Chemistry - A Molecular Science

OS

O

O

Example 5.11 Lewis structure of SO

2- 3

that shows all non-zero formal

CC

H H

H H

Example 5.12 Lewis structure of C

H 2

(^4)
The formal charges are determined as follows: FC
= 6 VE - [2 NB + S
1 /^2
(6 BE)] = +1 and FC
= 6 VE - [6 NB + O
1 /^2
(2 BE)] = -1.
Note the sum of the formal charges is +1 + 3
(-1) = -2, the charge on the ion. The negative
charge on the ion is distributed equ
ally among the three oxygen atoms.
Oxygen is more electronegative than sulf
ur, so the oxidation states are:
OX
= 6 VE - [2 NB + (0)(6 BE)] = +4 and OXS
= 6 VE - [6 NB + (1)(2 BE)] = -2. O
OX
= 6 VE - [2 NB + (0)(6 BE)] = +4 and OXS
= 6 VE - [6 NB + (1)(2 BE)] = -2. O
Because oxygen is very electronegative, t
he bonding electrons are almost always
assigned to it, which gives it a -2 oxidation st
ate in most of its compounds. The sum of the
oxidation states is +4 + [3
x(-2)] = -2, the charge on the ion.
Because oxygen is very electronegative, t
he bonding electrons are almost always
assigned to it, which gives it a -2 oxidation st
ate in most of its compounds. The sum of the
oxidation states is +4 + [3
x(-2)] = -2, the charge on the ion.
Example 5.12 Example 5.12 Draw the Lewis structure of C
H 2
, indicate all nonzero formal charges, and determine 4
Draw the Lewis structure of Cthe oxidation state of each atom.
H 2
, indicate all nonzero formal charges, and determine 4
the oxidation state of each atom.
ER = (2)(8) for C + (4)(2) for H = 24 electrons.
ER = (2)(8) for C + (4)(2) for H = 24 electrons.
VE = (2)(4) from C + (4)(1) from H = 12 valence electrons, or 6 pairs.
VE = (2)(4) from C + (4)(1) from H = 12 valence electrons, or 6 pairs.
SP =
1 /^2
(24 - 12) = 6 pairs must be shared. There are no lone pairs.
SP =
1 /^2
(24 - 12) = 6 pairs must be shared. There are no lone pairs.
Each H atom can have only one shared pair, so
the four C-H bonds require four shared
Each H atom can have only one shared pair, sopairs, which leaves two shared pairs for the C-C bond. All atoms have zero formal charge. The Lewis structure is shown in the margin.
the four C-H bonds require four shared
pairs, which leaves two shared pairs for the C-C bond. All atoms have zero formal charge. The Lewis structure is shown in the margin. Carbon is more electronegative than hydrogen, but the two carbons have the same electronegativities, so the C=C bonding
electrons are divided equally between the
carbons when determining the ox
idation states. Therefore,
Carbon is more electronegative than hydrogen, but the two carbons have the same electronegativities, so the C=C bonding
electrons are divided equally between the
carbons when determining the ox
idation states. Therefore,
OX
= 4 VE - [0 NB + 4 BEC
C-H

• 1 /^2
  (4 BE
  C=C
  )] = -2
  OX
  = 4 VE - [0 NB + 4 BEC
  C-H


• 
  

  1 /^2
  (4 BE
  C=C
  )] = -2
  OX
  = 1 VE - 0(2 BE) = +1. H
  OX
  = 1 VE - 0(2 BE) = +1. H
  Chapter 5 The Covalent Bond
  © by
  North
  Carolina
  State
  University