Question

Nessler’s reagent is:
A.\[{{K}_{2}}Hg{{I}_{4}}\]
B.\[{{K}_{2}}Hg{{I}_{4}}+KOH\]
C.\[{{K}_{2}}Hg{{I}_{2}}+KOH\]
D.\[{{K}_{2}}Hg{{I}_{4}}+Hg\]

Answer 1

Hint: We know that the Nessler’s reagent is used to detect ammonia. It is a pale solution which gets darker yellow when some amount of ammonia is added. A brown precipitate may form at higher concentration.

Complete answer:
First, think about the behavior of Nessler’s reagent and some reactions where Nessler’s reagent is used to study the complete behavior of it. Nessler’s reagent is used as a test for the presence of ammonia. Nessler’s reagent is more widely used for the determination of ammonium compounds. In the test the pale solution becomes deeper yellow in the presence of ammonia. It can also be prepared using anhydrous mercuric iodide and anhydrous potassium iodide.
Nessler’s reagent is a potassium tetraiodomercurate (II) which is an inorganic compound that consists of cations and anions. Here, cations are potassium ions and anions are tetraiodomercurate (II). Nessler’s reagent is an alkaline solution is ${{K}_{2}}Hg{{I}_{4}}$ along with $KOH.$ It is to note that we have to remember the chemical formula of Nessler’s reagent and also the complex formed when the reagent reacts with ammonia. A brown precipitate is formed due to this complex. The brown precipitate is known as Milan’s base. The colour of Nessler’s reagent is pale yellow. Nessler’s reagent is widely used for the determination of ammonia in the solution.

So, the correct answer is “Option B”.

Note:
Remember that the Nessler’s reagent is used for the qualitative analysis of ammonia. Qualitative analysis is basically defined as the chemical properties of an unknown substance that are determined by reacting the unknown substance with different reagents.