Determine the bond order in a molecule or ion with 10 valence electrons.

To determine the bond order of a molecule or ion with 10 valence electrons, we can follow these steps:

1. Break Down the Problem

• Identify how many bonds are present in the molecule or ion.
• Determine the number of bonding electrons and antibonding electrons.

2. Relevant Concepts

• Bond order (BO) can be calculated using the formula: $BO = \frac{N_b - N_a}{2}$ where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

3. Analysis and Detail

• We assume a hypothetical molecule or ion with 10 valence electrons.

• Common examples with 10 electrons include molecules like $N_2$ and $CO$, which have a bond order of 3.

  For example:

  • $N_2$ has 10 valence electrons distributed as follows:
    • Orbital configuration: $(1\sigma)^2 (1\sigma^*)^0 (2\sigma)^2 (2\sigma^*)^0 (2\pi)^4 (2\pi^*)^0$
    • Bonding electrons: 8
    • Antibonding electrons: 2

  Using the bond order formula: $BO = \frac{8 - 2}{2} = \frac{6}{2} = 3$

4. Verify and Summarize

• The calculated bond order is 3 for molecular systems with 10 valence electrons like $N_2$.
• The methodology remains consistent; the bond order indicates the stability of a bond—higher values represent more stable bonds.

Final Answer

The bond order in a molecule or ion with 10 valence electrons is typically 3, indicating a strong bonding interaction.