# Solutions 5: Coordination Complexes

These are solutions to select problems from homework set #5.

## S5.0

Yes, this complex ion will exhibit optical isomerism and is "chiral." ## S5.1

The crystal field theory describes the interaction between the ligands and the d and central atom electrons. Due to their repulsion and attraction different ligands produce a different splitting in the d orbitals. The change in energy levels due to the splitting is the energy the complex absorbs which can correlate to the wavelength of light. From Planck’s constant we know that $$\mathrm{E=h\nu}$$.

Using the speed of light $$\ce{c}$$ we know that $$\mathrm{c=\lambda\nu}$$ where $$\mathrm{\lambda}$$ is the wavelength and $$\mathrm{\nu}$$ is the frequency. We can solve for $$\mathrm{\nu=\dfrac{c}{\lambda}}$$. Then we can plug it into our previous equation to get $$\mathrm{E=\dfrac{hc}{\lambda}}$$

From the equation we know depending on the energy we can find the correlating wavelength that is absorbed by the compound and we see the color that is opposite of the absorbed wavelength on the color wheel. For example if a compound absorbed violet we would see yellow and if it absorbed green we would view it as red.

To review Crystal field theory, please visit "Crystal Field Theory".

## S5.2

$$\ce{Co^2+}$$ is a d7 ion.

• For strong field lignd field, the first 6 d-electrons will fill up the three stabilized orbitals and the final d-electron will be in the destabilized orbital. This will result in a single unpaired electron in the complex.
• for a weak field ligand field, the first 5 d-electrons will fill up all five orbitals (three stabilized and two destabilized). This is since the weak-field ligands have a weak crystal field splitting (hence the term) and the energy needed to put a second electron in an orbital is less than the spin pairing energy. The remaining two d-electrons will pair up with two in the stabilized orbitals and result in three unpaired electrons in the complex.

Drawing the d-orbital digrams for these two cases is left as an exercise for the student.

## S5.3

We will be looking at the metal center of these complex ions with respect to the crystal field theory. Diamagnetism is when the electrons in a complex ion fill up the valence shell, and paramagnetism is when they do not and leave a spin.

1. $$\ce{[Co(CN)6]^3-}$$: there are 6 of the ion $$\ce{CN-}$$ and the charge on the complex ion is -3. So the charge on the cobalt metal cation must be +3. We will be looking at $$\ce{Co^3+}$$ under a strong ligand field due to the strong attractive forces of the $$\ce{CN-}$$ ligands. Normally, $$\ce{Co}$$ has an electron configuration of [Ar] 4s23d7, so $$\ce{Co^3+}$$ will be [Ar] 3d6. The strong ligand field means that the electrons will fill the dxy,dxz, and dyz orbitals first before jumping up to the dz^2 and dx^2-y^2 orbitals (fill up bottom before top). Therefore, the 6 electrons in the valence shell will fill each orbital in the lower energy orbitals with two electrons, leaving no unpaired electrons and creating a Diamagnetic ion.

$$\ce{[MnI6]^4-}$$: there are 6 of the ion $$\ce{I-}$$ and the charge on the complex ion is -4. So the charge on the manganese metal cation must be +2. Looking at $$\ce{Mn^2+}$$ under weak ligand field due to the weak splitting of $$\ce{I-}$$ ligands. Normally $$\ce{Mn}$$ has the electron configuration of [Ar] 4s23d5, so $$\ce{Mn^2+}$$ will be [Ar] 3d5. The 5 electrons in the valence shell under the weak ligand field will pair each orbital (high and low energy) with one electron. Each electron is not paired, so therefore this ion is Paramagnetic.

1. $$\ce{[Mn(NO2)4]^2-}$$: there are four $$\ce{NO2-}$$ ligands with -1 charge and the total charge of the complex ion is -2. So the oxidation state of the manganese metal cation is +2. Normally $$\ce{Mn}$$ has the electron configuration of [Ar]4s23d5, so $$\ce{Mn^2+}$$ will be [Ar]3d5. The 5 electrons in the valence shell under a strong ligand field must follow the tetrahedral crystal field splitting diagram. In tetrahedral, there are two stabilized orbitals and three destabilized orbitals. In contrast to octaerl fields, tetrahedral field are weak ($$\Delta_t \approx \Delta_o$$). So each of the five orbitals will be filled with a single electron. Hence, there are five unpaired electrons in this complex ion.

## S5.4

$$\ce{Fe}$$ has an oxidation state of 2+ since $$\ce{NH3}$$ is neutral. This means that $$\ce{Fe}$$ has 6 electrons. If the complex is paramagnetic, then it is impossible to tell whether it is an octahedral or tetrahedral complex because in each geometry there are unpaired electrons. If the complex is diamagnetic, then it must be an octahedral complex because it cannot be diamagnetic as a tetrahedral complex. For review on paramagnetism and diamagnetism of complex coordinations, please visit "Ligand Field Bond Theory".

## S5.5

To answer this you should recognize that the cobalt nitrate will dissolve fully in solution and generate the hexaaqua ligated cobolt complex in solution. Many of these sorts of reactions are really ligand exchange reactions and you have to recognize the starting and end ligands to write the reaction properly.

$\ce{water + Co(NO3)2(s) \rightarrow Co(H2O)6(aq) + 2NO3-(aq)}$

1. $$\ce{2NaOH(aq) + Co(H2O)6 \rightarrow Co(OH)2(s) + 2Na^+ (aq) + 6H2O(l)}$$
2. $$\ce{Co(OH)2(s) + 4NH3(aq) \rightarrow Co(NH3)4^2+(aq) + 2OH- (aq)}$$

For review on the topic, visit "Chemical Reactions"

## S5.6

Out of the three different complexes, the $$\ce{[Co(en)3]^3+}$$ complex with the ($$\ce{en}$$) ligands has the largest $$\mathrm{K_f}$$ value (4.8 x 1048) due to the chelate effect (Table E4). The chelate effect shows the increased stability of chelates over complex with monodentate due to an increase in entropy associated with chelation.

Please review "Chelation" and "The Chelate Effect" for further information.

## S5.7

The crystal field theory splits the five degenerate d orbitals into different energy states. The change in energy between the two groups is related to a photon of energy. The absorption of a higher energy photon means that short wavelengths (with a high frequency) are being absorbed, therefore we see longer wavelengths of light such as red. The absorption of a low energy photon means that long wavelengths (with a low frequency) are being absorbed, therefore we see short wavelengths of light such as violet.

## S5.9

1. The $$\ce{Fe}$$ in $$\ce{[FeI6]^3-}$$ has an oxidation state of +3. Looking at the periodic table, $$\ce{Fe^3+}$$ is found to have 5 d-electrons. I is a weak ligand and therefore has a high spin. This has a coordination number of 6 so therefore is in an octahedral structure. The $$\ce{Co}$$ in $$\ce{[Co(CN)6]^3-}$$has an oxidation state of +3 meaning it has 6 d electrons. $$\ce{CN}$$ is a strong field ligand so it will have a low spin. Because of a coordination number of 6, it is in a octahedral structure. 2. The $$\ce{Fe}$$ in $$\ce{[FeBr6]-}$$ has an oxidation state of +5 and has 3 d electrons. $$\ce{Br}$$ is a weak field ligand and so it will have a high spin, but for the d3 . Coordination number of 6 will make it have an octahedral structure. ## S5.10

In principle, to solve this problem, the absolute crystal field stabilization energies must be calculated for the different geometries and the total number of d electrons (e.g. $$\ce{Cu}$$ has an oxidation state of +4 which means it is a d7 metal). However, this is a trick question of some sort, since the complex formula is given and the coordination number if 6. Therefore has an octahedral structure and you do not need to do the full stabilization calculations.
See Octahedral vs. tetrahedral Geometries for more details on determining tetrahedral vs. octhedral geometries.

## S5.11

Copper ions in aqueous solution are dressed by the solvent as hexaaquacopper(II) complex ions. The reaction due to the addition of $$NH_3$$ is a ligand exchange reaction.

$$\mathrm{\underset{Light\: Blue}{[Cu(H_2O)_4]^{2+}} + 4NH_3 \rightarrow \underset{Deep\: Blue}{[Cu(NH_3)_4]^{2+}} +4H_2O}$$

## S5.12

$$\ce{[Fe(en)3]^2+}$$ should have the largest overall $$\mathrm{K_f}$$ value because it consists of multiple polydentate ligands which have larger $$\mathrm{K_f}$$ values than monodentate. This is called the chelate effect (occurs in a ring).

## S5.13

1. $$\ce{Zn}$$ has oxidation number +2, so $$\ce{Zn^2+}$$ has electron configuration [Ar]3d10. There are zero unpaired electrons.
2. $$\ce{Cu}$$ has oxidation number +2, so $$\ce{Cu^2+}$$ has electron configuration [Ar]3d9. There is one unpaired electron. $$\ce{Mn}$$ has oxidation number +3, so $$\ce{Mn^3+}$$ has electrong configuration [Ar]3d4. There are four unpaired electrons. Therefore, this octahedral complex has more unpaired electrons than the tetrahedral complex.

## S5.14 1. Since $$\ce{Cu^2+}$$ has 9 electrons, the only way to arrange that in 5 d orbitals is to have one unpaired electron. Therefore it is paramagnetic.

## S5.15

1. Each has $$\ce{Fe^3+}$$ as the metal ion. The color difference is attributed to the fact that in one case $$\ce{Fe^3+}$$ is bonded to a strong field ligand, $$\ce{NH3}$$, and in the other it is bonded to a weaker field ligand, $$\ce{H2O}$$. When bonding to a strong field ligand, the oxtahedral complex has a large $$\mathrm{\Delta_\circ}$$ in its electron configuration for its d orbitals. This large $$\mathrm{\Delta_\circ}$$ makes it necessary to absorb a high energy photon, which has short wavelengths with a high frequency. If you absorb short wavelengths you will see a longer wavelength, thus $$\ce{[Fe(NH3)6]^3+}$$ is yellow. In contrast, when bonding to a weak field ligand, the oxtahedral comples has a small $$\mathrm{\Delta_\circ}$$ in its electron configuration for its d orbitals. This small $$\mathrm{\Delta_\circ}$$ makes means it is only necessary to absorb a low energy photon, which has long wavelengths with a low frequency. If a complex absorbs longer wavelengths, the observers will see a shorter wavelength color, thus $$\ce{[Fe(H2O)6]^3+}$$ is violet.
2. $$\ce{[Ni(CN)4]^2-}$$ is yellow because cyanide is a strong field ligand (via the spectralchemical series), and electrons require more energy to be promoted to the destabilized orbitals. Therefore, the electrons absorb shorter wavelength (in this case in the blue spectrum) and it reflect (do not absorb) the complementary yellow color (from the color wheel). $$\ce{[Ni(H2O)6]^2+}$$ is green because water is a weaker field ligand, and electrons require less energy photons to be promoted to the higher d orbital. Therefore, the electrons will absorb at longer wavelength (in this case from the red spectrum) and reflect the complementary color green.