Question

A on reaction with dil. \[H_2SO_4\] gives a colourless pungent gas that can turn \[Cr_2O_{7^{2
- }}/{H^ + }\]green. Green colour is due to formation of :
A. \[CrO_{4^{2 - }}\]
B. \[C{r^{ + 3}}\]
C. \[CrO_3\]
D. \[CrO_2Cl_2\]

Answer 1

Hint: In order to answer this question accurately you need to first have a brief understanding of what the equation of this asked question might be. That is going to be the key to your answer.Which means you should also know the gas which is colourless and pungent smelling according to this question. Let me just make it easy for you. The gas is \[SO_2\]( sulphur dioxide).

Complete step by step answer:
So, in case you haven’t understood already, this pungent gas \[SO_2\] reduces potassium dichromate
to green coloured chromium sulphate that is \[Cr_2(SO_4)_3\].The equation is:
\[K_2Cr_2O_7 + 2{H^ + } + 3SO_2 \to (Cr_2(SO_4)_3)(green) + 5H_2O + 2{K^ + }\]
Thus, the green colour is due to the formation of \[Cr(III)\] ion.
So, the correct answer is option (B) \[C{r^{ + 3}}\].

Additional information:
\[SO_2\] sulphur dioxide is a colourless gas with a pungent odour . It is a liquid when it’s under
pressure and it dissolves in water very easily. It is also considered as a toxic gas responsible for the
smell of burnt match sticks . It is an inorganic compound. Sulphur dioxide gas is oxidised to sulphuric
acid when passed through acidified potassium dichromate solution. The colour of the solution
changes from orange to green as potassium dichromate is reduced to chromic sulphate.


Note:
When sulphur dioxide is passed through an acidified \[K_2Cr_2O_7\] solution, the oxidation state
sulphur changes from \[ + 4\] to \[ + 6\] and that of chromium changes from \[ + 6\] to\[ + 3\].
The appearance of green colour is due to the reduction of chromium metal. This question is from the
inorganic section of chemistry. Making regular notes is the only way for you to fetch full marks from
this section.