
Which of the following is known as Nessler’s reagent?
A.An alkaline solution of \[Hg{I_2}\]
B.An aqueous solution of tetraiodomercurate $(II)$
C.An alkaline solution of tetraiodomercurate $(II)$
D.An aqueous solution of $HgC{l_2}$
Answer
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Hint:At 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.
Complete step by step answer:
An alkaline solution of tetraiodomercurate $(II)$ is known as Nessler’s agent. The chemical formula of Nessler’s agent is ${K_2}[Hg{I_4}]$. It is obtained by the reaction of $Hg{I_2}$ with excess of $KI$.
$Hg{I_2} + KI(excess) \to {K_2}[Hg{I_4}]$
So the correct answer is option C.
Additional Information:-Nessler’s reagent gives brown precipitate with ammonia. Nessler’s reagent is used for detecting the presence of $N{H_4}^ + $ ions. Potassium tetraiodomercurate $(II)$ is an organic compound consisting of potassium cations and the tetraiodomercurate $(II)$ anion. It is mainly used as Nessler’s reagent , a $0.09$ $mol/L$ solution of potassium tetraiodomercurate $(II)$ in potassium hydroxide. In aqueous solution the triodido complex adds iodide to give the tetrahedral tetraiodo dianion. In the presence of ammonia at high concentrations, the brown precipitate of nessler’s reagent may form. The formula for the brown precipitate is given as $3HgO.Hg{(N{H_3})_2}{I_2}$ and $N{H_2}.H{g_2}{I_3}$. This brown precipitate is called Millon's base. Millon’ s base is a yellow basic powder which is produced by the action of ammonia on mercuric oxide.
Note:
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.
Complete step by step answer:
An alkaline solution of tetraiodomercurate $(II)$ is known as Nessler’s agent. The chemical formula of Nessler’s agent is ${K_2}[Hg{I_4}]$. It is obtained by the reaction of $Hg{I_2}$ with excess of $KI$.
$Hg{I_2} + KI(excess) \to {K_2}[Hg{I_4}]$
So the correct answer is option C.
Additional Information:-Nessler’s reagent gives brown precipitate with ammonia. Nessler’s reagent is used for detecting the presence of $N{H_4}^ + $ ions. Potassium tetraiodomercurate $(II)$ is an organic compound consisting of potassium cations and the tetraiodomercurate $(II)$ anion. It is mainly used as Nessler’s reagent , a $0.09$ $mol/L$ solution of potassium tetraiodomercurate $(II)$ in potassium hydroxide. In aqueous solution the triodido complex adds iodide to give the tetrahedral tetraiodo dianion. In the presence of ammonia at high concentrations, the brown precipitate of nessler’s reagent may form. The formula for the brown precipitate is given as $3HgO.Hg{(N{H_3})_2}{I_2}$ and $N{H_2}.H{g_2}{I_3}$. This brown precipitate is called Millon's base. Millon’ s base is a yellow basic powder which is produced by the action of ammonia on mercuric oxide.
Note:
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.
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