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To ensure that group 4 cations are not precipitated with group 2, we add dil. $\ce{HCl}$ to ensure only group 2 sulfides of lower $K_\mathrm{sp}$ are precipitated.
For group 4, we add $\ce{NH4OH}$ along with $\ce{H2S}$ to shift the reaction in the forward direction since group 4 sulfides have higher $K_\mathrm{sp}$ which makes sense but why aren't group 2 sulfides of lower $K_\mathrm{sp}$ precipitated along with group 4?
Nomenclature aside, the full procedure calls for first making the acidic sulfide solution, precipitating the metals associated with that group, and then decanting the liquid and adding ammonia solution* to capture the ammonia-sulfide group. Thus the acid-sulfide precipitates don't stay in solution, rather they are already gone.
*Ammonia solution, not ammonium hydroxide. Ammonia dissolved in water is only a weak base, so the major species are $\ce{H2O}$ and $\ce{NH3}$, not $\ce{NH4^+}$ and $\ce{OH^-}$.
I get the confusion of grouping the cations. Some say there are 6 groups, other say 4 group but at the end of the day, it's the matter of convention.
Let's start with bunch of salts of unknown cations. You add dil. $\ce{HCl}$. If it forms insoluble chloride, then the cations present are $\ce{Ag+, Hg2^2+ and Pb^2+}$. If no precipitate is formed, add $\ce{H2S}$ (note the pH is already acidic). Sulfide having low $K_\mathrm{sp}$ will form precipitate out in acidic solution (hence the group reagent you are observing is dil. $\ce{\bf{HCl}}$ + $\ce{H2S}$). If not, then make the solution basic with ammonia solution + $\ce{NH4Cl}$ (this is your group reagent). Sulfides having larger $K_\mathrm{sp}$ values such as ($\ce{ZnS, NiS, CoS}$ and $\ce{MnS}$ precipitate. Also, at basic medium, $\ce{Al^3+, Fe^3+}$ and $\ce{Cr^3+}$ form insoluble hydroxides and are separated from the solution. now comes cations which formed soluble sulfides and chlorides. They can be separated by forming insoluble carbonates (note the solution is basic so the group reagent is $\ce{\bf{NH_3(aq.)+ NH_4Cl}}$ + $\ce{(NH4)2CO3}$). The cations that are left are determined through flame test.
So, in my opinion, the group can be done in following way:
\begin{array}{c|c} \mathbf{Group} & \mathbf{Cation} & \mathbf{Group~reagent} & \mathbf{Inference}\\\hline \text{Group~I} & \text{$\ce{Ag+, Hg2^2+, Pb^2+}$} & \text{dil. HCl} & \text{insoluble chlorides}\\ \text{Group~II} & \text{$\ce{Hg^2+, Pb^2+, Cu^2+, Bi^3+, Cd^2+, As^3+, Sb^3+ , Sn^4+}$} & \text{$\ce{dil. HCl + H2S}$} & \text{insoluble sulfide}\\ \text{Group~IIIA} & \text{$\ce{Ni^2+, Zn^2+, Mn^2+}$} & \text{ammonia soln. + $\ce{NH4Cl}$} & \text{insoluble sulfide}\\ \text{Group~IIIB} & \text{$\ce{Al^3+, Fe^3+, Co^2+, Cr^3+}$} & \text{ammonia soln. + $\ce{NH4Cl}$} & \text{insoluble hydroxides}\\ \text{Group~IV} & \text{$\ce{Ca^2+, Sr^2+, Ba^2+}$} & \text{$\ce{NH3(aq.) + NH4Cl + (NH4)2CO3}$} & \text{insoluble carbonates}\\ \text{Group~V} & \text{$\ce{Mg^2+, Na+, K+}$}& \text{flame test} & \text{color}\\ \text{Group~0} & \text{$\ce{NH4+}$}& \text{-} & \text{since it doesn't have any group reagent, I don't think it needs a separate group}\end{array}
This was my previous grouping, but I suppose I will add a comment on the updated table.
