Question

Which one amongst the following pairs of ions cannot be separated by ${H_2}S$ in dilute $HCl$ ?
 $B{i^{3 + }},\,S{n^{4 + }}$
 $A{l^{3 + }},\,H{g^{2 + }}$
 $Z{n^{2 + }},\,C{u^{2 + }}$
 $N{i^{2 + }},\,C{u^{2 + }}$

Answer 1

Hint: To answer this question we will first see which group each cation belongs to and whether it is precipitated by ${H_2}S$ in dilute $HCl$ . If the given pair of cations belong to the same group then they cannot be separated as both of them will get precipitated. Whereas if the pair of cations do not belong to the same group then one of them will get precipitated while the other does not. In this way we can decide whether the 2 cations can be separated or not.

Complete Step by Step Solution:
To find which pair of ions can be separated by ${H_2}S$ in dilute $HCl$ , we will first see which group the given cations belong to. $B{i^{3 + }},\,H{g^{2 + }},\,C{u^{2 + }},\,\& \,S{n^{4 + }}$ are the members of group number 2 while $Z{n^{2 + }},\,A{l^{3 + }}\,\& \,N{i^{2 + }}$ belong to group number 4.

Out of the 4 options given the question, the last 3 options have a cation pair of different groups. As a result when they react with ${H_2}S$ in dilute $HCl$ one of them precipitates while the other does not. Thus we can say that the 2 cations can be separated upon reaction with ${H_2}S$ in dilute $HCl$ .

In the option B. $H{g^{2 + }}$ will precipitate while $A{l^{3 + }}$ will not.
In the option C. $C{u^{2 + }}$ will precipitate while $Z{n^{2 + }}$ will not.
Similarly, in option D. $C{u^{2 + }}$ will precipitate and $N{i^{2 + }}$ will not.

Now, in option A. both the cations $\left( {B{i^{3 + }},\,S{n^{4 + }}} \right)$ are from group 2 and upon reaction with ${H_2}S$ in dilute $HCl$ will precipitate and thus are inseparable by this reaction. They both precipitate as their sulphides: $B{i_2}{S_3},\,Sn{S_2}$ and are black and yellow in colour respectively.

Note: While answering all such types of questions, always mention the group number to which the ion belongs to. Since, $B{i^{3 + }},\,S{n^{4 + }}$ were not separated by ${H_2}S$ in dilute $HCl$ , it does not mean that they cannot be separated at all. There are various other methods for separating cations and so $B{i^{3 + }},\,S{n^{4 + }}$ can be separated by some other test.