May 1: Ch. 11 -
INTERmolecular interactions - forces that act on particles of a substance
and allow them to interact with each other:
1) ion-dipole interactions (technically, INTERparticle rather then INTERmolecular)
Exist between ions of soluble ionic compounds and dipoles of polar solvents.
2) dipole-dipole interactions
Exist between permanent dipoles of polar molecules -
do not forget how to determine molecular polarity!
3) London Dispersion forces
Weak forces that exist between ALL molecules, both polar and nonpolar. However,
their existance is particularly important for nonpolar molecules (or atoms)
where no other intermolecular forces are possible. Also, for very large molecules
(or atoms) they become very important and often dominate over dipole-dipole
interactions.
4) Hydrogen bonding (to be covered next week)
!!!Intermolecular forces (their presence and relative strength) determine
the preferred state of matter compound
occurs in (the stronger the forces, the more condensed the phase!), and
other physical properties it has!!!
April 29: Ch. 10 continued: Dalton's
Law of partial pressures;
KMT of gases - know the implications of the postulates of the theory. Look
up figure 10.17.
Note, that the higher the temperature, the higher the average kinetic energy
of gas particle and hence their average speed!
Rates of Diffusion and Effusion (Graham's Law) - large molecules travel
slower than smaller ones.
The smaller the gas molecules, the larger the rate of effision and diffusion.
April 26: EXAM
3 is coming soon to the lecture theatre near you! For a sneak preview
click here
. Derivation
of the ideal gas law - introduction of TWO ideal gas constants.
It is your
choice which one you want to use, but remember that depending on that
choice, units of pressure and volume in the ideal gas law are predetermined!
Make sure that your units are consistently applied in calculations
in Ch. 10!!! LOTS of calculations similar
to the ones you will find on pp. 363 - 373 in BLB.
1 mol of any gas that behaves "ideally" takes up 22.41 L of volume
at STP conditions. See Figure 10.13.
When gases undergo changes due to changing, T, p, V conditions, the
number of moles of gas does not change. Equation 10.8 in BLB reflects that
fact. April 24:Ch.
9.5 - finished with some general conclusions and examples of molecules
with many central atoms. What needs a hybrid (on the central atom)?
(i) one electron pair per bond (regardless of its multiplicity)
(ii) a lone pair of electrons
(iii) a single electron in a radical Ch. 10 (you are
responsible for the first 8 units of Ch. 10 - study 10.3 on your own
)
Gases - the most disordered state of matter.
Pressure - know the units and conversion
between them.
April 22: Ch. 9.4 - 9.5 Valence
bond theory, continued.
The most important thing to realize about VB theory, is that its
power is in explaining what "must have occurred" on the central atom
in a particular molecule so it can provide bonding we are aware of (from
a Lewis structure)
If you draw a Lewis structure of a molecule, and it has 3 single
bonds and no lone pairs, YOU KNOW, that the central atom must have utilized
3 hybrid orbitals of the type sp2. How?
The number of AO's mixed equals the number of hybrid orbitals
resulting from the hybridization. In other words, what you put in (in
terms of the number), is what you get out. From 3 AO's you will get 3
hybrids!
Orbitals participating in hybridizations are: s, p, p, p, d, d,
etc... IN THAT ORDER.
If you need 2 hybrids, you will mix
together an s and a p orbital. The result will be
TWO sp hybrids! If you need
3 hybrids, you will mix together an s and two p orbitals.
The result will be THREE sp
2 hybrids! If you need 4
hybrids, you will mix together an s and three p orbitals. The result
will be FOUR sp3 hybrids! If you need 5
hybrids, you will mix together an s, three p, and one d orbital. The
result will be FIVE sp
3 d hybrids! If you need 6
hybrids, you will mix together an s, three
p, and two d orbitals. The result will be
SIX sp
3 d
2 hybrids! Remember - many atoms can have DIFFERENT
hybridizations in different molecules. You have to know in which molecule
you are determining the hybridization of an atom, or your guess may be
wrong.
For instance, C has an sp3
hybridization in methane, sp
2 in ethylene, and sp
in acetylene or carbon dioxide!
Count your bonds (and lone pairs, as they, too, require hybrid orbitals)
on the central atom BEFORE deciding on the type of hybridization of the
central atom. Be able to see similarities between
Table 9.1 (VSEPR predictions of VSEPR electron domain arrangements) and
Table 9.4 (VB predictions of hybrid orbital arrangements) April 19: Ch. 9.2 - 9.3: Distorsions from ideal
geometries due to differential repulsions:
LONE PAIRS of electrons require MORE SPACE than bonding pairs.
MULTIPLE BONDS require more space than SINGLE bonds.
Polarity of polyatomic molecules - depends on:
(i) polarities of individual bonds, and
(ii) molecular shape.
Some molecules, even though are made with polar bonds, are nonpolar,
becase due to the molecular shape, that is arrangement of BONDS in
3D space, bond polarities cancelled out!
Naming shapes of molecules with more than one central atom (you
can no longer do it with just one or two words)
Ch. 9.4 Valence bond theory - main assumptions -
know them.
Bonds are formed when two singly occupied orbitals (the most
common case) overlap with each other.
If an atom does not have orbitals with single electrons in them,
it has to be "prepared" for bonding, by undergoing HYBRIDIZATION.
April 17: Ch. 9.1-9.2:
examples, examples, examples. We have "derived" Tables 9.2 and 9.3.
April 15: Follow
the three steps to predict geometries using the VSEPR theory (p. 308
in BLB)
Table 9.1 - you will not have to memorize it IF you understand
how it came about.
Table 9.2 and Table 9.3 - watch for the gradual breaking
of symmetry as lone pairs replace bonds. Be able to draw ALL VSEPR geometries
(I called them in class VSEPR electron domain arrangements) for 2,
3, 4, 5, and 6 VSEPR electron domains and name them! You are responsible
for knowing the names of
VSEPR electron domain arrangements.
Also, as bonds get replaced by lone pair (on the central atoms),
the molecular shapes, though still based on the arrangements of VSEPR
electron domains, will have different names! For instance:
H
2 O
has 4 VSEPR electron domains (two single bonds and two lone pairs
of electrons) so the VSEPR electron
domain arrangement is TETRAHEDRAL, but because two domains were lone
pairs, the shape of a water molecule is BENT.
Can you tell the difference? April 12:
Homework 4 was handed out - download
it here
, Quiz 5 was administered. Ch. 8.9 continued:
calculations of reaction enthalpy from bond enthalpies.
Do problem 8-69 in BLB (it was not recommended before,
but I think it would be a good idea to look it up)
Ch. 9.1-9.2: You have to be proficient in drawing
Lewis structures in order to do well in this chapter.
VSEPR model works well for covalently bonded molecules
and ions containing main group elements.
VSEPR is based on MINIMIZATION OF REPULSIONS BETWEEN
DOMAINS OF ELECTRON DENSITY
Count as ONE VSEPR electron domain: any bond (regardless
whether it is single, double or triple), a lone pair of electrons
or a single electron (in a radical).
Eg. H2
O has 4 VSEPR electron domains
(two single bonds and two lone pairs of electrons), while CO
2 has only 2 (two
double bonds)
April 10:
Exceptions to the octet rule:
(ii) octet expansion - possible for larger atoms in groups IIIA-VIIIA
(Yes! the noble gases, too), in periods: 3, 4, 5, etc...
Condition necessary (we think, though the theoretical
jury is still out on this issue) is the presence of empty d-orbitals
in the valence shell of the atom, eg.: sulfur: [Ne] 3s2
3p 4 3d 0 .
Remeber that there is a limit to how much you want to expand
octet in pursuit of more optimal formal charges. The moment you
start putting more positive (less negative) formal charges on the
more electronegative atoms, you know you went too far! (iii) radicals: species with
unpaired electron(s), typically ones where the total number of
valence electrons is an odd number.
Ch. 8.9 - started: new terms: bond order, bond length,
bond enthalpy
The higher the bond order, the shorter the bond,
the stronger the bond! However, do not
compare apples with pears. Sometimes a single bond,
like an H-H bond is stronger than a double bond, eg. a N=N bond!
Bonds between smaller atoms tend to be stronger
than bonds between larger atoms.April 8:
Ch. 8.6 Lewis structures - continued:
CH3COOH, O3; Ch. 8.7 New terms: Resonance and
resonance structures; resonance hybrid. If there are 2 or
more different possibilities of accomodating electrons in a molecule
(or ion), several valid Lewis structures can be drawn. The one(s)
with the best (closes to zero) formal charges will be the best and
will contribute the most to the resonance hybrid.
Ch. 8.8 Exceptions to the octet rule:
(i) electron DEFICIENT species: eg.: BeCl
2 ,
AlCl 3
, BF3
If a molecule cannot internally satisfy octet
for its atoms, several molecules (two or more) may come together
and form dimers or polymers using the previously lone pairs of
electrons on Cl, or F to cross-link between different molecules. Can you show why forming
a double bond between F and B in BF
3 is not
an option?
