A) \[S{{n}^{4+}}\] is reducing while \[P{{b}^{4+}}\] is oxidizing
B) \[S{{n}^{2+}}\]is reducing while \[P{{b}^{4+}}\]is oxidizing
C) \[S{{n}^{2+}}\] is oxidising while \[P{{b}^{4+}}\]is reducing
D) \[S{{n}^{2+}}\]and\[P{{b}^{2+}}\]are both oxidising and reducing
Correct Answer: B
Solution :
Inability of \[n{{s}^{2}}\]electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect As a result, Pb(ll) is more stable than Pb(IV) Sn(IV) is more stable than Sn(ll) \[\therefore \]Pb(IV) is easily reduced to Pb(ll) \[\therefore \]Pb(IV) is oxidising agent Sn (ll) is easily oxidised to Sn(IV) \[\therefore \] Sn (ll) is reducing agent
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