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The coordination number refers to the number of binders that interact with the central metal ion. So, in the complex $\ce{[Cr(C2O4)3]^3-}$ chromium is bound to three oxalate ions.

Why its coordination number is considered 6? So, is it an incorrect definition of coordination number?

Can you help me?

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    $\begingroup$ Each oxalate ion has two binders. $\endgroup$ – Ivan Neretin Jan 14 at 12:03
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    $\begingroup$ A better way to think of it is that the coordination number is the number of points of contact between ligands and the metal. If a ligand like oxalate touches the metal at two places, it adds 2 to the coordination number instead of 1. $\endgroup$ – Andrew Jan 14 at 12:28
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This seems to be the case of "show, don't tell".

IUPAC's definition of coordination number (C.N.) which applies to inorganic complexes:

In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

Let's have a look at the one of the first determined structures of trioxalatochromate(III) inorganic salts [1]:

Crystal structure of trioxalatochromate(III)
Figure 1. Crystal structure of $\ce{[Cr(C2O4)3]^3-}$ anion (ICSD-109607). Color scheme: $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#8A99C7}{\Large\bullet}~\ce{Cr}$.

Here, there are three oxalate $\ce{C2O4^2-}$ ligands arranged in distorted octahedral environment (depicted with semi-transparent coordination polyhedron). Each oxalate is a bidentate chelating ligand, resulting in C.N. 6 of chromium center.

References

  1. van Niekerk, J. N.; Schoening, F. R. L. The Crystal Structure of Ammonium Trioxalatochromate(III), $\ce{(NH4)3[Cr(C2O4)3].2H2O}.$ Acta Cryst 1952, 5 (4), 499–505. DOI: 10.1107/S0365110X52001428.
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As Ivan Neretin told it, each group oxalate has two groups $\ce{-COO^-}$. So it may be bound to a central metal by two oxygen atoms, one per group $\ce{-COO^-}$.

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  • $\begingroup$ So, the binders aren't the whole molecule or ion, but the atoms bound to the metallic ion, I'm right? $\endgroup$ – user3204810 Jan 14 at 12:39

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