ch7- ionic bonds
DESCRIPTION
Ch7- Ionic Bonds. Valence electrons- electrons in the highest energy level (only s & p) - # of valence electrons corresponds to the group. 1 18 H 2 13 14 15 16 17 He - PowerPoint PPT PresentationTRANSCRIPT
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Ch7- Ionic Bonds
Valence electrons- electrons in the highest energy level (only s & p)
- # of valence electrons corresponds to the group. 1 18 H 2 13 14 15 16 17 He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar
lose electrons to gain electrons to form form positive ions negative ions called called cations. anions.
(A Negative ION)All atoms are trying to satisfy the octet rule
- get 8 electrons in outer energy level.
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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?
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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
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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?
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Ex 2) Write the electron configure & dot diagram for:
Al Br...
.. . ... .[Ne] 3s23p1 [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
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Ex 3) Same stuff for:
Atoms: Mg N
Ions:
Formula:
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Ex 3) Same stuff for:
Atoms:
Ions:
Formula:
Mg .. N... ..
[Ne]3s2 [He] 2s22p3
Mg+2 = [Ne]+2 N-3 = [He] 2s22p6 = [Ne]-3
Mg..N. ....
Mg..N.. ...
Mg
Ch7 HW#1
..
Mg3N2
Properties of Ionic Compounds - solids at room temp - don’t conduct electricity in solid state - will conduct electricity if
1 molten (liquid state) 2 dissolved in water
in both these cases the ions are free to move.
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen
2) Dot structures & electron configurations:K C Mg O
3) Gain or Lose? K C Mg O
4) E.C. for ions K+1 C+4 Mg+2 O-2
C-4
5) Why do nonmetals form anions when reacting?
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6
2) Dot structures & electron configurations:K C Mg O
3) Gain or Lose? K C Mg O
4) E.C. for ions K+1 C+4 Mg+2 O-2
C-4
5) Why do nonmetals form anions when reacting?
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6
2) Dot structures & electron configurations:K C Mg O
[Ar] 4s1 [He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4
3) Gain or Lose? K C Mg O
4) E.C. for ions K+1 C+4 Mg+2 O-2
C-4
5) Why do nonmetals form anions when reacting?
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6
2) Dot structures & electron configurations:K C Mg O
[Ar] 4s1 [He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4
3) Gain or Lose? K Lose 1 C gain or lose 4 elec Mg lose 2 O gain 2
4) E.C. for ions K+1 C+4 Mg+2 O-2
C-4
5) Why do nonmetals form anions when reacting?
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6
2) Dot structures & electron configurations:K C Mg O
[Ar] 4s1 [He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4
3) Gain or Lose? K Lose 1 C gain or lose 4 elec Mg lose 2 O gain 2
4) E.C. for ions K+1 [Ar] C+4 [He] Mg+2 [Ne] O-2 [Ne]2s22p4
C-4 [He] 2s22p6
[Ne][Ne]
5) Why do nonmetals form anions when reacting?
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Ch7 HW#11) How many valence electrons for:
a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6
2) Dot structures & electron configurations:K C Mg O
[Ar] 4s1 [He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4
3) Gain or Lose? K Lose 1 C gain or lose 4 elec Mg lose 2 O gain 2
4) E.C. for ions K+1 [Ar] C+4 [He] Mg+2 [Ne] O-2 [Ne]2s22p4
C-4 [He] 2s22p6
[Ne][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.
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6) Dot formula to determine formulas:a) potassium & iodine b) Ca & S
K I Ca S
c) Al & O d) Na & P
Al Na Na P
Na
7) Name:
O
O
O
Al
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6) Dot formula to determine formulas:a) potassium & iodine b) Ca & S
K I Ca S
KCl CaSc) Al & O d) Na & P
Al Na Na P
Na
7) Name:
O
O
O
Al
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6) Dot formula to determine formulas:a) potassium & iodine b) Ca & S
K I Ca S
KCl CaSc) Al & O d) Na & P
Al Al2O3 Na Na3P Na P
Na
7) Name:
Al
O
O
O
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6) Dot formula to determine formulas:a) potassium & iodine b) Ca & S
K I Ca S
KCl CaSc) Al & O d) Na & P
Al Al2O3 Na Na3P Na P
Na
7) Name:Potassium iodide Calcium sulfide
Aluminum oxide Sodium phosphate
Al
O
O
O
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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.)
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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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
9) Electron configurations:a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10
b) Gold (I) Au+1 Au: [Xe] 6s24f145d9 6s14f145d10
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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons9) Electron configurations:a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10
b) Gold (I) Au+1 Au: [Xe] 6s24f145d9 6s14f145d10
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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons9) 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
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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons9) 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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons9) 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
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Ch7 HW#2 8 – 12 8) Why do metals tend to form cations?
Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons9) 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
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12) Use dots to combine:a) Cu(I) & Cl:
Cu Cl
b) Cu(II) & Cl: Cu Cl
Cl
c) Fe (II) & O Fe O
d) Fe (III) & O
Fe
O
O
O
Fe
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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 O
d) Fe (III) & O
Fe
O
O
O
Fe
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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 O FeO
d) Fe (III) & O Fe2O3
Fe
O
O
O
Fe
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Ch7 Rev WS1. 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.
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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.
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5. Explain the electrical conductivity of melted and of aqueous solutions ofionic compounds, using the characteristics of ionic compounds.
6.Explain the physical properties of metals, using theory of metallic bonding.
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B. Questions22. Write electron dot structures for the following atoms:a. silicon b. rubidium c. barium d. tin e. iodine f. arsenic
Si Rb Ba Sn I As
23. complete the following table. outer electron outer electron formula of ion type of ion config of atom config of ion
Se
K
Ca
Br
N
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24. Use electron dot formulas to determine chemical formulas of the ioniccompound 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.
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Formulas of Ionic CompoundsElements 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?
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Example BRemember 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.
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3. How many valence electrons does the element gallium have?
4. Write the formula for the ion formed when nitrogen gains electronsto attain a noble gas configuration.
5. What is the formula for the compound formed when astatine andstrontium combine?
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Lab7.1 – Models
- due in 2 days
- Ch7 Rev WS due at beginning of period
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Ch8.1 – Covalent BondsSingle Covalent Bonds – one pair of electrons shared between 2 atoms. Ex1) H + H H H H-H
one shared pair Structural formula
Each dash represents one shared pair of electrons.
. . :
Ex2) F2
Ex3) H2O
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Ch8.1 – Covalent BondsSingle Covalent Bonds – one pair of electrons shared between 2 atoms. Ex1) H + H H H H-H
one shared pair Structural formula
Each dash represents one shared pair of electrons.
. . :
Ex2) F2 F + F F–F Unshared pairs of electrons (nonbonding pairs)
Take up more space then the shared pairs.
Ex3) H2O
:..... :
..
..... ..
.... : :
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Double Covalent Bonds – 2 shared pairs of electrons.Triple Covalent Bonds – 3 shared pairs of electron.
Ex4) O2
Ex5) N2
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Covalent CompoundsEx6) NH3
Ex7) CH4
Ex8) CO2
Ch8 HW # 1
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Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2
Cl Cl Br Br 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?
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Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2
Cl Cl Br Br I I
Cl–Cl Br – Br 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?
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Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2
Cl Cl Br Br I I
Cl–Cl Br – Br I – I
2) How many unshared pairs are in each halogen molecule3 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?
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Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2
Cl Cl Br Br I I
Cl–Cl Br – Br I – I
2) How many unshared pairs are in each halogen molecule3 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?
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Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2
Cl Cl Br Br I I
Cl–Cl Br – Br I – I
2) How many unshared pairs are in each halogen molecule3 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
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5) Dot structuresa) H2S H S
H H
b) PH3 P H H
c) ClF Cl F
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5) Dot structuresa) H2S H S H S H-S-H
H H H
b) PH3 P H H
c) ClF Cl F
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5) Dot structuresa) H2S H S H S H-S-H
H H H H H
b) PH3 P H P H P-H H H H
c) ClF Cl F
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5) Dot structuresa) H2S H S H S H-S-H
H H H H H
b) PH3 P H P H P-H H H H
c) ClF Cl F Cl F Cl-F
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Ch8.2 – More Covalent BondsEx) 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–
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c) SO3–2
d) NH4+
e) H30+
f) O3
Resonant Structures - double or triple bond that can jump around.Ch8 HW#2 6 – 8
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Lab8.1 – Conductivity
- due tomorrow
- Ch8 HW#2 due at beginning of period
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Ch8 HW #2 6 – 8 6) Dot structures for SO4
-2 & CO3-2
S O O O O
C O O O
7) Resonant structures for CO3-2
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8) Dot structures:
a) BF3 B F F F
b) O2 O O
c) NO2-1 N O O
d) F2 F F
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Ch8.3 – VESPR TheoryEx 1) Methane, CH4
C H H H H
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Ch8.3 – VESPR TheoryEx 1) Methane, CH4 H H
H C H or H-C-H 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
C H H H H
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Ex2) Ammonia, NH3
N H H H
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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
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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 shapeBond angle = 180˚
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Ex4) Carbon Dioxide, CO2
C O O
Ex5) Boron trifluoride, BF3
B F F F
Ch8 HW#3 9 – 11
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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 b. PCl3
C Cl Cl Cl Cl P Cl Cl Cl
c. SeCl2
Se Cl Cl
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11. Draw a. CO2 b. SiCl4 C O O Si Cl Cl Cl Cl
c. SO3 d. SCl2
S O O O S Cl Cl
e. CO f. I3+
C O I I I+
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Ch8 HW#41. Predict the electron dot structure, shape and bond angles a. silicon dioxide b. PH3 Si O O P H H H
c. sulfur dioxide d. N2O (1 N as central atom) S O O N N O
e. CH2O f. Dinitrogen tetroxide C H H O N N O O
O O
g. hydrogen peroxideH H O O
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Ch8.4 – PolarityNonpolar Bonds - electrons are shared equally between 2 atoms
Ex 1) Cl2
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Ch8.4 – PolarityNonpolar 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
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Ch8.4 – PolarityNonpolar 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 theweakest of all bonds.
(Extremely low melting & boiling points)
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Polar Bonds – Electrons are not shared equally between 2 atoms.
Ex2) HCl
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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
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Ex3) H2O
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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 calledHydrogen Bonding. (Helps explain higher melting and boiling points.)
Ex4) CO2
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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 calledHydrogen Bonding. (Helps explain higher melting and boiling points.)
Ex4) CO2
Contains 2 polar bonds, but the symmetry of the molecule cancels outthe polarity, making this molecule nonpolar.
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Ex5) BF3
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Ex5) BF3
Linear, trig planar, and tetrahedral molecules will be nonpolar IF allthe atoms attached to the central atom are the same.
Bent and pyramidal are always nonpolar.
Ex6) NaCl Na Cl
Ch8 HW#6 16 – 20
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Ex5) BF3
Linear, trig planar, and tetrahedral molecules will be nonpolar IF allthe 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
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Lab8.2 – Models
- due in 3 days
- Ch8 HW#6 due at beginning of period.
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CH8 HW#6 16 – 20 16) ID bonds a) H and Br b) K and CI(ionic, polar cov, nonpolar cov) c) C and O d) CI and F e) Li and O
f) Br and Br g) F and F
17)draw with details a)HF b)HOOH c) BrCl d) H2O
18) Not every molecule with polar bonds is nonpolar. Why is CCI4 nonpolar? CI
CI C CI
CI
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CH8 HW#6 16 – 20 16) ID bonds a) H and Br b) K and CI(ionic, polar cov, nonpolar cov) c) C and O d) CI and F e) Li and O
f) Br and Br g) F and F
17)draw with details a)HF b)HOOH c) BrCl d) H2Oa) H–F b) H–O
O–Hc) Br–CI d) O
H H18) Not every molecule with polar bonds is nonpolar. Why is CCI4 nonpolar? CI
CI C CI
CI
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19) Hydrogen bonding between 2 NH3’s & between NH3 & H2O
20)Rank Forcesstrongest middle weakest
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Ch7,8 Rev1. Define valence electron.2. Electron configs for N and N-3 .
3. Dot structures for correct bonding between Na and OIonic 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 higherboiling points than compounds with weak intermolecular forces?