${\mathrm{CrO}}_4^{2-}$(yellow) changes to ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$(orange) in pH=X and vice–versa in pH=Y. 

Determine X and Y 

${\mathrm{CrO}}_4^{-}$ is a yellow color and in acidic medium, it forms orange color solution due to the formation of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}.$ This implies that the reaction takes place at pH lower than 7.

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

The above reaction reversed in the basic medium where pH is greater than 7.

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

Therefore, X is 6 and Y is 8.