VSEPR Theory: Molecules with Lone Pairs Worksheet for 9th - 12th ... - Free Printable

Let’s work through each molecule step by step using the VSEPR theory and the instructions in the worksheet.

We’re given a table to fill out for 5 molecules: H₂O, H₃O⁺, OF₂, BF₃, OCl₂.

For each, we need to find:
- Total valence electrons
- Lewis structure (we’ll describe it)
- Number of bonding pairs
- Number of lone pairs on central atom
- Electron geometry
- Molecular shape
- Bond angle (approximate)
- Polar or nonpolar?

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1. H₂O (Water) — Already filled in as example.

Total valence e⁻: 8
Lewis: O in center, bonded to two H, two lone pairs on O
Bonding pairs: 2
Lone pairs on central atom: 2
Electron geometry: tetrahedral (4 electron domains)
Molecular shape: bent
Bond angle: 105° (less than 109.5° due to lone pair repulsion)
Polar? Yes — asymmetric, dipoles don’t cancel

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2. H₃O⁺ (Hydronium ion)

Step 1: Count valence electrons
H = 1 × 3 = 3
O = 6
Charge = +1 → subtract 1 electron
Total = 3 + 6 - 1 = 8 valence electrons

Step 2: Lewis structure
O is central. Bonded to 3 H atoms. That uses 3 bonds × 2 e⁻ = 6 e⁻.
Remaining electrons: 8 - 6 = 2 → that’s 1 lone pair on O.
But wait — O has 3 bonds and 1 lone pair → total 4 electron domains → correct.

Step 3: Bonding pairs = 3
Lone pairs on central atom = 1

Step 4: Electron geometry = tetrahedral (4 domains)

Step 5: Molecular shape = trigonal pyramidal (because one domain is lone pair)

Step 6: Bond angle ≈ 107° (slightly less than 109.5° due to lone pair)

Step 7: Polar? Yes — asymmetrical, dipoles don’t cancel

✔ Filled row:
| H₃O⁺ | 8 | O bonded to 3 H, 1 lone pair on O | 3 | 1 | tetrahedral | trigonal pyramidal | ~107° | polar |

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3. OF₂ (Oxygen difluoride)

Step 1: Valence electrons
O = 6
F = 7 × 2 = 14
Total = 6 + 14 = 20 valence electrons

Step 2: Lewis structure
O is central (less electronegative than F? Actually F is more electronegative, but O is usually central in such cases).
O bonded to two F atoms → 2 bonds = 4 e⁻ used.
Remaining: 20 - 4 = 16 e⁻ → distribute as lone pairs.
Each F needs 3 lone pairs (6 e⁻ each) → 2 F × 6 = 12 e⁻
Left: 16 - 12 = 4 e⁻ → 2 lone pairs on O.

So: O has 2 bonding pairs and 2 lone pairs → same as water!

Step 3: Bonding pairs = 2
Lone pairs on central atom = 2

Step 4: Electron geometry = tetrahedral

Step 5: Molecular shape = bent

Step 6: Bond angle ≈ 103°–105° (similar to water, maybe slightly less because F is more electronegative)

Step 7: Polar? Yes — bent shape, dipoles don’t cancel

✔ Filled row:
| OF₂ | 20 | O bonded to 2 F, 2 lone pairs on O | 2 | 2 | tetrahedral | bent | ~103° | polar |

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4. BF₃ (Boron trifluoride)

Step 1: Valence electrons
B = 3
F = 7 × 3 = 21
Total = 3 + 21 = 24 valence electrons

Step 2: Lewis structure
B is central. Bonded to 3 F atoms → 3 bonds = 6 e⁻ used.
Remaining: 24 - 6 = 18 e⁻ → each F gets 3 lone pairs (6 e⁻ each) → 3 × 6 = 18 → perfect.
Boron has only 3 bonds → no lone pairs. But boron is an exception — it can be stable with 6 electrons (incomplete octet).

Step 3: Bonding pairs = 3
Lone pairs on central atom = 0

Step 4: Electron geometry = trigonal planar (3 domains)

Step 5: Molecular shape = trigonal planar

Step 6: Bond angle = 120° (perfect trigonal planar)

Step 7: Polar? No — symmetrical, dipoles cancel out

✔ Filled row:
| BF₃ | 24 | B bonded to 3 F, no lone pairs on B | 3 | 0 | trigonal planar | trigonal planar | 120° | nonpolar |

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5. OCl₂ (Oxygen dichloride)

Step 1: Valence electrons
O = 6
Cl = 7 × 2 = 14
Total = 6 + 14 = 20 valence electrons

Step 2: Lewis structure
O is central. Bonded to two Cl atoms → 2 bonds = 4 e⁻ used.
Remaining: 20 - 4 = 16 e⁻
Each Cl needs 3 lone pairs (6 e⁻ each) → 2 × 6 = 12 e⁻
Left: 16 - 12 = 4 e⁻ → 2 lone pairs on O.

Same as OF₂ and H₂O!

Step 3: Bonding pairs = 2
Lone pairs on central atom = 2

Step 4: Electron geometry = tetrahedral

Step 5: Molecular shape = bent

Step 6: Bond angle ≈ 110°? Wait — actually, Cl is larger and less electronegative than F, so bond angle might be a bit larger than in OF₂. Typical value for OCl₂ is about 110.9°, but for school level, we can say ~110° or just “bent, ~105–110°”. Let’s use ~110° for accuracy.

Step 7: Polar? Yes — bent shape, dipoles don’t cancel

✔ Filled row:
| OCl₂ | 20 | O bonded to 2 Cl, 2 lone pairs on O | 2 | 2 | tetrahedral | bent | ~110° | polar |

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Now let’s compile all answers neatly:

Final Answer:

| Molecule | Total Valence Electrons | Lewis Structure Description | Bonding Pairs | Lone Pairs on Central Atom | Electron Geometry | Molecular Shape | Bond Angle | Polar or Nonpolar |
|----------|--------------------------|--------------------------------------------------|---------------|----------------------------|---------------------|------------------------|----------------|-------------------|
| H₂O | 8 | O bonded to 2 H, 2 lone pairs on O | 2 | 2 | tetrahedral | bent | 105° | polar |
| H₃O⁺ | 8 | O bonded to 3 H, 1 lone pair on O | 3 | 1 | tetrahedral | trigonal pyramidal | ~107° | polar |
| OF₂ | 20 | O bonded to 2 F, 2 lone pairs on O | 2 | 2 | tetrahedral | bent | ~103° | polar |
| BF₃ | 24 | B bonded to 3 F, no lone pairs on B | 3 | 0 | trigonal planar | trigonal planar | 120° | nonpolar |
| OCl₂ | 20 | O bonded to 2 Cl, 2 lone pairs on O | 2 | 2 | tetrahedral | bent | ~110° | polar |

This completes the table accurately based on VSEPR theory and electron counting.