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(A) The thermal decomposition of lead nitrate is given by:$2Pb(NO_3)_2 \xrightarrow{673 \ K} 2PbO + 4NO_2 + O_2$Here,$A$ is $NO_2$.On dimerisation,$NO

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$0$

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(A) The thermal decomposition of lead nitrate is given by:$2Pb(NO_3)_2 \xrightarrow{673 \ K} 2PbO + 4NO_2 + O_2$Here,$A$ is $NO_2$.On dimerisation,$NO_2$ forms $N_2O_4$ $(B)$:$2NO_2 \rightleftharpoons N_2O_4$The structure of $N_2O_4$ consists of two $NO_2$ units linked by a $N-N$ bond. The structure is $O_2N-NO_2$. It does not contain any bridged oxygen atoms (i.e.,no $N-O-N$ linkage).Therefore,the number of bridged oxygen atoms in $B$ is $0$.

The number of bridged oxygen atoms present in compound $B$ formed from the following reactions is
$Pb(NO_3)_2 \xrightarrow{673 \ K} A + PbO + O_2$
$A \xrightarrow{\text{Dimerise}} B$

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The number of bridged oxygen atoms present in compound $B$ formed from the following reactions is$Pb(NO_3)_2 \xrightarrow{673 \ K} A + PbO + O_2$$A \xrightarrow{\text{Dimerise}} B$