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?


closed as off-topic by Nilay Ghosh, Jon Custer, Mathew Mahindaratne, Mithoron, Poutnik Oct 9 at 14:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Oct 8 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:

$$\ce{[Cu(H2O)4]SO4 + 4KF -> K2[CuF4] + K2SO4 + 4H2O}$$

  • $\begingroup$ WP says CuF2 is only slightly soluble in water. Do we need to resort to complex formation to explain these results, and how is the complex less soluble than the slightly soluble binary fluoride anyway? $\endgroup$ – Oscar Lanzi Oct 8 at 15:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.