Problem 26 - Chemistry in 30 days

Question

Which one of the following pairs of species have the same bond order? (a) \(\mathrm{CN}^{-}\)and \(\mathrm{NO}^{+}\) (b) \(\mathrm{CN}\) and \(\mathrm{CN}^{+}\) (c) \(\mathrm{O}_{2}^{-}\)and \(\mathrm{CN}^{-}\) (d) \(\mathrm{NO}^{+}\)and \(\mathrm{CN}^{+}\)

Step-by-Step Solution

Answer

Pairs (a) \(\mathrm{CN}^{-}\) and \(\mathrm{NO}^{+}\); (b) \(\mathrm{CN}\) and \(\mathrm{CN}^{+}\) have the same bond order.

Step 1: Understand Bond Order

Bond order is a concept in molecular orbital theory, defined as half the difference between the number of bonding and anti-bonding electrons in a molecule. It provides insight into the stability and strength of a bond: \( \text{Bond Order} = \frac{1}{2} \times (\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}) \). Higher bond orders typically indicate stronger bonds.

Step 2: Calculate Bond Order for \(\mathrm{CN}^{-}\)

The electronic configuration of \(\mathrm{CN}^{-}\) is like \(\left[ 1s^2, 2s^2, 2p_\sigma^2, 2p_\pi^4, 2p_\sigma^2 \right]\). The bond order is \(\frac{1}{2}(8 - 2) = 3\).

Step 3: Calculate Bond Order for \(\mathrm{NO}^{+}\)

\(\mathrm{NO}^{+}\) has the electronic configuration of \(\left[ 1s^2, 2s^2, 2p_\sigma^2, 2p_\pi^4, 2p_\sigma^1 \right]\), similar to \( \mathrm{N}_2 \). The bond order is \(\frac{1}{2}(10 - 4) = 3\).

Step 4: Calculate Bond Order for \(\mathrm{CN}\)

For \(\mathrm{CN}\), the electronic configuration can be considered as the neutral version of \(\mathrm{CN}^{-}\), so it has \(13\) electrons. Bond order for \(\mathrm{CN}\) is \(\frac{1}{2}(7 - 2) = 2.5\).

Step 5: Calculate Bond Order for \(\mathrm{CN}^{+}\)

The \(\mathrm{CN}^{+}\) ion has one less electron than \(\mathrm{CN}\), resembling \(\mathrm{CO}\). Electronic configuration: \(\left[1s^2, 2s^2, 2p_\sigma^2, 2p_\pi^4, 2p_\sigma^1\right]\). The bond order is \(\frac{1}{2}(8 - 3) = 2.5\).

Step 6: Calculate Bond Order for \(\mathrm{O}_{2}^{-}\)

For \(\mathrm{O}_{2}^{-}\), the configuration includes one extra electron compared to \(\mathrm{O}_2\). Its bond order is \(\frac{1}{2}(9 - 5) = 2\).

Step 7: Compare Bond Orders

Compare the bond orders of all the calculated species. \(\mathrm{CN}^{-}\) and \(\mathrm{NO}^{+}\) both have a bond order of \(3\). \(\mathrm{CN}\) and \(\mathrm{CN}^{+}\) both have a bond order of \(2.5\).

Key Concepts

Molecular Orbital Theory

Molecular Orbital Theory provides a framework for understanding how electrons are distributed in molecules. In contrast to the localized view of chemical bonding in Lewis structures, molecular orbital theory considers electrons as delocalized over the entire molecule.
Molecular orbitals are formed by the linear combination of atomic orbitals from each atom contributing to the molecule. These combine to form bonding and antibonding molecular orbitals.

• Bonding Orbitals: Lower in energy, enhance electron density between nuclei, leading to attraction and bond formation.
• Antibonding Orbitals: Higher in energy, reduce electron density between nuclei, leading to repulsion and potentially bond destabilization.

Understanding this theory allows us to calculate bond order, a key factor in considering a molecule's stability and bond strength.

Stability of Molecular Species

The stability of a molecular species is often linked to its bond order. This number is derived from the Molecular Orbital Theory and is calculated using the formula: \[\text{Bond Order} = \frac{1}{2} (\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})\]A higher bond order generally correlates with a stronger and more stable bond.
For example, if a molecule like \( \mathrm{CN}^{-} \) has a bond order of 3, it indicates a very stable bond. The higher the bond order, the greater the overlap of orbitals, thus strengthening the bond. Conversely, a bond order of 0 suggests that no stable bond exists between atoms.
Molecular species with different bond orders can therefore be compared in terms of their stability, helping predict the likelihood of their persistence in a chemical environment.

Electronic Configuration

Electronic configuration is the arrangement of electrons in the molecular orbitals of a species.
Understanding electronic configurations is crucial to determining the bond order of a molecule.

• Configuration Examples: Both \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \) have a similar electronic structure, which leads to the same bond order, a hint of similar bond strength and stability.
• Electron Count: The electronic configuration changes when electrons are added or removed, such as moving from \( \mathrm{CN} \) to \( \mathrm{CN}^{+} \), altering the bond order and the stability of the molecule.

By determining the number of electrons in bonding and antibonding orbitals, the molecule's bond order can be calculated accurately.

Comparison of Bond Strengths

Bond strengths can be assessed through bond orders obtained from electronic configurations. In general, a higher bond order suggests stronger bonds.
As seen in the given examples:

• \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \) show a bond order of 3, indicating very strong bonds, comparable to nitrogen gas \( \mathrm{N}_2 \), known for its high stability.
• \( \mathrm{CN} \) and \( \mathrm{CN}^{+} \) both display a bond order of 2.5, indicating moderately strong bonds, but slightly less than \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \).
• \( \mathrm{O}_{2}^{-} \) with a bond order of 2, displays weaker bond characteristics compared to the aforementioned species, reflecting lesser stability.

The understanding of bond order, therefore, provides an intuitive approach to assess and compare molecular strength and predict chemical behavior.