$\mathrm{Ge}($ II)compounds are powerful reducing agents whereas $\mathrm{Pb}(\mathrm{IV})$ compounds are…

$\mathrm{Ge}($ II)compounds are powerful reducing agents whereas $\mathrm{Pb}(\mathrm{IV})$ compounds are strong oxidants. It is because
  1. $\mathrm{Pb}$ is more electropositive than Ge
  2. ionization potential of lead is less than that of Ge
  3. ionic radii of $\mathrm{Pb}^{2+}$ and $\mathrm{Pb}^{4+}$ are larger than those of $\mathrm{Ge}^{2+}$ and $\mathrm{Ge}^{4+}$
  4. of more pronounced inert pair effect in lead than in Ge

Solution

Ge(II) tends to acquire Ge (IV) state by loss of electrons. Hence it is reducing in nature. Pb(IV) tends to acquire $\mathrm{Pb}$ (II) O.S. by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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