How to Draw a Lewis Structure for SO4²⁻?

To draw the Lewis structure for the sulfate ion (SO₄^2-), follow these steps:

1. Count the total number of valence electrons:
   • Sulfur (S) has 6 valence electrons.
   • Each oxygen (O) has 6 valence electrons, and there are 4 oxygen atoms, so 4 × 6 = 24 electrons.
   • Add 2 more electrons for the 2- charge.
   • Total valence electrons = 6 + 24 + 2 = 32 electrons.
2. Place the least electronegative atom in the center:
   • Sulfur is less electronegative than oxygen, so it will be the central atom.
3. Connect the outer atoms to the central atom with single bonds:
   • Draw single bonds between sulfur and each of the four oxygen atoms. This uses 8 electrons (4 bonds × 2 electrons).
4. Distribute the remaining electrons:
   • Subtract the used electrons from the total: 32 – 8 = 24 electrons.
   • Place these electrons around the oxygen atoms to complete their octets. Each oxygen needs 6 more electrons (since they already have 2 from the single bond).
   • After placing 6 electrons on each oxygen, you will have used 24 electrons (4 oxygens × 6 electrons).
5. Check for octet completion:
   • Each oxygen now has 8 electrons (2 from the bond and 6 lone pairs).
   • Sulfur has 12 electrons around it (4 bonds × 2 electrons), which is acceptable because sulfur can expand its octet.
6. Add formal charges if necessary:
   • Calculate the formal charge for each atom to ensure the structure is correct.
   • For sulfur: Formal charge = 6 – 0 – 4 = +2.
   • For each oxygen: Formal charge = 6 – 6 – 1 = -1.
   • The overall charge of the ion is 2-, which matches the given charge.

Here is the final Lewis structure for SO₄^2-:

    O
    ||
O--S--O
    ||
    O

Each oxygen atom has two lone pairs of electrons, and the sulfur atom is double-bonded to two oxygen atoms and single-bonded to the other two.