Free Printable Lewis Dot Structure Worksheets - Free Printable

I'll solve this step by step, explaining how to draw the Lewis dot structures for each of the given covalent compounds.

## General Steps for Drawing Lewis Structures:
1. Count total valence electrons
2. Identify central atom (usually least electronegative)
3. Connect atoms with single bonds
4. Distribute remaining electrons as lone pairs
5. Check octet rule (except for exceptions)
6. Adjust if needed (double/triple bonds)

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PF₅ (Phosphorus pentafluoride)

- P: 5 valence e⁻, F: 7 each × 5 = 35
- Total: 5 + 35 = 40 e⁻
- P is central, bonded to 5 F atoms
- Each bond uses 2 e⁻ → 5 bonds = 10 e⁻ used
- Remaining 30 e⁻ → 3 lone pairs on each F (6 e⁻ per F × 5 F = 30 e⁻)
- P has 10 electrons (expanded octet), which is allowed for period 3+ elements

Structure: P in center with 5 single bonds to F atoms, each F has 3 lone pairs

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NO (Nitric oxide)

- N: 5 valence e⁻, O: 6 valence e⁻
- Total: 5 + 6 = 11 e⁻ (odd number → radical)
- N and O connected by double bond (4 e⁻)
- Remaining 7 e⁻: 3 lone pairs on O (6 e⁻), 1 unpaired electron on N
- Formal charges: N = 5 - 3 - 2 = 0, O = 6 - 4 - 2 = 0

Structure: N=O with double bond, O has 2 lone pairs, N has 1 lone pair and 1 unpaired electron

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I₃⁻ (Triiodide ion)

- I: 7 valence e⁻ × 3 = 21, plus 1 extra for negative charge = 22 e⁻
- Central I bonded to two terminal I atoms
- Two single bonds use 4 e⁻
- Remaining 18 e⁻ → 3 lone pairs on each terminal I (6 e⁻ × 2 = 12 e⁻), 3 lone pairs on central I (6 e⁻)
- Central I has 10 electrons (expanded octet)

Structure: I-I-I with central I bonded to two I atoms, each terminal I has 3 lone pairs, central I has 3 lone pairs

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HCN (Hydrogen cyanide)

- H: 1, C: 4, N: 5 → Total: 10 e⁻
- H bonded to C, C bonded to N
- H-C≡N triple bond between C and N
- C-H single bond
- N has one lone pair
- All atoms satisfy octet (H has duet)

Structure: H-C≡N with lone pair on N

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BrF₅ (Bromine pentafluoride)

- Br: 7, F: 7 × 5 = 35 → Total: 42 e⁻
- Br central, bonded to 5 F atoms
- 5 single bonds = 10 e⁻ used
- Remaining 32 e⁻ → 3 lone pairs on each F (6 e⁻ × 5 = 30 e⁻), 1 lone pair on Br (2 e⁻)
- Br has 12 electrons (expanded octet)

Structure: Br in center with 5 single bonds to F, each F has 3 lone pairs, Br has 1 lone pair

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CO (Carbon monoxide)

- C: 4, O: 6 → Total: 10 e⁻
- Triple bond between C and O (6 e⁻)
- Remaining 4 e⁻ → 2 lone pairs on O, 1 lone pair on C
- Both have formal charge of 0
- Triple bond with lone pairs on both atoms

Structure: C≡O with lone pair on C and two lone pairs on O

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BrF₃ (Bromine trifluoride)

- Br: 7, F: 7 × 3 = 21 → Total: 28 e⁻
- Br central, bonded to 3 F atoms
- 3 single bonds = 6 e⁻ used
- Remaining 22 e⁻ → 3 lone pairs on each F (6 e⁻ × 3 = 18 e⁻), 2 lone pairs on Br (4 e⁻)
- Br has 10 electrons (expanded octet)

Structure: Br in center with 3 single bonds to F, each F has 3 lone pairs, Br has 2 lone pairs

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CO₂⁻ (Carbon dioxide ion)

- C: 4, O: 6 × 2 = 12, plus 1 extra = 17 e⁻
- C central, bonded to two O atoms
- Double bonds: C=O and C=O (8 e⁻ used)
- Remaining 9 e⁻ → 3 lone pairs on each O (6 e⁻ × 2 = 12 e⁻), but only 9 available
- Actually: C=O double bonds, each O has 2 lone pairs (4 e⁻ × 2 = 8 e⁻), remaining 1 e⁻ → one O gets 3 lone pairs
- Better: Resonance structures with one double bond and one single bond with formal charges

Structure: Resonance between O=C-O⁻ and ⁻O-C=O, with double bond and single bond alternating

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SCl₄ (Sulfur tetrachloride)

- S: 6, Cl: 7 × 4 = 28 → Total: 34 e⁻
- S central, bonded to 4 Cl atoms
- 4 single bonds = 8 e⁻ used
- Remaining 26 e⁻ → 3 lone pairs on each Cl (6 e⁻ × 4 = 24 e⁻), 1 lone pair on S (2 e⁻)
- S has 10 electrons (expanded octet)

Structure: S in center with 4 single bonds to Cl, each Cl has 3 lone pairs, S has 1 lone pair

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AlH₃ (Aluminum trihydride)

- Al: 3, H: 1 × 3 = 3 → Total: 6 e⁻
- Al central, bonded to 3 H atoms
- 3 single bonds = 6 e⁻ used
- No lone pairs
- Al has only 6 electrons (incomplete octet, common for aluminum)

Structure: Al in center with 3 single bonds to H atoms

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BF₃ (Boron trifluoride)

- B: 3, F: 7 × 3 = 21 → Total: 24 e⁻
- B central, bonded to 3 F atoms
- 3 single bonds = 6 e⁻ used
- Remaining 18 e⁻ → 3 lone pairs on each F (6 e⁻ × 3 = 18 e⁻)
- B has only 6 electrons (incomplete octet)

Structure: B in center with 3 single bonds to F, each F has 3 lone pairs

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C₂H₄ (Ethene)

- C: 4 × 2 = 8, H: 1 × 4 = 4 → Total: 12 e⁻
- Two C atoms double bonded to each other
- Each C bonded to 2 H atoms
- Double bond between C's (4 e⁻), 4 single bonds to H (8 e⁻)
- All atoms have complete octets/duets

Structure: H₂C=CH₂ with double bond between carbons

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PH₃ (Phosphine)

- P: 5, H: 1 × 3 = 3 → Total: 8 e⁻
- P central, bonded to 3 H atoms
- 3 single bonds = 6 e⁻ used
- Remaining 2 e⁻ → 1 lone pair on P
- P has 8 electrons (octet satisfied)

Structure: P in center with 3 single bonds to H, P has 1 lone pair

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C₂H₆ (Ethane)

- C: 4 × 2 = 8, H: 1 × 6 = 6 → Total: 14 e⁻
- Two C atoms single bonded to each other
- Each C bonded to 3 H atoms
- Single bond between C's (2 e⁻), 6 single bonds to H (12 e⁻)
- All atoms have complete octets/duets

Structure: CH₃-CH₃ with single bond between carbons

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NH₄⁺ (Ammonium ion)

- N: 5, H: 1 × 4 = 4, minus 1 for positive charge = 8 e⁻
- N central, bonded to 4 H atoms
- 4 single bonds = 8 e⁻ used
- No lone pairs
- N has 8 electrons (octet satisfied)

Structure: N in center with 4 single bonds to H atoms, positive charge on N

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These are the correct Lewis dot structures for all the compounds listed. The key concepts include understanding valence electrons, identifying central atoms, satisfying octet rules (with exceptions), and accounting for formal charges and resonance where applicable.