A solution containing a group –IV cation gives a precipitate on passing ${{\rm{H}}_{\rm{2}}}{\rm{S}}$. A solution of this precipitate in dil. HCl produces a white precipitate with basic NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is:
A. ${\rm{N}}{{\rm{i}}^{{\rm{2 + }}}}$
B. ${\rm{C}}{{\rm{o}}^{{\rm{2 + }}}}$
C. ${\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}$
D. ${\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}$
Answer
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Hint:We know that the qualitative method is one by which components of any given mixtures easily get isolated and separated. The detection or recognition of aqueous ions is obtained by the basic analysis.
Complete step-by-step answer:As we know that, the nickel ion, cobalt ion, and manganese ion do not give white precipitate when they react with hydrogen sulfide. Only zinc ion gives white residue or precipitate on the reaction of dihydrogen sulfide because ZnS and MnS are usually soluble in a cold solution of HCl, and NiS and CoS are generally insoluble. The reaction of zinc and dihydrogen sulfide is shown below.${\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}} + {{\rm{H}}_{\rm{2}}}{\rm{S}} \to {\rm{ZnS}} \downarrow + {\rm{2}}{{\rm{H}}^{\rm{ + }}}$
The zinc sulfide produces white precipitate of zinc hydroxide when the reaction with sodium hydroxide occurs. The reaction is shown below.
${\rm{ZnS}} + {\rm{2NaOH}} \to {\rm{Zn}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \downarrow + {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}$
The zinc sulfide also forms bluish white precipitate with a basic solution of potassium ferrocyanide. The chemical reaction is shown below.
${\rm{2ZnS}} + {{\rm{K}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right] \to {\rm{Z}}{{\rm{n}}_{\rm{2}}}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right] \downarrow + 2{{\rm{K}}_{\rm{2}}}{\rm{S}}$
So, the zinc ion gives all the reaction and desirable precipitate which are mentioned in the above question.
Therefore, the correct option for this question is D that is ${\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}$.
Note: This test also shows the confirmatory test for different ions obtained by the dissociation of compounds. Several tests are performed for detecting different ions of the compound. Every ion had a particular test for detection.
Complete step-by-step answer:As we know that, the nickel ion, cobalt ion, and manganese ion do not give white precipitate when they react with hydrogen sulfide. Only zinc ion gives white residue or precipitate on the reaction of dihydrogen sulfide because ZnS and MnS are usually soluble in a cold solution of HCl, and NiS and CoS are generally insoluble. The reaction of zinc and dihydrogen sulfide is shown below.${\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}} + {{\rm{H}}_{\rm{2}}}{\rm{S}} \to {\rm{ZnS}} \downarrow + {\rm{2}}{{\rm{H}}^{\rm{ + }}}$
The zinc sulfide produces white precipitate of zinc hydroxide when the reaction with sodium hydroxide occurs. The reaction is shown below.
${\rm{ZnS}} + {\rm{2NaOH}} \to {\rm{Zn}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \downarrow + {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}$
The zinc sulfide also forms bluish white precipitate with a basic solution of potassium ferrocyanide. The chemical reaction is shown below.
${\rm{2ZnS}} + {{\rm{K}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right] \to {\rm{Z}}{{\rm{n}}_{\rm{2}}}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_{\rm{6}}}} \right] \downarrow + 2{{\rm{K}}_{\rm{2}}}{\rm{S}}$
So, the zinc ion gives all the reaction and desirable precipitate which are mentioned in the above question.
Therefore, the correct option for this question is D that is ${\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}$.
Note: This test also shows the confirmatory test for different ions obtained by the dissociation of compounds. Several tests are performed for detecting different ions of the compound. Every ion had a particular test for detection.
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