copyright©2000 by houghton mifflin company. all rights reserved. 1 bonding: general concepts...

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

1

Bonding:Bonding: General ConceptsGeneral Concepts

Chapter 8


2

OverviewTypes of chemical bonds, Electronegativity, Bond

polarity and Dipole Moments.The Ions: electron configurations, size, formula,

lattice energy calculations.Covalent bonds: model, bond energies, chemical

reactions.Lewis structure, exceptions to the octet rule,

resonance.Molecular structure models from Valence Shell

Electron Pair Model “VSEPR” for single and multiple bonds.


3

Bonds

Forces that hold groups of atoms together and make them function as a unit.

NaClNaCl – attraction is electrostatic since – attraction is electrostatic since NaNa++ and and ClCl-- are the are theStable forms for these elements. This is an example of Stable forms for these elements. This is an example of ““Ionic Bonding”Ionic Bonding”


4

Bond Energy

It is the energy required to break a bond.

It gives us information about the strength of a bonding interaction, as well as radius.


5

Bond Length

The distance where the system energy is a minimum.


6

Figure 8.1: (a) The interaction of two hydrogen atoms. (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms.


7

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


8

Covalent Bond

No electron transferElectrons are shared between two atoms,

positioned between the two nucleiExample: H2, O2, H2O, CO2, etc.


9

Ionic Bonds

Formed from electrostatic attractions of closely packed, oppositely charged ions.

Formed when an atom that easily loses electrons reacts with one that has a high electron affinity.


10

Li + F Li+ F -

The Ionic Bond

1s22s11s22s22p5 1s21s22s22p6[He][Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -


11

Cation is always smaller than atom from which it is formed.Anion is always larger than atom from which it is formed.

Ionic Compound


12

Ionic Bondsand

Coulomb’s Law

Q1 and Q2 = numerical ion charges

r = distance between ion centers (in nm)

E = 2.31 10 J nm (19 QQ r1 2 / )

Attractive forces are (-), repulsive are (+).


13

Covalent or Ionic?

Covalent and ionic are simply extreme cases.

Most molecules share electrons but “Unequally” due to the difference in electronegativity and electron affinity.

This will give rise to a dipole moment and the molecule becomes polar.


14

Electronegativity

The ability of an atom in a molecule to attract shared electrons to itself.

Linus Pauling simple model

Δ= (H X)actual (H X)expected

(H-H) + (X-X)(H-H) + (X-X)

22(H-X)experimental # (H-X)expected =

If = 0 => no polarity


15


16

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 to 0.1 Covalent

2 Ionic

0 < and <2 Polar Covalent


17


18

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; andthe 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

9.5


19

H F FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms. The molecule is called “Dipolar”.

electron richregion

electron poorregion e- riche- poor

+ -

Dipole MomentDipole Moment


20

(a) When no electric field is present, the molecules are randomly oriented.

Figure 8.2: The effect of an electric field on hydrogen fluoride molecules.

(b) When the field is turned on, the molecules tend to line up with their negative ends toward the positive pole and their positive ends toward the negative pole.


21


22

May exhibit dipole moment depending on their structure i.e. arrangement in space

Polyatomic MoleculesPolyatomic Molecules


23

Comparison of Ionic and Covalent Compounds


24

Figure 8.4: (a) The charge distribution in the water molecule. (b) The water molecule in an electric field.

V-ShapeV-Shape


25

Figure 8.5: (a) The structure and charge distribution of the ammonia molecule. The polarity of the N—H bonds occurs because nitrogen has a greater electronegativity than hydrogen. (b) The dipole moment of the ammonia

molecule oriented in an electric field.

Look for a “NET DIPOLE”Look for a “NET DIPOLE”


26

Look for a “NET DIPOLE”Look for a “NET DIPOLE”

NN

HH HHHH

Trigonal Pyramidal StructureTrigonal Pyramidal Structure


27


28

The carbon dioxide molecule CO2: The opposed bond polarities cancel out, and the carbon dioxide has no dipole moment:

Non-polar molecule

Note:Note: The C-O bond is polar, but the net dipole is Zero The C-O bond is polar, but the net dipole is ZeroExample: SOExample: SO33, CCl, CCl44, etc., etc.


29


30

Which of the following molecules have a dipole moment?H2O, CO2, SO2, and CH4

O HH

dipole momentpolar molecule

SO

O

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

H

H

HH

no dipole momentnonpolar molecule


31

Compounds

Two nonmetals react: They share electrons to achieve NGEC (Noble Gas Electron

Configurations) and form covalent bonds.A nonmetal and a representative group metal

react (ionic compound): The valence orbitals of the metal are emptied (cation) to achieve NGEC. The valence electron configuration of the nonmetal (anion) achieves NGEC.


32

Ions

Ionic compounds are always electrically neutral e.g. they have the same amount of +ve and –ve charges.

Common ions have noble gas configurations.


33

Electron Configurations of Cations and Anions

Na [Ne]3s1 Na+ [Ne]

Ca [Ar]4s2 Ca2+ [Ar]

Al [Ne]3s23p1 Al3+ [Ne]

Atoms lose electrons so that cation has a noble-gas outer electron configuration.

H 1s1 H- 1s2 or [He]

F 1s22s22p5 F- 1s22s22p6 or [Ne]

O 1s22s22p4 O2- 1s22s22p6 or [Ne]

N 1s22s22p3 N3- 1s22s22p6 or [Ne]

Atoms gain electrons so that anion has a noble-gas outer electron configuration.

Of Representative Elements


34

ns1

ns2

ns2

np1

ns2

np2

ns2

np3

ns2

np4

ns2

np5

ns2

np6

d1

d5 d10

4f

5f

Ground State Electron Configurations of the Elements


35

+1

+2

+3 -1-2-3

Cations and Anions Of Representative Elements


36

Notes on Ions

Hydrogen may form H+ or H- Tin forms Sn2+ and Sn4+.Transition metals exhibit a more

complicated behavior.


37

Example of Ionic Compounds

MgO magnesium oxide is formed of Mg2+ and O2-.

CaO formed from Ca2+ and O2-.Al2O3 is formed of 2Al3+ and 3O2-.


38

Figure 8.7: Sizes of ions related to positions of the elements on the periodic table.


39


40

Electronegativity increases

Ionization energy increases

Atomic size decreases

Metallic Character decreases

Ato

mic

siz

e in

crea

ses

Ch

emic

al a

ctiv

ity

incr

ease

s

Ioni

zati

on e

nerg

y de

crea

ses

Met

allic

cha

ract

er in

crea

ses

Ele

ctro

nega

tivi

ty in

crea

ses

GR

OU

PS

GR

OU

PS

PERIODS

PERIODIC TRENDS OF ELEMENTSPERIODIC TRENDS OF ELEMENTS


41

Isoelectronic Ions Contain the the same number of electrons

8O 1s22s22p4 O2- 1s22s22p6

9F 1s22s22p5 F- 1s22s22p6

11Na 1s22s22p63s1 Na+ 1s22s22p6

12Mg 1s22s22p63s2 Mg2+1s22s22p6

13Al 1s22s22p63s23p1 Al3+ 1s22s22p6

Which one you expect to have the smallest radius?Which one you expect to have the smallest radius?And why?And why?


42

Radii of Isoelectronic Ions

O2> F > Na+ > Mg2+ > Al3+

largest smallest13 protons vs. 10

electrons


43

Example

Choose the largest ion in each of the following groups:

1. Li+, Na+, K+, Rb+, Cs+

2. Ba2+ , Cs+ , I- , Te2-


44

To Reiterate

Cation is smaller than parent molecule.Anion is larger than parent molecule.Size increase down in a group (+ve or –ve).The larger the mass number the smaller the

size for isoelectronic cations and anions.


45

Electron Configurations of Cations of Transition Metals

When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.

Fe: [Ar]4s23d6

Fe2+: [Ar]4s03d6 or [Ar]3d6

Fe3+: [Ar]4s03d5 or [Ar]3d5

Mn: [Ar]4s23d5

Mn2+: [Ar]4s03d5 or [Ar]3d5


46

Why Compounds Exist?

The driving force behind the naturally occurring compounds (such as NaCl, H2O, etc.) is to yield a stable lower energy form.

A stable form is a an arrangement of atoms, held together by bonding that prevent decomposition.

This bonding energy when ions condense from gas phase into ionic solid is called Lattice Energy.


47

Lattice Energy

The change in energy when separated gaseous ions are packed together to form an ionic solid.

M+(g) + X(g) MX(s)

Lattice energy is negative (exothermic) from the point of view of the system.


48

Formation of an Ionic Solid1. Sublimation of the solid metal

M(s) M(g) [endothermic]

2. Ionization of the metal atoms

M(g) M+(g) + e [endothermic]

3. Dissociation of the nonmetal 1/2X2(g) X(g) [endothermic]

4. Formation of X ions in the gas phase:

X(g) + e X(g) [exothermic]

5. Formation of the solid (LATTICE) MX

M+(g) + X(g) MX(s) [quite exothermic]

Lattice Energy


49

The energy changes involved in the formation of solid

lithium fluoride from its elements.

From Gas From Gas to Solidto Solid

LatticeLatticeEnergyEnergy


50

The structure of lithium fluoride. Called also the NaCl structure where each ion is surrounded by 6 of the other ions

Applicable for all alkali-metals/halogen except the Cesium salts.


51

Lattice Energy = k( / )QQ r1 2

Q1, Q2 = charges on the ions (Na+, Cl-, Ca++, O--)

r = shortest distance between centers of the cations and anions (Ionic radii increases from top to bottom in a group)

The magnitude of Q’s and sizes of ions (ionic radii) will determine how strong is the lattice.


52

Comparison of the energy changes involved in the formation of solid sodium fluoride and solid magnesium oxide. Note all ions are isoelectronic


53

Lattice energy (E) increases as Q increases and/or

as r decreases.

Compd lattice energyMgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q= +2,-1

Q= +2,-2

r F < r Cl

Electrostatic (Lattice) Energy


54

Partial Ionic Character of the Covalent Bond

Introduce the concept of percent ionic character in a polar covalent bond

% ionic character = % ionic character =

Measured dipole of X-YMeasured dipole of X-Y

Calculated dipole of XCalculated dipole of X++YY--

x 100x 100


55

Experiments showed that none of the studied systems haveExperiments showed that none of the studied systems have100% ionic character despite large differences in electronegativity.100% ionic character despite large differences in electronegativity.

The reason is “Ion Pairing”The reason is “Ion Pairing”

during motion, ions approach closely and for a short periodduring motion, ions approach closely and for a short period of time their charges will be neutralized beforeof time their charges will be neutralized beforemoving away from each other. moving away from each other.

Ion Pairing


56

Molten NaCl conducts an electric current, indicating the presence of mobile Na+ and Cl- ions.

The more mobile the ions theThe more mobile the ions theStronger the current and theStronger the current and thebrighter the lamp!brighter the lamp!


57

The relationship between the ionic character of a covalent bond and the electronegativity difference of the bonded atoms.

>50%>50%IonicIonic

<50%<50%Non-IonicNon-Ionic

Note: These are “Molten” saltsNote: These are “Molten” salts and not “Aqueous” salts!and not “Aqueous” salts!


58

A Model for Covalent Bond

Models are attempts to explain how nature operates on the microscopic level based on experiences in the macroscopic world.


59

Fundamental Properties of Models

A model does not equal reality.

Models are oversimplifications, and are therefore often wrong.

Models become more complicated as they age.

We must understand the underlying assumptions in a model so that we don’t misuse it.


60

Generally, bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms

e.g. CH4 is 1652 KJ lower in energy than 1 mole of C and 4 mole of H.

Stability can be determined in terms if model called “Chemical Bond”

C-H bond energy = = 413 KJ/mol1652 KJ/mol1652 KJ/mol

44


61

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.

H2 (g) H (g) + H (g) H0 = 436.4 kJ

Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ

HCl (g) H (g) + Cl (g) H0 = 431.9 kJ

O2 (g) O (g) + O (g) H0 = 498.7 kJ O O

N2 (g) N (g) + N (g) H0 = 941.4 kJ N N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond


62

Using this model, one can determine other bond energies.Using this model, one can determine other bond energies.

CHCH33Cl is composed of Cl is composed of 3 C-H3 C-H bond and 1 C-Cl bond bond and 1 C-Cl bond

C-HC-H bond is known bond is known 413 KJ/mol

C-ClC-Cl bond energy = 1572 – 3(413) = 339 KJ/mol bond energy = 1572 – 3(413) = 339 KJ/mol


63

Average bond energy in polyatomic molecules

H2O (g) H (g) + OH (g) H0 = 502 kJ

OH (g) H (g) + O (g) H0 = 427 kJ

Average OH bond energy = 502 + 427

2= 464 kJ


64


65

Note : The shorter the bond the higher the bond energy

Bond Energy and Bond Length


66

Bond Energies and Enthalpy of Reaction

Bond breaking requires energy (endothermic).

Bond formation releases energy (exothermic).

H = D(bonds broken) D(bonds formed)

Energy required Energy released

ReactantsReactants ProductsProducts


67

Use bond energies to calculate the enthalpy change for:H2 (g) + F2 (g) 2HF (g)

H0 = BE(reactants) – BE(products)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

9.10


68

Localized Electron Model

A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.


69

Localized Electron Model

1. Description of valence electron arrangement (Lewis structure).

2. Prediction of geometry (VSEPR model).

3. Description of atomic orbital types used to share electrons or hold lone pairs.


70

Lewis Structure

Shows how valence electrons are arranged among atoms in a molecule.

Reflects central idea that stability of a compound relates to noble gas electron configuration.


71

1. Count valence electrons of all atoms and sum up.

2. In ions, add 1 for each negative charge and subtract 1 for each positive charge from the total valence electrons.

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

4. Complete an octet for all terminal atoms first except hydrogen which has duet.

5. If structure contains excess of electrons, put theses electrons on central atom and also try to form double and triple bonds as needed for octet.

Drawing Lewis Structures


72

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

Actual in 3DActual in 3D


73

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond


74

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 bonds double bonds

Triple bond – two atoms share three pairs of electrons

N N8e-8e-

N N

triple bondtriple bond

or


75

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


76

Carbon Dioxide COCarbon Dioxide CO22

oror

OO CC OO....

....

....

....

....

No octet yetNo octet yetOctetOctet

OO CC OO....

....

.... ....

BUTBUTDo not Break Do not Break

SymmetrySymmetry

OO CC OO....

....

....

....


77

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


78

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


79

Comments About the Octet Rule

2nd row elements C, N, O, F observe the octet rule.

2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive (BF3, BeH2, etc..).

3rd row and heavier elements can exceed the octet rule using empty valence d orbitals (SF6, PCl5, etc.).

When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.


80

Note:Note:For molecules that has several Third row (or For molecules that has several Third row (or higher) elements the extra electrons should be higher) elements the extra electrons should be placed on the central atom.placed on the central atom.

II33--

II II II....

....

.... ....

....

....

....

.... ........


81

Resonance

Occurs when more than one valid Lewis structure can be written for a particular molecule.

C=CC=C << C CC C < < C-CC-C……..

These are resonance structures. The actual structure is an average of the resonance structures. The value of the resonance bond is in between a single bond and double bond.


82

Resonance Structure of Ozone: O3

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?


83

How the charges in the previous examples were depicted?

Calculate the Formal charge on each atom.


84

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

Formal Charge


85

Two possible skeletal structures of formaldehyde

(CH2O)

H C O HH

C OH

Calculate and minimize Formal ChargesCalculate and minimize Formal Charges


86

H C O HC – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-



formal charge on C = 4 -2 - ½ x 6 = -1

formal charge on O = 6 -2 - ½ x 6 = +1


=1

2




-

-1 +1


87

C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-



HC O

H

formal charge on C = 4 -0 - ½ x 8 = 0

formal charge on O = 6 -4 - ½ x 4 = 0


=1

2




-

0 0


88

Formal Charge and Lewis Structures

1. For neutral molecules, a Lewis structure in which there are NO formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible (acceptable) than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1 HC O

H

0 0


89

Formal Charge

Not as good Better

CO O(-1) (0) (+1)

CO O(0) (0) (0)

Charges are minimizedCharges are minimized


90

VSEPR Model

The structure around a given atom is determined principally by minimizing electron pair repulsions.


91

Predicting a VSEPR Structure

1. Draw Lewis structure.

2. Put pairs as far apart as possible.

3. Determine positions of atoms from the way electron pairs are shared.

4. Determine the name of molecular structure from positions of the

atoms.


92


93

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


94

Cl ClBe

2 atoms bonded to central atom0 lone pairs on central atom


95

AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar


96


97


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB4 4 0 tetrahedral tetrahedral


98


99


100


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal


101


102


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral


103


104

Molecular structure of PCl6-

Octahedral GeometryOctahedral Geometry


105

Octahedral electron arrangement for Xe


106

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB2E 2 1trigonal planar

bent


107

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3E 3 1


tetrahedraltrigonal

pyramidal


108

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB3E 3 1 tetrahedraltrigonal

pyramidal

AB2E2 2 2 tetrahedral bent

H

O

H

V-ShapeV-Shape


109


110

bonding-pair vs. bondingpair repulsion

lone-pair vs. lone pairrepulsion

lone-pair vs. bondingpair repulsion

< <


111

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

See-SawSee-Saw


112

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

ClF

F

F


113

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

AB2E3 2 3trigonal

bipyramidallinear

I

I

I


114

Three possible arrangements of the electron pairs in the I3

- ion.

Least Lone Least Lone pair repulsionspair repulsions


115

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

Br

F F

FF

F


116

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

AB4E2 4 2 octahedral square planar

Xe

F F

FF


117

Possible electron-pair arrangements for XeF4.


118

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

See-saw / distortedtetrahedron


119

POLARITY OF COMPOUNDSPOLARITY OF COMPOUNDS

ALL MOLECULES WITHOUT LONE PAIRS WILL BE ALL MOLECULES WITHOUT LONE PAIRS WILL BE NON-POLARNON-POLAR

ABAB22, AB, AB33, AB, AB44, AB, AB55, AB, AB66


120

The molecular structure of methanol CH3OH. (a) The arrangement of electron pairs and atoms around the carbon atom. (b) The arrangement of bonding and lone pairs around the oxygen atom. (c) The molecular structure.

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