A) \[{{N}_{2}}O\,.\,FeS{{O}_{4}}\]
B) \[FeS{{O}_{4}}NO\]
C) \[N{{O}_{2}}\]
D) \[FeS{{O}_{4}}N{{O}_{2}}\]
Correct Answer: B
Solution :
In the qualitative analysis of nitrate a brown ring is formed due to the formation of\[[Fe(NO)S{{O}_{4}}]\]. This test is called ring test. Ring test Take a small amount of mixture in a test tube, shake with 10 mL of water and filter. To the filte rate add freshly prepared \[FeS{{O}_{4}}\] solution shake well. Add cone. \[{{H}_{2}}S{{O}_{4}}\]drop wise by the side of test tube. A brown ring \[Fe(NO)S{{O}_{4}}]\] appears at the junction of the two solutions. \[NaN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}NaHS{{O}_{4}}+HN{{O}_{3}}\] \[6FeS{{O}_{4}}+2HN{{O}_{3}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}\] \[3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2NO+4{{H}_{2}}O\] \[FeS{{O}_{4}}+NO\xrightarrow{{}}\underset{\begin{smallmatrix} brown\text{ }ring \\ nitroso\text{ }ferrous \\ sulphate \end{smallmatrix}}{\mathop{[Fe(NO)S{{O}_{4}}]}}\,\]
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