Cu SO4 on mixing with NH3 (1:4 ratio )...

`Cu SO_4 ` on mixing with ` NH_3 (1:4 ratio ) ` does not give test for ` Cu^(2+) ` ions but gives test for ` SO_4^(2-) ` ions why?

Text Solution

Verified by Experts

It is because when `NH_3` coordinates to `Cu^(2+)` ions and it forms the complex `[Cu(NH_3)_4]SO_4` copper ions are present in coordination sphere, therefore they are non-ionisable whereas `SO_(4)^(2-)` ions are counter ions which are ionisable.

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Which of the following complexes formed by Cu^(+2) ions is the most stable ?

`Cu^(+2)+4NH_3 hArr [Cu(NH_3)_4]^(+2)` log k=11.6

`Cu^(+2) + 4CN^(-) hArr [Cu(CN)_4]^(-2)` log k =27.3

`Cu^(+2)+2en hArr [Cu(en)_2]^(+2)` log k =15.4

`Cu^(+2)+4H_2O hArr [Cu(H_2O)_4]^(+2)` log k =8.9

Oxalic acid on heating with conc. H_(2)SO_(4) gives

CO only

`CO_(2)` only

`CO_(2)+H_(2)O`

`CO+CO_(2)+H_(2)O`

The oxidation number of N in NH4^(+) ion is

-3

-4