Solved Worksheet 1 Molecular Geometry & Polarity For each of ... - Free Printable

Explanation:
Let’s go through each molecule/ion step by step. For each, we need to:

1. Draw the Lewis structure (show valence electrons, bonds, lone pairs — obey octet rule where possible).
2. Use VSEPR theory to predict:
- Molecular geometry (shape of atoms only, ignoring lone pairs)
- Polarity (polar = net dipole ≠ 0; nonpolar = net dipole = 0)
3. Identify intermolecular forces present (only those that actually apply):
- London dispersion (always present, but weakest)
- Dipole–dipole (present in polar molecules)
- Hydrogen bonding (only if H is bonded directly to N, O, or F)
- Ion–dipole (only for ions dissolved in polar solvents — not relevant here unless specified; we’ll assume pure substances)

We’ll do one at a time.

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1. CH₃Br (bromomethane)
- C has 4 valence e⁻, 3H ×1 = 3, Br = 7 → total = 4+3+7 = 14 e⁻
- Central atom: C
- Bonds: C–H (3 bonds), C–Br (1 bond) → 4 single bonds = 8 e⁻ used in bonds
- Remaining 6 e⁻ → 3 lone pairs on Br
- No lone pairs on C → 4 bonding domains → tetrahedral electron geometry → tetrahedral molecular geometry
- C–Br bond is polar (Br more electronegative), and shape is not symmetric (H vs Br), so polar molecule
- Intermolecular forces: London dispersion, dipole–dipole (no H-bonding: H is bonded to C, not N/O/F)

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2. NI₃ (nitrogen triiodide)
- N: 5, I×3 = 3×7 = 21 → total = 26 e⁻
- Central: N
- Bonds: 3 N–I single bonds = 6 e⁻ used
- Remaining 20 e⁻ → 3 lone pairs on each I (3×6=18), plus 2 left → 1 lone pair on N
- So: 3 bonds + 1 lone pair on N → 4 electron domains → tetrahedral electron geometry → trigonal pyramidal molecular geometry
- Polar (lone pair pushes bonds down, asymmetrical) → polar
- Intermolecular forces: London dispersion, dipole–dipole (no H-bonding)

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3. H₂S
- H×2 = 2, S = 6 → total = 8 e⁻
- Central: S
- Bonds: 2 S–H → 4 e⁻ used
- Remaining 4 e⁻ → 2 lone pairs on S
- 2 bonds + 2 lone pairs = 4 domains → tetrahedral electron geometry → bent (V-shaped) molecular geometry
- Polar (bent + lone pairs → dipole doesn’t cancel) → polar
- Intermolecular forces: London dispersion, dipole–dipole (no H-bonding: H bonded to S, not N/O/F)

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4. CS₂
- C: 4, S×2 = 14 → total = 18 e⁻
- Central: C
- Double bonds: C=S (2 double bonds) → each double bond = 4 e⁻ → 8 e⁻ used
- Remaining 10 e⁻ → 2 lone pairs on each S (4×2 = 8), plus 2 left? Wait—better: each S needs 8 e⁻. In C=S, S shares 4 e⁻ in bond, so needs 4 more → 2 lone pairs per S = 8 e⁻. Total used: 8 (bonds) + 8 (lone pairs on S) = 16. But we have 18 e⁻ → mistake.

Let’s recalc properly:
Valence e⁻: C = 4, S = 6 each → 4 + 6 + 6 = 16 e⁻
Yes! 16 total.
C central, two double bonds: C=S and C=S → 2×4 = 8 e⁻ in bonds
Each S gets 2 lone pairs (4 e⁻ each) → 8 e⁻
Total = 8 + 8 = 16 ✓
No lone pairs on C → 2 bonding domains → linear electron & molecular geometry
Symmetric → dipoles cancel → nonpolar
Intermolecular forces: London dispersion only

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5. SO₂
- S: 6, O×2 = 12 → total = 18 e⁻
- Central: S
- Resonance: S double-bonded to one O, single-bonded to other with S having +1 formal charge, O⁻, or better: S has 1 lone pair, 2 bonds (one double, one single with resonance)
Standard Lewis: S with 1 lone pair, 2 bonding regions (but one is double bond) → 3 electron domains (2 bonds + 1 lone pair)
→ trigonal planar electron geometry → bent molecular geometry
Polar (asymmetric, lone pair) → polar
Forces: London, dipole–dipole

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6. CO₃²⁻
- C: 4, O×3 = 18, +2 charge → add 2 e⁻ → total = 24 e⁻
- Central: C
- Resonance: 3 equivalent C–O bonds (average 1.33), one double bond and two single bonds with delocalized charge
- Electron domains: 3 bonding regions, 0 lone pairs on C → trigonal planar geometry
Symmetric → even though bonds are polar, symmetry cancels dipole → nonpolar ion (note: ions are always charged, but *molecular polarity* refers to dipole moment; CO₃²⁻ has no permanent dipole due to symmetry)
But wait: for *molecular polarity*, polyatomic ions can be nonpolar if symmetric. Yes, CO₃²⁻ is nonpolar (dipole = 0).
Intermolecular forces: as an ion, in solid/solution it experiences ion–ion or ion–dipole, but for pure substance (e.g., in crystal), dominant is ionic bonding. However, the question asks for *intermolecular forces* — for ions, we usually say: ion–ion forces (if solid), but many textbooks list for ions: ionic forces (not technically intermolecular, but accepted here). Since the table includes NH₄⁺, likely they expect: for ions → ion–ion (or just “ionic”) forces. We’ll use ionic forces for CO₃²⁻ and NH₄⁺.

But let’s check standard approach in high school: For polyatomic ions, when considering forces *between ions*, it's ionic bonding. So for CO₃²⁻ (in a salt like Na₂CO₃), forces are ionic. Since the question says “intermolecular forces”, and includes ions, we’ll list: ionic forces.

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7. CCl₂O (phosgene)
- C: 4, Cl×2 = 14, O = 6 → total = 24 e⁻
- Central: C
- Structure: C double-bonded to O, single-bonded to two Cl
- No lone pairs on C → 3 bonding domains → trigonal planar geometry
- Polar? C=O is very polar, C–Cl also polar; geometry is trigonal planar but not symmetric (O vs two Cl) → net dipole → polar
- Forces: London, dipole–dipole

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8. OF₂ (oxygen difluoride)
- O: 6, F×2 = 14 → total = 20 e⁻
- Central: O
- Bonds: 2 O–F single bonds → 4 e⁻ used
- Remaining 16 e⁻ → 3 lone pairs on O (6 e⁻), and 3 lone pairs on each F (6×2 = 12) → total 6+12=18 + 4 = 22? Mistake.

Correct: 20 e⁻ total.
2 bonds = 4 e⁻
Each F needs 3 lone pairs = 6 e⁻ each → 12 e⁻
So far 4 + 12 = 16
Remaining 4 e⁻ → 2 lone pairs on O
So O has 2 bonds + 2 lone pairs = 4 domains → tetrahedral electron geometry → bent molecular geometry
Polar (bent, lone pairs, F more electronegative than O) → polar
Forces: London, dipole–dipole (no H-bonding: no H!)

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9. AsCl₃ (arsenic trichloride)
- As: 5, Cl×3 = 21 → total = 26 e⁻
- Central: As
- 3 As–Cl bonds = 6 e⁻
- Remaining 20 e⁻ → 3 lone pairs on each Cl (18 e⁻), 2 left → 1 lone pair on As
- So 3 bonds + 1 lone pair → tetrahedral electron geometry → trigonal pyramidal
- Polar → polar
- Forces: London, dipole–dipole

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10. NH₄⁺ (ammonium ion)
- N: 5, H×4 = 4, minus 1 e⁻ for +1 charge → total = 8 e⁻
- Central: N
- 4 N–H bonds = 8 e⁻ used
- No lone pairs on N → 4 bonding domains → tetrahedral geometry
- Symmetric → even though bonds are polar, dipoles cancel → nonpolar ion (no net dipole)
- As an ion: ionic forces (when in compound, e.g., NH₄Cl)

Now compile all:

| Molecule/ion | Electron Dot Structure (description) | Molecular Geometry | Polarity | Intermolecular Forces |
|--------------|----------------------------------------|--------------------|----------|------------------------|
| CH₃Br | C central, 3 H, 1 Br; Br has 3 lone pairs | Tetrahedral | Polar | London, dipole–dipole |
| NI₃ | N central, 3 I, 1 lone pair on N | Trigonal pyramidal | Polar | London, dipole–dipole |
| H₂S | S central, 2 H, 2 lone pairs on S | Bent | Polar | London, dipole–dipole |
| CS₂ | S=C=S, 2 lone pairs on each S | Linear | Nonpolar | London only |
| SO₂ | S central, 2 O, 1 lone pair on S | Bent | Polar | London, dipole–dipole |
| CO₃²⁻ | Resonance, 3 O around C, no lone pairs on C | Trigonal planar | Nonpolar (symmetric) | Ionic forces |
| CCl₂O | C central, 2 Cl, 1 O (double bond), no lone pairs on C | Trigonal planar | Polar | London, dipole–dipole |
| OF₂ | O central, 2 F, 2 lone pairs on O | Bent | Polar | London, dipole–dipole |
| AsCl₃ | As central, 3 Cl, 1 lone pair on As | Trigonal pyramidal | Polar | London, dipole–dipole |
| NH₄⁺ | N central, 4 H, no lone pairs | Tetrahedral | Nonpolar | Ionic forces |

Double-check polarity of CO₃²⁻ and NH₄⁺:
- CO₃²⁻: symmetric trigonal planar, bond dipoles cancel → no net dipole → nonpolar *molecule/ion* (though charged, dipole moment = 0)
- NH₄⁺: symmetric tetrahedral, bond dipoles cancel → no net dipole → nonpolar ion

Intermolecular forces for ions: In context of this worksheet, they expect “ionic” for ions. Some might say “ion–ion”, but “ionic forces” is standard shorthand.

Final Answer:
[See table below — filled completely]

Since the user expects the final answer in a specific format (just the completed table), and the instruction says “Provide only the final correct answer” after reasoning, here is the clean table:

Final Answer:
Molecule or ion | Electron Dot Structure | Molecular Geometry | Molecular Polarity | Intermolecular Forces
CH₃Br | C bonded to 3H and 1Br; Br has 3 lone pairs | Tetrahedral | Polar | London dispersion, dipole–dipole
NI₃ | N bonded to 3I; N has 1 lone pair | Trigonal pyramidal | Polar | London dispersion, dipole–dipole
H₂S | S bonded to 2H; S has 2 lone pairs | Bent | Polar | London dispersion, dipole–dipole
CS₂ | S=C=S; each S has 2 lone pairs | Linear | Nonpolar | London dispersion
SO₂ | S bonded to 2O (resonance); S has 1 lone pair | Bent | Polar | London dispersion, dipole–dipole
CO₃²⁻ | C bonded to 3O (resonance); C has no lone pairs | Trigonal planar | Nonpolar | Ionic forces
CCl₂O | C bonded to 2Cl and double-bonded to O; C has no lone pairs | Trigonal planar | Polar | London dispersion, dipole–dipole
OF₂ | O bonded to 2F; O has 2 lone pairs | Bent | Polar | London dispersion, dipole–dipole
AsCl₃ | As bonded to 3Cl; As has 1 lone pair | Trigonal pyramidal | Polar | London dispersion, dipole–dipole
NH₄⁺ | N bonded to 4H; N has no lone pairs | Tetrahedral | Nonpolar | Ionic forces