Worksheet on drawing covalent bonds, including instructions and examples for HCl and H2O.

It looks like you've uploaded a worksheet titled "Drawing Covalent Bonds", which is designed to teach students how to draw covalent bonding diagrams using electron dot structures (Lewis structures). The worksheet includes an introduction, step-by-step instructions, and two completed examples (A: HCl and B: H₂O).

However, the actual task or problem you want me to solve isn't fully visible in the image — it appears that the Activity section only shows examples A and B, but no additional problems are listed for you to complete. There may be more content below what's shown, or perhaps you're asking for help understanding how to apply these steps to other molecules.

Since your request says "solve the problem," I’ll assume you'd like:

1. A clear explanation of how to use the steps provided.
2. A demonstration of how to draw covalent bond diagrams for common molecules beyond the examples given (e.g., O₂, N₂, CH₄, etc.), following the same method.
3. Guidance on how to approach any missing problems if they exist.

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✔ Step-by-Step Guide to Drawing Covalent Bond Diagrams

Let’s go through the four steps from the worksheet with clear explanations and examples.

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#### 🔹 Step 1: Draw Electron Dot Diagrams (Lewis Structures)

Each element has a certain number of valence electrons. Use the periodic table:
- Hydrogen (H): 1 valence electron → `H•`
- Oxygen (O): 6 → `:Ö:` (with 4 dots around)
- Nitrogen (N): 5 → `:N·` (with 5 dots)
- Chlorine (Cl): 7 → `:Cl:` (with 7 dots)
- Carbon (C): 4 → `:C:` (with 4 dots)

> 💡 Remember: Dots represent unpaired valence electrons.

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#### 🔹 Step 2: Count Total Valence Electrons

Add up all valence electrons from all atoms in the molecule.

For example:
- H₂O: 2×H = 2 × 1 = 2; O = 6 → total = 8
- CH₄: C = 4; 4×H = 4×1 = 4 → total = 8
- O₂: 2×O = 2×6 = 12

> ⚠️ Exception: Hydrogen needs only 2 electrons (duet rule), others need 8 (octet rule).

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#### 🔹 Step 3: Determine Central Atom

The central atom is usually:
- The one with lowest electronegativity
- Or the one that appears only once in the formula
- Or the one with the highest valence

Examples:
- In H₂O, oxygen is central (only one O, and it can form 2 bonds)
- In CH₄, carbon is central
- In CO₂, carbon is central

If there are only two atoms (like O₂ or Cl₂), there is no central atom — they bond directly.

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#### 🔹 Step 4: Pair Up Electrons to Form Bonds

Use single lines (`—`) for shared pairs (bonds), and lone pairs as dots.

Rules:
- Each bond = 2 electrons shared
- Each atom should have 8 electrons (except H, which needs 2)
- Try single bonds first; if not enough electrons, make double/triple bonds
- Keep trying until all electrons are paired

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🧪 Let's Practice: Draw Covalent Bond Diagrams

We'll do three new molecules based on this method.

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#### Example 1: O₂ (Oxygen Gas)

| Element | Valence e⁻ | Electron Dot |
|--------|------------|--------------|
| O | 6 | `:Ö:` |

Step 1: Two oxygen atoms, each with 6 valence electrons
Step 2: Total valence e⁻ = 6 + 6 = 12
Step 3: No central atom — two O atoms bonded together
Step 4: Try single bond: O–O → each O has 2 electrons in bond, plus 6 lone electrons → 8 total?
- Single bond: 2 e⁻ shared → each O has 2 bonding e⁻ + 6 lone e⁻ = 8 → wait! But each O has only 6 electrons left after bonding?

Wait — let’s count:

- Single bond: O–O → each O gets 2 bonding electrons
- Each O has 6 lone electrons → total per O = 8
- But we used: 2 electrons in bond, 6 lone on each → total = 2 + 6 + 6 = 14 electrons? Too many!

Actually, total valence electrons = 12. So we can’t have 14.

So if we try a double bond:
- Double bond = 4 electrons shared (2 pairs)
- Each O shares 2 electrons → now each O has 4 bonding e⁻ + 4 lone e⁻ = 8
- Lone pairs: 4 electrons per O → 2 lone pairs
- Total electrons: 4 (bond) + 4 (lone on O1) + 4 (lone on O2) = 12 ✔

✔ Correct structure:
```
.. ..
:O=O:
.. ..
```

Or better written as:

```
:Ö=Ö:
```

With two lone pairs on each O.

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#### Example 2: N₂ (Nitrogen Gas)

Nitrogen has 5 valence electrons.

Total valence e⁻ = 5 + 5 = 10

Try single bond: N–N → each N has 2 bonding e⁻ + 3 lone e⁻ = 5 → not octet

Try double bond: N=N → each N has 4 bonding e⁻ + 3 lone e⁻ = 7 → still missing

Try triple bond: N≡N → each N has 6 bonding e⁻ + 1 lone pair (2 e⁻) = 8 ✔

Total electrons: 6 (bond) + 2 (lone on N1) + 2 (lone on N2) = 10 ✔

Structure:
```
:N≡N:
```

Each nitrogen has one lone pair.

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#### Example 3: CH₄ (Methane)

Carbon: 4 valence e⁻
Hydrogen: 1 each → 4×1 = 4
Total valence e⁻ = 4 + 4 = 8

Carbon is central (only one C)

Each H needs 1 bond (2 e⁻), C needs 4 bonds (8 e⁻)

Draw C in center, connect to 4 H atoms with single bonds.

Each bond = 2 e⁻ → 4 bonds = 8 e⁻ → perfect!

No lone pairs on C or H.

Structure:
```
H
|
H--C--H
|
H
```

In Lewis notation:
```
H
|
H—C—H
|
H
```

Each bond is a shared pair.

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✔ Summary of Key Rules

| Rule | Explanation |
|------|-------------|
| 🔹 Octet Rule | Atoms gain 8 electrons via bonding/lone pairs (except H, which needs 2) |
| 🔹 Shared Pairs | One pair = single bond; two = double; three = triple |
| 🔹 Central Atom | Usually the least electronegative or unique atom |
| 🔹 Trial & Error | Sometimes you need to try different arrangements |

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🛠️ How to Solve Any Problem Like This

1. Identify elements and their valence electrons
2. Add total valence electrons
3. Choose central atom
4. Connect atoms with single bonds first
5. Distribute remaining electrons as lone pairs
6. Check octets — if not satisfied, convert lone pairs into double/triple bonds

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❓ If You Have Specific Problems to Solve…

Please provide the missing part of the worksheet (e.g., compounds like NH₃, CO₂, F₂, etc.), and I’ll walk you through drawing their covalent bond diagrams step by step.

Alternatively, if you meant to ask for help interpreting the examples already shown (A and B), here's a quick recap:

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🔍 Recap of Examples Given

#### A. HCl
- H has 1 e⁻, Cl has 7 e⁻
- Share 1 pair → H–Cl bond
- Cl has 3 lone pairs (6 e⁻) + 1 bond = 8 e⁻
- H has 2 e⁻ → duet rule satisfied

#### B. H₂O
- O has 6 e⁻, 2 H atoms × 1 e⁻ = 2 → total 8 e⁻
- O bonded to two H atoms with single bonds
- O has 2 lone pairs (4 e⁻) + 2 bonds = 8 e⁻
- Each H has 2 e⁻ → satisfied

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