Consider the laboratory preparation of chlorine: $2\mathrm{KMnO}_4 + 16\,\mathrm{HCl} \rightarrow 2\,\mathrm{KCl} + 2\,\mathrm{MnCl}_2 + 5\,\mathrm{Cl}_2 + 8\,\mathrm{H_2O}$. Which species in this reaction is the OXIDISING agent?

1. Track oxidation states across the reaction. 2. Mn in $\mathrm{KMnO}_4$ is $+7$ (compute: $+1 + x + 4(-2) = 0 \Rightarrow x = +7$). In $\mathrm{MnCl}_2$, Mn is $+2$. So Mn went DOWN by 5 — REDUCED. 3. Cl in HCl is $-1$ (with H = $+1$). In $\mathrm{Cl}_2$, Cl is $0$. So Cl went UP — OXIDISED. 4. The species that contains the element being REDUCED is the OXIDISING AGENT (it caused the other species to be oxidised). So $\mathrm{KMnO}_4$ is the oxidising agent. 5. Option A reverses the roles: HCl is the REDUCING agent (it donates electrons to Mn). Option C is a salt by-product, neither oxidising nor reducing. Option D is the solvent. _Source: NCERT Class 11 Chemistry Part 2, Ch 7, §7.2 (oxidising/reducing agent definitions) + §7.3 examples with $\mathrm{KMnO}_4$, p. 3 + p. 8._