Practice Drawing Lewis Structures Worksheet for 9th - 12th Grade ... - Free Printable

Let’s go step by step to draw the Lewis structures for each compound. Remember:

- Count total valence electrons.
- Put least electronegative atom in center (except H).
- Connect atoms with single bonds first.
- Fill octets (duet for H) with lone pairs.
- If needed, make double/triple bonds to satisfy octets.
- For ions, add or subtract electrons and use brackets + charge.

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1. CaCl₂
Calcium is a metal — this is ionic! Not covalent → no Lewis structure like molecules. But if forced: Ca²⁺ and two Cl⁻ ions. Each Cl has 8 electrons (3 lone pairs + 1 bond? No — actually, it's just ions). So we write [Ca]²⁺ and [:Cl:]⁻ twice. But since it’s ionic, often not drawn as shared electrons. We’ll note that.

But wait — instruction says “draw Lewis structures” — so maybe they want ionic representation? Let’s do:
[Ca]²⁺ [:Cl:]⁻ [:Cl:]⁻

Actually, better to show each chloride ion separately with full octet and negative charge.

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2. OH⁻
Oxygen has 6 valence e⁻, H has 1, plus 1 extra from negative charge = 8 total.
Draw O–H single bond (2 e⁻ used). Remaining 6 e⁻ go on oxygen as 3 lone pairs. Oxygen now has 8 e⁻ (2 in bond + 6 lone), H has 2. Add bracket and minus sign.

Structure: [H–O:]⁻ with 3 lone pairs on O? Wait — O should have 3 lone pairs? Let’s count:
Bond uses 2 e⁻. Left: 6 e⁻ → 3 lone pairs on O. Yes. Total electrons around O: 2 (bond) + 6 (lone) = 8. Good.

So: [H - Ö :]⁻ (with three lone pairs on O — but one pair is shown as dots, others implied? Better to draw all.)

Actually standard way:
[:Ö - H]⁻ where Ö means O with 3 lone pairs (so 6 dots) plus bond to H.

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3. NH₃
N has 5 valence, each H has 1 → 5 + 3×1 = 8 e⁻.
Central N, bonded to 3 H atoms (3 bonds = 6 e⁻ used). Remaining 2 e⁻ → 1 lone pair on N.
Each H has 2 e⁻ (good), N has 8 (3 bonds ×2 + 2 lone = 8). Perfect.

Structure:
H
|
H – N :
|
H
(with two dots above N for lone pair)

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4. MgCl₂
Again, ionic compound. Mg²⁺ and two Cl⁻.
So: [Mg]²⁺ [:Cl:]⁻ [:Cl:]⁻
Each Cl has 8 electrons (3 lone pairs + negative charge).

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5. N₂
Two nitrogen atoms. Each N has 5 valence → total 10 e⁻.
Triple bond between them: N≡N. That uses 6 e⁻. Remaining 4 e⁻ → 2 lone pairs, one on each N.
Each N has 8 e⁻: 6 from triple bond + 2 from lone pair.

Structure: :N≡N:

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6. CO₂
C has 4, each O has 6 → 4 + 12 = 16 e⁻.
Central C, double bonds to each O: O=C=O. Each double bond = 4 e⁻ → 8 e⁻ used. Remaining 8 e⁻ → 4 lone pairs, 2 on each O.
Each O has 8 e⁻ (4 from double bond + 4 from lone pairs), C has 8 (from two double bonds).

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

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7. SO₄²⁻
S has 6, each O has 6 → 6 + 24 = 30, plus 2 for charge → 32 e⁻.
Central S, four O atoms. Try single bonds first: 4 bonds = 8 e⁻ used. Remaining 24 e⁻ → 12 lone pairs. Give each O 3 lone pairs (6 e⁻ each) → 4 O × 6 = 24 e⁻. Total used: 8 + 24 = 32.
But S only has 8 e⁻ (from 4 bonds) — okay? S can expand octet. But formal charges: each O has -1? Too many negatives. Better to make double bonds.

Standard structure: S with two double bonds and two single bonds (to O⁻). Or resonance.

Best: S central, double bond to two O, single bond to two O⁻. Then S has 12 e⁻? Allowed. Formal charge: S = 6 - 0 - 4 = +2? Wait.

Better calculation:
Total valence e⁻: 32.
If all single bonds: S connected to 4 O with single bonds → 8 e⁻ in bonds. Each O gets 3 lone pairs (6 e⁻) → 4×6=24. Total 32. Charges: each O has -1? No — if single bond, O has 7 e⁻? Wait no: in Lewis, each single-bonded O would have 3 lone pairs (6 e⁻) + 1 bond (2 e⁻ shared) → but for formal charge: O has 6 valence, owns 6 lone + 1 from bond = 7 → formal charge -1. Four O⁻ → total -4, S has +2? Not good.

Actual common structure: S with four bonds, but two are double bonds. So:
S double bond to two O (each double bond O has two lone pairs), single bond to two O⁻ (each with three lone pairs).
Electrons:
- Two double bonds: 4 e⁻ each → 8 e⁻
- Two single bonds: 2 e⁻ each → 4 e⁻
Total bonding e⁻: 12
Lone pairs:
- Double-bonded O: each has 2 lone pairs → 4 e⁻ per O → 8 e⁻ for two
- Single-bonded O: each has 3 lone pairs → 6 e⁻ per O → 12 e⁻ for two
Total lone e⁻: 8 + 12 = 20
Total e⁻: 12 + 20 = 32. Good.
Formal charges:
- S: valence 6, bonds: 4 (two double = 4 bonds? In terms of electron count: S shares 8 e⁻ in bonds → owns 4. Lone pairs: 0. Formal charge = 6 - 0 - 4 = +2
- Double-bonded O: valence 6, owns 4 lone + 2 from bond = 6 → FC 0
- Single-bonded O: valence 6, owns 6 lone + 1 from bond = 7 → FC -1
Total charge: +2 + 0 + 0 + (-1) + (-1) = 0? But ion is -2. Mistake.

Ah — I forgot: when you have single-bonded O with three lone pairs, that’s 6 + 1 = 7 electrons owned, so formal charge -1. Two of them → -2. S has formal charge +2? Then total 0. But we need -2. So wrong.

Correct approach: minimize formal charges. Best structure has S with no formal charge? Actually, sulfate is often drawn with S having expanded octet and formal charges minimized by resonance.

Standard accepted structure: S central, four equivalent S-O bonds with partial double bond character. But for Lewis, we draw two double bonds and two single bonds with charges, and indicate resonance.

So:
[O=S(-O⁻)(-O⁻)=O]²⁻ but with resonance so all bonds equal.

For drawing: put S in center, double bond to top O, double bond to right O, single bond to bottom O (with three lone pairs and negative charge), single bond to left O (same). Then enclose in brackets with 2-.

But formal charge: S = 6 - 0 - 4 (bonds) = +2? Bonds: each double bond counts as 2 bonds for formal charge? Standard formula: formal charge = valence - nonbonding - 1/2 bonding.

For S: valence 6, nonbonding 0, bonding electrons: in two double bonds (4 e⁻ each? No — each double bond is 4 electrons, so for S, in two double bonds, it shares 8 electrons, in two single bonds, shares 4 electrons? Total bonding electrons around S: 12? Then 1/2 of 12 = 6. Formal charge = 6 - 0 - 6 = 0. Oh! I see — I miscounted.

In Lewis structure, for formal charge:
Number of bonds doesn't matter; it's number of electrons assigned to atom.

Rule: formal charge = valence electrons - (number of lone pair electrons + half of bonding electrons)

For S in SO₄²⁻ with two double bonds and two single bonds:
- Bonding electrons: each double bond contributes 4 electrons to the bond, but for S, it "owns" half of them. So for one double bond, S owns 2 electrons (half of 4). Similarly, single bond: S owns 1 electron (half of 2).
So: two double bonds → S owns 2 × 2 = 4 electrons
Two single bonds → S owns 2 × 1 = 2 electrons
Total bonding electrons owned by S: 6
Lone pairs on S: 0
Formal charge = 6 - 0 - 6 = 0

For double-bonded O: valence 6, lone pairs: 4 electrons (two pairs), bonding electrons: 4 (double bond), so owns 2 from bond. Total owned: 4 + 2 = 6 → FC 0

For single-bonded O: valence 6, lone pairs: 6 electrons (three pairs), bonding electrons: 2, owns 1 from bond. Total owned: 6 + 1 = 7 → FC -1

So two single-bonded O each -1 → total -2, S 0, two double-bonded O 0 → overall -2. Perfect.

So structure: S with two double bonds to O (each with two lone pairs), and two single bonds to O⁻ (each with three lone pairs). Enclose in brackets, charge 2-.

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8. NO₃⁻
N has 5, three O have 18, plus 1 for charge → 24 e⁻.
Central N, three O. Try one double bond, two single bonds.
Bonding: one double (4 e⁻), two single (4 e⁻) → total 8 e⁻ in bonds.
Remaining 16 e⁻ → lone pairs.
Double-bonded O: two lone pairs (4 e⁻)
Each single-bonded O: three lone pairs (6 e⁻ each) → 12 e⁻
Total lone: 4 + 12 = 16. Good.
Formal charges:
N: valence 5, bonding electrons: in double bond owns 2, in two single bonds owns 1 each → total 4 owned. Lone pairs: 0. FC = 5 - 0 - 4 = +1
Double-bonded O: 6 - 4 - 2 = 0? Owned: 4 lone + 2 from bond = 6 → FC 0
Single-bonded O: 6 - 6 - 1 = -1 each
Total charge: +1 + 0 + (-1) + (-1) = -1. Good.
Resonance: the double bond rotates among the three O.

Structure: [O=N(-O⁻)-O⁻]⁻ with resonance arrows, or draw one version.

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9. BeF₂
Be has 2 valence, F has 7 each → 2 + 14 = 16 e⁻.
Be is central, bonded to two F. Single bonds: 4 e⁻ used. Remaining 12 e⁻ → 6 lone pairs, 3 on each F.
Be has only 4 e⁻ (two bonds) — but Be is exception, can have less than octet.
So: F–Be–F, each F has three lone pairs.

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10. C₂H₂ (different isomers possible)
Acetylene: H–C≡C–H. Triple bond between C, each C bonded to H.
Valence: C has 4 each → 8, H has 1 each → 2, total 10 e⁻.
Triple bond: 6 e⁻, two C–H bonds: 4 e⁻, total 10. Each C has 2 e⁻ from C–H and 6 from triple bond → 8 e⁻. H has 2. Good.

Is there another isomer? Like vinylidene? H₂C=C: but that would be carbene, unstable. Usually only acetylene is considered. But problem says "different isomers possible", so perhaps mention H₂C=C: but it's not stable. Or maybe they mean constitutional isomers? For C₂H₂, only one stable isomer: HC≡CH.

Perhaps they consider the linear structure only. I'll draw HC≡CH.

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11. CH₄
C has 4, H has 1 each ×4 → 8 e⁻.
Central C, four single bonds to H. Each bond 2 e⁻ → 8 e⁻ used. No lone pairs. C has 8 e⁻, each H has 2.

Structure: tetrahedral, but in 2D:
H
|
H – C – H
|
H

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12. Na₂SO₄
Ionic: two Na⁺ and SO₄²⁻. So draw [Na]⁺ [Na]⁺ and the sulfate ion as before.

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13. BrCl₃
Br has 7, Cl has 7 each ×3 → 7 + 21 = 28 e⁻.
Central Br, bonded to three Cl. Single bonds: 6 e⁻ used. Remaining 22 e⁻ → 11 lone pairs.
Give each Cl three lone pairs (6 e⁻ each) → 3×6=18 e⁻. Remaining 4 e⁻ → two lone pairs on Br.
Br has 3 bonds + 2 lone pairs = 5 electron domains → trigonal bipyramidal electron geometry, T-shaped molecular geometry.
Octet? Br has 10 e⁻ — allowed (expanded octet).

Structure: Br central, three Cl attached, two lone pairs on Br. Usually drawn with Cl at equatorial positions? Standard: T-shape, with lone pairs axial.

But for Lewis, just show Br with three single bonds to Cl, and two lone pairs on Br. Each Cl has three lone pairs.

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14. PCl₅
P has 5, Cl has 7 each ×5 → 5 + 35 = 40 e⁻.
Central P, five single bonds to Cl. Bonds use 10 e⁻. Remaining 30 e⁻ → 15 lone pairs, 3 on each Cl.
P has 10 e⁻ — expanded octet, ok.

Structure: P with five Cl around it, each Cl has three lone pairs.

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15. SF₆
S has 6, F has 7 each ×6 → 6 + 42 = 48 e⁻.
Central S, six single bonds to F. Bonds use 12 e⁻. Remaining 36 e⁻ → 18 lone pairs, 3 on each F.
S has 12 e⁻ — expanded octet.

Structure: octahedral, S with six F, each F has three lone pairs.

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16. Al³⁺
Aluminum ion. Lost 3 electrons. Valence shell empty? Or just [Al]³⁺ with no dots. Since it's ion, no bonds. Just write [Al]³⁺

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17. PO₄³⁻
P has 5, O has 6 each ×4 → 5 + 24 = 29, plus 3 for charge → 32 e⁻.
Similar to sulfate. Central P, four O. To minimize formal charges, make one double bond? But P can have expanded octet.

Standard: P with four single bonds to O, each O has three lone pairs. Then P has formal charge: valence 5, owns 4 from bonds (half of 8 bonding e⁻), lone pairs 0 → FC +1. Each O has FC -1? Four O⁻ → -4, total -3? +1 -4 = -3. Yes.

But usually drawn with double bond to reduce charge. However, phosphate is often represented with four equivalent P-O bonds with resonance.

For Lewis: [O=P(-O⁻)(-O⁻)(-O⁻)]³ but with resonance.

Formal charge if all single bonds: P FC = 5 - 0 - 4 = +1, each O FC = 6 - 6 - 1 = -1, total +1 -4 = -3. Good.

With one double bond: P double bond to one O, single to three O⁻.
P: bonding electrons: double bond owns 2, three single bonds own 3 → total 5 owned. FC = 5 - 0 - 5 = 0
Double-bonded O: FC 0
Single-bonded O: FC -1 each → three of them → -3
Total charge -3. Better.

So structure: P with one double bond to O (two lone pairs), three single bonds to O⁻ (each three lone pairs). Enclose in brackets, charge 3-.

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18. HNO₃
H, N, three O. Valence: H1, N5, O18 → 24 e⁻.
Structure: H attached to O, that O attached to N, N attached to two other O. One of those O double bonded.

Common: H–O–N(=O)–O⁻ but then charge? Better: H–O–N with double bond to one O, single bond to another O which has negative charge? But then formal charges.

Standard: H bonded to O, that O bonded to N, N double bonded to one O, single bonded to another O (which has negative charge? No).

Actually: nitric acid is HONO₂, with N central, bonded to OH, =O, and -O⁻ but then protonated? Confusing.

Correct Lewis: N central, bonded to three O: one is OH group (so O bonded to H), one is double bond O, one is single bond O with negative charge? But then total charge -1, but molecule neutral.

Valence electrons: 24.

Draw: H–O–N, then N double bond to O, and N single bond to O⁻. But then the O in H–O– has two lone pairs, and is single bonded to H and N.

Count electrons:
Bonds: H–O (2), O–N (2), N=O (4), N–O (2) → total bonding e⁻: 10
Lone pairs:
- O in H–O: two lone pairs? It has two bonds (to H and N), so needs two lone pairs to have 8 e⁻ → 4 e⁻
- Double-bonded O: two lone pairs → 4 e⁻
- Single-bonded O: three lone pairs → 6 e⁻
- N: no lone pairs
Total lone e⁻: 4 + 4 + 6 = 14
Total e⁻: 10 + 14 = 24. Good.

Formal charges:
N: valence 5, bonding electrons: in O–N single bond owns 1, in N=O owns 2, in N–O single bond owns 1 → total 4 owned. FC = 5 - 0 - 4 = +1
O in H–O: valence 6, owns 4 lone + 1 from H bond + 1 from N bond = 6 → FC 0
Double-bonded O: 6 - 4 - 2 = 0
Single-bonded O: 6 - 6 - 1 = -1
H: 1 - 0 - 1 = 0
Total charge: +1 (N) + (-1) (O) = 0. Good.

So structure: H–O–N(=O)–O⁻ but with the last O having three lone pairs and negative charge, and N having positive formal charge. Often written with resonance.

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19. BH₃
B has 3, H has 1 each ×3 → 6 e⁻.
Central B, three single bonds to H. Uses 6 e⁻. No lone pairs. B has 6 e⁻ — incomplete octet, but common for boron.

Structure: H–B–H, with another H, planar.

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20. BF₃
B has 3, F has 7 each ×3 → 3 + 21 = 24 e⁻.
Central B, three single bonds to F. Bonds use 6 e⁻. Remaining 18 e⁻ → 9 lone pairs, 3 on each F.
B has 6 e⁻ — incomplete octet.

Structure similar to BH₃, but F instead of H.

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Now, let's compile the final answers clearly.

Note: For ionic compounds, we represent as ions. For molecules, show bonds and lone pairs.

Since this is text-based, I'll describe each briefly, but for actual drawing, student should sketch.

But for answer, I'll list the key features.

Final Answer:

1. CaCl₂: [Ca]²⁺ [:Cl:]⁻ [:Cl:]⁻
2. OH⁻: [H–Ö:]⁻ (O has three lone pairs)
3. NH₃: H–N–H with one lone pair on N, and another H (pyramidal)
```
H
|
H–N:
|
H
```
4. MgCl₂: [Mg]²⁺ [:Cl:]⁻ [:Cl:]⁻
5. N₂: :N≡N:
6. CO₂: :Ö=C=Ö: (each O has two lone pairs)
7. SO₄²⁻: [O=S(-O⁻)(-O⁻)=O]²⁻ with two double bonds and two single bonds to O⁻, resonance
8. NO₃⁻: [O=N(-O⁻)-O⁻]⁻ with resonance, one double bond
9. BeF₂: F–Be–F, each F has three lone pairs
10. C₂H₂: H–C≡C–H
11. CH₄: Central C with four H, tetrahedral
12. Na₂SO₄: [Na]⁺ [Na]⁺ [SO₄]²⁻ (sulfate as in #7)
13. BrCl₃: Br central with three Cl and two lone pairs, T-shaped
14. PCl₅: P central with five Cl, trigonal bipyramidal
15. SF₆: S central with six F, octahedral
16. Al³⁺: [Al]³⁺
17. PO₄³⁻: [O=P(-O⁻)(-O⁻)(-O⁻)]³ or with resonance
18. HNO₃: H–O–N(=O)–O⁻ with formal charges (+ on N, - on single-bonded O)
19. BH₃: H–B–H with third H, trigonal planar, B has six electrons
20. BF₃: F–B–F with third F, each F has three lone pairs, B has six electrons

For drawings, ensure correct number of lone pairs and bonds. Use brackets for ions with charge outside.