Chem 2 AP Homework #10-2: pg.429 #18, 19, 21, 23, 24, 26, 27, 79
18
The bonds in beryllium hydride (BeH2) molecules are polar, and yet the dipole moment of
the molecule is zero. Explain.
BeH2 has a net dipole moment of 0 because it is linear, and the dipole moments of the individual
bonds, which have the same magnitude but opposite directions, cancel out.
H
19
Referring to Table 10.3, arrange the following molecules in order of increasing dipole
moment: H2O, H2S, H2Te, H2Se.
All four molecules have two bonds and two lone pairs (AB2E2) and therefore the bond angles are
not linear. Since electronegativity decreases going down a column (group) in the periodic table, the
electronegativity differences between hydrogen and the other Group 6 element will increase in the
order Te < Se < S < O, and the electronegativity of H is equal to that of Te. The dipole moments
will increase in the same order.
Te
H
21
H
Be
H
<
Se
H
S
<
H
H
H
<
O
H
H
List the following molecules in order of increasing dipole moment: H2O, CBr4, H2S, HF,
NH3, CO2.
CO2 = CBr4 (µ = 0 for both) < H2S < NH3 < H2O < HF
Br
O
23
O =
Br
C
<
C
Br
Br
S
H
H
<
N
H
H
H
<
O
H
H
< H
Which of the following molecules has the higher dipole moment?
Molecule (b) will have a higher dipole moment. In molecule (a), the trans arrangement cancels the
bond dipoles and the molecule is nonpolar.
H
Br
C
Br
C
H
C
Br
(a)
Br
H
C
(b)
H
F
2
24
HW 10-2: DIPOLE MOMENTS
The molecules shown in (b) and (d) are nonpolar. Due to the high symmetry of the molecules and
the equal magnitude of the bond moments, the bond moments in each molecule cancel one another.
The resultant dipole moment will be zero. For the molecules shown in (a) and (c), the bond moments
do not cancel and there will be net dipole moments. The dipole moment of the molecule in (a) is
larger than that in (c), because in (a) all the bond moments point in the same relative direction,
reinforcing each other (see Lewis structure below). Therefore, the order of increasing dipole moments
is:
(b) = (d) = 0 < (c) < (a).
Cl
Cl
Cl
Cl
=
26
Cl
<
Cl
Cl
<
Cl
Cl
Cl
(b)
(d)
(c)
(a)
Use valence bond theory to explain the bonding in Cl2 and HCl. Show how the atomic
orbitals overlap when a bond is formed.
In Cl2 there is an overlap between the 3p orbitals of the two chlorine atoms. In HCl the overlap is
between the 1s orbital of H and the 3p orbital of Cl:
Cl
3px
27
Cl
Cl
3px
H
1s
Draw a potential energy curve for the bond formation in F2.
–155 kJ/mol
142 pm
Cl
3px
HW 10-2: DIPOLE MOMENTS
79
3
Draw Lewis structures and give the other information requested for the following:
(a) SO3. The Lewis structure is:
O
O
S
O
O
The geometry is trigonal planar; the molecule is nonpolar.
O
S
O
(b) PF3 The Lewis structure is:
F
P
F
P
F
F
F
F
The molecule has trigonal pyramidal geometry. It is polar.
(c) F3SiH The Lewis structure is:
H
F
Si
H
F
Si
F
F
F
F
Net
Dipole
The molecule will be tetrahedral (AB4). Both fluorine and hydrogen are more electronegative
than silicon, but fluorine is the most electronegative element, so the molecule is polar (fluorine
side negative).
(d) SiH3– The Lewis structure is:
H
Si
H
Si
H
H
H
H
The ion has a trigonal pyramidal geometry (AB3E).
(e) Br2CH2 The Lewis structure and 3-dimensional structure are:
H
H
H
C
Br
Br
C
H
Br
Br
The molecule will be tetrahedral (AB4) but still polar. The negative end of the dipole moment
will be on the side with the two bromine atoms; the positive end will be on the hydrogen side.