Question

On passing $\text{S}{{\text{O}}_{2}}$ gas in acidified ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ solution, green colour is produced due to the formation of:
A. $\text{CrS}{{\text{O}}_{4}}$
B. ${{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}}$
C. $\text{C}{{\text{r}}_{2}}{{(\text{S}{{\text{O}}_{4}}\text{)}}_{3}}$
D. $\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$

Answer 1

Hint:> In this question firstly, we have to study the properties of sulphur dioxide and potassium dichromate. After it, we have to write a balanced chemical reaction in which potassium dichromate will be the oxidising agent.

Complete step by step answer:
- In the given question we have to identify the product formed by the reaction between sulphur dioxide and potassium dichromate.
- Sulphur dioxide gas is very poisonous and harmful and colourless in appearance.
- By volcanic activity the sulphur dioxide gas is released naturally and also artificially as a by-product of copper extraction.
- Whereas potassium dichromate is an orange to the red coloured inorganic compound which releases harmful gas when heated.
- Potassium dichromate is also a strong oxidising agent and used in many organic reactions.
- Oxidising agents are those which oxidise other molecules and reduce themselves.
- So, when the sulphur dioxide gas is passed in the acidified potassium dichromate then it yields chromium sulphate and water.
- The balanced reaction of the above process is:
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+3S{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+3{{H}_{2}}O$
- In the given reaction, the potassium dichromate is reduced to the chromium sulphate i.e. from +1 to +5 which is responsible for giving the green colour.
- Whereas sulphur dioxide is oxidised to the sulphate ion from +6 to +3.

Therefore, option C is the correct answer.

Note: In paints, inks, ceramic process, etc chromium is used and also it is a non-combustible inorganic compound. The oxidation of potassium in potassium dichromate and potassium sulphate is
${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
$2x+12-14=0$
$2x=2$
$x=+1$
In ${{K}_{2}}S{{O}_{4}}$
$2x-2-8=0$
$2x=10$
$x=+5$
Whereas the oxidation state of chromium in potassium dichromate and chromium sulphate is:
${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
$2+2x-14=0$
$2x=12$
$x=+6$
In $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$
$2x+3(-2)=0$
$2x=6$
$x=+3$