Lewis Structures Practice Worksheet - Free Printable

To solve this worksheet, we need to draw the Lewis structures for each compound. A Lewis structure shows how atoms are connected and where the electrons (both bonding and non-bonding) are located.

Here’s how we’ll approach each one:

1. Count total valence electrons.
2. Arrange atoms with single bonds first.
3. Add lone pairs to satisfy octets (or duet for hydrogen).
4. If needed, form double/triple bonds to complete octets.
5. Check formal charges if necessary.

Let’s go step by step for each compound.

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1) PBr₃

- Phosphorus (P) is in group 15 → 5 valence electrons.
- Bromine (Br) is in group 17 → 7 valence electrons each × 3 = 21.
- Total = 5 + 21 = 26 valence electrons.

Structure:
- P is central atom (less electronegative than Br).
- Connect each Br to P with a single bond → uses 3 bonds × 2 e⁻ = 6 e⁻.
- Remaining electrons: 26 - 6 = 20.
- Each Br needs 6 more electrons to complete octet (they already have 2 from bond) → 3 Br × 6 = 18 e⁻.
- Leftover: 20 - 18 = 2 e⁻ → put on P as a lone pair.

Final structure: P bonded to three Br atoms, each Br has 3 lone pairs, P has 1 lone pair.

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2) N₂H₂ (diazene)

- Nitrogen (N) → 5 valence electrons × 2 = 10.
- Hydrogen (H) → 1 valence electron × 2 = 2.
- Total = 12 valence electrons.

Structure:
- H atoms must be terminal (can only form 1 bond).
- So likely: H–N=N–H or H–N–N–H? Let’s try both.

Try H–N–N–H with single bonds:
- Bonds: 3 single bonds (H-N, N-N, N-H) → 6 e⁻ used.
- Remaining: 12 - 6 = 6 e⁻.
- Each N needs 6 more to complete octet? But they’re sharing only 2 electrons so far each → too few.

Better: Try double bond between nitrogens: H–N=N–H
- Bonds: 2 N-H bonds (4 e⁻) + 1 N=N double bond (4 e⁻) → total 8 e⁻ used.
- Remaining: 12 - 8 = 4 e⁻.
- Each N now has 3 bonds (so 6 electrons), needs 2 more → add 1 lone pair to each N → 4 e⁻ used.

Perfect! Structure: H–N=N–H, each N has one lone pair.

Note: This can also exist as cis/trans isomers, but for Lewis structure, just show connectivity and lone pairs.

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3) CH₃OH (methanol)

- Carbon (C) → 4
- Hydrogen (H) → 1 × 4 = 4 (three H on C, one H on O)
- Oxygen (O) → 6
- Total = 4 + 4 + 6 = 14 valence electrons

Structure:
- C is central for methyl group, O is attached to C and H.
- So: C bonded to 3 H and 1 O; O bonded to C and 1 H.
- Bonds: 3 C-H (6 e⁻) + 1 C-O (2 e⁻) + 1 O-H (2 e⁻) → total 10 e⁻ used.
- Remaining: 14 - 10 = 4 e⁻ → go to oxygen as 2 lone pairs.

Check: C has 4 bonds → good. O has 2 bonds + 2 lone pairs → 8 electrons → good. All H have 1 bond → good.

Structure:
H
|
H–C–O–H
|
H
(with two lone pairs on O)

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4) NO₂⁻ (nitrite ion)

- Nitrogen (N) → 5
- Oxygen (O) → 6 × 2 = 12
- Charge -1 → add 1 electron
- Total = 5 + 12 + 1 = 18 valence electrons

Structure:
- N is central.
- Try single bonds: N–O and N–O → 4 e⁻ used.
- Remaining: 18 - 4 = 14.
- Each O needs 6 more → 12 e⁻ → left with 2 e⁻ → put on N? But then N has only 4 electrons (2 bonds + 1 lone pair) → not stable.

Better: Make one double bond.

Option: One N=O double bond, one N–O single bond.

Bonds: 1 double (4 e⁻) + 1 single (2 e⁻) = 6 e⁻ used.
Remaining: 18 - 6 = 12 e⁻.

Assign:
- Double-bonded O: already has 4 e⁻ from bond → needs 4 more → 2 lone pairs.
- Single-bonded O: has 2 e⁻ from bond → needs 6 more → 3 lone pairs.
→ That’s 4 + 6 = 10 e⁻ used for oxygens.
Left: 12 - 10 = 2 e⁻ → put on N as 1 lone pair.

Now check formal charges:
- N: valence 5 - (2 bonds + 2 dots) = 5 - 4 = +1? Wait — actually, in Lewis terms: formal charge = valence - (number of bonds + lone electrons/2)? No — standard formula: FC = V - (L + B/2), where L = number of lone electrons, B = bonding electrons.

Actually simpler:
- For N: it has 1 lone pair (2 e⁻) and 3 bonds (double counts as 2, single as 1 → total 3 bonds → 6 bonding e⁻). Formal charge = 5 - (2 + 6/2) = 5 - (2+3) = 0.
- Double-bonded O: 6 - (4 + 4/2) = 6 - (4+2) = 0.
- Single-bonded O: 6 - (6 + 2/2) = 6 - (6+1) = -1.

So overall charge: 0 + 0 + (-1) = -1 → matches.

But resonance exists: the double bond can be with either oxygen. So we draw two resonance structures.

For simplicity, we can show one with brackets and charge.

Structure: [O=N–O]⁻ with lone pairs: double-bonded O has 2 lone pairs, single-bonded O has 3 lone pairs, N has 1 lone pair.

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5) C₂H₄ (ethylene)

- Carbon (C) → 4 × 2 = 8
- Hydrogen (H) → 1 × 4 = 4
- Total = 12 valence electrons

Structure:
- Two carbons connected.
- Each carbon bonded to 2 hydrogens.
- If all single bonds: C–C and 4 C–H → 5 bonds → 10 e⁻ used.
- Remaining: 2 e⁻ → not enough to give each C an octet (each would have only 3 bonds → 6 e⁻).

So make a double bond between carbons.

Bonds: 1 C=C double bond (4 e⁻) + 4 C–H single bonds (8 e⁻) → total 12 e⁻ → perfect.

Each carbon has 3 atoms attached (2 H and 1 C via double bond) → that’s 3 regions, but double bond counts as one region — wait, no: in terms of electrons, each carbon shares 4 electrons with other carbon (double bond) and 2 with each H → total 8 electrons per carbon → octet satisfied.

Structure: H₂C=CH₂

No lone pairs anywhere.

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Final Answer:

1) PBr₃: Central P with 3 single bonds to Br, each Br has 3 lone pairs, P has 1 lone pair.
2) N₂H₂: H–N=N–H, each N has 1 lone pair.
3) CH₃OH: C bonded to 3 H and 1 O; O bonded to H; O has 2 lone pairs.
4) NO₂⁻: Resonance structures with one N=O, one N–O⁻, N has 1 lone pair; overall charge -1.
5) C₂H₄: H₂C=CH₂, no lone pairs.