# Formation of green precipitation whilst adding potassium fluoride to aqueous solution of copper(II) sulfate [closed]

Aqueous copper sulfate solution (blue) gives a green precipitation with aqueous potassium fluoride. Explain these experimental results.

I've read answers like, $$\ce{H2O}$$ is a weak ligand and fluorine can replace it, but isn't $$\ce{H2O}$$ a stronger ligand than fluorine according to the spectrochemical series?

• Consider the possible products of a double displacement reaction. Which one is likely to be green? You may also want to read this WP article. – Oscar Lanzi Oct 8 '19 at 9:58

$$\ce{CuSO4}$$ exists as $$\ce{[Cu(H2O)4]SO4}$$ in solution. It is blue in colour due to the presence of $$\ce{[Cu(H2O)4]^{2+}}$$ ions.
Now if aqueous $$\ce{KF}$$ is added, the solution turns green due to formation of complexes $$\ce{[CuF4]^2-}$$ and $$\ce{[CuF6]^4-}$$(Initially copper(II) fluoride($$\ce{CuF2}$$) is formed but it formed complexe in present of water). The complete reaction is given by:
$$\ce{[Cu(H2O)4]SO4 + 4KF -> K2[CuF4] + K2SO4 + 4H2O}$$