Question

A metal nitrate reacts with $KI$ to give a black precipitate which in addition to excess $KI$ converts to an orange color solution. The cation of metal nitrate is:
A: $H{g^{2 + }}$
B: $B{i^{3 + }}$
C: $P{b^{2 + }}$
D: $C{u^ + }$

Answer 1

Hint: Salts that contain $NO_3^ - $ anions are called nitrates. Nitrates are commonly used in making explosives and fertilizers. Nitrates that are found naturally are present in vegetables, dairy products and meat.

Complete step by step answer:
In this question we have to find the metal nitrate that produces black precipitate on reacting with $KI$ and this precipitate when reacts with $KI$ again converts the solution orange.
Copper nitrate reacts with potassium iodide. The reaction that will take place when potassium iodide reacts with copper nitrate is,
$2Cu{\left( {N{O_3}} \right)_2} + 4KI\xrightarrow{{}}2CuI + 4KN{O_3} + {I_2}$
But precipitates formed are white in color. This means $C{u^ + }$ is not the required metal ion. As the metal ion we have to find in this question gives black precipitate.
Lead nitrate reacts with potassium iodide. The reaction that will take place when potassium iodide reacts with Lead nitrate is,
$Pb{\left( {N{O_3}} \right)_2} + 4KI\xrightarrow{{}}Pb{I_2} + KN{O_3}$
But precipitates formed are yellow in color. This means $P{b^{2 + }}$ is not the required metal ion. As the metal ion we have to find in this question gives black precipitate.
Bismuth nitrate reacts with potassium iodide. The reaction that will take place when bismuth iodide reacts with copper nitrate is,
$Bi{\left( {N{O_3}} \right)_3} + 3KI\xrightarrow{{}}Bi{I_3} + 3KN{O_3}$
Precipitates of bismuth iodide are black in color and the precipitate mentioned in the question are also black. This means $B{i^{3 + }}$ could be the answer. Now, further statement in the question states that when bismuth iodide precipitate will react with potassium iodide again, an orange solution will be formed. Reaction of bismuth iodide with potassium iodide is as follows,
$Bi{I_3} + KI\xrightarrow{{}}K\left[ {Bi{I_4}} \right]$
Complex that is formed in this reaction is orange in color (as required).
This means correct answer is option B that is $B{i^{3 + }}$.

Note:
Any compound possesses color because of valence electrons. In copper iodide, copper has inert gas configuration. This means copper doesn’t have any valence electrons. Due to which copper iodide precipitates are white colored.