Question

A brown ring is formed in the ring test for $NO_3^-$ ion. It is due to the formation of
(a) [Fe (H₂O)₅ (NO)]²⁺
(b) FeSO₄. NO₂
(c) [Fe (H₂O)₄ (NO)₂]²⁺
(d) FeSO₄. HNO₃

Answer 1

$\bold{OPTION \: (b) \: is \: correct}$

$\bold{BROWN \: RING \: TEST}$

This test is done for detecting the nitrate

$NO_3^{-1}$

EXPERIMENT

▶Take a test tube and mix solution of a nitrate salt or of dilute nitric acid.

▶Add freshly prepared solution of iron II sulphate in the testube. Only freshly prepared iron II sulphate is used because t on exposture to atmosphere is oxidised to iron III sulphate and test will not give results.

▶Add concentrate sulphuric acid slowly.

▶Cool the test tube.

OBSERVATION

▶A brown ring is formed at the junction of two liquids. Because concentrate sulphuric acid being heavier settles down and iron II sulphate layer remains above.

▶When the test tube is shaken the concentrate sulphuric acid may further mix with water and heat evolved may disturb the brown ring.

EQUATION :-

$6FeSO_4 +3H_2SO_4 + HNO_3 \longrightarrow 3Fe_2(SO_4)_3 + 4H_2O + 2NO$

$FeSO_4 + NO \longrightarrow FeSO_4.NO$

$\bold{NIROSO \: IRON \: II\: SULPHATE }$

Answer 2

FeSO₄. NO₂ is the answer...!!!