The pBlock Elements - Result Question 17
####18. It is because of inability of $n s^{2}$ electrons of the valence shell to participate in bonding that:-
[2017]
(a) $Sn^{2+}$ is oxidising while $Pb^{4+}$ is reducing
(b) $Sn^{2+}$ and $Pb^{2+}$ are both oxidising and reducing
(c) $Sn^{4+}$ is reducing while $Pb^{4+}$ is oxidising
(d) $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidising
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Solution:
- (d) Due to inert pair effect, $Pb$ (II) is more stable than $Pb(IV)$
$Sn(IV)$ is more stable than $Sn$ (II)
$\therefore \quad Pb(IV)$ is easily reduced to $Pb(II)$ and can acts as an oxidising agent whereas $Sn$ (II) is easily oxidised to $Sn(IV)$ and can acts as a reducing agent.
Inertness 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. It occurs due to poor shielding of the $n s^{2}$ electrons of the valence shell by the intervening $d$ and $f$ electrons.
