One resonance structure of the bromate ion is shown below.  What is the bond order of the bromine–oxygen bonds in BrO3—?

1. Break Down the Problem

To find the bond order of the bromine–oxygen bonds in the bromate ion $\text{BrO}_3^-$, we need to:

1. Identify the number of bonds between bromine and oxygen in the resonance structures.
2. Calculate the bond order based on the resonance contributions.

2. Relevant Concepts

The bond order can be calculated using the formula: $\text{Bond Order} = \frac{\text{Number of bonding electrons}}{\text{Number of bonds}}$

In resonance structures, the total number of bonds and the contribution of each bond must be taken into account.

3. Analysis and Detail

1. Resonance Structures of $\text{BrO}_3^-$: The bromate ion has three resonance structures with bromine connected to three oxygen atoms. In these structures:

   • There are one single bond and two double bonds in each resonance form.
   • In total, considering all resonance forms, each oxygen forms a bond with bromine.

2. Calculating Bond Order: Each resonance structure contributes differently to the overall bond character. Summarizing:

   • 2 bonds (double) + 1 bond (single) = 5 bonding interactions total over three resonance structures.
   • Since each resonance structure is equivalent, we consider the average bond character.

   Each oxygen thus has a bond order calculation: $\text{Bond Order} = \frac{\text{Number of total bonds}}{\text{Number of equivalent resonance structures}} = \frac{4 \text{ (total bonds)}}{3 \text{ (resonance forms)}}$ $= \frac{4}{3} \approx 1.33$

4. Verify and Summarize

The bond order calculations confirm that there are approximately $1.33$ bonds on average per Br–O bond due to resonance.

Final Answer

The bond order of the bromine–oxygen bonds in $\text{BrO}_3^-$ is approximately $1.33$.