1. Consider the well-known chromate–dichromate equilibrium in water:

$2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+ \rightleftharpoons \mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O}$

This is the more complete form of the given equilibrium.

2. We are asked when the equilibrium

$\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightleftharpoons 2\mathrm{CrO}_4^{2-}$

shifts to the right, i.e. when dichromate converts into chromate.

3. From the equilibrium expression above:

• Increasing $[\mathrm{H}^+]$ (acidic medium) shifts equilibrium to the left side of the given equation, favoring $\mathrm{Cr}_2\mathrm{O}_7^{2-}$.
• Decreasing $[\mathrm{H}^+]$ (basic medium) shifts equilibrium toward $\mathrm{CrO}_4^{2-}$.

4. Therefore, in a basic medium, the equilibrium shifts to the right.

5. Check options:

• A: weakly acidic medium — not correct, acidity favors dichromate
• B: basic medium — correct
• C: neutral medium — not the best answer
• D: acidic medium — incorrect

Hence, the correct option is B.