Covalent Bonding Worksheet featuring exercises on electron sharing, Lewis dot diagrams, and bond types for molecules like H₂, F₂, O₂, N₂, BF₃, NH₃, CO₂, CH₄, CO, and SO₂.

You're working on a Covalent Bonding Worksheet that asks you to draw Lewis dot diagrams for various molecules and circle the unpaired electrons that are shared to form covalent bonds. Let’s go through each problem step-by-step with clear explanations.

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1. H₂ (Hydrogen gas) — Already Solved

- Each H atom has 1 valence electron.
- They share their single electrons → forms a single bond (H–H).
- ✔ Correct: `H• + •H → H:H → H–H`

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2. F₂ (Fluorine gas) — Partially Solved

- Fluorine (F) is in group 17 → 7 valence electrons.
- Each F needs 1 more electron to complete octet.
- So, they share 1 electron each → forms a single bond.
- Lewis structure: Each F has 3 lone pairs + 1 bonding pair.

✔ Complete it like this:

```
..
:F : F:
.. ..
```

(Each F has 6 dots around it + 2 shared electrons = 8 total)

➡️ Circle the one unpaired electron from each F that gets shared.

---

3. O₂ (Oxygen gas) — Partially Solved

- Oxygen (O) has 6 valence electrons → needs 2 more to complete octet.
- Two O atoms share 2 electrons each → forms a double bond.
- Lewis structure: O=O with 2 lone pairs on each O.

✔ Draw it like this:

```
..
:O::O:
..
```

(Each O has 4 nonbonding electrons + 4 bonding electrons = 8 total)

➡️ Circle the 2 unpaired electrons from each O that get shared to form the double bond.

---

4. N₂ (Nitrogen gas) — “Is this a triple bond?”

- Nitrogen (N) has 5 valence electrons → needs 3 more to complete octet.
- Two N atoms share 3 electrons each → forms a triple bond.
- Lewis structure: N≡N with 1 lone pair on each N.

✔ Draw it like this:

```
:N:::N:
```

Or with dots:

```
.
:N≡N:
.
```

(Each N has 2 nonbonding electrons + 6 bonding electrons = 8 total)

➡️ Circle the 3 unpaired electrons from each N that get shared.

✔ Answer: YES, this is a triple bond.

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5. BF₃ (Boron trifluoride) — “You need 3 fluorine atoms here”

- Boron (B) has 3 valence electrons.
- Each Fluorine (F) has 7 → needs 1.
- B shares its 3 electrons with 3 F atoms → forms 3 single bonds.
- B ends up with only 6 electrons (incomplete octet — acceptable for B).

✔ Draw it like this:

```
..
:F:
|
:F – B – F:
|
:F:
```

(Each F has 3 lone pairs; B has no lone pairs.)

➡️ Circle the 3 unpaired electrons from B and one unpaired electron from each F that get shared.

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6. Ammonia (NH₃) — “How many hydrogen atoms are needed?”

- Nitrogen (N) has 5 valence electrons → needs 3 more.
- Each H provides 1 electron → so we need 3 H atoms.
- Forms 3 single bonds → N has 1 lone pair.

✔ Draw it like this:

```
H
|
H – N – H
|
(lone pair)
```

With dots:

```
H
|
H – N:
|
H
```

(N has 2 dots = lone pair)

➡️ Circle the 3 unpaired electrons from N and 1 from each H.

✔ Answer: 3 hydrogen atoms needed.

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7. Carbon dioxide (CO₂)

- Carbon (C) has 4 valence electrons.
- Each O has 6 → needs 2.
- C shares 2 electrons with each O → forms two double bonds: O=C=O.

✔ Draw it like this:

```
.. ..
:O :: C :: O:
.. ..
```

(Each O has 2 lone pairs; C has no lone pairs.)

➡️ Circle the 2 unpaired electrons from C and 2 from each O that get shared.

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8. Methane (CH₄) — “Careful here — 4 hydrogen atoms needed”

- Carbon (C) has 4 valence electrons → needs 4 more.
- Each H provides 1 → so 4 H atoms.
- Forms 4 single bonds → tetrahedral shape.

✔ Draw it like this:

```
H
|
H – C – H
|
H
```

(C has no lone pairs)

➡️ Circle the 4 unpaired electrons from C and 1 from each H.

✔ Answer: 4 hydrogen atoms needed.

---

9. Dihydrogen monoxide (H₂O) — “Most dangerous substance...” 😄

- Oxygen (O) has 6 valence electrons → needs 2.
- Each H provides 1 → so 2 H atoms.
- Forms 2 single bonds → O has 2 lone pairs.

✔ Draw it like this:

```
H
|
O:
|
H
```

(O has 4 dots = 2 lone pairs)

➡️ Circle the 2 unpaired electrons from O and 1 from each H.

✔ Answer: 2 hydrogen atoms needed.

*(Note: Dihydrogen monoxide is just water — H₂O — a joke about sensationalizing common chemicals.)*

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10. SO₂ (Sulfur dioxide) — “Hint: one pair of electrons from sulfur must be split up”

- Sulfur (S) has 6 valence electrons.
- Each O has 6 → needs 2.
- S can expand octet (it’s in period 3).
- Common structure: S double bonded to one O, single bonded to another O (with resonance), and S has 1 lone pair.

But the hint says: *“one pair of electrons from sulfur must be split up”* → meaning, one lone pair on S becomes unpaired to form a bond.

✔ One possible Lewis structure (with formal charges minimized):

```
..
O = S - O:
..
```

(S has 1 lone pair, one O has 3 lone pairs, other O has 2 lone pairs + 1 extra electron → negative charge)

Better to show resonance:

```
.. ..
:O - S = O: ↔ :O = S - O:
.. ..
```

In both, S has 1 lone pair and 2 bonds (total 8 electrons).

➡️ Circle the unpaired electrons involved in bonding. Since S starts with 6 electrons (3 lone pairs), you "split" one lone pair → now S has 2 unpaired electrons to form bonds with O atoms.

✔ Key point: S uses 2 electrons to form a double bond with one O, and 2 electrons to form a single bond with the other O (which gets an extra electron → negative charge). S keeps 1 lone pair.

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## ✔ Final Summary Table:

| # | Molecule | Bonds Formed | Notes |
|----|----------|--------------|-------|
| 1 | H₂ | Single | ✔ Done |
| 2 | F₂ | Single | Circle 1 unpaired e⁻ per F |
| 3 | O₂ | Double | Circle 2 unpaired e⁻ per O |
| 4 | N₂ | Triple | Circle 3 unpaired e⁻ per N → YES, triple bond |
| 5 | BF₃ | 3 singles | B shares 3 e⁻ with 3 F atoms |
| 6 | NH₃ | 3 singles | Needs 3 H atoms; N has 1 lone pair |
| 7 | CO₂ | 2 doubles | O=C=O; linear |
| 8 | CH₄ | 4 singles | Needs 4 H atoms; tetrahedral |
| 9 | H₂O | 2 singles | Needs 2 H atoms; bent shape |
|10 | SO₂ | 1 double + 1 single (resonance) | S splits a lone pair; has 1 lone pair left |

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Let me know if you want printable diagrams or want to check your own drawings! Happy bonding 😊