NOTES: CHEMICAL BONDING
Ionic Bonding
Example: calcium fluoride
CaF2
Metallic Bonding
http://www.drkstreet.com/resources/metallic-bonding-animation.swf
delocalized electrons:
“electron sea” model:
Alloys:
Covalent Bonding: (molecular compounds – non-metals ONLY!)
1. Diatomic Elements (BrINClHOF)
Cl2
O2
2. Nonpolar Covalent Bonds vs Polar Covalent Bonds:
F—F
H--F
3. Nonpolar Covalent vs Polar Covalent Molecules
Nonpolar vs. Polar Molecules: What’s
the difference?
2. Drawing Molecular Compounds
ELEMENT
Usual # of
BONDS
C
TRY: CF4, NI3, H2CO
N
O
H
Halogens**
3. Multiple Bonds
Bond
Pair of Electrons
Order
single
double
triple
Bond Length
Bond Strength
4. Drawing other compounds:
1. Put the most “needy” element in the center
2. Arrange atoms as symmetrically as possible
3. Count the total number of valence electrons available.
4. Count the total number of electrons each element needs for a complete octet. (8/element, EXCEPT:
H needs 2, Boron needs 6).
5. Subtract the total number available from the total number needed (#4 - #3) to get the number of
electrons in bonds. Divide by two to get the number of bonds.
6. Subtract the number of electrons in bonds from the total available to determine the number of
electrons in lone pair. Distribute the lone pair so that elements achieve their perfect outer shell.
RULES NOT TO BREAK: H can only bond 1x; oxygen can only bond 2x.
PRACTICE: Draw the Lewis Structure for each of the molecules/polyatomic ions below below: Then indicate
whether each molecule is polar or nonpolar.
F2CS
SO2
C2H2
SO3
O22-
ClO31-
PO43-
NH4+
HOMEWORK: 1. Draw Lewis diagrams for each species below.
2. Label each molecule as polar or nonpolar. The terms polar and nonpolar do not apply to ions.
a. H2O2
2H=2
2 O = 12
14-6 = 8
polar
b. HOBr
H=1
O=6
Br = 7
14 – 4 = 10
polar
c. NH3
N=5
3H = 3
8–6=2
Polar
d. CS2
C=4
2S = 12
16 – 4 = 12
nonpolar
e. C2F4
f. N2O (one of the Ns is central)
2C = 8
4 F = 28
36 – 10 = 26
Nonpolar
2N = 10
O=6
16-4=12
polar
g. SeO2
h. PO3 -3
Se = 6
2 O = 12
18 – 4 = 14
polar
P=5
3O= 18
-3 = 3
26 – 6 = 20
i. HCN
H=1
C=4
N=5
10 – 4 = 6
polar
j. CO32C=4
3 O = 18
2- = 2
24 – 6 = 18
k. NO2N=5
2 O = 12
-= 1
18 -4 = 14
l. N2O4
2 N = 10
4 O = 24
34 – 10 = 24
nonpolar