Question

In the reaction $HgC{{l}_{2}}+excess\,KI\to X\xrightarrow{N{{H}_{4}}Cl+NaOH}Y$, X and Y respectively are:
A.$Hg{{I}_{2}},N{{H}_{4}}I$
B.${{K}_{2}}Hg{{I}_{4}},{{(N{{H}_{4}})}_{2}}Hg{{I}_{4}}$
C.${{K}_{2}}Hg{{I}_{4}},H{{g}_{2}}N{{H}_{2}}OI$
D.$Hg{{I}_{2}},H{{g}_{2}}N{{H}_{2}}OI$

Answer 1

Hint: The idea of Nessler's Reagent is to be used in solving the question. Nessler's Reagent is prepared by mixing compounds of Hg and K and is used for detection of ammonium ions.

Complete answer:
In order to answer our question, we'd like to find out about Nessler's reagent. Potassium tetraiodomercurate(II) is a chemical compound which includes cations of cations and also the anion of tetraiodomercurate(II) . It's mainly called as Nessler's reagent, a concentration of $0.09mol\,{{L}^{-1}}$ solution consisting potassium tetraiodomercurate(II) in $2.5mol\,{{L}^{-1}}$ caustic potash, which can be used for detection of ammonia. it's prepared by crystallizing from a concentrated solution of mercuric iodide with iodide is that the monohydrate which is pale orange.[3] In solution this triodido complex adds iodide to relinquish the tetrahedral tetraiodo dianionIt is termed after the man Julius Nebler (Nessler), an alkaline solution of ${{K}_{2}}Hg{{I}_{4}}$ is termed Nessler's reagent. This pale solution becomes deeper yellow within the presence of ammonia. At higher concentrations, a brown precipitate may form. Now let us come to the question asked.
By using $KI,HgC{{l}_{2}},NaOH$ we can prepare the Nessler's Reagent. For this, we need to obtain mercuric iodide and the same is obtained by reaction of potassium iodide with mercuric chloride. The reaction is represented as:
     \[HgC{{l}_{2}}+2KI\to Hg{{I}_{2}}\downarrow +2KCl\]
Now with this orange coloured mercuric iodide, potassium iodide is reacted so that we get ${{K}_{2}}Hg{{I}_{4}}$.
     \[Hg{{I}_{2}}+2KI\to {{K}_{2}}Hg{{I}_{4}}\]
As said above, for detection of $N{{H}_{4}}^{+}$ions, Nessler's reagent can be used and a reaction illustrating the same is given below:
     \[{{K}_{2}}Hg{{I}_{4}}+NaOH+N{{H}_{4}}Cl\to H{{g}_{2}}N{{H}_{2}}OI+NaI+KCl+{{H}_{2}}O\]
So, X is ${{K}_{2}}Hg{{I}_{4}}$ and Y is $H{{g}_{2}}N{{H}_{2}}OI$ .

Hence we obtain our correct answer as option C.

Note:
It is to be noted that during the test for the presence of ammonium by using Nessler's reagent, the presence of the ion is confirmed as a brown ring gets formed. For example in this case, $H{{g}_{2}}N{{H}_{2}}OI$ forms as the brown precipitate.