Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) molecule, is a high- spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about \(10 \%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin? (e) CO is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

Step by step solution

01

(a) Number of unpaired electrons

Oxyhemoglobin is a low-spin Fe(II) complex with an \(\mathrm{O}_2\) molecule bound to the iron, while deoxyhemoglobin is a high-spin complex without the \(\mathrm{O}_2\) molecule. In both cases, we can assume the environment around the metal ion is octahedral. The Fe(II) ion has a d6 configuration. For a low-spin complex in an octahedral environment like oxyhemoglobin, electrons will occupy the lowest energy orbitals without any unpaired electrons due to strong ligand fields. Therefore, oxyhemoglobin has 0 unpaired electrons. In contrast, a high-spin complex like deoxyhemoglobin would have the electrons distributed across the orbitals with a larger number of unpaired electrons due to lower energy differences between the orbitals. In deoxyhemoglobin, there would be 4 unpaired electrons.

02

(b) Ligand in deoxyhemoglobin

In deoxyhemoglobin, the oxygen molecule is absent, and instead, a water molecule (H\(_2\)O) is coordinated to the iron.

03

(c) Different colors of hemoglobin and deoxyhemoglobin

The different colors of hemoglobin and deoxyhemoglobin can be explained by their different electronic structures and ligand coordination. The number of unpaired electrons and ligand identity can affect the energy levels and transitions between d-orbitals, which in turn influences the absorption of specific wavelengths of light that contributes to the color of the compounds. Oxyhemoglobin has a red color due to the low-spin configuration, which absorbs light more readily in the blue region of the visible spectrum. Deoxyhemoglobin, with its high-spin configuration and different ligand coordination, absorbs light in a different region of the spectrum, giving it a bluish cast.

04

(d) Equilibrium constants for CO and \(\mathrm{O}_2\) binding to hemoglobin

The fact that a 15-minute exposure to air containing 400 ppm of CO causes about \(10\%\) of the hemoglobin to convert into carboxyhemoglobin suggests that CO has a higher affinity for hemoglobin compared to \(\mathrm{O}_2\). Therefore, the equilibrium constant for binding CO (K\(_\text{CO}\)) is higher than the equilibrium constant for binding \(\mathrm{O}_2\) (K\(_\text{O2}\)). This means that hemoglobin has a higher tendency to bind to CO than \(\mathrm{O}_2\), causing the displacement of \(\mathrm{O}_2\) and formation of carboxyhemoglobin.

05

(e) Color of carboxyhemoglobin

CO is a strong-field ligand, causing the carboxyhemoglobin complex to have a low-spin configuration with fewer unpaired electrons. This would result in a different electronic structure compared to both hemoglobin and deoxyhemoglobin, causing the absorption of light at different wavelengths. Considering that oxyhemoglobin appears red due to its low-spin configuration, one might expect carboxyhemoglobin to also display a red color, possibly even more intensely red than oxyhemoglobin due to the stronger ligand field induced by CO.

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