Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin \(\mathrm{Fe}(\mathrm{II})\) 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) \(\mathrm{CO}\) is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

Step by step solution

01

A(a)Determining unpaired electrons

In low-spin complexes, the electrons try to fill in the lower-energy orbitals before starting to fill higher-energy ones. In high-spin complexes, electrons fill the orbitals more evenly. In the case of an octahedral low-spin \(\mathrm{Fe(II)}\) complex like oxyhemoglobin, the electron configuration will be \(t_{2g}^6\). It means that there would be 0 unpaired electrons for oxyhemoglobin (low-spin). In high-spin complexes, like deoxyhemoglobin, the electron configuration is \(t_{2g}^4 e_g^2\), and there would be 4 unpaired electrons.

02

A(b) Ligand in deoxyhemoglobin

In deoxyhemoglobin, the oxygen molecule (\(\mathrm{O}_{2}\)) is not coordinated to the iron. Instead, a water molecule (\(\mathrm{H}_{2}\mathrm{O}\)) typically occupies the coordination site in place of the \(\mathrm{O}_{2}\).

03

A(c) Explanation for color differences

The color of coordination compounds is due to the absorption of light in the visible region of the electromagnetic spectrum, which causes electronic transitions between the different energy levels of the d orbitals. The difference in color between oxyhemoglobin and deoxyhemoglobin can be attributed to the fact that they have different numbers of unpaired electrons and different electron configurations, which leads to different energy level splitting patterns and different absorbed wavelengths.

04

A(d) Relative equilibrium constants

The fact that 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 (carboxyhemoglobin) suggests that the equilibrium constant for the binding of CO to hemoglobin is much larger than the equilibrium constant for the binding of \(\mathrm{O}_{2}\) to hemoglobin. This means that CO binds to hemoglobin more strongly and efficiently than \(\mathrm{O}_{2}\) does.

05

A(e) Expected color of carboxyhemoglobin

As CO is a strong-field ligand, it will induce a high crystal field splitting of the d orbitals on the metal ion, which results in a low-spin complex. Since different ligands and different spin states imply different electron configurations and energy level splitting patterns, the color of carboxyhemoglobin is expected to be different from both oxy- and deoxyhemoglobin. Although we cannot predict the exact color of carboxyhemoglobin without more information, considering that CO forms a low-spin complex and leads to a different ligand field splitting, it could be expected to exhibit a different visible color in the spectrum.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!