Step 1: Write the reaction
Acidified \(K_2Cr_2O_7\) oxidises \(I^-\) to \(I_2\): \(Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O\).
Step 2: Oxidation state
In \(Cr_2O_7^{2-}\) chromium is \(+6\). The product \(Cr^{3+}\) has oxidation state \(+3\), so each Cr gains 3 electrons.
Step 3: Other options
\(+6\) is the starting value and \(+2\) or \(+4\) are not formed in this reaction.
Final Answer:
Chromium ends as Cr(III), oxidation state +3.
\[ \boxed{\text{(B)}\ +3} \]