2006 (A)
ANNOTATION:
6a) Glucose has all possibilities. Cyclohexane has nothing but London (default)
b) Glucose has hydrogen bonds that can bond to water molecules well, thus dissolving well. Cyclohexane
is the opposite. Also, glucose is polar. Cyclohexane is nonpolar, so it does not bond with h2o molecules as
well to dissolve.
c) i) 1—hydrogen and dipole dipole (H20 is involved!) (Intermolecular)
2—Covalent bonds within the molecule are broken (Intramolecular)
ii) DUH NO. BOILING IS PHYSICAL CHANGE NOT CHEMICAL! AKA. InTERmolecular forces are
broken. Water molecules itself are not separated into something different.
GUIDELINES:
2006 (B) NONE
2007 (A) NONE
2007 (B) NONE
2008 (A)
ANNOTATION:
6a) Pyridine is polar and can form hydrogen bonds. Benzene is nonpolar, cannot form hydrogen bond
with water. Blablabla~
b) Dimethyl ether has London dispersion and dipole dipole (weak). Ethanol has those IMFs of dimethyl
ether AND hydrogen bond that can be formed with water molecules, which are strong IMF which needs
high bp to break the bonds during boiling.
C) SO2: dipole dipole & London forces (weak!). Needs low mp to break bonds. SiO2: Network covalent,
strong bonds (much stronger than intermolecular forces), needs high mp to break bonds.
GUIDELINES
2008 (B) NONE
2009 (A) Question 6
ANNOTATION:
a) I) LOOK AT SIZE if they IMFs are the same. H2S: more electrons, more polarizable than H20
therefore stronger London dispersion.
ii) H2S: H—S bond is less polar than H—O bond (dipoledipole is all about polarity)—look at
electronegativity between S and O (S<O)—so H2S has weaker d-d force.
GUIDELINES
2009 (B) NONE
2010 (B) NONE
2010 (A) Question 5 ONLY d and f. OMIT e.
ANOTATION:
d) COURSE ITS FALSE! IT’S A TRICK. The bonds they tell you are INTRAMOLECULAR FORCES.
BOILING is about breaking INTERmolecular forces. !!!!
f) Ethanol has strong hydrogen bonds. Ethanethiol doesn’t. Ethanol can thus bond with water molecules
and dissolve, but ethanethiol cannot.
GUIDELINES
2011: (A)
5. Hydrazine is an inorganic compound with the formula N2H4 . The normal boiling point of N2H4 is 114°C,
whereas the normal boiling point of C2H6 is −89°C. Explain, in terms of the intermolecular forces present in
each liquid, why the boiling point of N2H4 is so much higher than that of C2H6.
ANNOTATION
N2H4: Has all 3 IMFs. C2H6 only has the weakest London. So it needs more energy to break the stronger
bonds of N2H4, which comes in higher bp.
GUIDELINES
N2H4 is a polar molecule with London dispersion
forces, dipole-dipole forces, and hydrogen bonding
between molecules, whereas C2H6 is nonpolar and
only has London dispersion forces between
molecules. It takes more energy to overcome the
stronger IMFs in hydrazine, resulting in a higher
boiling point.
1 point is earned for correct reference to the two
different types of IMFs.
1 point is earned for a valid explanation based on
the relative strengths of the IMFs.
2011 (B)
6. Use principles of molecular structure, intermolecular forces, and kinetic molecular theory to answer the
following questions. Ethyl methanoate, CH3CH2OCHO, is synthesized in the laboratory from ethanol, C2H5OH,
and methanoic acid, HCOOH , as represented by the following equation. C2H5OH(l) + HCOOH(l) ƨ
CH3CH2OCHO(l) + H2O(l)
Draw the complete Lewis electron-dot diagrams of a methanoic acid molecule and a water molecule in an
orientation that allows a hydrogen bond to form between them.
ANNOTATION
You see H2O, the easiest. So we identify water. Use your chapter 22 skills to recognize methanoic acid.
Now, only water hydrogen bonds with other substances. So it muz be h-bonding.
GUIDELINES
(Hydrogen Bonding Between Methanoic Acid and Water)
1 point is earned for a diagram showing a reasonable orientation between a methanoic acid molecule and a
water molecule.