Wet Test for Basic Radical Questions in English

(A) When $NaOH$ solution is added to $HgCl_2$,it forms a yellow precipitate of mercury$(II)$ oxide $(HgO)$.
The chemical reaction is: $HgCl_2 + 2NaOH \to HgO(s) + 2NaCl + H_2O$.
Thus,the correct option is $A$.

(D) $PbCl_2$ is sparingly soluble in cold water but becomes significantly more soluble in hot water. This behavior is characteristic of lead$(II)$ chloride,which is often used to separate lead from other metal ions in qualitative analysis.

(C) The $V^{-}A$ group (Group $15$ in modern $IUPAC$,but traditionally Group $V$ in qualitative analysis) contains $Bi^{3+}$.
When the black precipitate of $Bi_2S_3$ is dissolved in $50\% \ HNO_3$,it forms $Bi(NO_3)_3$.
Upon adding an excess of $NH_4OH$ to this solution,a white precipitate of $Bi(OH)_3$ is obtained.
Therefore,the correct option is $(C)$.

(D) The reaction between $NaCl$,$K_2Cr_2O_7$,and concentrated $H_2SO_4$ is known as the chromyl chloride test.
The chemical equation is: $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O$.
$CrO_2Cl_2$ (chromyl chloride) is evolved as an orange-red gas.

(B) Concentrated aqueous $NaOH$ is a strong base that reacts with amphoteric metal ions to form soluble complexes or salts,while it does not react with non-amphoteric metal ions like $Fe^{3+}$,which precipitate as hydroxides.
$Al^{3+}$ is amphoteric and reacts with $NaOH$ to form soluble sodium meta-aluminate ($NaAlO_2$ or $Na[Al(OH)_4]$).
$Fe^{3+}$ is not amphoteric and forms an insoluble brown precipitate of ferric hydroxide $(Fe(OH)_3)$ in the presence of $NaOH$.
Therefore,$NaOH$ can be used to separate $Al^{3+}$ from $Fe^{3+}$.

(A) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which has the chemical formula $K_2[HgI_4]$.
It is commonly used to detect the presence of ammonia $(NH_3)$ or ammonium ions $(NH_4^+)$ in a solution,forming a brown precipitate known as the iodide of Millon's base.

(A) When $H_2S$ gas is passed through a solution of $AgNO_3$,it forms silver sulfide $(Ag_2S)$,which is a black precipitate.
The chemical reaction is: $2AgNO_3 + H_2S \to Ag_2S \downarrow (\text{Black ppt.}) + 2HNO_3$.
Therefore,the correct option is $A$.

(B) The sulphides of $As$,$Sb$,and $Sn$ are generally soluble in yellow ammonium sulphide $(NH_4)_2S_x$. However,among the given options,$As_2S_3$ is specifically known to be soluble in ammonium carbonate $(NH_4)_2CO_3$ solution due to the formation of a soluble complex. $SnS$ and $Sb_2S_3$ are not soluble in $(NH_4)_2CO_3$,and $CdS$ is insoluble in both.

(B) When $SnS$ is dissolved in yellow ammonium sulphide $(NH_4)_2S_2$,it forms a soluble thio-salt: $SnS + (NH_4)_2S_2 \rightarrow (NH_4)_2SnS_3$.
Upon adding $HCl$ to this solution,the thio-salt decomposes to precipitate stannic sulphide $(SnS_2)$: $(NH_4)_2SnS_3 + 2HCl \rightarrow 2NH_4Cl + H_2S + SnS_2 \downarrow$.

(B) When $H_2S$ is passed through a solution containing an oxidising agent,the $H_2S$ is oxidised to colloidal sulphur $(S)$.
$H_2S + [O] \rightarrow H_2O + S \text{ (milky appearance)}$.
Therefore,the milkiness indicates the presence of an oxidising agent.

(B) $HNO_3$ is a strong oxidizing agent.
It oxidizes $Fe^{2+}$ ions to $Fe^{3+}$ ions.
If $Fe^{2+}$ is not oxidized to $Fe^{3+}$,it will not be completely precipitated as $Fe(OH)_3$ in the presence of $NH_4OH$ and $NH_4Cl$.

(A) In qualitative inorganic analysis,$Hg_2^{2+}$ belongs to Group-$I$ basic radicals.
When $HCl$ is added to an aqueous solution containing $Hg_2^{2+}$ ions,it reacts to form mercurous chloride $(Hg_2Cl_2)$,which is a white precipitate.
The chemical equation is: $Hg_2^{2+}(aq) + 2Cl^-(aq) \to Hg_2Cl_2(s) \downarrow$ (white precipitate).
Other ions like $Mg^{2+}$,$Zn^{2+}$,and $Cd^{2+}$ do not form insoluble chlorides with $HCl$.

(D) $Sb_2S_3$ (Antimony trisulphide) is an acidic sulphide.
It is insoluble in dilute acids but dissolves in alkaline solutions (like $NaOH$ or $KOH$) to form thio-salts,such as $Na_3SbS_3$ and $Na_3SbS_4$.

(B) The $IV^{th}$ group basic radicals $(Co^{2+}, Ni^{2+}, Mn^{2+}, Zn^{2+})$ are precipitated as their sulphides in an alkaline medium.
In the presence of $NH_4OH$,the dissociation of $H_2S$ is increased due to the removal of $H^+$ ions by $OH^-$ ions,which shifts the equilibrium $H_2S \rightleftharpoons 2H^+ + S^{2-}$ to the right.
This increases the concentration of $S^{2-}$ ions,allowing the solubility product of $IV^{th}$ group sulphides to be exceeded.
Therefore,$H_2S$ in the presence of $NH_4OH$ (alkaline medium) is the correct reagent.

(D) In qualitative inorganic analysis,the $II^{nd}$ group radicals (like $Cu^{2+}$,$Cd^{2+}$,$As^{3+}$,$Sb^{3+}$) are precipitated as sulphides by passing $H_2S$ gas in the presence of dilute $HCl$.
However,the question asks which radical is not precipitated in a $concentrated$ acid solution.
Actually,all these radicals ($Cu^{2+}$,$Cd^{2+}$,$As^{3+}$,$Sb^{3+}$) are precipitated as sulphides in the presence of $dilute$ $HCl$.
If the acid concentration is too high,the common ion effect (due to $H^+$) suppresses the dissociation of $H_2S$,reducing the concentration of $S^{2-}$ ions.
Among the given options,$CdS$ has the highest solubility product $(K_{sp})$ compared to $CuS$,$As_2S_3$,and $Sb_2S_3$.
Therefore,$Cd^{2+}$ is the most likely to remain in solution if the acid concentration is increased significantly,preventing the precipitation of $CdS$.

(A) In the presence of dilute $HCl$,$H_2S$ provides a low concentration of $S^{2-}$ ions due to the common ion effect.
This concentration is sufficient to exceed the solubility product $(K_{sp})$ of group $II$ metal sulphides (like $Bi_2S_3, SnS_2, HgS, CuS$) but not group $IV$ metal sulphides (like $ZnS, NiS$).
$Bi^{3+}$ and $Sn^{4+}$ both belong to group $II$ and will both precipitate as sulphides in dilute $HCl$,making them inseparable by this method.

(B) is the correct answer.
Both $Pb^{2+}$ and $Ag^+$ ions form white precipitates of their respective chlorides ($PbCl_2$ and $AgCl$) when treated with dilute $HCl$.
However,$PbCl_2$ is soluble in hot water,whereas $AgCl$ remains insoluble in hot water.
Therefore,hot water or hot dilute $HCl$ can be used to distinguish between them.

(A) The group reagent for the precipitation of group $II$ basic radicals in qualitative analysis is dil. $HCl + H_2S$.
In group $II$,the basic radicals are precipitated as their respective sulphides in an acidic medium provided by dil. $HCl$.

(B) In qualitative inorganic analysis,$Pb^{2+}$,$Ag^{+}$,and $Sn^{2+}$ are precipitated as chlorides in Group $I$ or $II$ analysis using $HCl$.
$Cu^{+}$ (cuprous ion) is generally unstable in aqueous solution and disproportionates into $Cu^{2+}$ and $Cu^0$.
However,among the given options,$Cu^{+}$ is not a standard cation encountered in the systematic group analysis scheme for precipitation by $HCl$ or $H_2S$ in the same manner as the others,as it is not stable in aqueous solution.
Therefore,$Cu^{+}$ is the ion that does not follow the standard precipitation behavior of the other listed ions.

(C) The reaction between lead$(II)$ acetate and hydrogen sulfide is as follows:
$Pb(CH_3COO)_2 + H_2S \to 2CH_3COOH + PbS \downarrow$
$PbS$ (Lead sulfide) is formed as a black precipitate.

(A) . $NH_4SCN$ reacts with $Fe^{3+}$ ions to form a blood-red colored complex,$[Fe(SCN)(H_2O)_5]^{2+}$.
$Fe^{2+}$ ions do not form this complex,thus allowing them to be distinguished.

(A) $Hg_2^{2+}$ ions react with dilute $HCl$ to form a white precipitate of mercury$(I)$ chloride $(Hg_2Cl_2)$.
$Hg_2^{2+} + 2Cl^{-} \longrightarrow Hg_2Cl_2 \downarrow \text{ (White ppt)}$
This precipitation occurs because the solubility product $(K_{sp})$ of $Hg_2Cl_2$ is very low,making it insoluble in dilute $HCl$.

(C) Lead sulphate $(PbSO_4)$ is insoluble in water and most acids,but it is soluble in a solution of ammonium acetate $(CH_3COONH_4)$.
This occurs because of the formation of a soluble complex or the displacement reaction:
$PbSO_4 + 2CH_3COONH_4 \rightarrow Pb(CH_3COO)_2 + (NH_4)_2SO_4$

(D) The precipitation of metal sulfides in the presence of $HCl$ depends on the solubility product $(K_{sp})$ of the sulfides.
$Cu^{2+}$,$Hg^{2+}$,and $Bi^{3+}$ belong to Group-$II$ of the qualitative analysis scheme,where their sulfides ($CuS$,$HgS$,$Bi_2S_3$) have very low $K_{sp}$ values and precipitate in the presence of $H_2S$ and $HCl$.
$Co^{2+}$ belongs to Group-$III$ (or $IV$ depending on the scheme),and its sulfide $(CoS)$ has a higher $K_{sp}$ compared to Group-$II$ sulfides.
Therefore,$CoS$ does not precipitate in the acidic medium provided by $HCl$.

(B) In qualitative inorganic analysis,the group reagent for analytical group $IV$ cations (such as $Co^{2+}$,$Ni^{2+}$,$Mn^{2+}$,$Zn^{2+}$) is $H_2S$ gas in the presence of $NH_4Cl$ and $NH_4OH$ (buffered alkaline medium). Therefore,the correct combination is $NH_4Cl + NH_4OH + H_2S$.

(D) In group $III$ analysis,$Fe^{3+}$ and $Cr^{3+}$ are precipitated as hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
$NH_4Cl$ provides $NH_4^+$ ions,which suppress the dissociation of $NH_4OH$ due to the common ion effect,thereby decreasing the concentration of $OH^-$ ions.
This ensures that only the hydroxides of group $III$ cations ($Fe^{3+}$,$Al^{3+}$,$Cr^{3+}$) precipitate,while preventing the precipitation of group $IV$ cations.
$Fe(OH)_3$ forms a reddish-brown precipitate,whereas $Cr(OH)_3$ forms a green precipitate,allowing for their differentiation.

(D) In the qualitative analysis of basic radicals,$H_2S$ gas is used as a group reagent for the precipitation of metal sulfides.
Specifically,$H_2S$ in the presence of dilute $HCl$ is used for group $II$ radicals (e.g.,$Cu^{2+}$,$Cd^{2+}$,$As^{3+}$).
$H_2S$ in the presence of $NH_4OH$ is used for group $IV$ radicals (e.g.,$Zn^{2+}$,$Mn^{2+}$,$Ni^{2+}$,$Co^{2+}$).
Therefore,$H_2S$ is used in group $II$ and group $IV$.

(B) $BaCO_3$ is a carbonate salt that is insoluble in water but reacts with dilute $HCl$ to form $BaCl_2$,$H_2O$,and $CO_2$ gas,resulting in a clear solution.
$ZnS$ is a metal sulfide that is insoluble in water but dissolves in dilute $HCl$ to form $ZnCl_2$ and $H_2S$ gas,resulting in a clear solution.
Both salts dissolve in dilute $HCl$ to form colourless solutions.
Reaction $1$: $BaCO_3 + 2HCl \to BaCl_2 + H_2O + CO_2 \uparrow$
Reaction $2$: $ZnS + 2HCl \to ZnCl_2 + H_2S \uparrow$
Therefore,the correct mixture is $BaCO_3$ and $ZnS$.

(C) $Cd^{2+}$ is not precipitated by $H_2S$ in a concentrated acid solution because the high concentration of $H^+$ ions suppresses the ionization of $H_2S$ due to the common ion effect.
Therefore,the solution must be diluted before passing $H_2S$ through it to precipitate $CdS$.

(C) $Na_2CO_3$ is a strong electrolyte that provides a high concentration of $CO_3^{2-}$ ions.
This high concentration of carbonate ions causes the precipitation of $Mg^{2+}$ ions as $MgCO_3$ along with the fifth group radicals $(Ba^{2+}, Sr^{2+}, Ca^{2+})$.
$(NH_4)_2CO_3$ is a weaker electrolyte and provides a lower concentration of $CO_3^{2-}$ ions,which is sufficient to exceed the solubility product $(K_{sp})$ of the fifth group carbonates but not $MgCO_3$.

(B) The addition of $NH_4Cl$ followed by $NH_4OH$ creates a buffer solution that suppresses the concentration of $OH^-$ ions due to the common ion effect of $NH_4^+$.
In this condition,the concentration of $OH^-$ is sufficient to exceed the solubility product $(K_{sp})$ of $III^{rd}$ group cations ($Al^{3+}$,$Fe^{3+}$,$Cr^{3+}$) but insufficient to precipitate $IV^{th}$ group cations ($Zn^{2+}$,$Ni^{2+}$,$Mn^{2+}$,$Co^{2+}$).
Therefore,only $Al(OH)_3$ and $Fe(OH)_3$ will precipitate as hydroxides.
The correct option is $B$.

(D) In an acidic medium,the ionization of $H_2S$ is suppressed due to the common ion effect of $H^+$ ions,resulting in a very low concentration of $S^{2-}$ ions.
Only the metal sulfide with the lowest solubility product $(K_{sp})$,which is $CuS$,will exceed its solubility product and precipitate.
$Cu^{2+}$ belongs to Group $II$ of the qualitative analysis scheme,while $Zn^{2+}$ and $Ni^{2+}$ belong to later groups that require higher $S^{2-}$ concentrations for precipitation.
Therefore,only $Cu^{2+}$ precipitates.

(B) When $Sb_2S_3$ is dissolved in yellow ammonium sulphide $(NH_4)_2S_2$,it forms a soluble thio-salt: $Sb_2S_3 + 2(NH_4)_2S_2 \to 2(NH_4)_3SbS_4$ (or similar complex).
Upon adding $HCl$ to this solution,the thio-salt decomposes to precipitate the higher sulphide of antimony,which is $Sb_2S_5$.
The reaction is: $2(NH_4)_3SbS_4 + 6HCl \to Sb_2S_5 + 3H_2S + 6NH_4Cl$.

(A) In a $0.3 \, M$ $HCl$ solution,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$ ions from $HCl$,which suppresses the dissociation of $H_2S$.
Only the sulfides of the Group $II$ basic radicals,which have very low solubility products $(K_{sp})$,will precipitate under these conditions.
Among the given ions,$Hg^{2+}, Cd^{2+},$ and $Cu^{2+}$ belong to Group $II$,while $Fe^{2+}$ and $Sr^{2+}$ do not precipitate in acidic medium.
Therefore,$Hg^{2+}, Cd^{2+},$ and $Cu^{2+}$ will precipitate as their respective sulfides.

(A) The reaction of $H_2S$ with metal salts leads to the formation of metal sulphides.
$CuCl_2 + H_2S \to CuS \text{ (Black ppt.)} + 2HCl$.
$CdCl_2 + H_2S \to CdS \text{ (Yellow ppt.)} + 2HCl$.
$ZnCl_2 + H_2S \to ZnS \text{ (White ppt.)} + 2HCl$.
$NaCl$ does not react with $H_2S$ to form a sulphide precipitate.
Therefore,the correct option is $A$.

(D) In qualitative inorganic analysis,the $Group \ I$ basic radicals are $Pb^{2+}$,$Hg_2^{2+}$,and $Ag^+$.
These ions form insoluble chlorides when treated with dilute $HCl$ due to their very low solubility product $(K_{sp})$ values.
$Hg^{2+}$ and $Cd^{2+}$ belong to $Group \ II$ and do not precipitate as chlorides with dilute $HCl$.
Therefore,the addition of dilute $HCl$ precipitates $Hg_2Cl_2$ and $PbCl_2$.

(A) The group $III$ radicals $(Fe^{3+}, Al^{3+}, Cr^{3+})$ are precipitated as hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
$NH_4Cl$ provides $NH_4^+$ ions,which suppress the ionization of $NH_4OH$ due to the common ion effect,ensuring that the concentration of $OH^-$ ions is low enough to prevent the precipitation of group $IV$ and $V$ radicals.
Any ammonium salt that provides $NH_4^+$ ions without introducing interfering anions can be used.
$NH_4NO_3$ is a suitable substitute because it provides $NH_4^+$ ions and does not interfere with the precipitation process.
$(NH_4)_2SO_4$ cannot be used because $SO_4^{2-}$ ions would precipitate $Ba^{2+}, Sr^{2+},$ and $Ca^{2+}$ as sulfates.

(A) The solubility of metal sulfides in $HNO_3$ depends on the solubility product $(K_{sp})$ of the sulfide and the oxidizing power of the acid.
$HgS$ has an extremely low $K_{sp}$ value $(10^{-52})$,which makes it insoluble in hot dilute $HNO_3$.
$PbS$,$CuS$,and $CdS$ have higher $K_{sp}$ values compared to $HgS$ and dissolve in hot dilute $HNO_3$ to form their respective nitrates and sulfur or $H_2S$ (which is then oxidized).
Therefore,$HgS$ does not dissolve.

(A) $Pb^{2+}$ ions can be precipitated by $HCl$ as $PbCl_2$ (Group $I$ basic radical).
$Pb^{2+}$ ions can also be precipitated by $H_2S$ in the presence of $HCl$ as $PbS$ (Group $II$ basic radical).
Therefore,$Pb^{2+}$ is the correct answer.

(C) The solubility of metal sulphates in water depends on the balance between hydration energy and lattice energy.
$PbSO_4$ is well-known to be insoluble in water due to its high lattice energy.
While $Bi_2(SO_4)_3$ is also sparingly soluble or insoluble due to its covalent character and high lattice energy,$PbSO_4$ is the standard textbook example of an insoluble sulphate in qualitative analysis.
Therefore,$PbSO_4$ is the most appropriate answer.

(B) The correct answer is $(B)$.
Mercurous chloride $(Hg_2Cl_2)$ reacts with ammonium hydroxide $(NH_4OH)$ to form a black precipitate of finely divided mercury $(Hg)$ and an amino-mercuric chloride complex.
The chemical equation is: $Hg_2Cl_2 + 2NH_4OH \to NH_2HgCl + Hg \text{ (black ppt)} + NH_4Cl + 2H_2O$.

(D) . The fifth group radicals $(Ba^{2+}, Sr^{2+}, Ca^{2+})$ are precipitated as carbonates using $(NH_4)_2CO_3$ in the presence of $NH_4Cl$ and $NH_4OH$.
When $(NH_4)_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions is kept low due to the common ion effect of $NH_4^+$.
However,if $Na_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions becomes very high.
This high concentration of $CO_3^{2-}$ causes $Mg^{2+}$ ions (which belong to the $VI^{th}$ group) to precipitate as $MgCO_3$ along with the $V^{th}$ group radicals,thus interfering with the analysis.

(B) The salt is light green,which is characteristic of $Fe^{2+}$ ions in aqueous solution.
When $H_2S$ is passed through a solution containing $Fe^{2+}$ ions (usually in the presence of a buffer or under specific conditions),it forms $FeS$,which is a black precipitate.
$FeS$ is a metal sulfide that is readily soluble in dilute $HCl$ because the reaction $FeS + 2HCl \rightarrow FeCl_2 + H_2S$ occurs,shifting the equilibrium.
Therefore,the metal ion present is $Fe^{2+}$.

(D) The separation of group $II$ cations is based on the solubility of their sulphides in yellow ammonium sulphide,$(NH_4)_2S_x$.
Group $IIA$ sulphides (e.g.,$HgS, PbS, Bi_2S_3, CuS, CdS$) are insoluble in yellow ammonium sulphide.
Group $IIB$ sulphides (e.g.,$As_2S_3, Sb_2S_3, SnS_2$) are soluble in yellow ammonium sulphide due to the formation of thio-salts.
In the given options,$CdS$ belongs to group $IIA$ (insoluble) and $As_2S_3$ belongs to group $IIB$ (soluble).
Therefore,yellow ammonium sulphide is used to separate $CdS$ and $As_2S_3$.

(B) In the qualitative analysis of basic radicals,$NH_4Cl$ and $NH_4OH$ (aqueous $NH_3$) are used as group reagents for Group-$III$ cations.
$NH_4Cl$ suppresses the dissociation of $NH_4OH$ due to the common ion effect,thereby reducing the concentration of $OH^-$ ions.
This low concentration of $OH^-$ is sufficient to exceed the solubility product $(K_{sp})$ of $Al(OH)_3$,$Fe(OH)_3$,and $Cr(OH)_3$,causing them to precipitate.
Therefore,$Al^{3+}$ is precipitated by these reagents.

(B) The reaction between $SnCl_2$ and $HgCl_2$ occurs in two steps:
$1$. $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 (\text{white ppt}) + SnCl_4$
$2$. $Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg (\text{gray ppt}) + SnCl_4$
Thus,the initial white precipitate of $Hg_2Cl_2$ turns gray due to the formation of metallic mercury $(Hg)$.

(C) The colourless crystalline salt $X$ is $MgSO_4$. It is soluble in dilute $HCl$.
When $NaOH$ solution is added to a solution of $MgSO_4$,it forms a white precipitate of magnesium hydroxide,$Mg(OH)_2$,which is insoluble in excess $NaOH$.
$MgSO_4 + 2NaOH \rightarrow Mg(OH)_2 \downarrow + Na_2SO_4$
In contrast,$Al^{3+}$,$Zn^{2+}$,and $Sn^{2+}$ salts form precipitates that are soluble in excess $NaOH$ due to the formation of soluble complex ions like $[Al(OH)_4]^-$,$[Zn(OH)_4]^{2-}$,or $[Sn(OH)_4]^{2-}$.

(A) Group $IV$ cations $(Zn^{2+}, Mn^{2+}, Ni^{2+}, Co^{2+})$ are precipitated as their sulfides in the presence of $NH_4OH$ and $H_2S$.
In this medium,the concentration of $S^{2-}$ ions is high due to the common ion effect and the basic medium,which exceeds the solubility product $(K_{sp})$ of group $IV$ sulfides.
Therefore,$H_2S$ must be highly ionised to provide a sufficient concentration of sulfide ions for precipitation.

(B) When aqueous $NaOH$ is added to a solution containing $Al^{3+}$ ions,a white gelatinous precipitate of $Al(OH)_3$ is formed.
$Al^{3+} + 3OH^{-} \to Al(OH)_3 \downarrow$ (white gelatinous precipitate)
This precipitate is amphoteric in nature and dissolves in excess $NaOH$ to form a soluble complex,sodium aluminate.
$Al(OH)_3 + OH^{-} \to [Al(OH)_4]^{-}$