Step 1: Recall the chromate-dichromate equilibrium.
In aqueous solution, chromate ($CrO_4^{2-}$, yellow) and dichromate ($Cr_2O_7^{2-}$, orange) exist in equilibrium that depends on pH.
Step 2: Write the equilibrium equation.
\[ Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O \]
Step 3: Apply Le Chatelier's principle in alkaline medium.
In alkaline medium, $OH^-$ concentration is high. By Le Chatelier's principle, the equilibrium shifts to the RIGHT to consume excess $OH^-$, forming more $CrO_4^{2-}$ (chromate). The solution turns yellow.
Step 4: What happens in acidic medium (for contrast).
In acidic medium, $H^+$ ions react with $OH^-$, shifting the equilibrium LEFT, converting chromate to dichromate (orange colour).
Step 5: Eliminate incorrect options.
$Cr^{3+}$ requires chemical reduction of Cr(VI), not just pH change. $CrO$ (Cr in +2 state) is not formed by pH adjustment. $CrO_3$ is the solid anhydride of chromic acid and is not a product in alkaline aqueous solution.
Step 6: State the final answer.
In alkaline medium, $Cr_2O_7^{2-}$ (orange dichromate) converts to $CrO_4^{2-}$ (yellow chromate). \[ \boxed{CrO_4^{2-}} \]