The number of bridged oxygen atoms present in compound B formed from the following reactions is $\mathrm{Pb}(\mathrm{NO}_{3})_{2} \xrightarrow{673 \mathrm{~K}} \mathrm{~A}+\mathrm{PbO}+\mathrm{O}_{2}$ $\mathrm{A} \xrightarrow{\text { Dimerise }} \mathrm{B}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · p-Block Elements

The number of bridged oxygen atoms present in compound B formed from the following reactions is

Pb (NO_{3})_{2} 673 K A + PbO + O_{2}

A Dimerise B

(A)0✓ Correct
(B)1
(C)2
(D)3

Answer: (A)

Solution

Lead nitrate decomposes at 673 K to give

NO_{2}

(compound A), PbO, and

O_{2}

NO_{2}

dimerizes to form

N_{2} O_{4}

(compound B), which has the structure O=N-N=O with no bridging oxygen atoms between the two nitrogen atoms.

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

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