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One pink solid (X) has the formula CoCl_3. 5NH_3. H_2O.A solution of this salt (X) in water is also pink and precipitates 3 moles AgCl on treatment with silver nitrate solution..When the pink solid (X) is heated, it loses 1 mole H_2O to yielda purple solid (Y) with the same molar ratio of NH_3:Cl:Co.The purple solid releases two of its chlorides rapidly on dissolution and after treatment with AgNO_3 solution.The coordination number of cobalt in complexes (X) and (Y) is six. The IUPAC name of pink (X) and purple (Y) complexes respectively are : |
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Answer» Tetraamminechloridoaquacobalt(III) chloride and pentaamminechloridocobalt(III) chloride `3Ag^(+)+3 Cl^(-)to3AgCldarr`(white) The following complex is obtained on HEATING and losing 1 mole of `H_2O` `underset(1:5:1)([Co(NH_3)_5Cl]Cl_2)("purple")overset(aq.)hArr[Co(NH_3)_5Cl]^(2+)+2Cl^(-)` `2Ag^(+)+2Cl^(-)to2AgCldarr`(white) So, pink (X) is `[Co(NH_3)_5(OH_2)]Cl_3` and IUPAC name would be pentaammineaquacobalt(III) chloride and purple (Y) is `[Co(NH_3)_5Cl]Cl_2` and IUPAC name would be pentaamminechloridocobalt(III)chloride. |
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