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 \(\mathrm{CN}^{+}\)

Short answer

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

Step-by-step solution

Determine Electrons in Molecular Orbitals

To find the bond order, we need to count the number of electrons in each species. For \( \mathrm{CN}^{-} \), there are 14 electrons. For \( \mathrm{NO}^{+} \), there are also 14 electrons. For \( \mathrm{CN}^{+} \), there are 12 electrons. For \( \mathrm{O}_2^{-} \), there are 17 electrons. From this, we will calculate the bond order for each species.

Calculate Bond Order

Bond order is calculated using the formula: \( \text{Bond Order} = \frac{1}{2}( \text{Number of bonding electrons - Number of antibonding electrons }) \).- For \( \mathrm{CN}^{-} \): configuration is similar to \( \mathrm{N}_2 \), bond order = 3.- For \( \mathrm{NO}^{+} \): similar to \( \mathrm{N}_2 \), bond order = 3.- For \( \mathrm{CN}^{+} \): with two fewer electrons compared to \( \mathrm{CN}^- \), bond order = 2.5.- For \( \mathrm{O}_2^{-} \): similar to \( \mathrm{O}_2 \) with an extra electron, bond order = 1.5.

Compare Bond Orders

From Step 2, we determine that the bond orders are:- \( \mathrm{CN}^{-} \) = 3- \( \mathrm{NO}^{+} \) = 3- \( \mathrm{CN}^{+} \) = 2.5- \( \mathrm{O}_2^{-} \) = 1.5By comparing these bond orders, the pair with the same bond order is \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \).

Key concepts

Bond Order

Bond order is a concept used to describe the stability of a bond between two atoms in a molecule. It is calculated from molecular orbital theory, which considers both bonding and antibonding orbitals that form when atoms combine to create a molecule.
The formula to calculate bond order is:\[ \text{Bond Order} = \frac{1}{2}( \text{Number of bonding electrons} - \text{Number of antibonding electrons} ) \]Here's why bond order matters:

• A higher bond order indicates a stronger, more stable bond.
• A bond order of 1 suggests a single bond, whereas a bond order higher than 1 often signifies double or triple bonds.
• A bond order of 0 implies no bond exists between the two atoms.

For example, in our exercise, both \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \) have a bond order of 3, suggesting a strong triple bond.

Electrons in Molecular Orbitals

To determine a molecule's bond order, it's essential to count its electrons. These electrons fill molecular orbitals, which are formed when atomic orbitals combine as atoms bond.
These orbitals are categorized as bonding or antibonding:

• Bonding orbitals: Lower energy spots where electrons are stable and contribute to bond formation.
• Antibonding orbitals: Higher energy places that electrons can fill, potentially destabilizing the bond.

Counting the electrons:

• For \( \mathrm{CN}^{-} \) and \( \mathrm{NO}^{+} \): both species have 14 electrons.
• \( \mathrm{CN}^{+} \) has 12 electrons, having lost an electron compared to \( \mathrm{CN}^{-} \).
• \( \mathrm{O}_2^{-} \) has 17 electrons, due to gaining an extra electron compared to \( \mathrm{O}_2 \).

At this stage, one might begin evaluating which molecular orbital each electron occupies, it's vital as it directly affects the bond order.

Bonding and Antibonding Electrons

Molecular orbitals can either stabilize or destabilize a molecule, depending on whether they are bonding or antibonding in nature. The electrons in these orbitals fundamentally define the bond's strength and order.
Bonding Electrons

• These electrons occupy lower-energy orbitals, pulling atoms together.
• They help increase bond order, reinforcing the bond.

Antibonding Electrons

• These are found in higher-energy orbitals, exerting a repulsive force between atoms.
• They tend to decrease bond order, making the bond weaker.

Calculating the difference between these electrons using our formula provides insight into the overall stability. Knowing how to differentiate and count these electrons is key to understanding molecular structure and stability in chemical bonds.