Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from

We need to determine the change in oxidation state of chromium when potassium dichromate acts as a strong oxidizing agent in acidic solution.

First, we find the oxidation state of Cr in $$K_2Cr_2O_7$$. Let the oxidation state be $$x$$. Then the equation $$2(+1) + 2x + 7(-2) = 0$$ leads to $$2 + 2x - 14 = 0$$, so $$2x = 12$$ and $$x = +6$$.

In acidic solution, the dichromate ion is reduced according to $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$. The oxidation state of chromium in the product $$Cr^{3+}$$ is $$+3$$.

Thus the oxidation state of chromium changes from $$+6$$ to $$+3$$. This matches Option 2, and the answer is $$\boxed{+6 \text{ to } +3}$$.