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Chapter 9: Problem 78

The transport of \(\mathrm{O}_{2}\) in the blood is carried out by hemoglobin. Carbon monoxide (CO) can interfere with \(\mathrm{O}_{2}\) transport because hemoglobin has a stronger affinity for CO than for \(\mathrm{O}_{2}\) If \(\mathrm{CO}\) is present, normal uptake of \(\mathrm{O}_{2}\) is prevented, depriving the body of needed \(\mathrm{O}_{2} .\) Using the molecular orbital model, write the electron configurations for \(\mathrm{CO}\) and for \(\mathrm{O}_{2} .\) From your configurations, give two property differences between \(\mathrm{CO}\) and \(\mathrm{O}_{2}\)

Short Answer

Expert verified
The electron configurations for CO and O鈧� based on the molecular orbital model are [蟽(1s)虏 蟽*(1s)虏 蟽(2s)虏 蟽*(2s)虏 蟽(2p)虏 蟺(2p)鈦碷 for CO and [蟽(1s)虏 蟽*(1s)虏 蟽(2s)虏 蟽*(2s)虏 蟽(2p)虏 蟺(2p)鈦� 蟺*(2p)虏] for O鈧�. Two property differences between CO and O鈧� based on these configurations are: 1. CO has a triple bond (bond order = 3) between its atoms, while O鈧� has a double bond (bond order = 2) between its atoms, making CO's bond stronger. 2. CO is diamagnetic (all electrons paired), while O鈧� is paramagnetic (unpaired electrons in 蟺*(2p) orbitals).

Step by step solution

01

Reminder on atomic orbitals for C and O

To build the molecular orbitals, let's recall the electron configurations of carbon (C) and oxygen (O) in their atomic forms: - Carbon has six electrons: 1s虏, 2s虏, and 2p虏. - Oxygen has eight electrons: 1s虏, 2s虏, and 2p鈦�.
02

Construct molecular orbitals for CO

Use molecular orbital theory to combine the atomic orbitals of carbon and oxygen to form molecular orbitals for CO: - The \(1s\) orbital from both C and O will form a sigma (蟽) bond and their antibonding counterpart (蟽*). - The \(2s\) orbital from both C and O will form another sigma bond (蟽) and their antibonding counterpart (蟽*). - The \(2p\) orbitals will form three bonding orbitals (蟽, 蟺, 蟺) and their antibonding counterparts (蟽*, 蟺*, 蟺*). Now, fill in the electrons in the molecular orbitals, using the Aufbau principle (from the lowest to a higher orbital) and remembering that CO has a total of 14 electrons: - 蟽(1s): 2 e鈦� - 蟽*(1s): 2 e鈦� - 蟽(2s): 2 e鈦� - 蟽*(2s): 2 e鈦� - 蟽(2p): 2 e鈦� - 蟺(2p): 4 e鈦� (2 electrons in each degenerate orbital) The electron configuration for CO is: [蟽(1s)虏 蟽*(1s)虏 蟽(2s)虏 蟽*(2s)虏 蟽(2p)虏 蟺(2p)鈦碷
03

Construct molecular orbitals for O鈧�

Follow the same procedure as for CO, but now for O鈧�, which contains 16 electrons: - 蟽(1s): 2 e鈦� - 蟽*(1s): 2 e鈦� - 蟽(2s): 2 e鈦� - 蟽*(2s): 2 e鈦� - 蟽(2p): 2 e鈦� - 蟺(2p): 4 e鈦� (2 electrons in each degenerate orbital) - 蟺*(2p): 2 e鈦� (1 electron in each degenerate orbital, as they have the same energy) The electron configuration for O鈧� is: [蟽(1s)虏 蟽*(1s)虏 蟽(2s)虏 蟽*(2s)虏 蟽(2p)虏 蟺(2p)鈦� 蟺*(2p)虏]
04

Identify two property differences based on electron configurations

1. Bond order: CO has a bond order of 3, which can be calculated as (number of bonding electrons - number of antibonding electrons) / 2 = (10 - 4) / 2 = 3. This means that CO has a triple bond between its atoms. On the other hand, O鈧� has a bond order of 2 (calculated as (10 - 6)/2), which implies that O鈧� has a double bond between its atoms. Therefore, CO has a stronger bond between its Carbon and Oxygen atoms than O鈧� has between its two Oxygen atoms. 2. Paramagnetism: Electrons in the 蟺*(2p) orbitals in O鈧� are unpaired, making O鈧� paramagnetic. In contrast, CO has all its electrons paired, so it is diamagnetic. This difference in magnetism is due to their distinct electron configurations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Electron configuration provides the arrangement of electrons within an atom or a molecule. It maps out which molecular orbitals these electrons occupy. For molecules like CO and O鈧�, we use the molecular orbital theory to determine how atomic orbitals from individual atoms combine to form molecular orbitals.

These molecular orbitals are filled with electrons according to the Aufbau principle鈥攁ll starting from the lowest energy level. For CO, with a total of 14 electrons, this results in an electron configuration of \[\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p)^2 \pi(2p)^4\].
While for O鈧�, which has 16 electrons, the configuration is \[\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p)^2 \pi(2p)^4 \pi^*(2p)^2\].
  • The difference in the electron count affects the properties of these molecules.
  • Understanding their electron configurations helps predict how they will interact with other molecules.

Bond Order
Bond order is an indicator of the strength and stability of a bond between two atoms within a molecule. It is defined as half the difference between the number of electrons in bonding and antibonding molecular orbitals.

For CO, the bond order is 3, calculated using the expression \((10 - 4) / 2 = 3\).
  • This signifies that CO has a triple bond, indicating a very strong bond between the carbon and oxygen atoms.
  • Triple bonds often mean higher bond energy and shorter bond lengths.
  • Such a triplet in CO allows it to have a high binding affinity, hence why CO can strongly attach to hemoglobin.

In contrast, O鈧� has a bond order of 2, \((10 - 6) / 2 = 2\), meaning it possesses a double bond.
  • A double bond suggests moderate bond strength and stability.
  • It helps oxygen remain reactive, fulfilling its role in biological processes like cellular respiration.
Paramagnetism
Paramagnetism in molecules is a result of unpaired electrons in their molecular orbitals. A molecule will be paramagnetic if even a single electron is unpaired.

O鈧� has unpaired electrons in the \(\pi^*(2p)\) orbitals, reflecting its paramagnetic nature.
  • This means that O鈧� is attracted to magnetic fields.
  • Paramagnetic properties arise because these unpaired electrons generate a magnetic field of their own.

On the other hand, CO is diamagnetic because all its electrons are arranged in pairs; no unpaired electrons exist.
  • Diamagnetic substances are repelled by magnetic fields.
  • CO's diamagnetism results in a lack of magnetic attraction, attributed to its full pairing of the molecular orbital electrons.
  • The absence of unpaired electrons in CO makes it more stable in terms of magnetic interaction.

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Most popular questions from this chapter

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. $$ \begin{array}{l}{\text { a. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{SO}_{3}}\end{array} $$ $$ \text {c} \mathrm{s}_{2} \mathrm{O}_{3}^{2-}\left[\begin{array}{c}{\mathrm{o}} \\\ {\mathrm{s}-\mathrm{s}-\mathrm{o}} \\ {\mathrm{o}} \\\ {\mathrm{o}}\end{array}\right]^{2-} $$ e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}+\)

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) can be produced from the reaction of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion \(\left(\mathrm{C}_{2}^{2-}\right)\)

Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule. a. \(\mathrm{BH}_{3}\) a. \(\mathrm{BH}_{3}\) b. \(\mathrm{N}_{2} \mathrm{F}_{2}\) c. \(\mathrm{C}_{4} \mathrm{H}_{6}\)

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) the hybrid orbital model?
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