
The following some statements about $HN{O_2}$ are:
i) It acts as both oxidising and reducing agent.
ii) It is a strong acid.
iii) Its anhydride in pure state exists as pale blue liquid
The correct combination is:
A. All are correct
B. i, iii are correct
C. ii, iii are correct
D. i, ii are correct
Answer
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Hint: Whether $HN{O_2}$ acts as a reducing or oxidising agent can be found by examining the oxidation number of nitrogen in the compound. The strength of an acid is determined using the acid dissociation constant. And lastly, the anhydride is formed by the removal of water from the compound.
Complete step by step answer:
First let us examine the oxidation number of nitrogen in $HN{O_2}$. As we know, hydrogen and oxygen always have oxidation numbers of $ + 1$ and $ - 2$ respectively. And since the compound has no charge, the sum of oxidation numbers should add up to zero. Taking the oxidation number of nitrogen as $x$, we get:
$ + 1 + x + ( - 2 \times 2) = 0$
$ \Rightarrow x = 4 - 1 = 3$
We know that nitrogen can exhibit oxidation numbers in the range of $ - 3$ to $ + 5$. Since the oxidation state of nitrogen in $HN{O_2}$ is $ + 3$, it can get reduced and lowered, meaning that the compound can get oxidized or reduced. Hence, statement (i) is correct.
Now let us move on to the second statement. $HN{O_2}$ dissociates in water to give:
$HN{O_2} \rightleftharpoons {H^ + } + NO_2^ - $
It is found that the dissociation constant for this reaction is quite low, indicating that $HN{O_2}$ is a weak acid. Hence, statement (ii) is wrong.
Moving on to the last statement, we know that an anhydride is produced when a compound is heated and loses water. In the case of $HN{O_2}$, we have:
$2HN{O_2}\xrightarrow{\Delta }{N_2}{O_3} + {H_2}O$
The anhydride formed, ${N_2}{O_3}$, exhibits a deep blue colour in solution. Hence, statement (iii) is correct.
As we have seen, only the first and third statements are correct.
So, the correct answer is Option B.
Note: The main reason why $HN{O_2}$ is a weak acid is because the conjugate base formed in water, the $NO_2^ - $ ion is a very unstable compound. Acids are considered to be strong only when their conjugate bases are stable in solution. Note that ${N_2}{O_3}$ is coloured as it absorbs light in the region of the visible spectrum and hence, appears blue.
Complete step by step answer:
First let us examine the oxidation number of nitrogen in $HN{O_2}$. As we know, hydrogen and oxygen always have oxidation numbers of $ + 1$ and $ - 2$ respectively. And since the compound has no charge, the sum of oxidation numbers should add up to zero. Taking the oxidation number of nitrogen as $x$, we get:
$ + 1 + x + ( - 2 \times 2) = 0$
$ \Rightarrow x = 4 - 1 = 3$
We know that nitrogen can exhibit oxidation numbers in the range of $ - 3$ to $ + 5$. Since the oxidation state of nitrogen in $HN{O_2}$ is $ + 3$, it can get reduced and lowered, meaning that the compound can get oxidized or reduced. Hence, statement (i) is correct.
Now let us move on to the second statement. $HN{O_2}$ dissociates in water to give:
$HN{O_2} \rightleftharpoons {H^ + } + NO_2^ - $
It is found that the dissociation constant for this reaction is quite low, indicating that $HN{O_2}$ is a weak acid. Hence, statement (ii) is wrong.
Moving on to the last statement, we know that an anhydride is produced when a compound is heated and loses water. In the case of $HN{O_2}$, we have:
$2HN{O_2}\xrightarrow{\Delta }{N_2}{O_3} + {H_2}O$
The anhydride formed, ${N_2}{O_3}$, exhibits a deep blue colour in solution. Hence, statement (iii) is correct.
As we have seen, only the first and third statements are correct.
So, the correct answer is Option B.
Note: The main reason why $HN{O_2}$ is a weak acid is because the conjugate base formed in water, the $NO_2^ - $ ion is a very unstable compound. Acids are considered to be strong only when their conjugate bases are stable in solution. Note that ${N_2}{O_3}$ is coloured as it absorbs light in the region of the visible spectrum and hence, appears blue.
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