Concept: Lewis Dot Structures

Lewis dot structures represent the valence electrons of atoms within a molecule, showing how these electrons are arranged to satisfy the octet rule (or duet rule for hydrogen). The central atom is usually the least electronegative atom (excluding hydrogen). Formal charge calculation helps in determining the most stable Lewis structure.

Why (D) is correct:

Let's analyze the steps to draw the Lewis structure for SO₂:

1. Count total valence electrons: Sulfur (Group 16) has 6 valence electrons. Oxygen (Group 16) has 6 valence electrons. So, for SO₂, total valence electrons = $6+2\times 6=18$.

2. Determine the central atom: Sulfur is less electronegative than oxygen, so sulfur is the central atom.

3. Draw single bonds: Connect the central sulfur atom to the two oxygen atoms with single bonds. This uses 4 electrons ($2\times 2$). Remaining electrons = $18-4=14$.

4. Distribute remaining electrons as lone pairs to terminal atoms: Each oxygen needs 6 more electrons to complete its octet. So, $2\times 6=12$ electrons are used. Remaining electrons = $14-12=2$.

5. Place remaining electrons on the central atom: The 2 remaining electrons are placed as a lone pair on the sulfur atom.

6. Check octets and formal charges: At this point, sulfur has 6 electrons (2 from each single bond + 2 lone pair), and each oxygen has 8 electrons. Sulfur does not have an octet. To satisfy the octet rule for sulfur, one lone pair from an oxygen atom must be converted into a double bond with sulfur.

7. Form double bonds to satisfy octets (if necessary) and minimize formal charges: If we form one S=O double bond, the structure becomes O=S-O. Let's calculate formal charges:

   • For O (double bond): $6-(4{\text{ lone pair e}}^{-}+\frac{1}{2}\times 4{\text{ bonding e}}^{-})=6-(4+2)=0$

   • For S: $6-(2{\text{ lone pair e}}^{-}+\frac{1}{2}\times 6{\text{ bonding e}}^{-})=6-(2+3)=+1$

   • For O (single bond): $6-(6{\text{ lone pair e}}^{-}+\frac{1}{2}\times 2{\text{ bonding e}}^{-})=6-(6+1)=-1$

   This structure (with one double bond and one single bond) is the most stable resonance structure as it minimizes formal charges compared to structures with only single bonds (where S would have a +2 formal charge and both O's -1). Sulfur can expand its octet, but for SO₂, the structure with one double bond and one single bond (and a lone pair on S) is generally preferred as it adheres to the octet rule for all atoms while minimizing formal charges. Option D correctly depicts this structure, showing one S=O double bond, one S-O single bond, a lone pair on sulfur, and appropriate lone pairs on oxygen atoms.

Option Analysis:

• A) This structure shows two single bonds and no lone pair on sulfur, giving sulfur only 4 electrons. This violates the octet rule for sulfur and is highly unstable.

• B) This structure shows two double bonds and no lone pair on sulfur. While sulfur can expand its octet, this structure would give sulfur 8 electrons (4 from each double bond) and 0 formal charge, and each oxygen would also have 0 formal charge. However, the presence of a lone pair on sulfur is characteristic of SO₂, and the structure with one double bond and one single bond (and a lone pair on S) is often considered more representative or one of the resonance forms.

• C) This structure shows two single bonds and two lone pairs on sulfur, giving sulfur 8 electrons. However, this would result in a formal charge of +2 on sulfur and -1 on each oxygen, which is less stable than option D.

• D) This structure shows one double bond, one single bond, and one lone pair on sulfur. This structure has formal charges of 0 on the double-bonded oxygen, +1 on sulfur, and -1 on the single-bonded oxygen. This is a valid resonance structure and is generally considered the most appropriate representation that minimizes formal charges while satisfying the octet rule for oxygen and allowing for a lone pair on sulfur.

Correct Answer: (D)