$Sn^{2+}$ is oxidising while $Pb^{4+}$ is reducing
$Sn^{2+}$ and $Pb^{2+}$ are both oxidising and reducing
$Sn^{4+}$ is reducing while $Pb^{4+}$ is oxidising
$Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidising
Hide Solution
Verified By Collegedunia
The Correct Option isD
Solution and Explanation
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
P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.