Q. It is because of inability of $ns^2$ electrons of the valence shell to participate in bonding that :
A) $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidising
B) $Sn^{2+}$ is oxidising while $Pb^{4+}$ is reducing
C) $Sn^{2+}$ and $Pb^{2+}$ are both oxidising and reducing
D) $Sn^{4+}$ is reducing while $Pb^{4+}$ is oxidising
Explanation
[Solution] The inert pair effect makes +2 more stable down the group: $Pb^{4+}$ is oxidising (goes to $Pb^{2+}$), while $Sn^{2+}$ is reducing (goes to $Sn^{4+}$). Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.