Problem 68 - Chemistry An Atoms First Approach

Question

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}^{+}, \mathrm{F}_{2},\) and \(\mathrm{F}_{2}^{-},\) predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

Step-by-Step Solution

Answer

The bond orders for F2, F2+, and F2- are 3, 2.5, and 2.5, respectively. F2 has no unpaired electrons, while F2+ and F2- each have one unpaired electron. The bond lengths for F2+ and F2- are equal and longer than the bond length in F2.

Step 1: Recall the Molecular Orbital Diagram for F2

First, we need to recall the molecular orbital (MO) diagram for F2 or any molecule with similar electronic configuration. In the case of F2, the atomic orbitals from each F atom mix to form molecular orbitals: the 1s orbitals mix to form 1σ and 1σ* (sigma and sigma-star) orbitals, while the 2s orbitals mix to form 2σ and 2σ*. Additionally, the 2p orbitals mix to form two pi orbitals (π2p since two of the 2p orbitals are perpendicular to the internuclear axis), and the remaining 2p orbital mixes with its partner from the other atom to form two more sigma orbitals (σ2p and σ2p*). In F2, there are a total of 18 electrons to be placed into the MO diagram derived from this information.

Step 2: Fill the Orbitals for Each Species

Start filling the molecular orbitals for each species (F2+, F2, and F2-) in increasing order of their energy levels, following Hund's rule and the Pauli exclusion principle. F2: 2 electrons in 1σ, 2 electrons in 1σ*, 2 electrons in 2σ, 4 electrons in π2p, 2 electrons in σ2p, and 2 electrons in π2p*, total of 14 valence electrons. F2+: Remove one electron from F2: 2 electrons in 1σ, 2 electrons in 1σ*, 2 electrons in 2σ, 4 electrons in π2p, 2 electrons in σ2p, and 1 electron in π2p*, total of 13 valence electrons. F2-: Add one electron to F2: 2 electrons in 1σ, 2 electrons in 1σ*, 2 electrons in 2σ, 4 electrons in π2p, 2 electrons in σ2p, and 3 electrons in π2p*, total of 15 valence electrons.

Step 3: Calculate Bond Orders, Unpaired Electrons, and Relative Bond Lengths

Bond Order = (Number of electrons in bonding MOs - Number of electrons in antibonding MOs) / 2 F2: Bond Order = (10 - 4) / 2 = 3 Unpaired Electrons = 0 Bond Length: Take it as a reference, since we will compare other species relative to it. F2+: Bond Order = (9 - 4) / 2 = 2.5 Unpaired Electrons = 1 Relative Bond Length: Since bond order is inversely proportional to bond length, F2+ has a longer bond length than F2. F2-: Bond Order = (10 - 5) / 2 = 2.5 Unpaired Electrons = 1 Relative Bond Length: F2- has the same bond order as F2+, so their bond lengths are also equal and longer than F2. In summary, the bond orders are 3 (F2), 2.5 (F2+), and 2.5 (F2-); there are no unpaired electrons in F2, and one unpaired electron in both F2+ and F2-. The relative bond lengths are the same for F2+ and F2-, and both are longer than the bond length in F2.

Key Concepts

Bond Order

Bond order is a crucial concept in Molecular Orbital Theory that helps in determining the strength and stability of a bond between two atoms in a molecule. It is calculated using the formula: \[\text{Bond Order} = \frac{\text{Number of electrons in bonding MOs} - \text{Number of electrons in antibonding MOs}}{2}\]The bond order can be thought of as the net number of chemical bonds present between a pair of atoms. A higher bond order typically indicates a stronger bond with a shorter bond length. In the case of \[\mathrm{F}_{2}^+\, \mathrm{F}_2, \text{ and } \mathrm{F}_{2}^-\] we have:

• For \(\mathrm{F}_2\): Bond Order = 3
• For \(\mathrm{F}_{2}^+\): Bond Order = 2.5
• For \(\mathrm{F}_{2}^-\): Bond Order = 2.5

Bond order helps in predicting other properties like bond length and magnetic behavior, which are useful for understanding the molecular structure.

Unpaired Electrons

The presence of unpaired electrons in a molecule is a key factor in determining its magnetic properties. Molecules with unpaired electrons are paramagnetic and tend to be attracted to an external magnetic field. In contrast, molecules without unpaired electrons are diamagnetic and are not attracted, or are even slightly repelled by a magnetic field. For \(\mathrm{F}_{2}\, \mathrm{F}_{2}^{+}, \text{ and } \mathrm{F}_{2}^{-}\), the number of unpaired electrons is calculated from their Molecular Orbital diagrams:

• \(\mathrm{F}_{2}\): 0 unpaired electrons
• \(\mathrm{F}_{2}^+\): 1 unpaired electron
• \(\mathrm{F}_{2}^-\): 1 unpaired electron

This means that both \(\mathrm{F}_{2}^+\) and \(\mathrm{F}_{2}^-\) exhibit some degree of magnetic behavior due to the presence of an unpaired electron, while \(\mathrm{F}_{2}\) does not.

Bond Length

Bond length is an important factor in molecular structure, representing the distance between the nuclei of two bonded atoms. The bond order is inversely related to bond length: as bond order increases, bond length generally decreases. This is because a higher bond order signifies more shared electron density between atoms, pulling them closer together. For the species in our example:

• \(\mathrm{F}_2\) has a bond order of 3, indicating the shortest bond length among the three molecules.
• Both \(\mathrm{F}_{2}^+\) and \(\mathrm{F}_{2}^-\) have a bond order of 2.5, so they have longer bond lengths compared to \(\mathrm{F}_2\).

Thus, understanding bond order allows you to infer the relative bond lengths, providing insight into the spatial arrangement of atoms within a molecule.