Question

In Nessler’s reagent, for the detection of ammonia the active species is:
(A) ${ Hg }_{ 2 }{ I }_{ 2 }$
(B) ${ Hg }^{ 2+ }$
(C) ${ HgI }_{ 2 }^{ 2- }$
(D) ${ HgI }_{ 4 }^{ 2- }$

Answer 1

Hint: Try to recall the reaction involved in the detection of ammonia with nessler's reagent to get an idea about solving this question and figuring out the correct option.

Complete step by step answer:
> Nessler's reagent - Potassium tetraiodomercurate(II) is an inorganic compound consisting of potassium cations and the tetraiodomercurate(II) anion. It is mainly used as Nessler's reagent.
> An alkaline solution of ${ K }_{ 2 }{ HgI }_{ 4 }$ is called Nessler's reagent.
We can use a 0.09 mol/L solution of potassium tetraiodomercurate(II) (${ K }_{ 2 }{ HgI }_{ 4 }$) in 2.5 mol/L potassium hydroxide, to detect ammonia or ammonium ions present in the solution. This pale solution becomes deeper yellow in the presence of ammonia. At higher concentrations, a brown precipitate may form.
${ NH }_{ 4 }^{ + }\quad +\quad 2[HgI_{ 4 }]^{ 2- }+\quad 4OH^{ - }\quad \rightarrow \quad HgO·Hg(NH_{ 2 })I\quad \downarrow \quad +\quad 7I^{ - }+\quad 3H_{ 2 }O$
Here, ${ HgI }_{ 4 }^{ 2- }$ is the active species in the detection of ammonia. Because it is responsible for reacting ${ NH }_{ 4 }^{ + }$ with and forming a precipitate.
$HgO·Hg(NH_{ 2 })I$ = Iodide of Millon's base, that forms brown precipitate.
So, the correct answer is option D.
Additional information: Nessler’s reagent was named after Julius Nessler who was a German chemist. Nessler's reagent provides a colorimetric measure of ammonia concentration.

Note: Preparation of Nessler’s reagent - We can prepare Nessler’s reagent by crystallizing it from a concentrated aqueous solution of mercuric iodide with potassium iodide is the monohydrate $KHgI_{ 3 }.H_{ 2 }O$, which is pale orange. In aqueous solution this triiodide complex adds iodide to give the tetraiodo dianion.