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Question: The pair(s) of ions where both the ions are precipitated upon passing \({{H}_{2}}S\) gas in presence...

The pair(s) of ions where both the ions are precipitated upon passing H2S{{H}_{2}}S gas in presence of dilute HCl, is(are) :
THIS QUESTION HAS MULTIPLE CORRECT OPTIONS
A. Ba+2,Zn+2B{{a}^{+2}},Z{{n}^{+2}}
B. Bi3+,Ke2+B{{i}^{3+}},K{{e}^{2+}}
C. Cu2+,Pb2+C{{u}^{2+}},P{{b}^{2+}}
D. Hg2+,Bi2+H{{g}^{2+}},B{{i}^{2+}}

Explanation

Solution

H2S{{H}_{2}}S dissociates to give the sulphide ion and HCl dissociates to give the chloride ion. The metal ion that can react with both the anions to form a compound will be the correct answer. So we need to consider both the gases to solve this question.

Complete answer:
-To solve this question, we need to see which of the following ions can be precipitated with the H2S{{H}_{2}}S gas. If the metals ions are able to form the precipitates of sulphides, then they are correct else they are wrong.
-One important thing to notice here is that we are given HCl also along with H2S{{H}_{2}}S gas. Now, HCl is an acid and it reacts with metals to liberate hydrogen gas and forms the respective metal chlorides.
-The formation of metal chlorides is not desired here. We need those metals only which can precipitate in the form of sulphides and not the chlorides. So we need to check for that option.
-The ions present on dissociation of the gases will be H+,S−2,Cl−{{H}^{+}},{{S}^{-2}},C{{l}^{-}} ions. Now they have to react with a set of metal ions so as to produce only the metal sulphides and not metal chlorides. Such sets of metal ions will be our answer.
-The effect shown here in this question is called the common ion effect. There are two different negative ions called anions present in the solution and the metals can react with both the ions. The tendency of reacting with the chloride ions is more as the sulphide ion is more stable than chloride ion.
-Only the group 2 elements are able to counterbalance this common ion effect and form the sulphides instead of forming the chlorides. Thus the solubility of chlorides is suppressed by the group 2 elements.
-As the solubility of chlorides decreases, there are more sulphide ions which can react with the metal ions to form the sulphides and precipitate in that form. That will get us our desired result.
-The radicals of group 3, 4 and 5 are not able to overcome the common ion effect. They also react and precipitate but not as sulphides but as chlorides. In the question, we are asked the sulphides and not the chlorides. So they will not be included in our answer.
-The metal ions of group 2 are Cu2+,Pb2+C{{u}^{2+}},P{{b}^{2+}}, Hg2+,Bi2+H{{g}^{2+}},B{{i}^{2+}}. The other ions do not belong to group 2 on losing their electrons and so they cannot form sulphides. They form the chlorides and precipitate in that form only. The reactions can be shown as
Cu2++H2S→CuS↓+H2 Pb2++H2S→PbS↓+H2 Hg2++H2S→HgS↓+H2 Bi2++H2S→BiS↓+H2 \begin{aligned} & C{{u}^{2+}}+{{H}_{2}}S\to CuS\downarrow +{{H}_{2}} \\\ & \text{P}{{\text{b}}^{2+}}+{{H}_{2}}S\to PbS\downarrow +{{H}_{2}} \\\ & H{{g}^{2+}}+{{H}_{2}}S\to HgS\downarrow +{{H}_{2}} \\\ & B{{i}^{2+}}+{{H}_{2}}S\to BiS\downarrow +{{H}_{2}} \\\ \end{aligned}

Therefore the correct options are C and D.

Note:
The sulphides are very less soluble in water and so they occur as precipitates. The compounds which are soluble in water dissolve in it and are not found as precipitates. All the sulphides except for the sulphides of the metals barium, calcium, magnesium, sodium, ammonium and potassium are insoluble in water.