Question

What occurs when a copper sulfate sample is treated successively with `(i)` water, and then `(ii)` `HBr`?

Answer 1

You form different coordination complexes of `Cu(II)`, each of which necessarily have a different colour.

Explanation:

And so we can write the reaction as....

`[Cu(OH_2)_6]^(2+) + 4Br^(-) rarr [CuBr_4]^(2-)`

You start with copper sulfate, a WHITE powder, and you dissolve some of this in water, and you form the `[Cu(OH_2)_6]^(2+)` complex, which has a beautiful, and distinctive blue colour, as seen in the LHS of the diagram.

Add quantities of bromide ligand to the solution, you form `[CuBr_4]^(2-)`, which has a distinctive green colour.