xCr₂O₇ ^2- + ySO₂ -----> Cr³⁺ + SO₄ ^2-
the x and y in the reaction are ?

Dear Student,

Let us consider the reaction is taking place in acidic condition. Balancing the reaction by ion electron method:
$$\begin{gathered} {\mathrm{xCr}}_2{{\mathrm{O}}_7}^{2-} + {\mathrm{ySO}}_2 \to {\mathrm{Cr}}^{3+} + {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Oxidation} : {\mathrm{SO}}_2 \to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} \to {\mathrm{Cr}}^{3+} \\ \mathrm{Step} 1: \mathrm{Balance} \mathrm{all} \mathrm{the} \mathrm{atoms} \mathrm{other} \mathrm{than} \mathrm{O} \\ \mathrm{Oxidation} : {\mathrm{SO}}_2 \to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} \to 2{\mathrm{Cr}}^{3+} \\ \mathrm{Step} 2: \mathrm{Balance} \mathrm{O} \mathrm{by} \mathrm{adding} {\mathrm{H}}_2\mathrm{O} \mathrm{to} \mathrm{O} \mathrm{deficient} \mathrm{side} \\ \mathrm{Oxidation} : {\mathrm{SO}}_2 + 2{\mathrm{H}}_2\mathrm{O}\to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} \to 2{\mathrm{Cr}}^{3+}+ 7{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Step} 3: \mathrm{Balance} \mathrm{the} \mathrm{hydrogen} \mathrm{atoms} \mathrm{by} \mathrm{adding} {\mathrm{H}}^{+} \\ \mathrm{Oxidation} : {\mathrm{SO}}_2 + 2{\mathrm{H}}_2\mathrm{O}\to {{\mathrm{SO}}_4}^{2-}+ 4{\mathrm{H}}^{+} \\ \mathrm{Reduction} : {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} + 14{\mathrm{H}}^{+} \to 2{\mathrm{Cr}}^{3+}+ 7{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Step} 4: \mathrm{Balance} \mathrm{the} \mathrm{charges} \mathrm{by} \mathrm{adding} \mathrm{electrons} \\ \mathrm{Oxidation} : {\mathrm{SO}}_2 + 2{\mathrm{H}}_2\mathrm{O}\to {{\mathrm{SO}}_4}^{2-}+ 4{\mathrm{H}}^{+} + 2{\mathrm{e}}^{-} \\ \mathrm{Reduction} : {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} + 14{\mathrm{H}}^{+} + 6{\mathrm{e}}^{-} \to 2{\mathrm{Cr}}^{3+}+ 7{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Adding} \mathrm{both} \mathrm{the} \mathrm{reaction} \mathrm{and} \mathrm{multiplying} \mathrm{each} \mathrm{half} \mathrm{reaction} \mathrm{by} \mathrm{the} \mathrm{smallest} \mathrm{whole} \mathrm{number} \mathrm{that} \mathrm{is} \mathrm{required} \mathrm{to} \mathrm{equalise} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{electron} \\ \mathrm{So}, \mathrm{we} \mathrm{mutiply} \mathrm{the} \mathrm{oxidation} \mathrm{half} \mathrm{by} 3 \mathrm{and} \mathrm{then} \mathrm{add} \mathrm{both} \mathrm{the} \mathrm{reactions}, \mathrm{and} \mathrm{we} \mathrm{get} \\ {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-} + 3{\mathrm{SO}}_2 + 2{\mathrm{H}}^{+}\to 2{\mathrm{Cr}}^{3+}+ 3{{\mathrm{SO}}_4}^{2-}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{So}, \mathrm{x}= 1 \mathrm{and} \mathrm{y}=3 \end{gathered}$$