Match List - I with List - II.
List-I (Colloid Preparation Method)   List-II (Chemical Reaction)
(a) Hydrolysis   (i) 2AuCl$$_3$$ + 3HCHO + 3H$$_2$$O $$\rightarrow$$ 2Au(sol) + 3HCOOH + 6HCl
(b) Reduction   (ii) As$$_2$$O$$_3$$ + 3H$$_2$$S $$\rightarrow$$ As$$_2$$S$$_3$$(sol) + 3H$$_2$$O
(c) Oxidation   (iii) SO$$_2$$ + 2H$$_2$$S $$\rightarrow$$ 3S(sol) + 2H$$_2$$O
(d) Double Decomposition   (iv) FeCl$$_3$$ + 3H$$_2$$O $$\rightarrow$$ Fe(OH)$$_3$$(sol) + 3HCl
Choose the most appropriate answer from the options given below:

We have to find, for every reaction given in List-II, the method of colloid preparation to which it belongs. The four methods mentioned in List-I are hydrolysis, reduction, oxidation and double decomposition. We shall compare each chemical equation with the characteristic feature of these methods one by one and obtain the correct matching.

First recall the identifying features of every method.

• In hydrolysis the salt of a weak base and a strong acid (or vice-versa) reacts with water. The metallic hydroxide produced remains as the sol.

• In reduction a metallic cation is reduced (its oxidation number decreases) to the free metal which appears in the colloidal form.

• In oxidation a species present in the solution is oxidised (its oxidation number increases) and the oxidised product forms the sol.

• In double decomposition two substances exchange their ions; an insoluble or sparingly soluble product separates out as the colloid.

Now we examine each reaction of List-II.

(iv) $$$\mathrm{FeCl_3 + 3H_2O \;\rightarrow\; Fe(OH)_3(sol) + 3HCl}$$$

Here ferric chloride reacts directly with water. We see water dissociates, chloride ions leave with hydrogen to form $$\mathrm{HCl}$$ and the basic hydroxide $$\mathrm{Fe(OH)_3}$$ stays in the colloidal state. This is exactly the definition of hydrolysis. So reaction (iv) corresponds to the hydrolysis method. Hence

$$ (a)\;{\text{Hydrolysis}} \;\longrightarrow\; (iv). $$

(i) $$$\mathrm{2AuCl_3 + 3HCHO + 3H_2O \;\rightarrow\; 2Au(sol) + 3HCOOH + 6HCl}$$$

Gold is present in the +3 oxidation state in $$\mathrm{AuCl_3}$$. On the product side free metallic gold $$\mathrm{Au}$$ is obtained; its oxidation number becomes 0. The oxidation number has decreased; hence gold ion has been reduced. The method used is therefore reduction. Thus

$$ (b)\;{\text{Reduction}} \;\longrightarrow\; (i). $$

(iii) $$$\mathrm{SO_2 + 2H_2S \;\rightarrow\; 3S(sol) + 2H_2O}$$$

Let us look at oxidation numbers:

• In $$\mathrm{SO_2}$$, sulphur is in +4 state. In elemental sulphur $$\mathrm{S}$$, it becomes 0, so sulphur in $$\mathrm{SO_2}$$ is reduced.

• In $$\mathrm{H_2S}$$, sulphur is -2; in elemental $$\mathrm{S}$$ it becomes 0, i.e. it is oxidised.

One species is oxidised, the other reduced, but the net formation of colloidal sulphur takes place because sulphide ion is being oxidised to free sulphur that remains in the dispersed phase. The essential step for colloid formation is the oxidation of $$\mathrm{H_2S}$$. Therefore this reaction is placed under the oxidation method. Consequently

$$ (c)\;{\text{Oxidation}} \;\longrightarrow\; (iii). $$

(ii) $$$\mathrm{As_2O_3 + 3H_2S \;\rightarrow\; As_2S_3(sol) + 3H_2O}$$$

Here arsenious oxide and hydrogen sulphide simply interchange the oxide and sulphide ions. No change in oxidation state of arsenic or sulphur occurs; the reaction proceeds through metathesis producing water and colloidal arsenic trisulphide, $$\mathrm{As_2S_3}$$, which is sparingly soluble. Such an ion-exchange reaction is termed double decomposition. Therefore

$$$ (d)\;{\text{Double Decomposition}} \;\longrightarrow\; (ii). $$$

Collecting all the matches we obtain

$$$ \begin{aligned} (a) &\;\rightarrow\; (iv),\\ (b) &\;\rightarrow\; (i),\\ (c) &\;\rightarrow\; (iii),\\ (d) &\;\rightarrow\; (ii). \end{aligned} $$$

This pattern exactly corresponds to Option B.

Hence, the correct answer is Option B.