Question

Which of the following precipitate is dirty white in color?
A. ${{CoS}}$
B. ${{ZnS}}$
C. ${{CuS}}$
D. ${{MnS}}$

Answer 1

Hint:Let’s initially know what the precipitate is. It is an insoluble compound which is formed when two solutions are reacted together. This insoluble compound gets settled in the solution. All of the given compounds are sulfides.

Complete answer:
We can see a precipitate through our naked eye. This is because precipitate appears in solid state. It is deposited at the bottom of the solution.
When double displacement reactions occur, precipitation is formed. Double displacement reactions occur due to the mutual exchange of ions. The ions are positive and negative ions. In such reactions, elements of different compounds are swapped.
All of the given compounds are solid compounds. There are no ions which are present in solid state. Thus the reactions occur only in aqueous solution form.
Mostly, all of the metals in the ${{3d}}$ block show different colors.
All sulfides are generally insoluble in water. ${{CoS}}$ and ${{CuS}}$ has a black color precipitate. While ${{MnS}}$ shows pink color. While ${{ZnS}}$ appears dirty white in color.
Thus the precipitate which is dirty white in color is ${{ZnS}}$.

Hence, the correct option is B.

Additional information:
D block elements generally have one or more unpaired number of electrons. d-d transition occurs when the unpaired electrons in the lower energy level get excited to higher energy level.

Note:
${{ZnS}}$ can be prepared by reacting ${{ZnC}}{{{l}}_2}$ with ${{N}}{{{a}}_2}{{S}}$. ${{ZnS}}$ is a white colored salt. It forms a dense white precipitate which gets deposited at the bottom. The chemical reaction is given below:
${{ZnC}}{{{l}}_2} + {{N}}{{{a}}_2}{{S}} \to {{ZnS}}$
We know the d block elements have the ability to form colored compounds because they have vacant or half-filled d orbitals.