A) \[N{{H}_{2}}=Hg-Cl\]
B) \[H{{g}_{2}}C{{l}_{2}}N{{H}_{3}}\]
C) \[Hg{{(N{{H}_{3}})}_{2}}C{{l}_{2}}\]
D) \[HgC{{l}_{2}}N{{H}_{3}}\]
Correct Answer: A
Solution :
\[HgC{{l}_{2}}+2N{{H}_{3}}\xrightarrow{{{H}_{2}}O}\] \[Hg+\underset{\begin{smallmatrix} mercuric \\ amino\text{ }chloride \end{smallmatrix}}{\mathop{N{{H}_{2}}HgCl}}\,\,\,+N{{H}_{4}}Cl\] \[\therefore \] \[HgC{{l}_{2}}\] on reaction with \[N{{H}_{4}}OH\] (or\[N{{H}_{3}}+{{H}_{2}}O\]) forms mercuric amino chloride.
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