heating cinnabar \(\displaystyle (\mathrm{HgS})\), the sulphide ore of mercury. The reactions are as follows: \[\begin{aligned} & 2 \mathrm{HgS}+3 \mathrm{O}_2 \xrightarrow{\text { Heat }} 2 \mathrm{HgO}+2 \mathrm{SO}_2 \\ & 2 \mathrm{HgO} \xrightarrow{\text { Heat }} 2 \mathrm{Hg}+\mathrm{O}_2 \end{aligned} \]
    Mercury is the one metal liquid at room temperature; its ore is cinnabar, \(\displaystyle \mathrm{HgS}\). Roasting in air first oxidises the sulphide; further heating decomposes the unstable oxide, releasing free mercury — a self-reduction, with no added reducing agent. \[2\mathrm{HgS(s)} + 3\mathrm{O_2(g)} \xrightarrow{\Delta} 2\mathrm{HgO(s)} + 2\mathrm{SO_2(g)} \] \[2\mathrm{HgO(s)} \xrightarrow{\Delta} 2\mathrm{Hg(l)} + \mathrm{O_2(g)} \]Answer: Metal: mercury (Hg); ore: cinnabar (HgS); roasting gives HgO, which decomposes on further heating to give Hg.

  2. Exercise 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\displaystyle \mathrm{N}_2
    (b)
    Li and O2\displaystyle \mathrm{O}_2
    (c)
    Al and Cl2\displaystyle \mathrm{Cl}_2
    (d)
    K and O2\displaystyle \mathrm{O}_2

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    NCERT’s answer
    (a)
    \(\displaystyle \mathrm{Mg}_3 \mathrm{~N}_2\)
    (b)
    \(\displaystyle \mathrm{Li}_2 \mathrm{O}\)
    (c)
    \(\displaystyle \mathrm{AlCl}_3\)
    (d)
    \(\displaystyle \mathrm{K}_2 \mathrm{O}\)
    Each metal carries the charge fixed by its group; the formula balances that charge against the non-metal's usual valency. \[\text{(a)}\quad 3\mathrm{Mg(s)} + \mathrm{N_2(g)} \rightarrow \mathrm{Mg_3N_2(s)} \] \[\text{(b)}\quad 4\mathrm{Li(s)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{Li_2O(s)} \] \[\text{(c)}\quad 2\mathrm{Al(s)} + 3\mathrm{Cl_2(g)} \rightarrow 2\mathrm{AlCl_3(s)} \] \[\text{(d)}\quad 4\mathrm{K(s)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{K_2O(s)} \]Answer: (a) \(\displaystyle \mathrm{Mg_3N_2}\) (b) \(\displaystyle \mathrm{Li_2O}\) (c) \(\displaystyle \mathrm{AlCl_3}\) (d) \(\displaystyle \mathrm{K_2O}\).

  3. Exercise 49

    What happens when
    (a)
    ZnCO3\displaystyle \mathrm{ZnCO}_3 is heated in the absence of oxygen?
    (b)
    a mixture of Cu2O\displaystyle \mathrm{Cu}_2 \mathrm{O} and Cu2S\displaystyle \mathrm{Cu}_2 \mathrm{S} is heated?

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    NCERT’s answer
    (a)
    It undergoes calcination. The chemical reaction can be given as \(\displaystyle \mathrm{ZnCO}_3 \xrightarrow{\text { Heat }} \mathrm{ZnO}+\mathrm{CO}_2\)
    (b)
    It undergoes auto reduction forming copper and sulphur dioxide
    \[2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \xrightarrow{\text { Heat }} 6 \mathrm{Cu}+\mathrm{SO}_2 \]
    (a)
    Heated with no oxygen present, the carbonate simply decomposes (calcination), driving off carbon dioxide and leaving the oxide — yellow while hot, white on cooling.
    \[\mathrm{ZnCO_3(s)} \xrightarrow{\Delta} \mathrm{ZnO(s)} + \mathrm{CO_2(g)} \]
    (b)
    The two copper compounds reduce each other on heating — no external reducing agent — freeing metallic copper; this self-reduction is the step actually used industrially.
    \[2\mathrm{Cu_2O(s)} + \mathrm{Cu_2S(s)} \xrightarrow{\Delta} 6\mathrm{Cu(s)} + \mathrm{SO_2(g)} \]
    Answer: (a) \(\displaystyle \mathrm{ZnO}\) + \(\displaystyle \mathrm{CO_2}\) form. (b) Metallic Cu + \(\displaystyle \mathrm{SO_2}\) form by self-reduction.

  4. Exercise 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?

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    NCERT’s answer
    (a)
    A is carbon, B is carbon monoxide and C is carbon dioxide
    (b)
    A belongs to Group - $\displaystyle 14$ of the Periodic Table
    Carbon is present in every carbohydrate, fat and protein we eat, so A is carbon. Limited oxygen gives the toxic monoxide; excess oxygen gives the dioxide that traps heat in the atmosphere. \[2\mathrm{C(s)} + \mathrm{O_2(g)}\ (\text{limited}) \rightarrow 2\mathrm{CO(g)} \] \[\mathrm{C(s)} + \mathrm{O_2(g)}\ (\text{excess}) \rightarrow \mathrm{CO_2(g)} \]Answer: A = carbon; B = \(\displaystyle \mathrm{CO}\) (toxic); C = \(\displaystyle \mathrm{CO_2}\) (global warming); A belongs to Group 14.

  5. Exercise 51

    Give two examples each of the metals that are good conductors and poor conductors of heat respectively.

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    NCERT’s answer
    (a)
    Good conductor : Ag and Cu
    (b)
    Poor conductor : Pb and Hg
    Thermal conductivity, \(\displaystyle k\), in decreasing order: \[k:\ \mathrm{Ag}\,(429) > \mathrm{Cu}\,(401) \gg \mathrm{Pb}\,(35) > \mathrm{Hg}\,(8)\ \ \mathrm{W\,m^{-1}\,K^{-1}} \]Answer: Good conductors of heat: \(\displaystyle \mathrm{Ag}\), \(\displaystyle \mathrm{Cu}\). Poor conductors of heat: \(\displaystyle \mathrm{Pb}\), \(\displaystyle \mathrm{Hg}\).

  6. Exercise 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\displaystyle 310 K (37\displaystyle 37°C)

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    NCERT’s answer
    Metal - Mercury (Hg); Non-metal - Bromine (Br) Two metals with melting points less than 310K are Cesium (Cs) and Gallium (Ga)
    Mercury is the metal and bromine the non-metal that stay liquid at ordinary room temperature. Gallium and caesium melt in the warmth of a hand — both below body temperature. \[T_{mp}(\mathrm{Ga}) \approx 303\ \mathrm{K}, \quad T_{mp}(\mathrm{Cs}) \approx 302\ \mathrm{K} \quad (\text{both} < 310\ \mathrm{K}) \]Answer: Metal: mercury (\(\displaystyle \mathrm{Hg}\)); non-metal: bromine (\(\displaystyle \mathrm{Br_2}\)). Metals with m.p. below $\displaystyle 310$ K: gallium and caesium.

  7. Exercise 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.

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    NCERT’s answer
    A — Ca; B— \(\displaystyle \mathrm{Ca}(\mathrm{OH})_2\); C — CaO \(\displaystyle \mathrm{Ca}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{aq})+\mathrm{H}_2(\mathrm{g})\) \(\displaystyle \mathrm{Ca}(\mathrm{OH})_2 \xrightarrow{\text{Heat}} \mathrm{CaO}+\mathrm{H}_2 \mathrm{O}\)
    \(\displaystyle A\) is a metal whose hydroxide is used for white-washing and whose oxide, on hydration, regenerates that same hydroxide -- this fixes \(\displaystyle A\) as calcium. \[\mathrm{Ca(s)} + 2\mathrm{H_2O(l)} \rightarrow \mathrm{Ca(OH)_2(aq)} + \mathrm{H_2(g)} \] \[\mathrm{Ca(OH)_2(s)} \xrightarrow{\Delta} \mathrm{CaO(s)} + \mathrm{H_2O(g)} \] \[\mathrm{CaO(s)} + \mathrm{H_2O(l)} \rightarrow \mathrm{Ca(OH)_2(aq)} \]

  8. Exercise 54

    An alkali metal A gives a compound B (molecular mass = 40\displaystyle 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.

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    NCERT’s answer
    A — Na; B — NaOH; C — \(\displaystyle \mathrm{NaAlO}_2\) \(\displaystyle 2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2\) \(\displaystyle \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{NaOH} \rightarrow 2 \mathrm{NaAlO}_2+\mathrm{H}_2 \mathrm{O}\)
    Molecular mass \(\displaystyle 40\) fixes \(\displaystyle B\) as \(\displaystyle \mathrm{NaOH}\) (\(\displaystyle 23+16+1\)), so \(\displaystyle A\) is sodium; \(\displaystyle \mathrm{Al_2O_3}\) is amphoteric and dissolves in the base to a soluble aluminate \(\displaystyle C\). \[2\mathrm{Na(s)} + 2\mathrm{H_2O(l)} \rightarrow 2\mathrm{NaOH(aq)} + \mathrm{H_2(g)} \] \[\mathrm{Al_2O_3(s)} + 2\mathrm{NaOH(aq)} \rightarrow 2\mathrm{NaAlO_2(aq)} + \mathrm{H_2O(l)} \]

  9. Exercise 55

    Give the reaction involved during extraction of zinc from its ore by
    (a)
    roasting of zinc ore
    (b)
    calcination of zinc ore

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    NCERT’s answer
    (a)
    \(\displaystyle 2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_2 \xrightarrow{\text { Heat }} 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_2(\mathrm{~g})\)
    (b)
    \(\displaystyle \mathrm{ZnCO}_3(\mathrm{~s}) \xrightarrow{\text { Heat }} \mathrm{ZnO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)
    Roasting heats a sulphide ore in excess air; calcination heats a carbonate ore with air excluded.
    (a)
    \[2\mathrm{ZnS(s)} + 3\mathrm{O_2(g)} \xrightarrow{\Delta} 2\mathrm{ZnO(s)} + 2\mathrm{SO_2(g)} \]
    (b)
    \[\mathrm{ZnCO_3(s)} \xrightarrow{\Delta} \mathrm{ZnO(s)} + \mathrm{CO_2(g)} \]

  10. Exercise 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.

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    NCERT’s answer
    \(\displaystyle \mathrm{M}=\mathrm{Cu}\); Black product— CuO \[2 \mathrm{Cu}+\mathrm{O}_2 \rightarrow 2 \mathrm{CuO} \]
    \(\displaystyle M\) does not displace \(\displaystyle \mathrm{H_2}\) from acids, so it lies below hydrogen in the reactivity series. Heated in air, it is oxidised to a black oxide. \[2\mathrm{Cu(s)} + \mathrm{O_2(g)} \xrightarrow{\Delta} 2\mathrm{CuO(s)} \] \[\text{reddish-brown Cu} \rightarrow \text{black CuO} \] Answer: \(\displaystyle M = \mathrm{Cu}\); the black product is \(\displaystyle \mathrm{CuO}\), formed when copper is heated in air.

  11. Exercise 57

    An element forms an oxide A2O3\displaystyle \mathrm{A}_2 \mathrm{O}_3 which is acidic in nature. Identify A as a metal or non-metal.

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    NCERT’s answer
    Since an oxide of element is acidic in nature, therefore, A will be a non-metal.
    Metal oxides are basic or amphoteric; only non-metal oxides are acidic. Since \(\displaystyle \mathrm{A_2O_3}\) is acidic, \(\displaystyle A\) is a non-metal.

  12. Exercise 58

    A solution of CuSO4\displaystyle \mathrm{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.

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    NCERT’s answer
    Fe is more reactive as compared to Cu. Therefore, Fe displaces Cu from \(\displaystyle \mathrm{CuSO}_4\) and forms \(\displaystyle \mathrm{FeSO}_4\). \[\mathrm{Fe}+\mathrm{CuSO}_4 \rightarrow \mathrm{FeSO}_4+\mathrm{Cu} \]
    Iron is more reactive than copper, so it displaces \(\displaystyle \mathrm{Cu}\) from \(\displaystyle \mathrm{CuSO_4}\) solution; the pot's own iron passes into solution as \(\displaystyle \mathrm{Fe^{2+}}\), pitting the metal, while copper deposits out. \[\mathrm{Fe(s)} + \mathrm{CuSO_4(aq)} \rightarrow \mathrm{FeSO_4(aq)} + \mathrm{Cu(s)} \]