Question

In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed? $$ \begin{array}{ll}{\text { a. } H_{2}} & {\text { c. } C_{2}^{2-}} \\ {\text { b. } B_{2}} & {\text { d. OF }}\end{array} $$

Short answer

c. \(C_{2}^{2-}\)

Step-by-step solution

Determine the molecular orbital electron configurations

For each of the diatomic molecules, we must determine the molecular orbital (MO) electron configuration, considering how it would change as an electron is removed. Start by identifying the valence electrons for each molecule.

Calculate bond orders for the original configurations

With the molecular orbital electron configurations, we can calculate the bond order for each molecule using the formula \(Bond Order = \frac{1}{2} (e_{b} - e_{a})\), where \(e_{b}\) and \(e_{a}\) are the number of bonding and antibonding electrons, respectively.

Recalculate bond orders after removing an electron

For each molecule, remove one electron from the molecular orbital with the highest energy and recalculate the bond order to determine how the bond strength is affected.

Compare the bond orders

Compare the bond orders before and after removing an electron to determine which molecule's bond strength is expected to weaken.

Choose the correct option

Based on the bond order analysis after removing an electron, identify which molecule fits the description of having a weakened bond strength. Let's go through each step for the given diatomic molecules: Step 1: MO electron configurations (original) - a. \(H_{2}\): 1σ² - b. \(B_{2}\): 1σ² 2σ² 1π² 2π² - c. \(C_{2}^{2-}\): 1σ² 2σ² 3σ² 1π^4 - d. \(OF\): 1σ² 2σ² 3σ² 1π^4 2π^2 Step 2: Bond orders (original configurations) - a. \(H_{2}\): BO = \(\frac{1}{2}(2-0)=1\) - b. \(B_{2}\): BO = \(\frac{1}{2}(6-4)=1\) - c. \(C_{2}^{2-}\): BO = \(\frac{1}{2}(8-4)=2\) - d. \(OF\): BO = \(\frac{1}{2}(10-6)=2\) Step 3: MO electron configurations after removing 1 electron and bond orders - a. \(H_{2}\): 1σ¹; BO = \(\frac{1}{2}(1-0)=0.5\) - b. \(B_{2}\): 1σ² 2σ² 1π² 2π¹; BO = \(\frac{1}{2}(6-5)=0.5\) - c. \(C_{2}^{2-}\): 1σ² 2σ² 3σ² 1π^3; BO = \(\frac{1}{2}(8-5)=1.5\) - d. \(OF\): 1σ² 2σ² 3σ² 1π^4 2π^1; BO = \(\frac{1}{2}(9-7)=1\) Step 4: Compare bond orders - a. \(H_{2}\): Original BO = 1; After removing 1 electron, BO = 0.5 - b. \(B_{2}\): Original BO = 1; After removing 1 electron, BO = 0.5 - c. \(C_{2}^{2-}\): Original BO = 2; After removing 1 electron, BO = 1.5 - d. \(OF\): Original BO = 2; After removing 1 electron, BO = 1 Step 5: Choose the correct option From the comparisons, it's clear that removing an electron from the \(C_{2}^{2-}\) molecule has the most significant effect on weakening the bond strength (from BO = 2 to BO = 1.5). So, the correct answer is: c. \(C_{2}^{2-}\)

Key concepts

Electron Configuration

The concept of electron configuration is a foundational topic in molecular orbital theory. In diatomic molecules, electron configuration describes the distribution of electrons among the molecular orbitals. Each molecule has a unique electron configuration based on its total number of electrons.
For example, in the hydrogen molecule (\(\text{H}_2\)), the simplest diatomic molecule, the electron configuration is expressed as 1σ², indicating that two electrons occupy the bonding sigma orbital. The electron configuration helps predict the molecule's chemical bonding and properties.

• In Boron (\(\text{B}_2\)), the configuration is 1σ² 2σ² 1π² 2π², showing that electrons fill according to energy levels.
• For more negative ions like C\(_2^{2-}\), extra electrons occupy anti-bonding or higher energy orbitals, which can change the bond properties.
• Oxygen fluoride (\(\text{OF}\)) has a more complex order due to its higher electron count, with both bonding and anti-bonding filled.

This distribution affects the molecule’s stability and can change significantly when electrons are added or removed.

Bond Order

Bond order is a key concept to understand the strength and stability of molecular bonds. It can be calculated using the formula: \(\text{Bond Order} = \frac{1}{2}(e_{b} - e_{a})\), where \(e_b\) is the number of bonding electrons and \(e_a\) is the number of antibonding electrons. The resulting number indicates how many chemical bonds exist between two atoms:
- A higher bond order implies a stronger bond with more shared electrons.
- Conversely, a lower bond order suggests a weaker bond which is less stable.
In our exercise:

• \(\text{H}_2\) originally has a bond order of 1, indicating a single bond.
• Upon removing an electron, this drops to 0.5, implying weak bonding.
• C\(_2^{2-}\) originally shows a bond order of 2, indicating a strong double bond, which decreases slightly to 1.5 with electron removal.

Thus, bond order changes are significant when an electron is added or removed, impacting bond strength and molecular stability.

Diatomic Molecules

Diatomic molecules consist of only two atoms. Often these two atoms are identical, forming simple molecules like hydrogen (\(\text{H}_2\)) or nitrogen (\(\text{N}_2\)). They are particularly useful in illustrating molecular orbital theory because their simplicity makes it easier to predict bonding patterns and molecular properties. For other molecules like \(\text{B}_2\) and \(\text{OF}\), despite containing different elements, the diatomic nature still allows us to analyze them using MO theory,

• These molecules have predictable forms, with electrons interacting within molecular orbitals accessible to valence electrons.
• Diatomic molecules often participate in forming gases and demonstrate fundamental chemistry principles.

Understanding diatomic molecules is crucial as they serve as a basis for comprehending more complex multi-atomic molecules.

Bond Strength

Bond strength refers to how strongly two atoms are held together within a molecule. It is directly tied to bond order—the greater the bond order, the stronger the bond. As calculated in our example, the bond strength of \(\text{H}_2\) is reduced when an electron is removed: its bond order falls from 1 to 0.5. This indicates less energy is required to break this bond now.
In the example of \(\text{C}_2^{2-}\), the bond order drops from 2 to 1.5 when an electron is removed, meaning the bond weakens, though it is still relatively strong with a bond order supporting a bond greater than a single bond.

• Higher bond orders indicate a stronger, shorter bond, with more shared electrons.
• Weaker bonds with lower bond orders are longer and have fewer electrons shared.

Understanding bond strength through bond order allows chemists to predict reactivity and stability of specific molecular arrangements.