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Solutions 6: Isomers of Coordination Complexes

  • Page ID
    10759
  • These are solutions to select problems from homework set #6.

    Q6.1

    \(\mathrm{Cu^{2+} ­ (aq) + CO_3^{2-} (aq) \rightarrow Cu(CO_3) (s)}\)

    \(\mathrm{Na_2CO_3 (s) \rightarrow Na^+ (aq) + CO_3^{2-}(aq)}\)

    The green color is most likely due to the unreacted Copper(II) carbonate.

    \(\mathrm{Cu(CO_3) (s) + 2 NH_3 (aq) \rightarrow [Cu(NH_3)_2]^{2+} (aq) + CO_3^{2-} (aq)}\)

    \(\ce{[Cu(NH3)2]^2+ (aq) + 2 H3O^+ (aq) \rightarrow [Cu(H2O)2]^2+ (aq) + 2 NH4+ (aq)}\)

    Q6.2

    The complex containing multiple bidentate ligands, \(\ce{[Ni(en)3]^2+}\), will have a larger formation constant, \(\mathrm{K_f}\). Despite the fact that each complex shares a common transition metal Nickel, we must note that all polydentate ligands have larger formation constants. This is known as the Chelate effect. As an empirical rule of thumb, the formation of a complex with a chelating agent, such as (en) will always be more thermodynamically stable than a complex formed with a monodentate ligand. This added stability can be linked back to entropy, which is dependent on disorder and randomness. With that in mind, the binding of a chelating agent with multiple teeth will ensure a smaller loss of entropy disorder. The result is a more stable reaction, thus a higher formation constant.

    Q6.3

    Not given

    Q6.4

    1. The \(\ce{Fe^4+}\)ion has a d4 electron configuration. Because \(\ce{CN^-}\) is a strong ligand, the lower energy orbitals will be filled first. The result is two unpaired electrons.
    1. In the octahedral \(\ce{[Ni(NO2)3(NH3)3]-}\) complex ion, \(\ce{Ni^2+}\) has a d8 electron configuration. Because these ligands are quite strong (see spectrochemical series), the lower energy orbitals will be entirely filled (using 6 electrons) and the resulting two electrons will then fill the destablized orbitals (as unpaired electrons). This results in a paramagnetic complex with two unpaired electrons

    The tetrahedral \(\ce{[Mn(SCN)4]}\) complex compound has strong ligands and will therefore have a high \(\Delta_t\). However, this does not matter much since the tdetrahedral splitting is weaker than the corresponding octahedral splitting (i.e., \(\Delta_t \approx 4/9\, \Delta_o\)). Thus tetrahedral complexes are typically high spin. The \(\ce{Mn^4+}\) ion has a d3 electron configuration with three unpaired electrons (two on the stabilized orbitals and one in the destabilized orbitals). Consequently, \(\ce{[Mn(SCN)4]}\) is also paragmagnetic, but with a greater magnetic moment than \(\ce{[Ni(NO2)3(NH3)3]-}\) since three electrons are unpaired.

    Q6.5

    1. First, we must find the electron configuration of \(\ce{Zn(II)}\), which will be 3d10. Automatically, we know that the structure will be diamagnetic and there will be no unpaired electrons. The next question is, will it be square planar or tetrahedral? Because \(\ce{Cl-}\) is a weak field ligand, we know that it will probably be a high-spin complex. This is helpful because tetrahedral complexes are generally high-spin. Also, because 3d transition metals prefer high spin complexes, we can assume that the probable structure will tetrahedral.
    2. Looking at the periodic table, we find that \(\ce{Ni(II)}\) will have an electron configuration of 3d8. If this wwere an octahedral complex, then we know expect complex to be paramagnetic. However, the coordination number of the complex is 4, narrowing our option to only square planar (diamagnetic) and tetrahedral (paramagnetic). For first-row atoms such as Ni2+, tetrahedral arrangement will be preferred if the ligands are large and weak-field. If the ligands are small (more rod-like like CN-) and are strong-field, the planar arrangement will be preferred. The cyanide ion is small and it is a strong-field ligand so tetracyanonickelate(II) will probably be square planar.

    For review on this topic, visit the page "Magnetic Properties of Coordination Complexes".

    Q6.6

    1. The second complex has the \(\ce{CN}\) ligand, which (according to the spectrochemical series) produces a greater splitting of d energy than \(\ce{NH3}\).Therefore, it is expected that \(\ce{­­­­­­­­­­­­­[Cu(CN)4]^2-}\) will absorb light of a shorter wavelength and higher energy than \(\ce{[Cu(NH3)4]^2+}\). Whichever color that the complex appears will merely be the direct opposite of what it absorbs. As a result, \(\ce{[Cu(NH3)4]^2+}\) absorbs orange and transmits/reflects light is blue. Likewise, \(\ce{­­­­­­­­­­­­­[Cu(CN)4]^2-}\) absorbs in the blue region and the transmitted light is orange.
    1. To solve, we must first take the ligand strengths into consideration. Our first complex, Copper(II) sulfate pentahydrate has a \(\ce{H2O}\) ligands which are stronger than \(\ce{Cl-}\) ligands in Nickel(II) chloride hexahydrate. This means that our first complex produces a greater splitting of the d energy. Therefore, we can expect Copper(II) sulfate pentahydrate to absorb light of a shorter wavelength than Nickel(II) chloride hexahydrate. Our copper complex will therefore absorb complementary orange and the transmitted light will be blue. Our Nickel complex will absorb complementary red and be seen as green.

    For review on this topic, visit the page "Colors of Coordination Complexes".