Answer:
In \[Pb{{O}_{2}}\] and \[Sn{{O}_{2}},\]
both lead and tin are present in +4-oxidation state. But due to stronger inert
pair effect, \[P{{b}^{2+}}\] ion is more stable than \[S{{n}^{2+}}\] ion. In
other words, \[P{{b}^{4+}}\]ions i.e. \[Pb{{O}_{2}},\]is more easily reduced to
\[P{{b}^{2+}}\]ions than \[S{{n}^{4+}}\]ions are reduced to \[S{{n}^{2+}}\]
ions. Thus, \[Pb{{O}_{2}}\] acts
as a stronger oxidising agent than \[Sn{{O}_{2}}\].
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