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Question

When $$KI$$ is added to acidified solution of sodium nitrite :


A
NO gas is liberated and I2 is set free
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B
N2 gas is liberated and HI is produced
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C
N2O gas is liberated and I2 is set free
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D
N2 gas is liberated and HOI is produced
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Solution

The correct option is A $$NO$$ gas is liberated and $$I_2$$ is set free

Reduction of the acid gives different products, depending on the reducing agent.

With $$I^-$$ and $$Fe^{2+}$$ ions, $$NO$$ is formed.

$$NaNO_2 + KI + H_2SO_4 \rightarrow H2O + I_2 + K_2SO_4 + NO + Na_2SO_4$$

$$2 NaNO_2 + 2 FeSO_4 + 2 H_2SO_4 \rightarrow Fe_2(SO_4)_3 + 2 NO + 2 H_2O + Na_2SO_4$$


Chemistry

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