What is the Lewis dot structure of sodium hyposulfite (Na2SO3)?
The Correct Answer and Explanation is:
Lewis Dot Structure of Sodium Sulfite (Na₂SO₃)
To draw the Lewis structure of sodium sulfite (Na₂SO₃), we first recognize that this compound is an ionic compound composed of two sodium ions (Na⁺) and one sulfite ion (SO₃²⁻). Sodium, as an alkali metal, donates one electron to become Na⁺ and does not contribute to the covalent bonding in the sulfite ion. Therefore, the focus is on the structure of the SO₃²⁻ ion.
Steps to Draw the Lewis Structure of SO₃²⁻:
- Count the total valence electrons:
- Sulfur (S) has 6 valence electrons.
- Each oxygen (O) atom has 6 valence electrons.
- The ion has a charge of -2, meaning 2 additional electrons.
Total:
= 6 (from S) + 3 × 6 (from O) + 2 (extra)
= 6 + 18 + 2
= 26 valence electrons
- Draw the skeletal structure:
Place sulfur (central atom) connected to three oxygen atoms with single bonds. - Distribute the remaining electrons:
After placing 3 single bonds (6 electrons), 20 electrons remain. Distribute these electrons to complete octets of the oxygen atoms. Each of the three oxygens gets 6 electrons (3 lone pairs), using 18 electrons. The remaining 2 electrons go on sulfur as a lone pair. - Check the formal charges:
This structure results in high formal charges. To reduce them, form one double bond between sulfur and one oxygen atom. The result: one S=O and two S–O⁻ bonds. - Enclose the structure in brackets with a 2⁻ charge, representing the sulfite ion.
Final Structure Summary:
- One sulfur atom at the center.
- One double bond to an oxygen atom.
- Two single bonds to two negatively charged oxygen atoms.
- The overall charge of the ion is 2⁻.
- Sodium ions (Na⁺) are not shown in the Lewis structure of the polyatomic ion.
This structure reflects the delocalized nature of the sulfite ion with resonance among the oxygen atoms.
