Identify the process in which change in the oxidation state is five:

Step 1 — Calculate oxidation state change in each option:

(1) $\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$: In $\mathrm{Cr_2O_7^{2-}}$: $2x + 7(-2) = -2 \Rightarrow x = +6$ In $\mathrm{Cr^{3+}}$: $x = +3$ Change = $6 - 3 = 3$ per Cr atom.

(2) $\mathrm{MnO_4^{-} \rightarrow Mn^{2+}}$: In $\mathrm{MnO_4^{-}}$: $x + 4(-2) = -1 \Rightarrow x = +7$ In $\mathrm{Mn^{2+}}$: $x = +2$ Change = $7 - 2 = \mathbf{5}$ ✓

(3) $\mathrm{CrO_4^{2-} \rightarrow Cr^{3+}}$: In $\mathrm{CrO_4^{2-}}$: $x + 4(-2) = -2 \Rightarrow x = +6$ Change = $6 - 3 = 3$

(4) $\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$: In $\mathrm{C_2O_4^{2-}}$: $2x + 4(-2) = -2 \Rightarrow x = +3$ In $\mathrm{CO_2}$: $x + 2(-2) = 0 \Rightarrow x = +4$ Change = $4 - 3 = 1$

Answer: Option (2) — $\mathrm{MnO_4^{-} \rightarrow Mn^{2+}}$, change = 5