We can draw three inequivalent Lewis structures for the bromate ion, BrO3-. The concepts of formal charge and electronegativity can help us choose the structure that is the best representation. 1. Assign formal charges to the elements in each of the structures below. Note: Count oxygen atoms starting from the left for each structure.
The Correct Answer and Explanation is:
To determine the best Lewis structure for the bromate ion (BrO₃⁻), we analyze three possible resonance structures using formal charges and electronegativity. Here’s how we determine the formal charges for each atom and then identify the best representation.
Step 1: Determine Total Valence Electrons
- Br (Group 17): 7 valence electrons
- O (Group 16): 3 × 6 = 18 valence electrons
- Extra electron for the negative charge: +1
- Total: 7 + 18 + 1 = 26 valence electrons
Possible Structures and Formal Charges
We’ll analyze three inequivalent Lewis structures that differ in the placement of double bonds and lone pairs.
Structure A: One Br=O double bond, two Br–O single bonds
(Starting from the left: O1 = double bonded; O2 & O3 = single bonded)
- O1 (double bonded): 6 valence – (4 nonbonding + 2 bonding) = 0
- O2, O3 (single bonded): 6 – (6 nonbonding + 1 bonding) = –1
- Br: 7 – (0 nonbonding + 6 bonding) = +1
➡️ Formal Charges:
Br = +1
O1 = 0
O2 = –1
O3 = –1
Structure B: Two Br=O double bonds, one Br–O single bond
(Left to right: O1 & O2 = double bonded; O3 = single bonded)
- O1 & O2 (double): 6 – (4 nonbonding + 2 bonding) = 0
- O3 (single): 6 – (6 nonbonding + 1 bonding) = –1
- Br: 7 – (0 nonbonding + 8 bonding) = 0
➡️ Formal Charges:
Br = 0
O1 = 0
O2 = 0
O3 = –1
Structure C: All Br–O single bonds
Each oxygen has a full octet, and Br has an expanded octet.
- All O atoms: 6 – (6 nonbonding + 1 bonding) = –1
- Br: 7 – (0 nonbonding + 6 bonding) = +3
➡️ Formal Charges:
Br = +3
Each O = –1
Best Structure? Structure B
- Structure A has –1 charges on two oxygens and +1 on Br, balancing to –1 overall.
- Structure B minimizes formal charges: only one oxygen has –1, and Br is neutral.
- Structure C has a high positive formal charge (+3) on Br, making it highly unstable.
Because formal charges are minimized in Structure B, and negative charges reside on the more electronegative oxygen atoms rather than bromine, Structure B is the best Lewis structure.
Conclusion
Using formal charge analysis and the concept of electronegativity, Structure B is the most stable and best represents the BrO₃⁻ ion. It maintains the correct number of valence electrons, places the negative charge on the more electronegative oxygen, and keeps formal charges close to zero.