Match each reaction in List-I with the oxidation state of the element named in List-II. List-I (P) Cu + dilute HNO3: nitrogen in the colourless gas…

P: $3\mathrm{Cu}+8\mathrm{HNO_3}(\text{dil})\to3\mathrm{Cu(NO_3)_2}+2\mathrm{NO}+4\mathrm{H_2O}$. The gas is $\mathrm{NO}$, nitrogen $+2$ (5), which becomes brown $\mathrm{NO_2}$ in air. Q: $\mathrm{Cu}+4\mathrm{HNO_3}(\text{conc})\to\mathrm{Cu(NO_3)_2}+2\mathrm{NO_2}+2\mathrm{H_2O}$, brown $\mathrm{NO_2}$, nitrogen $+4$ (1). R: $4\mathrm{Zn}+10\mathrm{HNO_3}\to4\mathrm{Zn(NO_3)_2}+\mathrm{NH_4NO_3}+3\mathrm{H_2O}$, the ammonium ion with nitrogen at $-3$ (4). S: $3\mathrm{Cl_2}+6\mathrm{OH^-}\xrightarrow{\Delta}5\mathrm{Cl^-}+\mathrm{ClO_3^-}+3\mathrm{H_2O}$, chlorate with chlorine at $+5$ (3). The value $+1$ (hypochlorite) is the product with cold dilute alkali and is not used. Option B interchanges P and Q, C assigns hypochlorite to S, and D uses wrong gases. The correct code is option A. Each List-II value is used at most once and each List-I entry has exactly one match.