question_answer

It is because of inability of \[n{{s}^{2}}\] electrons of the valence shell to participate in bonding that [NEET-2017]

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: A

Solution :

[a] Inability of 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)         Pb(IV) is easily reduced to Pb(ll)         Pb(IV) is oxidising agent Sn (ll) is easily oxidised to Sn(IV)         Sn (ll) is reducing agent