Question

Which of the following will form a reversible sol?
A.Gold sol
B.Sulphur sol
C.Rubber sol
D.Arsenous sulphide sol

Answer 1

Hint: Rubber is a polymeric substance and the polymer sols are stabilized by a phenomenon called steric stabilization normally, whereas inorganic sols are stabilized by electrostatic stabilization. Also, Gold sol and Sulphur sol are the type of multimolecular colloid.

Complete step by step answer:
When two constituents of the sol are separated by any means like evaporation, then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture, then that type of sol is known as reversible sol.
Examples of reversible sols are: gum, gelatin, starch, rubber.
Rubber is a polymeric substance and the polymer sols are stabilized by a phenomenon called steric stabilization normally. Branching in the polymeric chains of rubber provides enough steric hindrance to the polymeric particles which in turn resist coagulation in the liquid phase and thus the polymer particles just roam around the dispersing media freely and show lyophilic behavior.
Gold sol is a multimolecular colloid as a large number of molecules combine to form the colloidal particles. It is negatively charged and a lyophobic sol.
Sulphur sol is a type of multimolecular colloid as it consists of particles containing a thousand or more of Sulphur molecules.
Sulphur sol and gold sol, both are lyophobic sols and they are unstable and are built to colloidal size from small particles by the process of condensation method by chemical reactions.
$A{s_2}{S_3}$ (Arsenious sulphide) is a lyophobic colloid. It is formed due to the hydrolysis of Arsenious oxide $(A{s_2}{O_3})$ in boiled distilled water. Further ${H_2}S$ gas is passed through the solution.
S gas is passed through the solution. The reaction is as follows:
 $A{s_2}{O_3} + 3{H_2}O \to 2As{(OH)_3}$
 $2As{(OH)_3} + 3{H_2}S \to A{s_2}{S_3} + 6{H_2}O$
The particles in Arsenious sulphide colloidal solution are surrounded by HS ions which are formed due to the dissociation of ${H_2}S$ . Sulphide ion layer is surrounded by \[{H^ + }\] ions.

Therefore, the correct answer is option (C).

Note: The reduction of $AuCl_3$ with formaldehyde will give gold sol.
 $AuC{l_3} + HCHO + 3{H_2}O \to Au\left( {sol} \right) + 3HCOOH + 6HCl$
Oxidation of Sulphur sol is prepared by passing ${H_2}S$ into $S{O_2}$ .
 ${H_2}S + S{O_2} \to S\left( {sol} \right) + {H_2}O$