(a) $\overset{+2}{\mathrm{Cu}}\mathrm{O}_{(s)}^{-2} + \mathrm{H}_{\overset{0}{2(g)}}$ → $\overset{0}{\mathrm{Cu}_{(s)}} + \overset{+1}{\mathrm{H_2}}\mathrm{O}_{(g)}^{-2}$ O.N. of Cu decreases from + 2 in CuO to 0 in Cu but that of H increases from 0 in $\mathrm{H}_2$ to + 1 in $\mathrm{H_2O}$. Therefore, CuO is reduced to Cu but $\mathrm{H}_2$ is oxidised to $\mathrm{H_2O}$. Thus, this is a redox reaction. (b) $\overset{+3}{\mathrm{Fe_2}}\mathrm{O}_{\overset{-2}{3(s)}} + \overset{+2}{3\mathrm{C}}\mathrm{O}_{(g)}$ → $2\overset{0}{\mathrm{Fe}}_{(s)} + \overset{+4}{3\mathrm{C}}\mathrm{O}_{2(g)}$ O.N. of Fe decreases from +3 in $\mathrm{Fe_2O_3}$ to 0 in Fe while that of C increases from +2 in CO to +4 in $\mathrm{CO}_2$. Thus, $\mathrm{Fe_2O_3}$ is reduced and CO is oxidised. Thus, this is a redox reaction. (c) $\overset{+3}{4\mathrm{B}}\mathrm{Cl}_{\overset{-1}{3(g)}} + 3\overset{+1}{\mathrm{Li}}\overset{+3}{\mathrm{Al}}\mathrm{H}_{\overset{-1}{4(s)}}$ → $\overset{-3}{2\mathrm{B}}_2\mathrm{H}_{\overset{+1}{6(g)}} + 3\overset{+1}{\mathrm{Li}}\mathrm{Cl}_{(s)}^{-1} + 3\overset{+3}{\mathrm{Al}}\mathrm{Cl}_{\overset{-1}{3(s)}}$ O.N. of B decreases from +3 in $\mathrm{BCl}_3$ to –3 in $\mathrm{B_2H_6}$ while that of H increases from –1 in $\mathrm{LiAlH}_4$ to + 1 in $\mathrm{B_2H_6}$. Therefore, $\mathrm{BCl}_3$ is reduced while $\mathrm{LiAlH}_4$ is oxidised. Thus, this is a redox reaction. (d) $2\mathrm{K}_{(s)} + \mathrm{F}_{2(g)}$ → $2\mathrm{K}^{+}\mathrm{F}^{-}_{(s)}$ Each K atom has lost one electron to form $\mathrm{K}^{+}$ while $\mathrm{F}_2$ has gained two electrons to form two $\mathrm{F}^{-}$ ions. Therefore, K is oxidised while $\mathrm{F}_2$ is reduced. Thus, it is a redox reaction. (e) $\overset{-3}{4\mathrm{N}}\mathrm{H}_{\overset{+1}{3(g)}} + 5\mathrm{O}_{\overset{0}{2(g)}}$ → $\overset{+2}{4\mathrm{N}}\mathrm{O}_{(g)}^{-2} + \overset{+1}{6\mathrm{H}}_2\mathrm{O}_{(g)}^{-2}$ O.N. of N increases from –3 in $\mathrm{NH}_3$ to +2 in NO while that of O decreases from 0 in $\mathrm{O}_2$ to –2 in NO or $\mathrm{H_2O}$. Therefore, $\mathrm{NH}_3$ is oxidised while $\mathrm{O}_2$ is reduced. Thus, it is a redox reaction.