Identify what the reagent does

Acidified potassium dichromate paper contains $\mathrm{K_2Cr_2O_7}$ in acidic medium.

In acidic solution, dichromate ion is orange and gets reduced to $\mathrm{Cr^{3+}}$, which is green:

So, the paper turns green when it comes in contact with a reducing agent.

Check each option

Option A: $\mathrm{SO_2}$

Sulphur dioxide is a reducing agent. It reduces acidified dichromate to $\mathrm{Cr^{3+}}$.

The reaction is:

Hence, orange dichromate paper turns green.

So, A is correct.

Option B: $\mathrm{SO_3}$

Sulphur trioxide is not a reducing agent; it is the anhydride of sulphuric acid. It does not reduce dichromate to $\mathrm{Cr^{3+}}$.

So, B is incorrect.

Option C: $\mathrm{H_2S}$

Hydrogen sulphide is also a reducing agent and can reduce dichromate strongly. In acidic medium, it can reduce dichromate to green $\mathrm{Cr^{3+}}$.

Thus chemically, $\mathrm{H_2S}$ can also turn acidified dichromate green.

Option D: $\mathrm{CO_2}$

Carbon dioxide is neither reducing nor oxidizing under these conditions. No such color change occurs.

So, D is incorrect.

Exam-oriented conclusion

In standard qualitative chemistry, acidified potassium dichromate paper is classically used as a test for sulphur dioxide, which turns it from orange to green.

Although $\mathrm{H_2S}$ is also a reducing gas, the expected single-correct answer in this MCQ is:

$\boxed{\text{A: Sulphur dioxide}}$

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

Therefore, they agree.