CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 5 Mark Questions 2021
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CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 5 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Chemistry
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An amorphous solid 'A' burns in air to form a gas 'B' which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+ Identify the solid "A" and the gas "B" and write the reactions involved.
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On heating, lead (II) nitrate gives a brown gas 'A'. The gas 'A' on cooling changes to colourless solid 'B'. Solid 'B' on heating with NO changes to a blue solid 'C'. Identiy 'A', 'B' and 'C' and also write reactions involved and draw the structures of 'B' and 'C'.
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On heating compound(A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
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On heating compound (A)gives a gas (B)which is a constituent of air.This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives gives a compound (D) which is a part of acid rain. Identify compounds (A) to D and also give necessary equations of all the steps involved.
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An amorphous solid A burns in air to form a gas B which turns lime water milky. The gas is also produced as a byproduct during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe 3+ to Fe2+. Identify the solid 'A' and the gas 'B ' and write the reactions involved.
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CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 5 Mark Questions 2021 Answer Keys
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\('A'\quad is\quad { S }_{ 8 }.\quad 'B'is{ SO }_{ 2 }(g).\)
\({ S }_{ 8 }+{ 8O }_{ 2 }\overset { heat }{ \rightarrow } { 8SO }_{ 2 }(g)\)
'B' decolorizes
'B' turns lime water milky due to formation of
'B' is obtained by roasting of sulphideores
'B' reduces in aqueous solution. -
\(2Pb{ { { { (NO }_{ 3 }) } } }_{ 2 }\xrightarrow { heat } 2PbO(s)+{ NO }_{ 2 }+{ O }_{ 2 }\)
Brown (A)
\({ 2NO }_{ 2 }(g)\overset { cooling }{ \rightleftharpoons } { N }_{ 2 }{ O }_{ 4 }(s)\)
(B) Colourless
\({ N }_{ 2 }{ O }_{ 4 }+2NO\overset { heat }{ \underset { 250k }{ \rightleftharpoons } } { 2N }_{ 2 }{ O }_{ 3 }(s)\)
(C) Blue solid
Resonating Structures of \({ N }_{ 2 }{ O }_{ 4 }\) -
(i) \({ NH }_{ 4 }{ NO }_{ 2 }(s)\underrightarrow { heat } { N }_{ 2 }+{ 2H }_{ 2 }O\)
'A' 'B'
(ii) \({ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow 2{ NH }_{ 3 }(g)\)
'C' basic
(iii) \({ 4NH }_{ 3 }+{ 5O }_{ 2 }\longrightarrow 4NO+{ 6H }_{ 2 }O\)
\(\\ 2NO+{ O }_{ 2 }\longrightarrow { 2NO }_{ 2 }\)
'D' (part of acid rain)
\({ 3NO }_{ 2 }+{ H }_{ 2 }O\longrightarrow 2{ HNO }_{ 3 }+NO\) -
\((A)=NH_{ 4 }NO_{ 2 },\ (B)=N_{ 2 },\ (C)=NH_{ 3 },\ (D)=HNO_{ 3 }\)
\(\\ NH_{ 4 }NO_{ 2 }\overset { Heat }{ \longrightarrow } \underset { (B) }{ N_{ 2 } } +2H_{ 2 }O\)
\(\\ N_{ 2 }+3H_{ 2 }\overset { Catalyst }{ \longrightarrow } \underset { (C) }{ 2NH_{ 3 } } \)
\(\\ 4NH_{ 3 }+5O_{ 2 }\longrightarrow 4NO+6H_{ 2 }O\)
\(\\ 2NO+O_{ 2 }\longrightarrow 2NO_{ 2 }\)
\(\\ 3NO_{ 2 }+H_{ 2 }O\longrightarrow 2HNO_{ 3 }+NO\) -
(i) Since, the byproduct of roasting of sulphide ore is SO2' It also turns lime water milky. Therefore, gas 'B ' must be SO2.
(ii) As the gas B is obtained when amorphous solid 'A' burns in air therefore, amorphous solid 'A' must be sulphur, S8
\(\begin{aligned} &\mathrm{S}_{8}+8 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 8 \mathrm{SO}_{2}\\ &\text { (A) }\\ &2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2} \end{aligned}\)
(iii) Gas B reduces acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+ salts as shown below:
\(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 5 \mathrm{SO}_{4}^{2-}+4 \mathrm{H}^{+}+2 \mathrm{Mn}^{2+}\\ (Violet) (Colourless) \)
\(\\\underset{\text { (Yellow) }}{2 \mathrm{Fe}^{3+}}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\text { (Green) }}{2 \mathrm{Fe}^{2+}}+\mathrm{SO}_{4}^{2-}+4 \mathrm{H}^{+}\)
Thus, solid A is S8 and gas B is SO2
