Problem 71 - Chemistry An Atoms First Approach

Question

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. a. CO b. \(\mathrm{CO}^{+}\) c. \(\mathrm{CO}^{2+}\)

Step-by-Step Solution

Answer

The bond orders for CO, \(\mathrm{CO}^{+}\), and \(\mathrm{CO}^{2+}\) are 1, 1.5, and 2, respectively. CO and \(\mathrm{CO}^{+}\) are paramagnetic, while \(\mathrm{CO}^{2+}\) is diamagnetic. In terms of bond length, the order is CO > \(\mathrm{CO}^{+}\) > \(\mathrm{CO}^{2+}\), and for bond energy, the order is CO < \(\mathrm{CO}^{+}\) < \(\mathrm{CO}^{2+}\).

Step 1: Understand the Molecular Orbital Model

The molecular orbital model is a model that accounts for the formation of molecular bonds and provides an explanation for their electronic structure. The model considers the interaction between atomic orbitals to create molecular orbitals, which represent the probability distribution and energy levels of electrons in a molecule. We must first determine the electron configuration of the species that are given and then calculate the bond orders to determine the bond length and energy.

Step 2: Determine Electron Configurations of the Diatomic Species

Given the diatomic species listed, their electron configurations are as follows: a. CO: Carbon has 6 electrons, and Oxygen has 8 electrons, so CO has a total of 14 electrons. b. The \(\mathrm{CO}^{+}\) ion: CO has 14 electrons, and the positive charge means one electron has been removed. Therefore, the \(\mathrm{CO}^{+}\) ion has 13 electrons. c. The \(\mathrm{CO}^{2+}\) ion: CO has 14 electrons, and the double positive charge means two electrons have been removed. Therefore, the \(\mathrm{CO}^{2+}\) ion has 12 electrons.

Step 3: Calculate Bond Orders of the Diatomic Species

To calculate the bond order of a molecule, use the formula: \(Bond\,Order = \frac{1}{2} (Number\,of\,Electrons\,in\,Bonding\,Orbitals - Number\,of\,Electrons\,in\,Antibonding\,Orbitals)\) For CO, which has 14 electrons: Using the MO model, we can fill up the electrons in molecular orbitals: \(1\sigma^2, 1\sigma^*^2,2\sigma^2,2\sigma^*^2,3\sigma^2, 1\pi^4,1\pi^*^2\) Bond Order= \( \frac{1}{2} (8-6)=1\) For \(\mathrm{CO}^{+}\), which has 13 electrons: Fill up the electrons: \(1\sigma^2, 1\sigma^*^2,2\sigma^2,2\sigma^*^2,3\sigma^2, 1\pi^4,1\pi^*^1\) Bond Order= \( \frac{1}{2} (8-5)=1.5\) For \(\mathrm{CO}^{2+}\), which has 12 electrons: Fill up the electrons: \(1\sigma^2, 1\sigma^*^2,2\sigma^2,2\sigma^*^2,3\sigma^2, 1\pi^4,1\pi^*^0\) Bond Order= \( \frac{1}{2} (8-4)=2\)

Step 4: Determine Paramagnetism and Order

A species is paramagnetic if it has unpaired electrons, meaning they have unpaired spins that create a magnetic moment. From electron configurations: - CO: Even though it has paired electrons, CO is still considered paramagnetic. - \(\mathrm{CO}^{+}\): One unpaired electron makes this species paramagnetic. - \(\mathrm{CO}^{2+}\): All paired electrons make this species diamagnetic. Now, order them by increasing bond length and bond energy: The bond lengths are inversely proportional to bond orders. Thus, we have: CO (bond order = 1) > \(\mathrm{CO}^{+}\) (bond order = 1.5) > \(\mathrm{CO}^{2+}\) (bond order = 2) The bond energy is directly proportional to bond orders. Therefore, we have: CO (bond order = 1) < \(\mathrm{CO}^{+}\) (bond order = 1.5) < \(\mathrm{CO}^{2+}\) (bond order = 2)