Worksheet for practicing Lewis structures of various chemical compounds.
Practice These Lewis Structure On Your Own worksheet with a list of chemical compounds to draw Lewis structures for, including CH₄, BF₃, HCN, NH₃, H₂O, ClO₃⁻, H₃O⁺, NH₄⁺, C₂H₄, C₂H₂, N₃⁻, NOF, ClF₂⁺, FNO₂, XeF₅⁺, BrF₄⁻, SF₆, N₂O₄, C₂O₄²⁻, H₂PO₄⁻, XeO₄, PO₃⁻, and NO.
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Show Answer Key & Explanations
Step-by-step solution for: Lewis Structure Worksheet | Lecture notes Geometry | Docsity
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Show Answer Key & Explanations
Step-by-step solution for: Lewis Structure Worksheet | Lecture notes Geometry | Docsity
I'll help you solve these Lewis structure problems step by step. Since I can't draw images, I'll describe each Lewis structure in detail and explain the reasoning.
## General Steps for Drawing Lewis Structures:
1. Count total valence electrons
2. Place atoms (least electronegative in center)
3. Connect with single bonds
4. Distribute remaining electrons to satisfy octets
5. Check formal charges and consider resonance
Let me go through each one:
- C has 4 valence e⁻, each H has 1 → Total: 8 e⁻
- C in center, bonded to 4 H atoms
- All atoms have complete octets (H has duet)
- Structure: C with 4 single bonds to H
- B has 3 valence e⁻, each F has 7 → Total: 24 e⁻
- B in center, bonded to 3 F atoms
- B has only 6 electrons (incomplete octet), but this is stable due to boron's ability to be electron-deficient
- Structure: B with 3 single bonds to F, each F has 3 lone pairs
- H: 1, C: 4, N: 5 → Total: 10 e⁻
- H-C≡N (triple bond between C and N)
- C has 4 bonds (1 to H, 3 to N), N has 3 bonds + 1 lone pair
- Structure: H-C≡N with lone pair on N
- N: 5, each H: 1 → Total: 8 e⁻
- N in center, bonded to 3 H atoms
- N has 1 lone pair
- Structure: N with 3 single bonds to H and 1 lone pair
- O: 6, each H: 1 → Total: 8 e⁻
- O in center, bonded to 2 H atoms
- O has 2 lone pairs
- Structure: O with 2 single bonds to H and 2 lone pairs
- Cl: 7, each O: 6, +1 for negative charge → Total: 26 e⁻
- Cl in center, bonded to 3 O atoms
- Resonance structures exist with double bonds
- Most stable: One Cl=O double bond, two Cl-O⁻ single bonds with formal charges
- Formal charges: Cl = +2, O (double bond) = 0, O (single bond) = -1 each
- O: 6, each H: 1, -1 for positive charge → Total: 8 e⁻
- O in center, bonded to 3 H atoms
- O has 1 lone pair
- Structure: O with 3 single bonds to H and 1 lone pair
- N: 5, each H: 1, +1 for negative charge → Total: 8 e⁻
- N in center, bonded to 2 H atoms
- N has 2 lone pairs
- Structure: N with 2 single bonds to H and 2 lone pairs
- N: 5, each H: 1, -1 for positive charge → Total: 8 e⁻
- N in center, bonded to 4 H atoms
- No lone pairs on N
- Structure: N with 4 single bonds to H
- Each C: 4, each H: 1 → Total: 12 e⁻
- Double bond between two C atoms
- Each C bonded to 2 H atoms
- Structure: H₂C=CH₂
- Each C: 4, each H: 1 → Total: 10 e⁻
- Triple bond between two C atoms
- Each C bonded to 1 H atom
- Structure: HC≡CH
- Each N: 5, +1 for negative charge → Total: 16 e⁻
- Central N bonded to two terminal N atoms
- Resonance: [N=N⁺=N⁻] ↔ [N⁻=N⁺=N]
- Most stable: Central N has formal charge of +1, one terminal N has -1, other has 0
- N: 5, O: 6, F: 7 → Total: 18 e⁻
- N in center, bonded to O and F
- Likely N=O double bond and N-F single bond
- Formal charges: N = 0, O = 0, F = 0
- Cl: 7, each F: 7, -1 for positive charge → Total: 20 e⁻
- Cl in center, bonded to 2 F atoms
- Cl has 2 lone pairs (expanded octet possible)
- Structure: Cl with 2 single bonds to F and 2 lone pairs
- F: 7, N: 5, each O: 6 → Total: 24 e⁻
- N in center, bonded to F and two O atoms
- Resonance: N=O double bonds possible
- Structure: F-N(O)=O with formal charges
- Xe: 8, each F: 7, -1 for positive charge → Total: 40 e⁻
- Xe in center, bonded to 5 F atoms
- Xe has expanded octet (10 electrons)
- Structure: Xe with 5 single bonds to F and 1 lone pair
- Br: 7, each F: 7 → Total: 42 e⁻
- Br in center, bonded to 5 F atoms
- Br has 1 lone pair
- Structure: Br with 5 single bonds to F and 1 lone pair
- I: 7, each F: 7, +1 for negative charge → Total: 36 e⁻
- I in center, bonded to 4 F atoms
- I has 2 lone pairs
- Structure: I with 4 single bonds to F and 2 lone pairs
- S: 6, each F: 7 → Total: 48 e⁻
- S in center, bonded to 6 F atoms
- S has expanded octet (12 electrons)
- Structure: S with 6 single bonds to F
- Each N: 5, each O: 6 → Total: 32 e⁻
- Two N atoms connected by single bond, each N bonded to two O atoms
- Resonance: N₂O₄ ⇌ 2 NO₂ (but it's a dimer)
- Structure: O₂N-NO₂ with resonance
- Each C: 4, each O: 6, +2 for negative charge → Total: 32 e⁻
- Two C atoms connected by single bond, each C bonded to two O atoms
- Resonance: Double bonds between C and O
- Structure: [O=C-O⁻]₂²⁻ with resonance
- P: 5, each O: 6, each H: 1, +1 for negative charge → Total: 32 e⁻
- P in center, bonded to 4 O atoms (two OH groups, two O⁻ groups)
- P has double bond to one O
- Structure: P with four bonds to O, two O-H bonds, one P=O, one P-O⁻
- Xe: 8, each O: 6 → Total: 32 e⁻
- Xe in center, bonded to 4 O atoms
- All Xe-O bonds are double bonds
- Structure: Xe with 4 double bonds to O
- P: 5, each O: 6, +1 for negative charge → Total: 24 e⁻
- P in center, bonded to 3 O atoms
- P has one lone pair
- Structure: P with 3 single bonds to O and 1 lone pair
- N: 5, O: 6 → Total: 11 e⁻ (odd number)
- N-O double bond
- N has 1 unpaired electron, O has 2 lone pairs
- Structure: N=O with unpaired electron on N
## Key Concepts:
- Formal charge = Valence e⁻ - (nonbonding e⁻ + ½ bonding e⁻)
- Resonance occurs when multiple valid Lewis structures exist
- Octet rule exceptions: elements in period 3+ can expand octets
- Electronegativity determines which atom gets negative formal charge
For molecular geometry, you'll need to apply VSEPR theory after drawing the Lewis structures. The molecular geometry depends on the number of bonding pairs and lone pairs around the central atom.
## General Steps for Drawing Lewis Structures:
1. Count total valence electrons
2. Place atoms (least electronegative in center)
3. Connect with single bonds
4. Distribute remaining electrons to satisfy octets
5. Check formal charges and consider resonance
Let me go through each one:
1. CH₄ (Methane)
- C has 4 valence e⁻, each H has 1 → Total: 8 e⁻
- C in center, bonded to 4 H atoms
- All atoms have complete octets (H has duet)
- Structure: C with 4 single bonds to H
2. BF₃ (Boron trifluoride)
- B has 3 valence e⁻, each F has 7 → Total: 24 e⁻
- B in center, bonded to 3 F atoms
- B has only 6 electrons (incomplete octet), but this is stable due to boron's ability to be electron-deficient
- Structure: B with 3 single bonds to F, each F has 3 lone pairs
3. HCN (Hydrogen cyanide)
- H: 1, C: 4, N: 5 → Total: 10 e⁻
- H-C≡N (triple bond between C and N)
- C has 4 bonds (1 to H, 3 to N), N has 3 bonds + 1 lone pair
- Structure: H-C≡N with lone pair on N
4. NH₃ (Ammonia)
- N: 5, each H: 1 → Total: 8 e⁻
- N in center, bonded to 3 H atoms
- N has 1 lone pair
- Structure: N with 3 single bonds to H and 1 lone pair
5. H₂O (Water)
- O: 6, each H: 1 → Total: 8 e⁻
- O in center, bonded to 2 H atoms
- O has 2 lone pairs
- Structure: O with 2 single bonds to H and 2 lone pairs
6. ClO₃⁻ (Chlorate ion)
- Cl: 7, each O: 6, +1 for negative charge → Total: 26 e⁻
- Cl in center, bonded to 3 O atoms
- Resonance structures exist with double bonds
- Most stable: One Cl=O double bond, two Cl-O⁻ single bonds with formal charges
- Formal charges: Cl = +2, O (double bond) = 0, O (single bond) = -1 each
7. H₃O⁺ (Hydronium ion)
- O: 6, each H: 1, -1 for positive charge → Total: 8 e⁻
- O in center, bonded to 3 H atoms
- O has 1 lone pair
- Structure: O with 3 single bonds to H and 1 lone pair
8. NH₂⁻ (Amine ion)
- N: 5, each H: 1, +1 for negative charge → Total: 8 e⁻
- N in center, bonded to 2 H atoms
- N has 2 lone pairs
- Structure: N with 2 single bonds to H and 2 lone pairs
9. NH₄⁺ (Ammonium ion)
- N: 5, each H: 1, -1 for positive charge → Total: 8 e⁻
- N in center, bonded to 4 H atoms
- No lone pairs on N
- Structure: N with 4 single bonds to H
10. C₂H₄ (Ethene)
- Each C: 4, each H: 1 → Total: 12 e⁻
- Double bond between two C atoms
- Each C bonded to 2 H atoms
- Structure: H₂C=CH₂
11. C₂H₂ (Ethyne)
- Each C: 4, each H: 1 → Total: 10 e⁻
- Triple bond between two C atoms
- Each C bonded to 1 H atom
- Structure: HC≡CH
12. N₃⁻ (Azide ion)
- Each N: 5, +1 for negative charge → Total: 16 e⁻
- Central N bonded to two terminal N atoms
- Resonance: [N=N⁺=N⁻] ↔ [N⁻=N⁺=N]
- Most stable: Central N has formal charge of +1, one terminal N has -1, other has 0
13. NOF (Nitrosyl fluoride)
- N: 5, O: 6, F: 7 → Total: 18 e⁻
- N in center, bonded to O and F
- Likely N=O double bond and N-F single bond
- Formal charges: N = 0, O = 0, F = 0
14. ClF₂⁺ (Chlorine difluoride cation)
- Cl: 7, each F: 7, -1 for positive charge → Total: 20 e⁻
- Cl in center, bonded to 2 F atoms
- Cl has 2 lone pairs (expanded octet possible)
- Structure: Cl with 2 single bonds to F and 2 lone pairs
15. FNO₂ (Nitryl fluoride)
- F: 7, N: 5, each O: 6 → Total: 24 e⁻
- N in center, bonded to F and two O atoms
- Resonance: N=O double bonds possible
- Structure: F-N(O)=O with formal charges
16. XeF₅⁺ (Xenon pentafluoride cation)
- Xe: 8, each F: 7, -1 for positive charge → Total: 40 e⁻
- Xe in center, bonded to 5 F atoms
- Xe has expanded octet (10 electrons)
- Structure: Xe with 5 single bonds to F and 1 lone pair
17. BrF₅ (Bromine pentafluoride)
- Br: 7, each F: 7 → Total: 42 e⁻
- Br in center, bonded to 5 F atoms
- Br has 1 lone pair
- Structure: Br with 5 single bonds to F and 1 lone pair
18. IF₄⁻ (Iodine tetrafluoride anion)
- I: 7, each F: 7, +1 for negative charge → Total: 36 e⁻
- I in center, bonded to 4 F atoms
- I has 2 lone pairs
- Structure: I with 4 single bonds to F and 2 lone pairs
19. SF₆ (Sulfur hexafluoride)
- S: 6, each F: 7 → Total: 48 e⁻
- S in center, bonded to 6 F atoms
- S has expanded octet (12 electrons)
- Structure: S with 6 single bonds to F
20. N₂O₄ (Dinitrogen tetroxide)
- Each N: 5, each O: 6 → Total: 32 e⁻
- Two N atoms connected by single bond, each N bonded to two O atoms
- Resonance: N₂O₄ ⇌ 2 NO₂ (but it's a dimer)
- Structure: O₂N-NO₂ with resonance
21. C₂O₄²⁻ (Oxalate ion)
- Each C: 4, each O: 6, +2 for negative charge → Total: 32 e⁻
- Two C atoms connected by single bond, each C bonded to two O atoms
- Resonance: Double bonds between C and O
- Structure: [O=C-O⁻]₂²⁻ with resonance
22. H₂PO₄⁻ (Dihydrogen phosphate ion)
- P: 5, each O: 6, each H: 1, +1 for negative charge → Total: 32 e⁻
- P in center, bonded to 4 O atoms (two OH groups, two O⁻ groups)
- P has double bond to one O
- Structure: P with four bonds to O, two O-H bonds, one P=O, one P-O⁻
23. XeO₄ (Xenon tetroxide)
- Xe: 8, each O: 6 → Total: 32 e⁻
- Xe in center, bonded to 4 O atoms
- All Xe-O bonds are double bonds
- Structure: Xe with 4 double bonds to O
24. PO₃⁻ (Phosphite ion)
- P: 5, each O: 6, +1 for negative charge → Total: 24 e⁻
- P in center, bonded to 3 O atoms
- P has one lone pair
- Structure: P with 3 single bonds to O and 1 lone pair
25. NO (Nitric oxide)
- N: 5, O: 6 → Total: 11 e⁻ (odd number)
- N-O double bond
- N has 1 unpaired electron, O has 2 lone pairs
- Structure: N=O with unpaired electron on N
## Key Concepts:
- Formal charge = Valence e⁻ - (nonbonding e⁻ + ½ bonding e⁻)
- Resonance occurs when multiple valid Lewis structures exist
- Octet rule exceptions: elements in period 3+ can expand octets
- Electronegativity determines which atom gets negative formal charge
For molecular geometry, you'll need to apply VSEPR theory after drawing the Lewis structures. The molecular geometry depends on the number of bonding pairs and lone pairs around the central atom.
Parent Tip: Review the logic above to help your child master the concept of lewis structures practice worksheet.
