S. 8.3 part 4.
Lewis dot structures: Diagram representing the electron arrangement of ionic and covalent compounds.
Lewis dot structures for simple ionic compounds.
Steps:
1. Determine the charge of each ion.
2. Place single ion in the middle surrounded around the most abundant ion.
3. Place electrons around the ions.
Ex. 1: Lewis dot structure for CaBr2.
Ex. 2: Lewis dot structure for Li3N.
Questions:
- Pg. 183, # 85.
Lewis Structures of Covalent Compounds that obey the Octet Rule.
- Show how the Valence electrons are distributed in a molecule.
Satisfying Octet.
- All elements will have eight valence electrons.
Steps:
1) Count the total number of valence electrons (subtract one electron for every positive charge
and add one electron for every negative charge)
2) Atom with subscript value of 1 is the central atom and arrange other atoms around it.
3) Determine which atoms are bonding.
4) 2 electrons for every bond (2 e = 1 bond, 4 e = 2 bonds)
5) Use the remaining electrons to fill the elements valence orbitals.
6) If central atom is deficient (Less than 8 electrons) in electrons pull electrons to share from
bonding elements.
7) Replace electron pairs with dashed lines to indicate bonds.
Ex 1: Lewis dot structure for NH3
Steps:
1) N: 5 + 3H: 3 = 8 electrons.
2) Nitrogen is central atom, place the three hydrogen’s on three of Nitrogen’s four
sides.
3) Place two electrons in between bonding atoms, three hydrogen’s bonding with one
nitrogen.
4) Replace bonding electrons with dashed lines, bonding electrons equals 6.
5) Find remaining electrons to place around central atom by subtracting line 4 value
from line 1. 8 – 6 = 2. Place two electrons on the Nitrogens side with no
hydrogens.
6) Nitrogen has 8 valence electrons and each hydrogen has two valence electrons. All
is correct.
Ex. 2: Lewis dot structure for CNSteps:
1) C: 1 + N: 1 + 1 e-= 10 electrons.
2) Carbon and Nitrogen are placed horizontally beside each other.
3) Place two electrons in between bonding atoms.
4) Replace bonding electrons with dashed lines = 2 electrons.
5) Find remaining electrons to place around central atom by subtracting line 4 value
from line 1. 10 – 2 = 8. Place 4 electrons around carbon and nitrogen.
6) Both elements only have five electrons, place one electron from each element into
bonding area between elements.
7) Both elements have 7 electrons, place one electron from each element into bonding
area between elements.
8) Both elements now have full octets.
9) Enclose molecule in end brackets and place a superscript negative sign on the upper
right hand corner to represent the extra electron.
Ex. 3: Draw the Lewis dot structure for HOPO.
***Note*** Both 6a and 6b are correct. In nature there will be a mixture of both 6a and 6b with the
double bond changing back and forth.
Violation of the octet rule:
Electron deficient molecules: Molecule with one or more atoms that does not possess a full octet.
H, Be, B, and Al are octet rule exceptions.
- These electrons can only gain one electron in a covalent bond for every electron they can
contribute to the bond.
- Be has 2 valence electrons and share a maximum of 4 electrons.
- B and Al have 3 valence electrons and can share a maximum of 6 electrons.
Ex. 1: Draw the Lewis dot structure for BCl3.
All 24 electrons are used. Boron has a maximum of three bonds due to having 3 valence electrons, which
are the only electrons capable of bonding.
Elements lower than the second period (expanded Octet of Valence Electrons).
- Can attain more than an octet since electrons are placed in low – lying (lower level) d- and forbitals.
- Lewis dot structures apply.
Ex. 1: Draw the Lewis dot structure for SF6.
Ex. 2: Draw the Lewis Dot structure for H3PO4.
Questions:
- Pg. 188 # 86.
- Lewis dot structure quiz next class.