Question

In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed? a. \(\mathrm{H}_{2}\) c. \(\mathrm{C}_{2}^{2-}\) b. \(\mathrm{B}_{2}\) d. OF

Short answer

The bond strength would be expected to weaken as an electron is removed for \(\mathrm{H}_2\) (option a) and possibly for OF (option d), based on the changes in bond order of the molecules.

Step-by-step solution

Molecular Orbital Diagrams

To analyze the bond strength of each diatomic molecule, we must first understand their molecular orbital diagrams. These diagrams show the distribution of electrons in bonding and antibonding orbitals. Let's write out the electron configuration and molecular orbital diagrams for each molecule: a. \(\mathrm{H}_2\): 1s1 1s1 c. \(\mathrm{C}_2^{2-}\): 1s2 1s*2 2s2 2s*2 2p_x2 2p_y2 2p_z2 b. \(\mathrm{B}_2\): 1s2 1s*2 2s2 2s*2 2p_x2 2p_y1 2p_z1 d. OF: The atomic orbitals of O and F interact to form molecular orbitals, but we primarily need to focus on the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) since we are removing an electron.

Analyze Bond Order

Bond order is a measure of the number of bonding electrons minus the number of antibonding electrons, divided by two: Bond order = \(\frac{(number \ of \ bonding \ electrons) - (number \ of \ antibonding \ electrons)}{2}\) a. \(\mathrm{H}_2\): (2 - 0) / 2 = 1 c. \(\mathrm{C}_2^{2-}\): (10 - 6) / 2 = 2 b. \(\mathrm{B}_2\): (7 - 6) / 2 = 0.5 d. In OF, we need to look at whether either of the HOMO or LUMO would be disrupted by losing an electron.

Determine the Effect of Removing an Electron

Now let's see how removing an electron from each molecule affects the bond order: a. \(\mathrm{H}_2\) will become \(\mathrm{H}_2^+\) and the bond order reduces to (1 - 1) / 2 = 0. c. \(\mathrm{C}_2^{2-}\) will become \(\mathrm{C}_2^{-}\) and the bond order remains the same: (10 - 6) / 2 = 2. b. \(\mathrm{B}_2\) will become \(\mathrm{B}_2^+\) and the bond order increases to (7 - 5) / 2 = 1. d. In the case of OF, the HOMO is bonding and the LUMO is antibonding. Removing an electron will depend on whether we are removing an electron from the bonding HOMO or LUMO. If we remove from HOMO, the bond would weaken, but we cannot definitively say without more information.

Conclusion

Comparing the changes in bond order above, we can see that only the removal of an electron from \(\mathrm{H}_2\) and possibly OF (depending on the HOMO or LUMO) would weaken the bond. Therefore, we expect the bond strength to weaken for \(\mathrm{H}_2\) (option a) and possibly OF (option d) upon the removal of an electron.

Key concepts

Molecular Orbital Diagrams

Molecular Orbital (MO) Diagrams are invaluable tools for understanding the bonding in diatomic molecules. These diagrams depict how atomic orbitals combine to form bonding and antibonding molecular orbitals. A bonding orbital stabilizes the molecule because its electron density is found directly between the atoms, effectively gluing them together.

Antibonding orbitals, on the other hand, tend to have electron density found outside the inter-nuclear region, which can destabilize the bond if they are occupied. When comparing the MO diagrams of diatomic molecules, the position of electrons in these orbitals is critical to predicting properties like bond strength.

Let's review some helpful guidelines when drawing or interpreting these diagrams: Electrons fill the lowest energy molecular orbitals first, known as the Aufbau principle, and each orbital can hold a maximum of two electrons with opposite spins, reflecting Pauli's Exclusion Principle. Also, according to Hund's rule, electrons will not pair up in orbitals if not necessary, preferring to remain unpaired as they fill degenerate (energy-equal) orbitals. These principles help us to understand the electron distribution within a molecule and hence its chemical behavior.

Bond Order Calculation

Bond order tells us a lot about the strength and stability of a bond within a molecule. It is determined by calculating the difference between the number of electrons in bonding orbitals and antibonding orbitals, and then dividing by two, as reflected in the formula: \
\(Bond order = \frac{\text{(number of bonding electrons)} - \text{(number of antibonding electrons)}}{2}\).

A higher bond order implies a stronger, more stable bond, whereas a bond order of zero signifies a bond that does not exist or a molecule that is highly unstable. In simplifying this concept, think of bonding electrons as the 'strength' contributors and antibonding electrons as 'weakness' contributors. Thus, the presence of electrons in antibonding orbitals can indicate a point of vulnerability in the molecule's structure, which can be exposed when an electron is removed.

Electron Configuration

The electron configuration of a molecule can provide profound insights into its chemical behavior and properties. Electrons are found in various orbitals around the nucleus of an atom, and understanding their arrangement is vital. For diatomic molecules, this arrangement is depicted using molecular orbital theory which combines the atomic orbitals from each atom to predict the molecular electron configuration.

For example, the electron configuration of \
\(\mathrm{B}_2\), with a bond order of 0.5, reflects the presence of unpaired electrons, which contributes to its magnetic properties and relatively weak bond. In contrast, \
\(\mathrm{C}_2^{2-}\) has a higher bond order, indicating a more robust bond due to the paired-up electrons in bonding orbitals.

When removing an electron, it's critical to know which orbital the electron is coming from. Electrons from antibonding orbitals can increase bond order upon their removal, whereas losing an electron from a bonding orbital can reduce bond order and therefore bond strength. The electron configuration serves as the foundation for predicting how changes will affect the molecule, such as the removal or addition of an electron.