Problem 110 - Chemistry The Science in Context

Question

For which of the following diatomic molecules does the bond order increase with the loss of two electrons, forming the corresponding cation with a \(2+\) charge? a. \(\quad B_{2} \rightarrow B_{2}^{2+}+2 e^{-}\) b. \(C_{2} \rightarrow C_{2}^{2+}+2 e^{-}\) c. \(\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{2+}+2 \mathrm{e}^{-}\) d. \(\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{2+}+2 \mathrm{e}^{-}\)

Step-by-Step Solution

Answer

a. B₂ b. C₂ c. N₂ d. O₂ Answer: a. B₂

Step 1: Write the molecular electron configuration of the given molecules

First, we need to find the electron configuration for each molecule, using Molecular Orbital Theory. The atomic numbers for \(B\), \(C\), \(N\), and \(O\) are 5, 6, 7, and 8, respectively. Therefore, the electron configurations for these atoms are: Boron: \(1s^{2}2s^{2}2p^{1}\) Carbon: \(1s^{2}2s^{2}2p^{2}\) Nitrogen: \(1s^{2}2s^{2}2p^{3}\) Oxygen: \(1s^{2}2s^{2}2p^{4}\) In order to construct the molecular orbitals, we must combine the atomic orbitals for each diatomic molecule.

Step 2: Construct molecular orbitals and calculate bond order

Construct the molecular orbitals by combining the valence orbitals of two atoms. Keep in mind that the order of the molecular orbital energy levels changes after the nitrogen atom. For \(B\), \(C\), and \(N\) (\(1\sigma_g^2, 1\sigma_u^2, 2\sigma_g^2, 2\sigma_u^2, 3\sigma_g^2\)): - \(B_2 = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^4\) - \(C_2 = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^4\) - \(N_2 = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^6\) For \(O\), \(\sigma_g\) and \(\pi_u\) energies are swapped: - \(O_2 = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 \pi_u^4 3\sigma_g^2 3\sigma_u^2\) Then, calculate the bond orders for each molecule using the formula: Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2.

Step 3: Calculate bond orders for the cations

Remove two electrons from each molecule to form their corresponding cations. Calculate the bond order for each cation using the same formula: - \(B_2^{2+} = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^2\) - \(C_2^{2+} = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^2\) - \(N_2^{2+} = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 \pi_u^4\) - \(O_2^{2+} = 1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 \pi_u^2 3\sigma_g^2 3\sigma_u^2\)

Step 4: Compare bond orders of initial molecules and their cations

Calculate and compare the bond orders of the initial molecules and their cations to see if the bond order increases upon the loss of two electrons for each case: a. \(B_2: 1.5 \rightarrow B_2^{2+}: 2\) (Bond order increases) b. \(C_2: 2 \rightarrow C_2^{2+}: 2\) (Bond order stays the same) c. \(N_2: 3 \rightarrow N_2^{2+}: 2.5\) (Bond order decreases) d. \(O_2: 2 \rightarrow O_2^{2+}: 2.5\) (Bond order increases)

Step 5: Final answer

So, among given options - the bond order increases upon the loss of two electrons for \(B_2\). The answer is (a).

Key Concepts

Bond Order

Bond order is a concept derived from Molecular Orbital Theory that helps in understanding the strength of the chemical bond between two atoms in a molecule. It's calculated by taking the difference between the number of electrons in bonding orbitals and antibonding orbitals, divided by two. This is mathematically represented as:

• Bond order = \( \frac{\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals}}{2} \)

A higher bond order generally indicates a stronger, shorter bond. For example, in diatomic nitrogen (\(N_2\)), the bond order is 3, which corresponds to a very strong triple bond. It's essential to understand that changes in bond order, such as through the formation of cations, can significantly influence molecular properties. Therefore, monitoring how an electron loss or gain affects bond order is crucial in predicting molecular stability and reactivity.

Electron Configuration

Electron configuration in the context of Molecular Orbital Theory is vital in drawing conclusions about molecular stability and chemical properties. It describes the distribution of electrons in an atom or molecule across various orbitals. For diatomic molecules like \(B_2\), \(C_2\), \(N_2\), and \(O_2\), electrons are allocated to molecular orbitals derived from the constituent atomic orbitals.

• Each pair of electrons in molecular orbitals determines bonding between atoms.
• The order of energy levels for these molecular orbitals can vary, particularly from atom to atom.

For instance, up to the nitrogen molecule, the order typically is \(1\sigma_g, 1\sigma_u, 2\sigma_g, 2\sigma_u, 3\sigma_g\). In contrast, from \(O_2\) onwards, the energy arrangement switches to accommodate the increased number of electrons. Understanding how these electron configurations rearrange, especially when forming cations by the loss of electrons, reveals changes in molecular properties such as bond order and stability.

Cation Formation

Cation formation occurs when a molecule loses electrons, resulting in a positively charged species. This process is pivotal in molecular transformations and changes molecular properties significantly. When a diatomic molecule loses two electrons, it forms a \(2+\) cation, which can be notably different from its neutral state. Considering the example of \(B_2\) turning into \(B_2^{2+}\), the loss of two electrons leads to a rearrangement in the molecular orbital occupation. As a result, the bond order increases from 1.5 to 2.

• The molecule becomes more stable and the bond tighter.
• This direct relationship between electron loss and bond order increasing is not universal, as seen in \(C_2\) or \(N_2\) where such losses may not improve the bond order noticeably.

Thus, evaluating the effect of cation formation helps in predicting the nature and strength of the resulting molecular bond, underpinning much of the chemistry involving ionic species. It's an enlightening demonstration of how molecular orbital adjustments under changing charge states affect stability.