H 2 O - PowerPoint Presentation

Slide1

EXPLAINING VALENCE BOND THEORY

2F-1 (of

16)

1s atomicorbital w/ 1 e-

1s atomic orbital w/ 1 e-

H2

MOLECULAR ORBITAL

w/ 2 e-s

H

H

1s

1sSlide2

3p

y

3px

3pzEXPLAINING VALENCE BOND THEORY2F-2 (of 16)

3px atomicorbital w/ 1 e-

3px atomic orbital w/ 1 e-

Cl

2

MOLECULAR ORBITAL

w/ 2 e

-s

:

Cl ::

3s

3px

3pz

3s

3p

y

Cl

:

:

:Slide3

H

2

O

. .. O . . .2s

2py2px2pz2py

2px

2F-3

(of

16)Slide4

2p

x

2pz

H2O

. .. O . .

.2s2py

2py

2px

H

H

1s

1s

1s

1s

This predicts the shape of a water molecule to be bent with a 90º angleHowever, the molecule is bent with a 105º angle

Each bond is formed by combining a H 1s atomic orbital (with 1 e-) with an O 2p atomic orbital (with 1 e

-)

2F-4 (of 16)Slide5

1939

LINUS PAULING

Proposed that linear combinations of an atom’s valence atomic orbitals can produce another set of equivalent valence atomic orbitals

HYBRIDIZATION – The combining of 2 or more orbitals of different sublevels to make an equal number of HYBRID ORBITALS of equivalent energy (said to be DEGENERATE)

2F-5 (of 16)Slide6

E

2s

2p

hybridization

These 4 hybrid orbitals are called

sp

3

orbitals

sp

3

2F-6 (of 16)Slide7

2p

z

2p

x

2p

y2ssp3

sp

3

sp

3

sp

3hybridizationThe sp3 hybrid orbitals are arranged in a tetrahedron, with an angle of 109.5º between them

2F-7 (of 16)Slide8

sp

3

sp

3

sp3sp3This forms a bent molecule with a theoretical bond angle of 109.5º

Each bond is formed by combining an O sp3 atomic orbital (with 1 e-) with a H 1s atomic orbital (with 1 e-)2F-9 (of 16)Slide9

Each bond in H

2O is completely symmetrical around its internuclear axis

SIGMA BOND (σ) – A bond that is completely symmetrical around its

internuclear axisEach bond is named:σ(sp3+1s)2F-10 (of 16)Slide10

NH

3

. .

. N . .

2s2py2px2pz

2p

z

2p

x

2p

y2F-11 (of 16)Slide11

NH

3

. .

. N . .

2s2py2px2pz

2p

z

2p

x

H

H

1s

1s

This predicts the shape of an ammonia molecule to be trigonal pyramidal with a 90

º angle

However, the molecule is trigonal pyramidal with a 107º angle

2p

y

1s

.

H

2F-12

(of

16)Slide12

2F-13 (of

16)

E

2s2p

hybridization

sp

3Slide13

sp

3

sp

3

sp3sp3

This forms a trigonal pyramidal molecule with a bond angle of 109.5º Each sigma bond in NH3 is formed by combining a N sp3 atomic orbital (with 1 e-) with a H 1s atomic orbital (with 1 e

-)

Each bond is named:σ

(sp3+1s)2F-14

(of 16)Slide14

CH

4

. .

. C . 2s2py

2px2pz

.

.

C

. .

sp3sp3sp3sp3

E

2s

2p

hybridization

sp

3

2F-15

(of

16)Slide15

sp

3

sp

3

sp3sp3

This forms a tetrahedral molecule with a theoretical bond angle of 109.5º Each bond is named:

σ(sp3+1s)

2F-16 (of 16)

Central atoms with SN = 4 always undergo sp3 hybridization when bondingSlide16
Slide17

BH

3

2G-1 (of 18)

HH B HSN = 3Trigonal Planar

H

H

H

B

If the B undergoes sp

3

hybridization:

Central atoms with SN = 3 only need to hybridize 3 valance atomic orbitals

Trigonal Pyramidal

sp

3

sp

3

sp

3

sp

3Slide18

BH3

E

2s

2p

hybridization

sp2

2p

. .

.

B

2s

2p

y2px2pz

.

.

B

.

sp

2

sp

2

sp

2

2p

z

2G-2

(of

18)Slide19

2p

z

2p

x

2p

y

2s

hybridization

The angle between the sp

2 hybrid orbitals is 120º

sp

2sp22pz

sp2

The 2pz orbital is 90

º from the plane of the sp2 hybrid orbitals

2G-3 (of 18)Slide20

2p

z

2p

x

2p

y

2s

hybridization

sp

2

sp

22pzsp2

Rotating the top 90

º towards you

sp

2

sp

2

sp

2

2G-4

(of

18)Slide21

sp

2

sp

2

2pzsp2

2G-5 (of 19)

Rotating the top 90

º towards youSlide22

This forms a trigonal planar molecule with a bond angle of

120º

Each bond is named:

σ

(sp2

+1s)2G-8 (of 19)Slide23

BeH

2

H Be HSN = 2

LinearIf the Be undergoes sp3 hybridization:Central atoms with SN = 2 only need to hybridize 2 valance atomic orbitalsBent, 109.5º

H Be HIf the Be undergoes sp2 hybridization:Bent, 120º

2G-9 (of 19)Slide24

BeH2

E

2s

2p

hybridization

sp2p

. .

Be

2s

2py

2px

2p

z

.

. Be

sp

2p

y

sp

2p

z

2G-10

(of

19)Slide25

2p

z

2p

x

2p

y

2s

hybridization

The

sp

hybrid orbitals are linear, with an angle of 180º between them

sp2py2pzsp

The 2py and 2p

z orbitals are 90º from the

sp hybrid orbitals2G-11

(of 19)Slide26

sp

2p

y

2p

zsp

This forms a linear molecule with a bond angle of 180ºEach bond is named:σ(sp+1s)

2G-12 (of 19)Slide27

PH

5

5 + 5(1) = 10 valence e-s

HPHHHH

2G-13 (of 19)Central atoms with SN = 5 need to hybridize 5 valance atomic orbitalsSlide28

PH5

E

3s

3p

hybridization

sp3d3d

These 5 hybrid orbitals are called

3d

sp

3

d orbitals

2G-14

(of

19)Slide29

sp

3

d

sp

3dsp3dsp3d

sp3d

This forms a trigonal

bipyramidal molecule

Each bond is named:

σ(sp3d+1s)

PH

52G-15 (of 19)Slide30

SH6

H

SH

H

HHH6 + 6(1) = 12 valence e-s2G-16 (of 19)Central atoms with SN = 6 need to hybridize 6 valance atomic orbitalsSlide31

SH6

E

3s

3p

hybridization

sp3d23d

These 6 hybrid orbitals are called

3d

sp

3

d

2

orbitals

2G-17

(of

19)Slide32

sp

3

d

2

sp3d2sp3d

2sp3d2sp3d2sp3d2

SH6

This forms an octahedral molecule

Each bond is named:

σ

(sp

3

d2+1s)

2G-18 (of 19)Slide33

SN

Hybrid

.

23456spsp2sp3sp3

dsp3d2Hybrid Orbital GeometryLinearTrigonal PlanarTetrahedralTrigonal BipyramidalOctahedral

sp2

sp

2

sp

2

sp

3

sp

3

sp3sp3

sp3d

sp

3

d

sp

3

d

sp

3

d

sp

3

d

sp

3

d

2

sp

3

d

2

sp

3

d

2

sp

3

d

2

sp

3

d

2

sp

3

d

2

sp

sp

2G-19

(of

19)Slide34

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swfSlide35

HYDROCARBON – Molecules composed of only carbon and hydrogen

SATURATED HYDROCARBON – Hydrocarbons with only single bonds between the carbon atoms

The ending

-ane indicates only single bonds between the carbon atoms1 meth-2 eth-3 prop-4 but-5 pent-6 hex-

7 hept-8 oct-9 non-10 dec-The prefix tells the number of carbon atoms in the molecule2H-1 (of 10)Slide36

methane

:

CH

4HH C HHSN of C = 4 TetrahedralNonpolarEach bond: σ(sp3

+1s) HC HHH

 Hybridization = sp

3

sp

3

sp

3

sp

3

sp

3

2H-2 (of 10)Slide37

ethane

: C

2H6

H HH C C HH HSN of each C = 4

 Hybridization = sp3C-C bond: σ(sp3+sp3) Each C-H bond:

σ(sp3+1s) HC HH

H

H

H

C

2H-3 (of 10)

sp

3

sp

3

sp

3

sp

3Slide38

UNSATURATED HYDROCARBON – Hydrocarbons with at least one double or triple bond between carbon atoms

The ending

-ene indicates a double bond between carbon atoms

A double bond consists of a SIGMA BOND and a PI BONDPI BOND () – A bond that is only symmetrical upon a 180º rotation around its internuclear axis2H-4 (of 10)Slide39

C

C

ethene

: C2H4H HH C C H

SN of each C = 3  Hybridization = sp2

C-C bonds:

σ(sp2+sp2)

(2p+2p) Each C-H bond: σ(sp2+1s)

H

H

H

H

2H-5 (of 10)

sp

2

sp

2

2p

z

sp

2Slide40

The ending

-yne indicates a triple bond between carbon atoms

A triple bond consists of a 1 SIGMA BOND and 2 PI BONDS

2H-6 (of 10)Slide41

ethyne

: C

2H2

H C C HSN of each C = 2  Hybridization = sp

C-C bonds: σ(sp+sp) (2pz+2pz) (2py+2py)Each C-H bond: σ(sp+1s)

C

H

C

H

2H-7 (of 10)

sp

2p

y

2p

z

sp

Slide42

BOND ORBITAL MODELS (BOM’s) – Geometric

diagrams of the bonding in molecules

To draw a BOM for a molecule:

1 – Draw the Lewis structure and predict the hybridizations of each atom2 – Draw the correct geometries for the hybrid atomic orbitals of each atom, starting with atoms involved in double or triple bonds3 – Have atoms bond by overlapping atomic orbitals, and label each bond2H-8 (of 10)Slide43

C

C

H H C H H C H

H CDraw a BOM for CH3CHCH2

C

1

sp

3

C2sp2C3sp2

H

H

H

H

H

H

C

C

2

-C

3

bonds:

C

2

-H & C

3

-H bonds

:

C

1

-C

2

bond:

C

1

-H bonds:

2H-9 (of 10)

σ

(sp

2

+sp

2

)

(2p+2p)

σ

(sp

2

+1s

)

σ(sp3+sp

2)σ(sp3+1s)Slide44

H H C

C H C H C1sp2

C2spC3sp2Draw a BOM for CH2CCH2

C

C

H

H

H

C

C-C bonds:

C-H bonds:

H

2H-10 (of 10)

σ

(sp

2

+sp)



(

2p+2p)

σ

(sp

2

+1s)Slide45
Slide46

Valence Bond Theory

predicts many properties of moleculesbut it cannot predict

(1) magnetism of molecules (2) the e- arrangement of molecules with an odd number of e

-s (3) the e- arrangement in molecules with resonance (4) energies of excited states of molecules2I-1 (of 8)Slide47

MOLECULAR ORBITAL THEORY – Electrons in molecules exist in molecular orbitals

not restricted to regions of space between pairs of atoms, but are

treated as moving under the influence of all the nuclei in the molecule

2I-2 (of 8)

1932 ROBERT MULLIKEN AND FRIEDRICH HUNDProposed that molecular orbitals are not localized between pairs of atomsThe number of valence molecular orbitals (MO’s) in a molecule must equal the number of valence atomic orbitals (AO’s) from the atoms that make it upSlide48

H2

2 valence AO’s in the 2 H atoms,  2 valence MO’s in the molecule

This MO is the result of the addition of the 2 AO’s: 1s + 1s

This MO is the result of the subtraction of the 2 AO’s: 1s - 1s 2I-3 (of 8)

1s AO 1s AO

+++-Slide49

SIGMA MOLECULAR ORBITAL – An orbital that is completely symmetrical around its

internuclear axis

BONDING MOLECULAR ORBITAL – An orbital lower in energy than the original atomic orbitals because most of the e-

density is between the 2 nucleiANTIBONDING MOLECULAR ORBITAL – An orbital higher in energy than the original atomic orbitals because most of the e- density is outside the 2 nuclei2I-4 (of 8)

MO

1MO2σσ

b

*Slide50

MO Energy Level Diagram for H

2

E

atom 1atom 2molecule

1s1s

σ1sb

σ1s*

2I-5 (of 8)Slide51

Bond Order

= Bonding e

-s – Antibonding e-s

_______________________________________ 2Bond Order for H2 = 2 – 0 ______

2= 1 PARAMAGNETIC – A substance that is magnetic in a magnetic field Due to unpaired e-sDIAMAGNETIC – A substance that is never magnetic

Due to no unpaired e-sH2 is diamagnetic Electron configuration for H2 :(σ1sb)2

2I-6 (of

8)Slide52

He2

The valence AO’s of each He atom combine to form MO’s in the He2 molecule

MO Energy Level Diagram for He2

E

1s1s

σ1sbσ1s*

Bond Order for He

2

=

(2 – 2) / 2= 0  He2 does not exist

2I-7 (of 8)Slide53

Does He

2+ exist?If yes, give bond order, magnetism, and electron configuration notation

MO Energy Level Diagram for He2+

Bond Order for He2+= (2 – 1) / 2= ½  He2

+ does existHe2+ is paramagnetic Electron configuration for He2+ :(σ1sb)2 (σ1s

*)12I-8 (of 8)E

1s

1s

σ

1s

b

σ

1s

*Slide54
Slide55

Diatomic

Homonuclear Molecules of the 2nd Period

E

atom 1atom 2molecule Valence Orbitals

2s2p

2s2p?

2J-1 (of 16)Slide56

The 2 2s AO’s combine the same way as 2 1s AO’s, forming

a

s2sb MO and a

s2s* MO

2pz2px

2py

2p

z

2p

x

2p

yEach pair of p orbitals will combine to form a bonding MO and an antibonding MO2J-2 (of 16)Slide57

2p

x

2p

x

s2pxb

2px

2px

s2px*

2J-3 (of

16)-

+

--+-

++Slide58

2p

z

2p

z

PI MOLECULAR ORBITAL – An orbital that is symmetrical only upon a 180º rotation around the internuclear axisp2pzb

p

2pz

*

2pz2p

z

2J-4 (of

16)-

+

+

---++Slide59

The 2 2p

y AO’s combine the same way as 2 2pz AO’s, forminga

p2pyb MO and a p2p

y* MO 2J-5 (of 16)Slide60

E

atom 1

atom 2

molecule2s

2p2s

2p

s

2s

b

s

2s

*

s

2p

xb

s2p

x*

p2pz*

p2py*

p

2p

z

b

p

2p

y

b

Experimental data has determined the energies of the 8 MOs

2J-6 (of

16)Slide61

C2

Bond Order for

C2

= (6 – 2) / 2= 2 C2 is diamagneticElectron configuration for C2 :

(s2sb)2 (s2s*)2 (p2pb)42J-7 (of 16)

E2s

2p

2s

2p

s

2s

b

s

2s

*

s

2p

x

b

s

2p

x

*

p

2p

z

*

p

2p

y

*

p

2p

z

b

p

2p

y

bSlide62

F2

Bond Order for F

2

= (8 – 6) / 2= 1 F2 is diamagnetic Electron configuration for F2 :

(s2sb)2 (s2s*)2 (p2pb)4 (s2pb)2 (p2p*)4 2J-8 (of 16)

E

2s

2p

2s

2p

s

2s

b

s

2s

*

s

2p

x

b

s

2p

x

*

p

2p

z

*

p

2p

y

*

p

2p

z

b

p

2p

y

bSlide63

C

2

Bond Order

= 2 F2Bond Order= 1 Highest Bond Energy:

Longest Bond Length:C2F22J-9 (of 16)Slide64

Diatomic Heteronuclear Molecules of the 2

nd Period

2J-10 (of 16)Slide65

CN

Bond Order for

CN

= (7 – 2) / 2= 2½ CN is paramagnetic Electron configuration for CN :

(s2sb)2 (s2s*)2 (p2pb)4 (s2pb)1 N’s greater nuclear charge makes its AO energies lower that C’s2J-11 (of 16)

E

2s

2p

2s

2p

s

2s

b

s

2s

*

s

2p

x

b

s

2p

x

*

p

2p

z

*

p

2p

y

*

p

2p

z

b

p

2p

y

bSlide66

Benzene, C

6H6

C

CCCCC

H

H

H

H

H

H

C

CC

CC

C

H

H

H

H

H

H

Each C is sp

2

, and all atoms are planar

Each C has a p orbital perpendicular to the plane

2J-12

(of

16)

Molecular Orbital Theory can

also explain the

e

-

arrangement in molecules with resonanceSlide67

Resonance

structures assume that these p orbitals make distinct pi bonds

Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6

MO’s

2J-13

(of

16)Slide68

Resonance structures assume that these p orbitals make distinct pi bonds

Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6

MO’s

2J-14 (of 16)Slide69

2J-15

(of

16)

DELOCALIZED PI SYSTEM – A group of pi molecular orbitals spread out over more than 2 atoms

The e

-

s in the delocalized pi system

strengthen the bonds in

the

ring

Molecules possessing resonance

always bond

with delocalized pi systems, which are formed from

PARALLEL P ORBITALSSlide70

O N O

ONO

3-O N O

O

-

-

-

O N O

O

↔

↔

All atoms that can form a double bond in at least

1

resonance structure must

have parallel p orbitals to form the delocalized pi system

N

O

O

O

2J-16

(of

16)



m

ust be sp

2

N

O

O

OSlide71
Slide72

EMPIRICAL FORMULA CALCULATIONS

EMPIRICAL FORMULA – The simplest whole-number ratio of the atoms of different elements in a compound

C

6H6C1H1C6H12O6C1H2O1

H2OH2OMolecular Formula:Empirical Formula:2K-1 (of 8)Slide73

1) Assume you have 100 g of the compound

Find the empirical formula of a compound that is 75.0% carbon and 25.0% hydrogen by mass.



75.0 g C and 25.0 g H2) Calculate the moles of atoms of each element x 1 mol C ____________ 12.01 g C = 6.24

5 mol C 75.0 g Cx 1 mol H ____________ 1.01

g H = 24.75 mol H 25.0 g H3) Divide each number of moles by the smallest number of moles

6.245

mol C_______________

6.245

24.75

mol H

_______________ 6.245= 1.00 mol C= 3.96 mol H

4) The integer mole ratio must be the atom ratio:

CH4

2K-2 (of 8)Slide74

Find the empirical formula of a compound that is 90.0% carbon and 10.0% hydrogen by mass.

x 1

mol C

____________ 12.01 g C = 7.494 mol C 90.0 g Cx 1 mol

H ___________ 1.01 g H = 9.901 mol H 10.0 g H

7.494 mol C_______________ 7.4949.901

mol H

_______________ 7.494

= 1.000 mol

C= 1.321

mol H

If the moles of all elements are not within 0.1 moles of an integer, they must all be multiplied by an integer until they are integers 2 2= 2.00

mol C

= 2.64

mol H 3 3= 3.00 mol C

= 3.96 mol H

Empirical formula: C3H4

2K-3 (of 8)Slide75

Al

2(SO4)3

2 mol Al (26.98 g/mol) = 53.96 g 3

mol S (32.07 g/mol) = 96.21 g 12 mol O (16.00 g/mol) = 192.00 g 342.17 gMOLAR MASSES OF COMPOUNDSMOLAR MASS – The mass of one mole of molecules of a molecular substance, or one mole of formula units of an ionic substance

The mass necessary to have 1 mole of Al2(SO4)3 formula units2K-4 (of 8)= 342.2 gSlide76

CCl

4 1 mol C (12.01 g/mol) = 12.01 g

4 mol Cl (35.45 g/mol

) = 141.80 g 153.81 gCalculate the molar mass of carbon tetrachloride  153.81 g CCl4 = 1

mol CCl4Calculate the number of chlorine atoms in 1.00 g carbon tetrachloride.x 1 mol CCl4 __________________ 153.81 g CCl4

= 1.57 x 1022 atoms Cl 1.00 g CCl4x 4 mol Cl ______________ 1

mol CCl4

x 6.022 x 1023 atoms Cl

_____________________________ 1 mol Cl

2K-5 (of 8)Slide77

MOLECULAR FORMULA CALCULATIONS

MOLECULAR FORMULA – The actual number of the atoms of different elements in a molecule

C

1H1or C2H2Empirical Formula:Molecular Formula:C

1H1or C3H3or C4H4 etc.

2K-6 (of 8)Slide78

Find the molecular formula of a compound that is 43.7% phosphorus and 56.3% oxygen by mass, and has a molar mass of about 280 g/mol.

x 1

mol P

_____________ 30.97 g P = 1.411 mol P 43.7 g P

x 1 mol O ____________ 16.00 g O = 3.519 mol O 56.3 g O

1.411 mol P_______________ 1.4113.519

mol O

_______________ 1.411

= 1.00 mol P

= 2.49 mol O

Empirical formula: P

2O5Find the molar mass of the empirical formula 2 2= 2.00

mol P

= 4.98 mol

O2K-7 (of 8)Slide79

P

2O5 2 mol

P (30.97 g/mol) = 61.94 g 5 mol O (16.00 g/

mol) = 80.00 g 141.94 gDivide the compound’s actual molar mass by the empirical formula’s molar mass – it should be very close to an integer 280 g/mol________________141.94 g/mol

≈ 2 The molecular formula is 2 times the empirical formulaMolecular formula: P4O102K-8 (of 8)Slide80

REVIEW FOR TEST 2

Ionic BondingElectron Configuration Notation for IonsIon SizesLewis Structures for Ionic Compounds

Metallic BondingCovalent BondingValence Bond TheoryElectronegativity

Polar Covalent BondsNonpolar Covalent BondsLewis Structures for Covalent CompoundsSlide81

REVIEW FOR TEST 2

Bond OrderBond LengthBond EnergyFormal Charge

Resonance Steric NumberVSEPR Theory to Predict Molecular ShapesMolecular Polarity

HybridizationBond Orbital ModelsNaming BondsSlide82

REVIEW FOR TEST 2

Molecular Orbital TheoryBonding Molecular OrbitalsAntibonding Molecular OrbitalsSigma Molecular Orbitals

Pi Molecular OrbitalsMolecular Orbital Energy Level Diagrams Bond OrderParamagneticDiamagnetic

Electron Configuration NotationDelocalized Pi SystemsSlide83

REVIEW FOR TEST 2

Empirical Formula CalculationsMolar Masses of CompoundsMolecular Formula CalculationsReading

Experiments 6-10Chemical Separations Based on Solubilities

Solubility CurvesSpectrophotometric DeterminationsBeer’s Law