Wet Test for Basic Radical Questions in English

(B) $PbCl_2$ (Lead$(II)$ chloride) is known to be sparingly soluble in cold water but becomes significantly more soluble in hot water. This property is commonly used in qualitative inorganic analysis to separate lead ions from other metal ions.

(D) The test for zinc ions $(Zn^{2+})$ involves heating the substance with cobalt nitrate $(Co(NO_3)_2)$ on a charcoal cavity.
When zinc compounds are heated with $Co(NO_3)_2$,they form cobalt zincate $(CoZnO_2)$,which is known as Rinmann's green.
$ZnO + Co(NO_3)_2 \rightarrow CoZnO_2 + 2NO_2 + \frac{1}{2}O_2$.
Since $ZnO$,$3Zn(OH)_2 \cdot ZnCO_3$ (basic zinc carbonate),and $ZnSO_4$ all contain zinc,they can all produce $ZnO$ upon heating,which subsequently reacts to form Rinmann's green.
Therefore,all of these compounds can produce the green color.

(D) Mohr's salt is a double salt with the chemical formula $(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O$.
In an aqueous solution,it dissociates completely into its constituent ions:
$(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O (aq) \rightarrow 2NH_4^+ (aq) + Fe^{2+} (aq) + 2SO_4^{2-} (aq) + 6H_2O (l)$.
Since all these ions ($NH_4^+$,$Fe^{2+}$,and $SO_4^{2-}$) are present in the solution,the aqueous solution of Mohr's salt gives the test for all of them.

(A) $Ni^{+2}$ reacts with dimethylglyoxime $(DMG)$ in the presence of ammonium hydroxide to form a red precipitate of $[Ni(dmg)_2]$.
$Cu^{+2}$ reacts with excess ammonium hydroxide $(NH_4OH)$ to form a deep blue colored complex $[Cu(NH_3)_4]^{+2}$.

(D) The salt $A$ is $AgNO_3$.
$1$. Reaction with $Cu$: $2AgNO_3 + Cu \rightarrow Cu(NO_3)_2 + 2Ag$. The $Cu^{2+}$ ions formed in the solution impart a blue color.
$2$. Reaction with $HCl$: $AgNO_3 + HCl \rightarrow AgCl \downarrow + HNO_3$. $AgCl$ is a white precipitate.
$3$. Reaction with $Conc. H_2SO_4$: $AgNO_3 + H_2SO_4 \rightarrow HNO_3 + AgHSO_4$. The $HNO_3$ decomposes to release $NO_2$ gas,which appears as brown fumes.

(B) The chromyl chloride test is used for the detection of chloride ions $(Cl^-)$ in inorganic salts.
In this test,the chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$ to produce red vapors of chromyl chloride $(CrO_2Cl_2)$.
However,this test fails for certain chlorides such as $HgCl_2$,$SnCl_2$,$AgCl$,$PbCl_2$,and $SbCl_3$ because they are either covalent in nature or do not react to form the volatile chromyl chloride.
Therefore,$HgCl_2$ will not give a positive chromyl chloride test.

(D) The reaction $Fe(OH)_3 \downarrow + NaOH(\text{excess}) \longrightarrow \text{No reaction}$ indicates that $Fe(OH)_3$ is insoluble in excess $NaOH$ solution.
Since no products are formed,this corresponds to the category of 'no reaction'.
Therefore,option $D$ is correct.

(A) The reaction between $FeCl_3$ and excess $NH_3$ (aqueous ammonia) is a precipitation reaction.
$FeCl_3 + 3NH_3 + 3H_2O \longrightarrow Fe(OH)_3 \downarrow + 3NH_4Cl$
$Fe(OH)_3$ is a reddish-brown coloured precipitate.
Therefore,the reaction corresponds to the formation of a coloured precipitate.

(D) The reaction is: $2KCN + Pb(NO_3)_2 \longrightarrow Pb(CN)_2 \downarrow + 2KNO_3$.
In this reaction,$Pb(CN)_2$ is formed as a precipitate.
$Pb(CN)_2$ is a white-coloured precipitate.
Therefore,the correct assignment is $D$ for white precipitate.

(C) The reaction between $NiCl_2$ and $NaNO_3$ involves the exchange of ions: $NiCl_2 + 2NaNO_3 \longrightarrow Ni(NO_3)_2 + 2NaCl$.
Since all the products ($Ni(NO_3)_2$ and $NaCl$) are soluble in water,no precipitate is formed,and there is no observable chemical change.
Therefore,the mixture remains a clear/colourless solution (or simply a solution without precipitate).
Thus,the correct assignment is $C$.

(D) The reaction is: $(NH_4)_2SO_4 + Sr(OH)_2 \longrightarrow SrSO_4 \downarrow + 2NH_3 \uparrow + 2H_2O$.
In this reaction,$SrSO_4$ (Strontium sulfate) is formed as a precipitate.
$SrSO_4$ is known to be a white crystalline solid.
Therefore,the correct assignment for this reaction is $D$ (for white ppt.).

(B) Mohr's salt is $(NH_{4})_{2}SO_{4} \cdot FeSO_{4} \cdot 6H_{2}O$.
When Mohr's salt reacts with $NaOH$ solution,the ammonium ion $(NH_{4}^{+})$ reacts with the hydroxide ion $(OH^{-})$ to release ammonia gas $(NH_{3} \uparrow)$.
The reaction is: $(NH_{4})_{2}SO_{4} + 2NaOH \to Na_{2}SO_{4} + 2H_{2}O + 2NH_{3} \uparrow$.

(D) The reaction sequence describes the qualitative analysis of Group $V$ cations $(Ca^{2+}, Sr^{2+}, Ba^{2+})$.
$1$. When $NH_4Cl(s)$ and $(NH_4)_2CO_3$ solution are added,$Ca^{2+}, Sr^{2+},$ and $Ba^{2+}$ form carbonate precipitates $(Q = CaCO_3, SrCO_3, BaCO_3)$. $Mg^{2+}$ does not precipitate in the presence of $NH_4Cl$.
$2$. These carbonate precipitates are dissolved in $CH_3COOH$ to form their respective acetates.
$3$. Upon adding $(NH_4)_2C_2O_4$,$Ca^{2+}$ and $Sr^{2+}$ form oxalate precipitates $(R = CaC_2O_4, SrC_2O_4)$,whereas $Ba^{2+}$ remains in the filtrate because $BaC_2O_4$ is relatively more soluble.
Therefore,$Ba^{2+}$ forms precipitate $Q$ but does not form precipitate $R$ under these conditions.

(B) The reaction of $Cu^{2+}$ ions with potassium ferrocyanide $(K_4[Fe(CN)_6])$ produces a chocolate brown precipitate of copper$(II)$ hexacyanoferrate$(II)$,$Cu_2[Fe(CN)_6]$.
The reaction is: $2CuCl_2(aq.) + K_4[Fe(CN)_6](aq.) \rightarrow Cu_2[Fe(CN)_6](s) + 4KCl(aq.)$
The reaction of $Cl^-$ ions with silver nitrate $(AgNO_3)$ produces a white precipitate of silver chloride $(AgCl)$,which is insoluble in dilute $HNO_3$.
The reaction is: $CuCl_2(aq.) + 2AgNO_3(aq.) \rightarrow 2AgCl(s) + Cu(NO_3)_2(aq.)$
Therefore,$X$ is $CuCl_2$.

(D) $Fe^{2+}$ ions react with $NaOH$ to form a dirty green precipitate of $Fe(OH)_2$.
$Fe^{2+} (aq) + 2OH^- (aq) \rightarrow Fe(OH)_2 (s) \downarrow$ (Dirty green ppt.)
$Fe^{2+}$ ions react with $NH_3$ solution to form a dirty green precipitate of $Fe(OH)_2$.
$Fe^{2+} (aq) + 2NH_3 (aq) + 2H_2O (l) \rightarrow Fe(OH)_2 (s) \downarrow + 2NH_4^+ (aq)$
$Fe^{2+}$ ions react with $Na_2CO_3$ to form a precipitate of $FeCO_3$.
$Fe^{2+} (aq) + CO_3^{2-} (aq) \rightarrow FeCO_3 (s) \downarrow$
Since all the given reagents produce a precipitate with $FeSO_4$,the correct option is $D$.

(A) When excess $NaOH$ solution is added to a mixture of $Fe^{3+}$ and $Zn^{2+}$ ions:
$1$. $Fe^{3+}$ reacts with $NaOH$ to form a brown precipitate of $Fe(OH)_3$,which is insoluble in excess $NaOH$.
$2$. $Zn^{2+}$ reacts with $NaOH$ to form $Zn(OH)_2$,which then dissolves in excess $NaOH$ to form a soluble complex,$[Zn(OH)_4]^{2-}$.
$3$. Thus,the mixture can be separated by filtration,where $Fe(OH)_3$ remains as a precipitate and $[Zn(OH)_4]^{2-}$ passes into the filtrate.
Therefore,the correct pair is $Fe^{3+}(aq.), Zn^{2+}(aq.)$.

(D) The reaction between finely powdered $Fe$ and dilute $HCl$ is: $Fe + 2HCl(dil.) \to FeCl_2 + H_2\uparrow$. Here,$P$ is $FeCl_2$ and $Q$ is $H_2$.
$FeCl_2$ contains $Fe^{2+}$ ions.
$1$. With $AgNO_3$: $FeCl_2 + 2AgNO_3 \to 2AgCl(s) + Fe(NO_3)_2$. $AgCl$ is a white precipitate.
$2$. With $K_3[Fe(CN)_6]$: $3FeCl_2 + 2K_3[Fe(CN)_6] \to Fe_3[Fe(CN)_6]_2 + 6KCl$. This forms a blue precipitate (Turnbull's blue).
$3$. With $(NH_4)_2S$: $FeCl_2 + (NH_4)_2S \to FeS(s) + 2NH_4Cl$. $FeS$ is a black precipitate.
$4$. With $NH_4Cl + NH_4OH$: $Fe^{2+}$ ions do not form a precipitate with $NH_4OH$ in the presence of $NH_4Cl$ because the concentration of $OH^-$ ions is suppressed by the common ion effect of $NH_4^+$,which is insufficient to exceed the solubility product of $Fe(OH)_2$.

(D) $Fe^{2+}$ ions do not react with thiocyanate ions $(SCN^-)$ to form a colored complex. Therefore,statement $(d)$ is incorrect.
$(a)$ $Fe^{3+} + K_3[Fe(CN)_6] \to Fe[Fe(CN)_6]$ (Brown color solution).
$(b)$ $Fe^{2+} + K_3[Fe(CN)_6] \to Fe_3[Fe(CN)_6]_2$ (Turnbull's blue precipitate).
$(c)$ $Fe^{3+} + SCN^- \to [Fe(SCN)]^{2+}$ (Blood red color solution).
$(d)$ $Fe^{2+} + SCN^- \to \text{No reaction}$.

(A) In a neutral medium,$H_2S$ acts as a reducing agent for $Fe^{3+}$ ions.
The reaction is: $2Fe^{3+}(aq) + H_2S(g) \rightarrow 2Fe^{2+}(aq) + S(s) + 2H^+(aq)$.
Since $Fe^{2+}$ ions do not form a precipitate with $H_2S$ in a neutral medium,$Fe^{3+}$ does not yield a metal sulfide precipitate.
Conversely,$Cu^{2+}$,$Bi^{3+}$,and $Ag^+$ form their respective insoluble sulfides ($CuS$,$Bi_2S_3$,$Ag_2S$) in the presence of $H_2S$.

(D) The reaction $X_2S_n + H_2O \to X(OH)_n + H_2S$ indicates that the sulfide $X_2S_n$ undergoes hydrolysis to produce $H_2S$ gas $(Y)$.
$H_2S$ gas reacts with lead acetate $Pb(CH_3COO)_2$ to form a black precipitate of lead sulfide ($PbS$,$Z$).
This hydrolysis reaction occurs for cations whose hydroxides are insoluble in water,such as $Fe^{3+}$,$Al^{3+}$,and $Cr^{3+}$.
$Mg^{2+}$ does not form a stable sulfide $Mg_2S_3$ or $MgS$ that undergoes hydrolysis to produce $H_2S$ in this manner,as $MgS$ is typically prepared in dry conditions and $Mg(OH)_2$ is a weak base.
Therefore,$Mg^{2+}$ is the cation that does not follow this specific reaction pathway.

(B) $1$. $Pb^{2+} + 2KI \to PbI_2\downarrow$ (yellow ppt.). $PbI_2$ is soluble in excess $KI$ due to the formation of a soluble complex $[PbI_4]^{2-}$.
$2$. $Cu^{2+} + 2KI \to CuI_2 \to Cu_2I_2\downarrow + I_2$. $Cu_2I_2$ is insoluble in excess $KI$.
$3$. $Bi^{3+} + 3KI \to BiI_3\downarrow$ (black ppt.). $BiI_3$ is soluble in excess $KI$ due to the formation of $[BiI_4]^-$.
$4$. $Hg^{2+} + 2KI \to HgI_2\downarrow$ (red ppt.). $HgI_2$ is soluble in excess $KI$ due to the formation of $[HgI_4]^{2-}$.
Therefore,$Cu^{2+}$ is the cation that forms a precipitate which remains insoluble in excess $KI$.

(C) Potash alum is $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$. In aqueous solution,it provides $SO_4^{2-}$ ions.
Among the given cations,$Ba^{2+}$ reacts with $SO_4^{2-}$ ions to form a white precipitate of barium sulfate $(BaSO_4)$.
The reaction is: $Ba^{2+}(aq.) + SO_4^{2-}(aq.) \rightarrow BaSO_4(s) \downarrow$ (white precipitate).
Other cations like $Cu^{2+}$,$Zn^{2+}$,and $Ni^{2+}$ do not form precipitates with the sulfate ions present in potash alum under standard conditions.

(C) $Ni^{2+}$ ions react with dimethylglyoxime $(DMG)$ in the presence of ammonium hydroxide $(NH_4OH)$ to form a bright red precipitate of nickel dimethylglyoximate.
This reaction is a standard test for the detection of $Ni^{2+}$ ions in qualitative analysis.

(C) Lead sulfate $(PbSO_4)$ is generally insoluble in water and dilute acids. However,it is soluble in concentrated sulfuric acid due to the formation of hydrogen sulfate $(Pb(HSO_4)_2)$ and in solutions of ammonium acetate or ammonium tartrate due to the formation of soluble complex salts like lead acetate.

(B) $1$. The metal salt solution reacts with potassium chromate $(K_2CrO_4)$ in acetic acid to form a yellow precipitate. This indicates the presence of $Ba^{2+}$ or $Pb^{2+}$ ions,as both form yellow chromates ($BaCrO_4$ and $PbCrO_4$).
$2$. It forms a white precipitate with dilute $H_2SO_4$,which indicates the presence of $Ba^{2+}$,$Pb^{2+}$,or $Ca^{2+}$ (as $BaSO_4$,$PbSO_4$,and $CaSO_4$ are white).
$3$. It gives no precipitate with $NaCl$. $Pb^{2+}$ forms a white precipitate of $PbCl_2$ with $NaCl$,whereas $Ba^{2+}$ does not form a precipitate with $NaCl$ because $BaCl_2$ is soluble.
$4$. Therefore,the metal salt solution contains $Ba^{2+}$ ions,such as $Ba(NO_3)_2$.

(B) Copper sulphide $(CuS)$ is black in colour.
$Cu^{2+}$ is placed in group $II$ of inorganic qualitative analysis.
It is precipitated in the form of sulphide by passing $H_2S$ gas in the presence of dilute $HCl$.
The reaction is: $Cu^{2+} + H_2S \to \underset{\text{black}}{CuS} + 2H^{+}$

(C) Nessler's reagent,$K_2[HgI_4]$,is used for the detection and quantitative determination of ammonia (or $NH_4^+$) in solution.
It reacts with ammonia to form a brown precipitate of $Hg_2NI \cdot H_2O$ (often represented as $IHg-NH_2-Hg-O-I$ or similar structures),known as the iodide of Millon's base.
The reaction is: $NH_4^+ + 2[HgI_4]^{2-} + 4OH^- \rightarrow Hg_2NI \cdot H_2O + 7I^- + 3H_2O$.

(B) Silver ions $(Ag^+)$ and lead ions $(Pb^{2+})$ are both members of Group $I$ cations,which form insoluble chlorides with dilute $HCl$.
However,$PbCl_2$ is soluble in hot water,whereas $AgCl$ remains insoluble even in hot water.
Therefore,adding hot dilute $HCl$ solution allows for the distinction between silver and lead salts.

(B) The reagent is $BaCl_2$,which imparts a green colour to the flame.
$BaCl_2$ reacts with $K_2Cr_2O_7$ and conc. $H_2SO_4$ to form chromyl chloride $(CrO_2Cl_2)$,which is a red gas.
The chemical reaction is:
$2BaCl_2 + K_2Cr_2O_7 + 3H_2SO_4 \to K_2SO_4 + 2BaSO_4 + 2CrO_2Cl_2 \text{ (red gas)} + 3H_2O$

(A) The full form of $DMG$ is $Dimethylglyoxime$. It is a chelating agent commonly used in analytical chemistry to detect and quantify nickel ions $(Ni^{2+})$ by forming a red precipitate.

(A) The correct matches are:
$(i)$ Blood like red colour is due to the formation of the complex $[Fe(SCN)]^{2+}$,which corresponds to $(c)$.
$(ii)$ Sodium nitroprusside is the chemical compound $Na_2[Fe(CN)_5NO]$,which corresponds to $(d)$.
$(iii)$ Ammonium molybdate is $(NH_4)_2MoO_4$,which corresponds to $(b)$.
$(iv)$ Ferri-ferro cyanide is $Fe_4[Fe(CN)_6]_3$,which corresponds to $(a)$.
Therefore,the correct matching is $(i-c), (ii-d), (iii-b), (iv-a)$.

(A) The cation $Ca^{2+}$ does not form a precipitate with $CrO_4^{2-}$ because calcium chromate $(CaCrO_4)$ is relatively soluble in water compared to the chromates of $Sr^{2+}$,$Ba^{2+}$,and $Pb^{2+}$.
The solubility product $(K_{sp})$ values follow the order: $K_{sp}(BaCrO_4) < K_{sp}(SrCrO_4) < K_{sp}(CaCrO_4)$.
Since $CaCrO_4$ has a much higher $K_{sp}$ value,it remains in the solution.

(D) The carbonate of the bismuth ion $(Bi^{3+})$ is known to have a pale yellow color.
On the other hand,the carbonates of $Hg_{2}^{2+}$,$Sr^{2+}$,and $Li^{+}$ are white or colorless.

(D) $PbS$,$CuS$,$As_{2}S_{3}$,and $CdS$ are soluble in $50\% \ HNO_{3}$.
$HgS$ is insoluble in $50\% \ HNO_{3}$ (it requires aqua regia).
$Sb_{2}S_{3}$ is generally considered insoluble in $50\% \ HNO_{3}$ as it forms $Sb_{2}O_{5} \cdot xH_{2}O$ (antimonic acid) which is a white precipitate.
Therefore,the number of soluble sulphides is $4$.

(B) In qualitative analysis,metal ions are classified into groups based on their precipitation reactions:
$Mn^{2+}$ belongs to Group-$IV$ (precipitated as sulfide in alkaline medium).
$As^{3+}$ belongs to Group-$IIB$ (precipitated as sulfide in acidic medium).
$Cu^{2+}$ belongs to Group-$IIA$ (precipitated as sulfide in acidic medium).
$Al^{3+}$ belongs to Group-$III$ (precipitated as hydroxide in the presence of $NH_4Cl$ and $NH_4OH$).
Therefore,the correct matching is: $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(A) $1$. The brown fumes produced with concentrated $H_{2}SO_{4}$ and the formation of a dark brown ring with $FeSO_{4}$ indicate the presence of the nitrate ion $(NO_{3}^{-})$ in compound $X$.
$2$. The formation of a precipitate $Y$ when the solution of $X$ in dilute $HCl$ is treated with $H_{2}S$ gas indicates the presence of a Group-$II$ cation. Among the options,$Cu^{2+}$ is a Group-$II$ cation that forms a black precipitate $(CuS)$ with $H_{2}S$.
$3$. The precipitate $CuS$ $(Y)$ dissolves in concentrated $HNO_{3}$ to form $Cu(NO_{3})_{2}$.
$4$. Adding excess $NH_{4}OH$ to the $Cu(NO_{3})_{2}$ solution forms the deep blue complex $[Cu(NH_{3})_{4}]^{2+}$.
$5$. Therefore,compound $X$ is $Cu(NO_{3})_{2}$.

(B) In the presence of concentrated $HCl$,the medium is highly acidic.
$Al^{3+}$ and $Fe^{3+}$ do not form precipitates with $H_{2}S$ because their sulphides hydrolyze in water.
$Ni^{2+}$ and $Zn^{2+}$ require a basic medium to form precipitates with $H_{2}S$.
$Ca^{2+}$ and $Ba^{2+}$ form soluble sulphides.
Only $Cu^{2+}$ forms a precipitate $(CuS)$ in an acidic medium $(HCl + H_{2}S)$.
Therefore,the total number of cations precipitated is $1$.

(A) The qualitative analysis of $Ni^{2+}$ ions involves the addition of dimethylglyoxime $(DMG)$ in an ammoniacal (alkaline) medium.
This reaction results in the formation of a bright red precipitate of nickel dimethylglyoximate,$[Ni(DMG)_{2}]$.
The reaction is: $Ni^{2+} + 2DMG \xrightarrow{NH_4OH} [Ni(DMG)_{2}] \downarrow + 2H^+$.
Therefore,the reagent $(X)$ is dimethylglyoxime and the cation $(y^{2+})$ is $Ni^{2+}$.

(D) The correct answer is $D$.
When $H_2S$ gas is passed through an acidic solution,only group $II$ cations precipitate as their sulfides because the low concentration of $S^{2-}$ ions (due to the common ion effect of $H^+$) is sufficient to exceed the solubility product $(K_{sp})$ of group $II$ sulfides.
Among the given ions,$Cu^{2+}$ and $Pb^{2+}$ belong to group $II$.
$Al^{3+}$ belongs to group $III$ and $Ni^{2+}$ belongs to group $IV$,which do not precipitate in acidic medium.
The reactions are:
$Cu^{2+} + H_2S \longrightarrow CuS(s) + 2H^+$
$Pb^{2+} + H_2S \longrightarrow PbS(s) + 2H^+$

(D) The correct matching based on qualitative analysis of cations is:
$P (Pb^{2+}, Cu^{2+})$ belongs to Group-$II$,which uses $H_2S$ gas in the presence of dilute $HCl$.
$Q (Al^{3+}, Fe^{3+})$ belongs to Group-$III$,which uses $NH_4OH$ in the presence of $NH_4Cl$.
$R (Co^{2+}, Ni^{2+})$ belongs to Group-$IV$,which uses $H_2S$ in the presence of $NH_4OH$.
$S (Ba^{2+}, Ca^{2+})$ belongs to Group-$V$,which uses $(NH_4)_2CO_3$ in the presence of $NH_4OH$.
Therefore,the correct sequence is $P$ $\rightarrow i, Q$ $\rightarrow iii, R$ $\rightarrow iv, S$ $\rightarrow ii$.

(A) In qualitative inorganic analysis,cations are classified into groups based on their precipitation reactions with specific reagents.
$Zn^{2+}$,$Co^{2+}$,and $Ni^{2+}$ belong to group $IV$,which are precipitated as sulfides in the presence of $NH_4OH$ and $H_2S$.
$Fe^{3+}$ belongs to group $III$,which is precipitated as a hydroxide in the presence of $NH_4Cl$ and $NH_4OH$.
Therefore,$Fe^{3+}$ does not belong to group $IV$.

(C) The formation of $PbCrO_4$ (yellow precipitate),$PbI_2$ (yellow precipitate),and $PbSO_4$ (white precipitate) are standard confirmatory tests for $Pb^{2+}$ ions.
$Pb(NO_3)_2$ is a highly soluble,colourless compound in water and does not form a precipitate,therefore it cannot be used as a confirmatory test for $Pb^{2+}$ ions.

(B) In qualitative analysis,$(NH_4)_2CO_3$ is used as the group reagent for the $5^{th}$ group cations,which include $Ba^{2+}$,$Ca^{2+}$,and $Sr^{2+}$.
The chemical reaction for the precipitation of $Ba^{2+}$ is:
$Ba^{2+} + (NH_4)_2CO_3 \rightarrow BaCO_3 \downarrow + 2NH_4^{+}$
Thus,$Ba^{2+}$ is detected by obtaining a white precipitate of barium carbonate $(BaCO_3)$.

(B)

$PbSO_4$ (Lead sulphate)	White
$(NH_4)_2S$ (Ammonium sulphide)	Colorless/Soluble
$PbI_2$ (Lead iodide)	Bright yellow
$(NH_4)_3AsMo_{12}O_{40}$ (Ammonium arsinomolybdate)	Yellow

Therefore,the compound that is white in color is lead sulphate.

(C) The identification of cations in qualitative analysis is based on their group reagents:
$1$. Group $III$ cations (like $Al^{3+}$) are precipitated as hydroxides by $NH_4Cl + NH_4OH$ $(A-III)$.
$2$. Group $V$ cations (like $Sr^{2+}$) are precipitated as carbonates by $(NH_4)_2CO_3$ in the presence of $NH_4OH$ $(B-IV)$.
$3$. Group $IV$ cations (like $Mn^{2+}$) are precipitated as sulfides by $H_2S$ in the presence of $NH_4OH + NH_4Cl$ $(C-I)$.
$4$. Group $I$ cations (like $Pb^{2+}$) are precipitated as chlorides by dilute $HCl$ $(D-II)$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(D)

Group	Cations
Group $-II$	$Cu^{2+}$
Group $-III$	$Al^{3+}$
Group $-IV$	$Co^{2+}$
Group $-V$	$Ba^{2+}$
Group $-VI$	$Mg^{2+}$

The correct order of group number of ions is $\underset{(B)}{Cu^{2+}} < \underset{(A)}{Al^{3+}} < \underset{(D)}{Co^{2+}} < \underset{(C)}{Ba^{2+}} < \underset{(E)}{Mg^{2+}}$.
Therefore,the correct order is $B, A, D, C, E$.

(A) When a solution containing $Zn^{2+}$ ions is treated with $NH_4OH$ in the presence of $NH_4Cl$,$Zn(OH)_2$ is initially formed as a white precipitate.
Upon adding excess $NH_4OH$,the $Zn(OH)_2$ precipitate dissolves due to the formation of a soluble complex,$[Zn(NH_3)_4]^{2+}$.
The reaction is: $Zn(OH)_2 + 4NH_4OH \rightarrow [Zn(NH_3)_4](OH)_2 + 4H_2O$.
Thus,$Zn(OH)_2$ is the precipitate that dissolves in excess $NH_4OH / NH_4Cl$.

(A) The reaction of $Mg^{+2}$ ions with $NH_4OH$ and $Na_2HPO_4$ is a standard qualitative test for magnesium ions.
The chemical equation for the formation of the white crystalline precipitate is:
$Mg^{+2} + NH_4^+ + PO_4^{-3} \rightarrow Mg(NH_4)PO_4 \downarrow$
Thus,the formula of the white crystalline precipitate is $Mg(NH_4)PO_4$.