Step-by-step guide to drawing Lewis structures for common molecules, showing valence electrons and bonding patterns.

Let’s go through each molecule one by one and draw the correct Lewis structure step by step. We’ll count valence electrons, connect atoms with bonds, and add lone pairs to satisfy octets (or duet for hydrogen).

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a) HBr

- Hydrogen (H) has 1 valence electron.
- Bromine (Br) has 7 valence electrons.
- Total = 1 + 7 = 8 electrons

We connect H and Br with a single bond (uses 2 electrons).
That leaves 6 electrons → all go on Br as 3 lone pairs.

✔ Final structure:
H — Br: (with 3 lone pairs on Br)

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b) SCl₂

- Sulfur (S) has 6 valence electrons.
- Each Cl has 7 → 2 × 7 = 14
- Total = 6 + 14 = 20 electrons

Connect S to each Cl with a single bond → uses 4 electrons.
Left: 20 - 4 = 16 electrons → distribute as lone pairs.

Each Cl needs 3 lone pairs (6 electrons each) → 2 Cl × 6 = 12 electrons used.
Left: 16 - 12 = 4 electrons → put on S as 2 lone pairs.

✔ Final structure:
:Cl — S — Cl: (each Cl has 3 lone pairs, S has 2 lone pairs)

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c) PBr₃

- Phosphorus (P) has 5 valence electrons.
- Each Br has 7 → 3 × 7 = 21
- Total = 5 + 21 = 26 electrons

Connect P to each Br with a single bond → 3 bonds = 6 electrons used.
Left: 26 - 6 = 20 electrons.

Each Br needs 3 lone pairs (6 electrons) → 3 Br × 6 = 18 electrons.
Left: 20 - 18 = 2 electrons → put on P as 1 lone pair.

✔ Final structure:
Br with 3 lone pairs bonded to P, which also has 1 lone pair and two more Br atoms (same).

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d) C₂H₂ (acetylene)

- Each C has 4 → 2 × 4 = 8
- Each H has 1 → 2 × 1 = 2
- Total = 8 + 2 = 10 electrons

If we try single bonds: H—C—C—H → that’s 3 bonds = 6 electrons.
Left: 4 electrons → if we put them as lone pairs on carbons, each carbon only has 6 electrons total → not enough!

So we need multiple bonds. Try triple bond between carbons:

H—C≡C—H → that’s 1 bond from each H to C (2 bonds) + triple bond (3 bonds) = 5 bonds = 10 electrons → perfect!

No lone pairs needed — each carbon has 8 electrons (triple bond gives 6, plus 1 bond to H gives 2 more → 8 total).

✔ Final structure:
H — C ≡ C — H

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e) NO (nitric oxide)

- Nitrogen (N) has 5
- Oxygen (O) has 6
- Total = 5 + 6 = 11 electrons ← odd number! So it will be a radical.

Try single bond: N—O → uses 2 electrons. Left: 9.

Put 3 lone pairs on O (6 electrons), 1 lone pair and 1 unpaired electron on N → total 2+6+2+1=11.

But oxygen wants 8, nitrogen wants 8 — let’s try double bond:

N=O → uses 4 electrons. Left: 7.

Put 2 lone pairs on O (4 electrons), 1 lone pair and 1 unpaired electron on N → total 4+4+2+1=11.

Now check octets:
- O: double bond (4) + 2 lone pairs (4) = 8 ✔
- N: double bond (4) + 1 lone pair (2) + 1 unpaired = 7 ✘ → but this is okay because NO is a known radical.

Actually, the most stable structure has a double bond and the unpaired electron on nitrogen.

✔ Final structure:
:N = Ö· (double bond, O has 2 lone pairs, N has 1 lone pair and 1 unpaired electron)

Sometimes written as ·N=O: with dots showing the unpaired electron.

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f) O₃ (ozone)

- Each O has 6 → 3 × 6 = 18 electrons

Connect in a chain: O—O—O → 2 bonds = 4 electrons used.
Left: 14 electrons.

Give each end O 3 lone pairs (6 each) → 12 electrons.
Left: 14 - 12 = 2 → put on center O as 1 lone pair.

Now check octets:
- End O’s: 1 bond (2) + 3 lone pairs (6) = 8 ✔
- Center O: 2 bonds (4) + 1 lone pair (2) = 6 ✘ → not enough!

So we make a double bond with one side.

Try: O=O—O → now bonds use 6 electrons (double + single).
Left: 18 - 6 = 12.

Give left O (double-bonded): 2 lone pairs (4 electrons)
Right O (single-bonded): 3 lone pairs (6 electrons)
Center O: already has 3 bonds? Wait — better to think resonance.

Standard ozone structure: central O bonded to two others — one double, one single, and the single-bonded O has extra negative charge, central has positive, but we can show resonance.

Better way:

Draw O—O—O with one double bond and one single bond, and move lone pairs so that:

- One end O: double bond + 2 lone pairs → 8 electrons
- Other end O: single bond + 3 lone pairs → 8 electrons (but carries formal negative)
- Central O: 1 double + 1 single = 3 bonds → 6 electrons + 1 lone pair = 8? Wait — 3 bonds = 6 electrons, plus 1 lone pair = 2 → total 8 ✔

But formal charges: central O has 3 bonds → normally should have 2 → so +1 charge. Single-bonded O has 3 lone pairs + 1 bond → 7 electrons assigned → -1 charge.

Total charge: 0 → good.

And we show resonance: the double bond can be on either side.

✔ Final structure (one resonance form):
:Ö — O⁺ = Ö:⁻ ↔ :Ö:⁻ — O⁺ = Ö:

But often drawn with dashed lines or just one form with arrows.

For simplicity, we can write:

::O — O = O:: with lone pairs adjusted and note resonance.

Actually, standard drawing:

Left O: 3 lone pairs, single bond to center → formal charge -1
Center O: 1 lone pair, double bond to right, single to left → formal charge +1
Right O: 2 lone pairs, double bond → formal charge 0

Or vice versa.

✔ So final answer for ozone: two resonance structures where the double bond switches sides.

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Final Answer:

a) H—Br: (Br has 3 lone pairs)
b) :Cl—S—Cl: (each Cl has 3 lone pairs, S has 2 lone pairs)
c) Br₃P: (P has 1 lone pair, each Br has 3 lone pairs)
d) H—C≡C—H
e) ·N=O: (with N having 1 lone pair and 1 unpaired electron, O having 2 lone pairs)
f) Resonance structures of ozone: [::O⁻—O=O:: ↔ ::O=O⁺—O⁻::]