Lewis structure exercises | Learning Lab - Free Printable

Let’s go through each structure one by one and check if it follows the rules of Lewis structures — especially whether all atoms have complete octets (or duet for hydrogen), and whether formal charges make sense.

We’ll look at each lettered structure:

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a. SO₃ with double bonds to all three O atoms

- Sulfur has 6 valence electrons. In this structure, it’s forming 3 double bonds → that’s 6 bonding pairs = 12 electrons around sulfur.
- That’s allowed for sulfur because it’s in period 3 and can expand its octet.
- Each oxygen has 2 lone pairs + 1 double bond → 8 electrons → good.
✔ This is a valid Lewis structure.

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b. C bonded to two Cl and one O (double bond)

- Carbon: 4 valence electrons. Here it has 2 single bonds to Cl and 1 double bond to O → total 4 bonds → 8 electrons → good.
- Each Cl: 3 lone pairs + 1 single bond → 8 electrons → good.
- Oxygen: 2 lone pairs + 1 double bond → 8 electrons → good.
✔ Valid.

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c. H–F

- Hydrogen: 1 bond → 2 electrons → duet rule satisfied.
- Fluorine: 3 lone pairs + 1 bond → 8 electrons → good.
✔ Valid.

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d. SF₆

- Sulfur bonded to 6 fluorines → 6 single bonds → 12 electrons around sulfur → okay for expanded octet.
- Each F: 3 lone pairs + 1 bond → 8 electrons → good.
✔ Valid.

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e. S=C=S

- Central carbon: double bond to each S → 4 bonds → 8 electrons → good.
- Each sulfur: double bond + 2 lone pairs → 8 electrons → good.
✔ Valid.

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f. SeF₄ with one lone pair on Se

- Selenium: group 16 → 6 valence electrons.
- Bonds: 4 single bonds to F → uses 4 electrons.
- Lone pair: 2 electrons → total 6 electrons used from Se? Wait — actually, in Lewis terms, we count shared electrons too.
- Around Se: 4 bonds (8 electrons) + 1 lone pair (2 electrons) = 10 electrons → expanded octet → okay for Se (period 4).
- Each F: 3 lone pairs + 1 bond → 8 electrons → good.
✔ Valid.

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g. AlCl₃

- Aluminum: group 13 → 3 valence electrons.
- Forms 3 single bonds → 6 electrons around Al → incomplete octet.
- But aluminum often forms compounds with only 6 electrons (like AlCl₃) — it’s electron-deficient but still a real molecule.
- However, in strict Lewis structure rules, we usually expect octets unless it’s an exception like Be or B or Al.
- BUT — here’s the problem: each chlorine has 3 lone pairs and one bond → 8 electrons → fine.
- The issue is aluminum doesn’t have an octet — but that’s acceptable for Al in some cases.
⚠️ Some might say it’s “valid” as a real molecule, but strictly speaking, for a *complete* Lewis structure following octet rule, it’s not ideal. However, many textbooks accept AlCl₃ as drawn.

Wait — let’s hold off and compare with others.

Actually, looking again — this structure shows Al with only 3 bonds and no lone pairs → 6 electrons → violates octet rule. While it exists, in introductory chemistry, we often consider this invalid unless specified otherwise.

But let’s see what the question is asking — probably which ones are correctly drawn according to standard Lewis rules including octet.

Hold on — maybe we should check formal charges or other issues.

Actually, let’s move on and come back.

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h. O bonded to two Cl atoms

- Oxygen: 6 valence electrons. Here it has 2 single bonds → that’s 4 electrons from bonds, plus 2 lone pairs (4 electrons) → total 8 → good.
- Each Cl: 3 lone pairs + 1 bond → 8 electrons → good.
✔ Valid.

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i. SiBr₄

- Silicon: group 14 → 4 valence electrons → forms 4 single bonds → 8 electrons → good.
- Each Br: 3 lone pairs + 1 bond → 8 electrons → good.
✔ Valid.

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j. N≡N–O

Left N: triple bond to middle N → that’s 3 bonds → 6 electrons from bonds. Plus one lone pair → 2 more → total 8 → good.

Middle N: triple bond to left N (6 electrons) + single bond to O (2 electrons) → total 8 electrons → good.

Right O: single bond to N → 2 electrons from bond. Needs 6 more → should have 3 lone pairs → but in the diagram, it shows only 2 lone pairs? Let’s check:

In the image: :N≡N–Ö: ← wait, the right O has two dots above and two below? That’s 4 electrons → plus 2 from bond → only 6 electrons! ✘

Oxygen must have 8 electrons. Here it has only 6 → missing one lone pair.

So this structure is invalid.

Also, formal charge: left N has 5 valence - (3 bonds + 1 lone pair = 4 nonbonding? Wait:

Standard way: Formal charge = valence - (nonbonding + 1/2 bonding)

For right O: valence = 6, nonbonding = 4 (two lone pairs), bonding = 2 (one bond) → FC = 6 - (4 + 1) = +1? Not necessarily wrong, but the bigger issue is octet violation.

Oxygen has only 6 electrons → definitely invalid.

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Now back to g. AlCl₃

Aluminum has only 6 electrons → violates octet rule. In many contexts, especially introductory, this is considered an incomplete Lewis structure. Sometimes it’s shown with dative bonds or as a dimer, but as drawn, it’s not satisfying the octet rule.

Compare to BF₃ — same thing. Often marked as "incomplete" or "exception".

But let’s see — perhaps the question expects us to flag structures that break basic rules.

Another one: f. SeF₄ — we said it’s okay because Se can expand octet. Yes.

What about a. SO₃ — sometimes it’s drawn with resonance and one double bond, but here all double bonds — that’s also acceptable since S can handle 12 electrons.

Now, let’s list which ones are clearly invalid:

- j: Oxygen has only 6 electrons → invalid.
- g: Aluminum has only 6 electrons → violates octet → likely invalid for this context.

Are there any others?

Check c. HF — perfect.

b. COCl₂ — phosgene — correct.

d. SF₆ — correct.

e. CS₂ — correct.

h. Cl₂O — correct.

i. SiBr₄ — correct.

f. SeF₄ — correct (expanded octet).

a. SO₃ — correct.

So only g and j are problematic.

But wait — in g, is it possible that it’s meant to be AlCl₃ monomer? Even so, in Lewis dot terms, we usually require octets unless it's H, He, Li, Be, B, or sometimes Al — but even then, it’s noted as an exception.

However, looking at common textbook problems, when they ask “which Lewis structures are incorrect”, AlCl₃ as drawn (with only 3 bonds) is often flagged as having an incomplete octet.

Similarly, j is clearly wrong due to oxygen’s octet.

Is there another?

Look at f: SeF₄ — selenium has 4 bonds and 1 lone pair → 5 electron domains → trigonal bipyramidal — that’s fine. Electrons around Se: 4×2 (from bonds) + 2 (lone pair) = 10 — expanded octet — allowed.

No issue.

What about a: SO₃ — if you calculate formal charges:

S: valence 6, in structure: 0 lone pairs, 3 double bonds → bonding electrons = 12 → half is 6 → FC = 6 - 6 = 0.

Each O: valence 6, 4 nonbonding, 4 bonding → half bonding = 2 → FC = 6 - (4+2) = 0. So all zero — good.

Sometimes SO₃ is drawn with one double bond and two single bonds with charges, but this version is also acceptable.

So, final candidates for invalid: g and j

But let’s double-check j:

Structure: :N≡N–O:

The right O has only two lone pairs shown? In the image description: ":N≡N–Ö:" — assuming Ö means two dots above and two below? That’s four electrons, plus two from bond → six total. Should have three lone pairs → six nonbonding electrons.

Yes — missing one lone pair on oxygen.

Also, the middle nitrogen: triple bond to left N (6 electrons) + single bond to O (2 electrons) → 8 electrons — ok.

Left N: triple bond (6 electrons) + one lone pair (2 electrons) → 8 — ok.

Only oxygen is short.

So j is invalid.

For g: Al has three bonds → six electrons. No lone pairs. Octet not satisfied.

In many multiple-choice questions, this is considered incorrect unless specified.

Moreover, comparing to others — all others satisfy octet (or expanded where allowed).

So I think the intended answer is that g and j are incorrectly drawn.

But the question says: “You are an educational assistant helping a student solve homework problems.” and the user uploaded an image — but we don’t know the exact question.

Wait — re-reading the initial prompt: the user didn’t specify the question! They just said “solve the problem accurately” based on the image.

Looking at the structures, and typical such problems, the most common question is: “Which of these Lewis structures are incorrect?” or “Identify the errors.”

Given that, and our analysis, g and j have clear violations.

Perhaps f? No, Se can have expanded octet.

Another thought: in a, sulfur has 12 electrons — is that always accepted? Yes, for period 3+.

I think g and j are the main issues.

But let’s see if there’s a third.

What about c: HF — perfect.

All others seem fine.

Perhaps the question is to find which one has a mistake in electron count.

Maybe b: carbon has four bonds — good.

I think only g and j.

But to confirm, let’s think about formal charges or something else.

In j, besides oxygen’s octet, the formal charge on oxygen: valence 6, nonbonding 4, bonding 2 → FC = 6 - (4 + 1) = +1. Middle N: valence 5, nonbonding 0, bonding 8 → half is 4 → FC = 5 - 4 = +1. Left N: valence 5, nonbonding 2, bonding 6 → half is 3 → FC = 5 - (2+3) = 0. So total charge +2? But no indication of charge — so it should be neutral, but here it’s +2? That’s another error.

The molecule NNO should be neutral, but in this structure, formal charges sum to +2 (middle N +1, O +1, left N 0) — so it’s not balanced. Also, oxygen should have negative formal charge usually.

Correct Lewis for N₂O is [N=N=O] with charges or [N≡N-O]⁻ but with oxygen having three lone pairs and negative charge.

As drawn, it’s missing electrons on oxygen and has wrong formal charges.

Definitely invalid.

For g, AlCl₃, formal charge: Al valence 3, nonbonding 0, bonding 6 → half is 3 → FC = 3 - 3 = 0. Each Cl: 7 - (6 + 1) = 0. So formal charges are zero, but octet not satisfied. Still, in many curricula, it’s accepted as is, but strictly for Lewis structure completeness, it’s often marked as needing completion.

However, given that the other structures all satisfy octet (except expanded ones which are allowed), and g does not, I think it’s intended to be flagged.

Perhaps the answer is only j, but I doubt it.

Another idea: in f, SeF₄, selenium has 10 electrons — is that shown correctly? Yes, 4 bonds and one lone pair.

I think both g and j are incorrect.

But let’s see the options — the user didn’t provide choices, so we need to state which are wrong.

Perhaps the task is to identify the incorrect ones.

Since the instruction is to solve the problem, and based on standard practice, I'll go with g and j being incorrectly drawn.

Final decision:

Structures g and j are incorrectly drawn according to Lewis structure rules.

- g: Aluminum has only 6 electrons, violating the octet rule.
- j: Oxygen has only 6 electrons, violating the octet rule, and formal charges are unbalanced.

All others are valid.

So the answer should be g and j.

But to present it nicely.

Final Answer:
g and j