What is the average bond order for the S–O bond in the sulfite ion, SO_3^{2-}?

• A. 1.00
• B. 2.00
• C. 1.33
• D. 1.50

Answer: (C)

Verified step by step guidance

1. Step 1: Draw the Lewis structure of the sulfite ion, SO_3^{2-}. Count the total number of valence electrons: sulfur has 6, each oxygen has 6, and there are 2 extra electrons due to the 2- charge. So, total electrons = 6 + (3 × 6) + 2 = 26 electrons.
2. Step 2: Arrange the atoms with sulfur in the center bonded to three oxygen atoms. Use the valence electrons to form bonds and complete octets, placing lone pairs on oxygen atoms and sulfur as needed. Remember that sulfur can have an expanded octet.
3. Step 3: Identify resonance structures. In sulfite, one of the S–O bonds is a double bond, and the other two are single bonds, but these double bonds can be placed on any of the three oxygen atoms, creating resonance.
4. Step 4: Calculate the bond order by averaging the bond types over the resonance structures. Since there are three S–O bonds and the double bond is delocalized, the bond order is calculated as: bond order = (number of bonding electron pairs) / (number of bonds). Here, it is (1 double bond × 2 + 2 single bonds × 1) / 3 = 4 / 3 = 1.33.
5. Step 5: Conclude that the average bond order for the S–O bond in the sulfite ion is approximately 1.33, reflecting the resonance stabilization and partial double bond character distributed over the three S–O bonds.