Q.What is the effect of the following processes on the bond order in N2 and O2?

Concept: Bond Order Strength

Bond order = $\frac{1}{2}(\text{bonding electrons} - \text{antibonding electrons})$. Removing an electron from a bonding orbital decreases bond order; removing from an antibonding orbital increases it.

Reasoning

1. N$_2$ (14 e$^-$): MO configuration \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p}^4 \sigma_{2p}^2. Bond order = $\frac{10 - 4}{2} = 3$.
   • N$_2 \to$ N$_2^+$ removes one electron from the bonding $\sigma_{2p}$ orbital.
   • New bond order = $\frac{9 - 4}{2} = 2.5$ (decreases). …

The Concept: Bond Order and Molecular Orbital Diagrams

Bond order tells us how many net bonding pairs hold two atoms together. It's calculated as:

The key insight is that not all electrons are equal. Electrons in bonding molecular orbitals strengthen a bond; electrons in antibonding molecular orbitals weaken it. When you remove an electron (ionisation), the effect on bond order depends entirely on which type of orbital that electron came from.

For diatomic molecules of second-period elements, the molecular orbital filling order differs depending on whether the atoms are lighter (Li₂ to N₂) or heavier (O₂ to Ne₂). This is due to the energy crossover between the σ₂p and π₂p orbitals.

Now let's apply this to each case.

1. N₂ → N₂⁺ + e⁻

Step 1: Write the electron configuration of N₂

Nitrogen has 7 electrons per atom, so N₂ has 14 electrons total. Following the filling order for lighter molecules:

• σ₁s² (2 bonding)
• σ*₁s² (2 antibonding)
• σ₂s² (2 bonding)
• σ*₂s² (2 antibonding)
• π₂pₓ² π₂pᵧ² (4 bonding)
• σ₂p_z² (2 bonding)

That accounts for all 14 electrons. The last filled orbital is σ₂p_z, which is a bonding orbital.

Step 2: Calculate the initial bond order

Bonding electrons = 2 (σ₁s) + 2 (σ₂s) + 4 (π₂p) + 2 (σ₂p_z) = 10

Antibonding electrons = 2 (σ₁s) + 2 (σ₂s) = 4

This matches the known triple bond in N₂.

Step 3: Identify which electron is removed

The highest energy electron in N₂ is in the σ₂p_z bonding orbital. So N₂ → N₂⁺ removes one bonding electron.

Step 4: Calculate the new bond order

Bonding electrons become 9, antibonding remain 4.

2. O₂ → O₂⁺ + e⁻

Step 1: Write the electron configuration of O₂

Oxygen has 8 electrons per atom, so O₂ has 16 electrons total. For heavier molecules (O₂ to Ne₂), the σ₂p orbital fills before the π₂p orbitals:

• σ₁s² (2 bonding)
• σ*₁s² (2 antibonding)
• σ₂s² (2 bonding)
• σ*₂s² (2 antibonding)
• σ₂p_z² (2 bonding)
• π₂pₓ² π₂pᵧ² (4 bonding)
• π₂pₓ¹ π₂pᵧ¹ (2 antibonding) …