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.

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