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"Write Lewis structures that obey the octet rule for each of the following. Except for HCN and H₂CO, the first atom listed is the central atom. For HCN and H₂CO, carbon is the central atom. (a) HCN (b) PH₃ (c) CHCl₃ (d) NH₄⁺ (e) H₂CO (f) SeF₂ (g) CO₂ (h) O₂ (i) HBr"
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Step 1:
I'll solve this step by step, focusing on drawing Lewis structures that obey the octet rule for each molecule/ion.

(a) HCN (Hydrogen Cyanide)

Step 2:
: Identify the central atom

- Carbon is the central atom

Step 3:
: Count total valence electrons

- Total: $$1 + 4 + 5 = 10$$ valence electrons
- H: 1 electron - C: 4 electrons - N: 5 electrons

Step 4:
: Draw Lewis structure

H - C \equiv N:
- Single bond between H and C - Triple bond between C and N - Nitrogen has a lone pair (b) PH₃ (Phosphine)

Step 5:
: Identify the central atom

- Phosphorus (P) is the central atom

Step 6:
: Count total valence electrons

- Total: $$5 + (3 \times 1) = 8$$ valence electrons
- P: 5 electrons - H: 3 × 1 electron

Step 7:
: Draw Lewis structure

H - P:
- Three single bonds from P to H - One lone pair on P (c) CHCl₃ (Chloroform)

Step 8:
: Identify the central atom

- Carbon (C) is the central atom

Step 9:
: Count total valence electrons

- Total: $$4 + (3 \times 7) = 25$$ valence electrons
- C: 4 electrons - Cl: 3 × 7 electrons

Step 10:
: Draw Lewis structure

H - C(Cl)_{3}:
- Single bonds from C to three Cl atoms - Single bond from C to H - Chlorine atoms have three lone pairs each (d) NH₄⁺ (Ammonium ion)

Step 11:
: Identify the central atom

- Nitrogen (N) is the central atom

Step 12:
: Count total valence electrons

- Total: $$5 + (4 \times 1) - 1 = 8$$ valence electrons
- N: 5 electrons - H: 4 × 1 electron - Charge: + 1 (removes one electron)

Step 13:
: Draw Lewis structure

H H
| | - Four single bonds from N to H atoms (e) H₂CO (Formaldehyde)

Step 14:
: Identify the central atom

- Carbon (C) is the central atom

Step 15:
: Count total valence electrons

- Total: $$4 + 6 + (2 \times 1) = 12$$ valence electrons
- C: 4 electrons - O: 6 electrons - H: 2 × 1 electron

Step 16:
: Draw Lewis structure

H - C = O:
- Double bond between C and O - Two single bonds from C to H atoms (f) SeF₂ (Selenium difluoride)

Step 17:
: Identify the central atom

- Selenium (Se) is the central atom

Step 18:
: Count total valence electrons

- Total: $$6 + (2 \times 7) = 20$$ valence electrons
- Se: 6 electrons - F: 2 × 7 electrons

Step 19:
: Draw Lewis structure

F - Se - F:
- Two single bonds from Se to F atoms - Two lone pairs on Se (g) CO₂ (Carbon dioxide)

Step 20:
: Identify the central atom

- Carbon (C) is the central atom

Step 21:
: Count total valence electrons

- Total: $$4 + (2 \times 6) = 16$$ valence electrons
- C: 4 electrons - O: 2 × 6 electrons

Step 22:
: Draw Lewis structure

O = C = O
- Two double bonds between C and O atoms (h) O₂ (Oxygen molecule)

Step 23:
: Count total valence electrons

- Total: $$2 \times 6 = 12$$ valence electrons
- O: 2 × 6 electrons

Step 24:
: Draw Lewis structure

O \equiv O
- Double bond between O atoms (i) HBr (Hydrogen bromide)

Step 25:
: Draw Lewis structure

H - Br:
- Single bond between H and Br - Three lone pairs on Br

Final Answer

Lewis structures have been drawn for each molecule/ion, following the octet rule and showing all valence electrons.