Draw a Lewis structure for SO2 in which all atoms have a formal charge of zero. Do not consider ringed structures.
The Correct Answer and Explanation is :
To draw the Lewis structure of sulfur dioxide (SO₂) where all atoms have a formal charge of zero, follow these steps:
Step 1: Count the Valence Electrons
- Sulfur (S): Sulfur is in group 16, so it has 6 valence electrons.
- Oxygen (O): Oxygen is also in group 16, so each oxygen atom has 6 valence electrons.
- There are two oxygen atoms, so the total number of valence electrons is: 6(from S)+2×6(from O)=18 valence electrons.6 (\text{from S}) + 2 \times 6 (\text{from O}) = 18 \text{ valence electrons}.
Step 2: Draw a Skeleton Structure
- Place sulfur in the center and the two oxygen atoms around it. This is because sulfur is less electronegative than oxygen and can form more bonds.
- Connect each oxygen atom to the sulfur atom with a single bond (one pair of electrons between S and each O).
Step 3: Distribute Electrons
- After forming the bonds, each single bond uses 2 electrons. Since there are two bonds, this accounts for 4 electrons, leaving 14 electrons to be distributed.
Step 4: Complete Octets for Oxygen
- Place 6 electrons (as lone pairs) around each oxygen atom to complete their octet. After placing 6 electrons on each oxygen, all 12 of the remaining electrons are used up.
Step 5: Sulfur’s Electron Count
- Sulfur now has 4 electrons from the bonds with the two oxygen atoms and 4 electrons from the lone pairs (if any) directly on sulfur, meaning sulfur has 8 electrons in its valence shell.
Step 6: Minimize Formal Charges
- Formal charge is calculated using the formula: Formal charge=Valence electrons−(Nonbonding electrons+Bonding electrons2)\text{Formal charge} = \text{Valence electrons} – \left(\text{Nonbonding electrons} + \frac{\text{Bonding electrons}}{2}\right) Sulfur, with 6 valence electrons and 4 bonding electrons, and each oxygen with 6 valence electrons and 6 bonding electrons, will each have a formal charge of zero in this structure.
Final Structure
- The sulfur atom is in the center with double bonds to each oxygen atom. There are lone pairs on each oxygen atom.
- The structure is symmetrical, with both oxygens having an identical formal charge of zero.
Explanation (300 words)
In sulfur dioxide (SO₂), sulfur acts as the central atom because it can form more bonds than oxygen, which is a more electronegative atom. The goal in drawing the Lewis structure is to ensure that each atom obeys the octet rule, if possible, and that formal charges are minimized.
To achieve this, the structure starts with sulfur and the two oxygen atoms connected by single bonds. These single bonds account for four electrons. After these bonds are placed, the remaining electrons (14 in total) are distributed as lone pairs on the oxygen atoms. Oxygen, being more electronegative, prefers to complete its octet by having lone pairs.
Sulfur, while having more than 8 electrons, is capable of this due to its position in Period 3 of the periodic table, where it can accommodate more than 8 electrons in its valence shell. This allows sulfur to form double bonds with each oxygen to achieve an octet for both oxygen atoms. The formal charge on all atoms in this structure is zero, which is the ideal outcome in most molecules. This configuration satisfies the octet rule for the oxygen atoms and ensures minimal formal charge distribution across the molecule.
This structure demonstrates that sulfur dioxide has resonance, as the double bonds between sulfur and oxygen can shift, but the overall electron configuration remains the same. In this case, we represent the bonding in a way that minimizes formal charges and respects the octet rule for each oxygen atom while accommodating sulfur’s expanded valence shell.