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

Comment on the reactions of dihydrogen with (i) chlorine, (ii) sodium,

1.	Comment on the reactions of dihydrogen with (i) chlorine, (ii) sodium, and (iii) copper (II) oxide
Answer» Solution : (i) Dihydrogen reduces chlorine into chloride `(Cl^-)` ion and itself gets oxidised to `H^+` ion by chlorine to form hydrogen chloride. An electron pair is SHARED between H and Cl leading to the FORMATION of a covalent MOLECULE. (II) Dihydrogen is reduced by sodium to form NaH. An electron is transferred from Nato H leading to the formation of an ionic compound, `Na^(+)H^(-)`. (iii) Dihydrogen reduces copper(II) oxide to copper in zero oxidation state and itself gets oxidised to `H_2O`, which is a covalent molecule.