A solution prepared in conc. \[HCl\] sometimes gives white turbidity even in the absence of the first group. when water is added to it. It is due to the presence of
(A) $H{g^{ + 2}}$
(B) $S{b^{ + 3}}$
(C) $A{g^{ + 3}}$
(D) $S{b^{ + 3}}$or $B{i^{ + 3}}$ or both

Hint: Turbidity literally refers to the cloudiness that appears in the solution. While adding a compound to a solvent (mostly water) the compound dissolves completely and becomes clear known as the true solution. But sometimes, the particles remain suspended in the solvent making visibility opaque. This is known as turbidity. It appears when the solute is not dissolved completely in the solvent.

Complete Step by Step Solution:
Option A: $H{g^{ + 2}}$ belongs to group $1$and it is precipitated on addition of $HCl$. Hence there is no point of appearance of turbidity in the solution. Whereas as the precipitate of mercury chloride settles down, the solution becomes clear. Moreover, it is specified in the question that the first group is absent. Hence, option A is incorrect.

Option B: $S{b^{ + 3}}$reacts with chloride and produces $SbC{l_3}$precipitate. On adding water, the solution turns turbid.

Option C: silver belongs to group $1$ and is precipitated as silver chloride. As the solution is not of group $1$ so option C is ruled out.

Option D: similar to $Sb$, bismuth also precipitates as $BiC{l_3}$and turns turbid on addition of water.
Hence, the correct option is D.

Note: There are a number of methods to remove turbidity. Filtration is one of them. Another way could be by using the centrifuge machine. Centrifugation at a speed of around \[16500{\text{ }}rpm\]settles down the precipitate and above it the solution becomes clear which can be slowly poured into another beaker.