Hard Lewis Dot Structure Practice Problems

Why Lewis Dot Structures Get Harder Than They Should

Most students walk into Lewis structure problems thinking it'll be simple. Draw dots around atoms, connect them, move some around. Then they hit a problem like SCN^- or SO₃^2- and suddenly nothing makes sense.

The truth: Lewis structures aren't hard because the concept is complicated. They're hard because most textbooks skip the actual steps and expect you to figure out resonance, formal charge, and octet exceptions through osmosis.

This guide cuts the nonsense. Real problems. Real solutions. No "you've got this" garbage.

The Core Rules You Actually Need

Before touching any practice problem, you need these memorized:

• Valence electrons — count them for every atom in the molecule
• Octet rule — atoms want 8 electrons (except H, Li, Be, B which want fewer)
• Formal charge formula — FC = Valence e^- - (Nonbonding e^- + ½ Bonding e^-)
• Total formal charge — must match the ion's charge (or zero for neutral molecules)

If you don't know how to find valence electrons for any element in the periodic table, stop here and go learn that first. Everything else falls apart without this foundation.

Common Mistakes That Blow Your Answers

Mistake #1: Ignoring Formal Charge

Students pick the first Lewis structure they draw and call it done. Wrong. You need to minimize formal charges on each atom. A structure with a +2 on one atom and -1 on another is usually wrong when you could have +1 and 0.

Mistake #2: Getting Resonance Wrong

For polyatomic ions like NO₃^-, you need to draw all valid resonance structures. Then pick the best one using formal charge rules. Most students only draw one.

Mistake #3: Forgetting Octet Exceptions

Elements in period 3 and beyond can have expanded octets (12, 14, or even 16 electrons). Elements like sulfur and phosphorus regularly break the octet rule in valid structures.

Mistake #4: Counting Electrons Wrong

This sounds basic, but it's where most errors happen. Double bonds count as 4 electrons for octet purposes. Lone pairs count as 2. Don't double-count shared electrons.

Hard Practice Problems with Solutions

Problem 1: Sulfate Ion (SO₄^2-)

This one trips people up because of the expanded octet.

Step 1: Count total electrons
S = 6, O = 6 each × 4 = 24, plus 2 for the charge = 32 valence electrons

Step 2: Connect the skeleton
S goes in the center with four O atoms around it.

Step 3: Give every oxygen 3 lone pairs
That's 24 electrons used. 8 remain.

Step 4: Put remaining electrons on sulfur
Sulfur gets the last 8 electrons as lone pairs. This gives sulfur 12 electrons total (4 bonds × 2 + 4 lone pairs × 2 = 8 + 8 = 16... wait, let me redo this).

Actually: 4 S-O bonds = 8 electrons. 4 oxygens with 3 lone pairs each = 24 electrons. Total = 32. Perfect. Sulfur has 8 electrons around it from the bonds, plus the lone pairs you add. The structure has 2 double bonds and 2 single bonds to minimize formal charge.

Correct structure: S with 4 oxygens attached. Two double bonds, two single bonds. The two oxygens with single bonds each carry -1 formal charge. Sulfur has formal charge of 0.

Problem 2: Thiocyanate Ion (SCN^-)

Step 1: Count electrons
S = 6, C = 4, N = 5, plus 1 for the charge = 16 valence electrons

Step 2: Try the skeleton
Three possible arrangements: S-C-N, C-S-N, or S-N-C.

Step 3: Calculate formal charges for each arrangement

For S-C-N with a triple bond between C and N:

• S: FC = 6 - (2 + ½ × 2) = +5... no, that can't be right
• Let's draw it properly: S bonded to C, C triple bonded to N
• S has 2 lone pairs (4 e^-), C has none, N has 1 lone pair (2 e^-)
• S FC = 6 - (4 + ½ × 2) = 6 - 5 = +1
• C FC = 4 - (0 + ½ × 6) = 4 - 3 = +1
• N FC = 5 - (2 + ½ × 6) = 5 - 5 = 0
• Total = +2. Wrong.

Try S-C-N with double bond between S and C, triple between C and N:

• S FC = 6 - (4 + ½ × 4) = 6 - 6 = 0
• C FC = 4 - (0 + ½ × 8) = 4 - 4 = 0
• N FC = 5 - (2 + ½ × 6) = 5 - 5 = 0
• Total = 0. This works.

The correct structure is S=C=N^- with a double bond between S and C, triple bond between C and N, and a lone pair on sulfur.

Problem 3: Xenon Difluoride (XeF₂)

This one breaks rules students thought were solid.

Step 1: Count electrons
Xe = 8, F = 7 each × 2 = 14, minus nothing = 22 valence electrons

Step 2: Draw the skeleton
F-Xe-F in a straight line.

Step 3: Give fluorines their octets
Each F gets 3 lone pairs = 12 electrons used. 10 remain.

Step 4: Put remaining electrons on xenon
Xenon gets the remaining 10 electrons as 3 lone pairs on each side plus one in the middle. Wait, that gives Xe 10 electrons around it.

Xe has 3 bonds (counting the two Xe-F bonds as 6 electrons) plus 3 lone pairs on each end? No.

Structure: F-Xe-F with 3 lone pairs on each fluorine and 3 lone pairs on xenon. Xenon has 10 electrons around it (3 bonds + 6 lone pair electrons). This is an expanded octet situation — xenon can hold more than 8 electrons.

Comparison: Hard Polyatomic Ions

Getting Started: Your Systematic Approach

Follow these steps for every Lewis structure problem:

1. Count total valence electrons — add up all valence electrons from every atom, then add or subtract based on the ion's charge
2. Determine the central atom — almost always the least electronegative element (except H which is never central)
3. Draw the skeleton — connect atoms to the central atom with single bonds
4. Satisfy octets — give every atom 8 electrons (except H, Li, Be, B, and period 3+ elements)
5. Calculate formal charges — use FC = V - (LP + ½BP) for every atom
6. Minimize formal charges — if charges exist, adjust bonding (add/remove bonds, move lone pairs)
7. Check your work — total electrons match? All formal charges sum to the ion's charge?

When You're Stuck

Most Lewis structure problems go wrong at step 5. Students either skip formal charge calculation or don't know how to fix problems when formal charges are high.

If you have a +2 formal charge on one atom: form another bond to that atom (reduces FC by 1) or move a lone pair to form a double bond (reduces FC by 1 on the atom losing the pair, increases by 1 on the atom gaining).

If you have a -1 formal charge: remove a bond and give that electron to the negatively charged atom.

Practice with the ions in the table above. They're the ones that show up on exams. Get comfortable with formal charge math and expanded octets. Everything else is just variation on these themes.