7
$\begingroup$

How do you know the number of ligands surrounding a metal ion?

Example: the obtained complex of $\ce{Ni^2+}$ ion in $\ce{NH3}$ solution is written as $\ce{[Ni(NH3)6]^2+}$ but $\ce{Cu^2+}$ in $\ce{NH3}$ solution is $\ce{[Cu(NH3)4]^2+}$.

So how do you find the number for different metal ions like $\ce{Ag^2+}$?

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ The coordination number (number of ligands) changes depending on the type of ligand. In this case ammonia. $\endgroup$ – Ali Caglayan May 24 '15 at 19:48
  • $\begingroup$ I suspect you may have erroneously tried to use the silver dication $\ce{Ag^{2+}}$ as a standard example of transition metal ion by extrapolating the most common aqueous copper ion, $\ce{Cu^{2+}}$. In reality, $\ce{Ag^{2+}}$ is rather unstable, and silver is much more commonly found as a monocation, $\ce{Ag+}$. $\endgroup$ – Nicolau Saker Neto May 24 '15 at 22:53
  • 2
    $\begingroup$ @NicolauSakerNeto $\ce{Ag^{2+}}$ is perfectly stable in proper neighborhood. In fact $\ce{Ag+}$ disproportionate giving $\ce{Ag^{2+}}$ and metallic silver in some conditions. $\endgroup$ – permeakra May 24 '15 at 23:28
  • $\begingroup$ @permeakra Quite true, I didn't mean to imply it wasn't stable ever. To my understanding $\ce{Ag^{2+}}$ is stable in a narrower range of conditions, though, or at least doesn't overlap much with "common" conditions, for some reasonable definition of common. Coordination compounds in aqueous media with $\ce{Ag^{2+}}$ ions are relatively scarce, I would believe. $\endgroup$ – Nicolau Saker Neto May 24 '15 at 23:37
  • 2
    $\begingroup$ @NicolauSakerNeto you are incorrect in it. $\ce{Ag^{2+}}$ is quite stable in bipiridyl complexes and again, $\ce{Ag+}$ disproportionate in presence of some ligands pubs.acs.org/doi/abs/10.1021/ja00775a074 $\endgroup$ – permeakra May 24 '15 at 23:44
2
$\begingroup$

To my knowledge, there is no "rule" to predict the coordination number of a given metal ion. There are, however, guidelines, which are nicely summarized here.

  • Larger ions can accommodate more ligands
  • Bulky ligands will reduce the overall coordination number
  • Highly charged ions will tend to accept a greater number of Lewis Bases

With your example, we can make an educated guess as to the coordination number of Ag2+ using periodic trends. It is in the same group as copper and will be bigger since it is lower on the periodic table. Based on the three points above, I would predict that Ag2+ could have a coordination number greater than 4. It turns out that (distorted) octahedral coordination of the silver dication is not all that uncommon.

| improve this answer | |
$\endgroup$
5
$\begingroup$

1st-row $d$-element + 2nd-row $p$-element - usually octahedron except for $\ce{Ni(II),Cu(II),Co(I)}$ (often a square or a distorted octahedron with two weakly bound ligands) and $\ce{Zn}$ (usually a tetrahedron). 1st-row $d$-element + 3+-row $p$-element - usually a tetrahedron.

2nd/3rd row $d$-element are somewhat more complicated as they allow higher coordination number up to 9 ($\ce{ReH9^{2-}}$), but the situation is so much influenced by electron configuration, ligands and metal-metal interactions here that simple rules loose a lot of meaning.

In short, the coordination number is usually a compromise between 18-electron rule and size of ligands. For example, $\ce{V(CO)6}$ should dimerize, giving c.n. of 7 to vanadium, but can't do so because of size constrains. Somewhat similar happens with bulky alcoholats of titanium, that tend to include $\ce{Ti(IV)}$ in octahedral coordination, but lower c.n. is observable in case of tritox (tris-tert-butyl) alcoholat.

Though it is generally not recommended to guess the c.n., it is always better to google for some reference compounds.

| improve this answer | |
$\endgroup$

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.