Chemical Bonding I: The Covalent Bond Copyright © The McGraw-Hill Companies, Inc. Permission...

Chemical Bonding I:The Covalent Bond

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2

Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons thatparticpate in chemical bonding.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

3

Lewis Dot Symbols for the Representative Elements &Noble Gases

4

Li + F Li+ F -

The Ionic Bond

1s22s11s22s22p5 1s2 1s22s22p6

[He] [Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -

LiF

Ionic bond: the electrostatic force that holds ions together in an ionic compound.

5

Lattice energy (U) (LE) of an ionic compound is the energy required to break apart the ions in their lattice arrangement into the ions in the gas phase.

For example the energy to take NaCl from its lattice arrangement into the gas phase ions would be represented by NaCl(s) ---> Na+(g) + Cl-(g)

Electrostatic (Lattice) Energy

6

Born-Haber Cycle for Determining Lattice Energy

Hoverall = H1 + H2 + H3 + H4 + H5° °°°°°

8

A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F+

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond

9

Sigma () and Pi Bonds ()

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

How many and bonds are in the acetic acid (vinegar) molecule CH3COOH?

C

H

H

CH

O

O H bonds = 6 + 1 = 7

bonds = 1

10

8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e-

double bondsdouble bonds

Triple bond – two atoms share three pairs of electrons

N N8e- 8e-

N N

triple bondtriple bond

or

11

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond

12

H F FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms

electron richregion

electron poorregion e- riche- poor

+ -

13

Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.

Electron Affinity - measurable, Cl is highest

Electronegativity - relative, F is highest

X (g) + e- X-(g)

14

Variation of Electronegativity with Atomic Number

15

Covalent

share e-

Polar Covalent

partial transfer of e-

Ionic

transfer e-

Increasing difference in electronegativity

Classification of bonds by difference in electronegativity

Difference Bond Type

0 Covalent

2 Ionic

0 < and <2 Polar Covalent

16

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2.

Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic

H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent

N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent

17

1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.

2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Complete an octet for all atoms except hydrogen

4. If structure contains too many electrons, form double and triple bonds on central atom as needed.

Writing Lewis Structures

18

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)

5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

19

Write the Lewis structure of the carbonate ion (CO32-).

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

Step 5 - Too many electrons, form double bond and re-check # of e-

2 single bonds (2x2) = 41 double bond = 4

8 lone pairs (8x2) = 16Total = 24

20

A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O O+ -

OOO+-

O C O

O

- -O C O

O

-

-

OCO

O

-

-

What are the resonance structures of the carbonate (CO3

2-) ion?

21

Exceptions to the Octet Rule

The Incomplete Octet

H HBeBe – 2e-

2H – 2x1e-

4e-

BeH2

BF3

B – 3e-

3F – 3x7e-

24e-

F B F

F

3 single bonds (3x2) = 69 lone pairs (9x2) = 18

Total = 24

22

Exceptions to the Octet Rule

Odd-Electron Molecules

N – 5e-

O – 6e-

11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S – 6e-

6F – 42e-

48e-

S

F

F

F

FF

F

6 single bonds (6x2) = 1218 lone pairs (18x2) = 36

Total = 48

23Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chemical Bonding IIMolecular Geometry and

Hybridization of Atomic Orbitals

24

Valence shell electron pair repulsion (VSEPR) model:

Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs.

AB2 2 0

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

linear linear

B B

25

Cl ClBe

2 atoms bonded to central atom

0 lone pairs on central atom

26

AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

28


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB4 4 0 tetrahedral tetrahedral

30


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

32


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral

34

Predicting Molecular Geometry1. Draw Lewis structure for molecule.

2. Count number of lone pairs on the central atom and number of atoms bonded to the central atom.

3. Use VSEPR to predict the geometry of the molecule.

What are the molecular geometries of SO2 and SF4?

SO O

AB2E

bent

S

F

F

F F

AB4E

distortedtetrahedron

35

Dipole Moments and Polar Molecules

H F

electron richregion

electron poorregion

Dipole moment is defined as magnitude of charge (Q) times distance (r) between the charges. u = (Q)(r) Q charge in coulombs (C) r distance between charges in meters (m)

36

Behavior of Polar Molecules

field off field on

37

Which of the following molecules have a dipole moment? H2O, CO2, SO2, and CF4

O HH

dipole momentpolar molecule

SO

O

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

F

FFF

no dipole momentnonpolar molecule

38

Does CH2Cl2 have a dipole moment?

40

Bond Enthalpy Bond Length

H2

F2

436.4 kJ/mol

150.6 kJ/mol

74 pm

142 pm

Valence bond theory – bonds are formed by sharing of e- from overlapping atomic orbitals.

Overlap Of

2 1s

2 2p

How does Lewis theory explain the bonds in H2 and F2?

Sharing of two electrons between the two atoms.

41

Change in electron density as two hydrogen atoms approach each other.

42

Valence Bond Theory and NH3

N – 1s22s22p3

3 H – 1s1

If the bonds form from overlap of 3 2p orbitals on nitrogen with the 1s orbital on each hydrogen atom, what would the molecular geometry of NH3 be?

If use the3 2p orbitalspredict 90o

Actual H-N-Hbond angle is

107.3o

43

Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals.

1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals.

2. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

3. Covalent bonds are formed by:

a. Overlap of hybrid orbitals with atomic orbitals

b. Overlap of hybrid orbitals with other hybrid orbitals

44

Formation of sp3 Hybrid Orbitals

45

Formation of Covalent Bonds in CH4

46

Formation of sp Hybrid Orbitals

47

Formation of sp2 Hybrid Orbitals

48

# of Lone Pairs+

# of Bonded Atoms Hybridization Examples

2

3

4

5

6

sp

sp2

sp3

sp3d

sp3d2

BeCl2

BF3

CH4, NH3, H2O

PCl5

SF6

How do I predict the hybridization of the central atom?

1. Draw the Lewis structure of the molecule.

2. Count the number of lone pairs AND the number of atoms bonded to the central atom

Stoichiometry

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

50

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

51

The average atomic mass is the weighted

average of all of the naturally occurring

isotopes of the element.

52

Naturally occurring lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

(7.42 x 6.015) + (92.58 x 7.016)100

= 6.941 amu

Average atomic mass of lithium:

53

Average atomic mass (6.941)

54

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

Dozen = 12

Pair = 2

The Mole (mol): A unit to count numbers of particles

55

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

56

One Mole of:

C S

Cu Fe

Hg

57

x 6.022 x 1023 atoms K1 mol K

=

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K 1 mol K39.10 g K

x

8.49 x 1021 atoms K

58

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

SO2

59

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

60

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.

1Na 22.99 amu

1Cl + 35.45 amuNaCl 58.44 amu

For any ionic compound

formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu

1 mole NaCl = 58.44 g NaCl

NaCl

61

What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.97

8 O + 8 x 16.00310.18 amu

62

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.00%

63

Percent Composition and Empirical Formulas

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

nK = 24.75 g K x = 0.6330 mol K1 mol K

39.10 g K

nMn = 34.77 g Mn x = 0.6329 mol Mn1 mol Mn

54.94 g Mn

nO = 40.51 g O x = 2.532 mol O1 mol O

16.00 g O

64

Percent Composition and Empirical Formulas

K : ~~ 1.00.63300.6329

Mn : 0.63290.6329

= 1.0

O : ~~ 4.02.532

0.6329

nK = 0.6330, nMn = 0.6329, nO = 2.532

KMnO4

65

g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

g of O = g of sample – (g of C + g of H)

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

6.0 g C = 0.5 mol C

1.5 g H = 1.5 mol H

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

Combusted 11.5 g ethanol

Q. What is the Empirical formula of Ethanol

66

3 ways of representing the reaction of H2 with O2 to form H2O

A process in which one or more substances is changed into one or more new substances is a chemical reaction

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants products

67

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

68

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12

69


3. Start by balancing those elements that appear in only one reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

70


4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

71


5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

72

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Amounts of Reactants and Products

73

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

74

Limiting Reagent:

2NO + O2 2NO2

NO is the limiting reagent

O2 is the excess reagent

Reactant used up first in the reaction.

75

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

76

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

At this point, all the Al is consumed and Fe2O3 remains in excess.

77

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100%

Reaction Yield

78

1. What is the character of the bond in carbon monoxide?

a. ionicb. polar covalentc. non-polar covalentd. coordinate covalent

79

2. Which of the following is the bet explanation of the phenomenon of hydrogen bonding?a. Hydrogen has a strong affinity for holding onto

valence electronsb. Hydrogen can only hold two valence electronsc. Electron negative atoms disproportionately

carry shared pairs when bonded to hydrogend. Hydrogen bonds have ionic character

80

3. Which of the following has a formula weight between 74 and 75 grams per mole?a. KClb. C4H10Oc. (LiCl)2

d. BF3

81

4. In the reactoin shown, if 39.03 g of Na2S are reacted with 113.3 g of AgNO3, how much, if any, of either reagent will be left over once the reaction has gone to completion?

Na2S+ 2 AgNO3Ag2S + 2 NaNO3

a. 41.37 g AgNO3 b. 13.00 g Na2Sc. 26.00 g Na2Sd. 74.27 g AgNO3

82

What is the percent by mass of carbon in CO2?

a. 12%b. 27%c. 33 %d. 44 %

83

Sulfur dioxide oxidizes in the presence of O2 gas as per the reaction

2 SO2 + O2 2 SO3

Approximately how many grams of sulfur trioxide are produced by the complete oxidation of 1 mole of sulfur dioxide?

a. 1gb. 2gc. 80gd. 160 g

84

How many carbon atoms exist in 12 amu of 12C?

a. 1b. 12c. 6.02 x 1023

d. 7.22x1024

85

When an electron moves from a 2p to a 3s orbital, the atom containng that electron:

a. becomes a new isotopeb. becomes a new elementc. absorbs energyd. releases energy

87

5. One way to test for the presence of iron in solution is by adding potassium thiocyanate to the solution. The product when this reagent reacts with iron is FeSCN2+, which creates a dark red color in solution via the following net ionic equation.

Fe 3+ (aq) + SCN-FeSCN2+

How many grams of iron sulfate would be needed to produce 2 moles of

a. 400 gb. 800 gc. 200 gd. 500 g

Chemical Bonding I: The Covalent Bond Copyright © The McGraw-Hill Companies, Inc. Permission...

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