A) \[N{{H}_{4}}Cl\]
B) \[N{{H}_{4}}N{{O}_{3}}\]
C) \[N{{H}_{4}}N{{O}_{2}}\]
D) \[FeS{{O}_{4}}\]
Correct Answer: A
Solution :
\[N{{H}_{4}}Cl\xrightarrow{\Delta }\underset{Y}{\mathop{N{{H}_{3}}}}\,+\underset{Z}{\mathop{HCl}}\,\] |
\[N{{H}_{3}}+2{{K}_{2}}[Hg{{I}_{4}}]+3KOH\to {{H}_{2}}NHgO.\] |
\[HgI+7KI+{{H}_{2}}O\] |
Nessler's reagent brown ppt. iodide of Millon's base |
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