It is because of the 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.