Question

Nessler’s reagent contains
$A)$$H{g_2}{^{2{ + _{}}}_{}}$
$B)$$H{g^{2 + }}$
$C)$$Hg{I_4}^{2 - }$
$D)$$Hg{I_4}^ - $

Answer 1

Hint: Nessler’s reagent is an solution of mercury $(||)$ iodide $(Hg{I_2})$ in potassium iodide and potassium hydroxide named after the german physicist. It is utilized for the quantitative testing of ammonia.

Complete answer:
We know that the nessler’s reagent formula is potassium tetraiodomercurate$(||)$ $({K_2}[Hg{I_4}])$. It is an alkaline solution. In the presence of ammonia the color of the solution becomes pale yellow. A brown precipitate which is a derivative of millon’s base may form after increasing the concentration.
$N{H_4} + 2{[Hg{I_4}]_{}}^{2 - } + 4O{H^ - } \to HgO.Hg(N{H_2})I \downarrow + 7{I^ - } + 3{H_2}O$
- Nessler’s reagent has been utilized in determining ammonia in flow regime using scientific spectroscopy.
- Taking shape from a concentrated fluid arrangement of mercuric iodide with potassium iodide is the monohydrate $[K(Hg{I_3}).{H_2}O]$, which is pale orange. In aqueous arrangement this triodido complex adds iodide to give the tetrahedral tetraiodo dianion.
- As we already know it is a quantitative analysis. Quantitative analysis is the determination of the absolute or relative abundance of one, or few or specific substances present in the sample. When the absolute presence is known, the investigation of their supreme or relative abundance could help in deciding explicit properties. Knowing the piece of an example is vital, and few different ways have been created to make it possible as gravimetric and volumetric analysis.
And hence option C is the correct answer.

Note:
Remember the nessler's reagent is a solution of mercury iodide in potassium iodide and the color of the nessler’s reagent is yellow in color and the complex is responsible for the brown color precipitate.