Comment upon the reactions of dihydrogen with
(i) Chlorine
(ii) Sodium and
(iii) Copper (II) oxide

Explanation

(i) Dihydrogen reduces chlorine to `Cl^(-)` ion and itself gets oxidised to `H^(+)` ions. An electron pair is then shared between these two species to form a covalent melecules of hydrogen chloride.
`H_(2)(g) + Cl_(2)(g) to 2HCl (g)`
(ii) Sodium reduces dihydrogen to form hydride ion `(H^(-))` and itself gets oxidised to sodium ion `(Na^(+))` During this reaction, an electron is completely transferred from Na to H thereby ionic sodium hydride, `Na^(+) H^(-)`
`2Na(s) + H_(2)(g) overset(Delta)to 2Na^(+)H^(-)(s)`
(iii) Hydrogen reduces copper (II) oxide to copper metal. (zero oxidation state) while itself gets oxidized to form a covalent moleule of `H_(2)O` (in which H has an oxidation state of +1).
`overset(+2 -2)(CuO) (s) + overset(0)H_(2)(g) to overset(0)Cu(s) + overset(+1)H_(2) overset(-2)O(l)`