The number of bridged oxygen atoms present in compound B formed from the following reactions is {Pb(NO_3)_2 {673\,K} A…

Strategy: Trace the reaction sequence to find compound B, then count its bridging oxygen atoms.

Step 1: Identifying Compound A $\ce{Pb(NO3)2 ->[\ 673K\ ] A + PbO + O2}$ Lead(II) nitrate decomposes on heating. The gas A here must be $\ce{NO2}$ (brown gas): $\ce{2Pb(NO3)2 ->[\ \Delta\ ] 2PbO + 4NO2 + O2}$ So A = $\ce{NO2}$.

Step 2: Dimerisation of A to form B $\ce{A (NO2) ->[ ext{Dimerise}] B}$ $\ce{NO2}$ (brown gas) readily dimerises on cooling to form dinitrogen tetroxide: $\ce{2NO2 \rightleftharpoons N2O4}$ B = $\ce{N2O4}$

Step 3: Bridging Oxygen Atoms in N₂O₄ In $\ce{N2O4}$, the structure is $\ce{O2N-NO2}$ — a direct N–N bond with four oxygen atoms (2 on each N), and no bridging oxygen atoms between the two N atoms.

Number of bridging O atoms = 0

$\boxed{\text{Answer: (A) 0}}$