Worked example 1
Compare the two common Lewis structures of the sulfate ion SO₄ 2-.
Try it first: Count the 32 valence electrons and try the all single bond structure first.
0 of 3 steps revealed.
What you'll be able to do: Decide when expanding an octet is justified and weigh the formal charge argument against the octet argument.
Best after: Choosing the Best Lewis Structure with Formal Charge, Exceptions to the Octet Rule
Sulfate can be drawn with four single bonds or with two double bonds. Both appear in textbooks, and knowing why is the point of this lesson.
These are recommended, not required. You can start this lesson at any time.
With four S-O single bonds, sulfur has an octet but carries a formal charge of +2 while each oxygen carries -1. Converting two of those bonds to double bonds gives sulfur twelve electrons and drops its formal charge to zero, leaving only two oxygens at -1.
Rule one of formal charge ranking prefers the expanded structure. The octet rule prefers the strict one. Computational studies of sulfate show the S-O bonding is largely ionic with little d orbital participation, supporting the strict octet form, yet the expanded form remains standard in many textbooks because it reproduces the short observed S-O bond lengths.
SO₂ with one single and one double bond gives sulfur +1 and the single bonded oxygen -1. Drawing both bonds double gives every atom a formal charge of zero at the cost of ten electrons on sulfur. The same argument applies, and the same two answers appear in different sources.
State the structure, give the formal charges, and name the criterion. For example: minimising formal charge favours the expanded octet with two S=O bonds, while strict octet reasoning favours four single bonds with sulfur at +2. Naming the criterion earns credit either way.
A period-3 central atom can be drawn with more than eight electrons in some useful Lewis structures, but the drawing is a bookkeeping model of bonding, not evidence that literal d-orbital hybridization must occur. Second-period atoms such as N and O cannot exceed an octet in ordinary Lewis structures. Count electrons and check formal charges before choosing among valid drawings.
Only central atoms from period 3 downward, such as P, S, Cl, Br, I and Xe.
Four single bonds gives S at +2; two double bonds gives S at 0 with twelve electrons.
State whether you are minimising formal charge or preserving the octet.
Having several nonzero formal charges within one species, which usually costs energy.
FC = V - N - B/2
Compare the two common Lewis structures of the sulfate ion SO₄ 2-.
Try it first: Count the 32 valence electrons and try the all single bond structure first.
0 of 3 steps revealed.
Draw SO₂ so that all formal charges are zero, and state the cost.
Try it first: Count the 18 valence electrons.
0 of 3 steps revealed.
Why can PF₅ be drawn with five P-F bonds while NF₅ is not a comparable ordinary molecule?
Try it first: Compare the central atoms' periods and their valence-shell electron limits.
0 of 3 steps revealed.
Why it's wrong: Period 2 atoms cannot exceed eight electrons under any argument.
Check instead: Only expand a central atom in period 3 or below.
Why it's wrong: Both are defensible depending on the criterion applied.
Check instead: Name the criterion in your answer.
Why it's wrong: The valence budget is fixed, so a new bond must come from an existing lone pair.
Check instead: Recount bonding and non-bonding electrons after each change.
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For period 3 and heavier central atoms, expanding the octet can reduce formal charges dramatically, as in sulfate and sulfur dioxide. Modern calculations favour the strict octet form with charge separation, while many courses still teach the expanded form, so the defensible answer states the criterion being used.
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