Worksheet featuring Lewis dot structures for molecules like CH₄, H₂O, CO₂, and others, highlighting central atoms, bonding types, and octet rules.

Problem Overview:

The task involves constructing Lewis dot structures for a variety of molecules and ions, analyzing their bonding, resonance, and molecular geometry. The worksheet provides hints about the central atoms, bond types, and exceptions to the octet rule. Below, I will solve the problem by explaining how to construct Lewis structures for each molecule/ion and determine their properties.

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General Steps for Constructing Lewis Structures:

1. Determine the total number of valence electrons:
- Sum the valence electrons from all atoms.
- Add extra electrons for negative charges or subtract for positive charges.

2. Choose the central atom:
- Typically, the least electronegative atom is the central atom (except for hydrogen, which is never central).
- For polyatomic ions, consider the charge distribution.

3. Draw single bonds to connect atoms:
- Each bond represents 2 electrons.
- Ensure that all atoms achieve an octet (except for hydrogen, which needs 2 electrons).

4. Distribute remaining electrons as lone pairs:
- Place lone pairs on terminal atoms first, then on the central atom if needed.
- Fulfill the octet rule where possible.

5. Check for resonance:
- If multiple valid Lewis structures can be drawn, the molecule/ion exhibits resonance.

6. Determine the formal charge:
- Calculate formal charge for each atom to ensure the most stable structure.

7. Analyze molecular geometry:
- Use VSEPR theory to predict the shape based on electron pair repulsion.

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Detailed Solutions:

#### Row 1:
1. CH₄:
- Central atom: Carbon (C)
- Total valence electrons: \(4 + 4 \times 1 = 8\)
- Structure: C bonded to 4 H atoms with no lone pairs.
- Geometry: Tetrahedral

2. H₂O:
- Central atom: Oxygen (O)
- Total valence electrons: \(6 + 2 \times 1 = 8\)
- Structure: O bonded to 2 H atoms with 2 lone pairs on O.
- Geometry: Bent

3. CO₂:
- Central atom: Carbon (C)
- Total valence electrons: \(4 + 2 \times 6 = 16\)
- Structure: C double-bonded to each O atom.
- Geometry: Linear

4. N₂:
- Central atom: Nitrogen (N) (diatomic molecule)
- Total valence electrons: \(2 \times 5 = 10\)
- Structure: N triple-bonded to N with 1 lone pair on each N.
- Geometry: Linear

5. BeCl₂:
- Central atom: Beryllium (Be)
- Total valence electrons: \(2 + 2 \times 7 = 16\)
- Structure: Be bonded to 2 Cl atoms with no lone pairs on Be.
- Geometry: Linear (note: Be does not follow the octet rule)

#### Row 2:
6. BF₃:
- Central atom: Boron (B)
- Total valence electrons: \(3 + 3 \times 7 = 24\)
- Structure: B bonded to 3 F atoms with no lone pairs on B.
- Geometry: Trigonal planar (note: B does not follow the octet rule)

7. C₂H₄:
- Central atoms: Two carbon atoms (C=C double bond)
- Total valence electrons: \(2 \times 4 + 4 \times 1 = 12\)
- Structure: C=C double bond with H atoms attached.
- Geometry: Planar

8. C₂H₆:
- Central atoms: Two carbon atoms (C-C single bond)
- Total valence electrons: \(2 \times 4 + 6 \times 1 = 14\)
- Structure: C-C single bond with H atoms attached.
- Geometry: Tetrahedral

9. CO:
- Central atom: Carbon (C)
- Total valence electrons: \(4 + 6 = 10\)
- Structure: C triple-bonded to O with 1 lone pair on O.
- Geometry: Linear

10. O₂:
- Central atom: Oxygen (O) (diatomic molecule)
- Total valence electrons: \(2 \times 6 = 12\)
- Structure: O=O double bond with 2 lone pairs on each O.
- Geometry: Linear

#### Row 3:
11. NO:
- Central atom: Nitrogen (N)
- Total valence electrons: \(5 + 6 = 11\) (odd number)
- Structure: N bonded to O with 2 lone pairs on N and 2 lone pairs on O.
- Geometry: Linear

12. NO₂⁻:
- Central atom: Nitrogen (N)
- Total valence electrons: \(5 + 2 \times 6 + 1 = 18\)
- Structure: Resonance between two equivalent forms (N=O and N-O single bond).
- Geometry: Bent

13. NO₃⁻:
- Central atom: Nitrogen (N)
- Total valence electrons: \(5 + 3 \times 6 + 1 = 24\)
- Structure: Resonance among three equivalent forms (N=O and N-O single bond).
- Geometry: Trigonal planar

14. NH₃:
- Central atom: Nitrogen (N)
- Total valence electrons: \(5 + 3 \times 1 = 8\)
- Structure: N bonded to 3 H atoms with 1 lone pair on N.
- Geometry: Trigonal pyramidal

15. NH₄⁺:
- Central atom: Nitrogen (N)
- Total valence electrons: \(5 + 4 \times 1 - 1 = 8\)
- Structure: N bonded to 4 H atoms with no lone pairs.
- Geometry: Tetrahedral

#### Row 4:
16. O₃:
- Central atom: Oxygen (O)
- Total valence electrons: \(3 \times 6 = 18\)
- Structure: Resonance between two equivalent forms (O=O and O-O single bond).
- Geometry: Bent

17. ClF₃:
- Central atom: Chlorine (Cl)
- Total valence electrons: \(7 + 3 \times 7 = 28\)
- Structure: Cl bonded to 3 F atoms with 2 lone pairs on Cl.
- Geometry: T-shaped

18. SO₄²⁻:
- Central atom: Sulfur (S)
- Total valence electrons: \(6 + 4 \times 6 + 2 = 32\)
- Structure: S bonded to 4 O atoms with no lone pairs on S.
- Geometry: Tetrahedral

19. SF₆:
- Central atom: Sulfur (S)
- Total valence electrons: \(6 + 6 \times 7 = 48\)
- Structure: S bonded to 6 F atoms with no lone pairs on S.
- Geometry: Octahedral

20. SF₄:
- Central atom: Sulfur (S)
- Total valence electrons: \(6 + 4 \times 7 = 34\)
- Structure: S bonded to 4 F atoms with 1 lone pair on S.
- Geometry: See-saw

#### Row 5:
21. I₃⁻:
- Central atom: Iodine (I)
- Total valence electrons: \(2 \times 7 + 7 + 1 = 22\)
- Structure: Resonance between two equivalent forms (I-I single bond and I-I single bond with a shared lone pair).
- Geometry: Linear

22. XeCl₂:
- Central atom: Xenon (Xe)
- Total valence electrons: \(8 + 2 \times 7 = 22\)
- Structure: Xe bonded to 2 Cl atoms with 3 lone pairs on Xe.
- Geometry: Linear

23. PF₅:
- Central atom: Phosphorus (P)
- Total valence electrons: \(5 + 5 \times 7 = 40\)
- Structure: P bonded to 5 F atoms with no lone pairs on P.
- Geometry: Trigonal bipyramidal

24. CO₃²⁻:
- Central atom: Carbon (C)
- Total valence electrons: \(4 + 3 \times 6 + 2 = 24\)
- Structure: Resonance among three equivalent forms (C=O and C-O single bond).
- Geometry: Trigonal planar

25. BrF₅:
- Central atom: Bromine (Br)
- Total valence electrons: \(7 + 5 \times 7 = 42\)
- Structure: Br bonded to 5 F atoms with 1 lone pair on Br.
- Geometry: Square pyramidal

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Final Answer:

The detailed Lewis structures and analyses are provided above. The key takeaway is to carefully apply the steps for constructing Lewis structures, paying attention to exceptions like incomplete octets, resonance, and formal charges.

Final Answer:
\boxed{\text{See detailed solutions above}}