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

Explanation

(i) Dihydrogen reduces chlorine to chloride `(Cl^(-))` ion and itself gets oxidised to `H^(+)` ion in HCl. An electron pair is shared between H and Cl to form a covalent molecule of hydrogen chloride.
`H_2(g) + Cl_2(g) to 2HCl(g)`
(ii) Sodium reduces dihydrogen to form hydride `(H^-)` ions and itself gets oxidised to sodium `(Na^(+))` ion. During this reaction, an electron is transferred from Na to H leading to the formation of an ionic compound sodium hydride, `Na^(+)H^(-)`.
`2Na(s) + H_2(g) overset("Heat")to 2Na^+ H^(-) (s)`
(iii) Hydrogen reduces copper (II) oxide to copper metal (in zero oxidation state) and itself gets oxidised to `H_2 O`, which is a covalent molecules.
`overset(+1)(Cu)overset(-2)(O) + overset(0)H_2 (g) to overset(0)(Cu(s)) + overset(+1)(H_2O)overset(-2)((l))`