How many equivalent resonance structures can be drawn for the ion BrO_3^-?

• A. 3
• B. 2
• C. 1
• D. 4

Answer: (A)

Verified step by step guidance

1. Step 1: Draw the Lewis structure of the BrO_3^- ion. Start by counting the total number of valence electrons: bromine (Br) has 7, each oxygen (O) has 6, and add 1 extra electron for the negative charge. So, total electrons = 7 + 3×6 + 1 = 26 electrons.
2. Step 2: Connect the bromine atom to the three oxygen atoms with single bonds. This uses 6 electrons (3 bonds × 2 electrons each). Then, distribute the remaining electrons to complete the octets of the oxygen atoms first.
3. Step 3: After placing lone pairs on oxygens, check if bromine has an octet. Since bromine is in period 4, it can have an expanded octet. To satisfy the octet rule and minimize formal charges, form double bonds between bromine and some oxygen atoms by converting lone pairs on oxygen into bonding pairs.
4. Step 4: Identify all possible resonance structures by moving the double bonds among the three oxygen atoms while keeping the overall charge and connectivity the same. Each resonance structure differs only in the placement of the double bonds and lone pairs.
5. Step 5: Count the number of equivalent resonance structures. Because the three oxygen atoms are equivalent, the double bond can be placed with any one of the three oxygens, resulting in three equivalent resonance structures.