AP Chemistry worksheet on chemical bonding with Lewis structures and valence electron counts.

Let’s go step by step to solve this problem.

We are asked to:
1. Count the total number of valence electrons for each species (atom, ion, or molecule).
2. Draw the correct Lewis electron dot structure for each.

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Step 1: Understand valence electrons

Valence electrons = electrons in the outermost shell of an atom.

For main group elements (Groups 1–8), you can find valence electrons from the group number:

- Group 1 → 1 valence e⁻
- Group 2 → 2 valence e⁻
- Group 13 → 3 valence e⁻
- Group 14 → 4 valence e⁻
- Group 15 → 5 valence e⁻
- Group 16 → 6 valence e⁻
- Group 17 → 7 valence e⁻
- Group 18 → 8 valence e⁻ (except He, which has 2)

For ions:
- Add electrons for negative charges (e.g., F⁻ adds 1 electron)
- Subtract electrons for positive charges (e.g., K⁺ subtracts 1 electron)

For molecules:
Add up valence electrons from all atoms.

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Step 2: Go through each box and verify calculations and structures

I’ll check each one carefully.

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#### Row 1: Atoms

F (Fluorine)
Group 17 → 7 valence e⁻ ✔
Structure: F with 7 dots (3 pairs + 1 single) — shown correctly ✔

O (Oxygen)
Group 16 → 6 valence e⁻ ✔
Structure: O with 6 dots (2 pairs + 2 singles) — two versions shown, both acceptable ✔

K (Potassium)
Group 1 → 1 valence e⁻ ✔
Structure: K with 1 dot — correct ✔

Al (Aluminum)
Group 13 → 3 valence e⁻ ✔
Structure: Al with 3 dots — correct ✔

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#### Row 2: Ions

F⁻
F has 7, plus 1 for charge → 8 e⁻ ✔
Structure: [F]⁻ with 8 dots (4 pairs) — correct ✔

O²⁻
O has 6, plus 2 for charge → 8 e⁻ ✔
Structure: [O]²⁻ with 8 dots — correct ✔

K⁺
K has 1, minus 1 for charge → 0 e⁻ ✔
Structure: [K]⁺ with no dots — correct ✔

Al³⁺
Al has 3, minus 3 for charge → 0 e⁻ ✔
Structure: [Al]³⁺ with no dots — correct ✔

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#### Row 3: Molecules

F₂
Each F has 7 → 7+7=14 e⁻ ✔
Structure: F:F with 3 lone pairs on each F and 1 bond (2 shared e⁻) → total 14 e⁻ used ✔

H₂
Each H has 1 → 1+1=2 e⁻ ✔
Structure: H:H — single bond, 2 e⁻ total ✔

HF
H=1, F=7 → 8 e⁻ ✔
Structure: H-F with 3 lone pairs on F → total 8 e⁻ (2 in bond, 6 on F) ✔

NH₃
N=5, H×3=3 → 8 e⁻ ✔
Structure: N bonded to 3 H, with 1 lone pair on N → total 8 e⁻ ✔

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#### Row 4: More Molecules

CH₄
C=4, H×4=4 → 8 e⁻ ✔
Structure: C in center, bonded to 4 H — no lone pairs → 8 e⁻ used ✔

NF₃
N=5, F×3=21 → 26 e⁻ ✔
Structure: N bonded to 3 F, with 1 lone pair on N; each F has 3 lone pairs → let’s count:
Bonds: 3 bonds × 2 e⁻ = 6 e⁻
Lone pairs on N: 1 pair = 2 e⁻
Each F has 3 lone pairs → 3 F × 6 e⁻ = 18 e⁻
Total: 6 + 2 + 18 = 26 e⁻ ✔

SiF₄
Si=4, F×4=28 → 32 e⁻ ✔
Structure: Si in center, bonded to 4 F; each F has 3 lone pairs →
Bonds: 4 × 2 = 8 e⁻
Lone pairs on F: 4 F × 6 e⁻ = 24 e⁻
Total: 8 + 24 = 32 e⁻ ✔

C₂H₆ (Ethane)
C×2=8, H×6=6 → 14 e⁻ ✔
Structure: Two C bonded together, each C bonded to 3 H →
Bonds: 1 C-C bond (2 e⁻) + 6 C-H bonds (12 e⁻) = 14 e⁻ ✔

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#### Row 5: Last Set

MgH₂
Mg=2 (Group 2), H×2=2 → 4 e⁻ ✔
Structure: H-Mg-H — Mg has no lone pairs, each H shares 2 e⁻ → total 4 e⁻ ✔

LiH
Li=1, H=1 → 2 e⁻ ✔
Structure: Li:H — single bond, 2 e⁻ total ✔

AlH₃
Al=3, H×3=3 → 6 e⁻ ✔
Structure: Al bonded to 3 H — no lone pairs on Al → 3 bonds × 2 e⁻ = 6 e⁻ ✔

BH₃
B=3, H×3=3 → 6 e⁻ ✔
Structure: B bonded to 3 H — no lone pairs → 6 e⁻ used ✔
*(Note: BH₃ is often drawn as incomplete octet — that’s fine for boron)*

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✔ All answers in the worksheet are correct!

Final Answer:
All valence electron counts and Lewis structures provided in the worksheet are accurate and correctly drawn.