During the qualitative analysis of SO$$_3^{2-}$$ using dilute H$$_2$$SO$$_4$$, SO$$_2$$ gas is evolved which turns K$$_2$$Cr$$_2$$O$$_7$$ solution (acidified with dilute H$$_2$$SO$$_4$$):

We need to identify the color change when SO$$_2$$ gas is passed through acidified K$$_2$$Cr$$_2$$O$$_7$$ solution.

Reaction:

SO$$_2$$ is a reducing agent. It reduces the orange dichromate ion (Cr$$_2$$O$$_7^{2-}$$, where Cr is in +6 state) to green chromic ion (Cr$$^{3+}$$):

$$\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + 3\text{SO}_2 \to \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$$

The Cr$$^{3+}$$ ions in solution give a green color.

This color change from orange to green is a confirmatory test for the presence of SO$$_3^{2-}$$ ions (which produce SO$$_2$$ on treatment with acid).

The correct answer is Option 3: Green.