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Chemistry LibreTexts

24.E: Chemistry of Coordination Chemistry (Exercises)

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  • These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

    24.3: Nomenclature of Coordination Chemistry


    Write the name of the following complexes

    1. [CoCl3(NH3)3]
    2. [Co(ONO)3(NH3)3]
    3. [Fe(ox)2(H2O)2]-
    4. Ag2[HgI4]


    1. triamminetrichlorocobalt(III)
    2. triamminetrinitrito-O-cobalt(III); or triamminetrinitritocobalt(III)
    3. diaquadioxalatoferrate(III) ion
    4. silver(I) tetraiodomercurate(II)

    24.5: Color and Magnetism


    1. How many unpaired electrons are found in oxygen atoms ? 
    2. How many unpaired electrons are found in bromine atoms? 
    3. Indicate whether boron atoms are paramagnetic or diamagnetic. 
    4. Indicate whether F- ions are paramagnetic or diamagnetic. 
    5. Indicate whether Fe2+ ions are paramagnetic or diamagnetic.


    1. The O atom has 2s22p4 as the electron configuration. Therefore, O has 2 unpaired electrons.
    2. The Br atom has 4s23d104p5 as the electron configuration. Therefore, Br has 1 unpaired electron.
    3. The B atom has 2s22p1 as the electron configuration. Because it has one unpaired electron, it is paramagnetic.
    4. The F- ion has 2s22p6 has the electron configuration. Because it has no unpaired electrons, it is diamagnetic.
    5. The Fe2+ ion has 3d6 has the electron configuration. Because it has 4 unpaired electrons, it is paramagnetic. 

    24.6: Crystal Field Theory


    Describe crystal field theory in terms of its

    1. assumptions regarding metal–ligand interactions.
    2. weaknesses and strengths compared with valence bond theory.


    In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?


    Will the value of Δo increase or decrease if I ligands are replaced by NO2 ligands? Why?



    The value of Δo would increase because NO2− is a stronger ligand which means that it would have a larger split.


    Review oxidation state for more details.


    For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?


    How can CFT explain the color of a transition-metal complex?

    The color of a complex is determined by the wavelength it absorbed. For example, if a complex absorb red light, it shows green.

    How is the wavelength determined?

    It related to the structure of a complex. Based on CFT, if a complex has strong ligand field, it is low spin, which means the energy gap between each orbitals are high. The energy gap between orbitals relates to the wavelength of the complex absorbed: E = lamda * h

    Stronger the energy, shorter the wavelength absorbed, and the color maybe shows more red and orange-ish. 


    Do strong-field ligands favor a tetrahedral or a square planar structure? Why?


    For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.


    a. There are 6 ligands bonded to the central atom, so the complex is octahedral. 

    Since the complex has a negative 3 charge and the oxidation number of Cl is -1, Clgives the complex a -6 charge and Ti must be +3 for the complex to be -3 in total.

    Ti is the second transition metal in the first row of transition metals (atomic number 22). Ti3+ has the electron configuration of [Ar] dwhich means that there is one electron in the d orbital. It it unpaired (since there are no other electrons).

    Since Cl is pretty low in the Spectrochemical Series, it is a high energy ligand that would cause the complex to be high spin. However, since there is only one electron, whether it is high or low spin does not affect the number of paired/unpaired electrons.

    To see how the electron configuration would be for octahedral complexes, go to and scroll down to Electron Configurations in Octahedral Complexes

    b. There are 4 ligand bonded to the central atom, so the complex is either tetrahedral or square planar. In this case, it is tetrahedral. Since the overall charge is -2, and 4 Cl's give a -4 charge, Co's charge must be +2. 

    Co is the 7th transition metal in the first row of transition metals. Co2+ has an electron configuration of [Ar] dbecause transition metals lose electrons in the s orbital first, before they begin to lose in the d orbital. 

    As explained above, because of Cl this would be a high spin complex. In a high spin complex, all shells are half filled (not where the bottom ones are completely filled first before moving onto the top shells). Since there are 7 electrons and 5 shells, only two shells will be completely filled and there will be 3 unpaired electrons. 

    The image below is NOT how the electrons should look, but shows the shells. The only thing is that one more of the tetrahedral upper shell should have an unpaired electron (there are 8 electrons in the image instead of 7). You can ignore the sq planar side of the image since this complex was tetrahedral. 

    NOTE: I tried to attach an image here but could not do it. If you go to and scroll down to SAMPLE EXERCISE 24.9, you can see the image.

    1. [TiCl6]3−
    2. [CoCl4]2−


    For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

    1. [Cu(NH3)4]2+
    2. [Ni(CN)4]2−


    1. d9, square planar, neither high nor low spin, single unpaired electron

      First, we must determine the location of each of the ligands on the spectrochemical series. We can see that NH3 is slightly lower than CN, which is the greatest of the series. Since NH3 is only slighly lower but past the point that classifies is as a strong field, it is neither high or spin. 

      We then must verify the charge of Cu in the complex. Through calculations we come to the result that Cu=2+. In it's row, Cu is #11 so we subrtract 11-2=9. This tells us that the complex is a d^9 and square planar according the the 4 subscript. 

      Now, we enter the 9 electrons into the following diagram: 
      This diagram is specific for square planar complexes. We can classify this complex as weak field since it is weaker than [Ni(CN)4]^-2. If a complex is weak field, it has a small delta so we do NOT fill the bottom levels first. We fill each from Left to Right with one electron at a time. After we fill the diagram accordingly with the 9 electrons, the top level will be missing 1 electron and therefore will have a single unpaired electron. 
    2. d8, square planar, low spin, no unpaired electrons

      CN is highest on the spectrochemical series and therefore has a strong field, low spin, and large delta.
      We will fill out this diagram due to it's square planar shape. Nickel is #10 in it's row on the periodic table. After we solve for Nickel's charge, we find that it is +2. We then solve for d by 10-2=8. This proves that it is a d^8 complex with a large delta and low spin. If we fill out the diagram for a large delta, we fill in the lower orbitals fully before moving up to the higher ones. After we fill in the diagram accordingly, the only unfilled level will be the top one, dx^2-dy^2. This means that all of the other levels will be filled so no unpaired electrons will exist in the complex. 


    The ionic radii of V2+, Fe2+, and Zn2+ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.


    As we can see from the periodic table, all the mentioned metals, V, Fe, and Zn, are present in the 4th period in the periodic table. All these metals are 3d elements since electrons fill their 3d orbitals. As for the ionic radii, the similar ionic radii is due to the shielding effect, which is when the valence electrons are "shielded" from the attraction towards the nucleus. The atomic radii trend in the periodic table is that atomic radii decreases from left to right, but in the case of 3d elements, as we move from left to right, elements are added to the inner 3d orbital, thus pushing the outer 4s orbital. This explains why the atomic radii or ionic radii of these ions are so similar.