Lewis Structure Practice Problems - Free Printable

It looks like you've uploaded a list of molecules and ions for which you'd like Lewis structures drawn. However, I cannot view or process images directly — I can only analyze text-based input.

But since you’ve provided the text content of the problem (the list of compounds), I can help you solve all of these Lewis structure practice problems step by step.

Let’s go through each one systematically:

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General Steps to Draw Lewis Structures:

1. Count total valence electrons.
2. Identify central atom (usually least electronegative, except H).
3. Connect atoms with single bonds.
4. Distribute remaining electrons as lone pairs.
5. Check octets (or duet for H). Expand octet if needed (for elements in period 3+).
6. Use formal charges to assess best structure.

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Now let's solve each:

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a) ClF₂⁻ (Chlorine Difluoride Ion)

- Cl: 7 valence e⁻
- F: 7 × 2 = 14
- Add 1 e⁻ for negative charge → Total = 7 + 14 + 1 = 22 e⁻
- Central atom: Cl (less electronegative than F)
- Two Cl–F bonds: 2 × 2 = 4 e⁻ used
- Remaining: 22 – 4 = 18 e⁻ → place as lone pairs
- Each F gets 3 lone pairs (6 e⁻ each): 2 × 6 = 12 e⁻
- Remaining: 18 – 12 = 6 e⁻ → 3 lone pairs on Cl
- Cl has 2 bonds (4 e⁻) + 3 lone pairs (6 e⁻) = 10 e⁻ → expanded octet, OK for Cl
- Formal charge:
- Cl: 7 – (3 lone pairs = 6) – 2 bonds = 7 – 6 – 2 = –1 → matches charge
- F: 7 – 6 – 1 = 0

✔ Structure: Cl bonded to two F atoms, three lone pairs on Cl, three lone pairs on each F.

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b) ClF₄⁻ (Chlorine Tetrafluoride Ion)

- Cl: 7
- F: 7 × 4 = 28
- Add 1 e⁻ → total = 7 + 28 + 1 = 36 e⁻
- Central: Cl
- Four Cl–F bonds: 4 × 2 = 8 e⁻
- Remaining: 36 – 8 = 28 e⁻ → 3 lone pairs per F: 4 × 6 = 24 e⁻
- Left: 28 – 24 = 4 e⁻ → 2 lone pairs on Cl
- Cl has 4 bonds (8 e⁻) + 2 lone pairs (4 e⁻) = 12 e⁻ → expanded octet
- Formal charge on Cl: 7 – 4 – 4 = –1 → matches ion charge

✔ Structure: Cl with four F atoms bonded, two lone pairs on Cl, each F has three lone pairs.

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c) ClF₃ (Chlorine Trifluoride)

- Cl: 7
- F: 7 × 3 = 21
- Total = 28 e⁻
- Central: Cl
- Three Cl–F bonds: 6 e⁻
- Remaining: 22 e⁻ → 3 F atoms × 6 e⁻ = 18 e⁻
- Left: 4 e⁻ → 2 lone pairs on Cl
- Cl has 3 bonds + 2 lone pairs = 10 e⁻ → expanded octet
- Formal charge: Cl = 7 – 4 – 3 = 0 → neutral

✔ Structure: T-shaped (due to 2 lone pairs), Cl bonded to 3 F, each F has 3 lone pairs.

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d) ClF₅ (Chlorine Pentafluoride)

- Cl: 7
- F: 7 × 5 = 35
- Total = 42 e⁻
- Central: Cl
- Five Cl–F bonds: 10 e⁻
- Remaining: 32 e⁻ → 5 × 6 = 30 e⁻ for F lone pairs
- Left: 2 e⁻ → 1 lone pair on Cl
- Cl has 5 bonds (10 e⁻) + 1 lone pair (2 e⁻) = 12 e⁻ → expanded octet

✔ Square pyramidal geometry; Cl with 5 F and 1 lone pair.

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e) BeH₂ (Beryllium Hydride)

- Be: 2
- H: 1 × 2 = 2
- Total = 4 e⁻
- Central: Be
- Two Be–H bonds: 4 e⁻
- No lone pairs left
- Be has only 4 e⁻ → incomplete octet, but acceptable for Be
- H has 2 e⁻ each → stable

✔ Linear molecule; Be bonded to two H atoms, no lone pairs.

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f) AsO(OH)₃ (Arsenic Acid)

This is As(OH)₃O, but more commonly written as H₃AsO₄, so we assume it's AsO(OH)₃ meaning:

- Arsenic acid: As central, bonded to 4 O atoms: one double bond, three OH groups.

Valence electrons:
- As: 5
- O (double bond): 6
- 3 × O (in OH): 3 × 6 = 18
- 3 × H: 3 × 1 = 3
- Total = 5 + 6 + 18 + 3 = 32 e⁻

Structure:
- As bonded to:
- One O via double bond (no H)
- Three OH groups (each O bonded to H)
- Double bond: 4 e⁻
- Three single bonds to O: 3 × 2 = 6 e⁻
- Three O–H bonds: 3 × 2 = 6 e⁻
- Total bonding: 4 + 6 + 6 = 16 e⁻
- Remaining: 32 – 16 = 16 e⁻ → lone pairs
- Double-bonded O: 2 lone pairs (4 e⁻)
- Each OH oxygen: 2 lone pairs (4 e⁻ each) → 3 × 4 = 12 e⁻
- Total: 4 + 12 = 16 e⁻ → perfect

Formal charges:
- As: 5 – 0 – 4 = +1
- Double-bonded O: 6 – 4 – 2 = 0
- OH O: 6 – 4 – 2 = 0
- H: 0

But better to draw with one double bond and three single bonds, and assign formal charges accordingly.

✔ Final structure: As central, one double bond to O, three single bonds to O–H groups. As has +1 formal charge, double-bonded O has 0, others have 0.

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g) BF₄⁻ (Tetraborate Ion)

- B: 3
- F: 7 × 4 = 28
- Add 1 e⁻ → total = 3 + 28 + 1 = 32 e⁻
- Central: B
- Four B–F bonds: 8 e⁻
- Remaining: 24 e⁻ → 4 F atoms × 6 e⁻ = 24 e⁻
- All electrons used
- B has 4 bonds = 8 e⁻ → octet complete
- Formal charge: B = 3 – 0 – 4 = –1 → matches charge

✔ Tetrahedral; B bonded to four F, each F has 3 lone pairs.

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h) AsF₅ (Arsenic Pentafluoride)

- As: 5
- F: 7 × 5 = 35
- Total = 40 e⁻
- Central: As
- Five As–F bonds: 10 e⁻
- Remaining: 30 e⁻ → 5 × 6 = 30 e⁻ for F lone pairs
- As has 5 bonds = 10 e⁻ → expanded octet
- Formal charge: As = 5 – 0 – 5 = 0

✔ Trigonal bipyramidal geometry

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i) TeF₄ (Tellurium Tetrafluoride)

- Te: 6
- F: 7 × 4 = 28
- Total = 34 e⁻
- Central: Te
- Four Te–F bonds: 8 e⁻
- Remaining: 26 e⁻ → 4 F × 6 = 24 e⁻
- Left: 2 e⁻ → 1 lone pair on Te
- Te has 4 bonds + 1 lone pair = 10 e⁻ → expanded octet
- Formal charge: Te = 6 – 2 – 4 = 0

✔ See-saw geometry (trigonal bipyramidal electron geometry)

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j) I₃⁻ (Triiodide Ion)

- I: 7 × 3 = 21
- Add 1 e⁻ → total = 22 e⁻
- Central: middle I
- Two I–I bonds: 4 e⁻
- Remaining: 18 e⁻ → 3 terminal I atoms × 3 lone pairs = 9 × 2 = 18 e⁻
- But wait: each terminal I needs 3 lone pairs (6 e⁻), so 2 × 6 = 12 e⁻
- Middle I: after two bonds (4 e⁻), has 22 – 4 – 12 = 6 e⁻ → 3 lone pairs
- So: middle I has 2 bonds + 3 lone pairs = 10 e⁻ → expanded octet
- Formal charge:
- Terminal I: 7 – 6 – 1 = 0
- Central I: 7 – 6 – 2 = –1 → matches ion charge

✔ Linear shape; I–I–I with central I having 3 lone pairs

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k) F₃S–SF (Disulfur Tetrafluoride? Wait: F₃S–SF)

Wait: this seems like F₃S–SF, but likely meant to be F₃S–SF or SF₄?

Actually, F₃S–SF might be a typo or shorthand for SF₄ (sulfur tetrafluoride), but written as F₃S–SF.

Alternatively, it could be SF₄, which is sulfur tetrafluoride.

Assuming it's SF₄:

- S: 6
- F: 7 × 4 = 28
- Total = 34 e⁻
- Central: S
- Four S–F bonds: 8 e⁻
- Remaining: 26 e⁻ → 4 F × 6 = 24 e⁻
- Left: 2 e⁻ → 1 lone pair on S
- S has 4 bonds + 1 lone pair = 10 e⁻ → expanded octet
- Formal charge: S = 6 – 2 – 4 = 0

✔ See-saw geometry

If it's truly F₃S–SF, that would imply a disulfur compound: F₃S–S–F, but then we’d need more info.

But likely SF₄ is intended.

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l) TeCl₆ (Tellurium Hexachloride)

- Te: 6
- Cl: 7 × 6 = 42
- Total = 48 e⁻
- Central: Te
- Six Te–Cl bonds: 12 e⁻
- Remaining: 36 e⁻ → 6 × 6 = 36 e⁻ for Cl lone pairs
- Te has 6 bonds = 12 e⁻ → expanded octet
- Formal charge: Te = 6 – 0 – 6 = 0

✔ Octahedral geometry

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✔ Summary Table:

| Molecule | Total Valence e⁻ | Central Atom | Bonds | Lone Pairs on Central | Geometry |
|--------|------------------|-------------|-------|------------------------|----------|
| a) ClF₂⁻ | 22 | Cl | 2 | 3 | Bent (T-shaped?) |
| b) ClF₄⁻ | 36 | Cl | 4 | 2 | Square planar |
| c) ClF₃ | 28 | Cl | 3 | 2 | T-shaped |
| d) ClF₅ | 42 | Cl | 5 | 1 | Square pyramidal |
| e) BeH₂ | 4 | Be | 2 | 0 | Linear |
| f) AsO(OH)₃ | 32 | As | 4 | 0 (but one double bond) | Tetrahedral |
| g) BF₄⁻ | 32 | B | 4 | 0 | Tetrahedral |
| h) AsF₅ | 40 | As | 5 | 0 | Trigonal bipyramidal |
| i) TeF₄ | 34 | Te | 4 | 1 | See-saw |
| j) I₃⁻ | 22 | I | 2 | 3 | Linear |
| k) SF₄ | 34 | S | 4 | 1 | See-saw |
| l) TeCl₆ | 48 | Te | 6 | 0 | Octahedral |

> Note: For k), if it's not SF₄, please clarify. But F₃S–SF likely means SF₄.

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Would you like me to draw any of these structures using ASCII art or describe them visually? Or do you want formal charge calculations for any specific one?