Solved Name: Chemistry 63 EXPERIMENT #15 MOLECULAR SHAPE & | Chegg.com - Free Printable

Let’s go through each molecule one by one to check if the answers in the table are correct. We’ll look at:

1. # of valence electrons – total from all atoms.
2. Lewis Structure – how atoms are bonded and where lone pairs are.
3. Bond Polarity (P/NP) – is the bond between two different atoms? If yes, it’s polar (unless electronegativity difference is very small).
4. Shape of Molecule – based on VSEPR theory (electron domains around central atom).
5. Molecular Polarity (P/NP) – does the shape cancel out bond dipoles? If not, molecule is polar.

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1. H₂O

- Valence e⁻: H=1×2 = 2, O=6 → Total = 8 ✔
- Lewis: O in center, two H’s, two lone pairs on O ✔
- Bond polarity: O-H bonds are polar (O more electronegative) ✔
- Shape: Bent or “V-shape” (tetrahedral electron geometry, bent molecular) ✔
- Molecular polarity: Not symmetric → polar ✔
→ All correct.

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2. HBr

- Valence e⁻: H=1, Br=7 → Total = 8 ✔
- Lewis: H–Br with 3 lone pairs on Br ✔
- Bond polarity: H-Br is polar (Br more electronegative) ✔
- Shape: Linear (only two atoms) ✔
- Molecular polarity: Only one bond → polar ✔
→ All correct.

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3. NH₃

- Valence e⁻: N=5, H=1×3 = 3 → Total = 8 ✔
- Lewis: N in center, three H’s, one lone pair on N ✔
- Bond polarity: N-H bonds are polar ✔
- Shape: Trigonal pyramidal (tetrahedral electron geometry, one lone pair) ✔
- Molecular polarity: Not symmetric → polar ✔
→ All correct.

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4. CH₄

- Valence e⁻: C=4, H=1×4 = 4 → Total = 8 ✔
- Lewis: C in center, four H’s, no lone pairs ✔
- Bond polarity: C-H bonds are *almost* nonpolar (small electronegativity difference), but often considered nonpolar for simplicity ✔
- Shape: Tetrahedral ✔
- Molecular polarity: Symmetric → nonpolar ✔
→ All correct.

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5. CO₂

- Valence e⁻: C=4, O=6×2 = 12 → Total = 16 ✔
- Lewis: O=C=O, each O has two lone pairs ✔
- Bond polarity: C=O bonds are polar (O more electronegative) ✔
- Shape: Linear ✔
- Molecular polarity: Dipoles cancel → nonpolar ✔
→ All correct.

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6. H₂S

- Valence e⁻: H=1×2 = 2, S=6 → Total = 8 ✔
- Lewis: S in center, two H’s, two lone pairs on S ✔
- Bond polarity: S-H bonds are polar (S more electronegative than H) ✔
- Shape: Bent (“V-shape”) — similar to water ✔
- Molecular polarity: Not symmetric → polar ✔
→ All correct.

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7. H₂CO (formaldehyde)

- Valence e⁻: H=1×2 = 2, C=4, O=6 → Total = 12 ✔
- Lewis: C in center, double bond to O, single bonds to two H’s. O has two lone pairs. ✔
- Bond polarity: C=O and C-H bonds — C=O is very polar, C-H slightly polar → overall bonds are polar ✔
- Shape: Trigonal planar (C has 3 electron domains: double bond counts as one) ✔
- Molecular polarity: Not symmetric (O pulls harder) → polar ✔
→ All correct.

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8. CH₃Cl (chloromethane)

- Valence e⁻: C=4, H=1×3 = 3, Cl=7 → Total = 14 ✔
- Lewis: C in center, three H’s, one Cl. Cl has three lone pairs. ✔
- Bond polarity: C-Cl is polar, C-H slightly polar → overall polar bonds ✔
- Shape: Tetrahedral (C has 4 single bonds) ✔
- Molecular polarity: Not symmetric (Cl vs H) → polar ✔
→ All correct.

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9. CCl₄

- Valence e⁻: C=4, Cl=7×4 = 28 → Total = 32 ✔
- Lewis: C in center, four Cl’s, each Cl has three lone pairs ✔
- Bond polarity: C-Cl bonds are polar ✔
- Shape: Tetrahedral ✔
- Molecular polarity: Symmetric → dipoles cancel → nonpolar ✔
→ All correct.

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✔ Every row in your table is correctly filled out!

Final Answer:
All entries in the table are correct. No changes needed.