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Chapter 18: Tricks and Tips

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    Tips, tricks and hints

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    1. Some species are covalently bound to the metal while others are not.

      • As Werner concluded back in 1913, there are two kinds of bonding in a coordination compound. There is covalent bonding within the complex ion and ionic interactions between the complex ion and any counterions it might have.

        In solution, an ionic species break up into ions. That's always true. But a coordination complex is one ion -- it has many parts, but they are all bound together covalently, not ionically. When we dissolve a coordination complex the ligands bound to the metal remain bound to the metal. For example, when we dissolve [Co(NH3)5Cl]Br2 in water the following dissociation occurs:

        [Co(NH3)5Cl]Br2 (s) --> [Co(NH3)5Cl]2+ (aq) + 2 Br- (aq)

        Notice that the chloride ligands do not behave as ions. This is no different that saying that the oxygen atoms of sulfate remain bound to sulfur when NaSO4 is dissolved in water!

        Finally, notice that we always encase the complex ion in brackets when we write the formula. That helps make it clear we're dealing with one species containing several covalently bonded parts. Anything outside the brackets is treated like it is any "normal" ionic complex

    2. Use the CORRECT orbital splitting diagram

      • Students have a tendency to always write the octahedral field splitting diagram ("two up three down") when determining the electronic configuration of ions. However, recall that four coordinate complexes have different splitting diagrams.

        Tetrahedral have the "opposite" splitting diagram ("three up two down") and the splitting is much smaller than in octahedral complexes. Because this splitting is small in tetrahedral complexes, all tetrahedral complexes are high spin.

        When we recognize that a complex is four-coordinate remember that it could be tetrahedral or it could be square planar!! This can make an important difference in properties.

    3. The number of ligands and coordination number are NOT always the same.

      • It's easy to make this error. For example, [Co(NH3)5Cl]2+ has six ligands and it's six coordinate. But this is not always true.

        Coordination Number is the number of atoms coordinated to a complex. Because some ligands may coordinate through more than one atom, the number of ligands and the coordination number are not always the same!!

        A complex such as [Cr(en)3]3+ contains only 3 ligands, but it is octahedral!! That's because each en ligand is bidentate. Each en occupies two coordination sites around the metal. 3 ligands x 2 coordination sites = 6 coordinate!

        If you fail to recognize this difference you'll get into real trouble with a complex like [Ni(en)2Cl2]. This species has only four ligands, and it would be easy to use the wrong splitting diagram (such as square planar) instead of the correct one (octahedral). The coordination number of this species is SIX, not four.

    4. en can only bind in a cis fashion.

      • The ethylenediammine ligand can only bind in a cis fashion -- i.e. the two NH2 groups must always be 90 degrees apart. Never try to put the two ends of an en ligand 180 degrees apart -- it won't fit!

        By the way, recognizing this constraint makes it a lot easier to count geometric isomers when one of the ligands is en or a similar bidentate ligand.

    5. How do I know if this compound is colored?

      • When light hits a coordination complex, the electrons in a lower energy d-orbital may be excited/promoted to a higher energy orbital if the energy of the light is equal to the crystal field splitting energy.

        Visible "white" light is composed of many different frequencies/energies. If one of these is suitable for the promotion of an electron, the complex "consumes" that portion of the visible light, and so we see light that lacks certain frequencies = color. In the simplest form, we see the complement of the color (energy) that was absorbed as shown by the color wheel in Figure 18.13 of Chang.

        At our level of understanding, there are only two cases in which a compound is not colored:

        1. When there are no d-electrons to promote (a d0 configuration). For example, [Ti(H2O)6]4+ has no valence (i.e. d) electrons so we can not promote them! All wavelengths of our incident "white light" are transmitted or reflected equally and no color is observed.

        2. When there are 10 d electrons (a d10 configuration). When we promote an electron by absorbing light, we have to put it somewhere. All electron configurations except d10 have an empty higher energy orbital to accommodate an excited electron! We can't stuff more electrons into orbitals that are already filled.

        So, to recap ; If you have zero OR ten d (valence) electrons, the compound will be colorless. This does not depend on the geometry or crystal field splitting pattern (octahedral, tetrahedral, square planar etc.). Of course, if the compound is not d0 or d10 these factors would determine what color was observed!


    Chapter 18: Tricks and Tips is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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