Write balanced net ionic equation for the following reactions in acidic solution:

$H N O_{2} (a q) \to N O {_{3}}^{-} + N O (g)$

H N O_{2} (a q) \to H N O {_{3}}^{-} + 2 N O (g) + H_{2} O

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2 H N O_{2} (a q) \to 2 H N O {_{3}}^{-} + 2 N O (g) + 2 H_{2} O

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3 H N O_{2} (a q) \to N O {_{3}}^{-} + 2 N O (g) + H_{2} O + H^{+}

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None of the above

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Solution

The correct option is C $3 H N O_{2} (a q) \to N O {_{3}}^{-} + 2 N O (g) + H_{2} O + H^{+}$
The unbalanced chemical equation is as follows:
$H N O_{2} (a q) \to N O {_{3}}^{-} + N O (g)$ .
Balance all atoms except H and O.
$2 H N O_{2} (a q) \to N O {_{3}}^{-} + N O (g)$
The oxidation number of N changes from $+ 3$ to $+ 5$ .
The net change in the oxidation number is $2$ .
Thus, the increase in the oxidation number is balanced with the decrease in the oxidation number by multiplying $H N O_{2}$ and $N O$ with $2$ .
$3 H N O_{2} (a q) \to N O {_{3}}^{-} + 2 N O (g)$
Balance O atoms by adding $1$ water molecule to RHS.
$3 H N O_{2} (a q) \to N O {_{3}}^{-} + 2 N O (g) + H_{2} O$
Balance H atoms by adding $1$ $H^{+}$ to the RHS.
Hence, the balanced equation is as follows:
$3 H N O_{2} (a q) \to N O {_{3}}^{-} + 2 N O (g) + H_{2} O + H^{+}$ .

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