Systematic Procedure for Drawing Lewis Structures of Covalent Compounds ( handout #5 : Lewis Structures )

Overview of Lewis Structures for Covalent Molecules

Simple Concept:
- A Lewis structure is a 2D drawing that maps how atoms share valence electrons to form bonds.
- Core Goal: Atoms share electrons so each atom achieves a stable, full valence shell (resembling a noble gas).
Visual Key:
- Single Bond ( $-$ -): $1$ shared pair of electrons ( $2\,\text{e}^-$ ).
- Double Bond ( $=$ ): $2$ shared pairs of electrons ( $4\,\text{e}^-$ ).
- Triple Bond ( $\equiv$ ): $3$ shared pairs of electrons ( $6\,\text{e}^-$ ).
- Lone Pair ( $\cdot\cdot$ ): $2$ non-bonding valence electrons belonging exclusively to one atom.
Visual Diagram Breakdown:

         Lone Pair (Unshared Dots)
                 ..
             H - O - H
               /   \
        Single Bonds (Shared Lines)

Step 1: Determining Total Valence Electrons

Count Your Electron Budget:
- Sum the valence electrons from each atom in the molecule based on its periodic table group number.
- Elements in Group $1$ provide $1\,\text{e}^-$ , Group $14$ provide $4\,\text{e}^-$ , Group $15$ provide $5\,\text{e}^-$ , Group $16$ provide $6\,\text{e}^-$ , and Group $17$ provide $7\,\text{e}^-$ .
Adjust for Ionic Charge:
- Negative Charge (Anion): Add $1\,\text{e}^-$ for each negative charge unit.
- Positive Charge (Cation): Subtract $1\,\text{e}^-$ for each positive charge unit.
Visual Budget Example for Water ( $H_2O$ ):

  Hydrogen (H):  2 atoms x 1 e- = 2 e-
  Oxygen (O):    1 atom  x 6 e- = 6 e-
  ------------------------------------
  Total Electron Budget         = 8 e-

Step 2: Determining Atomic Arrangement

Place Atoms in the Molecular Framework:
- Central Atom: Position the least electronegative element in the middle (the element farthest from fluorine).
- Hydrogen Rule: Hydrogen ( $\text{H}$ ) can never be the central atom because it forms only one bond.
- Terminal (Outer) Atoms: Arrange the remaining atoms symmetrically around the central atom.
Visual Skeletal Arrangement:

  Water (H2O):           Carbon Dioxide (CO2):
    H   O   H                  O   C   O
  (O in center)              (C in center)

Step 3: Forming Initial Chemical Bonds

Connect Outer Atoms with Single Bonds:
- Draw a single line between the central atom and each surrounding outer atom.
- Each single bond costs $2\,\text{e}^-$ from your total electron budget.
- Subtract all bonding electrons to find your remaining pool of electrons.
Visual Single Bond Connection:

  Water (H2O):
    H - O - H
    Budget used: 2 single bonds x 2 e- = 4 e-
    Electrons remaining: 8 e- - 4 e-   = 4 e-

  Carbon Dioxide (CO2):
    O - C - O
    Budget used: 2 single bonds x 2 e- = 4 e-
    Electrons remaining: 16 e- - 4 e-  = 12 e-

Step 4: Satisfying Outer Atom Electron Shells

Distribute Remaining Electrons to Outer Atoms First:
- Place remaining electrons as lone pairs (dots) on outer atoms until each has a complete valence shell.
- Duet Rule: Hydrogen ( $\text{H}$ ) requires only $2\,\text{e}^-$ (fully satisfied by its single bond).
- Octet Rule: Carbon ( $\text{C}$ ), Nitrogen ( $\text{N}$ ), Oxygen ( $\text{O}$ ), and Halogens require an octet of $8\,\text{e}^-$ .
Visual Outer Shell Distribution:

  Water (H2O):
    H is satisfied with 2 e-.
    Place remaining 4 e- as lone pairs on central Oxygen:
        ..
    H - O - H
        ..
    (All 8 e- placed)

  Carbon Dioxide (CO2):
    Distribute 12 remaining e- to outer Oxygens:
    ..         ..
   :O - C - O:
    ..         ..
    (Both outer Oxygens now have 8 e-)

Step 5: Evaluating Central Atom Octets and Multiple Bonding

Inspect the Central Atom:
- Count the total electrons surrounding the central atom (bonding pairs plus lone pairs).
Outcome A — Complete Octet ( $8\,\text{e}^-$ ):
- If the central atom has $8\,\text{e}^-$ , the structure is complete (for example, $H_2O$ ).
Outcome B — Incomplete Octet (Fewer than $8\,\text{e}^-$ ):
- If the central atom has fewer than $8\,\text{e}^-$ , shift lone pairs from outer atoms inward to form double or triple bonds.
- Share electrons evenly from the outer atoms.
- Halogen Restriction: Never use electron pairs from halogens to create multiple bonds.
Visual Multiple Bond Transformation for Carbon Dioxide ( $CO_2$ ):

  Incomplete Central Octet (C has only 4 e-):
    ..         ..
   :O - C - O:
    ..         ..

  Move 1 lone pair inward from each Oxygen atom:
    ..       ..
    O = C = O
    ..       ..

  Completed Molecule Verification:
  - Left Oxygen:   2 bonds (4 e-) + 2 lone pairs (4 e-) = 8 e-
  - Central Carbon: 4 bonds (8 e-)                       = 8 e-
  - Right Oxygen:  2 bonds (4 e-) + 2 lone pairs (4 e-) = 8 e-
  All octets are fully satisfied.