Covalent Bonding E-Dot Worksheet - Covalent Bonding - Electron Dot ... - Free Printable

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

We are asked to draw electron dot structures (also called Lewis dot structures) for individual atoms and then for molecules. Electron dot structures show how valence electrons are arranged around atoms — both shared (in bonds) and unshared (lone pairs).

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Step 1: Draw electron dot structures for individual atoms

Valence electrons = group number for main group elements (for groups 1–2, 13–18). We’ll use the periodic table positions:

- H (Hydrogen) → Group 1 → 1 valence electron → `H·`
- C (Carbon) → Group 14 → 4 valence electrons → `·C·` with one dot on each side (top, right, bottom, left)
- N (Nitrogen) → Group 15 → 5 valence electrons → `·N·` with three single dots and one pair? Wait — better to place one dot per side first, then pair up. Standard way: N has 5 dots — one on each of four sides, and one extra paired with any one. Usually drawn as:
```
·
· N ·
·
```
But actually, we usually put one dot on top, right, bottom, left — that’s 4 — then the fifth goes next to one of them. So it looks like:
```
·
· N :
·
```
Actually, standard is: nitrogen has 3 single dots and 1 lone pair. Let me write it clearly:

Nitrogen: 5 valence electrons → typically drawn with one pair and three singles:
```
..
. N .
.
```
But in text, we can say: N with 5 dots — two together (a pair) and three separate.

To avoid confusion, let’s list them simply:

- H: 1 dot → `H·`
- C: 4 dots → one on each side → `·C·` (with dots top, right, bottom, left)
- N: 5 dots → one pair + three singles → e.g., pair on top, singles on right, bottom, left
- O: 6 dots → two pairs + two singles → e.g., pairs on top and right, singles on bottom and left
- F: 7 dots → three pairs + one single → e.g., pairs on top, right, bottom; single on left
- S: same as O (Group 16) → 6 valence electrons → two pairs + two singles
- Cl: same as F (Group 17) → 7 valence electrons → three pairs + one single

But since we’re drawing for atoms only (not molecules), we just show the valence electrons around the symbol.

So:

- H: •H
- C: •C• (dots on all 4 sides)
- N: :N• (pair on left, singles on top, right, bottom — or however, but total 5)
Actually, standard notation:

In many textbooks, they draw:

- H: H•
- C: •C• (four dots, one on each side)
- N: •N: (three dots and one pair — often pair on top, dots on other three sides)
Wait — let’s be precise.

Standard electron dot symbols:

- Hydrogen: H with one dot → H•
- Carbon: C with four dots, one on each side →
```
•
• C •
•
```
- Nitrogen: N with five dots — usually one pair and three singles → e.g.,
```
..
• N •
•
```
(That’s 2+1+1+1=5)

- Oxygen: O with six dots — two pairs and two singles → e.g.,
```
..
• O •
..
```
Wait — that would be 2+1+1+2=6? No — if you have two pairs and two singles, that’s 2×2 + 2×1 = 6. But in diagram, you place them on four sides. Typically:

Oxygen:
```
..
: O :
..
```
No — that’s 8 dots. Mistake.

Correct: oxygen has 6 valence electrons. So we place one dot on each of the four sides (that’s 4), then add two more as pairs on two of the sides. So:

Commonly drawn as:
```
..
• O •
..
```
Wait — that’s 2 (top) + 1 (right) + 2 (bottom) + 1 (left)? That’s 6? No — if top is a pair (2), right is single (1), bottom is pair (2), left is single (1) → total 6. Yes.

But in simple text, we can describe:

For atoms:

- H: 1 dot
- C: 4 dots (one on each side)
- N: 5 dots (one pair and three singles)
- O: 6 dots (two pairs and two singles)
- F: 7 dots (three pairs and one single)
- S: same as O → 6 dots
- Cl: same as F → 7 dots

Now, for the molecules, we need to draw:

- Dot structure: shows all valence electrons as dots
- Lewis structure: same as dot, but often uses lines for bonds
- Structural formula: uses lines for bonds, no dots for bonding electrons, but may show lone pairs

Actually, in many contexts, “Dot/Lewis/Structural” means:

- Dot: all electrons shown as dots
- Lewis: bonds as lines, lone pairs as dots
- Structural: bonds as lines, lone pairs sometimes omitted or shown

But the worksheet says: “diagram the electron dot, Lewis, and structural formulas”

Looking at the sample: N₂O – Sample Problem, and it lists “Dot”, “Lewis”, “Structural” as separate columns? In the table, for N₂O, it says:

“N₂O – Sample Problem
Dot Lewis Structural”

So probably, for each molecule, we are to provide three representations? But the column header says “Dot/Lewis/Structural” — maybe it’s asking for one diagram that combines them, or perhaps it’s implying to draw the Lewis structure which includes dots for lone pairs and lines for bonds.

Given that it's a high school level worksheet, and the instruction says “diagram the electron dot, Lewis, and structural formulas”, I think they want the Lewis structure (which shows bonds as lines and lone pairs as dots), and that serves as both Lewis and structural. Sometimes "structural" means just the connectivity without lone pairs, but here since it mentions “distribution of shared and lone pair electrons”, we must show lone pairs.

Moreover, in the sample N₂O, it lists “Dot”, “Lewis”, “Structural” separately, so perhaps for each molecule, we should provide three diagrams? But the table has only one column for “Dot/Lewis/Structural”. Looking back:

The table has:

| Molecule | Dot/Lewis/Structural | Molecule | Dot/Lewis/Structural |

And for N₂O, it says “Sample Problem” and then below it “Dot”, “Lewis”, “Structural” — suggesting that for N₂O, they expect three separate drawings. But for others, it’s not specified.

To be safe, and since the instruction says “diagram the electron dot, Lewis, and structural formulas”, I will assume that for each molecule, we need to provide the Lewis structure (which typically shows bonds as lines and lone pairs as dots), and that satisfies all three in common practice. For simplicity, and because it's a worksheet, I'll draw the Lewis structure with lines for bonds and dots for lone pairs.

Also, note: for some molecules, there might be multiple valid structures, but we’ll draw the most stable/common one.

Let’s tackle each molecule one by one.

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Molecule 1: CH₄ (Methane)

- Carbon has 4 valence electrons, each H has 1.
- Total valence electrons: 4 + 4×1 = 8
- Carbon is central atom, bonded to 4 H atoms.
- Each bond is a single bond (2 electrons shared).
- After forming 4 single bonds, carbon has used all 4 valence electrons, and each H has 2 electrons (duet).
- No lone pairs on carbon or hydrogens.

Lewis structure:

H
|
H - C - H
|
H

With no lone pairs. In dot form, we could show all electrons as dots, but Lewis structure uses lines for bonds.

Since the worksheet likely expects Lewis structure with lines and dots for lone pairs, and here no lone pairs, so just the structural formula with lines.

But to be thorough, let's describe:

- Dot structure: C in center, surrounded by 4 H, each H connected by a pair of dots (or line), and no other dots.
- Lewis: same as above, using lines for bonds.
- Structural: same.

So for CH₄, it's straightforward.

---

Molecule 2: CS₂ (Carbon disulfide)

- C: 4 valence, S: 6 each → total = 4 + 6×2 = 16 valence electrons
- Carbon is central (less electronegative than S)
- If we make single bonds: C-S-S, then C has two bonds, needs more. Better to make double bonds.
- Common structure: S=C=S
- Each double bond is 4 electrons, so two double bonds = 8 electrons used in bonding.
- Remaining electrons: 16 - 8 = 8 electrons → these go as lone pairs on S atoms.
- Each S in double bond has 2 bonds (4 electrons from bonds), so needs 4 more to complete octet → two lone pairs each.
- So each S has two lone pairs.

Lewis structure:

:S = C = S:

But with lone pairs: each S has two pairs, so:

.. ..
: S = C = S :
'' ''

In text:

Left S: two lone pairs (so four dots) and double bond to C
C: double bond to each S, no lone pairs
Right S: two lone pairs and double bond to C

Total electrons: each double bond is 4 electrons × 2 = 8, plus 4 lone pairs (each pair 2e) × 2 S × 2 pairs = 8 electrons? Wait: each S has two lone pairs → 4 electrons per S in lone pairs, so 8 total lone pair electrons, plus 8 bonding electrons = 16. Good.

---

Molecule 3: H₂O₂ (Hydrogen peroxide)

- H: 1 each ×2 = 2, O: 6 each ×2 = 12 → total 14 valence electrons
- Structure: H-O-O-H
- Bonds: O-O single bond, each O-H single bond → that's 3 single bonds = 6 electrons used
- Remaining electrons: 14 - 6 = 8 electrons → these are lone pairs.
- Each O is bonded to two atoms (one H and one O), so each O has two bonds → 4 electrons from bonds, needs 4 more to complete octet → two lone pairs each.
- So each O has two lone pairs.

Lewis structure:

H - O - O - H

With two lone pairs on each O.

So:

H - Ö - Ö - H (where Ö represents O with two lone pairs)

In dots:

Each O has : above and below, or something.

Text representation:

H - :O: - :O: - H

But the bonds are single, and each O has two lone pairs.

Yes.

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Molecule 4: N₂O (Sample Problem) — we don't need to solve, but for reference

It's given as sample, so skip.

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Molecule 5: CCl₄ (Carbon tetrachloride)

- C: 4, Cl: 7 each ×4 = 28 → total 32 valence electrons
- C central, bonded to 4 Cl
- Each bond single → 4 bonds × 2e = 8 electrons used
- Remaining: 32 - 8 = 24 electrons → lone pairs on Cl atoms
- Each Cl has one bond, so needs 6 more electrons → three lone pairs each
- 4 Cl × 3 pairs × 2e = 24 electrons → perfect

Lewis structure:

Cl
|
Cl - C - Cl
|
Cl

Each Cl has three lone pairs (six dots).

---

Molecule 6: CCl₃F (Trichlorofluoromethane)

- Similar to CCl₄, but one Cl replaced by F
- C: 4, Cl: 7×3=21, F:7 → total 4+21+7=32 valence electrons
- C central, bonded to 3 Cl and 1 F
- All single bonds → 4 bonds × 2e = 8e used
- Remaining 24e → lone pairs on halogens
- Each Cl and F has one bond, so each needs three lone pairs (6e)
- 4 atoms × 3 pairs × 2e = 24e → good

Structure:

Cl
|
Cl - C - F
|
Cl

Each halogen has three lone pairs.

---

Molecule 7: CH₃NH₂ (Methylamine)

- This is CH₃-NH₂
- Atoms: C, 3H (from CH₃), N, 2H (from NH₂) → total H:5, C:1, N:1
- Valence electrons: C:4, H:1×5=5, N:5 → total 4+5+5=14
- Structure: C bonded to 3H and to N; N bonded to C and 2H
- Bonds: C-H (3 bonds), C-N (1 bond), N-H (2 bonds) → total 6 single bonds = 12 electrons used
- Remaining: 14 - 12 = 2 electrons → these go as a lone pair on N
- Check octets: C has 4 bonds → 8e, good. Each H has 1 bond → 2e, good. N has 3 bonds (to C and two H) and one lone pair → 3×2 + 2 = 8e, good.

Lewis structure:

H
|
H - C - N - H
| |
H H

But N has a lone pair. So:

H
|
H - C - N - H
| |
H H

And on N, two dots (lone pair).

More accurately:

The carbon is tetrahedral, nitrogen is trigonal pyramidal.

In 2D:

H H
\ /
C - N - H
/ \
H H

But with lone pair on N.

Standard way:

Write as:

H
|
H-C-N-H
| |
H H

And add two dots on N.

Yes.

---

Molecule 8: Cl₂CO (Phosgene)

- Atoms: Cl, Cl, C, O
- Valence electrons: Cl:7×2=14, C:4, O:6 → total 24
- Carbon is central (least electronegative except H, but no H)
- Typically, C double bonded to O, and single bonded to two Cl
- Bonds: C=O double bond (4e), two C-Cl single bonds (4e) → total bonding electrons 8e
- Remaining: 24 - 8 = 16 electrons → lone pairs
- O in double bond has 4 electrons from bond, needs 4 more → two lone pairs
- Each Cl has one bond, needs 6 more → three lone pairs each
- So: O has two lone pairs (4e), each Cl has three lone pairs (6e each → 12e), total lone pair electrons: 4 + 12 = 16e → perfect

Lewis structure:

Cl
\
C = O
/
Cl

With two lone pairs on O, and three lone pairs on each Cl.

In text:

:Cl:
\
C = :O:
/
:Cl:

But with proper dots.

Each Cl has ::: around it, but in diagram, we show three pairs.

---

Now, to present the answers neatly.

Since the worksheet has a table, and for each molecule, we need to fill in the "Dot/Lewis/Structural" column, I will provide the Lewis structure for each, which shows bonds as lines and lone pairs as dots.

For the individual atoms in question 1, I'll list their electron dot symbols.

Let me compile the final answer.

First, for question 1: electron dot structures for atoms

- H: H•
- C: •C• (with dots on four sides)
- N: •N: (with a pair and three singles — specifically, often drawn with pair on top, and singles on left, right, bottom)
To be clear in text:

I'll describe them as:

- H: one dot
- C: four dots, one on each side of C
- N: five dots: one pair and three single dots
- O: six dots: two pairs and two single dots
- F: seven dots: three pairs and one single dot
- S: same as O — six dots: two pairs and two single dots
- Cl: same as F — seven dots: three pairs and one single dot

For molecules, I'll draw the Lewis structures using text diagrams.

Since this is text-based, I'll represent them as best as possible.

Final Answer format:

I need to provide the answers for both parts.

Part 1: Atom dot structures

Part 2: For each molecule, the Lewis structure (which covers dot, Lewis, and structural as per context)

Let me write it out.