Chemistry - A Molecular Science

H CA

H

H

O CB

OH

Example 5.8b Acetic acid, C

H 2

O 4

(^2)
HOOH
Example 5.9 Lewis structure of hydrogen peroxide
b) Use electron counting and the Lewis Stru
cture of acetic acid shown in the margin
to determine the oxidation states of the individual carbon atoms. CA
is assigned all of the C-H bonding electrons
(a = 1 because C is more electronegative)
and one of the two C-C bonding electrons (a
= ½ because the atoms are identical). BE = 7
for C

. CA

is assigned one electron from the C-C bond but none from the C=O or C-O B

bonds because O is more electronegative. BE = 1 for C

. Neither atom has any B

nonbonding electrons, so we can write the following: Atom

VE NB BE Oxid

ation State (OX)

CA

4

0

7

4 - [0 + 7] = -3

CB

4

0

1

4 - [0 + 1] = +3

One atom is –3, while one is +3. The averag

e of both carbon atoms is 0, the result

obtained in Part a.

Example 5.9

What are the oxidation states of the atoms in H

O 2

(hydrogen peroxide)? 2

The Lewis structure of hydrogen peroxide is shown in the margin. Oxygen is more electronegative than hydr

ogen, so the O-H bonding electrons are

assigned to the oxygen atom. The oxidat

ion state of each hydrogen atom is OX

= 1 - [0 + H

(0 x 2)] = +1. The two oxygen atoms have

identical electronegativities, so the O-O

bonding electrons are shared equally bet

ween the two oxygen atoms, but the O-H

bonding electrons are assigned to the oxygen.

The oxidation state of

each oxygen atom is

then determined to be -1. The oxidation state of oxygen is -1 in peroxides.

OX

= 6 VE - [4 NB + (1)(2 BEO

O-H

)+ (^

1 /^2

)(2 BE

O-O

)] = -1.

5.9

PRACTICE WITH LEWIS STRUCTURES

We conclude this chapter with several

examples of drawing Lewis structures and

determining formal char

ge and oxidation numbers.

Chapter 5 The Covalent Bond

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