question_answer 1)

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

Answer:

(i) Reaction of dihydrogen with chlorine is: \[\overset{0}{\mathop{{{H}_{2}}}}\,(g)+\overset{0}{\mathop{C{{l}_{2}}}}\,(g)\to \overset{+-}{\mathop{HCl}}\,\] In this reaction, hydrogen is oxidised and chlorine is reduced. Hydrogen chloride (H-Cl) is a polar covalent molecule. (ii) Re action of sodium metal with dihydrogen is: \[2\overset{0}{\mathop{Na}}\,+\overset{0}{\mathop{{{H}_{2}}}}\,\to 2\overset{+1}{\mathop{Na}}\,\overset{-1}{\mathop{H}}\,\] It is a redox reaction in which sodium is oxidised while hydrogen is reduced. This, reaction involves transfer of electron from sodium metal to hydrogen. (iii) Reaction of copper (II) oxide with dihydrogen is : \[\overset{+2}{\mathop{Cu}}\,\overset{-2}{\mathop{O}}\,+\overset{0}{\mathop{{{H}_{2}}}}\,\to \overset{0}{\mathop{Cu}}\,+\overset{+2}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{O}}\,\] It is al so a redox reaction in which copper is reduced while hydrogen is oxidised.