Draw the Lewis structure for the sulfite ion, SO3 2−. Which of the statements below is true for the Lewis structure of the sulfite ion?
The Correct Answer and Explanation is:
The Lewis structure for the sulfite ion (SO₃²⁻) can be drawn by following these steps:
- Determine the total number of valence electrons:
- Sulfur (S) is in Group 16, so it has 6 valence electrons.
- Oxygen (O) is also in Group 16, so each oxygen atom has 6 valence electrons.
- The sulfite ion has 3 oxygen atoms and a 2− charge, meaning it has 2 extra electrons. Total valence electrons = 6 (from sulfur) + 3 × 6 (from oxygen) + 2 (from the charge) = 24 valence electrons.
- Place the sulfur atom in the center:
Since sulfur is less electronegative than oxygen, it will be the central atom. - Draw bonds between the sulfur and the oxygen atoms:
Connect sulfur to each oxygen with a single bond. Each single bond accounts for 2 electrons, so using 3 bonds will use 6 electrons (3 bonds × 2 electrons). - Distribute the remaining electrons:
After placing the bonds, 18 electrons remain (24 – 6 used in bonds). These will be placed as lone pairs around the oxygen atoms. Since each oxygen needs 8 electrons to fulfill the octet rule, we place 6 electrons (3 lone pairs) on each oxygen atom. - Check for formal charges and octet rule:
The oxygen atoms now have 8 electrons each, but sulfur has only 6 electrons around it. To complete the octet for sulfur and minimize formal charges, one of the oxygen atoms will form a double bond with sulfur, using up two of its lone pairs and sharing two electrons with sulfur. This results in the following structure:
- Two oxygen atoms each have a single bond with sulfur and 3 lone pairs, carrying a formal charge of -1 each.
- One oxygen atom has a double bond with sulfur and 2 lone pairs, carrying no formal charge.
- Final structure:
The sulfite ion is a resonance structure with the negative charges delocalized across the two single-bonded oxygens. The formal charge on sulfur is 0, and the formal charges on the oxygen atoms are -1 (for the single-bonded oxygens).
True Statement:
The sulfite ion has resonance structures.
Explanation:
The sulfite ion, SO₃²⁻, has a resonance structure because the negative charges are delocalized between the three oxygen atoms. This means that, while one oxygen atom has a double bond with sulfur in one resonance form, another oxygen atom will form the double bond in a different resonance form, and so on. The negative charges on the oxygen atoms are spread over the three oxygens, stabilizing the ion.
In this case, the sulfite ion is an example of resonance where the actual structure is a hybrid of multiple possible configurations, rather than a single fixed arrangement. This delocalization of electrons helps distribute the negative charge and stabilize the ion.