Among the following, the correct statement is:

We are asked to examine all the four statements one by one and decide which one is correct.

First, let us recall the definitions of lyophilic and lyophobic sols. A lyophilic sol is a colloidal solution in which the dispersed phase has a strong affinity (literally “solvent-loving”) for the dispersion medium; such sols are very stable and are formed simply by mixing the components. Metal sulphide sols, for example $$As_2S_3$$ or $$CdS$$, are not of this type. They are difficult to prepare, readily precipitate on the addition of a small amount of electrolyte, and therefore belong to the lyophobic class. Hence the sentence “sols of metal sulphides are lyophilic” is wrong, so option A is false.

Now we consider Brownian movement. Brownian movement is the continuous, zig-zag motion shown by colloidal particles when observed under an ultramicroscope. The cause is the unbalanced bombardment of the colloidal particles by the much smaller, rapidly moving molecules of the dispersion medium. According to kinetic theory, the kinetic energy transferred in each collision is the same, so if the mass of the particle is smaller, its velocity must be larger, because $$\dfrac12 m v^2 = \text{constant}$$. Consequently the smaller the dispersed particles, the livelier and more conspicuous is their motion. Therefore the statement “Brownian movement is more pronounced for smaller particles than for bigger particles” is perfectly true, and option B is correct.

Next we turn to adsorption of gases on activated charcoal. A well-known empirical result is that the extent of physical adsorption of a gas on charcoal is roughly proportional to its critical temperature $$T_c$$, i.e. $$\text{Extent of adsorption} \;\propto\; T_c$$. The higher the critical temperature, the more easily the gas liquefies, and the greater is the van der Waals attraction between the gas molecules and the solid surface. The accepted experimental values of the critical temperatures are

$$T_c(H_2S) \approx 212^{\circ}\text{C} \;(485\;\text{K}),$$

$$T_c(Cl_2) \approx 144^{\circ}\text{C} \;(417\;\text{K}).$$

Because $$485\;\text{K} > 417\;\text{K}$$, hydrogen sulphide is more easily liquefied and is therefore expected to be adsorbed to a larger extent than chlorine. The sentence given in option C reads “one would expect charcoal to adsorb chlorine more than hydrogen sulphide”, which is the opposite of what the rule predicts, so option C is false.

Finally we examine the Hardy-Schulze law. The law states: The coagulating (precipitating) power of an ion for a given colloid increases with the valency of the ion that is opposite in charge to the colloidal particles. Thus a trivalent ion $$Al^{3+}$$ has enormously greater coagulating power than a divalent ion $$Ca^{2+}$$, which in turn is more powerful than a monovalent ion $$Na^{+}$$. The size of the ion does not enter the statement of the law at all. Hence “bigger the size of the ions, the greater is its coagulating power” is an incorrect restatement, and option D is false.

Summarising the discussion, only option B withstands scrutiny; the other three are incorrect.

Hence, the correct answer is Option B.