Question

Match the oxide of column I with their corresponding property (ies) in column II.

Answer 1

$\left(A\right)$ Oxidation state of $Pb$ in $P{b}_3{O}_4\Rightarrow 3x+(-8)=0$

$P{b}_3$ must cancel out the $-8$ and therefore, $Pb$ must have overall oxidation number of $+8$ or $+8/3$. However, oxidation numbers cannot be a fraction.

$P{b}_3{O}_4$ actually exists as a mixture of $PbO$ and $Pb{O}_2$ in the ratio $2:1$. The oxidation states of each are $+2$ and $+4$. It can act as both acid and base. Hence, it is amphoteric in nature.

$\left(B\right)$ Oxidation state of $Sn$ in $Sn{O}_2\Rightarrow x+(-4)=0\Rightarrow x=+4$

$Sn{O}_2$ can act as acid and base therefore is amphoteric in nature.

$\left(C\right)$ Oxidation state of $Pb$ in $Pb{O}_2\Rightarrow x+(-4)=0\Rightarrow x=+4$ It is a base in nature.

$\left(D\right)$ Oxidation state of $C$ in $C{O}_2\Rightarrow x+(-4)=0\Rightarrow x=+4$

It is acidic in nature.

$\left(E\right)$ Oxidation state of $Ge$ in $GeO\Rightarrow x+(-2)=0\Rightarrow x=+2$. It is amphoteric in nature.