Molecular Structure

1
Molecular Structure Intermolecular Forces

• Saturday Study Session 2
• 3rd Class

2
Lewis Structures

• Lewis structures are representations of
  molecules showing all electrons, bonding and
  nonbonding.

3
Draw the Lewis structure for CH2Cl2
OR
4
Draw the Lewis Structure for NO
5
Draw the Lewis Structure for XeF2

• Xe can have more than an octet of electrons!

6

• From the Lewis Structure we can determine
• Electron geometry
• Molecular geometry
• Hybrid Orbital
• Polarity
• Intermolecular bond

7
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.

8
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.

• The central atom in this molecule, A, has four
  electron domains.

9
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.

10
If all electron domains are bonds the molecular
shapes are like these below
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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.
12
Lets practice molecular geometry
13
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-

14
Bond Angles
120o
45o
15
Bond Angles for molecules without lone pairs of
electrons.
16
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.

17
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.

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Bond and Molecular POLARITY
19
Polar Covalent BondsIntramolecular

• 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.

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Polar Covalent Bonds

• \

• The greater the difference in electronegativity,
  the more polar is the bond.

21
Polarity

• Just because a molecule possesses polar bonds
  does not mean the molecule as a whole will be
  polar.

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Polarity of Molecules

• By adding the individual bond dipoles, one can
  determine the overall dipole moment for the
  molecule.

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Polarity
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Molecular Polarity

• Polar Molecules

• Non-Polar Molecules

• Must have some polar bonds
• (?EN gt 1.7)
• Overall net dipole
• Soluble in Water
• Look for
• Asymmetry
• -OH, -NH2 groups

• Polar or nonpolar bonds
• Dipoles cancel
• Insoluble in Water
• Look for
• Symmetrical molecule

25
Practice Molecular Polarity

• Which molecule is more polar?
• 1. CS2 or SF2
• 2. BH3 or NH3

NONPOLAR
POLAR
POLAR
NONPOLAR
26
Which is more polar?

• 3. Benzene

POLAR
OR
NONPOLAR
Glucose
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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)

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Inter vs. Intra molecular forces

• Inter (between molecules)

• Intra (inside molecules)

• London dispersion forces
• Dipole-dipole forces
• Hydrogen bonds

• Ionic bonds
• Covalent bonds
• Metallic bonds

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Dipole-Dipole forces
30
Hydrogen bonding
31
Which IMFs are present?

• 1. CF4
• 2. BF3
• 3. NH3
• 4. H2CS

• London dispersion forces
• 2. London dispersion forces,
• Dipole dipole forces
• 3. London dispersion forces, Hydrogen bonding
• 4. London dispersion forces,
• Dipole dipole forces
• (in water, weak H bonding)

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Effects of IMFs

• States of matter
• Phase changes
• Melting points
• Boiling points
• Vapor pressure

33
States of Matter
(Increasing)
Molecular Interactions ARE Intermolecular Forces
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Particles getting farther apart means they are
overcoming intermolecular forces by adding
energy.
35
Energy In
Energy Out
Must overcome IMFs
IMFs cause particles to congregate
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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 ?Hvap

37
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)
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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

39
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
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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.

41
MC Question 3 X CH3CH2CH2CH2CH3 Y
CH3CH2CH2CH2OH Z HOCH2CH2CH2OH 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 lt Y lt X B) Y
lt Z lt X C) Y lt X lt Z D) X lt Z lt Y E) X lt Y lt Z
42
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. Arent you glad
  you dont have to name all of them?

43
MC Question 4 Hydrogen Halide Normal Boiling
Points, C HF 19 HCl 85 HBr 67
HI 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.
44
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.

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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
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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.

47

•  MC Question 6
• Molecular iodine would be most soluble in  
• A) waterB) carbon tetrachloride C) vinegar
  (acetic acid and water)D) vodka (ethanol and
  water)E) equally soluble in all four

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Question 6 Answer

• Correct answer is B
• Molecular iodine is a nonpolar molecule.
• Carbon tetrachloride is the only nonpolar solvent
  listed.

49

• FR Question 1
• Explain each of the following in terms of the
  electronic structure and/or bonding of the
  compounds involved.
• At ordinary conditions,
• HF (normal boiling point 20C) is a liquid,
• whereas HCl (normal b.p. -114C) is a gas.

50
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.

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FR Question 2 (a) Identify the type(s) of
intermolecular attractive forces in (i) pure
glucose (ii) pure cyclohexane
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(b) Glucose is soluble in water but cyclohexane
is not soluble in water. Explain.
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FRQ 2 answer

• ALL molecules have London dispersion forces.
• Glucose has hydrogen bonding because its
  hydrogens are bonded to oxygen but cyclohexane
  does not. Cyclohexanes 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.

54
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.
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FRQ 3 Answer

• Process 1 requires overcoming London dispersion
  forces and hydrogen bonding.
• Process 2 requires overcoming much stronger
  covalent bonds.

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(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
57

• FR Question 4
• Explain each of the following in terms of atomic
  and molecular structures and/or intermolecular
  forces.
• Solid K conducts an electric current, whereas
  solid KNO3 does not.
• The normal boiling point of CCl4 is 77C, whereas
  that of CBr4 is 190C.
• Iodine has a greater boiling point than bromine
  even though the bond energy in bromine is greater
  than the bond energy in iodine

58
FRQ 4 Answer

1. 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.
2. CCl4 has a lower boiling point than CBr4 because
   CBr4 has more electrons and greater London
   dispersion forces.

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(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.
60

• 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.
• At 25C and 1 atm, F2 is a gas, whereas I2 is a
  solid.
• The melting point of NaF is 993C, whereas the
  melting point of CsCl is 645C.
• Ammonia, NH3 , is very soluble in water, whereas
  phosphine, PH3 , is only moderately soluble in
  water.

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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.

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(b) Coulombs 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 Coulombs 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.
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(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.