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Karnataka Board PUCPUC Science 2nd PUC Class 12

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’. Identify ‘A’, ‘B’ and ‘C’ and

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Question

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’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.

Long Answer
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Solution

The gas A is \[\ce{NO2}\]. The reactions are explained as under:

\[\ce{2Pb(NO3)2 ->[Heat][673 K] + 2PbO + \underset{(Brown colour)}{\underset{Nitrogen dioxide}{\underset{(A)}{4NO2}}} + O2}\]

\[\ce{2NO2 ->[Cooling] \underset{Solid (Colourless)}{\underset{(B)}{N2O4}}}\]

\[\ce{2NO + N2O4 ->[Heating] \underset{Dinitrogen trioxide}{\underset{(C)}{\underset{Blue solid}{2N2O3}}}}\]

Structure of B

Structure of C

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Chapter 7: The p-block Elements - Multiple Choice Questions (Type - I) [Page 100]

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NCERT Exemplar Chemistry Exemplar [English] Class 12
Chapter 7 The p-block Elements
Multiple Choice Questions (Type - I) | Q 71 | Page 100

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