Question

1. write lewis structures that obey the octet rule for each of the following: 3

a) chcl3 b) nh4+
c) h2co

Explanation:

Step1: Calculate total valence electrons for CHCl₃

C:4, H:1, 3Cl:3×7=21 → Total=4+1+21=26
Central atom: C. Bond C to H and 3Cl (4 bonds=8 e⁻ for C). Each Cl has 3 lone pairs (6 e⁻), H has 0.
Lewis structure: C at center, single bonds to H and 3Cl; each Cl has 3 lone pairs.

Step2: Calculate total valence electrons for NH₄⁺

N:5, 4H:4×1=4, minus 1 charge → Total=5+4-1=8
Central atom: N. Bond N to 4H (4 bonds=8 e⁻ for N). No lone pairs.
Lewis structure: N at center, single bonds to 4H; bracket with + charge.

Step3: Calculate total valence electrons for H₂CO

C:4, 2H:2×1=2, O:6 → Total=4+2+6=12
Central atom: C. Bond C to 2H (single bonds) and double bond to O (C has 8 e⁻). O has 2 lone pairs.
Lewis structure: C at center, single bonds to 2H, double bond to O; O has 2 lone pairs.

Answer:

a)

      Cl
      |
Cl — C — H
      |
      Cl
(Each Cl has 3 lone pairs)

b)

    [H]
    |
H — N — H⁺
    |
    H
(Brackets around structure with + charge)

c)

    O
    ||
H — C — H
(O has 2 lone pairs)