1 8/7/2015 George Mason University General Chemistry 211 Chapter 10 The Shapes (Geometry) of Molecules Acknowledgements Course Text: Chemistry: the Molecular

104/19/23

George Mason UniversityGeneral Chemistry 211

Chapter 10The Shapes (Geometry) of Molecules

Acknowledgements

Course Text: Chemistry: the Molecular Nature of Matter and Change, 7th edition, 2011, McGraw-

Hill Martin S. Silberberg & Patricia Amateis

The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

Lewis Electron-Dot Symbols A A Lewis electron-dot symbolLewis electron-dot symbol is a symbol in is a symbol in

which the electrons in the valence shell of an which the electrons in the valence shell of an atom or ion are represented by dots placed atom or ion are represented by dots placed around the letter symbol of the elementaround the letter symbol of the element

Note that the Note that the group (column) number group (column) number indicates the number of valence electronsindicates the number of valence electrons

04/19/23 2

.. .:

P: .

:

.SNa . .

. ..SiMg. . Al. .. Cl

:

: .: Ar:

:

::

Group I Group II Group VII Group VIIIGroup VI Group IV Group VGroup III

3s1 3s2 3s23p13s23p2 3s23p3 3s23p4 3s23p5 3s23p6

Lewis Electron-Dot Formulas A Lewis electron-dot formula is an illustration

used to represent the transfer of electrons during the formation of an ionic bond

The Magnesium has two electrons to give, whereas the Fluorines have only one “vacancy” each

Consequently, Magnesium can accommodate two Fluorine atoms

04/19/23 3

::

F.:

:

:F .: Mg. .

Mg[ F ]

:

:

:

:- 2+

[ F ]

:

:

:

:-

Lewis Structures The tendency of atoms in a molecule to have

eight electrons (ns2np6) in their outer shell (two for hydrogen) is called the octet rule

You can represent the formation of the covalent bond in H2 as follows:

This uses the Lewis dot symbols for the hydrogen atom and represents the covalent bond by a pair of dots

4

+H H H H

04/19/23

The Electron Probability Distribution for the H2 Molecule

504/19/23

Lewis Structures The shared electrons in H2 spend part of

the time in the region around each atom

In this sense, each atom in H2 has a helium (1s2) configuration

6

:H H

04/19/23

Lewis Structures The formation of a bond between H and Cl

to give an HCl molecule can be represented in a similar way

Thus, hydrogen has two valence electrons about it (as in He) and Cl has eight valence electrons about it (as in Ar)

7

:H

:

::Cl.H .

::

Cl :+

04/19/23

Lewis Structures Formulas such as these are referred to as

Lewis electron-dot formulas or Lewis structures

An electron pair is either a: bonding pair (shared between two

atoms) lone pair (electron pair that is not shared)

Hydrogen has no unbonded pairs

Chlorine has 3 unbonded pairs 8

::H Cl

::

04/19/23

Lewis Structures Rules for obtaining Lewis electron dot

formulas

Calculate the number of valence electrons for the molecule from:

group # for each atom (1-8)

add the charge of Anion

subtract the charge of a Cation

Put atom with the lowest group number and lowest electronegativity as the central atom

Arrange the other elements (ligands) around the central atom

904/19/23

Lewis Structures Rules for Lewis Dot Formulas

Distribute electrons to atoms surrounding the central atom to satisfy the octet rule for each atom

Distribute the remaining electrons as pairs to the central atom

If the Central atom is deficient in electrons to complete the octet; move electron pairs from surrounding atoms to complete central atom valence electron needs, that is form one or more double bonds (possibly triple bonds) around the central atom

04/19/23 10

Practice Problem Write a lewis structure for CCl2F2

Step 1: Arrange Atoms (Carbon is “Central Atom” because is has the lowest group number and lowest electronegativity

Step 2: Determine total number of valence electrons

1 x C(4) + 2 x Cl(7) + 2 x F(7) = 32

Step 3: Draw single bonds to central atom and subtract 2 e- for each single bond (4 x 2 = 8) 32 – 8 = 24 remaining

Step 4: Distribute the 24 remaining electrons in pairs around surrounding atoms (3 electron pairs around each Fluoride atom)

1104/19/23

Writing Lewis Dot Formulas The Lewis electron-dot formula of a

covalent compound is a simple two-dimensional representation of the positions of electrons in a molecule

Bonding electron pairs are indicated by either two dots or a dash

In addition, these formulas show the positions of lone pairs of electrons

04/19/23 12

Writing Lewis Dot Formulas The following rules allow you to write

electron-dot formulas even when the central atom does not follow the octet rule

To illustrate, draw the structure of:

Phosphorus Trichloride

3PCl04/19/23 13

Con’t on next slide

Writing Lewis Dot Formulas Step 1: Total all valence electrons in the

molecular formula. That is, total the group numbers of all the atoms in the formula

For a polyatomic anion, add the number of negative charges to this total

For a polyatomic cation, subtract the number of positive charges from this total

04/19/23 14

3PCl5 e-

(7 e-) x 3

P 3s23p3 Cl 3s23p5

(2+3) + 3x(2+5) = 5+21

26 total electrons


Writing Lewis Dot Formulas Step 2:

Arrange the atoms radially, with the least electronegative atom in the center

Place one pair of electrons between the central atom and each peripheral atom

PClCl

Cl

26 – 6 = 20 remaining

04/19/23 15


Writing Lewis Dot Formulas Step 3: Distribute the remaining electrons

to the peripheral atoms to satisfy the octet rule

16

P

Cl: :

Cl

::

::

Cl:

::

26 – (3 x 6 + 6) = 2 remaining

04/19/23


Writing Lewis Dot FormulasStep 4: Distribute any remaining electrons

(2) to the central atom. If the number of electrons on the central atom is less than the number of electrons required to complete the octet for that atom, use one or more electrons pairs from other atoms to form double or triple bonds

17

PClCl

Cl: :::

04/19/23

::

:::

:

Phosphorus has an octet of electronsNo double bonds required

Exceptions to the Octet Rule Although many molecules obey the octet

rule, there are exceptions where the central atom has more than eight electrons

Generally, if a nonmetal is in the third period or greater it can accommodate as many as twelve electrons, if it is the central atom

These elements have unfilled “d” subshells that can be used for bonding

1804/19/23

Exceptions to the Octet Rule For example, the bonding in phosphorus

pentafluoride, PF5, shows ten electrons surrounding the phosphorus

19

Total valence electrons 5 x 7 (F) + 5 (P) = 40

Distribute electrons to F atoms5 x 6 = 30

Establish bonding pairs5 x 2 = 10

Remaining electrons40 – 30 – 10 = 0

Phosphorus has “0” non-bonding pairs

04/19/23

: F :

:

F

FFP

F: ::

:::

:

::

:

::

Since Phosphorus is in Period 3, PF5 is a “hypervalent” moleculeThe Phosphorus utilizes electrons from other shells(vacant orbitals) to create a valence shell with more than 8 electrons

Exceptions to the Octet Rule In Xenon Tetrafluoride, XeF4, the Xenon atom

must accommodate two extra lone pairs

20

F :

::

: F

::

XeF :

::

: F

::

::

Total valence electrons 4 x 7 + 8 = 36

Distribute electrons to F atoms4 x 6 = 24

Establish bonding pairs4 x 2 = 8

Remaining electrons36 – 24 – 8 = 4

Add 2 non-bonding pairs to Xe

Xe violates “octet” rule

XeF4 is a “hypervalent” molecule and utilizes vacant “d” orbitals to create a valence shell with more than 8 electrons 04/19/23

Delocalized Bonding: Resonance The structure of Ozone, O3, can be

represented by two different Lewis electron-dot formulas

Experiments show, however, that both bonds are identical

21

O O

O:: :

::

:

OO

O :::

::

:

or

04/19/23

Ozone (O3)

Delocalized Bonding: Resonance

According to Resonance Theory, these two equal bonds are represented as one pair of bonding electrons spread over the region of all three atoms

This is called delocalized bonding, in which a bonding pair of electrons is spread over a number of atoms

22

OO

O

04/19/23

Ozone (O3)

Resonance & Bond Order Recall (Chap 9) – Bond Order

The number of electron pairs being shared by any pair of “Bonded Atoms” or

The number of electron pairs divided by the number of bonded-atom pairs

Ex. Ozone

04/19/23 23

Electron PairsBond Order =

Bonded - Atom Pairs

3

2

Electron PairsBond Order = = = 1.5

Bonded - Atom Pairs

Practice Problem

04/19/23 24

2

1

Electron PairsBond Order = = = 2

Bonded - Atom Pairs


Bonded - Atom Pairs1

= = 11


Bonded - Atom Pairs1

= = 11

C3H4

C3H6

Practice Problem

04/19/23 25

2

1


Bonded - Atom Pairs

Electron Pairs 1Bond Order = = = 1

Bonded - Atom Pairs 1

2

1


Bonded - Atom PairsElectron Pairs 1

Bond Order = = = 1Bonded - Atom Pairs 1

C4H4

C4H6

Practice Problem

04/19/23 26

Electron Pairs 9Bond Order = = = 1.5

Bonded - Atom Pairs 6Note : There are 6 equivalent C - C bonds in the aromatic Benzene molecule

C6H6

Formal Charge & Lewis Structures In certain instances, more than one feasible Lewis

structure can be illustrated for a moleculeFor example, H, C and N

The concept of “formal charge” can help discern which structure is the most likely

Formal Charge: An atom’s formal charge is:

Total number of valence electrons Minus all unshared electrons Minus ½ of its shared electrons

Formal Charges must sum to actual charge of species: Zero Charge for a Molecule Ionic Charge for an Ion

27

H C N CNHor: :

04/19/23

Formal Charge & Lewis Structures

When you can write several Lewis structures, choose the one having the least formal charge

28

H C N CNHor: :1 e- 4 e- 5 e- 1 e- 5 e- 4e-

“domain” electrons

group number

I IV V I V IV

-1+1

FCH: [1 - 0 - ½(2)] = 0

FCC: [4 - 0 - ½(8)] = 0

FCN: [5 - 2 - ½(6)] = 0

FCH: [1 - 0 - ½(2)] = 0

FCC: [4 - 2 - ½(6)] = -1

FCN: [5 - 0 - ½(8)] = +1

04/19/23

Note: HCN is a neutral molecule

Sum of Formal Charges in the preferred form (0) equals molecular charge (0)

FC: Total Valence e- – unshared e- – ½ shared

e-

Form I Form II

Preferred Form - Form I (Least Formal Charge)


29

Ozone

FCOA: [6 - 4 - ½(4)] = 0

FCOB: [6 - 2 - ½(6)] = +1

FCOC: [6 - 6 - ½(2)] = -1

FCOA: [6 - 6 - ½(2)] = -1

FCOB: [6 - 2 - ½(6)] = +1

FCOC: [6 - 4 - ½(4)] = 0

Both “Resonance” forms have the same formal chargeand thus, are identical

Note: Ozone (O3) is a neutral molecule Sum of Formal Charges (0) equals molecular charge (0)

04/19/23


04/19/23 30

BF F

F

BF F

F

FC B = 3 – 0 -(1/2 * 6)

= 0Even though B violates “Octet Rule”, this is the preferred form because it has “less” formal charge

FC B = 3 – 0 -(1/2 * 8) = -1FC F = 7 – 4 - (1/2 * 4) = +1

SOO

SOO

Boron TrifuorideBF3

Sulfur DioxideSO2

FC S = 6 – 2 – (1/2 * 6) = 1FC S = 6 – 2 – (1/2 * 8) = 0

Preferred Form (Less Formal Charge)

Resonance/Formal Charge – Nitrate Ion

04/19/23 31

Total Valence electrons - 3 x 6 (O) + 1 x 5 (N) + 1 (ion charge) = 24

Add 1 pair electrons between central atom and each other atom – 3 x 2 = 6

Add electrons to oxygen atoms to complete octet Nitrogen still missing 2 electrons to complete octet Borrow 2 electrons from one oxygen to form double bond Formal Charge – Nitrogen: 5 – (0 + ½*8) = 5 – 4 = +1 Formal Charge – Single bonded Oxygen: 6 – (6 + ½*2) = 6 – 7 = -1 x

2 = -2 Formal Charge – Double bonded Oxygen: 6 – (4 + ½*4) = 6 – 6 = 0 Net Charge of the ION is: +1 +(-2) + 0 = -1

Resonance/Formal Charge – Cyanate Ion

04/19/23 32

FCN = 5 – (6 + ½*2) = -2FCC = 4 – (0 + ½*8) = 0FCO = 6 – (2 + ½*6) = +1

FCN = 5 – (4 + ½*4) = -1FCC = 4 – (0 + ½*8) = 0FCO = 6 – (4 + ½*4) = 0

FCN = 5 – (2 + ½*6) = 0FCC = 4 – (0 + ½*8) = 0FCO = 6 – (6 + ½*2) = -1

Preferred Form:

Eliminate I – Higher formal charge on Nitrogen than Carbon & Oxygen Positive formal charge on Oxygen, which is more electronegative than NitrogenEliminate II – Forms II & III have the same magnitude of formal charges, but form III has a -1 charge on the more electronegative Oxygen atomForms II & III both contribute to the resonant hybrid of the Cyanate Ion, but form III is the more important

Note: Net formal charge in form III is same as ionic charge (-1)

Formal Charge vs Oxidation No

“Formal Charge” is used to examine resonance hybrid structures , whereas “Oxidation Number” is used to monitor “REDOX” reactions Formal Charge - Bonding electrons are assigned

equally to the atoms as if the bonding were “Nonpolar” covalent, i.e., each atom has half the electrons making up the bond Formal Charge = valence e- – (unbonded e- +

½ bonding e-) Oxidation Number - Bonding electrons are

transferred completely to the more electronegative atom, as if the bonding were “Ionic” Ox No. = valence e- – (unbonded e- + bonding

e-)04/19/23 33

Formal Charge vs Oxidation No

04/19/23 34

FC (-2) (0) (+1) (-1) (0) (0) (0) (0) (-1)

ON (-3) (+4) (-2) (-3) (+4) (-2) (-3) (+4) (-2)

Note: Oxidation Nos do not change from one resonance form to another (electronegativities remain same)

N 5 – (6 + ½ (1)) = -2C 4 – (0 + ½ (8)) = 0O 6 – (2 + ½ (6)) = +1

N 5 – (4 + ½ (4)) = -1C 4 – (0 + ½ (8)) = 0O 6 – (4 + ½ (4)) = 0

N 5 – (2 + ½ (6)) = 0C 4 – (0 + ½ (8)) = 0O 6 – (6 + ½ (2)) = -1

N 5 – (6 + 2) = -3C 4 – (0 + 0) = +4O 6 – (2 + 6) = -2

N 5 – (4 + 4) = -3C 4 – (0 + 0) = +4O 6 – (4 + 4) = -2

N 5 – (2 + 6) = -3C 4 – (0 + 0) = +4O 6 – (6 + 2) = -2

Note: Both Nitrogen (N) & Oxygen (O) are more electronegative than Carbon (C); thus, in the computation of Oxidation Number all the electrons are transferred to the N & O leaving C with no lone pairs and no bonded pairs

The Valence-Shell Electron Pair Repulsion Model (VSEPR)

04/19/23 35

The Valence-Shell Electron Pair Repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence shell electron pairs are arranged as far from one another as possible

Molecular geometry – The shape of a molecule is determined by the positions of atomic nuclei relative to each other, i.e., angular arrangement

Central Atom

Place atom with “Lower Group Number” in center(N in NF3 needs more electrons to complete octet)

If atoms have same group number (SO3 or ClF3), place the atom with the “Higher Period Number” in the center (Sulfur & Chlorine)

VSEPR Model of Molecular Shapes

The following rules and figures will help discern electron pair arrangements Select the Central Atom (Least

Electronegative Atom) Draw the Lewis structure Determine how many bonding electron

pairs are around the central atom Determine the number of non-bonding

electron pairs Count a multiple bond as “one pair” Arrange the electron pairs as far apart as

possible to minimize electron repulsions Note the number of bonding and lone pairs

04/19/23 36

VSEPR Model of Molecular Shapes To predict the relative positions of atoms around

a given atom using the VSEPR model, you first note the arrangement of the electron pairs around that central atom

Molecular Notation:A – The Central Atom (Least Electronegative atom)X – The Ligands (Bonding Pairs)a – The Number of LigandsE – Non-Bonding Electron Pairsb – The Number of Non-Bonding Electron Pairs

Double & Triple Bonds count as a “single” electron pair

The Geometric arrangement is determined by: sum (a + b)

04/19/23 37

AXaEb


04/19/23 38

Molecule

LewisStructur

e

ALxNy

Notation

Geometrice- Pair

Arrangement

3-DStructure

3-D View&

Isomers

BeH2 AX2E0

a = 2b = 0a + b = 2

Linear

BH3 AX3E0

a = 3b = 0a + b = 3

TrigonalPlanar

CH2Li2

AX4E0

orAX2X2E0

a = 4b = 0a + b = 4

Tetrahedral

OH2 AX2E2

a = 2b = 2a + b = 4

Tetrahedral

BiF5 AX5E0

a = 5b = 0a + b = 5

TrigonalBipyramidal


04/19/23 39

Molecule

LewisStructure

ALxNy

Notation

Geometrice- Pair

Arrangement

3-DStructure

3-D View&

Selected Isomers

SF5Cl

AX6E0

or AX5X1E0

a = 6b = 0a + b = 6 Octahedral

PCL4Br

AX5N0

orAX4X1E0

a = 5b = 0a + b = 5

TrigonalBipyramid

al

TeCl3Br

AX4E1

orAX3X1E1

b = 4a = 1a + b = 5

TrigonalBipyramid

al

SF4Cl2

AX6E0

orAX4X2E0

a = 6b = 0a + b = 6 Octahedral

XeF2 AX2E3

a = 2b = 3a + b = 5

TrigonalBipyramid

al

Arrangement of Electron Pairs About an Atom: Basic Shapes

04/19/23 40

CS2 HCN BeF2

NO2+


04/19/23 41

SO3 BF3 NO3− NO2

– CO32−

SO2 O3 PbCl2 SnBr2


04/19/23 42

CH4 SiCl4

SO42-

ClO4-

NH3 PF3 ClO3 H3O+

H2O OF2 SCl2


04/19/23 43

SF4, XeO2F2, IF4+,

IO2F2-

ClF3 BrF3

XeF2 I3- IF2

-

PF5 AsF5 SOF4


04/19/23 44

SF6

IOF5

BrF5 TeF5-

XeOF4

XeF4 ICl4

-

Electron Pair Arrangement

04/19/23 45

Electron Pair Arrangement

04/19/23 46

Linear Geometry Two electron pairs (linear arrangement)Two electron pairs (linear arrangement)

Double bonds count as a Double bonds count as a “single electron pair”

2 bonding pairs2 bonding pairs

0 non-bonding pairs0 non-bonding pairs

AXAXaaEEb b = a + b = 2 + 0 = 2 (Linear) = a + b = 2 + 0 = 2 (Linear)

Thus, according to the VSEPR model, the bonds Thus, according to the VSEPR model, the bonds are arranged linearly (bond angle = 180are arranged linearly (bond angle = 180oo))

Molecular shape of carbon dioxide is Molecular shape of carbon dioxide is linearlinear

04/19/23 47

::

::

Carbon is central atom because it has lower group number

Trigonal Planar Geometry Three electron pairs on Central atom

The three groups of electron pairs are arranged in a trigonal plane. Thus, the molecular shape of COCl2 is trigonal planar. The Bond angle is 120o

04/19/23 48

Central Atom - Carbon

3 bonding electron pairs

(double bond counts as 1 pair)

0 non-bonding electron pairs

a + b = 3 + 0 = 3

Trigonal PlanarCl

C

:

::

O

Cl :

::

: :

Trigonal Planar Geometry Effect of Double Bonds

Bond angles deviate from ideal angles when surrounding atoms and electron groups are not identical

A double bond has greater electron density and repels two single bonds more strongly than they repel each other

04/19/23 49

CH

H

O

122o

116o

Actual

C

H

H

O

120o

120o

Ideal

Trigonal Planar Geometry Effect of Lone Pairs

The molecular shape is defined only by the positions of the nuclei

When one of the three electron pairs in a trigonal planar molecule is a lone (non-bonding) pair, it is held by only one nucleus

It is less confined and exerts a stronger repulsive force than a bonding pair

This results in a decrease in the angle between the bonding pairs

04/19/23 50

The normal Trigonal Planar angle between the bonding pairs is

120o

Trigonal Planar Geometry Three electron pairs (Effect of ‘Lone’ pairs)

(trigonal planar arrangement)

Ozone has two bonding electron pairs about the central oxygen (double bond counts as 1 pair)

There is one non-bonding lone pair These groups have a:

Trigonal Planar arrangementAXaEb (a + b) = 2 + 1 = 3

Since one of the groups is a lone pair, the molecular geometry is described as bent or angular

04/19/23 51

O O

O:: :

:

:

:

SO3

BF3

NO3-

CO32-

<120o

Tetrahedral Geometry Four electron pairs

(Tetrahedral Arrangement)

Four electron pairs about the central atom lead to three different molecular geometries

a + b = 4 + 0 a + b = 3 + 1 a + b = 2 + 2

= 4 = 4 = 404/19/23 52

:Cl:

:::Cl:

:Cl

:: C Cl:

::

H

N

H

H :

:O

H

H :

Tetrahedral Geometry Molecular Geometries produced by

variable non-bonding electron pairs

04/19/23 53

:

O

H

H :

H

NH H

:

:Cl:

:

:Cl

::

C Cl:

::

:Cl: :

AX4

109o

AX4E107o

AX2E2

105o

CH4, SiCl4, SO42-, ClO4- PF3, ClO3

-, H3O+ OF2, SCl2

Note impact of non-bonding electron pairs on bond angle

107 o 105o107o

Trigonal Bipyramidal Five electron pairs

(trigonal bipyramidal arrangement)

This structure results in both 90o and 120o bond angles

04/19/23 54

: F :

::: F :

F :

::

: F

::

PF :

::

ASF5 SOF4

90o axial120o equatorial

Trigonal Bipyramidal Other molecular geometries are possible when

one or more of the electron pairs is a lone pair

04/19/23 55

XeO2F2 IF4+ IOF2

- ClF3 BrF3 XeF2 I3- IF2

-

S

F

F

F

F: Cl

F

F

:

:F Xe

F

F

::

:

<90o (ax)<120o (eq)

<90o (ax)180o

Octahedral Geometry Six electron pairs

(Octahedral arrangement)

This octahedral arrangement results in:

90o bond angles

04/19/23 56

F:

::

:F

::

SF:

::

:F

::

:F:

:

:F: : SF6 IOF5

90o

Other Geometries Six electron pairs

(octahedral arrangement)

04/19/23 57

square pyramidal square planar

BrF5 TeF5- XeOF4 XeF4 ICl4

-

Iodine violates octet ruleIodine is sp3d2 hybridizedIodine uses d orbitals

Noble gases not always inertXenon forms 6 electron domains

XeF

F :

F

F

:

I

F

F

F

:

F

F

<90o

90o

Practice ProblemIn the ICl4– ion, the electron pairs are arranged around the central iodine atom in the shape of

a. a tetrahedron

b. a trigonal bipyramid

c. a square plane

d. an octahedron

e. a trigonal pyramid

Ans: a

04/19/23 58

AXaEb

a + b = 4 + 0 = 4 (AX4 – Tetrahedral)

ICl

Cl

Cl

Cl

AX4

Dipole Moment The dipole moment () is a measure of the

degree of charge separation (molecular polarity) in a molecule The product of the magnitude of the charge

Q at either end of the molecular dipole times the distance r between the charges

= Q rExample: What is the dipole moment in

Debyes of a molecule with a:

Bond Length = 127 pm Electron Charge (e) 1.6x10-19 C 1 Debye = 3.34 x 10-30 C m

04/19/23 59

-12-19

-30

1 10 m 1Dμ = Qr = 1.60 10 C (127pm) ( ) ( ) = 6.08 D

pm 3.34 10 C •m)

Dipole Moment In the previous example the Dipole

Moment was calculated on the assumption that each of the atoms bore a full charge of 1 electronic unit , e, (+1 & -1).

That is: Q = 1 e = 1.60 x 10-19 Coulombs (C)

Such a molecule would be nearly ionic Atoms in asymmetric covalent molecules

would not exhibit full ionic charges, thus, the value of Q would be less than 1 e, depending on the relative electronegativity differences and the bond length

04/19/23 60

Dipole Moment andMolecular Geometry

04/19/23 61

The polarity of individual bonds within a molecule can be viewed as vector quantities

Thus, molecules that are perfectly symmetric have a zero dipole moment.

These molecules are considered nonpolar


04/19/23 62

However, molecules that exhibit any asymmetry in the arrangement of electron pairs would have a nonzero dipole moment. These molecules are considered polar

H

NH H

:

NH3 PF3 ClO3

H3O+


04/19/23 63

Formula Molecular Geometry Dipole Moment

AX Linear Can Be nonzero

AX2E0 Linear Zero

AX3E0 Trigonal Planar Zero

AX2E1 Trigonal Planar Bent Can Be nonzero

AX4E0 Tetrahedral Zero

AX3E1 Tetrahedral Trigonal Pyramidal Can Be nonzero

AX2E2 Tetrahedral Bent Can Be nonzero

AX5E0 Trigonal Bipyramidal Zero

AX4E1 Trigonal Bipyramidal SeeSaw Can Be nonzero

AX3E2 Trigonal Bipyramidal T-Shaped Can Be nonzero

AX2E3 Trigonal Bipyramidal Linear Can Be nonzero

AX6E0 Octahedral Zero

AX5E1 Octahedral Square Pyramidal Can Be nonzero

AX4E2 Octahedral Square Planar Zero

Practice ProblemThe Nitrogen atom would be expected to have the positive end of the dipole in the species

a. NH4+

b. Ca3N2

c. HCN

d. AlN

e. NO+

Ans: e

04/19/23 64

N is more Electronegative than HN is more EN than CaN is more EN than CN is more EN than AlO is more EN than Nitrogen

Practice ProblemWhich of the following molecules is polar? a. BF3 b. CBr4 c. CO2

d. NO2 e. SF6

Ans: d

04/19/23 65

NO O ● ●

● ● ● ●

● ●

● ●

● ●

●

The Lewis structures for BF3, CBr4, CO2, and SF6 do not have any non-bonding electrons on the central atomThe Lewis structure for NO2 shows one double bond and a lone non-bonding electron on the NitrogenThe VSEPR Molecular Geometry for NO2 is AX2E1 (a + b = 2 + 1 = 3) - Trigonal Planar

Formal Charge on N is 5 – 1 – ½ (6) = +1NO2 molecule is polar

NO O

● ●

● ●

● ●

● ●

● ●

● +

-N

O O

● ●

● ●

● ●

● ●

● ●

+●

-

Practice ProblemWhich of the following compounds is nonpolar? a. XeF2 b. HCl c. SO2

d. H2S e. N20

Ans: a

HCL is ionic and very polar SO2 has AX2E1 Trigonal Planar Bent geometry with

a dipole moment (polar) H2S has AX2E2 Tetrahedral Bent geometry and

with a dipole moment (polar) N2O has AX2E0 linear with asymmetric geometry.

Since oxygen is more EN than N, the molecule is polar

XeF2 has AX2E3 Trigonal Bypyramidal Geometry, but linear molecular geometry (nonpolar)

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Equation Summary

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Bonded - Atom Pairs

)- - -Formal Charge (FC) = Total Valence e - (Non - Bonding e + 1 / 2 Bonding e

)- - - Oxidation Number (ON) = Total Valence e - (Non - Bonding e + Bonding e

VSEPR Model - AXaEb

Geometric Configuration Determined by the sum (a + b)

Dipole Moment = = Q x r

Documents

1 8/7/2015 George Mason University General Chemistry 211 Chapter 10 The Shapes (Geometry) of Molecules Acknowledgements Course Text: Chemistry: the Molecular