Chemistry Worksheet Lewis Dot Structures Name: Block: 1. Draw ... - Free Printable

It looks like you've uploaded a chemistry worksheet titled "Ionic Bonding – KEY" that asks students to draw the dot diagrams for each element in the compound and show how they form ionic bonds. The worksheet includes various ionic compounds, and handwritten answers (in blue ink) are already provided.

Let’s go through this systematically and explain how to solve such problems correctly, using the examples from the worksheet.

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🔬 Objective:

For each ionic compound, draw Lewis dot structures (electron dot diagrams) for each atom involved, showing:
- Valence electrons.
- How electrons are transferred.
- Formation of ions (cations and anions).
- The resulting ionic compound.

We’ll analyze a few key examples from your image.

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✔ Example 1: NaCl (Sodium Chloride)

Step-by-step:

1. Sodium (Na):
- Atomic number: 11
- Electron configuration: [Ne] 3s¹ → 1 valence electron
- Dot diagram: `Na•`

2. Chlorine (Cl):
- Atomic number: 17
- Electron configuration: [Ne] 3s² 3p⁵ → 7 valence electrons
- Dot diagram: `:Cl:•` (with 7 dots around Cl)

3. Electron Transfer:
- Na donates its 1 valence electron to Cl.
- Na becomes Na⁺ (loses 1 e⁻)
- Cl becomes Cl⁻ (gains 1 e⁻)

4. Resulting Ions:
- Na⁺: No dots (0 valence electrons)
- Cl⁻: 8 dots (octet achieved)

✔ Your answer is correct:
```
Na• + :Cl:• → Na⁺ + [:Cl:]⁻
```

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✔ Example 2: CaO (Calcium Oxide)

1. Calcium (Ca):
- Group 2 → 2 valence electrons
- Dot diagram: `Ca:`

2. Oxygen (O):
- Group 16 → 6 valence electrons
- Dot diagram: `:Ö:` (6 dots)

3. Electron Transfer:
- Ca loses 2 electrons → becomes Ca²⁺
- O gains 2 electrons → becomes O²⁻

4. Resulting Ions:
- Ca²⁺: no dots
- O²⁻: 8 dots (full octet)

✔ Your answer shows:
```
Ca: + :Ö: → Ca²⁺ + [:Ö:]²⁻
```
✔️ Correct!

---

✔ Example 3: AlN (Aluminum Nitride)

1. Aluminum (Al):
- Group 13 → 3 valence electrons
- Dot diagram: `Al•••`

2. Nitrogen (N):
- Group 15 → 5 valence electrons
- Dot diagram: `:N:•`

3. Electron Transfer:
- Al loses 3 electrons → Al³⁺
- N gains 3 electrons → N³⁻

4. Resulting Ions:
- Al³⁺: no dots
- N³⁻: 8 dots

✔ Your answer shows:
```
Al••• + :N:• → Al³⁺ + [:N:]³⁻
```
✔️ Correct!

---

✔ Example 4: MgS (Magnesium Sulfide)

1. Mg: Group 2 → 2 valence e⁻ → `Mg:`
2. S: Group 16 → 6 valence e⁻ → `:S:`
3. Mg loses 2 e⁻ → Mg²⁺
4. S gains 2 e⁻ → S²⁻

✔ Your answer:
```
Mg: + :S: → Mg²⁺ + [:S:]²⁻
```
✔️ Correct!

---

✔ Example 5: FeCl₂ (Iron(II) Chloride)

Note: Iron can have multiple charges. Here it's Fe²⁺, since there are two Cl⁻ ions.

1. Fe: Transition metal → usually shown with 2 or 3 valence electrons depending on oxidation state. In Fe²⁺, it loses 2 electrons.
2. Cl: Each has 7 valence electrons.

So:
- Fe → loses 2 e⁻ → Fe²⁺
- Two Cl atoms each gain 1 e⁻ → two Cl⁻

✔ Your answer shows:
```
Fe + 2 Cl: → Fe²⁺ + 2 [:Cl:]⁻
```
✔️ Correct! (Note: Fe doesn't typically have a fixed dot structure, but we assume it transfers 2 electrons.)

---

✔ Example 6: CaF₂ (Calcium Fluoride)

- Ca: 2 valence e⁻ → loses both → Ca²⁺
- F: 7 valence e⁻ → gains 1 e⁻ → F⁻
- So, one Ca gives 2 e⁻ to two F atoms

✔ Your answer:
```
Ca: + 2 •F: → Ca²⁺ + 2 [:F:]⁻
```
✔️ Correct!

---

✔ Example 7: Li₂O (Lithium Oxide)

- Li: 1 valence e⁻ → loses 1 → Li⁺
- O: 6 valence e⁻ → needs 2 → gains 2 e⁻
- So, two Li atoms donate 1 e⁻ each to O

✔ Your answer:
```
2 Li• + :O: → 2 Li⁺ + [:O:]²⁻
```
✔️ Correct!

---

✔ Example 8: AgP (Silver Phosphide)

Wait — AgP is unusual. Silver typically forms Ag⁺, phosphorus forms P³⁻.

But phosphide ion is P³⁻, so AgP would be Ag⁺P³⁻ — not common.

However, silver often forms Ag⁺, and phosphorus can be P³⁻ in ionic compounds.

So:
- Ag: loses 1 e⁻ → Ag⁺
- P: gains 3 e⁻ → P³⁻

But this requires 3 Ag⁺ and 1 P³⁻ → formula should be Ag₃P, not AgP.

⚠️ Error here:

Your answer shows:
```
Ag• + •P• → Ag⁺ + P³⁻
```
This implies a 1:1 ratio, which is incorrect.

✔ Correct answer:
- Ag₃P: three Ag atoms each lose 1 e⁻ → 3 Ag⁺
- One P gains 3 e⁻ → P³⁻

So:
```
3 Ag• + •P• → 3 Ag⁺ + [:P:]³⁻
```

✘ Your answer is incorrect for AgP — it should be Ag₃P.

---

✔ Example 9: Ba(NO₃)₂ (Barium Nitrate)

This is not a simple binary ionic compound — it contains polyatomic ion NO₃⁻.

- Ba: Group 2 → Ba²⁺
- NO₃⁻: nitrate ion (already charged)

So:
- Ba loses 2 e⁻ → Ba²⁺
- Two NO₃⁻ ions are needed to balance charge

But the question says "draw the dot diagrams for each element".

So you need to show:
- Ba: loses 2 e⁻ → Ba²⁺
- N and O in NO₃⁻: already bonded covalently, but overall charge is -1

You’re showing:
```
Ba: → Ba²⁺
N and O → NO₃⁻
```
But you're not drawing full Lewis structure for NO₃⁻.

The nitrate ion has resonance — but in basic ionic bonding, you just show:
- Ba²⁺
- Two NO₃⁻ ions

Your answer shows:
```
Ba: → Ba²⁺
N and O → NO₃⁻
```
But you didn’t draw the actual dot structure of NO₃⁻.

✔ Better way:
- Show Ba losing 2 e⁻
- Show NO₃⁻ as a unit with 8 electrons around O and formal charges

But since this is likely a beginner-level worksheet, your simplified version may be acceptable.

Still, it’s incomplete — the polyatomic ion should be drawn properly.

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✔ Example 10: Al₂(SO₄)₃ (Aluminum Sulfate)

Again, polyatomic ion SO₄²⁻

- Al: loses 3 e⁻ → Al³⁺
- SO₄²⁻: sulfate ion

So:
- 2 Al³⁺ → total +6 charge
- 3 SO₄²⁻ → total -6 charge

You showed:
```
Al: → Al³⁺
S and O → SO₄²⁻
```

But again, SO₄²⁻ is a polyatomic ion — you should represent it with proper Lewis structure (including double bonds and formal charges), or at least label it.

In a simplified format, it’s okay to write:
```
2 Al: + 3 SO₄²⁻ → 2 Al³⁺ + 3 SO₄²⁻
```

But you didn’t show the transfer — instead, you drew arrows from Al to S and O, which is misleading.

✘ This is not accurate — electrons are transferred to the ion, not directly to S or O.

✔ Better: Show Al losing 3 e⁻, and SO₄²⁻ accepting them (but it's already charged).

So, your representation is flawed.

---

✔ Example 11: ZnI₂ (Zinc Iodide)

- Zn: Group 12 → loses 2 e⁻ → Zn²⁺
- I: Group 17 → gains 1 e⁻ → I⁻
- So, one Zn gives 2 e⁻ to two I atoms

Your answer:
```
Zn: + 2 I: → Zn²⁺ + 2 [:I:]⁻
```
✔️ Correct!

---

✔ Example 12: AlCl₃ (Aluminum Chloride)

- Al: 3 valence e⁻ → loses 3 → Al³⁺
- Cl: 7 valence e⁻ → gains 1 → Cl⁻
- So, one Al gives 3 e⁻ to three Cl atoms

Your answer:
```
Al••• + 3 Cl: → Al³⁺ + 3 [:Cl:]⁻
```
✔️ Correct!

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## 📌 Summary of Corrections:

| Compound | Your Answer | Correct? | Notes |
|--------|------------|---------|------|
| NaCl | ✔️ | Yes | Good |
| CaO | ✔️ | Yes | Good |
| AlN | ✔️ | Yes | Good |
| MgS | ✔️ | Yes | Good |
| FeCl₂ | ✔️ | Yes | Good |
| CaF₂ | ✔️ | Yes | Good |
| Li₂O | ✔️ | Yes | Good |
| AgP | ✘ | No | Should be Ag₃P, not AgP |
| Ba(NO₃)₂ | ⚠️ | Partial | Need to show NO₃⁻ properly |
| Al₂(SO₄)₃ | ⚠️ | Incorrect | Misleading electron transfer to S/O |
| ZnI₂ | ✔️ | Yes | Good |
| AlCl₃ | ✔️ | Yes | Good |

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## ✔ Final Advice:

When solving these:
1. Identify charges of ions.
2. Balance electrons transferred.
3. For binary ionic compounds, show electron transfer clearly.
4. For polyatomic ions, treat them as units — don’t break them into individual atoms unless asked.
5. Use correct formulas — e.g., AgP should be Ag₃P.

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Would you like me to create a corrected version of this worksheet with proper Lewis structures and explanations?