Ch7- Ionic Bonds
Valence electrons- electrons in the highest energy level
(only s & p)
- # of valence electrons corresponds to the group.
1
H
2
Li Be
Na Mg
lose electrons to
form positive ions
called cations.
18
13 14 15 16 17 He
B C N O F Ne
Al Si P S Cl Ar
gain electrons to form
negative ions called
anions.
(A Negative ION)
All atoms are trying to satisfy the octet rule
- get 8 electrons in outer energy level.
Ex1) Write the electron configurations and dot diagrams for:
Na
Cl
Write the e.c. & dot diagram for the ions:
Na+ :
Cl- :
What is the chemical formula for sodium chloride?
Ex1) Write the electron configurations and dot diagrams for:
Na
Cl
[Ne] 3s1
[Ne] 3s2 3p5
Write the e.c. & dot diagram for the ions:
Na+ : [Ne] or [Na]+
Cl- : [Ne] 3s2 3p6 or [Cl]-
What is the chemical formula for sodium chloride? NaCl
Ex 2) Write the electron configure & dot diagram for:
Al
Br
Write the e.c. & d.d. for the ions:
Al+3
Br-1
What is the chemical formula for aluminum bromide?
Ex 2) Write the electron configure & dot diagram for:
.
Al. .
[Ne] 3s23p1
..Br. ..
..
[Ar] 4s23d104p5
Write the e.c. & d.d. for the ions:
Al+3 = [Ne]+3
Br-1 = [Ar] 4s23d104p6 or [Kr]-1
What is the chemical formula for aluminum bromide?
AlBr3
Ex 3) Same stuff for:
Atoms:
Mg
N
Ions:
Formula:
Ex 3) Same stuff for:
Atoms:
Ions:
Mg..
.
Mg.
. .
Mg
[Ne]3s2
Mg+2 = [Ne]+2
. ..
N..
[He] 2s22p3
N-3 = [He] 2s22p6 = [Ne]-3
Formula: Mg3N2
. ..
N. .
.. ..
N. Properties of Ionic Compounds
- solids at room temp
.
Mg.
- don’t conduct electricity in solid state
- will conduct electricity if
1 molten (liquid state)
Ch7 HW#1
2 dissolved in water
in both these cases the ions are free to move.
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
2) Dot structures & electron configurations:
K
3) Gain or Lose?
K
C
C
Mg
O
Mg 
O
4) E.C. for ions
K+1
C+4
C-4
Mg+2
5) Why do nonmetals form anions when reacting?
O-2
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
1
4
2
6
2) Dot structures & electron configurations:
K
3) Gain or Lose?
K
C
C
Mg
O
Mg 
O
4) E.C. for ions
K+1
C+4
C-4
Mg+2
5) Why do nonmetals form anions when reacting?
O-2
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
1
4
2
6
2) Dot structures & electron configurations:
K
[Ar] 4s1
3) Gain or Lose?
K
C
[He] 2s22p2
Mg
[Ne] 3s2
O
[Ne] 2s22p4
C
Mg 
O
4) E.C. for ions
K+1
C+4
C-4
Mg+2
5) Why do nonmetals form anions when reacting?
O-2
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
1
4
2
6
2) Dot structures & electron configurations:
K
[Ar] 4s1
3) Gain or Lose?
K  Lose 1
C
[He] 2s22p2
Mg
[Ne] 3s2
C  gain or lose 4 elec
O
[Ne] 2s22p4
Mg  lose 2 O  gain 2
4) E.C. for ions
K+1
C+4
C-4
Mg+2
5) Why do nonmetals form anions when reacting?
O-2
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
1
4
2
6
2) Dot structures & electron configurations:
K
[Ar] 4s1
3) Gain or Lose?
K  Lose 1
C
[He] 2s22p2
Mg
[Ne] 3s2
C  gain or lose 4 elec
O
[Ne] 2s22p4
Mg  lose 2 O  gain 2
4) E.C. for ions
K+1 [Ar]
C+4 [He]
Mg+2 [Ne]
C-4 [He] 2s22p6
[Ne]
5) Why do nonmetals form anions when reacting?
O-2 [Ne]2s22p4
[Ne]
Ch7 HW#1
1) How many valence electrons for:
a) Potassium b) Carbon
c) Magnesium d) Oxygen
1
4
2
6
2) Dot structures & electron configurations:
K
[Ar] 4s1
3) Gain or Lose?
K  Lose 1
C
[He] 2s22p2
Mg
[Ne] 3s2
C  gain or lose 4 elec
O
[Ne] 2s22p4
Mg  lose 2 O  gain 2
4) E.C. for ions
K+1 [Ar]
C+4 [He]
Mg+2 [Ne]
C-4 [He] 2s22p6
[Ne]
5) Why do nonmetals form anions when reacting?
Have 4 or more electron in outer energy level
Easier to gain electrons to satisfy octet rule.
O-2 [Ne]2s22p4
[Ne]
6) Dot formula to determine formulas:
a) potassium & iodine
b) Ca & S
K
I
Ca
c) Al & O
Al
Al
7) Name:
S
d) Na & P
O
O
O
Na
Na
Na
P
6) Dot formula to determine formulas:
a) potassium & iodine
b) Ca & S
K
I
Ca
KCl
c) Al & O
Al
Al
7) Name:
S
CaS
d) Na & P
O
O
O
Na
Na
Na
P
6) Dot formula to determine formulas:
a) potassium & iodine
b) Ca & S
K
I
Ca
KCl
c) Al & O
CaS
d) Na & P
Al
O
Al
O
O
7) Name:
S
Al2O3
Na
Na
Na
Na3P
P
6) Dot formula to determine formulas:
a) potassium & iodine
b) Ca & S
K
I
Ca
KCl
c) Al & O
S
CaS
d) Na & P
Al
O
Al
O
Al2O3
Na
Na
Na
Na3P
P
O
7) Name:
Potassium iodide
Calcium sulfide
Aluminum oxide
Sodium phosphate
Ch7.2 – Metallic Compounds
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
- have free floating valence electrons
- good conductors of heat and electricity
- if electrons enter one end, others will exit other end.
(basis of electric circuits)
- ductility and malleability caused by electrons shielding cations
from each other, even when the metal is smashed or bent.
(ionic solids shatter because like charges get pressed
together, then repel.)
Metallic Structures: (common, simple ones)
Body-centered cubes (BCC)
- every atom has 8 neighbors
- Na,K,Fe,Cr,W
Face-centered cubic (FCC)
- every atom has 12 neighbors
- Cu,Ag,Au,Al,Pb
Hexagonal Closest Packing (HCP)
-12 neighbors, but diff shape
- Mg,Zn,Cd
Quiz tomorrow:
1.Balance equation:
2.Mass-mass: 10 grams of ____ reacts with excess ____.
How much ____ produced?
3.Density: Givens: grams and cm3
4.Temp conversion: ˚C = ___K
CH7 HW#2 8-12
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
b) Gold (I) Au+1
Au: [Xe] 6s24f145d9
c) Cadmium Cd+2
Cd: [Kr] 5s24d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
10) Ductile Malleable 11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
b) Manganese Mn [Ar] 4s23d5
c) Iron
Fe: [Ar] 4s23d6
[Ar] 4s13d10
6s14f145d10
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
[Ar] 4s13d10
b) Gold (I) Au+1
Au: [Xe] 6s24f145d9
c) Cadmium Cd+2
Cd: [Kr] 5s24d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
10) Ductile Malleable 11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
b) Manganese Mn [Ar] 4s23d5
c) Iron
Fe: [Ar] 4s23d6
6s14f145d10
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
[Ar] 4s13d10
b) Gold (I) Au+1
c) Cadmium Cd+2
Cu+1: [Ar] 3d10
Au: [Xe] 6s24f145d9
6s14f145d10
Au+1: [Xe] 4f145d10
Cd: [Kr] 5s24d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
10) Ductile Malleable 11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
b) Manganese Mn [Ar] 4s23d5
c) Iron
Fe: [Ar] 4s23d6
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
[Ar] 4s13d10
Cu+1: [Ar] 3d10
b) Gold (I) Au+1
Au: [Xe] 6s24f145d9
6s14f145d10
Au+1: [Xe] 4f145d10
c) Cadmium Cd+2 Cd: [Kr] 5s24d10
Cd+2 [Kr] 4d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
Hg+2: [Xe] 4f145d10
10) Ductile Malleable 11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
b) Manganese Mn [Ar] 4s23d5
c) Iron
Fe: [Ar] 4s23d6
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
[Ar] 4s13d10
Cu+1: [Ar] 3d10
b) Gold (I) Au+1
Au: [Xe] 6s24f145d9
6s14f145d10
Au+1: [Xe] 4f145d10
c) Cadmium Cd+2 Cd: [Kr] 5s24d10
Cd+2 [Kr] 4d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
Hg+2: [Xe] 4f145d10
10) Ductile - a metal can be drawn into a wire
Malleable - a metal can be pounded into a sheet (or shape)
11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
b) Manganese Mn [Ar] 4s23d5
c) Iron
Fe: [Ar] 4s23d6
Ch7 HW#2 8 – 12
8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons
9) Electron configurations:
a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9
[Ar] 4s13d10
Cu+1: [Ar] 3d10
b) Gold (I) Au+1
Au: [Xe] 6s24f145d9
6s14f145d10
Au+1: [Xe] 4f145d10
c) Cadmium Cd+2 Cd: [Kr] 5s24d10
Cd+2 [Kr] 4d10
d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10
Hg+2: [Xe] 4f145d10
10) Ductile - a metal can be drawn into a wire
Malleable - a metal can be pounded into a sheet (or shape)
11) Electron configurations for +3 charged:
a) Chromium Cr: [Ar] 4s23d4
Cr+3: [Ar] 3d3
b) Manganese Mn [Ar] 4s23d5
Mn+3: [Ar] 3d4
c) Iron
Fe: [Ar] 4s23d6
Fe+3: [Ar] 3d5
12) Use dots to combine:
a) Cu(I) & Cl:
Cu Cl
b) Cu(II) & Cl:
Cu
Cl
Cl
c) Fe (II) & O
Fe
d) Fe (III) & O
O
Fe
O
Fe
O
O
12) Use dots to combine:
a) Cu(I) & Cl:
Cu Cl
CuCl
b) Cu(II) & Cl:
Cu
Cl
CuCl2
Cl
c) Fe (II) & O
Fe
d) Fe (III) & O
O
Fe
O
Fe
O
O
12) Use dots to combine:
a) Cu(I) & Cl:
Cu Cl
CuCl
b) Cu(II) & Cl:
Cu
Cl
CuCl2
Cl
c) Fe (II) & O
Fe
d) Fe (III) & O
O
FeO
Fe
O
Fe
O
O
Fe2O3
Ch7 Rev WS
1. Use the periodic table to find the number of valence electrons in an atom.
a. sodium
b. carbon
c. phosphorus
Draw the electron dot formulas of these representative elements:
K
AI
O
CI
2. Describe the formation of a cation from an atom of a metallic element,
using the octet rule and the importance of noble-gas electron configurations.
Describe the formation of the sodium ion using an electron dot structure.
3. Describe the formation of an atom of a non metallic element.
Describe the formation of the chloride ion using an electron dot structure.
4. List the characteristics of an ionic bond.
5. Explain the electrical conductivity of melted and of aqueous solutions of
ionic compounds, using the characteristics of ionic compounds.
6.Explain the physical properties of metals, using theory of metallic bonding.
B. Questions
22. Write electron dot structures for the following atoms:
a. silicon b. rubidium c. barium d. tin e. iodine f. arsenic
Si
Rb
Ba
Sn
23. complete the following table.
outer electron
outer electron
config of atom config of ion
Se
K
Ca
Br
N
I
formula of ion
As
type of ion
24. Use electron dot formulas to determine chemical formulas of the ionic
compound formed when the fallowing elements combine.
a. strontium and fluorine
b. magnesium and chlorine
c. sodium and oxygen
25. Same for aluminum and nitrogen and then name it.
Formulas of Ionic Compounds
Elements exchange electrons in ionic bonds . Cations form from metals
by the loss of valence electrons. Anions form from nonmetals by the gain
of electrons. Ionic bonds form as the result of oppositely charged ions
attracting one another. An ionic compound always contains at least
one positive ion (cation) and one negative ion (anion.) These ions must
combine in such a way as to produce a neutral compound. The formula unit
of an ionic compound represents the smallest sample of an ionic compound
that has the composition of that compound. This formula unit will reflect
the balance of charges of the compound’s ions. The use of electron dot
formulas is a helpful tool in predicting formulas of ionic compounds.
This worksheet will help to show you how to write formulas of various
ionic compounds.
1. How many valence electrons does the element iodine have?
What is the formula for iodine’s most stable ion?
Example B
Remember that metals lose valence electrons, and nonmetals gain
electrons in order to achieve electron configurations resembling those
of noble gases. Sufficient numbers of atoms of each element must be
included in the formula so that the number of electrons lost by one
element is equal to the number of electrons gained by the other
element. Sulfur needs two electrons to fill its octet of electrons. Sodium
has only one electron to lose.
2. Determine the formula of the ionic compound formed when barium
and phosphorus combine.
3. How many valence electrons does the element gallium have?
4. Write the formula for the ion formed when nitrogen gains electrons
to attain a noble gas configuration.
5. What is the formula for the compound formed when astatine and
strontium combine?
Lab7.1 – Models
- due in 2 days
- Ch7 Rev WS due at beginning of period
Ch8.1 – Covalent Bonds
Single Covalent Bonds – one pair of electrons shared between 2 atoms.
Ex1) H. + H.
H:H
one shared pair
Ex2) F2
Ex3) H2O
H-H
Structural formula
Each dash represents
one shared pair of electrons.
Ch8.1 – Covalent Bonds
Single Covalent Bonds – one pair of electrons shared between 2 atoms.
Ex1) H. + H.
H:H
one shared pair
Ex2) F2
..
..
:F
... + :F...
H-H
Structural formula
Each dash represents
one shared pair of electrons.
.. ..
:..
F–F
..:
Unshared pairs of electrons
(nonbonding pairs)
Take up more space then the shared pairs.
Ex3) H2O
Double Covalent Bonds – 2 shared pairs of electrons.
Triple Covalent Bonds – 3 shared pairs of electron.
Ex4)
O2
Ex5) N2
Covalent Compounds
Ex6)
NH3
Ex7)
CH4
Ex8)
CO2
Ch8 HW # 1
Ch8 HW#1 1 – 5
1) Dot structures for diatomic molecules
a) Chlorine, Cl2
b) Bromine, Br2
Cl
Cl
Br
Br
c) Iodine, I2
I
I
2) How many unshared pairs are in each halogen molecule
3) Why necessary to form double & triple bond sometimes?
4) How many electrons does each atom contribute in:
Double Bond?
Triple Bond?
Ch8 HW#1 1 – 5
1) Dot structures for diatomic molecules
a) Chlorine, Cl2
b) Bromine, Br2
Cl
Cl
Cl–Cl
Br
Br
Br – Br
c) Iodine, I2
I
I
I–I
2) How many unshared pairs are in each halogen molecule
3) Why necessary to form double & triple bond sometimes?
4) How many electrons does each atom contribute in:
Double Bond?
Triple Bond?
Ch8 HW#1 1 – 5
1) Dot structures for diatomic molecules
a) Chlorine, Cl2
b) Bromine, Br2
Cl
Cl
Cl–Cl
Br
Br
Br – Br
c) Iodine, I2
I
I
I–I
2) How many unshared pairs are in each halogen molecule
3 pairs per atom, 6 pairs per molecule
3) Why necessary to form double & triple bond sometimes?
4) How many electrons does each atom contribute in:
Double Bond?
Triple Bond?
Ch8 HW#1 1 – 5
1) Dot structures for diatomic molecules
a) Chlorine, Cl2
b) Bromine, Br2
Cl
Cl
Cl–Cl
Br
Br
Br – Br
c) Iodine, I2
I
I
I–I
2) How many unshared pairs are in each halogen molecule
3 pairs per atom, 6 pairs per molecule
3) Why necessary to form double & triple bond sometimes?
For stability, to satisfy the octet rule
4) How many electrons does each atom contribute in:
Double Bond?
Triple Bond?
Ch8 HW#1 1 – 5
1) Dot structures for diatomic molecules
a) Chlorine, Cl2
b) Bromine, Br2
Cl
Cl
Cl–Cl
Br
Br
Br – Br
c) Iodine, I2
I
I
I–I
2) How many unshared pairs are in each halogen molecule
3 pairs per atom, 6 pairs per molecule
3) Why necessary to form double & triple bond sometimes?
For stability, to satisfy the octet rule
4) How many electrons does each atom contribute in:
Double Bond? 2 electrons per atom
Triple Bond? 3 electrons per atom
5) Dot structures
a) H2S
H
H
b) PH3
P
c) ClF
Cl
S
H
H
H
F
5) Dot structures
a) H2S
H
H
b) PH3
P
c) ClF
Cl
S
H
H
H
F
H S
H
H-S-H
5) Dot structures
a) H2S
H
H
b) PH3
P
c) ClF
Cl
S
H
H
H
F
H S
H
H
P H
H
H-S-H
H
P-H
H
5) Dot structures
a) H2S
H
H
b) PH3
P
c) ClF
Cl
S
H
H
H
F
H S
H
H
P H
H
Cl F
H-S-H
H
P-H
H
Cl-F
Ch8.2 – More Covalent Bonds
Ex) Write electron dot structures for:
a) carbon monoxide, CO
Coordinate covalent bond - one atom contributes a pair
of electrons to the bond.
b) Hydroxide, OH–
c) SO3–2
d) NH4+
e) H30+
f) O3
Resonant Structures - double or triple bond that can jump around.
Ch8 HW#2 6 – 8
Lab8.1 – Conductivity
- due tomorrow
- Ch8 HW#2 due at beginning of period
Ch8 HW #2 6 – 8
6) Dot structures for SO4-2 & CO3-2
S
O
O
O
C
O
O
O
7) Resonant structures for CO3-2
O
8) Dot structures:
a) BF3
B
F
b) O2
O
O
c) NO2-1
N
O
d) F2
F
F
F
O
F
Ch8.3 – VESPR Theory
Ex 1) Methane, CH4
C
H H H H
Ch8.3 – VESPR Theory
Ex 1) Methane, CH4
C
H H H H
H
H C H
H
H
or H-C-H
H
- Electrons pairs seek maximum separations in 3 dimensions
Instead of a 2-D cross, they get greater separation
in a 3-D tetrahedral shape.
H
C
bond angle
Tetrahedral bond angles = 109.5˚
H
Ex2) Ammonia, NH3
N
H
H
H
Ex2) Ammonia, NH3
N
H
H
H
Pyramidal shape
Bond angles = 107˚
- Unshared pairs of electrons take up more space,
forces shared pairs closer together.
Ex3) Water, H2O
H
H
O
Ex2) Ammonia, NH3
N
H
H
H
Pyramidal shape
Bond angles = 107˚
- Unshared pairs of electrons take up more space,
forces shared pairs closer together.
Ex3) Water, H2O
H
H
O
Bent shape
Bond angle = 180˚
Ex4) Carbon Dioxide, CO2
C
O
O
Ex5) Boron trifluoride, BF3
B
F
F
F
Ch8 HW#3 9 – 11
Ch8 HW#3 9 – 11
9. BF3 is trigonal planar. Add a fluoride ion, F–, in a coordinate
covalent bond, what is its shape?
B
F
F
F
F–
10. Use VSEPR to draw:
a. CCl4
C
Cl
Cl
c. SeCl2
Se
Cl
Cl
Cl
b. PCl3
Cl
P
Cl
Cl
Cl
11. Draw
a. CO2
C
O
b. SiCl4
O
Si
c. SO3
S
O
Cl
d. SCl2
O
O
S
e. CO
f. I3+
C
I
O
Cl
Cl
Cl
I
I+
Cl
Cl
Ch8 HW#4
1. Predict the electron dot structure, shape and bond angles
a. silicon dioxide
b. PH3
Si O
O
P
H
H
H
c. sulfur dioxide
S
O
O
d. N2O (1 N as central atom)
N
N
O
e. CH2O
C H
f. Dinitrogen tetroxide
N
N
O
H
O
O
O
g. hydrogen peroxide
H
H
O
O
O
Ch8.4 – Polarity
Nonpolar Bonds - electrons are shared equally between 2 atoms
Ex 1) Cl2
Ch8.4 – Polarity
Nonpolar Bonds - electrons are shared equally between 2 atoms
Ex 1) Cl2
Cl
Cl
Cl – Cl
All of HNOFClBrI is nonpolar. But sometimes the random motion of the
shared electrons causes temporary polarity. This induces polarity in the
neighboring molecules.
Cl – Cl
Cl – Cl
Ch8.4 – Polarity
Nonpolar Bonds - electrons are shared equally between 2 atoms
Ex 1) Cl2
Cl
Cl
Cl – Cl
All of HNOFClBrI is nonpolar. But sometimes the random motion of the
shared electrons causes temporary polarity. This induces polarity in the
neighboring molecules.
δ– Cl – Cl δ+ δ– Cl – Cl δ+
This type of bond between molecules
is called Dispersion Forces. It is the
weakest of all bonds.
(Extremely low melting & boiling points)
Polar Bonds – Electrons are not shared equally between 2 atoms.
Ex2) HCl
Polar Bonds – Electrons are not shared equally between 2 atoms.
Ex2) HCl
H
Cl
+ H – Cl –
This arrangement is called a Dipole.
The Dipole – Dipole Bond between molecules is a little stronger.
+ H – Cl - + H – Cl higher melting & boiling points
Ex3) H2O
Ex3) H2O
Water contains 2 polar bonds, and overall is a polar molecule.
Its polarity attracts other water molecules.
When hydrogen is involved in polar bonds, it is called
Hydrogen Bonding. (Helps explain higher melting and boiling points.)
Ex4) CO2
Ex3) H2O
Water contains 2 polar bonds, and overall is a polar molecule.
Its polarity attracts other water molecules.
When hydrogen is involved in polar bonds, it is called
Hydrogen Bonding. (Helps explain higher melting and boiling points.)
Ex4) CO2
Contains 2 polar bonds, but the symmetry of the molecule cancels out
the polarity, making this molecule nonpolar.
Ex5) BF3
Ex5) BF3
Linear, trig planar, and tetrahedral molecules will be nonpolar IF all
the atoms attached to the central atom are the same.
Bent and pyramidal are always nonpolar.
Ex6) NaCl
Na
Cl
Ch8 HW#6 16 – 20
Ex5) BF3
Linear, trig planar, and tetrahedral molecules will be nonpolar IF all
the atoms attached to the central atom are the same.
Bent and pyramidal are always nonpolar.
Ex6) NaCl
Na
Cl
Na+
Cl–
Ionic!
Ch8 HW#6 16 – 20
Lab8.2 – Models
- due in 3 days
- Ch8 HW#6 due at beginning of period.
CH8 HW#6 16 – 20
16) ID bonds a) H and Br
(ionic, polar cov,
nonpolar cov)
c) C and O
d) CI and F
f) Br and Br
b) K and CI
e) Li and O
g) F and F
17)draw with details a)HF
b)HOOH
c) BrCl
18) Not every molecule with polar bonds
is nonpolar. Why is CCI4 nonpolar?
CI
CI
C
CI
CI
d) H2O
CH8 HW#6 16 – 20
16) ID bonds a) H and Br
(ionic, polar cov,
nonpolar cov)
c) C and O
d) CI and F
f) Br and Br
b) K and CI
e) Li and O
g) F and F
17)draw with details a)HF
a) H–F
c)
b)HOOH
c) BrCl
b)
H–O
O–H
Br–CI
O
d)
H
18) Not every molecule with polar bonds
is nonpolar. Why is CCI4 nonpolar?
CI
CI
C
CI
CI
d) H2O
H
19) Hydrogen bonding between 2 NH3’s & between NH3 & H2O
20)Rank Forces
strongest
middle
weakest
Ch7,8 Rev
1. Define valence electron.
2. Electron configs for N and N-3 .
3. Dot structures for correct bonding between Na and O
Ionic or covalent?
4. Describe diff between ionic and covalent bonds.
5. Dot structure shape and polarity for OF2.
6. Resonant structures for SO2.
7. Why do compounds with strong intermolecular forces have higher
boiling points than compounds with weak intermolecular forces?