Target: “so42- lewis structure” (sulfate ion, SO42−)
The sulfate ion is a polyatomic anion with sulfur as the central atom bonded to four oxygens. A correct SO42− Lewis structure must (1) use the correct total number of valence electrons, (2) satisfy octets for oxygen, and (3) assign reasonable formal charges while accounting for resonance.
Step 1: Count total valence electrons
Sulfur (Group 16) contributes 6 valence electrons, each oxygen contributes 6, and the 2− charge adds 2 more electrons.
| Species | Valence electrons per atom | Number of atoms | Total |
|---|---|---|---|
| S | 6 | 1 | 6 |
| O | 6 | 4 | 24 |
| Charge contribution | 2− adds 2 electrons | 2 | |
| Total valence electrons | 32 | ||
Total: \[ 6 + 4\cdot 6 + 2 = 32 \]
Step 2: Choose a skeleton and place single bonds
Sulfur is the central atom (less electronegative than oxygen and capable of bonding to four atoms). Begin with four single bonds: S—O to each oxygen.
Four single bonds use \(4 \times 2 = 8\) electrons, leaving: \[ 32 - 8 = 24 \]
Step 3: Complete octets on the oxygens
Each oxygen already has one bond (2 shared electrons) and needs 6 more electrons (three lone pairs) to complete an octet. For four oxygens, that is \(4 \times 6 = 24\) electrons, exactly the number remaining.
After placing these lone pairs, sulfur has 8 electrons around it (four bonds), and each oxygen has an octet.
Step 4: Check formal charges and improve the structure
Formal charge is computed by: \[ \mathrm{FC} = V - \left(N + \frac{B}{2}\right) \] where \(V\) is the atom’s valence electrons, \(N\) is its nonbonding electrons, and \(B\) is its bonding electrons.
For the “all single bonds” drawing (four S–O single bonds), each oxygen has three lone pairs and one bond.
| Atom | Structure features | \(V\) | \(N\) | \(B\) | Formal charge |
|---|---|---|---|---|---|
| O (each) | 3 lone pairs, 1 single bond | 6 | 6 | 2 | \(6 - \left(6 + \frac{2}{2}\right) = -1\) |
| S | 4 single bonds | 6 | 0 | 8 | \(6 - \left(0 + \frac{8}{2}\right) = +2\) |
This gives four oxygens at \(-1\) each (total \(-4\)) and sulfur at \(+2\), for a net \(-2\), which is correct. However, formal charges are more separated than necessary.
A commonly preferred Lewis depiction lowers formal-charge separation by converting lone pairs on two oxygens into bonding pairs, forming two S=O double bonds. In that depiction:
- Two oxygens are double-bonded to sulfur and have formal charge \(0\).
- Two oxygens remain single-bonded with formal charge \(-1\) each.
- Sulfur has formal charge \(0\).
| Atom type | Bonding and lone pairs | \(V\) | \(N\) | \(B\) | Formal charge |
|---|---|---|---|---|---|
| O in S=O (2 atoms) | 2 lone pairs, 1 double bond | 6 | 4 | 4 | \(6 - \left(4 + \frac{4}{2}\right) = 0\) |
| O in S–O (2 atoms) | 3 lone pairs, 1 single bond | 6 | 6 | 2 | \(6 - \left(6 + \frac{2}{2}\right) = -1\) |
| S (central) | 2 double bonds + 2 single bonds | 6 | 0 | 12 | \(6 - \left(0 + \frac{12}{2}\right) = 0\) |
Visualization: resonance-aware Lewis picture of SO42−
Geometry from electron domains (VSEPR)
Around sulfur there are four bonding regions (four S–O connections). With four electron domains and no lone pairs on sulfur in the skeletal description, the electron-domain geometry is tetrahedral, consistent with the sulfate ion’s approximately tetrahedral shape.
Answer
The so42- lewis structure (SO42−) is drawn with sulfur in the center bonded to four oxygens inside brackets with an overall 2− charge; a standard lowest-formal-charge depiction shows two S=O double bonds and two S–O− single bonds, and six equivalent resonance forms imply a resonance hybrid with all S–O bonds equivalent (bond order 1.5) in a tetrahedral arrangement.