Question

The white precipitate (P) is :
A. ${ Cu }_{ 2 }{ I }_{ 2 }$
B. ${ Cu }{ I }_{ 2 }$
C. ${ K }_{ 2 }{ [CuI }_{ 4 }{ ] }$
D. None of the above

Answer 1

Hint: Precipitation reaction: When during a chemical reaction, an insoluble solid product is formed which can be separated from the solution by filtration, then the solid product is known as precipitate and the reaction is known as precipitation reactions.
Complete answer:
The following reaction of the process will take place;
${ CuCO }_{ 3 }{ +2HCl\rightarrow { CuCl }_{ 2 } }{ +CO }_{ 2 }{ +H }_{ 2 }{ O }$
When copper carbonate reacts with hydrochloric acid then copper (II) chloride along with carbonic acid will be formed, which quickly decomposes into carbon dioxide and water. (it decomposes nearly quickly).
This solution becomes blue due to the presence of copper ${ (II) }$ ions.

${ CuCl }_{ 2 }{ +2KI\rightarrow Cu }{ I }_{ 2 }{ +2KCl }$

${ 2CuI }_{ 2 }{ \rightarrow }{ Cu }_{ 2 }{ I }_{ 2 }{ + }{ I }_{ 2 }$

When copper chloride reacts with potassium iodide and copper iodide will be formed along with potassium chloride. Here, ${ Cu }^{ 2+ }$ changes into ${ Cu }^{ 1+ }$, so it becomes stable now as it has fully filled ${ 3d }^{ 10 }$ electronic configuration. The white precipitate of ${ Cu }_{ 2 }{ I }_{ 2 }$ will be formed.
So, the correct option is A.

Additional Information:
Some properties of a precipitation reaction:
The precipitation reaction can take place between ions present in the aqueous solution to form the product.
These are also known as ionic reactions since the ions take part in the reaction actively to form the product.
These reactions undergo aqueous solution in an ionic state.
These reactions depend on the concentration of the solution, temperature, buffer solution, and so on.
These reactions can be used in wastewater treatment.
These reactions help to determine a particular element present in the solution.
These are used to extract magnesium from seawater.

Note: The possibility to make a mistake is that the white precipitate of ${ Cu }_{ 2 }{ I }_{ 2 }$ will be formed only when copper chloride reacts with an excess of ${ KI }$ but if ${ KI }$ is not present in excess then ${ Cu }{ I }_{ 2 }$ will be formed.