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Comment upon reactions of dihydrogen with (i) Chlorine ,(ii) Sodium and (iii) Copper(II) oxide. |
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Answer» Solution :(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 oxidizedto 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)` |
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