The equilibrium $$Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$$ is shifted to the right in :

We need to determine in which medium the equilibrium $$Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$$ shifts to the right.

Write the complete ionic equilibrium

The interconversion between dichromate and chromate ions in aqueous solution involves $$H^+$$ and $$OH^-$$ ions. The complete equation is:

$$Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$$

Alternatively, in terms of $$H^+$$:

$$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$$

Apply Le Chatelier's Principle

The forward reaction (dichromate to chromate) consumes $$OH^-$$ ions. According to Le Chatelier's Principle:

- In a basic medium (excess $$OH^-$$): The high concentration of $$OH^-$$ drives the equilibrium to the right, favouring the formation of yellow $$CrO_4^{2-}$$ (chromate) ions.

- In an acidic medium (excess $$H^+$$): The $$OH^-$$ concentration is low, so the equilibrium shifts to the left, favouring the orange $$Cr_2O_7^{2-}$$ (dichromate) ions.

Conclusion

This is why dichromate solutions appear orange in acidic conditions and turn yellow (chromate) in basic conditions. The equilibrium shifts to the right in a basic medium.

The correct answer is Option (2): a basic medium.