Answer the following question from the passage below:

The process of accumulation of molecular species at the surface of solid or liquid is called adsorption. It may be physisorption or chemisorption. Low temperature is favorable for physisorption. Amount of gas adsorbed by a solid depends upon nature of gas. Easily liquifiable gases are readily adsorbed. Catalyst increases the rate of reaction. Catalysis may be homogenous or heterogenous Enzymes are biological catalysts. They work at optimum pH and temperature. Solutions are of three types i.e. true solutions, colloids and suspensions. Depending on the interaction between dispersed phase and dispersion medium colloids may be Lyophilic and Lyophobic. Depending on type of particles there may multimolecular, macro-molecular and associated colloids. Sol. gel, emulsions are also various types of colloids.

The correct order of rate of adsorption of different gases  by 1g activated charcoal is:

A. \(H_2 < CH_4 < SO_2\)

B. \(H_2 < SO_2 < CH_4\)

C. \(CH_4 < SO_2 < H_2\)

D. \(SO_2 < CH_4 < H_2\)

E. \(SO_2 < H_2 < CH_4\)

Choose the correct answer from the options given below:

A only

The correct answer is option 2. A only.

The order of adsorption for the gases by 1g of activated charcoal should indeed be: \(A. H_2 < CH_4 < SO_2\)

Strength of intermolecular forces: As you mentioned, the key factor is the strength of interaction between the gas molecules and the activated charcoal surface.

Polarizability and size: While these play a role, for these specific molecules, the impact of intermolecular forces is more significant.

Here is the breakdown of the molecules:

\(H_2\): Smallest molecule with the weakest intermolecular forces (London dispersion forces).

\(CH_4\): Slightly larger than \(H_2\), but still has weak intermolecular forces (London dispersion forces).

\(SO_2\): Larger than both, but the key difference lies in its polarity. \(SO_2\) is a polar molecule with a bent shape, allowing it to interact more effectively with the surface through dipole-dipole interactions and even hydrogen bonding with the hydroxyl groups on activated charcoal.

Therefore, due to the stronger interaction with the surface, \(SO_2\) will be adsorbed the most, followed by \(CH_4\) and then \(H_2\). This aligns with option A only.