Molecular Structure &
Intermolecular Forces
Saturday Study Session #2
3rd Class
Lewis Structures
Lewis structures are representations of molecules
showing all electrons, bonding and nonbonding.
Draw the Lewis structure for CH2Cl2
OR
Draw the Lewis Structure for NO+
Draw the Lewis Structure for XeF2
• *Xe can have more than an octet of electrons!
From the Lewis Structure we can
determine:
Electron geometry
Molecular geometry
Hybrid Orbital
Polarity
Intermolecular bond
Molecular Shapes
• The shape of a molecule plays an important role
in its reactivity.
• By noting the number of bonding and nonbonding
electron pairs, we can easily predict the shape of
the molecule.
Electron Domains
• The central atom in
this molecule, A, has
four electron
domains.
• We can refer to the
electron pairs as electron
domains.
• In a double or triple bond,
all electrons shared
between those two atoms
are on the same side of the
central atom; therefore,
they count as one electron
domain.
Valence-Shell Electron-Pair Repulsion
Theory (VSEPR)
“The best arrangement of a given number of
electron domains is the one that minimizes the
repulsions among them.”
If all electron domains are bonds the molecular
shapes are like these below:
If some of the electron domains are unshared pairs of
electrons then the molecular shapes are as indicated
in this chart and the one on the following slide.
Let’s practice molecular
geometry
Molecular Geometry answers
1. CH2Cl2
2. NO+
3. XeF2
1. Tetrahedral
2. Linear
3. Linear, but its
electron geometry
is octahedral due
to 3 unshared
pairs of e-
Bond Angles
120o
45o
Bond
Angles for
molecules
without
lone pairs
of
electrons.
Nonbonding Pairs and Bond Angle
• Nonbonding pairs are physically
larger than bonding pairs.
• Therefore, their repulsions are
greater; this tends to decrease
bond angles in a molecule.
Multiple Bonds and Bond Angles
• Double and triple
bonds place greater
electron density on
one side of the central
atom than do single
bonds.
• Therefore, they also
affect bond angles.
Bond and Molecular POLARITY
Polar Covalent Bonds
Intramolecular
• Though atoms often form
compounds by sharing
electrons, the electrons
are not always shared
equally.
• Fluorine pulls harder on the electrons it shares
with hydrogen than hydrogen does.
• Therefore, the fluorine end of the molecule
has more electron density than the hydrogen
end.
Polar Covalent Bonds
• \
The greater the
difference in
electronegativity,
the more polar is
the bond.
Polarity
• Just because a molecule
possesses polar bonds
does not mean the
molecule as a whole will
be polar.
Polarity of Molecules
By adding the
individual bond
dipoles, one can
determine the overall
dipole moment for the
molecule.
Polarity
Molecular Polarity
Polar Molecules
•
•
•
•
Non-Polar Molecules
Must have some polar • Polar or nonpolar
bonds
bonds
(∆EN > 1.7)
• Dipoles cancel
Overall net dipole
•
Insoluble
in
Water
Soluble in Water
• Look for
Look for
– Symmetrical
– Asymmetry
molecule
– -OH, -NH2 groups
Practice Molecular Polarity
• Which molecule is more polar?
1. CS2 or SF2
NONPOLAR
POLAR
2. BH3 or NH3
NONPOLAR
POLAR
Which is more polar?
3. Benzene
POLAR
NONPOLAR
OR
Glucose
Molecules Stick Together
• All molecules have some attractive forces for
each other.
• Polar molecules have more types of attractive
forces than do nonpolar molecules.
• These attractive forces
are called INTERMOLECULAR
FORCES (IMF)
Inter vs. Intra molecular forces
Inter (between molecules)
• London dispersion forces
• Dipole-dipole forces
• Hydrogen bonds
Intra (inside molecules)
• Ionic bonds
• Covalent bonds
• Metallic bonds
Dipole-Dipole forces
Hydrogen bonding
Which IMFs are present?
1. CF4
1. London dispersion forces
2. BF3
2. London dispersion forces,
Dipole – dipole forces
3. NH3
4. H2CS
3. London dispersion forces,
Hydrogen bonding
4. London dispersion forces,
Dipole – dipole forces
(in water, weak H bonding)
Effects of IMFs
• States of matter
• Phase changes
–Melting points
–Boiling points
–Vapor pressure
States of Matter
(Increasing)
Molecular Interactions ARE Intermolecular Forces
Particles getting farther apart means they are overcoming intermolecular forces by
adding energy.
Energy In
Energy Out
Must
overcome
IMFs
IMFs cause
particles to
congregate
Effects of IMFs on properties
• Greater IMFs = higher melting and boiling
points and lower vapor pressure.
• Greater Molar Mass = more electrons =
greater IMFs
• Volatile substances have high VP due to low
IMFs
• Greater IMFs = high ΔHvap
MC Question 1
In which of the following processes are
covalent bonds broken?
A) I2(s) → I2(g)
B) CO2(s) → CO2(g)
C) NaCl(s) → NaCl(l)
D) C(diamond) → C(g)
E) Fe(s) → Fe(l)
Question 1 Answer
• Correct answer is D
• Diamond is a covalently bonded network
crystal. In order to form a gas the covalent
bonds must be broken.
• A and B the molecules remain intact, only IMF
are “broken”
• C is held together with ionic bonds
• E is held together with metallic bonds
MC Question 2
The structural isomers C2H5OH and CH3OCH3
would be expected to have the same values for
which of the following? (Assume ideal behavior.)
A) Gaseous densities at the same temperature and
pressure
B) Vapor pressures at the same temperature
C) Boiling points
D) Melting points
E) Heats of vaporization
Question 2 Answer
• Correct answer is A
• Density is a function of mass and volume.
Isomers have the same molecular mass, same
volume can be assumed.
• All other answers the values change according
to differences in IMFs.
• C2H5OH has hydrogen bonding but CH3OCH3
does not.
MC Question 3
X: CH3–CH2–CH2–CH2–CH3
Y: CH3–CH2–CH2–CH2–OH
Z: HO–CH2–CH2–CH2–OH
Based on concepts of polarity and hydrogen bonding,
which of the following sequences correctly lists the
compounds above in the order of their increasing
solubility in water?
A) Z < Y < X
B) Y < Z < X
C) Y < X < Z
D) X < Z < Y
E) X < Y < Z
Question 3 Answer
• Correct answer is E
• The pure hydrocarbon butane X, is the least
polar, thus has the lowest solubility in water.
• The presence of an –OH group on butanol Y,
makes it more soluble than butane, but less
soluble than the 1,3-propanediol
• Z, that contains two –OH groups. Aren’t you
glad you don’t have to name all of them?
MC Question 4
Hydrogen Halide Normal Boiling Points, °C
HF
HBr
+19
– 67
HCl
HI
– 85
– 35
The relatively high boiling point of HF can be correctly
explained by which of the following?
A) HF gas is more ideal.
B) HF is the strongest acid.
C) HF molecules have a smaller dipole moment.
D) HF is much less soluble in water.
E) HF molecules tend to form hydrogen bonds.
Question 4 Answer
• Correct answer is E
• Hydrogen bonding occurs when a H is bound
to a “highly electronegative atom” (F, N or O)
• The bonded H is attracted to an unshared
electron pair or another highly electronegative
atom on a neighboring molecule.
MC Question 5
Which of the following gases deviates most
from ideal behavior? (Ideal gases assume no
interparticle attractions)
A) SO2
B) Ne
C) CH4
D) N2
E) H2
Question 5 Answer
• Correct answer is A
• Deviations occur due to molecular volume (larger
molecules have more mass as well) and attractive
forces.
• The more electrons present, the more polarizable
a molecule, thus the greater the London
dispersion (induced dipole-induced dipole)
attractive forces become.
• SO2 has a higher molecular mass, more electrons
and is more polarizable than the other answer
choices.
MC Question 6
Molecular iodine would be most soluble
in:
A) water
B) carbon tetrachloride
C) vinegar (acetic acid and water)
D) vodka (ethanol and water)
E) equally soluble in all four
Question 6 Answer
• Correct answer is B
• Molecular iodine is a nonpolar molecule.
• Carbon tetrachloride is the only nonpolar
solvent listed.
FR Question 1
Explain each of the following in terms of the
electronic structure and/or bonding of the
compounds involved.
(a)At ordinary conditions,
HF (normal boiling point = 20°C) is a liquid,
whereas HCl (normal b.p. = -114°C) is a gas.
FRQ 1 Answer
• HF exhibits hydrogen bonding but HCl does
not. Both molecules have dispersion forces
(HCl slightly greater than HF) but the hydrogen
bonds are stronger in HF (F very highly
electronegative) and require more energy to
overcome to allow HF molecules to leave the
liquid state and enter the gaseous state.
FR Question 2
(a) Identify the type(s) of intermolecular attractive
forces in:
(i) pure glucose
(ii) pure cyclohexane
(b) Glucose is soluble in water but cyclohexane is
not soluble in water. Explain.
FRQ 2 answer
• ALL molecules have London dispersion forces.
• Glucose has hydrogen bonding because its
hydrogens are bonded to oxygen but
cyclohexane does not. Cyclohexane’s
hydrogens are bonded only to carbon.
• Glucose can also form hydrogen bonds with
water increasing its solubility, while hexane
can not hydrogen bond with water.
FR Question 3
Consider the two processes represented below.
Process 1: H2O(l) → H2O(g)
∆H° = +44.0 kJ
mol
Process 2: H2O(l) → H2(g) + 1/2 O2(g) ∆H° = +286 kJ
mol
(i) For each of the two processes, identify the
type(s) of intermolecular or intramolecular
attractive forces that must be overcome for the
process to occur.
FRQ 3 Answer
• Process 1 requires overcoming London
dispersion forces and hydrogen bonding.
• Process 2 requires overcoming much stronger
covalent bonds.
(ii) Indicate whether you agree or disagree with the
statement in the box below. Support your answer with a
short explanation.
When water boils, H2O molecules break
apart to form hydrogen molecules and
oxygen molecules
Water Boiling: H2O(l) → H2O(g)
molecules remain intact
Water molecules breaking apart: H2O(l) → H2(g) + 1/2 O2(g)
This process requires electrolysis
FR Question 4
Explain each of the following in terms of atomic
and molecular structures and/or intermolecular
forces.
(a) Solid K conducts an electric current, whereas
solid KNO3 does not.
(b) The normal boiling point of CCl4 is 77°C,
whereas that of CBr4 is 190°C.
(c) Iodine has a greater boiling point than bromine
even though the bond energy in bromine is
greater than the bond energy in iodine
FRQ 4 Answer
(a) Solid K has metallic bonds with a mobile sea
of electrons allowing current to flow. The
electrons in KNO3 are localized in ionic and
covalent bonds and are not allowed to move
throughout the material.
(b) CCl4 has a lower boiling point than CBr4
because CBr4 has more electrons and greater
London dispersion forces.
(c) Bond energies measure the
strength of the covalent bonds in the
diatomic molecules I2 and Br2. The
boiling points depend upon
intermolecular forces. Both molecules
have only London dispersion forces.
Iodine has more electrons per
molecule than bromine causing the
higher boiling point for iodine.
FR Question 5
Use appropriate chemical principles to account for
each of the following observations. In each part,
your response must include specific information
about both substances.
(a) At 25°C and 1 atm, F2 is a gas, whereas I2 is a
solid.
(b) The melting point of NaF is 993°C, whereas the
melting point of CsCl is 645°C.
(c) Ammonia, NH3 , is very soluble in water,
whereas phosphine, PH3 , is only moderately
soluble in water.
FRQ 5 Answer
(a) F2 molecules are smaller and have fewer
electrons than I2 molecules. The only
intermolecular forces present in both molecules
are London dispersion forces. Fluorine has fewer
IMF allowing the molecules to overcome the
attractive forces at 25oC and move apart into the
gas phase, while iodine molecules with more
electrons have greater London forces keeping
the molecules very close together as a solid.
(b) Coulomb’s Law states that the magnitude of the attractive
forces between two charged particles is equal to the product of
the charges of the particles divided by the square of the distance
between the particles.
NaF and CsCl are both ionic compounds. In order to melt their
ionic bonds must be broken. The bonds in NaF are stronger than
the bonds in CsCl because Na and F are smaller atoms than Cs and
Cl respectively. This means that their nuclei are closer the other
atoms’ electron cloud, increasing the strength of the attractive
forces between atoms. Also the charges of Na and Cs are the
same as are the charges of F and Cl. Using Coulomb’s Law in both
cases would result in the same number in the numerator of his
equation but with a smaller value in the denominator for NaF
than for CsCl resulting in stronger forces for NaF and thus a
higher melting point.
(c) Ammonia can form hydrogen bonds with
water but phosphine can not. The ability to form
hydrogen bonds along with the polarity that
both molecules exhibit increases the solubility of
ammonia in water as compared to that of
phosphine.