Exercício Resolvido - Quimica Geral (223)

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Chapter 10: Chemical Bonding I: Basic Concepts 
446 
129. (D)
 (a) The two bond dipole moments can be added 
geometrically, by placing the head of one at the tail 
of the other, as long as we do not change the 
direction or the length of the moved dipole. The 
resultant molecular dipole moment is represented by 
the arrow drawn from the tail of one bond dipole to 
the head of the other. This is shown in the figure to 
the right. The 52.0" angle in the figure is one-half of 
the 104" bond angle in water. The length is given as 
1.84 D. We can construct a right angled triangle by 
bisecting the 76.06 angle. The 
right angled triangle has a hypotenuse O * H bond dipole and the two other 
angles are 526 and 386. The side opposite the bisected 76.0 6 angles is ½ (1.84 D) 
0.92 D. We can calculate the bond dipole using: sin 38.06 
0.92D
O-H bond dipole
 
0.61566, hence O*H bond dipole 1.49 D. 
(b) For H S2 , we do not know the bond angle. We shall 
represent this bond angle as 2:. Using a similar 
procedure to that described in part (a), above, a 
diagram can be constructed and the angle 2 : 
calculated as follows: 
cos : 
1
(0.93D)
2
0.67 D
 0.694 : 46.056 
or 2: 92.1 6 
The H*S*H angle is approximately 926.5
(c)
sin (<) = 
0.25 D
1.87 D
<5= 7.6o
Molecule and associated 
individual bond dipoles
Relationship between
dipole moment(molecular) 
and bond dipoles(Vector 
addition)
Geometric Relationship
1.87 D
0.23 D
C
Cl <
o
Mathematical solution:
1.87 D
0.23 D
C
H
ClCl
Cl
(90 + <)
o
<
o
=C-H + 3(=C-Cl) 1.04 D
0.30 D + 3(x) 1.04 D
x 0.25 D =C-Cl
C
H
ClCl
Cl
0.3 D
1.87 D
1.87 D
1.87 D
The H-C-Cl bond angle is (90 + <)o = 90o + 7.6 o = 97.6o 
52o
52o
1.
84
D
76o
52o
52o
104o
38o
38o
0.
93
D
0.67 D
0
.6
7
D
90
o
:
0
.6
7
D
:
:
: