Question

Which of the following solutions does not change its colour on passing ozone through it?
A. Starch iodide solution
B. Alcohol solution of benzidine
C. Acidic solution of ${K_2}C{r_2}{O_7}$
D. Acidified solution of $FeS{O_4}$

Answer 1

Hint: The inorganic molecule ozone, often known as trioxygen, has the chemical formula \[{O_3}.\] It's a pale blue gas with a pronounced unpleasant odour. It is a far less stable allotrope of oxygen than the diatomic allotrope \[{O_2}\] , breaking down to \[{O_2}\] in the lower atmosphere.

Complete answer: Ozone is a strong oxidising agent because an atom of nascent oxygen that is more reactive than oxygen decomposes fast to yield. Ozone breaks down into nascent oxygen atoms. Sulphide is converted to sulphate by ozone oxidation.
The iodine reacts with the starch in the paper, resulting in a purple stain. The amount of ozone in the air determines the intensity of the purple colour. So, starch iodide solution, alcohol solution of benzidine and acidified solution of $FeS{O_4}$ changes its colour on passing ozone through it as all these solutions have a tendency to get oxidized.
But the acidified solution of ${K_2}C{r_2}{O_7}$ ($Cr$ in $ + 6$ oxidation state) is itself an oxidizing agent so it does not react with ozone to change its colour.
So, the correct option is: (C) Acidic solution of ${K_2}C{r_2}{O_7}$

Note:
Because ozone's disintegration into oxygen releases heat, it is thermodynamically less stable than oxygen. Ozone is made up of three molecules of oxygen and is thus in an unstable state, therefore it gives up one molecule of oxygen to return to a diatomic state.