Lewis Structure of Sulfate: Octet vs Formal Charge

Ask three chemistry teachers to draw the sulfate ion and you may get two different answers. One draws sulfur with four single bonds and a +2 charge sitting on it. Another draws two of the bonds as double bonds so that every charge except two vanishes. Both drawings appear in respected textbooks, and both have defenders. The ion itself is not confused at all: it is a perfectly tetrahedral SO₄²⁻ with four identical S–O bonds. The disagreement is about which cartoon describes it best.

The problems below work through both structures step by step, then test the reasoning on related species. By the end you should be able to draw either version, defend it, and say honestly where each one falls short. If you need a refresher on the bookkeeping first, see how to calculate formal charge and the general method in Lewis dot structures.

Problem 1: Count the electrons and draw the octet structure

Task: Draw a Lewis structure of SO₄²⁻ in which every atom obeys the octet rule, and find the formal charge on each atom.

Step 1: count valence electrons.

Source	Electrons
S (group 16)	6
4 × O (group 16)	4 × 6 = 24
2− charge	+2
Total	32

Step 2: skeleton. Sulfur is the less electronegative atom, so it goes in the middle with the four oxygens around it. Four single bonds use 8 electrons, leaving 24.

Step 3: complete the outer atoms. Each oxygen needs three lone pairs (6 electrons) to reach an octet. Four oxygens × 6 = 24. All electrons are now placed, and sulfur has exactly eight electrons in four bonding pairs.

Step 4: formal charges. Formal charge = valence electrons − nonbonding electrons − ½(bonding electrons).

• Sulfur: 6 − 0 − ½(8) = +2
• Each oxygen: 6 − 6 − ½(2) = −1

Check: +2 + 4(−1) = −2, which matches the ion’s charge. This is structure A, the octet structure. It obeys the octet rule everywhere but carries large, separated formal charges.

Problem 2: Minimise the formal charges

Task: Starting from structure A, convert lone pairs into double bonds to reduce the formal charges as far as possible. How many electrons surround sulfur now?

Move one lone pair from an oxygen into a new S=O bond. That oxygen now has two lone pairs and a double bond: 6 − 4 − ½(4) = 0. Sulfur now has 10 bonding electrons: 6 − 0 − ½(10) = +1.

Do it once more with a second oxygen:

• Sulfur: 6 − 0 − ½(12) = 0
• The two double-bonded oxygens: 6 − 4 − ½(4) = 0
• The two single-bonded oxygens: −1 each

Check: 0 + 0 + 0 − 1 − 1 = −2. Correct.

This is structure B, the expanded-octet structure. Sulfur is surrounded by 12 electrons. Only two formal charges remain, and they sit on oxygen, the more electronegative element, which the formal-charge guidelines like.

Why stop at two double bonds? A third would give sulfur −1 and leave one oxygen at −1, pushing negative charge onto the less electronegative atom. So structure B is where formal-charge minimisation ends.

Problem 3: Resonance and the average bond

Task: How many equivalent resonance forms does structure B have, and what average S–O bond order does it predict? Compare with structure A.

Structure B places its two double bonds on two of the four oxygens. The number of ways to pick 2 oxygens out of 4 is 4! / (2! × 2!) = 6. All six are equivalent drawings, and the real ion is a blend of them.

Average bond order in B: there are 6 bonds’ worth of electron pairs (two doubles + two singles = 2×2 + 2×1 = 6) shared over 4 bonds, so 6 ÷ 4 = 1.5. Each negative charge is spread over all four oxygens: −2 ÷ 4 = −½ per oxygen.

Structure A has only one arrangement, so no resonance is needed to make the bonds equal. Its bond order is 1 for every bond, and each oxygen carries −1.

What experiment says: all four S–O bonds in sulfate are the same length, about 149 pm. Both models agree the bonds are equivalent. They disagree on why the bonds are shorter and stronger than a plain single bond would suggest. Structure B says partial double-bond character. Structure A, taken together with modern calculations, says strong electrostatic attraction between S(+) and O(−) that tightens each bond. More on that in the discussion below.

Problem 4: Oxidation number, formal charge and shape

Task: Give the oxidation number of sulfur, and predict the shape and bond angle of SO₄²⁻ using VSEPR. Does the choice between A and B change your answers?

Oxidation number. Oxidation numbers treat every bond as fully ionic, giving the electrons to the more electronegative atom. Each O is −2: x + 4(−2) = −2, so x = +6. This does not depend on how you draw the bonds, because a double bond to oxygen gives both of its pairs to oxygen anyway.

Notice the three different “charges” on sulfur:

Model	Charge on S
Oxidation number (fully ionic)	+6
Formal charge, structure A	+2
Formal charge, structure B	0

None of these is the real charge on the atom; each is an accounting convention. Calculated partial charges on sulfur fall between the extremes, positive and substantial.

Shape. Sulfur has four bonding domains and no lone pairs in either structure (a double bond counts as one domain). Four domains give a tetrahedral shape with bond angles of 109.5°. The choice between A and B changes nothing here, which is reassuring. See VSEPR molecular geometry for the full method.

Problem 5: Sulfuric acid, where the bonds really do differ

Task: Draw the octet and minimum-formal-charge structures of H₂SO₄ (two OH groups on sulfur). Which bonds would you expect to be longest?

Valence electrons: 2(1) + 6 + 4(6) = 32, the same as sulfate because the two protons replace the 2− charge.

Octet version. Four S–O single bonds, each OH oxygen bonded to one H. Formal charges: S = +2; the two terminal O = −1 each; the two OH oxygens = 6 − 4 − ½(4) = 0; H = 0. Total 0. Correct.

Minimum-formal-charge version. Make both terminal oxygens double-bonded. Now every atom has a formal charge of 0, and sulfur carries 12 electrons.

Bond lengths. Here, unlike in sulfate, the bonds are genuinely different. The S–OH bonds are measurably longer than the bonds to the two terminal oxygens. Both pictures predict this: structure B says “single versus double”; structure A says the terminal oxygens carry −1 and are pulled in hard by the +2 sulfur, while the OH oxygens are neutral and not. Once again the two pictures agree on the observable result and disagree on the story.

Problem 6: The isoelectronic series

Task: SiO₄⁴⁻, PO₄³⁻, SO₄²⁻ and ClO₄⁻ all have 32 valence electrons. Find the central atom’s formal charge in the octet structure of each, and the number of double bonds needed to bring that formal charge to zero.

In every octet structure the central atom has four bonding pairs and no lone pairs, so its formal charge is (group valence electrons) − 4.

Ion	Central atom valence e⁻	Formal charge (octet)	Double bonds to reach 0	Electrons on central atom then
SiO₄⁴⁻	4	0	0	8
PO₄³⁻	5	+1	1	10
SO₄²⁻	6	+2	2	12
ClO₄⁻	7	+3	3	14

Each double bond lowers the central formal charge by one. For silicate the octet structure already has zero on silicon, so nobody draws silicate with double bonds. For perchlorate, formal-charge minimisation demands a chlorine with 14 electrons around it, which many students (reasonably) find alarming.

The table exposes the real question. Formal-charge rules push you towards more and more electrons around the central atom as you move across the period. Is that physically meaningful, or is the rule being stretched past what it was built for?

What modern bonding theory says

The expanded-octet picture was once justified by saying that sulfur uses its empty 3d orbitals to hold the extra electrons. Detailed quantum-chemical calculations over recent decades have mostly abandoned that explanation. The 3d orbitals of sulfur are too high in energy and too diffuse to take a large part in bonding. They do contribute a little, mainly as a mathematical refinement that improves the calculation, but not as genuine extra bonding orbitals holding four extra electrons.

What the calculations do show is that the S–O bonds are highly polar, with a large positive charge on sulfur and negative charge on the oxygens. That fits structure A better. The extra shortening and strengthening of the bonds, beyond a plain single bond, is then explained by the strong electrostatic pull between the charged atoms, plus some back-donation of oxygen lone-pair density towards sulfur through orbitals other than 3d. So a fair summary is:

• Structure A describes the charge distribution more honestly.
• Structure B, as a set of six resonance forms, conveys that the bonds are stronger and shorter than simple single bonds.
• Neither is a literal photograph. The real ion has four equivalent bonds, tetrahedral geometry and significant ionic character in each S–O bond.

This is not a settled “one right drawing” question, and it is worth saying so plainly in an answer if the context allows.

What will your exam expect?

Here is the awkward part: exam boards and textbooks are not consistent. Some mark schemes reward the expanded-octet structure because it minimises formal charge. Others accept or prefer the octet structure with charges on every atom. Some accept both if the working is clear. A few courses avoid the question entirely by asking only for shape and bond angle.

Practical advice:

1. Check your syllabus and past mark schemes. Whatever your specification shows is the version to learn for the exam.
2. Show your formal charges. Either structure, fully labelled, demonstrates understanding. An unlabelled structure is ambiguous.
3. Don’t rely on “sulfur uses its 3d orbitals” as the reason unless your course explicitly teaches it; it is the weakest part of the old explanation.
4. Remember what is certain: 32 valence electrons, tetrahedral, 109.5°, four equivalent bonds, oxidation state +6.

For more practice on awkward central atoms, try the Lewis structures practice set and the discussion of blended drawings in resonance structures.

Key takeaways

• SO₄²⁻ has 32 valence electrons; drawing it with four single bonds gives S a formal charge of +2 and each O −1.
• Converting two lone pairs to S=O bonds gives the expanded-octet structure: S formal charge 0, 12 electrons on S, six resonance forms, average bond order 1.5.
• All four S–O bonds are equal, about 149 pm, and the ion is tetrahedral (109.5°) whichever drawing you use.
• Modern calculations show little d-orbital participation and strongly polar S–O bonds, which favours the octet picture for describing charge, while the resonance picture captures the bond shortening.
• Oxidation number (+6), formal charge (+2 or 0) and real partial charge are different quantities.
• Exam expectations differ: follow your syllabus, and label formal charges clearly.