Question

It is because of inability of $ns^{2}$ electrons of the valence shell to participate in bonding that:-

• 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

Answer 1

Answer: (D)

$Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidising

Answer 2

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, $Pb ( II )$ is more stable than $Pb ( IV )$ $Sn ( IV )$ is more stable than $Sn ( II )$ $\therefore Pb ( IV )$ is easily reduced to $Pb ( II )$ $\therefore Pb ( IV )$ is oxidising agent $Sn ( II )$ is easily oxidised to $Sn ( IV )$ $\therefore Sn ( II )$ is reducing agent