Lewis structure drawing — step by step with formal charge minimization

Question

Draw the Lewis structure of $\text{SO}_4^{2-}$ (sulphate ion). How do we use formal charge minimization to find the best structure?

(CBSE 11 + JEE Main + NEET)

Solution — Step by Step

S: 6 electrons, each O: 6 electrons, charge: 2 extra electrons.

Total = $6 + 4(6) + 2 = \mathbf{32}$ electrons = 16 pairs.

S is the central atom (least electronegative non-hydrogen). Connect each O to S with a single bond.

4 bonds used = 8 electrons spent. Remaining: $32 - 8 = 24$ electrons.

Each O needs 3 more lone pairs to complete its octet: $4 \times 6 = 24$ electrons. All remaining electrons used.

Check S: it has only 4 bonds (8 electrons around it). For a structure with only single bonds, each O has formal charge $= 6 - 6 - 1 = -1$ , and S has formal charge $= 6 - 0 - 4 = +2$ . The total is $+2 + 4(-1) = -2$ . Correct charge, but formal charges are large.

S can expand its octet (it has $d$ -orbitals). Convert two lone pairs from two oxygens into double bonds:

Now: 2 oxygens are doubly bonded (formal charge = $6 - 4 - 2 = 0$ ), 2 oxygens are singly bonded (formal charge = $6 - 6 - 1 = -1$ ), S has formal charge = $6 - 0 - 6 = 0$ .

Total charge: $0 + 2(0) + 2(-1) = -2$ . Correct and formal charges are minimized.

Best Lewis structure: S in centre with 2 double bonds and 2 single bonds to O, charge of $-1$ on each singly-bonded O.

flowchart TD
    A["Lewis Structure Algorithm"] --> B["Step 1: Count total valence electrons"]
    B --> C["Step 2: Identify central atom - least electronegative"]
    C --> D["Step 3: Draw single bonds to terminal atoms"]
    D --> E["Step 4: Complete octets on terminal atoms"]
    E --> F["Step 5: Place remaining electrons on central atom"]
    F --> G{"Central atom has octet?"}
    G -- Yes --> H["Calculate formal charges"]
    G -- No --> I["Form double/triple bonds from lone pairs"]
    I --> H
    H --> J{"Formal charges minimized?"}
    J -- No --> K["Try more double bonds if central atom can expand octet"]
    K --> H
    J -- Yes --> L["Best Lewis Structure"]

Why This Works

Lewis structures show how valence electrons are distributed in a molecule. The octet rule (8 electrons around each atom) mimics the stability of noble gas configurations. Formal charge tells us how “happy” each atom is — lower formal charges mean more realistic electron distribution.

Formal charge = (valence electrons) - (non-bonding electrons) - (bonding electrons / 2). The best structure is the one where formal charges are closest to zero, and any negative formal charges are on the more electronegative atoms.

Alternative Method

For simple molecules, skip formal charge and use the shortcut: the number of bonds formed by an atom = 8 minus its valence electrons (for elements in the second period). C forms 4 bonds, N forms 3, O forms 2, F forms 1. This gives the correct structure for most common molecules.

For JEE, know that elements in the third period and below (S, P, Cl, etc.) can expand their octet using $d$ -orbitals. This is why $\text{SF}_6$ has 12 electrons around S, and $\text{PCl}_5$ has 10 around P. Second period elements (C, N, O, F) NEVER exceed 8 electrons.

Common Mistake

Students forget to add or subtract electrons for charged species. For $\text{NH}_4^+$ , subtract 1 electron (positive charge = electron lost). For $\text{SO}_4^{2-}$ , add 2 electrons (negative charge = electrons gained). Getting the total electron count wrong makes the entire Lewis structure incorrect.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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