Select all correct statements about the brown ring test for nitrate (take NO as $NO X^{+}$ in the complex).

ABromide and iodide interfere because conc. $H X_{2} SO X_{4}$ liberates $Br X_{2}$ or $I X_{2}$ whose colour masks the ring.
B $Fe X^{2 +}$ reduces nitrate to NO in acid and is itself oxidised to $Fe X^{3 +}$ .
CNitrite does not interfere because $Fe X^{2 +}$ cannot reduce it to NO.
DNO replaces one water of $[Fe (H X_{2} O) X_{6}] X^{2 +}$ to give the brown $[Fe (H X_{2} O) X_{5} NO] X^{2 +}$ with Fe in the +1 state.

Answer: A, B, D

3 Fe X^{2 +} + NO X_{3} X^{-} + 4 H X^{+} 3 Fe X^{3 +} + NO + 2 H X_{2} O

then

[Fe (H X_{2} O) X_{6}] X^{2 +} + NO [Fe (H X_{2} O) X_{5} NO] X^{2 +} + H X_{2} O

with Fe +1 for

NO X^{+}

Br X_{2}

and

I X_{2}

colours mask the ring. Nitrite is reduced to NO even more easily and gives a brown colour itself, so it does interfere and the last statement is false.

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Select all correct statements about the brown ring test for nitrate (take NO as NOX+ in the complex).

Answer: A, B, D

More questions on Chromyl chloride, brown ring and ring tests

Select all correct statements about the brown ring test for nitrate (take NO as $NO X^{+}$ in the complex).