1 Answer | Add Yours
The reaction of `Cu^(2+)` ion with `NH_3` draws some complication since `NH_3` has lone pairs and is considered as electron donor. On the other hand, `Cu^(2+)` has vacant orbitals which can accommodate `NH_3` molecules as ligands.
The first reaction would has limited amount of `NH_3` molecules which can form `Cu(OH)_2` . The reaction can be written as:
`Cu^(2+) + 2 NH_3 + 2 H_2O -> Cu(OH)_2 _(s) + 2 NH_4^(+)`
For the second reaction, there is an excess of `NH_3` which apparently can produce coordination and the product would be `Cu[(NH_3)_4]^(2+)` . The reaction equation can be written as:
`Cu^(2+) _(aq) + 4 NH_3 (excess) -> Cu[(NH_3)_4]^(2+) _(aq)`
In the first reaction, precipitates of copper hydroxide would be present. Copper hydroxide is a blue compound which is slightly soluble in water and has a `k_sp` value of `2.20*10^(-20)` . On the second reaction, the product formed is aqueous which means that they are soluble in the solution.
We’ve answered 317,632 questions. We can answer yours, too.Ask a question