Question

It is because of inability of ${\rm{n}}{{\rm{s}}^2}$ electrons of the valence shell to participate in bonding that

Accepted Answer

${\rm{S}}{{\rm{n}}^{2 + }}$ is reducing while ${\rm{P}}{{\rm{b}}^{{\rm{4 + }}}}$ is oxidising

Answer

${\rm{S}}{{\rm{n}}^{4 + }}$ is reducing while ${\rm{P}}{{\rm{b}}^{{\rm{4 + }}}}$ is oxidising

Answer

${\rm{S}}{{\rm{n}}^{2 + }}$ is oxidising while ${\rm{P}}{{\rm{b}}^{{\rm{4 + }}}}$ is reducing

Answer

${\rm{S}}{{\rm{n}}^{2 + }}{\rm{ and P}}{{\rm{b}}^{{\rm{2 + }}}}$ are both oxidising and reducing

NEET UG CHEMISTRY MCQ: 14 TH GROUP ELEMENTS (P-BLOCK)

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