N₂ (Nitrogen molecule)

Here are the Lewis structures for N$_2$, O$_2$, Cl$_2$, and CO$_2$:

• N$_2$ (Nitrogen molecule)

  1. Total valence electrons: Each nitrogen atom has 5 valence electrons. So, $2 \times 5 = 10$ valence electrons.
  2. Draw a single bond between the two nitrogen atoms: N-N (2 electrons used, 8 remaining).
  3. Distribute the remaining electrons to satisfy octets. To achieve an octet, each N needs 6 more electrons. This requires forming a triple bond.
  $$\boxed{:\text{N}\equiv\text{N}:}$$

  Each nitrogen atom has 6 electrons from the triple bond and 2 electrons from its lone pair, totaling 8 electrons.

• O$_2$ (Oxygen molecule)

  1. Total valence electrons: Each oxygen atom has 6 valence electrons. So, $2 \times 6 = 12$ valence electrons.
  2. Draw a single bond between the two oxygen atoms: O-O (2 electrons used, 10 remaining).
  3. Distribute the remaining 10 electrons as lone pairs. If 6 electrons are placed on each oxygen, that's 12 electrons, which is too many. Place 6 on one O and 4 on the other.
  4. The oxygen with only 6 electrons (2 from bond + 4 from lone pairs) needs 2 more. Form a double bond by moving a lone pair from the other oxygen.
  $$\boxed{:\text{O}=\text{O}:}$$

  Each oxygen atom has 4 electrons from the double bond and 4 electrons from its two lone pairs, totaling 8 electrons.

• Cl$_2$ (Chlorine molecule)

  1. Total valence electrons: Each chlorine atom has 7 valence electrons. So, $2 \times 7 = 14$ valence electrons.
  2. Draw a single bond between the two chlorine atoms: Cl-Cl (2 electrons used, 12 remaining).
  3. Distribute the remaining 12 electrons as lone pairs to satisfy octets. Each chlorine atom needs 6 more electrons.
  $$\boxed{:\text{Cl}-\text{Cl}:}$$

  Each chlorine atom has 2 electrons from the single bond and 6 electrons from its three lone pairs, totaling 8 electrons.

• CO$_2$ (Carbon dioxide)

  1. Total valence electrons: Carbon has 4, and each oxygen has 6. So, $4 + (2 \times 6) = 16$ valence electrons.
  2. Carbon is the central atom. Draw single bonds to the oxygen atoms: O-C-O (4 electrons used, 12 remaining).
  3. Distribute the remaining 12 electrons as lone pairs to the outer oxygen atoms first. Each oxygen needs 6 more electrons. :O-C-O: (6 electrons on each oxygen, 12 used. Central carbon has 4 electrons).
  4. The central carbon atom only has 4 electrons. To achieve an octet, it needs 4 more electrons. Move one lone pair from each oxygen atom to form double bonds with the carbon atom.
  $$\boxed{:\text{O}=\text{C}=\text{O}:}$$

  Each oxygen atom has 4 electrons from the double bond and 4 electrons from its two lone pairs, totaling 8 electrons. The central carbon atom has 4 electrons from each double bond, totaling 8 electrons.