Practice worksheet for drawing Lewis structures of common molecules and ions.
Lewis structures practice worksheet featuring various chemical compounds including BCl₃, NF₃, PH₃, BaI₂, Li₂O, AlH₃, CBr₄, NBr₃, CH₂Cl₂, CaBr₂, H₂Se, NO₂⁻, NO₂Cl, SCF₂, SO₃, P₂, ClO₂⁻, BrO₃⁻, ClO₄⁻, SiS₂, S₃, and NH₄⁺.
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Show Answer Key & Explanations
Step-by-step solution for: Solved You In the worksheet below you will draw the Lewis | Chegg.com
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Show Answer Key & Explanations
Step-by-step solution for: Solved You In the worksheet below you will draw the Lewis | Chegg.com
Here are the Lewis structures for the compounds and ions listed in your practice sheet.
General Rules Used:
1. Count Valence Electrons: Add up electrons from all atoms. Add 1 for every negative charge, subtract 1 for every positive charge.
2. Central Atom: Usually the least electronegative atom (except Hydrogen, which is never central).
3. Connect Atoms: Draw single bonds between the central atom and outer atoms.
4. Fill Octets: Give outer atoms 8 electrons (lone pairs). Put any remaining electrons on the central atom.
5. Check Central Atom: If it doesn't have 8 electrons, make double or triple bonds by sharing lone pairs from outer atoms.
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1. BCl₃ (Boron Trichloride)
* Valence e⁻: B(3) + 3×Cl(7) = 24 e⁻
* Structure: Boron is an exception to the octet rule; it is stable with only 6 valence electrons.
* Drawing: B in center, single bonded to three Cl atoms. Each Cl has 3 lone pairs. B has 0 lone pairs.
2. NF₃ (Nitrogen Trifluoride)
* Valence e⁻: N(5) + 3×F(7) = 26 e⁻
* Structure: N in center, single bonded to three F atoms.
* Drawing: N has 1 lone pair. Each F has 3 lone pairs.
3. SBr₂ (Sulfur Dibromide)
* Valence e⁻: S(6) + 2×Br(7) = 20 e⁻
* Structure: S in center, single bonded to two Br atoms.
* Drawing: S has 2 lone pairs. Each Br has 3 lone pairs.
4. PH₃ (Phosphine)
* Valence e⁻: P(5) + 3×H(1) = 8 e⁻
* Structure: P in center, single bonded to three H atoms.
* Drawing: P has 1 lone pair. H atoms have no lone pairs.
5. BaI₂ (Barium Iodide)
* Type: Ionic Compound (Metal + Nonmetal).
* Structure: Barium loses 2 electrons to become Ba²⁺. Iodine gains 1 electron to become I⁻.
* Drawing: [Ba]²⁺ and two separate [ :I: ]⁻ ions (each Iodine has 8 dots around it).
6. Li₂O (Lithium Oxide)
* Type: Ionic Compound.
* Structure: Two Lithiums lose 1 electron each to become Li⁺. Oxygen gains 2 electrons to become O²⁻.
* Drawing: Two [Li] ions and one [ :Ö: ]²⁻ ion (Oxygen has 8 dots).
7. AlH₃ (Aluminum Hydride)
* Valence e⁻: Al(3) + 3×H(1) = 6 e⁻
* Structure: Al in center, single bonded to three H atoms.
* Drawing: Aluminum is an exception like Boron; it is stable with 6 electrons here. No lone pairs on Al.
8. CBr₄ (Carbon Tetrabromide)
* Valence e⁻: C(4) + 4×Br(7) = 32 e⁻
* Structure: C in center, single bonded to four Br atoms.
* Drawing: Carbon has 0 lone pairs. Each Br has 3 lone pairs.
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9. NBr₃ (Nitrogen Tribromide)
* Valence e⁻: N(5) + 3×Br(7) = 26 e⁻
* Structure: Similar to NF₃. N in center, single bonded to three Br atoms.
* Drawing: N has 1 lone pair. Each Br has 3 lone pairs.
10. CH₂Cl₂ (Dichloromethane)
* Valence e⁻: C(4) + 2×H(1) + 2×Cl(7) = 20 e⁻
* Structure: C in center, bonded to two H's and two Cl's.
* Drawing: Carbon has 0 lone pairs. Each Cl has 3 lone pairs. H's have none.
11. CaBr₂ (Calcium Bromide)
* Type: Ionic Compound.
* Structure: Ca becomes Ca²⁺. Two Br become Br⁻.
* Drawing: [Ca]² and two separate [ :Br: ]⁻ ions (each Br has 8 dots).
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12. H₂Se (Hydrogen Selenide)
* Valence e⁻: Se(6) + 2×H(1) = 8 e⁻
* Structure: Se in center, single bonded to two H atoms.
* Drawing: Se has 2 lone pairs.
13. NO₃¹ (Nitrate Ion)
* Valence e⁻: N(5) + 3×O(6) + 1(charge) = 24 e⁻
* Structure: N in center. To satisfy the octet for Nitrogen, you need one double bond.
* Drawing: N double bonded to one O, single bonded to two O's.
* The double-bonded O has 2 lone pairs.
* The single-bonded O's have 3 lone pairs each.
* Put brackets around the whole structure with a "-1" charge outside.
14. NO₂Cl (Nitryl Chloride)
* Note: The prompt says "(N center)".
* Valence e⁻: N(5) + 2×O(6) + Cl(7) = 24 e⁻
* Structure: N in center. N needs 4 bonds to be stable here.
* Drawing: N double bonded to both Oxygen atoms, single bonded to Cl.
* Both O's have 2 lone pairs.
* Cl has 3 lone pairs.
* N has 0 lone pairs.
15. SCF₂ (Thiocarbonyl difluoride)
* Valence e⁻: S(6) + C(4) + 2×F(7) = 24 e
* Structure: Carbon is usually central. It forms a double bond with Sulfur to get enough electrons.
* Drawing: C in center, double bonded to S, single bonded to two F's.
* S has 2 lone pairs.
* Each F has 3 lone pairs.
* C has 0 lone pairs.
16. SO₃ (Sulfur Trioxide)
* Valence e⁻: S(6) + 3×O(6) = 24 e⁻
* Structure: S in center. Sulfur can expand its octet (have more than 8 electrons) to minimize formal charge.
* Drawing: S double bonded to all three Oxygen atoms.
* Each O has 2 lone pairs.
* S has 0 lone pairs (12 electrons total around S).
* *(Note: Some classes teach resonance with one double bond and two single bonds. Check with your teacher if they want the expanded octet version above or the resonance version).*
17. P₂ (Diatomic Phosphorus)
* Valence e⁻: 2×P(5) = 10 e⁻
* Structure: Similar to N₂. They share 3 pairs of electrons.
* Drawing: P P (Triple bond). Each P has 1 lone pair.
18. ClO₂⁻¹ (Chlorite Ion)
* Valence e⁻: Cl(7) + 2×O(6) + 1(charge) = 20 e⁻
* Structure: Cl in center.
* Drawing: Cl single bonded to two O's.
* Cl has 2 lone pairs.
* Each O has 3 lone pairs.
* Brackets with "-1" charge.
19. BrO₃⁻¹ (Bromate Ion)
* Valence e⁻: Br(7) + 3×O(6) + 1(charge) = 26 e⁻
* Structure: Br in center. Like Chlorate, Bromine expands its octet.
* Drawing: Br double bonded to two O's, single bonded to one O.
* Double-bonded O's have 2 lone pairs.
* Single-bonded O has 3 lone pairs.
* Br has 1 lone pair.
* Brackets with "-1" charge.
20. ClO₄⁻¹ (Perchlorate Ion)
* Valence e⁻: Cl(7) + 4×O(6) + 1(charge) = 32 e⁻
* Structure: Cl in center. Expands octet to minimize charge.
* Drawing: Cl double bonded to three O's, single bonded to one O.
* Double-bonded O's have 2 lone pairs.
* Single-bonded O has 3 lone pairs.
* Cl has 0 lone pairs.
* Brackets with "-1" charge.
21. SiS₂ (Silicon Disulfide)
* Valence e⁻: Si(4) + 2×S(6) = 16 e⁻
* Structure: Linear, similar to CO₂.
* Drawing: Si in center, double bonded to both Sulfur atoms (S=Si=S).
* Each S has 2 lone pairs.
* Si has 0 lone pairs.
22. S₃ (Trisulfur)
* Valence e⁻: 3×S(6) = 18 e⁻
* Structure: Bent shape, similar to Ozone (O₃).
* Drawing: Central S double bonded to one S and single bonded to the other.
* Central S has 1 lone pair.
* Double-bonded end S has 2 lone pairs.
* Single-bonded end S has 3 lone pairs.
* *(Resonance exists here, flipping the double bond side).*
23. NH₄⁺¹ (Ammonium Ion)
* Valence e⁻: N(5) + 4×H(1) - 1(charge) = 8 e⁻
* Structure: N in center, single bonded to four H atoms.
* Drawing: N has 0 lone pairs.
* Brackets with "+1" charge.
General Rules Used:
1. Count Valence Electrons: Add up electrons from all atoms. Add 1 for every negative charge, subtract 1 for every positive charge.
2. Central Atom: Usually the least electronegative atom (except Hydrogen, which is never central).
3. Connect Atoms: Draw single bonds between the central atom and outer atoms.
4. Fill Octets: Give outer atoms 8 electrons (lone pairs). Put any remaining electrons on the central atom.
5. Check Central Atom: If it doesn't have 8 electrons, make double or triple bonds by sharing lone pairs from outer atoms.
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Part 1: Basic Molecules
1. BCl₃ (Boron Trichloride)
* Valence e⁻: B(3) + 3×Cl(7) = 24 e⁻
* Structure: Boron is an exception to the octet rule; it is stable with only 6 valence electrons.
* Drawing: B in center, single bonded to three Cl atoms. Each Cl has 3 lone pairs. B has 0 lone pairs.
2. NF₃ (Nitrogen Trifluoride)
* Valence e⁻: N(5) + 3×F(7) = 26 e⁻
* Structure: N in center, single bonded to three F atoms.
* Drawing: N has 1 lone pair. Each F has 3 lone pairs.
3. SBr₂ (Sulfur Dibromide)
* Valence e⁻: S(6) + 2×Br(7) = 20 e⁻
* Structure: S in center, single bonded to two Br atoms.
* Drawing: S has 2 lone pairs. Each Br has 3 lone pairs.
4. PH₃ (Phosphine)
* Valence e⁻: P(5) + 3×H(1) = 8 e⁻
* Structure: P in center, single bonded to three H atoms.
* Drawing: P has 1 lone pair. H atoms have no lone pairs.
5. BaI₂ (Barium Iodide)
* Type: Ionic Compound (Metal + Nonmetal).
* Structure: Barium loses 2 electrons to become Ba²⁺. Iodine gains 1 electron to become I⁻.
* Drawing: [Ba]²⁺ and two separate [ :I: ]⁻ ions (each Iodine has 8 dots around it).
6. Li₂O (Lithium Oxide)
* Type: Ionic Compound.
* Structure: Two Lithiums lose 1 electron each to become Li⁺. Oxygen gains 2 electrons to become O²⁻.
* Drawing: Two [Li] ions and one [ :Ö: ]²⁻ ion (Oxygen has 8 dots).
7. AlH₃ (Aluminum Hydride)
* Valence e⁻: Al(3) + 3×H(1) = 6 e⁻
* Structure: Al in center, single bonded to three H atoms.
* Drawing: Aluminum is an exception like Boron; it is stable with 6 electrons here. No lone pairs on Al.
8. CBr₄ (Carbon Tetrabromide)
* Valence e⁻: C(4) + 4×Br(7) = 32 e⁻
* Structure: C in center, single bonded to four Br atoms.
* Drawing: Carbon has 0 lone pairs. Each Br has 3 lone pairs.
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Part 2: Lewis Structure B Practice
9. NBr₃ (Nitrogen Tribromide)
* Valence e⁻: N(5) + 3×Br(7) = 26 e⁻
* Structure: Similar to NF₃. N in center, single bonded to three Br atoms.
* Drawing: N has 1 lone pair. Each Br has 3 lone pairs.
10. CH₂Cl₂ (Dichloromethane)
* Valence e⁻: C(4) + 2×H(1) + 2×Cl(7) = 20 e⁻
* Structure: C in center, bonded to two H's and two Cl's.
* Drawing: Carbon has 0 lone pairs. Each Cl has 3 lone pairs. H's have none.
11. CaBr₂ (Calcium Bromide)
* Type: Ionic Compound.
* Structure: Ca becomes Ca²⁺. Two Br become Br⁻.
* Drawing: [Ca]² and two separate [ :Br: ]⁻ ions (each Br has 8 dots).
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Part 3: More Complex Molecules & Ions
12. H₂Se (Hydrogen Selenide)
* Valence e⁻: Se(6) + 2×H(1) = 8 e⁻
* Structure: Se in center, single bonded to two H atoms.
* Drawing: Se has 2 lone pairs.
13. NO₃¹ (Nitrate Ion)
* Valence e⁻: N(5) + 3×O(6) + 1(charge) = 24 e⁻
* Structure: N in center. To satisfy the octet for Nitrogen, you need one double bond.
* Drawing: N double bonded to one O, single bonded to two O's.
* The double-bonded O has 2 lone pairs.
* The single-bonded O's have 3 lone pairs each.
* Put brackets around the whole structure with a "-1" charge outside.
14. NO₂Cl (Nitryl Chloride)
* Note: The prompt says "(N center)".
* Valence e⁻: N(5) + 2×O(6) + Cl(7) = 24 e⁻
* Structure: N in center. N needs 4 bonds to be stable here.
* Drawing: N double bonded to both Oxygen atoms, single bonded to Cl.
* Both O's have 2 lone pairs.
* Cl has 3 lone pairs.
* N has 0 lone pairs.
15. SCF₂ (Thiocarbonyl difluoride)
* Valence e⁻: S(6) + C(4) + 2×F(7) = 24 e
* Structure: Carbon is usually central. It forms a double bond with Sulfur to get enough electrons.
* Drawing: C in center, double bonded to S, single bonded to two F's.
* S has 2 lone pairs.
* Each F has 3 lone pairs.
* C has 0 lone pairs.
16. SO₃ (Sulfur Trioxide)
* Valence e⁻: S(6) + 3×O(6) = 24 e⁻
* Structure: S in center. Sulfur can expand its octet (have more than 8 electrons) to minimize formal charge.
* Drawing: S double bonded to all three Oxygen atoms.
* Each O has 2 lone pairs.
* S has 0 lone pairs (12 electrons total around S).
* *(Note: Some classes teach resonance with one double bond and two single bonds. Check with your teacher if they want the expanded octet version above or the resonance version).*
17. P₂ (Diatomic Phosphorus)
* Valence e⁻: 2×P(5) = 10 e⁻
* Structure: Similar to N₂. They share 3 pairs of electrons.
* Drawing: P P (Triple bond). Each P has 1 lone pair.
18. ClO₂⁻¹ (Chlorite Ion)
* Valence e⁻: Cl(7) + 2×O(6) + 1(charge) = 20 e⁻
* Structure: Cl in center.
* Drawing: Cl single bonded to two O's.
* Cl has 2 lone pairs.
* Each O has 3 lone pairs.
* Brackets with "-1" charge.
19. BrO₃⁻¹ (Bromate Ion)
* Valence e⁻: Br(7) + 3×O(6) + 1(charge) = 26 e⁻
* Structure: Br in center. Like Chlorate, Bromine expands its octet.
* Drawing: Br double bonded to two O's, single bonded to one O.
* Double-bonded O's have 2 lone pairs.
* Single-bonded O has 3 lone pairs.
* Br has 1 lone pair.
* Brackets with "-1" charge.
20. ClO₄⁻¹ (Perchlorate Ion)
* Valence e⁻: Cl(7) + 4×O(6) + 1(charge) = 32 e⁻
* Structure: Cl in center. Expands octet to minimize charge.
* Drawing: Cl double bonded to three O's, single bonded to one O.
* Double-bonded O's have 2 lone pairs.
* Single-bonded O has 3 lone pairs.
* Cl has 0 lone pairs.
* Brackets with "-1" charge.
21. SiS₂ (Silicon Disulfide)
* Valence e⁻: Si(4) + 2×S(6) = 16 e⁻
* Structure: Linear, similar to CO₂.
* Drawing: Si in center, double bonded to both Sulfur atoms (S=Si=S).
* Each S has 2 lone pairs.
* Si has 0 lone pairs.
22. S₃ (Trisulfur)
* Valence e⁻: 3×S(6) = 18 e⁻
* Structure: Bent shape, similar to Ozone (O₃).
* Drawing: Central S double bonded to one S and single bonded to the other.
* Central S has 1 lone pair.
* Double-bonded end S has 2 lone pairs.
* Single-bonded end S has 3 lone pairs.
* *(Resonance exists here, flipping the double bond side).*
23. NH₄⁺¹ (Ammonium Ion)
* Valence e⁻: N(5) + 4×H(1) - 1(charge) = 8 e⁻
* Structure: N in center, single bonded to four H atoms.
* Drawing: N has 0 lone pairs.
* Brackets with "+1" charge.
Parent Tip: Review the logic above to help your child master the concept of lewis structures practice worksheet.
