Question

State whether the given statement is true or false:
$N{{O}_{2}}$ is a mixed anhydride of two oxoacids.

Answer 1

Hint: The term mixed anhydride suggests that it is a double acid anhydride. So we have to look at how nitrogen dioxide, $N{{O}_{2}}\text{ }$reacts with water. Remember the variable oxidation states of nitrogen, apply that concept.

Complete step by step answer:
The compounds that on reaction with water form a mixture of two different acids are called mixed anhydrides. If we remove water from an oxoacid, the corresponding oxide is called its anhydride.
When $N{{O}_{2}}\text{ }$ is added with water, there is no acid where the oxidation state of nitrogen is +4.
So, $N{{O}_{2}}\text{ }$ is dissociated into$HN{{O}_{3}}$ and $HN{{O}_{2}}$.
The reaction occurs as follows:
$N{{O}_{2}}\text{ + }{{H}_{2}}O\text{ }\rightleftarrows \text{ }HN{{O}_{3}}\text{ + }HN{{O}_{2}}$
It is not entirely a pure acid anhydride because of its abnormal oxidation state and thus disproportionate into two acids, nitric acid and nitrous acid.
So the statement is true.

Additional information:
Oxide of phosphorous acid (${{H}_{3}}P{{O}_{4}}$) is ${{P}_{2}}{{O}_{5}}$, where oxidation state of phosphorous of the oxide is +5 which is possible because it has vacant p-orbital. So the product of hydrolysis of the oxide is only one oxyacid which is a simple redox reaction.

Note: We have to check the oxidation state of the metal cation of the anhydride of the corresponding oxoacid. If the oxidation state exists, then there will be a simple redox reaction. If not, then we have to perform disproportionation reactions in order to maintain the oxidation.