Wet Test for Basic Radical Questions in English

(D) When water is added to a solution containing $Sb^{3+}$ or $Bi^{3+}$ ions in concentrated $HCl$,hydrolysis occurs.
$Sb^{3+} + H_2O + Cl^{-} \to SbOCl \text{ (white turbidity)} + 2H^+$
$Bi^{3+} + H_2O + Cl^{-} \to BiOCl \text{ (white turbidity)} + 2H^+$
This reaction occurs because the dilution decreases the concentration of $HCl$,shifting the equilibrium towards the formation of insoluble oxychlorides. Thus,the correct option is $(D)$.

(D) The reagent used to test for $Fe^{2+}$ ions is potassium ferricyanide,$K_3[Fe(CN)_6]$.
When $K_3[Fe(CN)_6]$ reacts with $Fe^{2+}$ ions,it forms a deep blue precipitate known as Turnbull's blue.
The chemical reaction is: $3Fe^{2+} + 2[Fe(CN)_6]^{3-} \to Fe_3[Fe(CN)_6]_2$ (Turnbull's blue).

(A) In the systematic qualitative analysis of basic radicals,the $V$ group radicals $(Ca^{2+}, Ba^{2+}, Sr^{2+})$ are precipitated as carbonates using $(NH_4)_2CO_3$ in the presence of $NH_4Cl$ and $NH_4OH$.
$Mg^{2+}$ does not precipitate with $(NH_4)_2CO_3$ in the presence of $NH_4Cl$ and $NH_4OH$ because the concentration of carbonate ions is suppressed by the common ion effect,and $MgCO_3$ is relatively more soluble.
Therefore,only $Ca^{2+}, Ba^{2+}, Sr^{2+}$ are precipitated.

(C) The sulfides of $As^{3+}, As^{5+}, Sb^{3+}, Sb^{5+}, Sn^{2+},$ and $Sn^{4+}$ are soluble in yellow ammonium sulfide $( (NH_4)_2S_x )$.
These sulfides react to form soluble thio-complexes (polysulphide complexes).
Among the given options,$SnS$ is soluble in yellow ammonium sulphide,while $CuS, CdS,$ and $PbS$ are insoluble.

(B) Concentrated $NaOH$ solution is used to separate metal ions based on the amphoteric nature of their hydroxides.
$Al^{3+}$ forms a soluble complex $[Al(OH)_4]^-$ with excess $NaOH$,while $Fe^{3+}$ forms an insoluble precipitate $Fe(OH)_3$ which does not dissolve in excess $NaOH$.
Therefore,a mixture of $Al^{3+}$ and $Fe^{3+}$ can be separated using $NaOH$.

(C) In an acidified solution,$H_2S$ provides a low concentration of $S^{2-}$ ions due to the common ion effect of $H^+$.
$Ag^+$ and $Cu^{2+}$ belong to Group $II$ of the qualitative analysis scheme,which precipitate as sulfides in acidic medium.
$Zn^{2+}$ belongs to Group $III$ (or $IV$ depending on the scheme) and requires a basic medium to precipitate as $ZnS$.
Therefore,both $Ag^+$ and $Cu^{2+}$ will form precipitates ($Ag_2S$ and $CuS$).
Since the question asks which forms a precipitate and both $Ag$ and $Cu$ are options,the most accurate answer in a multiple-choice context where only one is selected is often $Cu^{2+}$ or $Ag^+$. Given the options,$Cu^{2+}$ is a classic example of Group $II$ precipitation.

(D) When $H_2S$ gas is passed through an aqueous solution containing $Cu^{2+}$,$Pb^{2+}$,and $Zn^{2+}$ ions,all these metal ions form their respective insoluble sulfides.
$Cu^{2+} + H_2S \rightarrow CuS \downarrow + 2H^+$
$Pb^{2+} + H_2S \rightarrow PbS \downarrow + 2H^+$
$Zn^{2+} + H_2S \rightarrow ZnS \downarrow + 2H^+$
Since all three ions form precipitates with $H_2S$,the correct option is $(D)$.

(C) $H_2S$ is a weak acid that dissociates as $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
When $HCl$ is added,the concentration of $H^+$ ions increases significantly.
According to Le Chatelier's principle,this common ion effect suppresses the ionization of $H_2S$,thereby decreasing the concentration of $S^{2-}$ ions.
This ensures that only the solubility product $(K_{sp})$ of $II$ group metal sulfides is exceeded,allowing for their selective precipitation.

(D) In qualitative analysis,$Mn^{2+}$,$Ni^{2+}$,and $Cd^{2+}$ are precipitated as sulfides by $H_2S$ gas in the presence of $NH_3$ (or $NH_4OH$).
$Cd^{2+}$ is a group $II$ radical,while $Mn^{2+}$ and $Ni^{2+}$ are group $IV$ radicals.
$Ca^{2+}$ belongs to group $V$ and is not precipitated by $H_2S$ gas.
It is typically precipitated as $CaCO_3$ by the addition of $(NH_4)_2CO_3$ in the presence of $NH_4Cl$ and $NH_4OH$.

(B) In the qualitative analysis of basic radicals,$II$ group cations form sulfides upon passing $H_2S$ gas in the presence of dilute $HCl$.
The sulfides of $Pb^{2+}$,$Hg^{2+}$,and $Cu^{2+}$ are black in color.
However,$CdS$ (Cadmium sulfide) is yellow in color.
Therefore,if a black precipitate is obtained,the mixture does not contain $Cd^{2+}$.

(B) The reaction of bismuth nitrate with potassium iodide $(KI)$ produces a black precipitate of bismuth$(III)$ iodide $(BiI_3)$.
$Bi(NO_3)_3(aq) + 3KI(aq) \xrightarrow{\quad} BiI_3(s) + 3KNO_3(aq)$
Upon adding an excess of $KI$,the black precipitate dissolves to form a soluble complex,potassium tetraiodobismuthate$(III)$,which is orange in color.
$BiI_3(s) + KI(aq) \xrightarrow{\quad} K[BiI_4](aq)$
Thus,the cation is $Bi^{3+}$.

(A) In qualitative analysis,$Hg^{2+}$ and $Pb^{2+}$ both belong to Group $II$ and form black precipitates ($HgS$ and $PbS$ respectively) when $H_2S$ is passed in the presence of dilute $HCl$. Therefore,they cannot be distinguished by this reagent alone.

(B) The salt is $BaS_2O_3$ (Barium thiosulfate).
$1$. Reaction with dil. $HCl$: $BaS_2O_3 + 2HCl \to BaCl_2 + SO_2 \uparrow + S \downarrow$. The $SO_2$ gas has a pungent smell,and the sulfur $(S)$ forms a yellow precipitate.
$2$. Flame test: $Ba^{2+}$ ions impart a grassy green color to the flame.
$3$. Reaction with potassium chromate: $Ba^{2+} + K_2CrO_4 \to BaCrO_4 \downarrow + 2K^+$. $BaCrO_4$ is a yellow precipitate.

(C) The reaction with $HCl$ indicates the presence of $Pb^{2+}$,$Ag^+$,or $Hg_2^{2+}$ as they form white chloride precipitates.
When $NH_4OH$ is added to a mercurous salt $(Hg_2^{2+})$,it undergoes disproportionation to form a black precipitate of metallic mercury $(Hg)$ and amino-mercuric chloride $(Hg(NH_2)Cl)$.
The reaction is: $Hg_2^{2+} + 2NH_4OH \to Hg(l) + Hg(NH_2)Cl(s) + NH_4^+ + 2H_2O$.
This black precipitate does not dissolve in excess $NH_4OH$.

(A) The salt is $AgNO_3$. The reactions are as follows:
$1$. With excess ammonia: $AgNO_3 + 2NH_3 \to [Ag(NH_3)_2]NO_3$ (White precipitate).
$2$. With dilute $NaCl$: $AgNO_3 + NaCl \to AgCl \downarrow$ (White precipitate) $+ NaNO_3$.
$3$. With $H_2S$: $2AgNO_3 + H_2S \to Ag_2S \downarrow$ (Black precipitate) $+ 2HNO_3$.

(C) The correct answer is $C$.
$1$. When $H_2S$ is passed through the solution,the formation of a black precipitate indicates the presence of $Pb^{2+}$ ions,forming $PbS$ (lead sulfide).
$Pb^{2+} + H_2S \xrightarrow{\text{acidic}} PbS \downarrow \text{ (black ppt.)} + 2H^+$
$2$. The black precipitate $(PbS)$ dissolves in hot $HNO_3$ to form lead nitrate:
$3PbS + 8HNO_3 \to 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$
$3$. Upon adding conc. $H_2SO_4$ to the solution containing $Pb^{2+}$,a white precipitate of lead sulfate $(PbSO_4)$ is formed:
$Pb(NO_3)_2 + H_2SO_4 \to PbSO_4 \downarrow \text{ (white ppt.)} + 2HNO_3$

(A) The reaction of calomel $(Hg_2Cl_2)$ with ammonium hydroxide $(NH_4OH)$ is a disproportionation reaction.
$Hg_2Cl_2 + 2NH_4OH \to Hg + Hg(NH_2)Cl + NH_4Cl + 2H_2O$.
The product $Hg(NH_2)Cl$ is known as amino mercuric chloride,which is a black precipitate.
Therefore,the correct option is $A$.

(C) The correct answer is $(C)$.
When silver nitrate $(AgNO_3)$ is mixed with common salt $(NaCl)$,it forms a white precipitate of silver chloride $(AgCl)$.
The chemical reaction is: $AgNO_3(aq) + NaCl(aq) \to AgCl(s) + NaNO_3(aq)$.
This white precipitate of $AgCl$ is soluble in dilute ammonium hydroxide $(NH_4OH)$ due to the formation of a soluble complex: $AgCl(s) + 2NH_4OH(aq) \to [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.

(B) The correct statement is that from a mixed precipitate of $AgCl$ and $AgI$,ammonia solution dissolves only $AgCl$.
$AgCl$ is soluble in aqueous ammonia due to the formation of the complex $[Ag(NH_3)_2]Cl$,whereas $AgI$ is much less soluble and does not dissolve significantly in dilute ammonia.

(B) The correct answer is $(b)$ $CdS$.
$CdS$ (Cadmium sulphide) is yellow in colour.
$CuS$ (Copper sulphide) is black in colour.
$ZnS$ (Zinc sulphide) is white in colour.
$CoS$ (Cobalt sulphide) is black in colour.

(C) When $6 \ M \ NaOH$ is added in excess to a solution containing $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$:
$1$. $Fe^{3+}$ reacts with $OH^-$ to form a reddish-brown precipitate of $Fe(OH)_3$,which is insoluble in excess $NaOH$.
$2$. $Zn^{2+}$ reacts with $OH^-$ to form $Zn(OH)_2$,which dissolves in excess $NaOH$ to form the soluble complex $[Zn(OH)_4]^{2-}$.
$3$. $Cu^{2+}$ reacts with $OH^-$ to form a blue precipitate of $Cu(OH)_2$,which is insoluble in excess $NaOH$.
However,in the context of standard qualitative analysis,$Fe^{3+}$ is separated as $Fe(OH)_3$ from the others. Given the options,$6 \ M \ NaOH$ is the correct reagent to distinguish $Fe^{3+}$ (as a precipitate) from the amphoteric $Zn^{2+}$ (as a soluble complex).

(D) and $(c)$ are correct.
$Fe^{2+}$ ions react with potassium ferricyanide $(K_3[Fe(CN)_6])$ to form a blue precipitate known as Turnbull's blue,which is potassium ferric ferrocyanide,$KFe[Fe(CN)_6]$.
$Fe^{3+}$ ions react with potassium thiocyanate $(KSCN)$ to form a blood-red complex,$[Fe(SCN)(H_2O)_5]^{2+}$,or simply $Fe(SCN)_3$ in solution,which is a characteristic test for ferric ions.

(B) $Ni^{2+}$ reacts with dimethylglyoxime $(DMG)$ in the presence of ammonium hydroxide to form a cherry-red precipitate of nickel-dimethylglyoxime complex,$[Ni(DMG)_2]$.
This reaction is highly specific for the identification of $Ni^{2+}$ ions.

(C) Lead sulfate $(PbSO_4)$ is insoluble in water. It dissolves in an ammoniacal solution of ammonium acetate due to the formation of a soluble complex,lead acetate.

(A) In qualitative inorganic analysis,$Pb^{2+}$ ions belong to Group-$I$ and Group-$II$.
$Pb^{2+}$ reacts with $HCl$ to form $PbCl_2$ (white precipitate) in Group-$I$.
$Pb^{2+}$ also reacts with $H_2S$ in the presence of $HCl$ to form $PbS$ (black precipitate) in Group-$II$.

(B) The salt is $Pb(NO_3)_2$. The reactions are as follows:
$1$. With ammonia solution: $Pb(NO_3)_2 + 2NH_4OH \to Pb(OH)_2 \downarrow (\text{white}) + 2NH_4NO_3$
$2$. With dilute $NaCl$: $Pb(NO_3)_2 + 2NaCl \to PbCl_2 \downarrow (\text{white}) + 2NaNO_3$
$3$. With $H_2S$: $Pb(NO_3)_2 + H_2S \to PbS \downarrow (\text{black}) + 2HNO_3$

(D) In qualitative inorganic analysis,$Cu^{2+}$,$Hg^{2+}$,and $Bi^{3+}$ belong to Group-$II$ cations. These cations precipitate as their respective sulfides $(CuS, HgS, Bi_2S_3)$ in the presence of $H_2S$ gas in an acidic medium $(HCl)$.
$Co^{2+}$ belongs to Group-$IV$ cations. Group-$IV$ sulfides (like $CoS, NiS, MnS, ZnS$) are soluble in acidic medium $(HCl)$ due to their higher solubility product $(K_{sp})$ values compared to Group-$II$ sulfides.
Therefore,$CoS$ will not precipitate in the presence of $HCl$.

(B) The color of metal sulfides is a key identification feature in qualitative inorganic analysis.
$ZnS$ (Zinc sulfide) is white.
$CdS$ (Cadmium sulfide) is bright yellow.
$NiS$ (Nickel sulfide) is black.
$PbS$ (Lead sulfide) is black.
Therefore,the correct option is $CdS$.

(D) In qualitative analysis,$H_2S$ gas in the presence of dilute $HCl$ (acidic medium) provides a low concentration of $S^{2-}$ ions due to the common ion effect.
This low concentration of $S^{2-}$ is sufficient to exceed the solubility product $(K_{sp})$ of group $II$ sulfides like $CuS$,but not enough to precipitate group $III$ or $IV$ sulfides like $NiS$ or $ZnS$.
Therefore,only $Cu^{2+}$ ions will form a precipitate of $CuS$ in an acidic medium.

(C) Concentrated $NaOH$ is used to separate metal ions based on the amphoteric nature of their hydroxides.
$Fe^{3+}$ forms a brown precipitate of $Fe(OH)_3$ which is insoluble in excess $NaOH$.
$Cr^{3+}$ forms $Cr(OH)_3$ which is amphoteric and dissolves in excess $NaOH$ to form a soluble complex,$[Cr(OH)_4]^-$.
Therefore,$Cr^{3+}$ and $Fe^{3+}$ can be separated using concentrated $NaOH$.

(D) The $Fe^{+2}$ ion reacts with potassium ferricyanide,$K_3[Fe(CN)_6]$,to form a deep blue precipitate known as Turnbull's blue.
The chemical reaction is: $3Fe^{+2} + 2[Fe(CN)_6]^{-3} \rightarrow Fe_3[Fe(CN)_6]_2$ (Turnbull's blue).
Therefore,$K_3[Fe(CN)_6]$ is the specific reagent used to identify $Fe^{+2}$ ions.

(A) When $HCl$ is added to an aqueous solution containing $Ag^{+}$ ions,it reacts to form silver chloride $(AgCl)$,which is a white precipitate.
The chemical reaction is: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s) \downarrow$ (white precipitate).
Other ions like $Mg^{2+}$,$Zn^{2+}$,and $Cd^{2+}$ do not form insoluble chlorides under these conditions.

(A) $Fe^{3+}$ ions react with $NH_4SCN$ (Ammonium thiocyanate) to form a blood-red colored complex,$[Fe(SCN)(H_2O)_5]^{2+}$,whereas $Fe^{2+}$ ions do not produce this color.
Therefore,$NH_4SCN$ is used to distinguish between $Fe^{2+}$ and $Fe^{3+}$ ions.

(A) In qualitative analysis,the solubility of metal sulfides in dilute $HNO_3$ depends on the solubility product $(K_{sp})$ of the sulfide and the oxidizing power of the acid.
$HgS$ (mercuric sulfide) has an extremely low $K_{sp}$ value $(10^{-52})$,which makes it insoluble in hot dilute $HNO_3$.
Other sulfides like $PbS$,$CuS$,and $CdS$ have higher $K_{sp}$ values and are soluble in hot dilute $HNO_3$ because the acid oxidizes the $S^{2-}$ ion to elemental sulfur,shifting the equilibrium forward.

(A) In qualitative inorganic analysis,$Pb^{2+}$,$Ag^{+}$,$Sn^{2+}$,and $Cu^{2+}$ are all classified as Group $I$ or Group $II$ cations.
$Ag^{+}$ is precipitated as $AgCl$ by $HCl$ (Group $I$).
$Pb^{2+}$ is precipitated as $PbCl_2$ by $HCl$ (Group $I$) and as $PbS$ by $H_2S$ (Group $II$).
$Cu^{2+}$ and $Sn^{2+}$ are precipitated as sulfides by $H_2S$ in the presence of $HCl$ (Group $II$).
However,the question implies which ion is not precipitated by the combination of $HCl$ and $H_2S$ under standard conditions. All listed ions are precipitated by these reagents in their respective groups. Given the standard options,there is a technical error in the question as all these ions are precipitated by these reagents. If we assume the question asks for an ion that does not belong to these groups,none of the options fit. However,if this is a multiple-choice question where one must be selected,it is likely flawed. Based on standard analytical chemistry,all these ions are precipitated.

(A) In qualitative analysis,$H_2S$ in the presence of $0.3 \, M \, HCl$ is used to precipitate Group-$II$ cations.
Group-$II$ cations include $Hg^{2+}, Pb^{2+}, Bi^{3+}, Cu^{2+}, Cd^{2+}, As^{3+}, Sb^{3+},$ and $Sn^{4+}$.
Among the given ions $(Hg^{2+}, Cd^{2+}, Sr^{2+}, Fe^{2+}, Cu^{2+})$,the ions $Hg^{2+}, Cd^{2+},$ and $Cu^{2+}$ belong to Group-$II$.
$Sr^{2+}$ belongs to Group-$V$ and $Fe^{2+}$ belongs to Group-$III$,which do not precipitate under these acidic conditions.
Therefore,$Hg^{2+}, Cd^{2+},$ and $Cu^{2+}$ will precipitate as their respective sulfides.

(C) When lead acetate $(Pb(CH_3COO)_2)$ reacts with hydrogen sulfide $(H_2S)$,it forms lead sulfide $(PbS)$ as a precipitate.
The chemical reaction is: $Pb(CH_3COO)_2 + H_2S \rightarrow PbS \downarrow + 2CH_3COOH$.
Lead sulfide $(PbS)$ is a black-colored precipitate.

(A) In the qualitative analysis of inorganic salts,cations are classified into different groups based on their solubility products and the common ion effect.
Group $IV$ cations $(Co^{2+}, Ni^{2+}, Mn^{2+}, Zn^{2+})$ are precipitated as their sulfides.
The reagent used for Group $IV$ is $H_2S$ gas in the presence of an ammoniacal solution (buffered with $NH_4Cl$ and $NH_4OH$).
This provides a high concentration of $S^{2-}$ ions,which is necessary to exceed the solubility product of Group $IV$ sulfides.

(B) When $SnCl_2$ is added to $HgCl_2$,a white precipitate of $Hg_2Cl_2$ is formed first.
$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 (white) + SnCl_4$
On adding excess $SnCl_2$,the white precipitate turns grey/black due to the formation of metallic mercury.
$Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg (grey/black) + SnCl_4$
Therefore,the precipitate changes from white to grey/black.

(B) When $H_2S$ gas is passed through a solution containing $Zn^{2+}$ ions,it forms $ZnS$ (Zinc sulfide),which is a white precipitate.
$Zn^{2+} (aq) + H_2S (g) \rightarrow ZnS (s) + 2H^+ (aq)$
Therefore,the solution contains $Zn^{2+}$ ions.

(B) The sulfides of group $IIA$ (like $As_2S_3$,$Sb_2S_3$,and $SnS_2$) are soluble in ammonium sulfide $(NH_4)_2S$ or ammonium carbonate $(NH_4)_2CO_3$ due to the formation of thio-salts.
$As_2S_3 + 3(NH_4)_2CO_3 \rightarrow (NH_4)_3AsS_3 + (NH_4)_3AsO_3 + 3CO_2$.
$CdS$ is a group $IIB$ sulfide and is insoluble in $(NH_4)_2CO_3$.

(A) When $H_2S$ gas is passed through a solution containing metal ions,it forms metal sulfides.
$AgNO_3$ reacts with $H_2S$ to form $Ag_2S$,which is a black precipitate.
$2Ag^+ + H_2S \rightarrow Ag_2S(s) + 2H^+$.
Copper nitrate $(Cu(NO_3)_2)$ also forms a black precipitate of $CuS$ $(Cu^{2+} + H_2S \rightarrow CuS(s) + 2H^+)$.
However,in standard qualitative analysis,$Ag_2S$ is a classic example of a black precipitate formed in acidic media. Both $A$ and $D$ are technically correct,but $AgNO_3$ is the most common textbook example for this reaction.

(B) Silver $(Ag^+)$ and lead $(Pb^{2+})$ ions are both members of Group-$I$ in qualitative inorganic analysis.
They are precipitated as chlorides by adding dilute $HCl$.
To distinguish between them,the precipitate is treated with hot water.
Lead chloride $(PbCl_2)$ is soluble in hot water,whereas silver chloride $(AgCl)$ is insoluble.
Therefore,hot dilute $HCl$ or hot water is used to separate and distinguish these ions.

(A) The reaction of $Mg^{2+}$ ions with $NH_4OH$ and $Na_2HPO_4$ is a standard test for magnesium.
The chemical equation is:
$MgSO_4 + NH_4OH + Na_2HPO_4 \rightarrow Mg(NH_4)PO_4 \downarrow + Na_2SO_4 + H_2O$.
The white crystalline precipitate formed is magnesium ammonium phosphate,$Mg(NH_4)PO_4$.

(A, B, C, D) Aqueous $FeCl_3$ solution is yellow in color.
When $K_4[Fe(CN)_6]$ is added to $FeCl_3$,it forms a blue-colored precipitate of Prussian blue,$Fe_4[Fe(CN)_6]_3$.
When $H_2S$ is added,it reduces $Fe^{3+}$ to $Fe^{2+}$ and forms a milky white precipitate of sulfur.
When $NH_4CNS$ or $KCNS$ is added,it forms a blood-red colored complex,$[Fe(CNS)(H_2O)_5]^{2+}$.
Since all the given options result in a color change,this is a multiple-correct type question. However,in standard testing contexts,$NH_4CNS$ or $KCNS$ are the most common reagents used to test for $Fe^{3+}$ ions via a distinct color change.

(A) In qualitative analysis,the precipitation of group $II$ sulfides depends on the concentration of $H_2S$ and the acidity of the solution.
$HgS$ (mercuric sulfide) has a very low solubility product $(K_{sp})$,which makes it the least soluble among the given sulfides.
When an acidic solution containing $Hg^{2+}$ ions is treated with $H_2S$,$HgS$ precipitates.
Upon dilution,the concentration of $H^+$ ions decreases,but $HgS$ remains precipitated due to its extremely low $K_{sp}$ value,ensuring complete precipitation compared to others like $PbS$,$CdS$,or $CuS$ which may redissolve or require specific conditions.

(B) In the systematic qualitative analysis of cations,the group reagents are used to precipitate specific groups of cations.
Group $IV$ cations ($Co^{2+}$,$Ni^{2+}$,$Mn^{2+}$,$Zn^{2+}$) are precipitated as their sulfides.
The group reagent for Group $IV$ is $H_2S$ gas in the presence of $NH_4Cl$ and $NH_4OH$ (ammoniacal medium).
Therefore,the correct combination is $NH_4Cl + NH_4OH + H_2S$.

(B) In qualitative inorganic analysis,the group $III$ cations are precipitated as hydroxides using $NH_4Cl$ and $NH_4OH$ (aqueous $NH_3$).
$Al^{3+}$ belongs to group $III$ and forms $Al(OH)_3$ precipitate in the presence of these reagents.
$Ca^{2+}$ is a group $V$ cation,$Mg^{2+}$ is a group $VI$ cation,and $Zn^{2+}$ is a group $III$ cation but it is usually precipitated with $H_2S$ in the presence of $NH_4OH$ (group $III$ $B$). However,$Al^{3+}$ is the standard representative for group $III$ $A$ precipitation by $NH_4OH$.

(B) The reaction of $BaS_2O_3$ with dilute $HCl$ produces $SO_2$ gas (pungent smell) and elemental sulfur (yellow crystals).
$BaS_2O_3 + 2HCl \to BaCl_2 + SO_2 \uparrow + S \downarrow + H_2O$
$Ba^{2+}$ ions impart a characteristic apple-green color to the flame.
$Ba^{2+}$ reacts with potassium chromate $(K_2CrO_4)$ to form a yellow precipitate of barium chromate $(BaCrO_4)$:
$Ba^{2+} + K_2CrO_4 \to BaCrO_4 \downarrow + 2K^+$
Thus,the salt is $BaS_2O_3$.