Wet Test for Basic Radical Questions in English

(B) The reaction of $Hg^{2+}$ with $KI$ produces a red precipitate of $HgI_2$:
$Hg^{2+} + 2KI \longrightarrow HgI_2 \downarrow \text{(red)} + 2K^+$
In excess $KI$,$HgI_2$ dissolves to form a colourless soluble complex,potassium tetraiodomercurate$(II)$:
$HgI_2 + 2KI \longrightarrow K_2[HgI_4]$
Additionally,$Hg^{2+}$ reacts with cobalt$(II)$ thiocyanate to form a deep blue crystalline precipitate of mercury$(II)$ tetrathiocyanatocobaltate$(II)$,$Hg[Co(SCN)_4]$:
$Hg^{2+} + [Co(SCN)_4]^{2-} \longrightarrow Hg[Co(SCN)_4] \downarrow \text{(deep blue)}$
Thus,the metal ion is $Hg^{2+}$.

(A) The reaction described is the Lauth's violet or Methylene Blue test for sulfides.
$1.$ $X$ is $Na_2S$ because it provides $S^{2-}$ ions which react with $p$-amino-$N,N$-dimethylaniline in the presence of an oxidizing agent $(Y)$ to form methylene blue.
$2.$ $Y$ is $FeCl_3$ (an oxidizing agent). $FeCl_3$ reacts with potassium hexacyanoferrate$(II)$ $(K_4[Fe(CN)_6])$ to form Prussian blue $(Fe_4[Fe(CN)_6]_3)$,which is an intense blue precipitate. This precipitate dissolves in excess reagent to form a soluble complex.
$3.$ $FeCl_3$ reacts with potassium hexacyanoferrate$(III)$ $(K_3[Fe(CN)_6])$ to form $Fe[Fe(CN)_6]$ (iron$(III)$ hexacyanoferrate$(III)$),which gives a brown coloration.
Therefore,$X = Na_2S$,$Y = FeCl_3$,and $Z = Fe[Fe(CN)_6]$.
The correct option is $(D, C, B)$.

(A) In an acidified aqueous solution,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$ ions from the acid.
Only the sulphides of Group $II$ cations,such as $Cu^{2+}$ and $Hg^{2+}$,have a sufficiently low solubility product $(K_{sp})$ to precipitate under these conditions.
The sulphides of $Mn^{2+}$ and $Ni^{2+}$ (Group $III$ and $IV$ cations) have higher $K_{sp}$ values and remain in the solution.
Therefore,$CuS$ and $HgS$ are precipitated.
Hence,the correct option is $A$.

(D) $1$. Flame test: Both $Mn^{2+}$ and $Cu^{2+}$ do not show a characteristic green flame color that distinguishes them uniquely in this context (often $Cu^{2+}$ is associated with green,but $Mn^{2+}$ does not show a characteristic green flame). Thus,$(A)$ is incorrect.
$2$. Precipitation with $H_2S$ in acidic medium: $CuS$ has a very low solubility product $(K_{sp})$,so it precipitates in acidic medium. $MnS$ has a higher $K_{sp}$ and does not precipitate in acidic medium. Thus,$(B)$ is correct.
$3$. Precipitation with $H_2S$ in basic medium: Both $CuS$ and $MnS$ precipitate in basic medium. Thus,$(C)$ is incorrect.
$4$. Reduction potential: The standard reduction potential $E^0$ for $Cu^{2+}/Cu$ $(+0.34 \ V)$ is higher than that of $Mn^{2+}/Mn$ $(-1.18 \ V)$. Thus,$(D)$ is correct.
Therefore,the correct options are $(B)$ and $(D)$.

(A) $1$. The addition of $NaCl$ to the solution containing nitrates of $X$ and $Y$ produces white precipitates of $AgCl$ and $PbCl_2$.
$2$. $PbCl_2$ is soluble in hot water,while $AgCl$ is not. Thus,the residue $P$ is $AgCl$ and the hot filtrate $Q$ contains $PbCl_2$.
$3$. $AgCl$ (residue $P$) is soluble in aqueous $NH_3$ to form $[Ag(NH_3)_2]Cl$ and in excess sodium thiosulfate to form $Na_3[Ag(S_2O_3)_2]$.
$4$. The hot solution $Q$ containing $Pb^{2+}$ ions reacts with $KI$ to form a yellow precipitate of $PbI_2$ $(Pb^{2+} + 2I^- \rightarrow PbI_2 \downarrow)$.
$5$. Therefore,$X$ is $Ag$ and $Y$ is $Pb$.

(C) To form a white precipitate with both dilute $NaOH$ and dilute $HCl$,the cations must satisfy the following conditions:
$1$. With dilute $HCl$,the cation must form a white precipitate (e.g.,$Pb^{2+}$ or $Ag^+$).
$2$. With dilute $NaOH$,the cation must form a white precipitate (e.g.,$Pb^{2+}$,$Zn^{2+}$,or $Bi^{3+}$).
Analyzing the options:
- $Pb(NO_3)_2$: Forms white $PbCl_2$ with $HCl$ and white $Pb(OH)_2$ with $NaOH$.
- $Zn(NO_3)_2$: Forms soluble $ZnCl_2$ with $HCl$ and white $Zn(OH)_2$ with $NaOH$.
- $Bi(NO_3)_3$: Forms soluble $BiCl_3$ with $HCl$ and white $Bi(OH)_3$ with $NaOH$.
- $AgNO_3$: Forms white $AgCl$ with $HCl$ but forms brown $Ag_2O$ with $NaOH$.
- $Hg(NO_3)_2$: Forms soluble $HgCl_2$ with $HCl$ and yellow $HgO$ with $NaOH$.
For the mixture to give white precipitates only with both reagents:
- Mixture $A$ ($Pb(NO_3)_2$ and $Zn(NO_3)_2$): $Pb^{2+}$ gives white $PbCl_2$ and $Pb(OH)_2$; $Zn^{2+}$ gives white $Zn(OH)_2$. This works.
- Mixture $B$ ($Pb(NO_3)_2$ and $Bi(NO_3)_3$): $Pb^{2+}$ gives white $PbCl_2$ and $Pb(OH)_2$; $Bi^{3+}$ gives white $Bi(OH)_3$. This works.
- Mixture $C$ ($AgNO_3$ and $Bi(NO_3)_3$): $Ag^+$ gives brown $Ag_2O$ with $NaOH$. Incorrect.
- Mixture $D$ ($Pb(NO_3)_2$ and $Hg(NO_3)_2$): $Hg^{2+}$ gives yellow $HgO$ with $NaOH$. Incorrect.
Thus,mixtures $A$ and $B$ are correct.

(D) Ammoniacal $H_2S$ is the group reagent for the fourth group of cationic radicals in qualitative analysis.
$Fe^{3+}$ and $Al^{3+}$ belong to the third group and precipitate as hydroxides,$Fe(OH)_3$ and $Al(OH)_3$,in the presence of $NH_4OH$ and $NH_4Cl$.
$Mg^{2+}$ does not precipitate with $H_2S$ in the presence of $NH_4OH$.
$Zn^{2+}$ belongs to the fourth group and reacts with $H_2S$ in the presence of $NH_4OH$ to form a white precipitate of $ZnS$.

(A, D) $1.$ The precipitate $P$ is $PbCl_2$,which is formed by the reaction of $Pb^{2+}$ with $HCl$. $PbCl_2$ is known to be soluble in hot water. Thus,$P$ contains $Pb^{2+}$.
$2.$ The filtrate $Q$ contains $Cr^{3+}$. When treated with $H_2S$ in an ammoniacal medium,it forms a green precipitate $R$ of $Cr(OH)_3$. When $Cr(OH)_3$ is treated with $H_2O_2$ in an aqueous $NaOH$ medium,it undergoes oxidation to form a yellow solution $S$ containing $Na_2CrO_4$ (sodium chromate).
Reaction: $2Cr(OH)_3 + 4NaOH + 3H_2O_2 \longrightarrow 2Na_2CrO_4 + 8H_2O$.

(C) The sulphides $PbS$,$CuS$,$HgS$,$Ag_2S$,$NiS$,$CoS$,and $Bi_2S_3$ are black in colour.
$MnS$ is buff coloured.
$SnS_2$ is yellow coloured.
Therefore,the total number of black coloured sulphides is $7$.

(B) In qualitative analysis,the group-$II$ cations are precipitated as sulfides by passing $H_2S$ gas in the presence of dilute $HCl$.
The group-$II$ cations include $Cu^{2+}, Pb^{2+}, Hg^{2+}, Bi^{3+}, Cd^{2+}, As^{3+}, Sb^{3+},$ and $Sn^{4+}$.
In option $(C)$,both $Cu^{2+}$ and $Pb^{2+}$ belong to group-$II$ and will precipitate as sulfides.
In option $(D)$,both $Hg^{2+}$ and $Bi^{3+}$ belong to group-$II$ and will precipitate as sulfides.
Therefore,both $(C)$ and $(D)$ are correct pairs.

(B) The reaction sequence is as follows:
$1$. $PbS$ (Galena) reacts with $HNO_3$ to form lead$(II)$ nitrate: $3PbS + 8HNO_3 \rightarrow 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$. Thus,$A = PbS$.
$2$. $Pb(NO_3)_2$ reacts with $H_2SO_4$ to form lead$(II)$ sulfate precipitate: $Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4(s) + 2HNO_3$. Thus,$B = PbSO_4$.
$3$. $PbSO_4$ is dissolved in ammonium acetate and acetic acid to form lead$(II)$ acetate,which then reacts with $K_2CrO_4$ to form a yellow precipitate of lead$(II)$ chromate: $PbSO_4 + 2CH_3COONH_4 \rightarrow Pb(CH_3COO)_2 + (NH_4)_2SO_4$ and $Pb(CH_3COO)_2 + K_2CrO_4 \rightarrow PbCrO_4(s) + 2CH_3COOK$. Thus,$C = PbCrO_4$.
Therefore,$A = PbS$,$B = PbSO_4$,and $C = PbCrO_4$.

(D) The reaction of $K_4[Fe(CN)_6]$ with specific cations produces characteristic precipitates:
$1$. $Cu^{2+} + K_4[Fe(CN)_6] \rightarrow Cu_2[Fe(CN)_6] \downarrow$ (Reddish-brown precipitate)
$2$. $Fe^{3+} + K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$ (Prussian blue precipitate)
$3$. $Zn^{2+} + K_4[Fe(CN)_6] \rightarrow K_2Zn_3[Fe(CN)_6]_2 \downarrow$ (White/Bluish-white precipitate)
$4$. $Ca^{2+} + K_4[Fe(CN)_6] \rightarrow K_2Ca[Fe(CN)_6] \downarrow$ (White precipitate)
$Ba^{2+}, NH_4^{+},$ and $Mg^{2+}$ do not form characteristic precipitates with $K_4[Fe(CN)_6]$.
Thus,there are $4$ such cations.

(B) $CoS$ reacts with aqua-regia to form $CoCl_2$ (Compound $B$).
When $CoCl_2$ reacts with $KNO_2$ in the presence of acetic acid $(CH_3COOH)$,it forms a yellow precipitate of potassium cobaltinitrite $(K_3[Co(NO_2)_6])$.
$3CoS + 2HNO_3 + 6HCl \rightarrow 3CoCl_2 + 3S + 2NO + 4H_2O$
$CoCl_2 + 7KNO_2 + 2CH_3COOH \rightarrow K_3[Co(NO_2)_6] \downarrow (\text{yellow}) + 2KCl + 2CH_3COOK + NO + H_2O$

$(D)$ The colours of the given inorganic sulphides are as follows $ : $
$1$. $(NH_4)_2S$ is colourless.
$2$. $PbS$ is black.
$3$. $CuS$ is black.
$4$. $As_2S_3$ is yellow.
$5$. $As_2S_5$ is yellow.
Therefore, $As_2S_3$ and $As_2S_5$ are the yellow coloured sulphides.

(C) : $2 CuSO_4 + K_4[Fe(CN)_6] \xrightarrow{CH_3COOH} Cu_2[Fe(CN)_6] + 2 K_2SO_4$ (Chocolate brown precipitate). This is correct.
$B$: $4 FeCl_3 + 3 K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12 KCl$ (Prussian blue precipitate). This is correct.
$C$: $3 ZnCl_2 + 2 K_4[Fe(CN)_6] \xrightarrow{NH_4OH} K_2Zn_3[Fe(CN)_6]_2 + 6 KCl$ (White or bluish white precipitate). This is correct.
$D$: $MgCl_2$ does not form a blue precipitate with $K_4[Fe(CN)_6]$. This is incorrect.
$E$: $BaCl_2$ does not form a white precipitate with $K_4[Fe(CN)_6]$. This is incorrect.
Therefore,the correct tests are $A, B$ and $C$.

(B) When an ammonium salt is treated with sodium hydroxide $(NaOH)$,it releases ammonia gas $(NH_3)$:
$NH_4^{+} + OH^{-} \longrightarrow NH_3 \uparrow + H_2O$
Ammonia gas $(NH_3)$ reacts with Nessler's reagent $(K_2[HgI_4] + KOH)$ to form a brown-coloured precipitate known as the iodide of Millon's base:
$2[HgI_4]^{2-} + NH_3 + 3OH^{-} \longrightarrow HgO \cdot Hg(NH_2)I + 7I^{-} + 2H_2O$
Thus,$X = NH_3$ and $Y = K_2HgI_4 + KOH$.

(C) The classification of cations into groups for qualitative analysis is based on their solubility products and the reagents used:
$1$. $Pb^{2+}$ belongs to $Group-I$ (precipitated as $PbCl_2$).
$2$. $Al^{3+}$ belongs to $Group-III$ (precipitated as $Al(OH)_3$ in the presence of $NH_4Cl$ and $NH_4OH$).
$3$. $Co^{2+}$ belongs to $Group-IV$ (precipitated as $CoS$ in the presence of $NH_4Cl$,$NH_4OH$,and $H_2S$).
$4$. $Mg^{2+}$ belongs to $Group-VI$ (remains in solution after other groups are precipitated and is identified as $Mg_2P_2O_7$).
Therefore,the correct matching is:

$A. Co^{2+}$	$III. Group-IV$
$B. Mg^{2+}$	$IV. Group-VI$
$C. Pb^{2+}$	$I. Group-I$
$D. Al^{3+}$	$II. Group-III$

Thus,the correct option is $C$.

(D) In qualitative inorganic analysis,cations are classified into groups based on their reaction with specific group reagents.
Group-$I$ cations $(Pb^{2+}, Ag^+, Hg_2^{2+})$ are precipitated as chlorides using dil. $HCl$.
Group-$II$ cations $(Cu^{2+}, As^{3+}, etc.)$ are precipitated as sulfides using $H_2S$ in the presence of dil. $HCl$.
Group-$III$ cations $(Fe^{3+}, Al^{3+}, Cr^{3+})$ are precipitated as hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
Group-$IV$ cations $(Zn^{2+}, Mn^{2+}, Ni^{2+}, Co^{2+})$ are precipitated as sulfides using $H_2S$ in the presence of $NH_4OH$.
Group-$V$ cations $(Ba^{2+}, Sr^{2+}, Ca^{2+})$ are precipitated as carbonates using $(NH_4)_2CO_3$ in the presence of $NH_4OH$ and $NH_4Cl$.
Option $D$ is incorrect because $Ba^{2+}$ and $Ca^{2+}$ belong to Group-$V$,not Group-$IV$,and the reagent listed is also incorrect.

(D) The reaction of $Zn^{2+}$ ions with potassium ferrocyanide $(K_4[Fe(CN)_6])$ produces a white or blue-white precipitate of zinc ferrocyanide,$Zn_2[Fe(CN)_6]$.
The chemical equation is: $2Zn^{2+} + K_4[Fe(CN)_6] \rightarrow Zn_2[Fe(CN)_6] + 4K^+$.
Therefore,the ion $A$ is $Zn^{2+}$.

(D) . Lake Test is used to identify $Al^{3+}$ ions.
$B$. Nessler's Reagent is used to identify $NH_4^{+}$ ions.
$C$. Potassium thiocyanate is used to identify $Fe^{3+}$ ions.
$D$. Brown Ring Test is used to identify $NO_3^{-}$ ions.
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(A) In inorganic qualitative analysis,the cations are classified into groups based on their precipitation reactions with specific group reagents:

$B. Cu^{2+}$	Group $II$
$A. Al^{3+}$	Group $III$
$D. Co^{2+}$	Group $IV$
$C. Ba^{2+}$	Group $V$
$E. Mg^{2+}$	Group $VI$

Arranging them in increasing order of group number gives the sequence: $B, A, D, C, E$.

(A) In the qualitative analysis of cations,the $III^{rd}$ group consists of $Fe^{+3}$,$Al^{+3}$,and $Cr^{+3}$.
Before adding the group reagent ($NH_4OH$ in the presence of $NH_4Cl$),it is necessary to ensure that all iron is present in the $Fe^{+3}$ state.
If iron is present as $Fe^{+2}$,it will not precipitate as $Fe(OH)_3$ effectively with the $III^{rd}$ group reagent.
Therefore,concentrated $HNO_3$ is added to oxidize $Fe^{+2}$ to $Fe^{+3}$ according to the reaction: $6Fe^{+2} + 2HNO_3 + 6H^+ \rightarrow 6Fe^{+3} + 2NO + 4H_2O$.
Thus,both Statement-$1$ and Statement-$2$ are correct.

(A) $Pb^{2+}$ ion is capable of precipitating $Cl^{-}$,$I^{-}$,and $SO_{4}^{2-}$ ions as insoluble salts.
$Pb^{2+} + 2Cl^{-} \longrightarrow PbCl_{2} \downarrow$ (White precipitate)
$Pb^{2+} + 2I^{-} \longrightarrow PbI_{2} \downarrow$ (Yellow precipitate)
$Pb^{2+} + SO_{4}^{2-} \longrightarrow PbSO_{4} \downarrow$ (White precipitate)
While $Ba^{2+}$ precipitates $SO_{4}^{2-}$,it does not form insoluble precipitates with $Cl^{-}$ and $I^{-}$ in aqueous solutions.

(A) When $BiCl_{3}$ is diluted with water,it undergoes hydrolysis to form a white precipitate of bismuth oxychloride $(BiOCl)$.
The chemical reaction is: $BiCl_{3} + H_{2}O \rightleftharpoons BiOCl \downarrow + 2HCl$
Here,$BiOCl$ is the white precipitate.

(A) The reaction is the Chromyl Chloride test: $4Cl^{-} + Cr_{2}O_{7}^{2-} + 6H^{+} \rightarrow 2CrO_{2}Cl_{2} + 3H_{2}O$.
The deep red vapours consist of $CrO_{2}Cl_{2}$ (Chromyl chloride).
Statement $A$ is false because the vapours do not contain $Cl_{2}$ gas.
When these vapours are passed into $NaOH$,a yellow solution of sodium chromate is formed: $CrO_{2}Cl_{2} + 4OH^{-} \rightarrow CrO_{4}^{2-} + 2Cl^{-} + 2H_{2}O$.
This solution,when treated with lead acetate,gives a yellow precipitate of lead chromate: $CrO_{4}^{2-} + Pb^{2+} \rightarrow PbCrO_{4} \downarrow$ (yellow).

(B) The chromyl chloride test is used to detect the presence of chloride ions.
When a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$,red vapors of chromyl chloride $(CrO_2Cl_2)$ are evolved.
When these vapors are passed into a solution of sodium hydroxide $(NaOH)$,a yellow solution of sodium chromate $(Na_2CrO_4)$ is formed.
Upon adding lead acetate $(CH_3COO)_2Pb$ to this yellow solution,a yellow precipitate of lead chromate $(PbCrO_4)$ is obtained.

(B) The reaction sequence is as follows:
$ZnSO_4$ $\xrightarrow{NH_4OH} Zn(OH)_2 \downarrow \text{ (White ppt., } X)$ $\xrightarrow{NH_4OH \text{ (excess)}} [Zn(NH_3)_4]^{2+} \text{ (Clear solution, } Y)$ $\xrightarrow{H_2S} ZnS \downarrow \text{ (Ppt., } Z)$
Therefore,the metal $M$ is $Zn$ and the precipitate $Z$ is $ZnS$.

(B) The group reagent consisting of $NH_4Cl$ and aqueous $NH_3$ $(NH_4OH)$ is used for the precipitation of Group $III$ cations in qualitative inorganic analysis.
These cations are $Al^{3+}$,$Fe^{3+}$,and $Cr^{3+}$.
Among the given options,$Al^{3+}$ belongs to Group $III$ and will be precipitated as $Al(OH)_3$.

(C) $NH_{3}$ is used in the preparation of Tollen's reagent $([Ag(NH_{3})_{2}]^{+})$.
$NH_{3}$ is also used as a group reagent for the analysis of $III$ group basic radicals (as $NH_{4}OH$).
$NH_{3}$ is not used as a group reagent for the analysis of $IV$ group basic radicals,where $H_{2}S$ in the presence of $NH_{4}OH$ is used.
Nessler's reagent is $K_{2}[HgI_{4}]$ in $KOH$,which is used to test $NH_{4}^{+}$ ions,but $NH_{3}$ is not a component of the reagent itself.

(D) Magneson reagent is a dye used in the qualitative analysis of $Mg^{2+}$ ions.
In an alkaline medium,$Mg^{2+}$ ions form a blue-coloured lake (precipitate) with the magneson reagent.
This reaction is used as a confirmatory test for magnesium,where $Mg(OH)_2$ adsorbs the dye to form a deep blue precipitate.

(C) When a concentrated solution of $Co(NO_3)_2$ reacts with $NaNO_2$ in $50\%$ acetic acid,the hexanitritocobaltate$(III)$ complex ion,$[Co(NO_2)_6]^{3-}$,is formed.
Upon adding a salt containing the potassium ion $(K^+)$,a yellow precipitate of potassium hexanitritocobaltate$(III)$,$K_3[Co(NO_2)_6]$,is produced.
This reaction is a standard qualitative test for the identification of potassium ions.
Therefore,metal $M$ is potassium $(K)$.

(A) The sequence of reactions is as follows:
$1$. $2AgNO_{3} + Na_{2}S_{2}O_{3} \rightarrow Ag_{2}S_{2}O_{3} \downarrow (X) + 2NaNO_{3}$
$2$. $Ag_{2}S_{2}O_{3} + 3Na_{2}S_{2}O_{3} \rightarrow 2Na_{3}[Ag(S_{2}O_{3})_{2}] (Y)$
$3$. $2Na_{3}[Ag(S_{2}O_{3})_{2}] + H_{2}O \xrightarrow{\Delta} Ag_{2}S (Z) + Na_{2}SO_{4} + 2Na_{2}S_{2}O_{3} + H_{2}SO_{4}$
Thus,$X = Ag_{2}S_{2}O_{3}$,$Y = Na_{3}[Ag(S_{2}O_{3})_{2}]$,and $Z = Ag_{2}S$.

(A) Test $1$: The reaction with aqueous $NaOH$ and gentle heating produces $NH_3$ gas (which turns damp red litmus blue),indicating the presence of ammonium ions $(NH_4^+)$.
Test $2$: The reaction with dilute $HCl$ produces $SO_2$ gas (which turns lime water milky and acidified $K_2Cr_2O_7$ paper green),indicating the presence of sulphite ions $(SO_3^{2-})$.
Carbonate ions would turn lime water milky but would not change the color of acidified $K_2Cr_2O_7$.
Therefore,compound $X$ contains ammonium and sulphite ions.

(C) In the qualitative analysis of cations,$Ba^{2+}$ and $Ca^{2+}$ belong to Group $V$.
They are precipitated by the addition of $(NH_{4})_{2}CO_{3}$ in the presence of $NH_{4}Cl$ and $NH_{4}OH$.
Both $Ba^{2+}$ and $Ca^{2+}$ react with the carbonate ion $(CO_{3}^{2-})$ to form their respective insoluble carbonates,$BaCO_{3}$ and $CaCO_{3}$.

(B) $1$. An apple green flame test is a characteristic feature of $Ba^{2+}$ ions.
$2$. $Ba^{2+}$ ions react with $K_{2}CrO_{4}$ in an acetic acid medium to form a yellow precipitate of barium chromate $(BaCrO_{4})$.
$3$. The formation of a canary yellow precipitate upon heating the sodium carbonate extract with conc. $HNO_{3}$ and ammonium molybdate is a confirmatory test for the presence of phosphate ions $(PO_{4}^{3-})$.
$4$. Therefore,the salt contains $Ba^{2+}$ as the cation and $PO_{4}^{3-}$ as the anion.

(D) In qualitative inorganic analysis,metal ions are classified into groups based on their precipitation reactions.
$Pb^{2+}$ and $Cu^{2+}$ belong to Group $II$ and precipitate as sulfides in an acidic medium.
$Fe^{3+}$ belongs to Group $III$ and precipitates as $Fe(OH)_3$ in the presence of $NH_4OH$ and $NH_4Cl$.
$Mn^{2+}$ belongs to Group $IV$ and precipitates as $MnS$ when $H_2S$ is passed through an alkaline solution (in the presence of $NH_4OH$).
Therefore,$Mn^{2+}$ is the ion that precipitates under the given conditions.
The electronic configuration of Manganese $(Mn)$ is $[Ar] 3d^5 4s^2$. It can lose all $7$ valence electrons to achieve its highest oxidation state of $+7$,as seen in compounds like $KMnO_4$.