Question

$Pb{O_2}$ is an amphoteric oxide and so, $Pb{O_2}$ is soluble in $NaOH$ and also in $HCl$. State whether the statement is true or false.

Answer 1

We have the statement in the question and we need to conclude whether the above statement is true or false. To check the statement we must confirm that $Pb{O_2}$ is an amphoteric oxide. So we should know the basics of salts. If $Pb{O_2}$ is an amphoteric oxide then we will be able to conclude the statement.

Complete step by step answer:
First we will discuss the term amphoteric. An amphoteric compound or molecules is the compound which can react with both acids and base. Most of the elements form oxides and hydroxides. For example $A{l_2}{O_3}$ is an amphoteric oxide. Most of the metals form amphoteric oxides or hydroxides.
Now we will consider $Pb{O_2}$ (lead oxide). We will react with lead oxide with both acids and bases one by one and we will note the observation.
First we will react with lead (II) oxide with a base Sodium hydroxide. The product of the reaction will be sodium plumbate (IV) and water. The reaction takes place in a boiling solution. The reaction can be written as given below.
$Pb{O_2} + 2NaOH\xrightarrow{{}}N{a_2}Pb{O_3} + {H_2}O$
Now we will react with lead (II) oxide with an acid hydrogen chloride. When lead (II) oxide reacts with an acid the product formed is lead tetrachloride and water. The reaction can be written as given below.
$Pb{O_2} + 4HCl\xrightarrow{{}}PbC{l_4} + 2{H_2}O$
We can conclude that $Pb{O_2}$ (lead oxide) is an amphoteric oxide as it reacts with both acids as well as bases. From the above chemical reactions we can also conclude that $Pb{O_2}$ (lead oxide) is soluble in $NaOH$ and $HCl$.

Therefore, the statement is true.

Note: In the reactions mentioned in the answer we can easily observe that the oxidation state of lead is $+ 4$. $PbC{l_4}$ is soluble in water and it has a tetrahedral configuration, with lead as the central atom.