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It is because of inability of ns^(2) ele...

It is because of inability of `ns^(2)` electrons of the valence shell to participate in bonding that:

A

`Sn^(2+)` is oxidizing while `Pb^(4+)` is reducing

B

`Sn^(+2)` and `Pb^(2+)` are both oxidizing and reducing

C

`Sn^(4+)` is reducing while `Pb^(4+)` is oxidizing

D

`Sn^(2+)` is reducing while `Pb^(4+)` is oxidizing

Text Solution

Verified by Experts

The correct Answer is:
D
[In English-(d), in Hindi -(a)]
`{:(" "Sn^(+2)" "to " "Sn^(+4)),(" (R.A)"" "Sn^(+2) lt Sn^(+4)"Stability order"),(""Pb^(+4)" " to " "Pb^(+2)),("(O.A)"" " Pb^(+2) gt Pb^(+4) "Stability order(inert pair effect)"):}`
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