11
$\begingroup$

$\ce{CuCl2}$ is green in colour while $\ce{CuCl}$ is white. What is the reason for this? Is it because the first compound exists as a complex ion and has d-d transitions?

$\endgroup$
  • 7
    $\begingroup$ Both should be considered as complex ions in solid state. $\endgroup$ – Jan Feb 4 '16 at 16:35
13
$\begingroup$

The reason is indeed $\text d$-to-$\text d$ transitions. The d orbitals of both $\ce{CuCl}$ and $\ce{CuCl2}$ have two possible energy levels (based on crystal field theory).

$\ce{Cu(II)}$ is a $\text{d}^{9}$ metal, and $\ce{CuCl2}$ exists as a (distorted) octahedral around the copper atom. Based on crystal field theory, the d orbitals of an octahedral metal center can be represented as shown below. In this case, because $\ce{Cu(II)}$ is $\text{d}^{9}$, the three lower energy suborbitals are fully filled in the ground state, and but the top two suborbitals only contain a combined three electrons. Therefore, the energy required to excite that electron is $\Delta _o$. When that electron returns to the ground state, it gives off a different wavelength of light than it absorbed (in this case, the light emitted is apparently not visible light).

Octahedral splitting

$\ce{Cu(I)}$, on the other hand, is a $\text{d}^{10}$ metal ion, so each of the d suborbitals is fully filled. In order for an electron to be excited in this situation, it would have to leave the 3d orbital (and the $n=3$ shell altogether) and enter the $n=4$ shell. This is rather large energy jump compared to the intraorbital jump an electron in $\ce{CuCl2}$ has to make. Therefore, $\ce{CuCl}$ does not readily absorb visible light as $\ce{CuCl2}$ does, so it appears colorless. That is, until it see moist air.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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