Wet Test for Basic Radical Questions in English

(C) The test for sulfide ions $(S^{2-})$ involves the addition of sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ to an alkaline solution containing the sulfide ions.
This reaction produces a deep violet or purple coloration due to the formation of the complex ion $[Fe(CN)_5NOS]^{4-}$.
The reaction is: $S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}$.

(C) The solubility product $(K_{sp})$ values for alkaline earth metal oxalates are very low.
Specifically,the $K_{sp}$ values for $CaC_2O_4$,$SrC_2O_4$,and $BaC_2O_4$ are all quite small,meaning they are all sparingly soluble in water.
When an oxalate solution (such as ammonium oxalate) is added to a mixture containing $Ba^{2+}$,$Sr^{2+}$,and $Ca^{2+}$ ions,all three ions will form insoluble white precipitates of their respective oxalates.
Therefore,all three ions precipitate.

(C) The group-$II$ basic radicals are precipitated as their sulfides.
$H_2S$ is a weak electrolyte and dissociates as: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
In the presence of dilute $HCl$,the concentration of $H^+$ ions increases due to the common ion effect.
According to Le Chatelier's principle,this suppresses the dissociation of $H_2S$,thereby lowering the concentration of $S^{2-}$ ions.
This low concentration of $S^{2-}$ is sufficient to exceed the solubility product $(K_{sp})$ of group-$II$ sulfides but not that of group-$IV$ sulfides,allowing for selective precipitation.

(A) Calomel is $Hg_2Cl_2$. When it reacts with ammonium hydroxide $(NH_4OH)$,it undergoes a disproportionation reaction.
The reaction is: $Hg_2Cl_2 + 2NH_4OH \rightarrow Hg + Hg(NH_2)Cl + NH_4Cl + 2H_2O$.
The black precipitate formed is a mixture of finely divided mercury $(Hg)$ and amino mercuric chloride $(Hg(NH_2)Cl)$.

(A) In the analysis of group $III$ radicals,$NH_4Cl$ is added to suppress the dissociation of $NH_4OH$ by the common ion effect,ensuring that the concentration of $OH^-$ ions is sufficient only to precipitate group $III$ hydroxides (like $Fe(OH)_3$,$Al(OH)_3$,$Cr(OH)_3$) and not group $IV$ or $V$ hydroxides. $NH_4NO_3$ acts as a strong electrolyte and provides the common $NH_4^+$ ion,serving the same purpose as $NH_4Cl$.

(C) When $Na_2O_2$ is added to the mixture,$Fe^{2+}$ is oxidized to $Fe^{3+}$,which precipitates as $Fe(OH)_3$ (brown residue).
$Al^{3+}$ forms soluble $[Al(OH)_4]^-$ and $Cr^{3+}$ is oxidized to $CrO_4^{2-}$ (yellow filtrate).
Thus,the filtrate is yellow due to the presence of chromate ions,and the residue is brown due to the presence of ferric hydroxide.

(B) $1$. The salt must contain $Cl^-$ ions to form a white precipitate of $AgCl$ with $AgNO_3$. This eliminates option $D$.
$2$. The salt must contain $Ba^{2+}$ ions to form a white precipitate of $BaSO_4$ with dilute $H_2SO_4$. This eliminates options $A$ and $C$.
$3$. $BaCl_2$ contains both $Cl^-$ and $Ba^{2+}$ ions.
$4$. $Ba^{2+}$ ions impart a characteristic apple-green color to the flame in the flame test.
$5$. Therefore,$BaCl_2$ is the correct salt.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,$K_2[HgI_4]$.
It is primarily used to detect the presence of ammonium ions $(NH_4^+)$ or ammonia $(NH_3)$.
When it reacts with ammonium ions in a basic medium,it forms a brown precipitate known as the iodide of Millon's base,which is $HgO \cdot Hg(NH_2)I$.

(D) In qualitative analysis,$H_2S$ is used to precipitate metal sulfides.
$Zn^{2+}$ and $Mn^{2+}$ both belong to Group-$IV$ of the basic radical analysis.
When $H_2S$ is passed through an alkaline solution (buffered with $NH_4OH$ and $NH_4Cl$),both $Zn^{2+}$ and $Mn^{2+}$ precipitate as $ZnS$ and $MnS$ respectively.
Since both precipitate under the same conditions,they cannot be separated by this method.
In contrast,other pairs listed belong to different groups or have different solubility products,allowing for separation.

(B) Mercurous chloride $(Hg_2Cl_2)$ reacts with ammonium hydroxide $(NH_4OH)$ to form a black precipitate of metallic mercury $(Hg)$ and amino mercuric chloride $(Hg(NH_2)Cl)$.
The reaction is: $Hg_2Cl_2 + 2NH_4OH \rightarrow Hg(NH_2)Cl + Hg + NH_4Cl + 2H_2O$.
The black color is due to the formation of finely divided metallic mercury $(Hg)$.

(C) Both $Pb^{2+}$ and $Ba^{2+}$ ions form yellow precipitates with $K_2CrO_4$ in acetic acid and white precipitates with dilute $H_2SO_4$.
However,lead salts $(Pb^{2+})$ form precipitates with $NaCl$ $(PbCl_2)$ or $NaI$ $(PbI_2)$,whereas barium salts $(Ba^{2+})$ do not form precipitates with these reagents.
Therefore,the given salt must be a barium salt.
When sodium carbonate $(Na_2CO_3)$ is added to a barium salt solution,it forms a white precipitate of barium carbonate $(BaCO_3)$.

(A) The reaction of magnesium ions with disodium hydrogen phosphate in the presence of ammonium hydroxide is a standard test for magnesium ions.
The chemical equation is: $Mg^{2+} + Na_2HPO_4 + NH_4OH \to Mg(NH_4)PO_4 \downarrow + 2Na^+ + H_2O$
The white crystalline precipitate formed is magnesium ammonium phosphate,$Mg(NH_4)PO_4$.

(A) The reaction of calomel $(Hg_2Cl_2)$ with ammonium hydroxide $(NH_4OH)$ is a disproportionation reaction.
The reaction is as follows:
$Hg_2Cl_2 + 2NH_4OH \rightarrow Hg(NH_2)Cl + Hg + NH_4Cl + 2H_2O$.
In this reaction,a black precipitate consisting of a mixture of mercury $(Hg)$ and amino mercury$(II)$ chloride $(Hg(NH_2)Cl)$ is formed.

(D) Sulfide ions $(S^{2-})$ react with sodium nitroprusside to form a deep violet-colored complex.
The chemical reaction is: $Na_2[Fe(CN)_5NO] + S^{2-} \rightarrow [Fe(CN)_5NOS]^{4-} + 2Na^+$.
This test is a standard qualitative test for sulfide ions.

(B) The correct answer is $B$.
Sometimes,yellow turbidity is observed upon passing $H_2S$ gas even when second group radicals are absent.
This occurs because certain fourth group radicals (such as $Zn^{2+}$,$Mn^{2+}$,$Ni^{2+}$,$Co^{2+}$) may precipitate as sulphides if the concentration of $S^{2-}$ ions in the solution is sufficiently high,leading to the formation of a colloidal suspension or turbidity.

(B) In an acidic medium,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$ ions from $HCl$ on the dissociation of $H_2S$.
$Pb^{2+}$ and $Cu^{2+}$ belong to Group-$II$ of the qualitative analysis scheme,which are precipitated as sulphides in an acidic medium because their solubility products $(K_{sp})$ are very low.
$Zn^{2+}$ and $Ni^{2+}$ belong to later groups (Group-$III$ and $IV$) and require a higher concentration of $S^{2-}$ ions for precipitation,which is not achieved in an acidic medium.
Therefore,only $PbS$ and $CuS$ will precipitate.
Thus,the correct option is $(B)$.

(C) When ammonia $(NH_4OH)$ is added to a solution containing $Fe^{3+}$ ions,a red (reddish-brown) precipitate of ferric hydroxide $(Fe(OH)_3)$ is formed.
This precipitate is soluble in dilute $HCl$.
$Fe^{3+} + 3NH_4OH \rightarrow Fe(OH)_3 \downarrow + 3NH_4^+$
$Fe(OH)_3 + 3HCl \rightarrow FeCl_3 + 3H_2O$

(C) The correct answer is $(C)$.
Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which is represented as $K_2HgI_4 + KOH$.

(B) The reaction sequence describes the chromyl chloride test.
$1$. $A$ is a dichromate salt (e.g.,$K_2Cr_2O_7$),which is orange.
$2$. When $A$ reacts with $B$ (a chloride salt,$NH_4Cl$) and concentrated $H_2SO_4$,it produces $C$ (Chromyl chloride,$CrO_2Cl_2$),which is a red gas.
$3$. $CrO_2Cl_2$ reacts with $NaOH$ and $AgNO_3$ to form $Ag_2CrO_4$ (Red precipitate,$D$).
$4$. $Ag_2CrO_4$ dissolves in $NH_3$ to form a yellow solution.
$5$. $B$ $(NH_4Cl)$ reacts with $NaOH$ upon heating to release $F$ ($NH_3$ gas),which gives white fumes with $HCl$ due to the formation of $NH_4Cl$ smoke.
Therefore,$B$ is $NH_4Cl$ and $F$ is $NH_3$.

(C) $1$. When $H_2S$ gas is passed through the acidified solution,$Cu^{2+}$ and $Cd^{2+}$ ions precipitate as $CuS$ and $CdS$ (Group $II$ cations). The filtrate contains $Cr^{3+}$ and $Fe^{3+}$.
$2$. Boiling with $HNO_3$ ensures $Fe^{2+}$ is oxidized to $Fe^{3+}$.
$3$. Upon adding $NH_4Cl$ and excess $NaOH$ (or $NH_4OH$),$Fe^{3+}$ precipitates as $Fe(OH)_3$ (brown precipitate).
$4$. $Cr^{3+}$ reacts with excess $NaOH$ to form a soluble complex,$[Cr(OH)_4]^-$,which remains in the filtrate.
$5$. Therefore,the filtrate contains $Cr^{3+}$ ions and $Na^+$ ions (from the added $NaOH$). Thus,the filtrate gives a test for sodium and chromium ions.

(A) The color of $Hg_2I_2$ (mercurous iodide) is yellow-green,but it is often categorized as yellow. However,$BiI_3$ (bismuth triiodide) is known to be black in color. $Ag_2CrO_4$ (silver chromate) is brick red. $BaCrO_4$ (barium chromate) is yellow. All the given options are actually correct in terms of their characteristic colors. If this is a multiple-choice question where one must be incorrect,there might be a typo in the question source. Based on standard chemical properties,all these matches are correct.

(C) $1$. The borax bead test for $Cr^{3+}$ in a reducing flame produces a green-colored bead due to the formation of $Cr_2O_3$.
$2$. When $Cr^{3+}$ compounds are treated with $H_2O_2$ in an alkaline medium,they are oxidized to chromate ions $(CrO_4^{2-})$.
$3$. The chromate ions react with lead$(II)$ acetate $(Pb(OAc)_2)$ to form a yellow precipitate of lead chromate $(PbCrO_4)$.
$4$. Therefore,the metal ion present in compound $A$ is $Cr^{3+}$.

(B) $1$. The reaction $X$ $\xrightarrow[\Delta]{NaOH} Y$ $\xrightarrow{HCl} \text{White fumes}$ indicates that $Y$ is ammonia $(NH_3)$ gas,which reacts with $HCl$ to form $NH_4Cl$ (white fumes). This confirms that $X$ contains the ammonium ion $(NH_4^+)$.
$2$. The reaction $X$ $\xrightarrow{CaCl_2} Z \text{ (White ppts.)}$ $\xrightarrow{KMnO_4} \text{Colourless solution}$ indicates that $Z$ is calcium oxalate $(CaC_2O_4)$,which is a white precipitate. Oxalate ions $(C_2O_4^{2-})$ react with $KMnO_4$ in an acidic medium to form a colourless solution ($CO_2$ and $Mn^{2+}$).
$3$. Combining these,the salt $X$ is ammonium oxalate,$(NH_4)_2C_2O_4$.

(B) $1$. When $H^{+}/H_2S$ is added,$Pb^{2+}$,$Cu^{2+}$,and $Cd^{2+}$ precipitate as their respective sulfides ($PbS$,$CuS$,$CdS$) because they belong to Group $II$ of the qualitative analysis scheme.
$2$. The remaining ions in solution $(X)$ are $Al^{3+}$ and $Zn^{2+}$.
$3$. When excess $NaOH$ is added to solution $(X)$ containing $Al^{3+}$ and $Zn^{2+}$,both form soluble complexes.
$4$. $Al^{3+} + 4OH^{-} \rightarrow [Al(OH)_4]^{-}$ (soluble aluminate).
$5$. $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_4]^{2-}$ (soluble zincate).
$6$. Thus,a clear solution is obtained containing two radicals as soluble complexes.

(C) $1$. $Fe(OH)_3$ forms a red-brown precipitate. (Correct)
$2$. $Cu_2[Fe(CN)_6]$ forms a chocolate-brown precipitate. (Correct)
$3$. $BaSO_4$ is a white precipitate,not green. (Incorrect)
$4$. $ZnS$ forms a greenish-white precipitate. (Correct)
Therefore,the incorrect match is $BaSO_4$ $\rightarrow$ Green.

(D) $1$. Ion $A$ reacts with conc. $H_{2}SO_{4}$ to produce a gas $G_{1}$. Ion $B$ reacts with $NaOH$ to produce a gas $G_{2}$.
$2$. The reaction between $G_{1}$ and $G_{2}$ produces white fumes.
$3$. In option $D$,if $A = Cl^{-}$,then $Cl^{-} + H_{2}SO_{4} \rightarrow HCl(g) (G_{1})$.
$4$. If $B = NH_{4}^{+}$,then $NH_{4}^{+} + NaOH \rightarrow NH_{3}(g) (G_{2}) + H_{2}O$.
$5$. The reaction between $HCl(g)$ and $NH_{3}(g)$ produces $NH_{4}Cl(s)$,which appears as white fumes.
$6$. Therefore,the correct ions are $Cl^{-}$ and $NH_{4}^{+}$.

(B) $1$. $X$ gives a green flame test,which is characteristic of $Ba^{2+}$ ions.
$2$. $X$ reacts with $K_2CrO_4$ to form a yellow precipitate $T$. $BaCrO_4$ is a yellow precipitate.
$3$. $X$ reacts with $dil. HCl$ to form a yellow precipitate $Y$ and a pungent smelling gas $Z$.
$4$. $BaS_2O_3 + 2HCl \rightarrow BaCl_2 + S(s) + SO_2(g) + H_2O$. However,$BaS_2O_3$ is not the correct fit for the yellow precipitate $Y$ in this context.
$5$. Considering $PbS_2O_3 + 2HCl \rightarrow PbCl_2(s) + S(s) + SO_2(g) + H_2O$. $PbCl_2$ is white.
$6$. Re-evaluating: $BaS_2O_3$ is the correct choice as $Ba^{2+}$ gives a green flame test. The yellow precipitate $T$ is $BaCrO_4$. The reaction with $HCl$ produces $SO_2$ (pungent gas) and the precipitate $Y$ is sulfur (often appearing yellow).

(C) The metallic salt $A$ reacts with $K_2CrO_4$ in the presence of acetic acid to give a yellow precipitate $B$,which indicates that $A$ contains $Ba^{2+}$ ions. $Sr^{2+}$ and $Ca^{2+}$ do not form a precipitate with $K_2CrO_4$ in acetic acid.
The reactions are as follows:
$1. Ba^{2+} + K_2CrO_4 \xrightarrow{CH_3COOH} BaCrO_4 \downarrow (\text{Yellow ppt } B)$
$2. Ba^{2+} + dil. H_2SO_4 \rightarrow BaSO_4 \downarrow (\text{White ppt } C)$
$3. Ba^{2+} + Na_2CO_3 \rightarrow BaCO_3 \downarrow (\text{White ppt } D)$
Thus,the white precipitate $D$ is barium carbonate $(BaCO_3)$.

(C) $1.$ $Fe^{3+}$ reacts with potassium ferricyanide $(K_3[Fe(CN)_6])$ to form a brown-colored complex.
$2.$ $Fe^{2+}$ reacts with potassium ferricyanide $(K_3[Fe(CN)_6])$ to form Turnbull's blue precipitate.
$3.$ $Fe^{3+}$ reacts with potassium thiocyanate $(KSCN)$ to form a blood-red colored complex,$[Fe(SCN)(H_2O)_5]^{2+}$.
$4.$ $Fe^{2+}$ does not give a brown color with ammonium thiocyanate; it typically shows no reaction or a very faint color change compared to $Fe^{3+}$.
Therefore,statements $(1)$,$(2)$,and $(3)$ are correct.

(B) The reaction sequence describes the chromyl chloride test and related chemistry:
$1$. $(T)$ is $K_2Cr_2O_7$ (orange-red,but often associated with dichromate chemistry).
$2$. $K_2Cr_2O_7 + NH_4Cl + conc. H_2SO_4 \rightarrow CrO_2Cl_2$ (Chromyl chloride,$V$,red gas).
$3$. $CrO_2Cl_2 + NaOH \rightarrow Na_2CrO_4$ (yellow solution),which with $AgNO_3$ gives $Ag_2CrO_4$ (red ppt.,$W$).
$4$. $Ag_2CrO_4 + NH_3 \rightarrow [Ag(NH_3)_2]^+$ (soluble complex,$X$).
$5$. $Ag_2CrO_4 + dil. HCl \rightarrow AgCl$ (white ppt.,$Y$).
$6$. $NH_4Cl + NaOH \xrightarrow{\Delta} NH_3$ (gas,$Z$,gives white fumes with $HCl$).

(D) The reaction of alkaline earth metal ions with oxalate ions $(C_2O_4^{2-})$ in the presence of acetic acid $(CH_3COOH)$ leads to the formation of insoluble oxalates.
$Ca^{2+} + C_2O_4^{2-} \rightarrow CaC_2O_4 \downarrow$ (White ppt.)
$Sr^{2+} + C_2O_4^{2-} \rightarrow SrC_2O_4 \downarrow$ (White ppt.)
$Ba^{2+} + C_2O_4^{2-} \rightarrow BaC_2O_4 \downarrow$ (White ppt.)
All three cations ($Ca^{2+}$,$Sr^{2+}$,and $Ba^{2+}$) form insoluble oxalates in a weakly acidic medium like acetic acid.

(D) Aqueous $(NH_4)_2S$ provides $S^{2-}$ ions.
$Al^{3+}$ is precipitated as $Al(OH)_3$ due to hydrolysis: $2Al^{3+} + 3S^{2-} + 6H_2O \longrightarrow 2Al(OH)_3 \downarrow + 3H_2S \uparrow$.
$Pb^{2+}$ and $Cu^{2+}$ form insoluble metal sulfides: $Pb^{2+} + S^{2-} \longrightarrow PbS \downarrow$ and $Cu^{2+} + S^{2-} \longrightarrow CuS \downarrow$.
$K^{+}$ salts are generally soluble in water,so $K_2S$ remains in the solution.

(D) The reddish-brown fumes evolved on treating salt $A$ with dil. $H_2SO_4$ and warming indicate the presence of nitrite $(NO_2^-)$ ions,which produce $NO_2$ gas.
The reaction is: $2NO_2^- + 2H^+ \rightarrow H_2O + NO + NO_2$ (reddish-brown).
The blue colour produced upon adding $C$,starch,and acetic acid is the characteristic test for iodine $(I_2)$.
This implies that $C$ is an oxidizing agent that oxidizes $I^-$ (if present in $A$) or provides $I_2$ to react with starch.
However,the question asks what $C$ may contain to produce the blue colour with the salt solution of $A$.
Since $A$ contains $NO_2^-$,and the test for $I^-$ involves oxidation to $I_2$ (which turns starch blue),$C$ must be an oxidizing agent like $NO_2^-$ itself or a reagent that facilitates the reaction.
Given the options,$I^-$ is the species that reacts to form $I_2$,but the question asks what $C$ contains to produce the blue colour.
Actually,the reaction of $NO_2^-$ with $I^-$ in acidic medium produces $I_2$.
Thus,$C$ must contain $I^-$.

(C) The correct formulas for the ammonium phosphomolybdate and ammonium arsenomolybdate precipitates are $(NH_4)_3[P(Mo_{12}O_{40})]$ and $(NH_4)_3[As(Mo_{12}O_{40})]$ respectively.
Both of these are yellow precipitates.
$MnO(OH)_2$ is a brown precipitate formed during the test for manganese.
$K_3[Co(NO_2)_6]$ is a yellow precipitate formed during the test for cobalt.
In the provided options,the chemical formulas for the arsenomolybdate and phosphomolybdate complexes were written incorrectly as $(NH_4)_3[As(Mo_3O_{10})_4]$ and $(NH_4)_3[P(Mo_3O_{10})_4]$.
Since the question asks for the incorrect match and the provided formulas in options $C$ and $D$ are chemically inaccurate representations of the standard yellow precipitates,they are technically incorrect. However,assuming the intent was to identify the standard test results,all listed colours are correct for their respective compounds. Given the options,if we must choose an incorrect match based on the provided stoichiometry,options $C$ and $D$ are chemically flawed.

(A) Both $Zn^{2+}$ and $Al^{3+}$ ions form white precipitates with $NaOH$ and $NH_4OH$.
However,$Zn(OH)_2$ is soluble in excess of both $NaOH$ and $NH_4OH$ due to the formation of soluble complexes like $[Zn(OH)_4]^{2-}$ and $[Zn(NH_3)_4]^{2+}$.
$Al(OH)_3$ is soluble in excess $NaOH$ to form $[Al(OH)_4]^-$,but it is insoluble in excess $NH_4OH$.
Therefore,$NH_4OH$ can be used to distinguish between them because $Al(OH)_3$ will remain as a precipitate while $Zn(OH)_2$ will dissolve.
$NaOH$ is less effective for distinction because both hydroxides dissolve in excess $NaOH$.

(D) Both $Al^{3+}$ and $Fe^{3+}$ form precipitates with $NaOH$ solution.
$Al^{3+} + 3OH^- \rightarrow Al(OH)_3 \downarrow$ (white precipitate)
$Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow$ (brown precipitate)
$Al(OH)_3$ is amphoteric in nature and dissolves in excess $NaOH$ to form a soluble complex,$Na[Al(OH)_4]$,whereas $Fe(OH)_3$ is basic and does not dissolve in excess $NaOH$.
Therefore,$Al^{3+}$ and $Fe^{3+}$ can be separated using concentrated aqueous $NaOH$.

(B) $(1)$ The reaction of $NH_4^+$ with Nessler's reagent ($K_2HgI_4$ in $KOH$) gives a brown precipitate of Iodide of Millon's base,not blue. So,$(1)$ is incorrect.
$(2)$ $Ni^{2+}$ reacts with dimethylglyoxime $(dmg)$ in the presence of $OH^-$ to form a rosy red precipitate of $[Ni(dmg)_2]$. This is correct.
$(3)$ $FeCl_3$ reacts with $NH_4OH$ to form a reddish brown precipitate of $Fe(OH)_3$,not green. So,$(3)$ is incorrect.
$(4)$ $CuSO_4$ reacts with excess $NH_4OH$ to form a deep blue solution of $[Cu(NH_3)_4]SO_4$. This is correct.
Therefore,the correct observations are $(2)$ and $(4)$.

(D) $1$. $Sodium$ nitroprusside $(Na_2[Fe(CN)_5NO])$ reacts with $S^{2-}$ ions to form a purple or violet complex,$[Fe(CN)_5NOS]^{4-}$. This is correct.
$2$. The brown ring complex is $[Fe(H_2O)_5NO]SO_4$,where $Fe$ is in the $+1$ oxidation state. This is correct.
$3$. $Fe^{3+}$ ions react with $SCN^-$ to form a blood-red complex $[Fe(SCN)(H_2O)_5]^{2+}$. Basic ferric acetate also gives this test. This is correct.
$4$. Tollen's reagent $([Ag(NH_3)_2]^+)$ is used to detect aldehydes by forming a silver mirror. $H_3PO_4$ is a phosphoric acid and does not reduce Tollen's reagent to form a silver mirror. This is incorrect.

(B) $1$. $Cu^{2+}$ reacts with $K_4[Fe(CN)_6]$ to form $Cu_2[Fe(CN)_6]$,which is a chocolate-brown precipitate.
$2$. In the presence of $HCl$,the concentration of $S^{2-}$ ions is suppressed due to the common ion effect. $CuS$ $(K_{sp} \approx 10^{-36})$ precipitates,but $NiS$ $(K_{sp} \approx 10^{-21})$ does not precipitate because its ionic product remains lower than its $K_{sp}$. Thus,the statement that both give black precipitate is incorrect.
$3$. $Fe^{3+}$ reacts with $SCN^-$ to form $[Fe(SCN)]^{2+}$,which is blood-red.
$4$. $Cu^{2+}$ in a reducing flame forms metallic copper,which gives a red-coloured bead.

(D) Group $IV$ cations include $Co^{2+}, Ni^{2+}, Mn^{2+},$ and $Zn^{2+}$.
When $H_2S$ is passed through a basic solution of $Zn^{2+}$,a white precipitate of $ZnS$ is formed.
This precipitate dissolves in dilute $HCl$ to form $ZnCl_2$: $ZnS + 2HCl \rightarrow ZnCl_2 + H_2S$.
$ZnCl_2$ reacts with $NaOH$ to form a white precipitate of $Zn(OH)_2$: $Zn^{2+} + 2OH^{-} \rightarrow Zn(OH)_2 \downarrow$ (White).
With potassium ferrocyanide $(K_4[Fe(CN)_6])$,$Zn^{2+}$ ions form a bluish-white precipitate of $K_2Zn_3[Fe(CN)_6]_2$: $3Zn^{2+} + 2K^{+} + 2[Fe(CN)_6]^{4-} \rightarrow K_2Zn_3[Fe(CN)_6]_2 \downarrow$ (Bluish-white).

(B) The presence of $Fe^{3+}$ ions is indicated by the blood-red color formed with $CNS^{-}$ ions: $Fe^{3+} + 3CNS^{-} \to Fe(CNS)_3$ (Ferric thiocyanate,blood-red coloration).
$Fe^{3+}$ ions react with $K_4[Fe(CN)_6]$ to form a blue precipitate (Prussian blue): $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \to Fe_4[Fe(CN)_6]_3$.
The chromyl chloride test is specific for the presence of chloride ions $(Cl^{-})$,where $X$ reacts with $K_2Cr_2O_7$ and concentrated $H_2SO_4$ to evolve reddish-brown vapors of chromyl chloride $(CrO_2Cl_2)$.
Since the salt $X$ contains both $Fe^{3+}$ and $Cl^{-}$ ions,the salt is $FeCl_3$.

(C) In qualitative inorganic analysis,cations are divided into groups based on their precipitation reactions.
$Pb^{2+}$ (Group $II$),$Cu^{2+}$ (Group $II$),and $As^{3+}$ (Group $II$) are precipitated as sulfides by $H_2S$ in the presence of $dil. \ HCl$ due to the low solubility product of their sulfides.
$Co^{2+}$ belongs to Group $IV$ and is not precipitated by $H_2S$ in the presence of $dil. \ HCl$ because the concentration of $S^{2-}$ ions is too low due to the common ion effect of $H^+$.
$Co^{2+}$ is precipitated as $CoS$ only in the presence of $NH_4OH$ and $H_2S$ (Group $IV$ reagent).

(A) Hydrated cobalt$(II)$ salts,such as $CoCl_2 \cdot 6H_2O$,are pink in colour due to the presence of the $[Co(H_2O)_6]^{2+}$ complex.
Upon heating,the water of crystallization is lost,and the salt converts to anhydrous cobalt$(II)$ chloride $(CoCl_2)$,which is blue in colour.
Therefore,the presence of the $Co^{2+}$ cation is most likely.

(A) In the analysis of group $V$ cations,$(NH_4)_2CO_3$ is used as a group reagent in the presence of $NH_4Cl$ and $NH_4OH$.
$NH_4Cl$ suppresses the ionization of $(NH_4)_2CO_3$ due to the common ion effect,keeping the concentration of $CO_3^{2-}$ low enough to exceed only the solubility product $(K_{sp})$ of $Ca^{2+}$,$Ba^{2+}$,and $Sr^{2+}$ carbonates.
If $Na_2CO_3$ is used instead,it is a strong electrolyte and dissociates completely to provide a high concentration of $CO_3^{2-}$ ions.
This high concentration of $CO_3^{2-}$ exceeds the $K_{sp}$ of $MgCO_3$ as well,causing $Mg^{2+}$ ions to precipitate along with the group $V$ cations.

(B) The reaction of $Fe^{3+}$ with $SCN^{-}$ produces a blood-red complex $[Fe(SCN)]^{2+}$.
$Fe^{2+}$ does not form this complex,making statement $B$ incorrect.
$Na_2ZnO_2$ reacts with $H_2S$ to form a white precipitate of $ZnS$.
$Cu^{2+}$ reacts with excess $NH_3$ to form the deep blue $[Cu(NH_3)_4]^{2+}$ complex.

(A) In an acidified aqueous solution,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$.
Only the sulphides of group $II$ cations $(Cu^{2+}, Hg^{2+})$ have a solubility product $(K_{sp})$ low enough to be exceeded by the ionic product at this low $S^{2-}$ concentration.
Therefore,$CuS$ and $HgS$ precipitate,while $Mn^{2+}$ and $Ni^{2+}$ remain in the solution.

(B) The reaction between sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ and sulfide ions $(S^{2-})$ is a characteristic test for sulfide ions.
The reaction is: $Na_2[Fe(CN)_5NO] + Na_2S \rightarrow Na_4[Fe(CN)_5(NOS)]$.
This reaction results in the formation of a violet-colored complex known as sodium thionitroprusside.

(A) $(i) \, Ni^{2+}$ reacts with Dimethylglyoxime $(DMG)$ to form a rosy red precipitate of $[Ni(DMG)_2]$. Thus,$(i) - D$.
$(ii) \, Ag^{+}$ reacts with Sodium thiosulphate $(Na_2S_2O_3)$ to form a soluble complex $Na_3[Ag(S_2O_3)_2]$. Thus,$(ii) - A$.
$(iii) \, Cu^{2+}$ reacts with Ammonia $(NH_3)$ to form a deep blue complex $[Cu(NH_3)_4]^{2+}$. Thus,$(iii) - C$.
$(iv) \, S^{2-}$ reacts with Sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ to form a purple colored complex $Na_4[Fe(CN)_5(NOS)]$. Thus,$(iv) - B$.
Therefore,the correct matching is $(i)-D, (ii)-A, (iii)-C, (iv)-B$.