Chapter 8

The Basics
of Chemical Bonding

Chemistry, 7th Edition

International Student Version
Brady/Jespersen/Hyslop

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved.
.

Chapter in Context

Describe the necessary conditions for bond
formation

Examine the factors involved in ionic bonding

Write electron configurations of ions

Write Lewis symbols for atoms and ions

Describe covalent bonds, the octet rule, and
multiple bonds

Understand the energetics of bond formation

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 2
.

1
 Chapter in Context, cont’d

Learn trends in electronegativity

Examine how electronegativity affects bond
polarity and the reactivity of elements

Draw Lewis structures of molecules

Calculate and use formal charges

Draw and understand resonance structures

Classify organic compounds and identify
functional groups

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 3
.

Chemical Bonds

Attractive forces that hold atoms together
in complex substances

Molecules and ionic compounds
Why study?

Changes in these bonding forces are the
underlying basis of chemical reactivity

During reaction

Break old bonds

Form new bonds

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 4
.

2
 Two Classes of Bonds

Covalent bonding

Occurs in molecules

Sharing of electrons

Ionic Bonding

Occurs in ionic solid

Electrons transferred from one atom to another

Simpler

We will look at this first

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 5
.

Ionic Bonds

Result from attractive forces between
oppositely charged particles

Na+ Cl –

Metal - nonmetal bonds are ionic because:

Metals have

Low ionization energies

Easily lose electrons to be stable

Non-metals have

Very exothermic electron affinities

Formation of lattice stabilizes ions
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 6
.

3
 Ionic Compounds

Formed from metal and nonmetal
Na + Cl Na+ + Cl− NaCl(s)

e−

Ionic Bond

Attraction between + and – ions in ionic
compound

Why does this occur? Why is e – transferred?

Why Na+ and not Na2+ or Na–?

Why Cl– and not Cl2– or Cl+?

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 7
.

Ionic Compounds
Ionic crystals:

Exist in 3-dimensional array
of cations and anions called
a lattice structure

Ionic chemical formulas:

Always written as empirical
formula

Smallest whole number ratio
of cation to anion
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 8
.

4
 Energetics

Must look at energy of system to answer
these questions

For any stable compound to form from its
elements

Potential energy of system must be lowered.

Net decrease in energy ∆Hf° < 0 (negative)

What are factors contributing to energy
lowering for ionic compound?

Use Hess’s Law to determine

Conservation of energy

Envision two paths
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 9
.

Lattice Energy

Amount that PE of system decreases when
one mole of solid salt is formed from its gas
phase ions

Energy released when ionic lattice forms

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 10
.

5
 Two Paths to Evaluate Energy
1. Single step
Na(s) + ½Cl2(g) → NaCl(s) ∆Hf°= – 411.1 kJ/mol

2. Stepwise path
Na(s) → Na(g) ∆Hf°(Na, g) = 107.8 kJ/mol
½Cl2(g) → Cl(g) ∆Hf°(Cl, g) = 121.3 kJ/mol
Na(g) → Na (g) + e
+ – IE(Na) = 495.4 kJ/mol
Cl(g) + e → Cl (g)
– – EA(Cl) = –348.8 kJ/mol
Na+(g) + Cl–(g) → NaCl(s) ∆Hlattice = –787 kJ/mol

Na(s) + ½Cl2(g) → NaCl(s) ∆Hf°= –411 kJ/mol

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 11
.

Lattice Energy

Always ∆HLattice= – = exothermic

∆HLattice gets more exothermic (larger negative
value) as ions of opposite charge are brought closer
together

Ions tightly packed with opposite charged ions next
to each other q q
∆H Lattice ∝ + −
d

Any increase in PE due to ionizing atoms is more
than met by decrease in PE from formation of
crystal lattice. Even for +2 and –2 ions

Therefore, forming ionic solids is an overall
exothermic process and they are stable compounds
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 12
.

6
 Your Turn!
Assuming that the separation between
cations and anions in the lattice is nearly
identical, which species would have the
greatest lattice energy?
A. sodium chloride
B. calcium chloride
C. calcium nitride Highest charges
D. sodium oxide (+2 and -3)
E. calcium oxide

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 13
.

Your Turn!
Which combination below will have the most
negative lattice energy?

A. Low-charge ions separated by large distances

B. Low-charge ions separated by small distances
C. High-charged ions separated by large distances
D. High-charged ions separated by small distances
As q increases, ∆H increases
q q
∆H Lattice ∝ + −
d As d increases, ∆H decreases
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 14
.

7
 Why do Metals form Cations and
Nonmetals form Anions?
Metal Nonmetal

Left hand side of
Right hand side of
Periodic Table Periodic Table

IE small and positive
IE large and positive

Little energy required to
Difficult to remove e–
remove electrons
EA large and negative

EA small and negative
But easy to add e–
or positive
Exothermic—large amount

Not favorable to attract an of energy given off
electron to it
PE of system decreases

Least expensive, energy-
Least expensive, energy-
wise, to form cation wise, to form anion
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 15
.

Electron Configurations of Ions

Review of electron configurations:
Follow Aufbau ordering
Electron configuration based on “filling” an
atom with electrons. Follows order in the
periodic table

Reflect electron energy level
Electron configuration based on increasing
values of n and then in any given energy level
by increasing values of ℓ. Helpful format for
explaining how ions form

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 16
.

8
 Electron Configurations of Ions

How electronic structure affects types of ions
formed
e.g.,
Na 1s22s22p63s1 = [Ne] 3s1
Na+ 1s2 2s22p6 = [Ne]
IE1 = 496 kJ/mol small not too difficult
IE2 = 4563 kJ/mol large ~10 x larger very difficult

Can remove first electron, as doesn't cost too much

Can’t remove second electron, as can't regain lost
energy from lattice

Thus, Na2+ doesn’t form
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 17
.

Electron Configurations of Ions

e.g., Ca [Ar] 4s2
Ca2+ [Ar]
IE1 small = 590 kJ/mol not too difficult
IE2 small = 1140 kJ/mol not too difficult
IE3 large = 4940 kJ/mol too difficult

Can regain by lattice energy ~2000 kJ/mole if
+2, –2 charges

But third electron is too hard to remove

Can’t recoup required energy through lattice
formation

Therefore Ca3+ doesn't form
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 18
.

9
 Electron Configurations of Ions

Stability of noble gas core above or below
the valence electrons effectively limits the
number of electrons that metals lose

Ions formed have noble gas electron
configuration

True for anions and cations
e.g., Cl 1s22s22p63s23p5 = [Ne]3s2 3p5
Cl– 1s22s22p63s23p6 = [Ar]

Adding another electron

Requires putting it into next higher n shell

Energy cost too high
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 19
.

e.g.,
O 1s22s22p4
O– 1s22s22p5 EA1 = –141 kJ/mol
O2– 1s22s22p6 = [Ne] EA2 = +844 kJ/mol
EAnet = +703 kJ/mol
endothermic

Energy required to form cation is more than
made up for by the increase in ∆HLattice
caused by higher –2 charge
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 20
.

10
 Electron Configurations of Ions
Generalization:

When ions form

Atoms of most representative elements (s and p
block)

Tend to gain or lose electrons to obtain nearest
Noble gas electron configuration

Except He (two electrons), all noble gases have
eight electrons in highest n shell
Octet Rule

Atoms tend to gain or lose electrons until
they have achieved outer (valence) shell
containing octet of eight electrons
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 21
.

Octet Rule

Works well with

Group 1A and 2A
metals

Al

Non-metals

H and He can't obey

Limited to 2 electrons
in the n = 1 shell

Doesn't work with

Transition metals

Post transition metals
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 22
.

11
 Your Turn!
What is the correct electron configuration for Cs
and Cs+?
A. [Xe] 6s1, [Xe]
B. [Xe] 6s2, [Xe] 6s1
C. [Xe] 5s1, [Xe]
D. [Xe] 6s1, [Xe] 6s2
E. [Xe] 6p2, [Xe] 6p1

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 23
.

Your Turn!
What charges are aluminum, phosphorus,
sulfur, strontium, and rubidium most likely to
have when they become ions?
A. -3, +3, +2, -2, -1, respectively
B. -5, -3, -2, +2, +1, respectively
C. +3, -3, -2, +2, +1, respectively
D. +13, +3, -2, +2, +2, respectively
E. +3, +5, -2, -2, +1, respectively

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 24
.

12
 Transition Metals

First electrons are lost from outermost s orbital

Lose electrons from highest n first, then ℓ
e.g., Fe [Ar] 3d 64s2
Fe2+ [Ar] 3d 6 loses 4s electrons first
Fe3+ [Ar] 3d 5 then loses 3d electrons

Extra stability due to half-filled d subshells

Consequences

M 2+ is a common oxidation state as two
electrons are removed from the outer ns shell

Ions of larger charge result from loss of d
electrons
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 25
.

Post Transition Metals

e.g.,
Sn [Kr] 4d 105s25p2
Sn2+ [Kr] 4d 105s2

Neither has noble gas electron configuration

Have emptied 5p subshell
Sn4+ [Kr] 4d 10

Does have empty 5s subshell

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 26
.

13
 Transition Metals

Not easy to predict which ions form and
which are stable

But ions with exactly filled or half-filled d
subshells are extra stable and therefore tend
to form

Mn2+ [Ar]3d 5

Fe3+ [Ar]3d 5

Zn2+ [Ar]3d 10

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 27
.

Predicting Cation Configurations

Consider Bi, whose aufbau configuration is:
[Xe]6s 2 4f 14 5d 10 6p3. What ions are expected?

Rewrite configuration: [Xe]4f 14 5d 10 6s 2 6p 3

Bi3+ and Bi5+

Consider Fe, whose aufbau configuration is:

[Ar]4s 2 3d 6. What ions are expected?

Rewrite configuration: [Ar]3d 6 4s 2

Fe2+ and Fe3+
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 28
.

14
 Predicting Anion Configurations
Non-metals gain electrons to become
isoelectronic with next larger noble gas

O: [He]2s 22p 4 + 2? e– → ? O2– : [He]2s 2 2p 6

? e– → ? N3– : [He]2s 2 2p 6
N: [He]2s 22p 3 + 3

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 29
.

Your Turn!
What are the correct electron configurations for
Cu and Cu2+ ?

A. [Ar] 3d 94s2, [Ar] 3d 9

B. [Ar] 3d 104s1, [Ar] 3d 84s1
C. [Ar] 3d 104s1, [Ar] 3d 9
D. [Ar] 3d 94s2, [Ar] 3d 104s1
E. [K] 3d 94s2, [Ar] 3d 9

Filled and half-filled orbitals are particularly

stable
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 30
.

15
 Your Turn!
What are the correct electron configurations for
zirconium(II) and zirconium(IV) ions?

A. [Kr] 5d 2 [Kr] 4d 1
B. [Ar] 4d 25s2 [Ar] 5s2
C. [Kr] 4d 2 [Kr]
D. [Kr] 4d 65s2 [Kr] 4d 6
E. [Rb] 4d 2 [Rb]

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 31
.

Lewis Symbols

Electron bookkeeping method

Way to keep track of e–’s

Write chemical symbol surrounded by dots for each
e–
Group # 1A 2A 3A 4A
Valence e–'s 1 2 3 4
e– conf'n ns1 ns2 ns2np1 ns2np2

Li Be B C
Na Mg Al Si
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 32
.

16
 Lewis Symbols
Group # 5A 6A 7A 8A
Valence e-'s 5 6 7 8
e- conf'n ns2np3 ns2np4 ns2np5 ns2np6
He
N O F Ne
P S Cl Ar
For the representative elements
Group number = number of valence e–’s
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 33
.

Lewis Symbols

Can use to diagram electron transfer in ionic
bonding

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 34
.

17
 Covalent Compounds

Form individual separate molecules

Atoms bound by sharing electrons

Do not conduct electricity

Often low melting point
Covalent Bonds

Shared pairs of electrons between two
atoms

Two H atoms come together, why?

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 35
.

Covalent Bond

Attraction of valence electrons of one atom
by nucleus of other atom

Shifting of electron density

As distance between nuclei decreases,
probability of finding either electron near
either nucleus increases

Pulls nuclei closer together

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 36
.

18
 Covalent Bond

As nuclei get
close

Begin to repel
each other

Both have
high positive
charge

Final internuclear distance between two atoms in

bond

Balance of attractive and repulsive forces

Bond forms since there is a net attraction
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 37
.

Covalent Bond

Two quantities characterize this bond
Bond Length (bond distance)

Distance between 2 nuclei = rA + rB
Bond Energy

Also bond strength

Amount of energy released when bond formed
(decreasing PE) or

Amount of energy must put in to “break” bond

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 38
.

19
 Your Turn!
Which species is most likely covalently bonded?
A. CsCl
B. NaF
C. CaF2
D. CO
E. MgBr2

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 39
.

Your Turn!
What is the primary driving force behind the
formation of covalent bonds?
A. The energy released when two electrons
attract each other
B. The energy gained when two electrons attract
each other
C. The energy released when nuclei attract
electrons
D. The energy released by two nuclei attracting
each other
E. The energy gained by the repulsion between
nuclei
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 40
.

20
 Your Turn!
What force(s) limit(s) the distance between nuclei
in a covalent bond?

A. The repulsive forces between nuclei

B. The attractive forces between nuclei
C. The attractive forces between electrons
D. The attractive forces between nuclei and
electrons
E. The octet rule

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 41
.

Lewis Structures

Molecular formula drawn with Lewis Symbols

Method for diagramming electronic structure
of covalent bonds

Uses dots to represent electrons

Covalent bond

Shared pair of electrons

Each atom shares electrons so has complete
octet ns 2np 6

Noble gas electron configuration

Except H which has complete shell with 2
electrons
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 42
.

21
Octet rule and covalent bonding
When atoms form covalent bonds, they tend
to share sufficient electrons so as to achieve
outer shell having eight electrons

Indicates how all atoms in molecule are attached
to one another

Accounts for ALL valence electrons in ALL atoms
in molecule
Let’s look at some examples
Noble Gases: eight valence electrons

Full octet ns 2np 6

Stable monatomic gases

Don’t form compounds
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 43
.

Lewis Structures
Diatomic Gases:

H and Halogens
H2

H· + ·H → H:H or HH

Each H has two electrons through
sharing

Can write shared pair of electrons as a
line ()
: or  signify a covalent bond
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 44
.

22
 Lewis Structures
Diatomic Gases:
F2
F + F FF F F

Each F has complete octet

Only need to form one bond to complete
octet

Pairs of electrons not included in covalent
bond are called lone pairs

Same for rest of halogens: Cl2, Br2, I2
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 45
.

Lewis Structures
Diatomic Gases:
HF
H + F HF H F

Same for HCl, HBr, HI

Molecules are diatomics of atoms that need
only one electron to complete octet

Separate molecules

Gas in most cases because very weak
intermolecular forces
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 46
.

23
 Your Turn!
How many electrons are required to complete
the octet around nitrogen?
A. 2
B. 3
C. 1
D. 4
E. 6

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 47
.

Lewis Structures
Many nonmetals form more than one covalent bond

C N O
Needs 4 electrons Needs 3 electrons Needs 2 electrons
Forms 4 bonds Forms 3 bonds Forms 2 bonds
H
H N H O H
H C H
H H
H H
H C H H N H O H
H H H
methane ammonia water
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 48
.

24
 Multiple Bonds

Single Bond

Bond produced by sharing one pair of
electrons between two atoms

Many molecules share more than one

pair of electrons between two atoms

Multiple bonds

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 49
.

Double Bonds

Two atoms share two pairs of electrons
e.g., CO2

O C O O C O O C O
Triple bond

Three pairs of electrons shared between
two atoms
e.g., N2

N N N N N N
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 50
.

25
 Your Turn!
Which species is most likely to have multiple
bonds?
A. CO
B. H2O
C. PH3
D. BF3
E. CH4

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 51
.

Electronegativity and Bond

Polarity

Two atoms of same
element form bond

Equal sharing of
electrons

Two atoms of different

elements form bond

Unequal sharing of
electrons

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 52
.

26
 Why?

One atom usually attracts
electrons more strongly
than the other
Result

Unbalanced distribution
of electron density within bond

Electron cloud tighter around Cl in HCl

Slight positive charge around H

Slight negative charge around Cl

This is not a complete transfer of an electron
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 53
.

Electronegativity and Bond

Polarity

Leads to concept of
partial charges
δ+ δ–
H——Cl
δ+ on H = +0.17
δ– on Cl = –0.17

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 54
.

27
 Polar Covalent Bond

Also known as a polar bond

Bond that carries partial + and – charges at
opposite ends

Bond is dipole

Two poles or two charges involved
Polar Molecule

Molecule has partial positive and negative
charges at opposite ends of a bond

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 55
.

Dipole Moment

Quantitative measure of extent to which
bond is polarized

Dipole moment = Charge on either end ×
distance between them

µ=q×r

Units = debye (D)

1 D = 3.34 × 10–30 C m (Coulomb meter)

The size of the dipole moment or the
degree of polarity in the bond depends on
the differences in abilities of bonded atoms
to attract electrons to themselves
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 56
.

28
 Your Turn!
Which situation below results in the largest
dipole moment?
A. +1 and -1 charges separated by 6 Å
B. +1 and -1 charges separated by 8 Å
C. +2 and -2 charges separated by 4 Å
D. +2 and -2 charges separated by 6 Å
E. +2 and -2 charges separated by 8 Å
µ=q×r dipole moment increases as
both q and r increase.

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 57
.

Dipole Moments and Bond Lengths

for Some Diatomic Molecules

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 58
.

29
 Electronegativity (EN)

Relative attraction of atom for electrons in
bond

Quantitative basis

Table of electronegativities – Fig. 8.9

Difference in electronegativity
= estimate of bond polarity
∆EN = |EN1 – EN2|
e.g., N—H Si—F
δ– δ+ δ+ δ–

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 59
.

Electronegativity Table

Linus Pauling

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 60
.

30
 Trends in Electronegativity

EN increases from left to right across period
as Zeff increases

EN decreases from top to bottom down group
as n increases
Ionic and Covalent Bonding

Are the two extremes of bonding

Actual is usually somewhere in between.

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 61
.

Your Turn!
Which of the following species has the least
polar bond?
A. HCl
B. HF
C. HI
D. HBr

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 62
.

31
 Using Electronegativities

Difference in electronegativity
∆EN = EN A – EN B
Measure of ionic character of
bond

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 63
.

Using Electronegativities

Nonpolar Covalent Bond

No difference in electronegativity

Ionic Character of bond

Degree to which bond is polar

∆EN > 1.7 means mostly ionic

> 50% ionic

More electronegative element almost completely
controls electron

∆EN < 0.5

Means almost purely covalent

Nonpolar: < 5% ionic

0.5 < ∆EN < 1.7 polar covalent
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 64
.

32
 Result

Elements in same region of periodic table

i.e., two nonmetals

Have similar electronegativities

Bonding more covalent

Elements in different regions of periodic

table

i.e., metal and nonmetal

Have different electronegativities

Bonding predominantly ionic
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 65
.

Reactivities of Elements Related

to Electronegativities

Parallels between EN and its reactivity

Tendency to undergo redox reactions

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 66
.

33
 Reactivities of Elements Related
to Electronegativities
Metals

Low EN: easy to oxidize (Groups 1A and 2A)

High EN: hard to oxidize (Pt, Ir, Rh, Au, Pd)

Reactivity decreases across row as
electronegativity increases
Nonmetals

Oxidizing power increases across row as EN
increases

Oxidizing power decreases down a column as
electronegativity decreases
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 67
.

Your Turn!
Predict the type of bonding (covalent or ionic)
in the following: magnesium chloride, carbon
tetrachloride, iron(III) oxide, sulfur dioxide,
carbon disulfide
A. ionic, ionic, ionic, covalent, covalent
B. covalent, ionic, covalent, ionic, ionic
C. ionic, covalent, covalent, covalent, ionic
D. ionic, covalent, ionic, covalent, covalent

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 68
.

34
 Drawing Lewis Structures

Very useful

Way of diagramming structure

Used to describe structure of molecules

Can be used to make reasonably accurate
predictions of shapes of molecules

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 69
.

Drawing Lewis Structures

Not all molecules obey the octet rule

Holds rigorously for second row elements like
C, N, O, and F

B and Be sometimes have less than octet
BeCl2, BCl3

2nd row can never have more than eight
electrons

3rd row and below, atoms often exceed octet

Why?

n = 3 shell can have up to 18 electrons as
now have d orbitals in valence shell
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 70
.

35
Method for Drawing Lewis Structure

1. Decide how atoms are bonded

Skeletal structure = arrangement of atoms

Central atom

Usually given first

Usually least electronegative
2. Count all valence electrons (all atoms)
3. Place two electrons between each pair of
atoms

Draw in single bonds

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 71
.

Method for Drawing Lewis Structure

4. Complete octets of terminal atoms
(atoms attached to central atom) by
adding electrons in pairs
5. Place any remaining electrons on central
atom in pairs
6. If central atom does not have octet

Form double bonds

If necessary, form triple bonds

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 72
.

36
 Example: SiF4
Skeletal Structure
1 Si = 1 × 4e– = 4 e– F
4 F = 4 × 7e– = 28 e– F Si F
Total = 32 e–
single bonds – 8 e– F
24 e– Complete terminal
atom octets
F lone pairs – 24 e–
0 e– F
F Si F
F
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 73
.

Example: H2CO3

CO32– oxoanion, so C central, and O s
around, H+ attached to two O s
1 C = 1 × 4e– = 4 e– O
3 O = 3 × 6e– = 18 e– H O C O H
2 H = 2 × 1e– = 2 e–
Total = 24 e– O
single bonds – 10 e–
H O C O H
14 e–
O lone pairs – 14 e–

But C only has 6 e–
0 e–
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 74
.

37
 Example: H2CO3 (cont.)

Too few electrons O

Must convert one of H O C O H
lone pairs on O to
second bond to C

Form double bond
between C and O O
H O C O H

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 75
.

Example: N2F2
2 N = 2 × 5e– = 10 e– Skeletal Structure
2F = 2 × 7e– = 14 e–
F N N F
Total = 24 e–
Complete terminal
single bonds – 6 e–
atom octets
18 e–
F lone pairs – 12 e– F N N F
6 e–
Put remaining electrons
N electrons – 6 e–
on central atom
0 e–
F N N F
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 76
.

38
 Example: N2F2 (cont.)

Not enough electrons to complete octets

on nitrogen

Must form double bond between nitrogen
atoms to satisfy both octets

F N N F F N N F

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 77
.

Expanded Octets
Elements after Period 2 in the Periodic Table

Are larger atoms

Have d orbitals

Can accept 18 electrons

For Lewis structures

Follow same process as before but add extra
electrons to the central atom

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 78
.

39
 Example: PCl5
1 P = 1 × 5 e– = 5 e– Cl
5 Cl = 5 × 7e– = 35 e– Cl Cl
P
Total = 40 e–
Cl Cl
single bonds – 10 e–
30 e–
Cl
Cl lone pairs – 30 e–
0 e– Cl Cl
P

P has 10 electrons Cl Cl

Third period element

Can expand its shell
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 79
.

Electron Deficient Structures

Boron often has six electrons around it

Three pairs

Beryllium often has four electrons around it

Two pairs

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 80
.

40
 Example: BBr3
1 B = 1 × 3 e– = 3 e–
Br
3 Br = 3 × 7e– = 21 e–
Total = 24 e– B
single bonds – 6 e– Br Br
18 e–
Br lone pairs – 18 e– Br
0 e–

B has only six electrons B

Does not form double bond Br Br

Has incomplete octet
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 81
.

Your Turn!
How many electrons does NO2 have and how
many are in the structure below?

A. 16, 18
B. 17, 18
C. 18, 18
D. 16, 16
E. 15, 16

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 82
.

41
 Your Turn!
What type of bond exists between the C and O
atoms in CH2O (i.e., single, double, etc.)?
O
A. single bond
C
B. double bond
H H
C. triple bond
D. There is no bond between the C and the O
because the oxygen is bonded to a H:
C-H-H-O

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 83
.

If more than one Lewis structure

can be drawn, which is correct?

Experiment always decides

Concepts such as formal charge and

resonance help to make predictions

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 84
.

42
 Example: H2SO4 O

1 S = 1 × 6e– = 6 e– H O S O H
4 O = 4 × 6e– = 24 e– O
2 H = 2 × 1e– = 2 e– O
Total = 32 e–
single bonds – 12 e– H O S O H
20 e– O
O lone pairs – 20 e– O
0 e–
H O S O H
• n = 3, has empty d orbitals
• Could expand its octet O
• Could write structure with double bonds.
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 85
.

How Do We Know Which is Accurate?

Experimental evidence

In this case bond lengths from X-ray data

S—O bonds (no H attached) are shorter 142 pm

S—O—H, S—O longer 157 pm

Indicates that two bonds are shorter than the
other two

Structure with S=O for two O’s without H’s is
more accurate

Preferred Lewis structure

Even though it seems to violate octet rule
unnecessarily
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 86
.

43
 Formal Charge (FC)

Apparent charge on atom

Bookkeeping method

Does not represent real charges
FC = # valence e–
– [# unshared e– + ½ (# bonding e–)]
FC = # valence e–
– [# bonds to atom + # unshared e–]

Indicate formal charges by placing them in
circles around atoms

Sum of FC on all atoms must equal the
overall charge on ion or molecule
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 87
.

FC = #valence e– − [#bonds to atom

+ # unshared e– ]
-1 Structure 1
O FCS = 6 – (4 + 0) = 2
+2 FCH = 1 – (1 + 0) = 0
H O S O H FCO(s) = 6 – (1 + 6) = –1
O -1 FCO(d) = 6 – (2 + 4) = 0

O Structure 2
FCS = 6 – (6 + 0) = 0
H O S O H FCH = 1 – (1 + 0) = 0
FCO(s) = 6 – (2 + 4) = 0
O
FCO(d) = 6 – (2 + 4) = 0
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 88
.

44
 H2SO4

No formal charges on any atom in structure 2
Conclusion:

When several Lewis structures are possible

Those with smallest formal charges

Most stable

Preferred
Most Stable Lewis Structure
1. Lowest possible formal charges are best
2. All FC ≤ |1|
3. Any negative FC on most electronegative element
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 89
.

Example: CO2
1 C = 1 × 4e– = 4 e–
2 O = 2 × 6e– = 12 e–
Total = 16 e–
single bonds – 4 e–
12 e–
O lone pairs – 12 e–
0 e–

C only has four
electrons

Need two extra bonds
Which of these is correct?
to O to complete octet
Need another criteria

3 ways you can do this
Come back to this
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 90
.

45
 CO2 Which Structure is Best

Use formal charge to +1 −1
determine which
structure is best
FCC = 4 – (4 + 0) = 0
FCO(single) = 6 – (1 + 6) = –1
FCO(double) = 6 – (2 + 4) = 0
FCO(triple) = 6 – (3 + 2) = +1
-1 +1

Central structure best

All FC’s = 0
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 91
.

Can Use Formal Charges to Explain

Boron Chemistry

BCl3

Why doesn’t a double bond form here?

FCB = 3 – 0 – 3 = 0

FCCl = 7 – 6 – 1 = 0

All FCs = 0 so the molecule has the best
possible structure

It doesn’t need to form double bond

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 92
.

46
 Your Turn!
What is the formal charge on Xe for the
following?

A. +2, +4
B. +2, +3
C. +4, 0
D. +4, +2

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 93
.

Your Turn!
What is the formal charge on each atom in the
structure below?

O N S

A. O: -2 N: 0 S: -2
B. O: 0 N: 0 S: 0
C. O: -1 N: 0 S: -1
D. O: -2 N: +1 S: +2
E. O: -1 N: 0 S: 0

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 94
.

47
 Resonance: Explaining Multiple
Equivalent Lewis Structures

Can use formal charge to decide between two
different Lewis structures

Need an explanation of equivalent structures

The resonance concept provides the way to
interpret equivalent structures

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 95
.

Resonance: When a Single Lewis

Structure Fails
Example: NO3–
1 N = 1 × 5e– = 5 e–
3 O = 3 × 6e– = 18 e–
–1 charge = 1 e–
Total = 24 e–
single bonds – 6 e–
18 e–
O lone pairs –18 e–
0 e–
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 96
.

48
 Example: NO3–

Lewis structure predicts
one bond shorter than
other two
Experimental observation:
• All three N—O bond lengths are same
• All shorter than N—O single bonds
• Have to modify Lewis Structure
• Electrons cannot distinguish O atoms
• Can write two or more possible structures
simply by moving where electrons are
• Changing placement of electrons
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 97
.

What are Resonance Structures?

Multiple Lewis structures for single molecule

No single Lewis structure is correct

Structure not accurately represented by any one Lewis
structure

Actual structure = “average” of all possible structures

Double headed arrow between resonance structures
used to denote resonance

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 98
.

49
 Resonance Structures

Lewis structures assume electrons are
localized between 2 atoms

In resonance structures, electrons are
delocalized

Smeared out over all atoms

Can move around entire molecule to give
equivalent bond distances
Resonance Hybrid

Way to depict
resonance delocalization
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 99
.

Example: CO32–

1 N = 1 × 5e– = 5 e–
3 O = 3 × 6e– = 18 e–
−1 charge = 1 e–
Total = 24 e–
single bonds – 6 e–
18 e–
O lone pairs –18 e–
0 e–
C only has 6 electrons so
a double bond is needed
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 100
.

50
 Three Equivalent Resonance
Structures

All have same net formal charges on C and O’s

FC = –1 on singly bonded O’s

FC = O on doubly bond O and C

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 101
.

Resonance Structures Not Always

Equivalent

Two or more Lewis Structures for same compound
may or may not represent electron distributions of
equal energy
How Do We Determine Which are Good
Contributors?
1. All octets are satisfied
2. All atoms have as many bonds as possible
3a. FC ≤ 1
3b. Any negative charges are on the more
electronegative atoms.

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 102
.

51
Drawing Good Resonance Structures
1. All must be valid Lewis Structures
2. Only electrons are shifted

Usually double or triple bond and lone pair

Nuclei can't be moved

Bond angles must remain the same
3. Number of unpaired electrons, if any, must remain
the same
4. Major contributors are the ones with lowest
potential energy (see above)
5. Resonance stabilization is most important when
delocalizing charge onto two or more atoms
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 103
.

Example: NCO−

1 C = 1 × 4e– = 4 e–
1 N = 1 × 5e– = 5 e–
1 O = 1 × 6e– = 6 e–
–1 charge = 1 e–
Total = 16 e–
single bonds – 4 e–
12 e–
lone pairs –12 e–
0 e–

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 104
.

52
 Example: NCO–
FCN = 5 – 2 – 3 = 0 −1

FCC = 4 – 0 – 4 = 0 Best
FCO = 6 – 6 – 1 = –1

FCN = 5 – 4 – 2 = –1 −1

FCC = 4 – 0 – 4 = 0 OK
FCO = 6 – 4 – 2 = 0

FCN = 5 – 6 – 1 = –2 −2 +1
FCC = 4 – 0 – 4 = 0 Not
Accept-
FCO = 6 – 2 – 3 = +1 able
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 105
.

Resonance Stabilization

Actual structure is more stable than either

single resonance structure

For benzene

The extra stability is ~146 kJ/mol

Resonance energy

Extra stabilization energy from resonance
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 106
.

53
 Your Turn!
Which of the following is false?
A. Not all resonance structures for a given
molecule are equivalent
B. Resonance structures indicate that the
molecule rapidly switches between multiple
electron distributions
C. Resonance structures are necessary when a
single Lewis structure is not sufficient to
depict the electron distribution in a molecule
D. Formal charges are calculated the same way
in resonance structures as in other structures
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 107
.

Your Turn!
Which of the structures below exhibit resonance
(hint: there is more than one)?
A. NO2
B. H2O
C. N3–
D. N2O (nitrogen is central atom)
E. CH3CH2CH

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 108
.

54
 Coordinate Covalent Bonds
Ammonia

Normal covalent bonds

One electron from each atom shared
between the two
H

3H + N N
H H

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 109
.

Coordinate Covalent Bond

Ammonium Ion

H+ has no electrons

N has lone pair

Can still get 2 electrons shared between
them
H H +

N + H+ H N H
H H
H
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 110
.

55
 Coordinate Covalent Bond

Both electrons of shared pair come from just
one of two atoms

Once bond formed, acts like any other
covalent bond

Can't tell where electrons came from after
bond is formed

Useful in understanding chemical reactions

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 111
.

Coordinate Covalent Bond

Especially boron (electron deficient

molecule) reacts with nitrogen compounds
that contain lone pair of electrons

H Cl H Cl

N + B H N B Cl
H H Cl Cl
H Cl

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 112
.

56
 Learning Check
Example: Draw the best Lewis structure for
the following
HClO4

XeF4

I3–

BrF5
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 113
.

Carbon Compounds

Carbon-containing compounds

Exist in large variety

Mostly due to multiple ways in which C can form
bonds

Functional groups

Groups of atoms with similar bonding

Commonly seen in C compounds

Molecules may contain more than one

functional group
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 114
.

57
 Important Compounds of Carbon
Alkanes e.g.,

Hydrocarbons CH4 methane

Only single bonds CH3CH3 ethane
CH3CH2CH3 propane
Isomers

Same molecular formula H

Different physical properties H C H

Different connectivity H H
(structure) H C C C H
H H H H
H H H
H C C C C H
butane iso-butane
H H H H
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 115
.

Hydrocarbons
H C C H ethylene

Alkenes (ethene)

Contain at least one H H
H H
double bond
H C C C C H
butene
H H H H

H C C H acetylene

Alkynes (ethyne)
H H

Contain at least one
triple bond H C C C C H
butyne
H H
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 116
.

58
 Oxygen Containing Organics

Alcohols H H H

Replace H with OH H C O H C C O
H H H H H
methanol ethanol

Ketones H O H

Replace CH2 with H C C C H
C=O
H H

Carbonyl group
acetone
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 117
.

Carbonyl Group

Carbon in hydrocarbon with double
bond to oxygen

Aldehydes

Ketones

Carboxylic acids

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 118
.

59
 Oxygen Containing Organics

Aldehydes H O

At least one atom
H C C H
attached to C=O is H
H
acetaldehyde

Organic Acids H O

Contains carboxyl H C C O
group
H H

COOH
acetic acid

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 119
.

Nitrogen Containing Organics

Amines

Derivatives of NH3 with H’s replaced by alkyl
groups

H H H H H
H C N H H C N C H
H H H
methylamine dimethylamine

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 120
.

60
 Your Turn!
How many isomers are there of butanol?
A. none
B. 2
C. 3
D. 6
E. 4

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 121
.

Your Turn!
Which molecule contains a carbonyl group
(hint: draw the Lewis structures to help
decide)?
A. C2H6O CH3-O-CH3
B. N(CH3)3 amine
C. CH3CCH alkyne (CH3-C≡H)
D. CH4O alcohol (CH3-O-H)
E. CH2O aldehyde O
C
H H
Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved. 122
.

61