Comment on the reactions of dihydrogen with $(i)$ chlorine,$(ii)$ sodium,and $(iii)$ copper $(II)$ oxide.

(N/A) $(i)$ Dihydrogen reacts with chlorine to form hydrogen chloride $(HCl)$. The reaction is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$. Here,dihydrogen is oxidised to $H^+$ and chlorine is reduced to $Cl^-$,forming a covalent molecule.
$(ii)$ Dihydrogen reacts with sodium to form sodium hydride $(NaH)$. The reaction is: $2Na(s) + H_2(g) \rightarrow 2NaH(s)$. Here,an electron is transferred from $Na$ to $H$,forming an ionic compound,$Na^+H^-$.
$(iii)$ Dihydrogen reduces copper $(II)$ oxide to copper metal. The reaction is: $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$. Dihydrogen acts as a reducing agent,getting oxidised to water $(H_2O)$.