Question

Photochemical smog is ……. in character while classical smog is ……. in character.
A. oxidizing, reducing
B. reducing, oxidizing
C. oxidizing, oxidizing
D. reducing, reducing

Answer 1

Hint: The oxidizing and reducing nature depends on the ability to lose or gain electrons by a compound. So we can solve this question if we know the compounds present in photochemical smog and classical smog.

Complete step by step solution:
-Dispersion is a system where particles of certain phases are dispersed into a medium of another phase. There are three types of dispersions named solutions, colloids and suspensions. Solution is a homogeneous mixture of a material which is dispersed into another material such that they get completely mixed with one another.
-A colloid is a heterogeneous mixture of particles which are small but the particles will not settle down if left for a particular period of time. It means that the particles will not undergo sedimentation.
-Suspension is a mixture of large particles which undergo the process of sedimentation when left for a particular period of time. It is also a heterogeneous dispersion. They can be physically separated from solutions and colloids.
-Smog is a colloidal solution in which the dispersed phase in solid and the dispersion medium is gas. Photochemical smog has high concentrations of ozone and $HN{{O}_{3}}$ while classical smog has high concentrations of sulphur-dioxide.
-These gases present in smog classify them as reducing or oxidizing agents. Oxidizing agents are those which reduce themselves to oxidize the other compound present in the solution. Reducing agents are those which oxidize themselves to reduce the other compound present in the solution.
-Ozone and nitric acids are very good oxidizing agents as they can easily lose electrons and reduce themselves from 0 to -2 oxidation state and +5 to -3,-1,0,+1,+3 oxidation states respectively. So photochemical smog is also called oxidizing smog.
-Sulphur-dioxide present in the classical smog is a very good reducing agent as it can easily gain electrons to oxidize itself and attain the oxidation state of +6 by converting into $S{{O}_{3}}$. The oxidation state of $S{{O}_{2}}$ is +4 only.

Therefore the correct option is A.

Note: Our eyes cannot distinguish between colloids and solutions. This is because the dispersed particles can be too small to be visible by the naked eye. They are distinguished based on a phenomenon called Tyndall Effect. It is difficult to classify the substances as solution, colloid and suspension.