# Why ionic radii of Cu2+ is less than Zn2+?

Fully filled orbital has more effective nuclear charge than incompletely filled orbital. So atomic or ionic radii of elements or ions having fully filled orbitals should be less than that of elements having incompletely filled orbitals (e.g. $\ce{F}$ and $\ce{Ne}$).

Now $\ce{Cu^2+}$ has a $\ce{d^9}$ configuration and $\ce{Zn^2+}$ has a $\ce{d^10}$ configuration. So, $\ce{Zn^2+}$ has full filled d orbital and is thus stable. According to my assumption above, $\ce{Zn^2+}$ should have less ionic radii than $\ce{Cu^2+}$. But it is found that $\ce{Zn^2+}$ has more ionic radii than $\ce{Cu^2+}$.

Ionic radii of $\ce{Cu^2+}$ is $\ce{73}$ pm and that of $\ce{Zn^2+}$ is $\ce{74}$ pm. Why is this so?

(source)

• are you sure that Zn2+ will have $d^{10}$ configuration coz then there will be no electron in the 4s sub-shell – manshu Dec 27 '15 at 17:49
• @manshu Of course it will. It is the cation’s most stable electronic configuration. – Jan Dec 27 '15 at 17:53
• @Jan Then why does nickel have $4s^2 3d^8$ configuration. even if both have same number of electrons – manshu Dec 27 '15 at 18:00
• $d^{10}$ shell is very rigid while $d^9$ shell is polarisable and, since copper is typically square planar, it moves slightly from the ligands, allowing tighter association. – permeakra Dec 27 '15 at 19:27
• @manshu chemistry.stackexchange.com/a/4500/7475 Just because the electron count is the same does not mean the species are the same — zinc(II) has two more protons in its nucleus. – Jan Dec 27 '15 at 21:43

$\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ are the orbitals with higher energy. As it can be seen, they are located between the metal atom's center and the ligands. As such, electrons on said orbitals have clear effect on metal-ligand distance, so for low-spin state the observed ionic radius monotonously descreases up to 6 electrons (fully occupied lower orbitals) and then monotonously increases. For the high-spin state, the increases are also located on steps when an electron is added to higher orbital, but in this case the electrons are added on d4-d5 and d8-d10 configurations, so the increase happens in two steps. For the 2nd and 3rd rows, only few last steps on the curve are observed as most elements in these rows tend to form metal-metal bonds in low oxidation states and thus their true ionic radii are unknown.
And now, why are these dependences of limited use. Consider the graph of $\ce{M-Cl}$ distances for chlorides of the first d-row below. The data was obtained from the open crystallography database. No correlation with the plot above can be established, despite most of the chlorides except for $\ce{ZnCl2}$ having their metal in sixfold coordination. This is because said chlorides are not isostructural. Layered $\ce{MoS2}$ and $\ce{CdI2}$ structures together with chains of squares and 3D-lattices occur. Furthermore, for iron two different structures (layered and rutile-like ones) were found with, obviously, different $\ce{M-Cl}$ distances (the layered one is plotted. As a fun fact, the longer distance is observed in the higher-pressure rutile modification). An obvious conclusion for the plot is that only for isosctructural fragments we can conclude something based on electron structure of the central metal atom.