Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
1
The keys are done quickly. I do make mistakes. If you think you found an error, please bring it to my attention so
I can correct it for others. Thank you.
Problem 1 (p 2) – What is the total nuclear charge and effective nuclear charge for each of the atoms below? How
does this affect the electron attracting ability of an atom?
H
Ztotal =
core electrons =
Zeffective =
Li
Na
+1 +3
+11
2e-
K
B
Al
C
+19 +5 +13 +6
Si
N
P
O
+14 +7 +15 +8
S
+1
+1 +3
+3
+4
+4 +5
Cl
+16 +9 +17
2e- 10e- 18e- 2e- 10e- 2e- 10e- 2e- 10e- 2e-
+1 +1
F
+5 +6
Br
I
He
+35 +53 +2
Ne
Ar
+10
+18
10e- 2e- 10e- 28e- 46e- 0e+6 +7
+7
+7
+7
+2
2e-
10e-
+8
+8
Atoms with higher Zeff tend to hold onto their valence elctrons tighter. Atoms with the same Zeff, but lower valence
shells also hold tighter because the valence electrons are closer to the same Zeff.
Problem 2 (p 2) - Which atom in each pair below probably requires more energy to steal away an electron (also
called ionization potential IP)? Why? Are any of the comparisons ambiguous? Why? Check your answers with
the data in the I.P. table on page 7.
a. C vs N
b. N vs O
c. O vs F
d. F vs Ne
Since all comparisons are in a row, the answers use essentially the same reasoning.
C
N
Both C and N are in the same row and use the same n=2 valence shell. N has a
Zeff = +5 and holds onto its electrons tighter than C with a Zeff = +4.
N
O
Both N and O are in the same row and use the same n=2 valence shell. O has a
Zeff = +6 and holds onto its electrons tighter than N with a Zeff = +5.
O
F
Both O and F are in the same row and use the same n=2 valence shell. F has a
Zeff = +7 and holds onto its electrons tighter than O with a Zeff = +6.
F
Ne
Both F and Ne are in the same row and use the same n=2 valence shell. Ne has a
Zeff = +8 and holds onto its electrons tighter than F with a Zeff = +7.
Problem 3 (p 4) – Write and atomic configuration for H, B, C, N, O, F, Ne, S, Cl, Br, I.
hydrogen: 1s1
boron: 1s2, 2s2, 2p1
carbon: 1s2, 2s2, 2p2
nitrogen: 1s2, 2s2, 2p3
oxygen: 1s2, 2s2, 2p4
fluorine: 1s2, 2s2, 2p5
neon: 1s2, 2s2, 2p6
sulfur: 1s2, 2s2, 2p6, 3s2, 3p4
chlorine: 1s2, 2s2, 2p6, 3s2, 3p5
bromine: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p5
Iodine: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2, 5p5
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
2
Problem 4 (p 5) - Explain the atomic trends in ionization potential in a row (Na vs Si vs Cl) and in a column (F vs
Cl vs Br).
These three elements (Na vs Si vs Cl) are all in the same row. The Zeff increases across a row (in the same shell) so
the rightmost element should hold onto its electron the tightest and have the highest ionization potential
(Na = 118 > Si = 189 > Cl = 300 all kcal/mole).
These three elements (F vs Cl vs Br) are all in the same column. The Zeff is +7 for all of them. Fluorine’s valence
electrons are in the n=2 shell (closest to the +7 Zeff), chlorine’s in the n=3 and bromine’s in n=4 (farthest from the +7
Zeff). Being closer to the Zeff should produce a stronger attraction, thus fluorine has the highest ionization potential
(F = 402 > Cl = 300 > Br = 273 all kcal/mole).
Problem 5 (p 6) - Explain the atomic trends in atomic radii in a row (Li vs C vs F) and in a column (C vs Si vs Ge).
Across a row Zeff increases, in the same shell and the valence electrons are held tighter and have smaller radii (Li =
167, C = 67, F = 42). In a column, each atom has the same Zeff, but there are additional layers, so larger radii are
found (C = 67, Si = 111, Ge = 125).
Valence electrons have been lost in cations so there is a deficiency of electron density. Across a row, the higher
positive Zeff should pull in the core electron tighter and produce a smaller radius (Li+ = 90, Be+2 = 59 and B+3 = 41).
Valence electrons have been added in anions so there is an excess of electron density, which repel one another
(found on the right side, in a row). There is greater excess on the nitrogen (with a lower Zeff = +5), than the oxygen,
(Zeff = +6), than the fluorine, (Zeff = +7). The nitrogen electron cloud expands the most and has the largest radius
(N-3 = 132, O-2 = 126 and F-- = 119).
Cation and anion radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom]
Li +1 = 90
Na+1 = 116
K +1
= 152
+1
Rb = 166
+2
Be
Mg +2
Ca +2
Sr +2
= 59
= 86
= 114
= 132
3d elements
4d elements
B+3
Al +3
Ga +3
In +3
= 41
= 68
= 76
= 94
C=
Si =
Ge =
Sn =
N -3
P
As
Sb
= 132
=
=
=
O -2 = 126
S -2 = 170
Se -2 = 184
Te -2 = 207
F -1 = 119
Cl -1 = 167
Br -1 = 182
I -1 = 206
Problem 6 (p 6) - In the tables above:
a. Explain the cation distances compared to the atomic distances.
Each time an electron is lost from an element the positive charge goes up and the element holds on tighter
to its remaining electrons. Also, electron/electron repulsion goes down as each electron is lost. This
should contract the electron cloud more and more.
b. Explain the anion distances compared to the atomic distances.
Each time an electron is gained to an element the negative charge goes up and there is more and more
electron/electron repulsion. This should expand the electron cloud more and more.
c. Explain the cation distances in a row (Li = 90A, Be = 59A, B = 41A, A = angstrom).
Similar to a. Each time an electron is lost from an element the positive charge goes up and the element
holds on tighter to its remaining electrons. Also, electron/electron repulsion goes down as each electron is
lost. This should contract the electron cloud more and more.
d. Explain the anion distances in a row (N = 132A, O = 126A, F = 119A, A = angstrom).
On the right side, in a row, there is an excess of electrons, which repel each other. There is greater excess
on the nitrogen, Zeff = +5, than the oxygen, Zeff = +6, than the fluorine, Zeff = +7. The nitrogen electron
cloud expands the most and has the largest radius.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
3
Problem 7 (p 7) Classify each bond type below as pure covalent, polar covalent or ionic according to our simplistic
guidelines above. If a bond is ionic, rewrite it showing correct charges.
b
a
c
e
d
C
C
C
C
N
O
C
f
+2
Mg Br
Br
F
C
Br
ionic
pure covalent
polar covalent
polar covalent
polar covalent
X = 2.5 - 2.5 = 0
X = 3.0 - 2.5 = 0.5
X = 3.5 - 2.5 = 1.0
X = 4.0 - 2.5 = 1.5
h
i
g
C
Na
O
Cl
polar covalent
X = 3.0 - 2.5 = 0.5
o
mp = 711 C
bp = decomposes
(bp indicates ionic)
l
k
j
Na
H
X = 3.0 - 1.2 = 1.8
H
F
H
C
C
C
C
polar covalent
ionic
pure covalent
X = 2.5 - 2.2 = 0.3
X = 3.2 - 0.9 = 2.3
mp = 801oC
bp = 1413oC
(bp indicates ionic)
polar covalent
X = 3.5 - 2.2 = 1.3
n
m
C
C
O
C
N
pure covalent
pure covalent
X = 2.5 - 2.5 = 0
X = 2.5 - 2.5 = 0
p
o
N
X = 4.0 - 2.2 = 1.8
mp = -84oC
bp = +20oC
(bp indicates molecular)
q
Na
O
r
N
H
C
H
Cl
ionic
polar covalent
X = 3.0 - 2.5 = 0.5
s
N
O
polar covalent
polar covalent
X = 3.5 - 2.5 = 1.0
t
H
H
N
X = 3.0 - 2.5 = 0.5
O
O
ionic
polar covalent
X = 3.0 - 1.0 = 2.0
X = 3.5 - 3.0 = 0.5
mp = 390oC
bp = 430oC
(bp indicates ionic)
mp = 318oC
bp = 1388oC
(bp indicates ionic)
polar covalent
polar covalent
X = 3.2 - 2.5 = 0.7
X = 3.0 - 2.2 = 0.8
v
u
Li
X = 3.5 - 0.9 = 2.6
pure covalent
X = 3.5 - 3.5 = 0.0
Cl
O
w
H
H
N
Na
ionic
pure covalent
X = 3.5 - 3.2 = 0.3
X = 3.0 - 0.9 = 2.1
mp = 210oC
bp = 400oC
(bp indicates ionic)
Problem 8 (p 12) - Predict the formula for the combination of the following pairs of ions. What kinds of melting
points would be expected for these salts?
KF
KNO2
KNO3
K2O
KHCO3
K3PO4
K2CO3
BaF2
Ba(NO2)2
Ba(NO3)2
BaO
Ba(HCO3)2
Ba3(PO4)2
BaCO3
ZnF2
Zn(NO2)2
Zn(NO3)2
ZnO
Zn(HCO3)2
Zn3(PO4)2
ZnCO3
AlF3
Al(NO2)3
Al(NO3)3
Al2O3
Al(HCO3)3
AlPO4
Al2(CO3)3
All of these salts have very high melting points.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
4
Problem 9 (p 13) – Supply dipole arrows to any polar bonds above (according to our arbitrary rules). Make sure
they point in the right direction.
Only the polar bonds have been redrawn below. If a molecule is completely nonpolar, it is not shown.
H
H
H
C
N
H
H
H
H
H
C
O
H
H
H
H
C
F
H
N
H
O
O
F
H
N
O
F
H
N
O
H3C
N
C
H2
C
C
N
H
C
H
H
N
O
N
H
H
H
C
H
H
N
H
H
C
O
H
H
F
F
F
H
H
H
H
H
N
O
C
H
C
N
H
H
C
C
N
H
H2C
H2C
O
NH
H2C
H2C
C
H2
Problem 10 (p 19) - The following pairs of molecules have the same formula; they are isomers. Yet, they have
different melting points. Match the melting points with the correct structure and provide an explanation for the
difference. Hint: single bonds can rotate and pi bonds tend to be rigid, fixed shapes.
H3C
a.
H3C
C
C
mp = -32oC
CH3
H2 C
C
C
b.
H
mp = -126oC
c.
mp = -105oC
mp = -185oC
d.
C6H14
mp = -95oC
C6H12
mp = +6oC
mp = +79oC
mp = -88oC
Molecules with rigid, symmetrical shapes tend to pack tighter (closer) in their crystal lattice structures. Closer
contact makes for stronger intermolecular forces of attraction, meaning it is harder to break down the lattice
structure, meaning a higher melting point (all other things being equal). In each part the molecules have the same
formula (or almost = c) and similar weak dispersion forces. The compound with the more regular shape and higher
melting point is the first one in a, b and d, while the second structure has a more regular shape in part c.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
5
Problem 11 (p 23) – Provide an explanation for the different boiling points in each column. (µ = dipole moment,
X = difference in electronegativities)
4B
5B
6B
CH4
mp = -182oC
bp = -164oC
 = 0.0 D
 = 0.3
SiH4
mp = -185oC
bp = -112oC
 = 0.0 D
 = 0.3
GeH4
mp = -165oC
bp = -88oC
 = 0.0 D
 = 0.2
SnH4
mp = -146oC
bp = -52oC
 = 0.0 D
 = 0.2
NH3
mp = -78oC
bp = -33oC
 = 1.42 D
 = 0.8
PH3
mp = -132oC
bp = -88oC
 = 0.0 D
 = 0.0
AsH3
H2O
mp = 0oC
bp = +100oC
 = 1.80 D
 = 1.2
7B
H2S
mp = -82oC
bp = -60oC
 = 1.0 D
 = 0.4
H2Se
mp = -111oC mp = -66oC
bp = -41oC
bp = -62oC
=?D
 = 0.0 D
 = 0.4
 = 0.0
SbH3
H2Te
mp = -88oC mp = -49oC
bp = -17oC
bp = -2oC
 = 0.0 D
=?D
 = 0.2
 = 0.1
8B
100
He
mp = -272oC
50
bp = -269oC
Temp
 = 0.0 D
(oC)
 = NA
0
Ne
HCl
mp = -114oC mp = -249oC
-50
bp = -85oC bp = -246oC
 = 1.0 D
 = 0.0 D
-100
 = 1.0
 = NA
Ar
HBr
-150
mp = -87oC mp = -189oC
o
o
bp = -67 C bp = -185 C
 = 0.0 D
 = 0.8 D
-200
 = NA
 = 0.8
HI
mp = -51oC
bp = -35oC
 = 0.4 D
 = 0.5
Kr
mp = -157oC
bp = -153oC
 = 0.0 D
 = NA
boiling points (oC)
H2O
HF
mp = -84oC
bp = +20oC
 = 1.86 D
 = 1.8
HF
H2Te
SbH3
HI
H2Se
NH3
H2S
HCl
PH3
AsH3
SnH4
HBr
GeH4
SiH4
Kr
CH4
Ar
Ne
-250
He
-300
row
2
row
3
row
5
row
4
Answer: Each dotted line represents the boiling points in the hydrides in a column in the periodic table, plus the Noble gases.
In the Noble gases there is a continuning trend towards higher boiling point as the main atom gets larger with more and
more polarizable electron clouds (stronger dispersion forces). Where bonds to hydrogen become polar, hydrogen bonding
becomes important, leading to stronger attractions for neighbor molecules and higher boiling points deviate from the trends
observed due to dispersion forces. Deviations are clearly seen with NH3, H2O and HF. Melting points are often less predictable
and are not plotted.
a
H3C
H2
C
b
CH3
 = 0.08 D
bp = -42oC
mp = -188oC
H2O sol. = 40mg/L
O
H3C
CH3
 = 1.30 D
bp = -22oC
mp = -141oC
H2O sol. = 71g/L
H
g
c HC
3
=?D
bp = +138oC
mp = -78oC
H2O sol. = 22g/L
O
H
d
H
O
 = 1.69 D
bp = +65oC
mp = -98oC
H2O sol. = miscible
O
H f
H
O
 = 1.69 D
bp = +78oC
mp = -114oC
H2O sol. = miscible
h
H
=?D
bp = +157oC
mp = -50oC
H2O sol. = 6g/L
e
 = 1.68 D
bp = +98oC
mp = -126oC
H2O sol. = ?
O
 = 1.66 D
bp = +118oC
mp = -90oC
H2O sol. = 73g/L
H
i
O
O
=?D
bp = +195oC
mp = -16oC
H2O sol.  0
Structure a, b and c are similar in size. Structure c only has dispersion forces, while structure b has dispersion forces and polar bonds,
while structure c has dispersion forces, polar bonds and hydrogen bonds. Intermolecular forces of interaction increase in order of
dispersion forces < polar bonds < hydrogen bonds, so c has the highest bp, followed by b and lowest is a. Compound d is like c except
it has an extra methyl, thus greater dispersion forces, which is similar for e, f, g, h and i. All have increasing boiling points because of
their extra dispersion forces (on top of their hydrogen bonds).
Problem 12 (p 24) – Provide an explanation for the different boiling points in each series. In part b, an H-F bond is
expected to be more polar than an H-O bond, so why does HF boil lower than H2O? Think of an "H bond" donor
site as a hand that can grab and a lone pair as a handle that can be held onto. If you only had one hand, it really
doesn't make any difference how many handles are available to grab hold of and if you only have one handle, it
doesn’t make any difference how many hands you have. But, if you have two hands and two handles a much
stronger grip is possible with twice the handholds.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
6
a. and b
H
H
H
C
=0D
bp = -161 oC
mp = -182oC
no polarity, only weak dispersion forces ot attraction f or neighboring molecules
H
NH2
H
H
hydrogen bonding is possible, but only 2 per molecule on average, also, N-H is less
polar than O-H
N
H
H
= 1.47 D
bp = -33 oC
mp = -78oC
NH3
HO
OH
H
= 1.78 D
bp = +100 oC
mp = 0oC
H
hydrogen bonding is possible, 4 per molecule on average, OH is
our most polar 'typical' bond
O
H
H
OH2
H2O
F
hydrogen bonding is possible, H-F is our most polar 'typical' H bond, but
only 2 per molecule on average, so lower bp than water
H
F
H
= 1.85 D
bp = +20 oC
mp = -84oC
F
H
c.
H
H3 C
H 3C
CH3
H
N
H3C
F
 = 1.85 D
bp = -78 oC
mp = -142oC
 = 1.78 D
bp = +65 oC
mp = -97oC
 = 1.47 D H
bp = -7 oC
mp = -93oC
=0D
bp = -89 oC
mp = -183oC
H3 C
O
H bonding is the most important factor affecting bp's in these molecules. CH3OH makes stronger H bonds than CH3NH2 and the other
two do not have H bonds. While CH3F has the most polar bond, fluorine is so electronegative that it does not share its lone pairs much
and is not very polarizable. Ethane has no polarity and only has weak dispersion forces to form attractions with neighbor molecules, so
has the lowest bp.
d.
H2
C
H2
C
H3C
C
H2
 = 1.2 D
bp = +48 oC
mp = -83oC
NH2
H3C
CH3
CH3
N
 = 1.0 D H
bp = +36 oC
mp = ??oC
H3C
N
 = 0.6 D
bp = +3 oC
mp = -117oC
donate H bond
2
1
0
accept H bond
1
1
1
CH3
zero
H bonds
More H bonds possible on the
first structure, so stronger attractions
for neighbor molecules and a higher
bp. The third structure does not have
any polarized protons (N-H) to
hydrogen bond with.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
H2
C
e.
H3C
H2
C
CH3
H3 C
C
H2
f.
O
H
O
 = 2.33 D
bp = -19 oC
mp = -92 oC
C
H2
 = 1.69 D
bp = +195 oC
mp = -13oC
H
H bonding forms stronger
connections to neighbor
molecules and requires more
energy to remove a molecule
from the liquid (higher bp).
The third structure can H bond
at two ends, the second structure
at one end and the first
structure cannot H bond at all.
H
 = 58oC
O
H
CH2
C
H3 C
 = 1.78 D
bp = +65 oC
mp = -97oC
O
2
O
CH3
H
O
1
C
H
H
 = 1.22 D
bp = +97 oC
mp = -126oC
0
 = 84oC
H2
C
H
C
H2
 = 0.08 D
bp = 0 oC
mp = -136oC
H bonding
possibilities =
O
7
H
 = 26oC
O
C
H3C
 = 1.69 D
bp = +78 oC
mp = -114 oC
 = 2.7 D
bp = +20 oC
mp = -123 oC
O
H3C
CH3
 = 2.91 D
bp = +56 oC
mp = -95 oC
CH
H3 C
CH3
 = 1.66 D
bp = +82 oC
mp = -89 oC
In each part we are comparing two similar structures, one with a polar C=O bond with a polar C-O bond and H bonding possibilities.
The alcohols, with H bonding, have stronger attractions for the neighbor molecules, thus higher bp's. The differences are larger on the
smaller moleules, where the H bonding represents a larger fraction of the total interactions.
Problem 13 (p 25) – Provide an explanation for the different boiling points in each series.
a.
F
bp = +85 oC
Cl
I
Br
bp = +155 oC
bp = +130 oC
bp = +188 oC
Similar molecules. The highest boiling points go with the largest dispersion forces (greatest polarizability, I > Br > Cl > F)
b.
OH
bp = -0.5 oC
O
bp = +5 oC
bp = +78 oC
Structure 3 has H bonding and polarity (strongest attractions), structure 2 has polar bonds and structure 1 only has dispersion forces
(weakest attractions)
c.
O
bp = -47 oC
HO
H2N
O
bp = +210 oC
bp = +101 oC
bp = +20 oC
O
O
C
O
bp = -78 oC
(sublimes)
Structure 1 only has weak dispersion forces, structure 2 had polar bonds (stronger), structure 3 has H bonds too (stronger), structure 4 has
the strongest polarity and H bonds with both N-H bonds (highest). Finally CO2 has polar bonds, but its symmetry cancels out its polarity.
The oxygen atoms hold so tightly to their electrons that they are less polarizable than structure 1 and sublimes at a lower temperature.
d.
N
o
bp = -42 oC
bp = -23 C
bp = +82 oC
Structure 1 (propane) only has weak dispersion forces, and has a bent shape that prevents closer contact with neighbor molecules.
Structure 2 (propyne) only has weak dispersion forces, but has a linear shape that allows closer contact with neighbor molecules, so
stronger attractions. Structure 3 has the shape of propyne, and relatively strong polarity which makes stronger attractions and a higher bp.
Chem 201/Beauchamp
H2
C
e.
H3 C
H3 C
CH3
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
bp = -89 oC
H2
C
H2
C
CH3
H2
C
H
C
H3 C
CH3
CH3
H3 C
CH3
bp = +36 oC
8
C
o
bp = +30 C
CH3
CH3
bp = +10 oC
More carbons have greater contact surface area so stronger dispersion forces (C2 < C5). Straight chains can approach closer than when
branches are present (branches tend to push neighbor molecules away = weaker dispersion forces). More branches weaken those attractions
even more (D < C < A).
e.
CH4
bp = -164 oC
CH3Cl
CH2Cl2
bp = -24 oC
bp = +40 oC
CCl4
CHCl3
bp = +61 oC
bp = +77 oC
Chlorine is a larger atom (than C) and has polarizable lone pairs. More chlorines = more dispersion forces = higher bp. B, C and D have
polarity too. A and E are not polar, but E has four chlorine atoms and larger dispersion forces dominate here (highest bp).
H2
C
H2
C
f.
H3 C
H3C
CH3
H2
C
H
H
H3C
N
O
bp = +78 oC
bp = +17 oC H
Bp's follow H bonding possibilities. Structure 3 has the strongest H bonds because O-H is more polar. Structure 2 has some H bonding
possibilitiies so a higher bp than structure 1 which only has dispersion forces. All 3 molecues are similar in size, so comparisons are fair.
bp = -42 oC
g.
H2
C
H3 C
C
H2
hexane, C6H14
mp = -91 oC
bp = +69 oC
H2
C
H2
C
C
H2
H2C
CH3
CH2
H2C
cyclohexane, C6H12
H
H
CH2
C
H2
benzene, C6H6
o
mp = +6 C
bp = +81 oC
mp = +5 oC
bp = +80 oC
H
C
C
C
C
H
C
C
H
H
All 3 structures only have weak dispersion forces and are approximately the same size. The rings are more regular in shape and probably
pack closer in the solid, thus the much higher mp's (similar to one another) and are also closer in the liquid with no branches wildly swinging
around to bump neighbor molecules farther away, thus the higher bp's too (again, rings similar to one another).
Problem 14 (p 25) – Match the given boiling points with the structures below and give a short reason for your
answers. (-7oC, +31 oC, +80 oC, +141oC, 1420oC)
O
O
KCl
2-butanone
MW = 72 g/mol
+80 oC
2-methyl-1-butene
OH
propanoic acid
potassium chloride
2-methylpropene
MW = 70 g/mol
MW = 74 g/mol
MW = 74.5 g/mol
MW = 56 g/mol
o
+31 C
+141oC
1420oC
-7oC
Structure 4 has ionic bonds, by far the strongest forces to overcome and has an extremely high bp. Structure 3 has strong H bonding
possiblities and the second high bp. Structure 1 has polar C=O bond so higher than structures 2 and 5, which only have weak dispersion
forces. Structure 2 is larger and has a larger surface area and more dispersion forces than structure 5, so has a higher bp.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
9
Problem 15 (p 28) – Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DMSO) that
makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use
DMSO (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the
difference from the methanol picture.
= cations are smaller because they have lost electrons and hold onto the remaining electrons tighter.
= anions are larger because they have extra electron density that is repulsive and expands the electron clouds
polar H
H3C
H
H3C
CH3
O
H
H
O
H
H3C
O
H3C
H
O
O
CH3
S
S
H3C
H3C
CH3
S
S
CH3
CH3
O
H
H3C
O
O
O
H
strong bond dipole
O
H3C
O
CH3
S
S
O
O
H3C
H3C
CH3
CH3
CH3
Methanol is good at solvating positive and negative charge.
DMSO is very good at solvating positive charge
but it is poor at solvaing n egative charge.
Problem 16 (p 29) –
a. Hexane (density = 0.65 g/ml) and water (density = 1.0 g/ml) do not mix. Which layer is on top?
b. Carbon tetrachloride (density = 1.59 g/ml) and water (density = 1.0 g/ml) do not mix. Which layer is on top?
b
H2
C
H2
C
hexane
H3C
C
H2
O
water
H
 = 1.8 D
H
O
CH3
C
H2
=0D
Differences in polarity keep
these two from mixing and
water is more dense than
hexane and sinks to the bottom.
H
water
H
Cl
Cl
C
carbon tetrachloride
Cl
Cl
=0D
 = 1.8 D
Differences in polarity keep
these two from mixing and
carbon tetrachloride is
more dense than water
and sinks to the bottom.
Problem 17 (p 29) - The melting point of NaCl is very high ( 800oC) and the boiling point is even higher
( 1400oC). Does this imply strong, moderate or weak forces of attraction between the ions? Considering your
answer, is it surprising that NaCl dissolves so easily in water? Why does this occur? Consider another chloride salt,
AgCl. How does your analysis work here? What changed?
Lattic structure - depends on
the size and charge of the ions.
Each ion is surrounded on many sides by oppositely charged ions. To introduce
the disorder of a liquid (melt) or a gas (boil) requires a very large input of energy
(mp indicates the amount of energy required to break down the ordered lattice
structure and boiling point indicates the amount of energy required to remove an ion
pair from the influence of all neighbors. Ionic bonds (ionic attractions on all sides)
can only be bronke at great expense in energy.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
10
However, many water molucules working together can break down this structure by solvating both the cations and
anions (solvation energy). When solvation energy > lattice energy, the salt will dissolve (NaCl). When
solvation energy < lattice energy the salt will not dissolve (AgCl).
AgCl
NaCl
OH2
OH2
H2O
HO
H
H
OH
lattice
energy
solvation
energy
lattice
energy
solvation
energy
H2O
Small difference between
two very large energy values.
DG =
dissolves
DG =
does not dissolve
Problem 18 (p 29) – a. Which solvent do you suspect would dissolve NaCl better, DMSO or hexane? Explain your
choice? b. Which solvent do you suspect would dissolve NaCl better, methanol or benzene? Explain your
choice?
a
H3C
S
H3C
O
CH3
H3 C
S
H3C
O
O
O
CH3
O
H3C
S
very strong interactions
O
S
CH3
S
H3C
CH3
S
CH3
much weaker interactions,
but still possible.
CH3
Hexane cannot help dissolve ions at all because it has no
polarity. The salt sinks to the bottom (if more dense).
OK overall
b
polar H
H3C
CH3
H
O
O
H3C
CH3
O
H
O
H
H
O
H3 C
H
O
H
H3 C
H
O
Methanol is good at solvating positive and negative charge.
CH3
Benzene is not good at dissolving ions at all because it
has no polarity, although its pi electrons can weakly
interact with cations. The salt sinks to the bottom
(if more dense).
Problem 19 (p 30) – The terms "hydrophilic" and "hydrophobic" are frequently used to describe structures that mix
well or poorly with water, respectively. Biological molecules are often classified in a similar vein as water soluble
(hydrophilic) or fat soluble (hydrophobic). The following list of well known biomolecules are often classified as fat
soluble or water soluble. Examine each structure and place it in one of these two categories. Explain you reasoning.
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
11
O
b
a
O
c
N
H
N
OH
O
P
OH
O
O
N
vitamin A
N
vitamin B2
(riboflavin)
Hydrophobic - mostly nonpolar
OH
O
vitamin B6
(pyridoxine)
OH
HO
N
H
OH
Hydrophilic - lots of
"OH" and polarity.
Hydrophilic - lots of
"OH" and polarity
and charge.
OH
Hydrophilic - lots of
"OH" and polarity
and charge. N
O
d
NH2
O
O
e
N
HO
O
P
O
O
P
O
O
N
N
O
HO
O
O
ATP
O
P
HO
f
Hydrophilic - lots of
"OH" and polarity.
vitamin C
(ascorbic acid)
OH
vitamin D2
OH
OH
HO
O
Hydrophobic - mostly nonpolar
Problem 20 (p 31) – a. Carbohydrates are very water soluble and fats do not mix well with water. Below, glucose is
shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point
out why each is classified in the manner indicated.
O
HO
HO
O
HO
OH =
HO
HO
HO
OH
glucose (carbohydrate)
O
O
O
O
OH
OH
O
O
typical saturated triglyceride (fat)
Chem 201/Beauchamp
Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key
12
b. All of the “OH” groups in glucose can be methylated. What do you think this will do to the solubility of
glucose? Why? One of these structures is soluble in carbon tetrachloride the other one is not. Which one is it and
why?
OCH3
OH
methylation
reaction
O
HO
HO
O
H3CO
H3CO
OH
OCH3
OCH3
methylated glucose
OH
glucose
The methylation reaction converts the alcohol groups (OH) to ether groups (OCH3) and the ability
to hydrogen bond goes down, as does water solubility. However, the ability to dissolve in nonpolar
carbon tetrachloride goes up, so B will be much more soluble in CCl4.
Problem 21 (p 31) – Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your
body to solubilize the fats, so that they can be absorbed and transported in the aqueous blood. The major bile salt
glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good
compromise structure that can mix with both the fat and aqueous blood. Use the ‘rough’ 3D drawings below to
help your reasoning, or better yet, build models to see the structures for yourself (though it’s a lot of work).
OH
H
H
O
H
synthesized in many,
many steps in the body
H
H
N
H
O
cholesterol
HO
HO
OH
H
blood
H2O
H2O
OH
blood
HO
HO
OH
O2C
CO2
nonpolar
fats
inside
blood
H2O
H2O
OH
blood
H2O
O2C
CO2
HO
OH
HO
blood
OH
OH
blood
H2O
blood
H2O
HO
H2O
blood
glycolate
(bile salt)
O
Glycolate has a nonpolar, hydrophobic
face that can cover the inside of a fat ball
and a hydorphilic face that can point
outward toward the aqueous blood, which
allows fats to be transported throughout
the body to reach fat storage cells and other
essential locations.
H2O
HO
blood
H