Question

$Ge\left(II\right)$ compounds are powerful reducing agents where as $Pb\left(IV\right)$ compounds are strong oxidants. It can be due to:

• A. $Pb\text{is more electronegative than}Ge$
• B. $\text{Ionisation potential of lead is less than that of}Ge$
• C. $\text{Ionic radii of}P{b}^{2+}andP{b}^{4+}\text{are larger than those of}G{e}^{2+}andG{e}^{4+}$
  
• D. $\text{More pronounced inert pair effect in lead than in}Ge$

Answer 1

The correct option is D $\text{More pronounced inert pair effect in lead than in}Ge$

The pair of electrons in valence $s-$orbital is reluctant to take part in bond formation due to poor shielding effect of $d-andf-$ electron in heavier elements. This is known as inert pair effect.
It is generally observed in post transition metal compounds.
More pronounced inert pair effect is in lead than in $Ge$.
So, $Pb$ tends to be in $+2$ oxidation state where as $Ge$ is stable in $+4$ oxidation state.
Hence, $Pb\left(IV\right)$ compounds are reduced to $Pb\left(II\right)$ and acts as strong oxidants.
$Ge\left(II\right)$ compounds are oxidised to $Ge\left(IV\right)$ and acts as powerful reducing agents.