The oxidation state of chromium in the final product formed in the reaction between $$KI$$ and acidified $$K_2 Cr_2 O_7$$ solution is:

The reaction between $$KI$$ and acidified $$K_2Cr_2O_7$$ yields a chromium-containing product whose oxidation state must be determined.

In acidic medium, $$K_2Cr_2O_7$$ acts as a strong oxidizing agent, oxidizing $$I^-$$ to $$I_2$$, while $$Cr_2O_7^{2-}$$ (with Cr in the +6 oxidation state) undergoes reduction.

The balanced redox equation under these conditions is:

$$Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$$

From this equation, chromium is reduced from +6 in $$Cr_2O_7^{2-}$$ to +3 in $$Cr^{3+}$$, which subsequently forms $$Cr_2(SO_4)_3$$ or $$CrCl_3$$ depending on the acid employed. Each chromium atom thus gains 3 electrons.

Hence, the oxidation state of chromium in the product is Option C) +3.