Question

An aqueous solution of an inorganic compound (X) gives the following reactions.
(i) With an aqueous solution of $\text{BaC}{{\text{l}}_{\text{2}}}$, a precipitate insoluble in dilute HCl is obtained.
(ii) Addition of excess of KI gives a brown precipitate which turns white on addition of excess hypo solution.
(iii) With an aqueous solution of potassium ferrocyanide a chocolate colored precipitate is obtained.
A. $\text{CuS}{{\text{O}}_{\text{4}}}$
B. $\text{BaS}{{\text{O}}_{\text{4}}}$
C. BaCl
D. NaI

Answer 1

Hint: To give the correct option, we have to do the reaction in terms of reactivity series. The answer of this question is blue in colour when hydrated and when it is dehydrated, the colour is white.

Step by step answer:
Let us take the first part of the question, in which an inorganic compound X is reacting with an aqueous solution of $BaCl_2$. Let us guess our first option as inorganic compound X. We should note that, when $CuSO_4$ reacts with $BaCl_2$, double decomposition reaction happens. Barium sulphate and copper chloride will be formed. It’s a general principle of double decomposition reactions that we should remember that the least soluble of the four possible compounds always determines the outcome, although everything may stay in solution together until the solvent starts to evaporate. Then the least soluble one crystallizes out first.
$\text{BaC}{{\text{l}}_{\text{2}}}\text{ }+\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }\to \text{ BaS}{{\text{O}}_{\text{4}}}\text{ }+\text{ CuC}{{\text{l}}_{\text{2}}}$
So, from the above reaction we can say that the inorganic compound X in this reaction is $CuSO_4$.
Let us take the second part of the question in which addition of KI to inorganic compound X gives a brown precipitate which turns white in addition to excess hypo solution.
Copper sulphate reacts with potassium iodide to form cuprous iodide and iodine.
$\text{2CuS}{{\text{O}}_{\text{4}}}+\text{4KI}\to \text{C}{{\text{u}}_{\text{2}}}{{\text{I}}_{\text{2}}}\downarrow +{{\text{I}}_{\text{2}}}+\text{2}{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{4}}$
We should know that it is a type of redox reaction and not precipitation reaction or double decomposition reaction. We should note that, it involves oxidising agent like $C{{u}^{\text{2}}}^{+}$and$\text{2}{{\text{I}}^{-}}$. We should note that $C{{u}^{\text{2}}}^{+}$is reduced to$\text{C}{{\text{u}}^{+}}$, which will get transformed from blue to a white precipitate while $\text{2}{{\text{I}}^{-}}$ oxidises to ${{\text{I}}_{\text{2}}}$ that is from colourless to yellow or brown.
In the third part of the question, inorganic compound (X) reacts with an aqueous solution of potassium ferrocyanide a chocolate coloured precipitate is obtained.
We should note that, potassium ferrocyanide is ${{K}_{\text{4}}}\left[ Fe{{\left( CN \right)}_{\text{6}}} \right]$. Let us now react with copper sulphate with potassium ferrocyanide.

& \text{2}C{{u}^{\text{2}}}^{+}+{{K}_{\text{4}}}\left[ Fe{{\left( CN \right)}_{\text{6}}} \right]\to C{{u}_{\text{2}}}\left[ Fe{{\left( CN \right)}_{\text{6}}} \right]+\text{4}{{K}^{+}} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Red-brown\left( chocolate \right)ppt \\\ \end{aligned}$$ From the above discussion and reaction, we can say that option A is the correct answer. Inorganic compound(X) is $CuSO_4$. Note: We should note that colours of precipitates help to identify compounds. We can decide which ions (cations or anions) are in the compound by comparing colours of different precipitates. Let us compare AgCl and AgBr precipitates. $$\text{A}{{\text{g}}^{+}}$$ion is common in both compounds. But their colours are different due to different halide ions. AgCl is a white precipitate and AgBr is a light yellow precipitate.