NEET Solved Paper 2017 Question 30
Question: It is because of inability of $ ns^{2} $ electrons of the valence shell to participate in bonding that
Options:
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
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Answer:
Correct Answer: B
Solution:
- Inability of $ ns^{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