Salt Analysis

Only $\mathrm{Cu}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ca}^{2+} \& \mathrm{Zn}^{2+}$ form precipitate with $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

To determine the number of white coloured salts among the listed compounds, we look at their common color characteristics:

(a) $\mathrm{SrSO}_4$ - Generally, most sulfate salts are white unless the cation imparts a color. Strontium sulfate is indeed white.

(b) $\mathrm{Mg}\left(\mathrm{NH}_4\right)\mathrm{PO}_4$ - Magnesium ammonium phosphate is also known to be white. This is typical for many phosphates with colorless cations.

(c) $\mathrm{BaCrO}_4$ - Barium chromate is well-known for its yellow color, a characteristic of many chromates.

(d) $\mathrm{Mn}(\mathrm{OH})_2$ - Manganese hydroxide can have a faint pink to white color, but in this context, we consider it white for counting purposes.

(e) $\mathrm{PbSO}_4$ - Lead sulfate is white, consistent with many lead salts not influenced by strong coloring anions.

(f) $\mathrm{PbCrO}_4$ - Lead chromate is distinguished by its yellow color, again a feature of chromate salts.

(g) $\mathrm{AgBr}$ - Silver bromide is known for its pale yellow to virtually white appearance, but it is more accurately described as pale yellow.

(h) $\mathrm{PbI}_2$ - Lead iodide exhibits a bright yellow color, typical for many iodides with heavy metal cations.

(i) $\mathrm{CaC}_2\mathrm{O}_4$ - Calcium oxalate is white, as most insoluble oxalates are.

(j) $\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3\mathrm{COO}\right)\right]$ - This complex containing iron is brown red due to the presence of iron ions.

Summarizing, the white salts among the given compounds are $\mathrm{SrSO}_4$, $\mathrm{Mg}\left(\mathrm{NH}_4\right)\mathrm{PO}_4$, $\mathrm{Mn}(\mathrm{OH})_2$, $\mathrm{PbSO}_4$, and $\mathrm{CaC}_2\mathrm{O}_4$, making a total of 5 white colored salts.

All the following metal ions will respond to flame test.

$\mathrm{Sr}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Cu}^{2+}$

$ \mathrm{KI}+\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\frac{1}{2} \mathrm{I}_2 $

Moles of KI required $=2$

$\begin{gathered}\mathrm{Zn}^{2+}+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2 \\ \mathrm{OR} \\ \mathrm{Zn}^{2+}+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]\end{gathered}$

Univalent metal means alkali metal. A is binary compound so A can be KO or KO₂.

But here A react completely with sulphur.

$\therefore$ A must be KO₂.

To check A and B are right or not we have to check with their molar ratio.

Here molar ratio of KO₂ and S = 2 : 1

Moles of 1.422 g KO₂ = ${{1.422} \over {71}} = 0.02$

Moles of 0.321 g S = ${{0.321} \over {32}} = 0.01$

$\therefore$ Molar ratio = 0.02 : 0.01 = 2 : 1

$\therefore$ A and B are right choice.

When K⁺ is the monovalent cation and Al⁺³ is the trivalent cation then double salt C is $ = {K_2}S{O_4}.A{l_2}{(S{O_4})_3}.24{H_2}O$

(i) Gas A turned lime water milky, which indicates the presence of carbon dioxide. Therefore, gas A is carbon dioxide (CO₂).

(ii) Gas B, which condenses to a colourless liquid: This is most likely water (H₂O).

(iii) $Y$ when dissolved in water yields an alkaline solution and the solution on treatment with $\mathrm{BaCl}_2$ solution forms a white ppt, of $Z$. The compound $Z$ on treatment with acid gives effervescence of $\mathrm{CO}_2$ so $Z$ and hence $Y$ must be a carbonate, $\mathrm{CO}_3^{2-}$. We can thus write $Y$ as $M \mathrm{CO}_3$ or $M_2 \mathrm{CO}_3$.

(iv) $X$ on being heated yields a carbonate $Y$, hence $\mathrm{CO}_{2(g)}$ i.e. $A$ and another gas $B$, hence it must be a bicarbonate, $\mathrm{HCO}_3^{-}$.


$4.4 \mathrm{~g}$ of $\mathrm{CO}_2$ is given by $M\left(\mathrm{HCO}_3\right)=16.8 \mathrm{~g}$

$44 \mathrm{~g}$ of $\mathrm{CO}_2$ is given by $M\left(\mathrm{HCO}_3\right)$

$ =\frac{16.8}{4.4} \times 44=168 \mathrm{~g} $

Because 2 molecules of $M\left(\mathrm{HCO}_3\right)$ are involved in the reaction so molecular weight of

$M\left(\mathrm{HCO}_3\right)=168 / 2=84$

Let the atomic weight be $M$.

Molecular weight of $M\left(\mathrm{HCO}_3\right)=M+1+12+48$$=M+61$

$ \therefore M+61=84 \text { or } M=84-61=23 $

Thus the metal must be $\mathrm{Na}$ and so the given salt $X$ is $\mathrm{Na}\left(\mathrm{HCO}_3\right)$.

$\mathrm{FeCl}_3$ is a salt of strong acid and weak base. In water, it hydrolyses slowly producing brown ppt of $\mathrm{Fe}(\mathrm{OH})_3$.

$ \mathrm{FeCl}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{HCl}+\mathrm{Fe}(\mathrm{OH})_3 \downarrow $(Brown)

Let's analyze the given information step-by-step to identify the compounds.

(i) Dissolution in dilute sulfuric acid without gas evolution: This suggests that compound A is likely a metal compound that forms a soluble sulfate. The lack of gas rules out carbonates and many other reactive compounds.

(ii) Decolorization of KMnO₄: Potassium permanganate ($KMnO_4$) is a strong oxidizing agent. Its decolorization indicates that compound A, or a species derived from it in the acidic solution, is readily oxidized. This points towards a compound containing a metal in a lower oxidation state, capable of being oxidized to a higher oxidation state.

(iii) Strong heating producing pungent gases (B and C) and brown residue (D): The pungent gases suggest the presence of sulfur and/or nitrogen. The brown residue is indicative of a metal oxide. The fact that gases are produced implies that compound A might be a metal salt containing an anion which decomposes upon heating.

(iv) Gases B and C turn a dichromate solution green: Dichromate ($Cr_2O_7^{2-}$) solutions are orange. Their reduction to a green solution indicates that gases B and C are reducing agents. The green color is characteristic of chromium(III) ions ($Cr^{3+}$). This strongly suggests that one of the gases (B or C) is $SO_2$, which is capable of reducing dichromate.

(v) The green solution gives a white precipitate (E) with barium nitrate: The formation of a white precipitate with barium nitrate is a classic test for sulfate ions ($SO_4^{2-}$). The reaction is:

$Ba^{2+}(aq) \ + \ SO_4^{2-}(aq) \ \rightarrow \ BaSO_4(s)$

Therefore, gas B or C must be sulfur dioxide ($SO_2$), which is oxidized to sulfate ions ($SO_4^{2-}$) upon reaction with the dichromate solution.

(vi) Residue D gives a magnetic substance (E) upon reduction: This strongly suggests that residue D is a metal oxide of a transition metal, and the magnetic substance E is the corresponding metal. The reduction reaction can be generally represented as:

$Metal Oxide \ + \ Reducing \ Agent \ \rightarrow \ Metal$

Considering all the observations, a plausible scenario is that compound A is ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$). This is a light green crystalline solid. Let's check if it fits the observations:

(i) & (ii): Ferrous sulfate dissolves in dilute sulfuric acid and gets oxidized by $KMnO_4$.

(iii): Upon strong heating, ferrous sulfate decomposes. The gases produced include sulfur dioxide ($SO_2$ - gas B) and sulfur trioxide ($SO_3$ - gas C). Both have pungent smells. The brown residue (D) is ferric oxide ($Fe_2O_3$).

(iv): $SO_2$ reduces dichromate to $Cr^{3+}$.

(v): $SO_2$ is oxidized to $SO_4^{2-}$, which forms a white precipitate of barium sulfate ($BaSO_4$) with barium nitrate.

(vi): Reduction of $Fe_2O_3$ on charcoal produces elemental iron (Fe - E), which is magnetic.

Therefore, the compounds are:

A: Ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$)

B: Sulfur dioxide ($SO_2$)

C: Sulfur trioxide ($SO_3$)

D: Ferric oxide ($Fe_2O_3$)

E: Iron (Fe) (or Barium Sulfate ($BaSO_4$), depending on which step (v) or (vi) is considered)

(i) The compound $C$ produced by heating $A$ is white in colour and changes to yellow on heating, thus compound $C$ may be $\mathrm{ZnO}$. $C$ with dil. $\mathrm{HCl}$ and $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right.$ ]gives white ppt. This confirms that the compound $C$ must be $\mathrm{ZnO}$.

$ \begin{aligned} & A \stackrel{\Delta}{\longrightarrow} { } \underset{C}{\mathrm{ZnO}}+B \text { (gas) } \\\\ & \underset{C} {\mathrm{ZnO}}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{O} \\\\ & 2 \mathrm{ZnCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow 4 \mathrm{KCl}+\mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] \downarrow (White ppt) \end{aligned} $

(ii) The gas $B$ turns lime water milky and milkiness disappear with continuous passage of gas. Hence, the gas is $\mathrm{CO}_2$ and compound $A$ in $\mathrm{ZnCO}_3$.

$ \begin{aligned} \underset{C}{\mathrm{CO}_2}+\mathrm{Ca}(\mathrm{OH})_2 & \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CaCO}_3 \downarrow \\\\ \mathrm{CaCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} & \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \\\\ \underset{A} {\mathrm{ZnCO}_3} & \stackrel{\Delta}{\longrightarrow} \underset{C}{\mathrm{ZnO}}+\underset{B}{\mathrm{CO}_2} \end{aligned} $

(iii) The solution of $A$ gives white ppt of $\mathrm{ZnS} D$ with $\mathrm{NH}_4 \mathrm{OH}$ and excess of $\mathrm{H}_2 \mathrm{~S}$.

$ \begin{aligned} & \mathrm{ZnCO}_3+\mathrm{HCl} \longrightarrow \underset{C}{\mathrm{CO}_2 \uparrow}+\mathrm{ZnCl}_2 \\\\ & \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{~S} \stackrel{\mathrm{NH}_4 \mathrm{OH}}{\longrightarrow} 2 \mathrm{HCl}+{\underset{D}{\mathrm{ZnS}} \downarrow \text { (white) }} \end{aligned} $

(iv) The solution of $A$ also gives initially a white ppt $E$ with $\mathrm{NaOH}$, which dissolve in excess of reagent.

$ \begin{aligned} & \mathrm{ZnCl}_2+2 \mathrm{NaOH} \longrightarrow \underset{E \text { (white) }} {\mathrm{Zn}(\mathrm{OH})_2} \downarrow+2 \mathrm{NaCl} \\\\ & \mathrm{Zn}(\mathrm{OH})_2+2 \mathrm{NaOH} \longrightarrow \underset{ \text { (Soluable) }}{\mathrm{Na}_2\left[\mathrm{Zn}(\mathrm{OH})_4\right]} \end{aligned} $

As nitric acid is an oxidising agent, it oxides H₂S to S.