how to draw lewis structure, Summaries of Chemistry

Chem 1211k, Topic module 7: How to draw lewis structures

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Drawing Lewis Structure
Unit 3_Module 7
- We know that most atoms are square structure where there is 1s +3p orbitas = 4
sides
- Exception H is a circle because it only has 1s
Ex. C2H4
1.
- Carbon-> Group 14 -> 4 valence electrons
- Hydrogen -> group 1 -> 1 valence electron
o Since we have 2 carbons = 8 electrons
o We have 4 hydrogens -> 4 electrons
o Total = 12 valence electrons = 8+4
2. Draw
- Carbon (4 valence electrons)
• C •
- Hydrogen has 1 valence electron
H•
o Then you draw 2 carbons (each with 4 dots)
o And 4 hydrogens (each with 1 dot)
3. Determine which one is the central atom, and which one is the terminal atom?”
o Core atom has two or more unpaired electrons. Therefore, they can form two
or more bonds and link two or more atoms together like a chain.
o Terminal atoms only have one unpaired electron. Therefore, they can only
form one bond and, as such, will end a chain of atoms
i. In this example the core atom is C
ii. The terminal atom is H
- Start bond core atoms
b. C-C
c. Now each C has 3 unpaired electrons left
4. Attach the terminal atoms
o We know that each H has 1 unpaired electron
o We can attach 2 H to each C
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Drawing Lewis Structure Unit 3_Module 7

  • We know that most atoms are square structure where there is 1s +3p orbitas = 4 sides
  • Exception H is a circle because it only has 1s Ex. C2H
  • Carbon-> Group 14 - > 4 valence electrons
  • Hydrogen - > group 1 - > 1 valence electron o Since we have 2 carbons = 8 electrons o We have 4 hydrogens - > 4 electrons o Total = 12 valence electrons = 8+
  1. Draw
  • Carbon (4 valence electrons)
    • C •

  • Hydrogen has 1 valence electron H• o Then you draw 2 carbons (each with 4 dots) o And 4 hydrogens (each with 1 dot)
  1. Determine “which one is the central atom, and which one is the terminal atom?” o Core atom has two or more unpaired electrons. Therefore, they can form two or more bonds and link two or more atoms together like a chain. o Terminal atoms only have one unpaired electron. Therefore, they can only form one bond and, as such, will end a chain of atoms i. In this example the core atom is C ii. The terminal atom is H
  • Start bond core atoms b. C-C c. Now each C has 3 unpaired electrons left
  1. Attach the terminal atoms o We know that each H has 1 unpaired electron o We can attach 2 H to each C

------à o Since we still have an extra unpaired electron on each C o We will have to form aa double bound i. We made the double bound because: “too many unpaired electrons, and they are adjacent atoms - > form double bonds” Ex #2: CO

  1. Count valence electrons o C = 4 VE o O = 6 VE o Total = 4+6+6=16 electrons
  2. Draw structure o C -
  • C •
  • o O ••
  • O • •• § Observations: O2 has 2 lone pairs, 2 unpaired electrons
  1. Core vs. terminal atom o C-> 4 unpaired = core o C-> 2 unpaired = can be core but is usually a terminal § Structure starts like this: OCO
  1. Are there any unpaired electrons a. if so, it can be double or triple bond
  2. is there a charge in the compound? a. Add/ remove electrons Tips:
  • Period 2 elements (C, N ,O ,F) follow the octet rule (8VE) (lack d orbital do not have room to hold more than 8 electrons)
  • Hydrogen full with 2 VE
  • Period 3+ can expand Octet o P, Cl ,S can often have 10 or 12 e o Use d orbital for extra bonding