Chemistry - A Molecular Science

Chapter 5 The Covalent Bond

Example 5.13

Draw the Lewis structure of acetone (C

H 3

O), indicate all nonzero formal charges, 6

and determine the oxidation state of each atom. The oxygen atom is attached to the central carbon atom, and there are no O-H bonds.

ER = 3(8) for C + 1(8) for O + 6

(2) for H = 44 electrons.

VE = 3(4) from C + 1(6) from O + 6(1) from H = 24 valence electrons.

SP = ½

(44 - 24) = 10 pairs must be shared.

CC

O

C

(a) Skeletal structure

H CH H

CC

H

H

H

O

(b) Lewis Structure

The skeleton formed by the carbon and oxygen

atoms that is consistent with the given

information, which is shown in the margin figure labeled Example 5.13a, contains three shared pairs. The skeleton with

the six C-H bonds will require another six pairs, for a total

of nine pairs. Ten shared pairs are required,

so there must be a double bond, which

cannot be drawn to a hydrogen atom. Theref

ore, there is either a C=C or C=O double

bond. We use the fact that

carbon always has four bonds to decide where the double

bond must be. If the double bond is a C=C bond,

then one of the terminal atoms can have

only two C-H bonds (1 C=C + 2 C-H = 4 bonds), and the central carbon will have four bonds (1 C-C + 1 C-O + 1 C=C). Only three C-H bonds can be drawn to the other carbon atom, so the sixth H would have to be bound to O, which is a violation of the given information. We conclude that the double bond must be a C=O bond to arrive at the structure shown as the margin figure labeled Example 5.13b. The oxidation states are determined by again dividing the C-C bonding electrons between the carbon atoms and assigning the C-H bonding

electrons to the more electronegative

carbon atom. The C=O bonding electrons are assigned to the oxygen. Therefore, the oxidation states of the carbon atoms can be determined to be OX

= 4 VE - [0 NB + 6BEC

C-H

+ ½

(2 BE

C-C

)] = -3 (terminal carbons)

OX

= 4 VE - [0 NB + ½C

(4 BE

C-C

) + 0(4 BE

C=O

)] = +2 (central carbon)

Thus, the three carbon atoms have an average

oxidation state of (-3 - 3 +2)/3 = -4/3.

Using the method of Section 4.4, we would determine the oxidation state of carbon in C^3

H^6

O as 3OX

+ 6(+1) + 1(-2) = 0, which also yielC

ds -4/3. Thus, the method in Section

4.4 determines the average oxidat

ion state of each atom in a molecule, while the method

presented in this chapter determines the ox

idation state of each individual atom.

The oxygen is assigned all four bonding

electrons, so its oxidation state is

OX

= 6 VE - [4 NB + 4 BE] = 6 - 8 = -2 O

No bonding electrons are assigned to the hydrogen, so its oxidation state is

OX

= 1 VE - [0 NB + 0 BE] = +1 H

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