Question

In which of the pairs the precipitates are red and black coloured respectively and both are soluble in excess KI solution?

(A) $Hg{I_2},H{g_2}{I_2}$

(B) $Hg{I_2},Bi{I_3}$

(C) $C{u_2}{I_2},AgI$

(D) $Cd{I_2},Pb{I_2}$

Answer 1

In order to know which precipitate is in what colour we must know the characteristic of all the precipitates given. We should know what kind of reaction will take place when it reacts with the given KI solution.

Complete step by step answer:

$Hg{I_2}$ is a molecular formula of Mercury (II) Iodide or Mercury diiodide. $Hg{I_2}$ is the precipitate which is red in colour. When mercury (II) iodide is added to the excess of potassium iodide, it will form a complex ${K_2}[Hg{I_4}]$ . The formed complex is soluble in water. The reaction is given below:

$Hg{I_2} \downarrow + 2KI \to {K_2}[Hg{I_4}]$

Similarly, $Bi{I_3}$ is the molecular formula of an inorganic compound called Bismuth (III)iodide or Bismuth triiodide. $Bi{I_3}$ is the precipitate that is grey-black in colour. When Bismuth (III)iodide is added to an excess of KI, then it will form a complex ${K_2}[Bi{I_5}]$. The formed complex of bismuth is also soluble in water. The reaction is given below:

$Bi{I_3} \downarrow + 2KI \to {K_2}[Bi{I_5}]$

Therefore, the pair the precipitates are red and black coloured respectively and both are soluble in excess KI solution is $Hg{I_2}, Bi{I_3}$.

Hence, the correct answer is option (B) $Hg{I_2},Bi{I_3}$.

Additional information:

Uses of Mercury (II) iodide are given:

- In order to detect ammonia, $Hg{I_2}$ can be used.

- Used in manufacturing Blister ointment.

- Mercury (II) Iodide is used for preparing Nessler’s reagent.

- It is used in imaging devices.

Note: Mercury (II) iodide has various colours like red, yellow and orange. But red is the most stable one and the remaining two are meta-stable. This is because Mercury (II) iodide can be crystallized from various solvents and forms three polymorphs red, yellow and orange.