CH 3- CHEMICAL BONDING Jun 2014 Pt1.pdf

7/21/2019 CH 3- CHEMICAL BONDING Jun 2014 Pt1.pdf

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Chemical Bonding

Chapter 3

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


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Chemical Bonding

3.0 Introduction

3.1 Lewis Structure

3.2 Ionic Bond

3.3 Covalent Bond

3.4 Molecular Geometry and VSEPR theory

3.5 Valence bonds theory and hybridizations

3.6 Intermolecular Forces

3.7 Metallic Bonding

2


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3.0 Introduction to Chemical Bonds

Ionic Bonds

When a metal atom loses electrons it becomes

a cation

metals relatively easy to remove electrons

from them When a nonmetal atom gains electrons it

becomes an anion

Nonmetals are easier to add electrons to

these atoms

The oppositely charged ions are then attracted

to each other, resulting in an ionic bond


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5

Li + F Li+ F -

The Ionic Bond

1s22s1 1s22s22p5 1s2 1s22s22p6

[He] [Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -

LiF

Ion ic bond:the electrostatic force that holds ions together in

an ionic compound.


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3.0 Introduction to Chemical BondsCovalent Bonds

6

Covalent bond:chemical bond in which two

or more electrons are shared by two atoms.

Covalent compounds are compounds that

contain only covalent bonds.

In a covalent bond, each electron in a shared

pair is attracted to the nuclei of both atoms.

This attraction holds the two atoms together.

H + H H:H or H-H


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8

8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bondtwo atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e- double bonds

Triple bondtwo atoms share three pairs of electrons

N N

8e-8e-

N N

triple bond

or


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3.0 Introduction to Chemical Bonds

Three Models of Chemical Bonding.


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Topic

3.1 Lewis Structure

G.N. Lewis (1875-1946)


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3.1 Lewis Structure

Lewis Theory- using valence electrons to explain

bonding of atoms

Using Lewis Theory, we can draw models called

Lewis Structures or known as Electron-DotStructures.

Lewis structure represent the valence electrons of

main-group elements as dots surrounding the

symbol for the particular element.

13


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Valence Electrons

Because chemical bonding involves the transferor sharing of electrons between two or more

atoms, valence electrons are most important in

bonding

Lewis theory focuses on the behavior of thevalence electrons

3.1 Lewis Structure


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Determining the Number of Valence

Electrons in an Atom The column number on the Periodic Table will tell you

how many valence electrons a main group atom has Transition Elements all have two valence electrons. Why?

3.1 Lewis Structure


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Lewis Electron-Dot Symbols

Nitrogen, N, is in Group 5A, therefore has 5 valence electrons.

To draw the Lewis symbol for any main-groupelement:

Note the A-group number, which gives the

number of valence electrons.

Place one dot at a time on each of the four sidesof the element symbol.

Keep adding dots, pairing them, until all are

used up.

N

N

N

N

or or or

3.1 Lewis Structure


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Lewis Electron-Dot Symbols for Elements

in Periods 2 and 3.

3.1 Lewis Structure


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19

Lewis Dot Symbols for the Representative Elements &

Noble Gases


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20

1. Draw skeletal structure of compound showing

what atoms are bonded to each other. Put leastelectronegative element in the center.

2. Count total number of valence e-. Add 1 for each

negative charge. Subtract 1 for each positivecharge.

3. Complete an octet for all atoms excepthydrogen.

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

needed.

Writing Lewis Structures

E l 1


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Example 1

Write the Lewis structure for nitrogen trifluoride (NF3) in which

all three F atoms are bonded to the N atom.

NF3is a colorless, odorless, unreactive

gas.


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E l 1


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Example 1

Step 3: We draw a single covalent bond between N and each

F, and complete the octets for the F atoms. We place

the remaining two electrons on N:

Because this structure satisfies the octet rule for all the atoms,

step 4 is not required.

CheckCount the valence electrons in NF3(in bonds and in

lone pairs). The result is 26, the same as the total number of

valence electrons on three F atoms (3 7 = 21) and one N

atom (5).

E l 2


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Example 2

Write the Lewis structure for nitric acid (HNO3) in which the

three O atoms are bonded to the central N atom and the

ionizable H atom is bonded to one of the O atoms.

HNO3is a strong electrolyte.

E l 2


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Example 2

Solut ionWe follow the procedure already outlined for writing

Lewis structures.

Step 1: The skeletal structure of HNO3is

Step 2: The outer-shell electron configurations of N, O, and H

are 2s2

2p3

, 2s2

2p4

, and 1s1

, respectively. Thus, thereare 5 + (3 6) + 1, or 24, valence electrons to

account for in HNO3.

E l 2


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Example 2

Step 3: We draw a single covalent bond between N and each

of the three O atoms and between one O atom and

the H atom. Then we fill in electrons to comply withthe octet rule for the O atoms:

Step 4: We see that this structure satisfies the octet rule for

all the O atoms but not for the N atom. The N atom

has only six electrons. Therefore, we move a lone

pair from one of the end O atoms to form another

bond with N.


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Topic

3.2 Ionic Bond


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3.2 Ionic Bond

Lewis Structures of Ions

Cations have Lewis symbols withoutvalence

electrons

electrons lost in the cation formation

Anions have Lewis symbols witheight valence

electrons

electrons gained in the anion formation

Example:


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The Octet Rule for Ionic Bonding

Metals form cations by losing enough electrons

until they get the same electron configuration as

the previous noble gas

Nonmetals form anions by gaining enough

electrons until they get the same electron

configuration as the next noble gas

The noble gas electron configurations are very

stable because they have 8 electrons (an octet) in

their outermost shells (except He)

3.2 Ionic Bond


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Stable Electron Arrangements

and Ion Charge

3.2 Ionic Bond


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Lewis Theory and Ionic Bonding

Lewis symbols can be used to represent the

transfer of electrons from metalatom to nonmetal

atom, resulting in ions that are attracted to each

other and therefore bond

+

3.2 Ionic Bond


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Orbital diagram

Another Way to Depict Electron Transfer in

the Formation of Li+and F-.

Electron configuration

Li 1s2

2s1

+ F 1s2

2p5

Li+

1s2

+ F-

1s2

2s2

2p6

Li

1s 2p

2s

1s 2p

2s

F

+

1s 2p2s

Li+

1s 2p

2s

F-

3.2 Ionic Bond


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Predicting Ionic Formulas

Using Lewis Symbols Electrons are transferred until the metal loses all

its valence electrons and the nonmetal has an

octet

Numbers of atoms are adjusted so the total

number of electrons lost balance up with number

of electrons gained

Lewis structure Formula

Li2O

3.2 Ionic Bond


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CHECKPOINT 1:

Depicting Ion Formation

SOLUTION:

Use partial orbital diagrams and Lewis symbols to

depict the formation of Na+and O2ions from the

atoms, and determine the formula of the compoundformed.

2Na+ +

2-O

Na

O

Na

3.2 Ionic Bond


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CHECKPOINT 2

Using Lewis theory to predict chemical formulas of ionic

compounds

Predict the formula of the compound that forms between

calcium and chlorine.

Draw the Lewis dot symbolsof the elements.

Ca

Cl

Transfer all the valence electrons

from the metalto the nonmetal,

adding more of each atom as you go,until all electrons are lost from the

metal atoms and all

nonmetal atoms have eight electrons.

CaCl

Cl

Ca2+

CaCl2

3.2 Ionic Bond


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Periodic Trends in Lattice Energy

Latt ice energyis the energy required to separate 1 mol of an

ionic solid into gaseous ions.

Lattice energy is a measure of the strength of the ionic bond.

3.2 Ionic Bond

Lattice energy is affected by ionic sizeand ionic charge.

As ionic size increases, lattice energy decreases. Lattice

energy therefore decreases down a group on the periodictable.

As ionic charge increases, lattice energy increases.


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Lattice Energy vs. Ion Size

3.2 Ionic Bond


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Lattice Energy vs Ion Charge

Lattice Energy =

910 kJ/mol

Lattice Energy =

3414 kJ/mol

3.2 Ionic Bond


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3.2 Ionic Bond


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Example 3: Order the following ionic compounds in

order of increasing magnitude of lattice energy:

CaO, KBr, KCl, SrO

First examine the ion charges

and order by sum of the charges

Ca2+& O2-,

K+& Br,

K+& Cl,

Sr2+& O2

(KBr, KCl) < (CaO, SrO)

Then examine the ion sizes of

each group and order by

radius; larger < smaller

(KBr, KCl) same cation,

Br> Cl(same Group)

KBr < KCl < (CaO, SrO)

(CaO, SrO) same anion,

Sr2+> Ca2+(same Group)

KBr < KCl < SrO < CaO


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1. High melting and boiling point Lewis theory predicts ionic compounds should have high

melting points and boiling points because breaking down

the crystal should require a lot of energy the stronger the attraction (larger the lattice energy), the higher the

melting point

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Properties of Ionic Compounds3.2 Ionic Bond

NaCl (s) NaCl (liquid)


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CHECKPOINT 3

Which ionic compound below has the highest

melting point? KBr (734 C)

CaCl2(772 C)

MgF2(1261 C)

KBr

CaCl2

MgF2


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Properties of Ionic Compounds

2. Hard and brittle crystalline solid Lewis theory implies that the positions of the ions

in the crystal lattice are critical to the stability ofthe structure

Lewis theory predicts that moving ions out ofposition should therefore be difficult, and ionicsolids should be hard

hardness is measured by rubbing two materials

together and seeing which streaks or cuts the other the harder material is the one that cuts or doesnt

3.2 Ionic Bond


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Lewis theory implies that if the ions are displacedfrom their position in the crystal lattice, that repulsiveforces should occur

This predicts the crystal will become unstable and

break apart. Lewis theory predicts ionic solids will bebrittle.

Ionic solids are brittle. When struck they shatter.

+ - + + + ++ + + +- -

--

--

--

+ - + + + ++ + + +

- -

-

-

-

-

-

-

+ - + + + ++ + + +- - - - - - - -

3.2 Ionic Bond


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3. Good electrical conductor in liquid state

Lewis theory implies that, in the ionic solid, the ions

are locked in position and cannot move around

does not conduct electricity Lewis theory implies that, in the liquid state or when

dissolved in water,the ions will have the ability to

move around

conduct electricity

4. Many ionic solids are soluble in waterbut not inorganic solvents

3.2 Ionic Bond

Example 4


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Example 4

StrategyWe use electroneutrality as our guide in writing

formulas for ionic compounds, that is, the total positive charges

on the cations must be equal to the total negative chargeson the anions.

Solut ionThe Lewis dot symbols of Al and O are

Because aluminum tends to form the cation (Al3+) and oxygen

the anion (O2) in ionic compounds, the transfer of electrons is

from Al to O. There are three valence electrons in each Al

atom; each O atom needs two electrons to form the O2ion,

which is isoelectronic with neon.

Example 4


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Example 4

Thus, the simplest neutralizing ratio of Al3+to O2is 2:3; two

Al3+ions have a total charge of +6, and three O2ions have a

total charge of 6. So the empirical formula of aluminum oxideis Al2O3, and the reaction is

CheckMake sure that the number of valence electrons (24) is

the same on both sides of the equation. Are the subscripts in

Al2O3reduced to the smallest possible whole numbers?


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3.3 Covalent Bond

Lewis theory implies that another way atoms can

achieve an octet of valence electrons is to share

their valence electrons with other atoms

The shared electrons would then count toward

each atoms octet (Octet rule applied)

The sharing of valence electrons is called

covalent bonding


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....OSO.. .... .... .. ..

Bonding Pairs and Lone Pairs

Bonding pairs Lone pairs

Electrons that are shared by atoms are calledbonding pairs

Electrons that are not shared by atoms butbelong to a particular atom are called lone pairs

ornonbonding pairs

3.3 Covalent Bond


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Bonding Pairs and Lone Pairs

3.3 Covalent Bond

Generally, elements follow certain common

bonding patterns

The # of bonds = Group # or 8 Group #

C = 4 bonds & 0 lone pairs, N = 3 bonds & 1 lone pair,

O= 2 bonds & 2 lone pairs, H and halogen = 1 bond,

Be = 2 bonds & 0 lone pairs,

B = 3 bonds & 0 lone pairs

B C N O F

3 3 C l B d


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Single Covalent Bonds

F

F

F

F

HH O

HH O

F F

When two atoms share one pair of electrons it iscalled a single covalent bond One atom may use more than one single bond to

fulfill its octet

H only duet

3.3 Covalent Bond

3 3 C l t B d


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Double Covalent Bond

When two atoms share two pairs of electrons the

result is called a double covalent bond

four electrons

OO

O

O

3.3 Covalent Bond

3 3 C l t B d


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Triple Covalent Bond

When two atoms share three pairs of electrons theresult is called a triple covalent bond

six electrons

N

N

N N

3.3 Covalent Bond

3 3 C l t B d


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Lewis Structures

3.3 Covalent Bond

Lewis theory predicts that atoms will be most stablewhen they have their octet of valence electrons.

It does not require that atoms have the same number

of lone pair electrons they had before bonding.

First use the octet rule

3 3 C l t B d


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The Steps in Converting a Molecular

Formula into a Lewis Structure.Molecular

Formula

Atom

placement

Place atom with lowest

EN in center.

Step 1

Add A-group numbers.Step 2

Sum of

valence e-

Draw single bonds, and

subtract 2e-for each bond.

Step 3

Remaining

valence e-

Lewis

structure

Step 4 Give each atom8e-

(2e-for H).

3.3 Covalent Bond

3 3 C l t B d


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Writing Lewis Structures for NF3

Atom

placement

Sum of

valence e-

1 x N = 1 x 5 = 5e-

3 x F = 3 x 7 = 21 e-

Total = 28 e-

Molecular

FormulaN has a lower EN than F, so N is placed in the center.

Lewis

structure

Remaining

valence e-

3.3 Covalent Bond

3 3 C l t B d


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Writing Lewis Structures for Molecules

with One Central AtomExample 4 : Write a Lewis structure for CCl2F2

Step 1: Carbon has the lowest EN and is the

central atom. The other atoms areplaced around it.

Step 2: [1 x C(4e-)] + [2 x F(7e-)] + [2 x

Cl(7e-)] = 32 valence e-

Step 3-4: Add single bonds, then give

each atom a full octet.

3.3 Covalent Bond

3 3 C l t B d


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Writing Lewis Structures for Molecules with

More than One Central AtomExample 5 :Write the Lewis structure for

methanol (molecular formula CH4O)

Step 1: Place the atoms relative to each

other. H can only form one bond, so Cand O must be central and adjacent

to each other.

Step 2: [1 x C(4e-)] + [1 x O(6e-)] + [4 x

H(1e-)] = 14 valence e-


each atom (other than H) a full

octet.

3.3 Covalent Bond

3 3 C l t B d


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Multiple Bonds

If there are not enough electrons for the centralatom

to attain an octet, a multiple bond is present.

Step 5: If the central atom does not have a full octet,change a lone pair on a surrounding atom into another

bonding pair to the central atom, thus forming a

multiple bond.

3.3 Covalent Bond

3 3 C l t B d


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Writing Lewis Structures for Molecules with

Multiple BondsExample 6: Write Lewis structures for the

following: Ethylene (C2H4)PLAN: After following steps 1 to 4 we see that the central

atom does not have a full octet. We must thereforeadd step 5, which involves changing a lone pair to a

bonding pair.

SOLUTION:

C2H4has 2(4) + 4(1) = 12 valence e-

. H can have only one bond peratom.

3.3 Covalent Bond

3 3 Co alent Bond


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Step 1: Place the atoms relative to each

other. H can only form one bond, so C

must be central and N adjacent to

carbon

Step 2: [1 x C(4e-)] + [1 x N(5e-)] + [1 x

H(1e-)] = 10 valence e-


each atom (other than H) a full

octet.

Example 7: Write Lewis structure for HCN

NH C

H C N

3.3 Covalent Bond

3 3 Covalent Bond


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Writing Resonance Structures

SOLUTION:

Example 8 : Write resonance structures for the nitrate ion, NO3

and find the bond order.

Nitrate has [1 x N(5e-)] + [3 x O(6e-)] + 1e-] = 24 valence e-

After Steps 1-4:

3.3 Covalent Bond

A resonance structureis one of two or more Lewis structures

for a single molecule that cannot be represented accurately by

only one Lewis structure.

3 3 Covalent Bond


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Step 5. Since N does not have a full octet, we change alone pair from O to a bonding pair to form a double bond.

Writing Resonance Structures

3.3 Covalent Bond

3 3 Covalent Bond


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Example 9: Writing Lewis structures of

molecules, HNO3

1. Write skeletal structure H always terminal

in oxyacid, H outside attached to Os make least electronegative atom

central N is central not H

2. Count valence electrons sum the valence electrons for each

atom

add one electron for each charge

subtract one electron for each + charge

ONOH

O

N = 5H = 1

O3= 36 = 18

Total = 24 e

3.3 Covalent Bond

3 3 Covalent Bond


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3. Attach atom together with pairs of electrons, and

subtract from the total

dont forget, a line represents 2 electrons

Electrons

Start 24

Used 8Left 16ONOH

O..

..

.. ....

.

.

.

.

.

.

.

.

.

.

.

.

.

.

3.3 Covalent Bond


molecules, HNO3


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4. Complete octets, outside-in

H is already complete with 2

1 bond

and re-count electrons

N = 5H = 1

O3= 36 = 18

Total = 24 e

ElectronsStart 24

Used 8

Left 16

ElectronsStart 16

Used 16

Left 0

ONOH

O

....

..

....

.

.

.

.

.

.


molecules, HNO3



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5. If all octets complete, giveextra electrons to the centralatom

elements with dorbitals canhave more than eight electrons

Period 3 and below

6. If central atom does not haveoctet, bring in electrons from

outside atoms to share follow common bonding patterns if

possible

ONOH

O

ONOH

O

..

..

..

..

.

.

.

.

.

.

..

..

..

..

.

.

.

.

.

.

..

ONOH

O

..

...

.

.

.

.

.

..

..


molecules, HNO3

CHECKPOINT 4


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CHECKPOINT 4

Draw Lewis Structures of the Following

CO2

SeOF2

NO2

H3PO4

SO32

P2H4


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CO2

SeOF2

NO2

H3PO

4

SO32

P2H4

16 e

26 e

18 e

26 e

32 e

14 e

Two possible skeletal structures of formaldehyde (CH2O)


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70

H C O HH

C O

H

An atoms formal chargeis 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

totalnumber

ofbonding

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.

C 4 e- 2 single bonds (2x2) = 4-1 +1


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71

H C O H

C4 e

O6 e-

2H2x1 e-

12 e-

2 single bonds (2x2) = 4

1 double bond = 4

2 lone pairs (2x2) = 4

Total = 12

formal charge

on C= 4 - 2- x 6 = -1

formal charge

on O= 6 - 2- x 6 = +1

formal charge

on an atom in

a Lewis

structure

=1

2

total number

of bonding


of valence

electrons in

the free atom

-total number

of nonbonding

electrons-

1 1

C 4 e- 2 single bonds (2x2) = 4H 0 0


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72

C4 e

O6 e-

2H2x1 e-

12 e-


1 double bond = 4

2 lone pairs (2x2) = 4

Total = 12

HC O

H

formal charge

on C= 4 - 0- x 8 = 0

formal charge

on O= 6 - 4- x 4 = 0

formal charge

on an atom in

a Lewis

structure

=1

2

total number

of bonding


of valence

electrons in

the free atom

-total number

of nonbonding

electrons-

0 0

Example 10


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StrategyThe Lewis structure for the carbonate ion was

developed as below:

The formal charges on the atoms can be calculated using the

given procedure.

Solut ionWe subtract the number of nonbonding electrons and

half of the bonding electrons from the valence electrons of each

atom.

Example 10


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The C atom: The C atom has four valence electrons and there

are no nonbonding electrons on the atom in the

Lewis structure. The breaking of the double bondand two single bonds results in the transfer of four

electrons to the C atom. Therefore, the formal

charge is 4 4 = 0.

The O atom in C=O: The O atom has six valence electrons and

there are four nonbonding electrons on

the atom. The breaking of the double

bond results in the transfer of two

electrons to the O atom. Here the formal

charge is 6 4 2 = 0.

Example 10


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The O atom in CO: This atom has six nonbonding electrons

and the breaking of the single bond

transfers another electron to it.Therefore, the formal charge is

6 6 1 = 1.

Thus, the Lewis structure for with formal charges is

CheckNote that the sum of the formal charges is 2, the same

as the charge on the carbonate ion.

Formal Charge and Lewis Structures


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76

g

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 than those with small formal charges.

3. Among Lewis structures having similar distributions offormal charges, the most plausible structure is the one in

which negative formal charges are placed on the more

electronegative atoms.

Example 11


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Formaldehyde (CH2O), a liquid with a disagreeable odor,

traditionally has been used to preserve laboratory specimens.

Draw the most likely Lewis structure for the compound.

Example 11


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StrategyA plausible Lewis structure should satisfy the octet

rule for all the elements, except H, and have the formal charges

(if any) distributed according to electronegativity guidelines.

Solut ionThe two possible skeletal structures are

Example 11


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First we draw the Lewis structures for each of these possibilities

To show the formal charges, we follow the procedure before. In

(a) the C atom has a total of five electrons (one lone pair plus

three electrons from the breaking of a single and a double

bond). Because C has four valence electrons, the formal

charge on the atom is 4 5 = 1. The O atom has a total offive electrons (one lone pair and three electrons from the

breaking of a single and a double bond). Since O has six

valence electrons, the formal charge on the atom is 6 5 = +1.

Example 11


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In (b) the C atom has a total of four electrons from the breaking

of two single bonds and a double bond, so its formal charge is

4 4 = 0. The O atom has a total of six electrons (two lonepairs and two electrons from the breaking of the double bond).

Therefore, the formal charge on the atom is 6 6 = 0. Although

both structures satisfy the octet rule, (b) is the more likely

structure because it carries no formal charges.

CheckIn each case make sure that the total number of

valence electrons is 12. Can you suggest two other reasons

why (a) is less plausible?

CHECKPOINT 5


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CO2

SeOF2

NO2

H3PO4

SO32

P2H4

CHECKPOINT 5

Assign formal charges

CHECKPOINT 5 (answers)


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CO2

SeOF2

NO2

all 0

Se = +1

H3

PO4

SO32

P2H4

CHECKPOINT 5 (answers)

P = +1

rest 0

S = +1

all 0

3.3 Covalent Bond


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Exceptions to Octet Rule

Some elements violate the octet rule1. Incomplete octet

Be forms twobonds with no lone pairs in its compounds

B. Al forms threebonds with no lone pairs in itscompounds

2. Expanded octet

many elements may end up with more than eight valence

electrons in their structure if they can use their empty dorbitals for bonding

3. Odd number electron speciese.g., NO

- Will have one unpaired electron, free radical & very

reactive

3.3 Covalent Bond

3.3 Covalent Bond


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Molecules with Electron-Deficient Atoms

B and Be are

commonly electron-

deficient.


3.3 Covalent Bond

Exceptions to the Octet Rule


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85


The Incomplete Octet

H HBeBe2e-

2H2x1e-

4e-

BeH2

BF3

B3e-

3F3x7e-

24e-

F B F

F


9 lone pairs (9x2) = 18Total = 24

3.3 Covalent Bond


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Expanded Valence Shells

An expanded valence shell is only possible for nonmetals

from Per iod 3 or higherbecause these elements have

available d o rbi tals.


Odd-Electron SpeciesA molecule with an

odd number of

electrons is called a

free radical.



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87

Odd-Electron Molecules

N5e-O6e-

11e-

NO N O

The Expanded Octet

(central atom with principal quantum number n > 2)

SF6S6e-

6F42e-

48e-S

F

F

F

FF

F

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

Total = 48

Example 13


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Draw the Lewis structure for aluminum triiodide (AlI3).

AlI3has a tendency to

dimerize or form two units as

Al2I6.

Example 13


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StrategyWe follow the procedures used in Examples 11 and

12 to draw the Lewis structure and calculate formal charges.

Solut ionThe outer-shell electron configurations of Al and I are

3s23p1and 5s25p5, respectively. The total number of valence

electrons is 3 + 3 7 or 24. Because Al is less electronegative

than I, it occupies a central position and forms three bonds with

the I atoms:

Note that there are no formal charges on the Al and I atoms.

Example 13


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CheckAlthough the octet rule is satisfied for the I atoms, there

are only six valence electrons around the Al atom.

Thus, AlI3is an example of the incomplete octet.

Example 14


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Draw the Lewis structure for phosphorus pentafluoride (PF5), in

which all five F atoms are bonded to the central P atom.

PF5is a reactive

gaseous compound.

Example 14


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StrategyNote that P is a third-period element. We follow the

procedures given in Examples 11 to draw the Lewis structure

and calculate formal charges.

Solut ionThe outer-shell electron configurations for P and F

are 3s2

3p3

and 2s2

2p5

, respectively, and so the total number ofvalence electrons is 5 + (5 7), or 40.

Phosphorus, like sulfur, is a third-period element, and therefore

it can have an expanded octet.

Example 14


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The Lewis structure of PF5is

Note that there are no formal charges on the P and F atoms.

CheckAlthough the octet rule is satisfied for the F atoms,there are 10 valence electrons around the P atom, giving it an

expanded octet.

Example 15


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Draw a Lewis structure for the sulfate ion in which all

four O atoms are bonded to the central S atom.

Example 15


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StrategyNote that S is a third-period element. We follow the

procedures given in Example 11 to draw the Lewis structure

and calculate formal charges.

Solut ionThe outer-shell electron configurations of S and O

are 3s23p4and 2s22p4, respectively.

Step 1: The skeletal structure of is

Example 15

St 2 B h O d S G 6A l d h


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Step 2: Both O and S are Group 6A elements and so have

six valence electrons each. Including the two

negative charges, we must therefore account for atotal of 6 + (4 6) + 2, or 32, valence electrons in

Step 3: We draw a single covalent bond between all thebonding atoms:

Example 15

N t h f l h th S d O t


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Next we show formal charges on the S and O atoms:

Note that we can eliminate some of the formal charges for

by expanding the S atoms octet as follows:

Example 15


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At this stage of learning, you should realize that bothrepresentations are valid Lewis structures and you

should be able to draw both types of structures.

One helpful rule is that in trying to minimize formal

charges by expanding the central atoms octet, only

add enough double bonds to make the formal charge

on the central atom zero.

Example 15

Th th f ll i t t ld i f l h


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Thus, the following structure would give formal charges on

S(2) and O(0) that are inconsistent with the electronegativities

of these elements and should therefore not be included torepresent the ion.

Example 16

D L i t t f th bl d


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Draw a Lewis structure of the noble gas compound xenon

tetrafluoride (XeF4) in which all F atoms are bonded to the

central Xe atom.

Example 16

St t N t th t X i fifth i d l t W f ll th


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StrategyNote that Xe is a fifth-period element. We follow the

procedures in Example 11 for drawing the Lewis structure and

calculating formal charges.

Solut ion

Step 1: The skeletal structure of XeF4is

Step 2: The outer-shell electron configurations of Xe and F

are 5s25p6and 2s22p5, respectively, and so the total

number of valence electrons is 8 + (4 7) or 36.

Example 16

St 3 W d i l l t b d b t ll th


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Step 3: We draw a single covalent bond between all the

bonding atoms. The octet rule is satisfied for the F

atoms, each of which has three lone pairs. The sumof the lone pair electrons on the four F atoms (4 6)

and the four bonding pairs (4 2) is 32. Therefore,

the remaining four electrons are shown as two lone

pairs on the Xe atom:

We see that the Xe atom has an expanded octet.

There are no formal charges on the Xe and F atoms.

3.3 Covalent Bond


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Determine BestLewis Structures

BestLewis structure have:

All atoms obey the octet rule

Zero or fewer formal charges

smaller formal charges

negative formal charge on the moreelectronegative atom

Like charges on adjacent atoms are notdesirable

Sum of formal charges equal zero, or equalthe ionic charge for polyatomic ions

CHECKPOINT 6


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CHECKPOINT 6

Show the resonance forms of NCO-and

predict which resonance structure is the

most stable structure.

104

Answer:


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NCOhas 3 possible resonance forms:

N C O

A

N C O

B

N C O

C

N C O N C O N C O

formal charges

2 0 +1 1 0 0 0 0 1

Forms B and C have negative formal charges on N and O; thismakes them more preferred than form A.

Form C has a negative charge on O which is the more

electronegative element, therefore C contributes the most to the

resonance hybrid.

3.3 Covalent Bond


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Properties of Covalent Compounds

1) Low melting points and boiling points

involves breaking the attractions between the molecules,

but not the bonds between the atoms

the covalent bonds are strong, but the attractions

between the moleculesare generally weak

intermolecular forces

2) Do not conduct electricity in the solid or liquid state

there are no charged particles (no ions) around to allow

the material to conduct

3.3 Covalent BondP ti f C l t C d


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3) Electronegativity The ability of an atom to attract bonding electrons to itself is

called electronegativity

Increases across period (left to right) and

Decreases down group (top to bottom) fluorine is the most electronegative element

francium is the least electronegative element

opposite of atomic size trend

The larger the difference in electronegativity, the more polarthe bond

negative end toward more electronegative atom

Properties of Covalent Compounds

The Electronegativities of Common Elements


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108

3.3 Covalent Bond


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Polar Covalent Bonding

Covalent bonding between unlike atoms results in unequalsharingof the electrons

The more electronegative atom pulls the shared electronscloser to its side

one end of the bond has larger electron density than theother

The result is a polar covalent bond

bond polarity

the end with the larger electron density gets a partialnegative charge ()

The other end that is electron deficient gets a partialpositive charge (+)

3.3 Covalent Bond


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Polar Covalent Bonding


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B d P l it

3.3 Covalent Bond


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HF

H Fd+ d-

H F

EN = 2.1 EN = 4.0

Bond Polarity

Classification of bonds by difference in electronegativity


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113

Covalent

share e-

Polar Covalent

partial transfer of e-

Ionic

transfer e-

Increasing difference in electronegativity

Difference Bond Type

0 Covalent

2 Ionic

0 < and


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Classify the following bonds as ionic, polar covalent, or

covalent:

(a) the bond in HCl

(b) The bond in KF

(c) the CC bond in H3CCH3

Excercise


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StrategyWe follow the 2.0 rule of electronegativity difference.

Solut ion(a) The electronegativity difference between H and Cl is 0.9,

which is appreciable but not large enough (by the 2.0 rule)

to qualify HCl as an ionic compound. Therefore, the bond

between H and Cl is polar covalent.(b) The electronegativity difference between K and F is 3.2,

which is well above the 2.0 mark; therefore, the bond

between K and F is ionic.

(c) The two C atoms are identical in every respectthey arebonded to each other and each is bonded to three other H

atoms. Therefore, the bond between them is purely

covalent.

3.3 Covalent Bond


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In covalent bonding, the more electronsin two atoms share,

the stronger the bond

Bond strength is measured by how much energy needed

to break the bond

triple bonds are stronger than double bonds, and double

bonds are stronger than single bonds

The more electrons two atoms share, the shorter the bond

Bond length is determined by measuring the distance

between the nuclei of bonded atoms

In general, triple bonds are shorter than double bonds,

and double bonds are shorter than single bonds

4) Covalent Bond Strength and Bond Length

3.3 Covalent Bond


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Trends in Bond Strength

In general, the more electrons two atoms share,the stronger the covalent bond must be comparing bonds between like atoms

CC (837 kJ) > C=C (611 kJ) > CC (347 kJ) CN (891 kJ) > C=N (615 kJ) > CN (305 kJ)

In general, the shorter the covalent bond, thestronger the bond must be comparing similar types of bonds

BrF (237 kJ) > BrCl (218 kJ) > BrBr (193 kJ)

bonds get weaker down the column

bonds get stronger across the period

3.3 Covalent Bond


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Bond Lengths

The distance between the nuclei ofbonded atoms is called the bond

length Because the actual bond length

depends on the other atoms aroundthe bond we often use the average

bond length averaged for similar bonds from many

compounds

3.3 Covalent Bond


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Bond Lengths


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121


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122

Copyright The McGraw-Hill Companies, Inc. Permission required for

Chemical Bonding II:Molecular Geometry and

Hybridization of Atomic

Orbitals

Molecular Geometry


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Molecular Geometry

3-dimensional arrangement of atoms in amolecules

Molecular geometry affects its physical and

chemical properties We often describe the shape of a molecule

with terms that relate to geometric figures

The geometric figures also havecharacteristic angles that we call bondangles

Using Lewis Theory toPredict


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PredictMolecular Shapes

In general, bond lengths and bondangles are determined by

experiment By knowing the number of electrons

surrounding a central atoms (Lewisstructure), we can simply predict the

overall geometry of the molecules.

Lewis theory says that these regions of


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125

electron groups should repel each other

This idea can then be extended topredict the shapes of molecules

the position of atoms surrounding a

central atom will be determined by

where the bonding electron groupsare

the positions of the electron groups

will be determined by trying tominimize repulsions between them

VSEPR Theory


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The geometric arrangement of electron pairs around acentral atom by minimize the electrostatic repulsionbetween electron pairs.

Electron groups around the central atom of a molecule willbe most stable when they are separated as far apart aspossiblewe call this valence shell electron pair

repulsiontheory (VSEPR) The resulting geometric arrangement will allow us to predict

the shapes and bond angles in the molecule

VSEPR formula: ABaEb

A = central atom; B = surrounding atoms

E = lone pairs on central atom; a, b = integers (1,2,3)

Electron Groups


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The Lewis structure predicts the number ofvalence electron pairs around the central atom(s)

Each lone pair of electrons constitutes oneelectron group on a central atom

Each bonding pair constitutes one electron group

on a central atom regardless of whether it is single, double, or

triple

O N O

there are three electron

groups on Nthree lone pair

one single bond

one double bond

Valence shel l electro n pair repu lsion(VSEPR) model:


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128

Predict the geometry of the molecule from the

electrostatic repulsions between the electron(bonding and nonbonding) pairs.

AB2 2 0

Class

# of atoms

bonded to

central

atom

# lone

pairs on

central

atom

Arrangement

of electron

pairs

Molecular

Geometry

linear linear

B B

# of atoms

bonded to

# lone

pairs on Arrangement

VSEPR


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129

AB2 2 0 linear linear

Class

bonded to

central

atom

pairs on

central

atom

g

of electron

pairs

Molecular

Geometry

AB3 3 0trigonal

planar

trigonal

planar

Boron Trifluoride


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130

# of atoms

bonded to

# lone

pairs on Arrangement Molecular

VSEPR


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131

AB2 2 0linear linear

Class

bonded to

central

atom

pairs on

central

atom

Arrangement

of electron

pairs

Molecular

Geometry

AB3 3 0trigonal

planar

trigonal

planar

AB4 4 0tetrahedral tetrahedral

Methane


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132

# of atoms

bonded to# lone


VSEPR


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133

AB2 2 0linear linear

Class

central

atom

pairs on

central

atom

g

of electron

pairs

Molecular

Geometry

AB3 3 0trigonal

planar

trigonal

planar


AB5 5 0trigonal

bipyramidal

trigonal

bipyramidal

Phosphorus Pentachloride


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134

# of atoms

bonded to

# lone

pairs on Arrangement

VSEPR


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135

AB2 2 0 linear linear

Class

bonded to

central

atom

pairs on

central

atom

of electron

pairs

Molecular

Geometry

AB3 3 0trigonal

planar

trigonal

planar


AB5 5 0trigonal

bipyramidal

trigonal

bipyramidal

AB6 6 0 octahedraloctahedral

Sulfur Hexafluoride


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136


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137


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138

bonding-pair vs. bonding-

pair repulsionlone-pair vs. lone-pair

repulsion

lone-pair vs. bonding-

pair repulsion> >

# of atoms

bonded to

# lone


VSEPR


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139

Class central

atom

p

central

atom

g

of electron

pairs

Geometry

AB3 3 0trigonal

planar

trigonal

planar

AB2E 2 1trigonal

planarbent


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# of atoms

bonded to

t l

# lone

pairs onArrangement

of electron Molecular

VSEPR


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141

Class

central

atomcentral

atom

of electron

pairsMolecular

Geometry


AB3E 3 1tetrahedral

trigonal

pyramidal

AB2E2 2 2tetrahedral bent

# of atoms

bonded to# lone

pairs onArrangement


VSEPR


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142

Class

central

atomcentral

atom

of electron

pairs Geometry

AB5 5 0 trigonalbipyramidal

trigonal

bipyramidal

AB4E 4 1 trigonal

bipyramidal

distorted

tetrahedron

Cl

# of atoms

bonded to

central

t

# lone

pairs on

central

t

Arrangement

of electron

pairsMolecular

G t


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143

Class atom atom pairs Geometry

AB5 5 0trigonal

bipyramidal

trigonal

bipyramidal

AB4E 4 1trigonal

bipyramidal

distorted

tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

# of atoms

bonded to

central

# lone

pairs on

central

Arrangement



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144

Class atom atom pairs Geometry

AB5 5 0 trigonalbipyramidal

trigonal

bipyramidal

AB4E 4 1 trigonalbipyramidal

distorted

tetrahedron

AB3E2 3 2 trigonalbipyramidal

T-shaped

AB2E3 2 3 trigonalbipyramidal

linear

# of atoms

bonded to

# lone

pairs on

t l

Arrangement

f l t M l l

VSEPR


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145

Classcentral

atom

central

atom

of electron

pairsMolecular

Geometry

AB6 6 0octahedraloctahedral

AB5E 5 1octahedral

square

pyramidal

# of atoms

bonded to

central

# lone

pairs on

central

Arrangement

of electronMolecular

Geometry


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146

Class atom atom pairsGeometry

AB6 6 0octahedraloctahedral

AB5E 5 1octahedral

square

pyramidal

AB4E2 4 2octahedral square

planar


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147


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Predicting Molecular


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149

gGeometry

1. 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.

Example 1Use the VSEPR model to predict the geometryof the following molecules and ions:


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of the following molecules and ions:

(a) AsH3

(b) OF2

(c)

(d)

(e) C2H4

Example 1StrategyThe sequence of steps indetermining molecular geometry is as follows:


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determining molecular geometry is as follows:

Solut ion

(a) The Lewis structure of AsH3is

There are four electron pairs around thecentral atom; therefore, the electron pairarrangement is tetrahedral


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Example 1Recall that the geometry of a molecule is determined onlyby the arrangement of atoms (in this case the O and F


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by the arrangement of atoms (in this case the O and Fatoms). Thus, removing the two lone pairs leaves us with

two bonding pairs and a bent geometry, like H2O. Wecannot predict the FOF angle accurately, but we know that itmust be less than 109.5 because the repulsion of thebonding electron pairs in the OF bonds by the lone pairson O is greater than the repulsion between the bonding

pairs.

(c) The Lewis structure of is

Example 1There are four electron pairs around the central atom;therefore, the electron pair arrangement is tetrahedral.


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Because there are no lone pairs present, thearrangement of the bonding pairs is the same as theelectron pair arrangement. Therefore, has atetrahedral geometry and the ClAlCl angles are all109.5.

(d) The Lewis structure of is

There are five electron pairs around the central I atom;therefore, the electron pair arrangement is trigonalbipyramidal. Of the five electron pairs, three are lonepairs and two are bonding pairs.

Example 1Recall that the lone pairs preferentially occupy theequatorial positions in a trigonal bipyramid. Thus,

i h l i l i h li


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removing the lone pairs leaves us with a linear geometryfor, that is, all three I atoms lie in a straight line.

(e) The Lewis structure of C2H4is

The C=C bond is treated as though it were a single bondin the VSEPR model. Because there are three electron

pairs around each C atom and there are no lone pairspresent, the arrangement around each C atom has atrigonal planar shape like BF3, discussed earlier.

Example 1Thus, the predicted bond angles in C2H4are all 120.


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Comment

(1) The ion is one of the few structures for which the

bond angle (180) can be predicted accurately eventhough the central atom contains lone pairs.

(2) In C2H4, all six atoms lie in the same plane. The overallplanar geometry is not predicted by the VSEPR model, but

we will see why the molecule prefers to be planar later. Inreality, the angles are close, but not equal, to 120because the bonds are not all equivalent.

Predict the geometries of the following species using the

CHECKPOINT 7


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Predict the geometries of the following species using theVSEPR method:

a) PCl3

b) H2O

c) CHCl3d) ClF3

e) TeCl4

trigonal pyramidal

bent shape

Tetrahedral

t-shape

seesaw

CHECKPOINT 8


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Which of the following species are tetrahedral?SiCl4, SeF4, XeF4, Cl4, CdCl4

2-

Answer: SiCl4 Cl4 and CdCl42-

Dipole Moments and Polar

Molecules


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H F

electron rich

regionelectron poorregion

d+ d-

ris the distance between charges

1 D = 3.36 x 10-30C m

Dipole moment, m, is ameasure of bond polarity

a dipole is a material with a + and end

it is directly proportional to the sizeof the partial charges and to thedistance between the charges

Generally, the more electronstwo atoms share and the largerthe atoms are, the larger thedipole moment

A dipole arrow is usedto the polarity of a bond

m= Qx rQis the charge

* Note to students: You will not be

tested on DP calculation

Behavior of Polar Molecules


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160

field off field on

Molecular Geometry and Polarity

3.3 Covalent Bond


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y y

For a molecule to be polar it must1. have polar bonds

electronegativity differencetheory

2. have an unsymmetrical shape and an overall DP

vector addition

Polarity affects the intermolecular forces ofattraction

therefore boiling points and solubilities Nonbonding pairs affect molecular geometry and

also molecular polarity

Molecule Polarity3.3 Covalent Bond


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The HCl bond is polar. The bonding electrons

are pulled toward the Cl end of the molecule. The

net result is a polar molecule.

Molecule Polarity

3.3 Covalent Bond


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The OC bond is polar. The bonding electrons arepulled equally toward both O ends of the molecule. The 2

dipole cancell out each other. The net result is a nonpolar

molecule.

Molecule Polarity

3.3 Covalent Bond


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Molecule Polarity

The HO bond is polar. Both sets of bonding electrons are pulled

toward the O end of the molecule. A net dipole moment (dotted dipole

arrow) appears. The net result is a polar molecule.

Predicting Polarity of Molecules

3.3 Covalent Bond


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1. Draw the Lewis structure and determine themolecular geometry

2. Determine whether the bonds in the moleculeare polar or non-polar

3. Determine whether the polar bonds add

together to give a net dipole moment

Predict whether NH3is a polar molecule


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1. Draw the Lewis structureand determine themolecular geometry

a) eight valence electrons

b) three bonding + one lonepair = trigonal pyramidalmolecular geometry



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2. Determine if the bonds arepolar

a) Find electronegativitydifference EN

b) if the bonds are not polar,we can stop here anddeclare the molecule will benonpolar

ENN = 3.0

ENH = 2.1

3.0 2.1 = 0.9

therefore the bonds

are polar covalent

Predict whether NH is a polar molecule


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3) Determine whetherthe polar bonds add

together to give anet dipole momenta) vector addition

b) generally, asymmetricshapes result inuncompensatedpolarities and a netdipole moment

The HN bond is polar. All

the sets of bondingelectrons are pulled toward

the N end of the molecule.

The net result is a polar

molecule.

Bond moments and resultant dipole moments

in NH3and NF3.


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169


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170

Example 2

Predict whether each of the following


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gmolecules has a dipole moment:

(a) BrCl

(b) BF3(trigonal planar)

(c) CH2Cl2(tetrahedral)

Example 2Strategy

Keep in mind that the dipole moment of a


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p pmolecule depends on both the difference in

electronegativities of the elements present andits geometry.

A molecule can have polar bonds (if the

bonded atoms have differentelectronegativities), but it may not possess adipole moment if it has a highly symmetricalgeometry.

Example 2Solut ion

(a) Because bromine chloride is diatomic, it has ali t Chl i i


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linear geometry. Chlorine is more

electronegative than bromine. so BrCl is polarwith chlorine at the negative end

Thus, the molecule does have a dipole moment.In fact, all diatomic molecules containingdifferent elements possess a dipole moment.

Example 2(b) Because fluorine is more electronegative than

boron, each BF bond in BF3(boron trifluoride) ispolar and the three bond moments are equal


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polar and the three bond moments are equal.However, the symmetry of a trigonal planar shapemeans that the three bond moments exactly cancelone another:

An analogy is an object that is pulled in the

directions shown by the three bond moments. If theforces are equal, the object will not move.Consequently, BF3has no dipole moment; it is anonpolar molecule.

B-F

Example 2(c) The Lewis structure of CH2Cl2(methylenechloride) is


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This molecule is similar to CH4in that it hasan overall tetrahedral shape. However,because not all the bonds are identical,there are three different bond angles: HCH,HCCl, and ClCCl. These bond angles areclose to, but not equal to, 109.5.

Example 2Because chlorine is more electronegative thancarbon, which is more electronegative thanhydrogen the bond moments do not cancel and


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hydrogen, the bond moments do not cancel and

the molecule possesses a dipole moment:

Thus, CH2Cl2is a polar molecule.

Example 3Explain why CO2is nonpolar, but OCS is polar.


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Ans: In CO2the two bond moments point inopposite directions and are of equalmagnitude. Therefore, they cancel. In OCS,even though the two bond moments point in

opposite directions, they are not of the samemagnitude and do not cancel.

CHECKPOINT 9


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List the following molecules in order ofdecreasing dipole moment: H2O, HF, CO2

Answer: HF, H2O,CO2

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