• # question_answer 7) Why do the following reactions proceed differently? $P{{b}_{3}}{{O}_{4}}+8HCl\to 3PbCl+C{{l}_{2}}+4{{H}_{2}}O$and $P{{b}_{3}}{{O}_{4}}+4HN{{O}_{3}}\to 2Pb{{(N{{O}_{3}})}_{2}}+Pb{{O}_{2}}+2{{H}_{2}}O.$

$P{{b}_{3}}{{O}_{4}}$ is a mixture of $PbO$ and $Pb{{O}_{2}}$in 2 : 1 ratio. $PbO$has stable oxidation state of +2 while in $Pb{{O}_{2}}$, its oxidation state is +4. Thus, $Pb{{O}_{2}}$ acts as an oxidising agent and oxidises $HCl$into$C{{l}_{2}}$. $PbO$on the other hand is a basic oxide which forms a salt $(PbC{{l}_{2}})$ with $HCl$ The reaction, $P{{b}_{3}}{{O}_{4}}+8HCl\to 3PbC{{l}_{2}}+C{{l}_{2}}+4{{H}_{2}}O$ may be divided into two parts. $2PbO+4HCl\to 2PbC{{l}_{2}}+2{{H}_{2}}O$(Acid base reaction) $Pb{{O}_{2}}+4HCl\to PbC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O$(Redox reaction) In the second reaction, $\text{HN}{{\text{O}}_{\text{3}}}$acts as oxidising agent only once the reaction may not occur between $Pb{{O}_{2}}$ and $HN{{O}_{3}}$. $PbO$being a basic oxide forms salt with$HN{{O}_{3}}$. $2PbO+4HN{{O}_{3}}\to 2Pb(N{{O}_{3}})+2{{H}_{2}}O$ ($Pb{{O}_{2}}$remain unaffected.)