Download Solutions to Problem Assignment 6

Solutions to Problem Assignment 6 (Intermolecular Forces)
1. For molecules to be held together by London Dispersion Forces, it must be
possible to INDUCE an electric dipole moment in the molecule with an electric
field. Although such induced moments vary in magnitude, in principle, an electric
field can induce an electric dipole moment in any molecule, even in a
homonuclear diatomic molecule such as H2. For example, a uniform electric field
pointing along the internuclear direction, with the electrical potential difference
shown, would produce a small polarization of the electron distribution in the H–H
bond.
polarity of the electric field
No electric field
H
H
–
+
δ+
H
δ–
H
With electric field
aligned along
the H-H
internuclear axis
with the polarity
shown
For molecules to be held together by dipole-dipole interactions, the molecule
must possess a PERMANENT electric dipole moment. For example, the
heteronuclear diatomic molecule, HCl, possesses a permanent electric dipole
moment but the homonuclear diatomic molecule, H2 , does not.
For molecules to be held together by a hydrogen bond, the molecule must
contain a H atom bonded to an E L E C T R O N E G A T I V E atom -- the
electronegative atom should also have unshared pair(s) of electrons -- this is
usually true of electronegative atoms.
The relative strengths of such intermolecular interactions are:
hydrogen-bond > dipole-dipole > London dispersion
(a) CH4: London forces ONLY. CH4 has no permanent electric dipole moment
and H is NOT bonded to a very electronegative atom.
(b) CH3Cl: London forces AND dipole-dipole interactions. Chloromethane has
a permanent electric dipole moment pointing along the C – Cl bond.
(c) CH2Cl2: London forces AND dipole-dipole interactions. Dichloromethane
has a permanent electric dipole moment pointing along the line bisecting the ∠ Cl
– C – Cl.
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(d) CHCl3 : London forces AND dipole-dipole interactions. Trichloromethane
has a permanent electric dipole moment pointing along the line parallel with the
H – C axis.
(e) CCl4 : London forces ONLY. Tetrachloromethane has no permanent
electric dipole moment. The C – Cl bonds are polar, but all the bond moments
cancel due to the tetrahedral structure of CCl4 .
: S:
: :
: :
H
H
2. (a)
Since S is more electronegative than H, each S – H
bond is polarized with the bond moments directed as shown. Since H2S is a
bent molecule the vectorial sum of the bond dipole moments will produce a nonzero total dipole moment. Since the permanent dipole moment is NON-ZERO,
H2 S will show dipole-dipole interactions.
O
C
O
Since O is more electronegative than C,
(b)
each C – O bond is polarized with the bond moments directed as shown. Since
CO2 is linear, the bond moments cancel exactly, so that CO2 has a ZERO total
permanent dipole moment. Thus, CO 2 will N O T show dipole-dipole
interactions.
F
F
C
F
F
(c)
Since F is more electronegative than C, each C – F bond is polarized, and each
bond has an associated bond dipole moment. However, because of the
TETRAHEDRAL arrangement of the C – F bonds, the bond moments cancel
exactly, so that CF4 has a ZERO total permanent dipole moment. Thus, CF4 will
NOT show dipole-dipole interactions.
:
C
O
(d)
CO will show dipole-dipole interactions.
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3. Since Na+F– and Al2 O3 (2 [Al3+ ] 3 [O2– ]) form extended ionic lattices and
since the force between ion pairs are MUCH stronger than hydrogen-bonding
forces > dipole-dipole forces > London dispersion forces, NaF and Al2 O3 will
have the highest boiling points. Since Al2O3 has ions with a greater charge than
either of the ions in Na +F–, the boiling point of Al2O3 will be greater than that of
NaF.
Of the remaining molecules, none of them is capable of forming hydrogen-bonds.
Only one of them, NO, has a permanent electric dipole moment and so it is the
only molecule which can have dipole-dipole intermolecular interactions; NO will
have a higher boiling point than either Ar or CH4.
The only non-vanishing intermolecular interaction in the case of Ar and CH4 is a
London dispersion interaction. Since the electron distribution in CH4 is spread
out over a greater volume than that in the Ar ATOM, it is easier to INDUCE a
dipole moment in CH4 than it is in Ar. Thus, the London dispersion
intermolecular force between a pair of CH4 molecules is greater than the London
dispersion force between a pair of Ar atoms. Thus, it is easier to separate a pair
of Ar atoms.
The order of INCREASING boiling point is:
Ar < CH4 < NO < NaF < Al2O3 .