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Ch7- Ionic Bonds ns- electrons in the highest energy level nce electrons corresponds to the group. 18 13 14 15 1 B C N Al Si P S lose electrons to gain electrons to form form positive ions negative ions called called cations. anions. (A N egative ION ) atoms are trying to satisfy the octet rule get 8 electrons in outer energy level.

Ch7- Ionic Bonds

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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 - PowerPoint PPT Presentation

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Page 1: Ch7- Ionic Bonds

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.

Page 2: Ch7- Ionic Bonds

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?

Page 3: Ch7- Ionic Bonds

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

Page 4: Ch7- Ionic Bonds

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?

Page 5: Ch7- Ionic Bonds

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

Page 6: Ch7- Ionic Bonds

Ex 3) Same stuff for:

Atoms: Mg N

Ions:

Formula:

Page 7: Ch7- Ionic Bonds

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.

Page 8: Ch7- Ionic Bonds

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?

Page 9: Ch7- Ionic Bonds

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?

Page 10: Ch7- Ionic Bonds

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?

Page 11: Ch7- Ionic Bonds

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?

Page 12: Ch7- Ionic Bonds

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?

Page 13: Ch7- Ionic Bonds

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.

Page 14: Ch7- Ionic Bonds

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

Page 15: Ch7- Ionic Bonds

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

Page 16: Ch7- Ionic Bonds

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

Page 17: Ch7- Ionic Bonds

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

Page 18: Ch7- Ionic Bonds

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.)

Page 19: Ch7- Ionic Bonds

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

Page 20: Ch7- Ionic Bonds

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

Page 21: Ch7- Ionic Bonds

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

Page 22: Ch7- Ionic Bonds

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

Page 23: Ch7- Ionic Bonds

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

Page 24: Ch7- Ionic Bonds

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

Page 25: Ch7- Ionic Bonds

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

Page 26: Ch7- Ionic Bonds

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

Page 27: Ch7- Ionic Bonds

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

Page 28: Ch7- Ionic Bonds

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

Page 29: Ch7- Ionic Bonds

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.

Page 30: Ch7- Ionic Bonds

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.

Page 31: Ch7- Ionic Bonds

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.

Page 32: Ch7- Ionic Bonds

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

Page 33: Ch7- Ionic Bonds

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.

Page 34: Ch7- Ionic Bonds

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?

Page 35: Ch7- Ionic Bonds

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.

Page 36: Ch7- Ionic Bonds

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?

Page 37: Ch7- Ionic Bonds

Lab7.1 – Models

- due in 2 days

- Ch7 Rev WS due at beginning of period

Page 38: Ch7- Ionic Bonds

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

Page 39: Ch7- Ionic Bonds

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

:..... :

..

..... ..

.... : :

Page 40: Ch7- Ionic Bonds

Double Covalent Bonds – 2 shared pairs of electrons.Triple Covalent Bonds – 3 shared pairs of electron.

Ex4) O2

Ex5) N2

Page 41: Ch7- Ionic Bonds

Covalent CompoundsEx6) NH3

Ex7) CH4

Ex8) CO2

Ch8 HW # 1

Page 42: Ch7- Ionic Bonds

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?

Page 43: Ch7- Ionic Bonds

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?

Page 44: Ch7- Ionic Bonds

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?

Page 45: Ch7- Ionic Bonds

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?

Page 46: Ch7- Ionic Bonds

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

Page 47: Ch7- Ionic Bonds

5) Dot structuresa) H2S H S

H H

b) PH3 P H H

c) ClF Cl F

Page 48: Ch7- Ionic Bonds

5) Dot structuresa) H2S H S H S H-S-H

H H H

b) PH3 P H H

c) ClF Cl F

Page 49: Ch7- Ionic Bonds

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

Page 50: Ch7- Ionic Bonds

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

Page 51: Ch7- Ionic Bonds

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–

Page 52: Ch7- Ionic Bonds

c) SO3–2

d) NH4+

e) H30+

f) O3

Resonant Structures - double or triple bond that can jump around.Ch8 HW#2 6 – 8

Page 53: Ch7- Ionic Bonds

Lab8.1 – Conductivity

- due tomorrow

- Ch8 HW#2 due at beginning of period

Page 54: Ch7- Ionic Bonds

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

Page 55: Ch7- Ionic Bonds

8) Dot structures:

a) BF3 B F F F

b) O2 O O

c) NO2-1 N O O

d) F2 F F

Page 56: Ch7- Ionic Bonds

Ch8.3 – VESPR TheoryEx 1) Methane, CH4

C H H H H

Page 57: Ch7- Ionic Bonds

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

Page 58: Ch7- Ionic Bonds

Ex2) Ammonia, NH3

N H H H

Page 59: Ch7- Ionic Bonds

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

Page 60: Ch7- Ionic Bonds

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˚

Page 61: Ch7- Ionic Bonds

Ex4) Carbon Dioxide, CO2

C O O

Ex5) Boron trifluoride, BF3

B F F F

Ch8 HW#3 9 – 11

Page 62: Ch7- Ionic Bonds

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

Page 63: Ch7- Ionic Bonds

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+

Page 64: Ch7- Ionic Bonds

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

Page 65: Ch7- Ionic Bonds

Ch8.4 – PolarityNonpolar Bonds - electrons are shared equally between 2 atoms

Ex 1) Cl2

Page 66: Ch7- Ionic Bonds

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

Page 67: Ch7- Ionic Bonds

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)

Page 68: Ch7- Ionic Bonds

Polar Bonds – Electrons are not shared equally between 2 atoms.

Ex2) HCl

Page 69: Ch7- Ionic Bonds

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

Page 70: Ch7- Ionic Bonds

Ex3) H2O

Page 71: Ch7- Ionic Bonds

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

Page 72: Ch7- Ionic Bonds

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.

Page 73: Ch7- Ionic Bonds

Ex5) BF3

Page 74: Ch7- Ionic Bonds

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

Page 75: Ch7- Ionic Bonds

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

Page 76: Ch7- Ionic Bonds

Lab8.2 – Models

- due in 3 days

- Ch8 HW#6 due at beginning of period.

Page 77: Ch7- Ionic Bonds

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

Page 78: Ch7- Ionic Bonds

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

Page 79: Ch7- Ionic Bonds

19) Hydrogen bonding between 2 NH3’s & between NH3 & H2O

20)Rank Forcesstrongest middle weakest

Page 80: Ch7- Ionic Bonds

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?