How to Calculate Formal Charge (and Pick the Best Lewis Structure)

Many molecules and ions can be drawn with more than one valid Lewis structure. Which one is closest to reality? Formal charge is a simple bookkeeping tool that helps you decide. It assigns a charge to each atom as if the electrons in every bond were shared exactly equally, then compares that with the atom’s normal number of valence electrons. The structure with the smallest, most sensible formal charges is usually the best. This guide gives you the method, then works through examples that come up again and again in exams.

The formula

Formal charge = (valence electrons) − (non-bonding electrons) − ½(bonding electrons)

An equivalent and quicker version:

Formal charge = (valence electrons) − (dots) − (lines)

where “dots” are the electrons in lone pairs on that atom (count each electron, so a lone pair is 2) and “lines” are the number of bonds to that atom (each line counts as 1).

• Valence electrons: the number of outer electrons in the free atom — from the group number (C 4, N 5, O 6, halogens 7, H 1).
• The sum of the formal charges in a structure always equals the overall charge of the molecule or ion. This is a great check.

Step-by-step method

1. Draw a valid Lewis structure with the correct total number of valence electrons (see Lewis dot structures).
2. For each atom, count its lone-pair electrons and its bonds.
3. Apply the formula to each atom.
4. Check that the formal charges add up to the overall charge.
5. If several structures are possible, compare them using the rules below.

Rules for choosing the best structure

When comparing possible Lewis structures, prefer the one where:

1. Formal charges are as close to zero as possible (fewer and smaller formal charges).
2. Any negative formal charge sits on the more electronegative atom.
3. Adjacent atoms don’t carry the same sign of formal charge.
4. Atoms in period 2 (C, N, O, F) never exceed an octet.

Worked example 1: carbon dioxide, CO₂

Total valence electrons: C (4) + 2 × O (6) = 16.

Structure A: O=C=O (each O has two lone pairs)

• C: 4 − 0 − 4 = 0
• each O: 6 − 4 − 2 = 0
• Sum: 0 ✓

Structure B: O≡C–O (the triple-bonded O has one lone pair; the single-bonded O has three)

• C: 4 − 0 − 4 = 0
• O (triple): 6 − 2 − 3 = +1
• O (single): 6 − 6 − 1 = −1
• Sum: 0 ✓

Best: A, with all formal charges zero. This matches experiment: both C–O bonds in CO₂ are the same length (116 pm), consistent with two double bonds.

Worked example 2: ammonium, NH₄⁺

Total valence electrons: N (5) + 4 × H (1) − 1 (positive charge) = 8, so four N–H bonds and no lone pairs.

• N: 5 − 0 − 4 = +1
• each H: 1 − 0 − 1 = 0
• Sum: +1 ✓ (matches the ion’s charge)

So the positive charge “belongs” formally to nitrogen. (In reality it’s spread over the whole ion; formal charge is bookkeeping, not a measured charge.) Compare ammonia, NH₃: N has 5 − 2 − 3 = 0.

Worked example 3: the thiocyanate ion, SCN⁻

Total valence electrons: S (6) + C (4) + N (5) + 1 (negative charge) = 16. Carbon is central.

Structure A: [S=C=N]⁻ (S has two lone pairs, N has two lone pairs)

• S: 6 − 4 − 2 = 0
• C: 4 − 0 − 4 = 0
• N: 5 − 4 − 2 = −1

Structure B: [S–C≡N]⁻ (S has three lone pairs, N has one)

• S: 6 − 6 − 1 = −1
• C: 4 − 0 − 4 = 0
• N: 5 − 2 − 3 = 0

Structure C: [S≡C–N]⁻ (S has one lone pair, N has three)

• S: 6 − 2 − 3 = +1
• C: 0
• N: 5 − 6 − 1 = −2

Structure C is clearly worst (charges of +1 and −2). A and B both have a single −1 charge. By rule 2, the negative charge should be on the more electronegative atom — nitrogen (3.04) rather than sulfur (2.58) — so A is slightly favoured. In reality both A and B contribute: the true structure is a resonance hybrid (see resonance structures). This is also why thiocyanate can bond to metals through either S or N.

Worked example 4: carbon monoxide, CO

Total valence electrons: 4 + 6 = 10.

Structure A: C=O (a double bond uses 4 electrons; the remaining 6 go as two lone pairs on O and one lone pair on C)

• C: 4 − 2 − 2 = 0, but carbon has only 6 electrons around it — an incomplete octet
• O: 6 − 4 − 2 = 0
• Sum: 0 ✓

Structure B: C≡O (each atom has one lone pair)

• C: 4 − 2 − 3 = −1
• O: 6 − 2 − 3 = +1
• Sum: 0 ✓, and both atoms have full octets.

Here the rules conflict. Structure A has zero formal charges but leaves carbon with an incomplete octet; structure B gives full octets but puts a negative charge on the less electronegative atom. B is the accepted structure, because full octets for period 2 atoms usually take priority. It fits the evidence: CO has a very short bond (113 pm) and a very high bond enthalpy (1,077 kJ mol⁻¹), consistent with a triple bond. The −1 formal charge on carbon also helps explain why CO binds to metals — including the iron in haemoglobin — through its carbon atom (see carbon monoxide poisoning).

This example is a useful reminder: count electrons carefully and check the octet rule before trusting the formal charges.

Worked example 5: the sulfate ion, SO₄²⁻

Total valence electrons: S (6) + 4 × O (6) + 2 = 32.

Structure A: all single bonds (S bonded to four O; each O has three lone pairs)

• S: 6 − 0 − 4 = +2
• each O: 6 − 6 − 1 = −1
• Sum: +2 − 4 = −2 ✓

Structure B: two S=O and two S–O⁻ (S has 12 electrons around it)

• S: 6 − 0 − 6 = 0
• double-bonded O: 6 − 4 − 2 = 0
• single-bonded O: −1 each
• Sum: −2 ✓

Formal charge rules favour B, which is why many textbooks draw sulfate with two double bonds and an “expanded octet” on sulfur. However, modern calculations suggest sulfur doesn’t really use d orbitals to exceed its octet, and structure A with strongly polar S–O bonds (plus resonance) is a better description of the bonding. Exam boards differ — follow your syllabus. This case shows the limits of formal charge: it’s a helpful guide, not a law.

Worked example 6: dinitrogen monoxide, N₂O

Total valence electrons: 2 × N (5) + O (6) = 16. The arrangement is N–N–O (nitrogen in the centre).

Structure A: N=N=O

• N (end): 5 − 4 − 2 = −1
• N (centre): 5 − 0 − 4 = +1
• O: 6 − 4 − 2 = 0

Structure B: N≡N–O

• N (end): 5 − 2 − 3 = 0
• N (centre): +1
• O: 6 − 6 − 1 = −1

Both have one +1 and one −1. B puts the negative charge on oxygen, the more electronegative atom, so B is favoured — but A also contributes to the resonance hybrid.

Common mistakes

• Counting lone pairs instead of lone-pair electrons. A lone pair contributes 2 to the “dots” count.
• Counting bonding electrons instead of bonds in the quick formula. Use either ½ × bonding electrons or the number of lines — not both.
• Forgetting the overall charge when counting total electrons: add electrons for negative ions, subtract for positive ions.
• Treating formal charge as real charge. It’s bookkeeping; real charge distribution depends on electronegativity (see bond polarity).
• Ignoring the octet rule for period 2 atoms in the rush for zero formal charges.

Practice

Calculate the formal charge on each atom and choose the better structure:

1. Hydroxide ion, OH⁻.
2. Nitrate ion, NO₃⁻ (one N=O and two N–O).
3. Ozone, O₃ (O=O–O).

Answers:

1. O: 6 − 6 − 1 = −1; H: 0. Sum −1 ✓.
2. N: 5 − 0 − 4 = +1; double-bonded O: 0; each single-bonded O: −1. Sum −1 ✓. (Three equivalent resonance structures.)
3. Central O: 6 − 2 − 3 = +1; double-bonded end O: 0; single-bonded end O: 6 − 6 − 1 = −1. Sum 0 ✓. (Two resonance structures; both O–O bonds are actually equal.)

Key takeaways

• Formal charge = valence electrons − lone-pair electrons − bonds.
• Formal charges must add up to the overall charge.
• Prefer structures with charges near zero, negative charges on electronegative atoms, and full octets for period 2 atoms.
• Formal charge is a bookkeeping tool; when structures are close, the real molecule is often a resonance hybrid.

Test yourself further with Lewis structures practice.