Draw the Lewis structure for the bromate ion (BrO3-) with minimized formal charges. How many TOTAL likely resonance structures exist for BrO3-? A) 1 B) 2 C) 3 D) 4 E) 5
The Correct Answer and Explanation is:
The correct answer is: C) 3
Explanation:
The bromate ion (BrO₃⁻) consists of a central bromine atom bonded to three oxygen atoms and carries an overall negative charge. To understand the number of likely resonance structures, we first consider its Lewis structure with minimized formal charges.
Step-by-step Lewis Structure:
- Count valence electrons:
- Br (Group 17): 7 valence electrons
- O (Group 16): 6 valence electrons × 3 = 18
- Extra electron (due to −1 charge): 1
- Total = 7 + 18 + 1 = 26 valence electrons
- Skeleton structure: Place Br in the center bonded to 3 oxygen atoms.
- Assign bonds and lone pairs:
- Form single bonds from Br to each O → uses 6 electrons.
- Distribute remaining 20 electrons to complete octets on O atoms.
- This would leave Br with only 6 bonding electrons, which is insufficient because Br can expand its octet (period 4 element).
- Create double bonds:
- To minimize formal charges, convert one of the Br–O single bonds into a double bond. This makes one oxygen neutral (0 formal charge) and the others carry -1 formal charges. Br now has a formal charge of 0.
- Resulting resonance structures:
- The double bond between Br and O can be placed in three different locations (with any one of the three O atoms).
- Each of these forms a distinct resonance structure because the location of the double bond shifts while the atoms and connectivity remain the same.
Conclusion:
There are three equivalent resonance structures for BrO₃⁻ with minimized formal charges, each with one double bond and two single bonds, shifting among the three oxygen atoms. Hence, the total number of likely resonance structures is 3.
Correct choice: C) 3.
