$\ce{CuCl2}$ 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?

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    $\begingroup$ Both should be considered as complex ions in solid state. $\endgroup$
    – Jan
    Commented Feb 4, 2016 at 16:35

1 Answer 1


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.


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