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(i) A black coloured compound (B) is formed on passing hydrogen sulphide through the solution of a compound (A) in NH_(4)OH. (ii) (B) on treatment with hydrochloric acid and potassium chlorate gives (A). (iii) (A) on treatment with potassium cyanide gives a buff coloured precipitate which dissolves in excess of this reagent forming a compound (C). (iv) The compound (C) is changed into a compound (D) when its aqueous solution is boiled: (v) The solution of (A) was treated with excess of sodium bicarbonate and then with bromine water. On cooling and shaking for some time, a green colour of compound (E) is formed. No change is observed on heating. Identify (A) to (E) and give chemical equations for the reactions at steps (l) to (iv) |
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Answer» Solution :The FORMATION of black coloured compound (B) by passing `H_(2)S` through the alkaline solution of the compound indicate that this asalt of the group TV TMCats (`CO^(2+)` or `Ni^(2+)`). However, the given reactions ESPECIALLY, reaction (ili), indicates that compound (A) is a cobalt salt `(CoCl_(2))` which explains all the given reactions. (i) `UNDERSET((A))(CoCl_(2))+2NH_(4)OH+H_(2)Stounderset((B))(CoS)+2NH_(4)Cl+2H_(2)O` (ii) `CoS+2HCl+underset((from KClO_(3)))([O])toCoCl_(2)+H_(2)S` `2KlCO_(3)to2Kcl+3O_(2)` (iii) `CoCl_(2)+2KCNtoCo(CN)_(2)darr+2Kcl` `Co(CN)_(2)+4KCNtoK_(4)[Co(CN)_(6)]` (iv) `2K_(4)[Co(CN)_(6)]+[O]+H_(2)Oto2K_(3)[Co(CN)_(6)]+2KOH` (v) `CoCl_(2)+6NaHCO_(3)toNa_(4)[Co(CO_(3))_(3)]+2NaCl+3CO_(2)+3H_(2)O` `2Na_(4)[Co(CO_(3))_(3)]+2NaHCO_(3)+Oto2Na_(3)[Co(CO_(3))_(3)]+2Na_(2)CO_(3)+H_(2)O` |
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