Covalent Bonding Worksheet featuring chemical structures and electron dot notation for various molecules.

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

We are given a table with molecules on the left and their Lewis structures (with electron dots) on the right. Our job is to check if each Lewis structure correctly shows all valence electrons for that molecule — including bonding pairs and lone pairs.

We’ll go row by row, count total valence electrons from atoms, then check if the drawing matches.

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Row 1: H₂O

Atoms: 2 H + 1 O
Valence electrons: H has 1 each → 2×1 = 2; O has 6 → total = 8

In the drawing:
- Two single bonds (O–H), each bond = 2 electrons → 4 electrons used in bonds
- Oxygen has two lone pairs → 4 more electrons
Total shown: 4 (bonds) + 4 (lone pairs) = 8 ✔ Correct!

But wait — look at the drawing provided: it shows O with only ONE lone pair? Let me recheck...

Actually, looking again — in the image, H₂O is drawn as:

H – O – H
:

That’s only 2 electrons as lone pair on O? That would be wrong.

Wait — no, actually in standard notation, “:” means one lone pair (2 electrons). But oxygen needs TWO lone pairs (4 electrons) to have 8 total.

So if the drawing shows only one lone pair (:), that’s only 2 electrons → total electrons shown = 4 (bonds) + 2 (lone pair) = 6 ✘ Missing 2 electrons.

BUT — let’s double-check the actual image description you gave. You wrote:

> H₂O → H–O–H with : under O

If “:” is meant to represent two dots (one lone pair), then yes — missing one more lone pair.

However, sometimes in shorthand, people write “:” meaning two dots, but oxygen should have four dots (two lone pairs).

Looking back at your original text:

You wrote for H₂O:
“H–O–H” and below O there’s “:”

That likely means only 2 electrons shown as lone pair → incorrect.

But wait — maybe it's a typo in how we’re reading it? Let’s assume the drawing intended to show two lone pairs. In many textbooks, water is drawn with two lone pairs on oxygen.

Given that this is a worksheet asking to identify errors, and other rows clearly have mistakes, perhaps H₂O is correct? Let’s hold off and check others first.

Actually — let’s count properly based on what’s written.

In your transcription:

For H₂O:
Structure: H–O–H with “:” under O → that’s 2 electrons as lone pair.

Total valence electrons needed: 8
Used in bonds: 2 bonds × 2e = 4e
Lone pairs shown: 2e
Total shown: 6e → missing 2e → ERROR.

So H₂O is incorrectly drawn — oxygen should have two lone pairs (four dots), not one.

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Row 2: N₂

Atoms: 2 N
Each N has 5 valence electrons → total = 10

Drawing: N≡N with “:” on each N? Wait — you wrote:

“N≡N” and then “:N≡N:” ? Actually you said:

> N₂ → :N≡N:

That would mean triple bond (6 electrons) plus one lone pair on each nitrogen (2e each → 4e total) → 6+4=10 ✔ Correct!

Yes — nitrogen gas is N≡N with one lone pair on each N. So this is correct.

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Row 3: NH₃

Atoms: 1 N + 3 H
Valence: N=5, H×3=3 → total=8

Drawing: You wrote “H–N–H” with another H and “..” above N?

Actually you transcribed:

> NH₃ → H–N–H with H attached and “..” above N

Assuming it’s pyramidal: N bonded to three H, and one lone pair (“..”) on N.

Bonds: 3 single bonds → 6 electrons
Lone pair: 2 electrons
Total: 8 ✔ Correct!

Yes — ammonia has 3 bonds and 1 lone pair on nitrogen → perfect.

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Row 4: HF

Atoms: H + F
Valence: H=1, F=7 → total=8

Drawing: H–F with “:::” around F? You wrote:

> HF → H–F with “:::” — probably meaning three lone pairs on F?

Fluorine in HF: single bond (2e), and three lone pairs (6e) → total 8e ✔ Correct!

Yes — fluorine always has 3 lone pairs when bonded once.

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Row 5: CCl₄

Atoms: 1 C + 4 Cl
Valence: C=4, Cl×4=28 → total=32

Drawing: Central C with four Cl atoms, each Cl has “:::” (three lone pairs)?

Each Cl–C bond = 2e → 4 bonds = 8e
Each Cl has 3 lone pairs → 6e per Cl → 4×6=24e
Total: 8 + 24 = 32 ✔ Correct!

Carbon has no lone pairs — good, since it’s making 4 bonds.

This is correct.

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Row 6: CO₂

Atoms: 1 C + 2 O
Valence: C=4, O×2=12 → total=16

Drawing: You wrote “O=C=O” — double bonds to each oxygen.

Each double bond = 4e → two double bonds = 8e
Now, each oxygen should have two lone pairs (since double-bonded oxygen typically has two lone pairs).

So each O: 2 lone pairs = 4e → two O’s = 8e
Total: 8 (bonds) + 8 (lone pairs) = 16 ✔ Correct!

But wait — in your transcription, did they draw lone pairs?

You wrote for CO₂: “O=C=O” — no mention of lone pairs.

Ah — here’s the issue! If the drawing shows ONLY “O=C=O” with NO lone pairs on oxygens, then:

Bonds: 8e
Lone pairs: 0e → total shown = 8e ✘ Missing 8 electrons!

Oxygen MUST have lone pairs. Each oxygen in CO₂ has two lone pairs.

So if the diagram doesn’t show them, it’s WRONG.

Looking back at your original text:

> CO₂ → O=C=O

No dots mentioned → so likely missing lone pairs → ERROR.

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Row 7: CH₄

Atoms: 1 C + 4 H
Valence: C=4, H×4=4 → total=8

Drawing: C in center, four H’s, no lone pairs.

Bonds: 4 single bonds → 8e
No lone pairs → total 8e ✔ Correct!

Methane is perfectly drawn like this.

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Row 8: Cl₂

Atoms: 2 Cl
Valence: 7×2=14

Drawing: Cl–Cl with “:::” on each Cl? You wrote:

> Cl₂ → Cl–Cl with “:::” — assuming three lone pairs per Cl

Bond: 2e
Each Cl has 3 lone pairs → 6e each → 12e
Total: 2 + 12 = 14 ✔ Correct!

Yes — chlorine molecule has single bond and three lone pairs per atom.

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Now let’s summarize which ones are INCORRECT:

From above:

- H₂O: Drawn with only one lone pair on O → should have two → ✘ Incorrect
- CO₂: Drawn as O=C=O with no lone pairs → each O needs two lone pairs → ✘ Incorrect
- All others seem correct.

Wait — let’s recheck H₂O.

In your initial transcription, you wrote for H₂O:

> H–O–H with “:” under O

If “:” represents two electrons (one lone pair), then yes — missing one more lone pair.

But sometimes in quick sketches, “:” might imply two dots, but still — oxygen needs four dots.

Similarly, for CO₂ — definitely missing lone pairs.

Are there any others?

What about N₂? We said :N≡N: — that’s correct.

NH₃ — correct.

HF — correct.

CCl₄ — correct.

CH₄ — correct.

Cl₂ — correct.

So only H₂O and CO₂ are wrong.

But let’s make sure about H₂O.

Standard Lewis structure for water:

H
\
O:
/
H

With two lone pairs on O — so four dots.

If the drawing shows only two dots (one pair), it’s missing two electrons.

Same for CO₂ — must show lone pairs on oxygens.

Therefore, the molecules with incorrect Lewis structures are:

→ H₂O
→ CO₂

Final Answer:
H₂O and CO₂