When $K I$ is added to acidified solution of sodium nitrite :

N O

gas is liberated and

I_{2}

is set free

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N_{2}

gas is liberated and

H I

is produced

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N_{2} O

gas is liberated and

I_{2}

is set free

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N_{2}

gas is liberated and

H O I

is produced

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Solution

The correct option is A $N O$ gas is liberated and $I_{2}$ is set free
Reduction of the acid gives different products, depending on the reducing agent.
With $I^{-}$ and $F e^{2 +}$ ions, $N O$ is formed.
$N a N O_{2} + K I + H_{2} S O_{4} \to H 2 O + I_{2} + K_{2} S O_{4} + N O + N a_{2} S O_{4}$
$2 N a N O_{2} + 2 F e S O_{4} + 2 H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + 2 N O + 2 H_{2} O + N a_{2} S O_{4}$

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