April 30:
EXAM 3 is on May 2, 2002. Please, bring a LARGE SCANTRON SHEET!!!!!!!!!!!!!!
Homework 5 was handed out (due on May 7) Ch. 10 continued - units 7 and 8
(handout on KMT)
New terms: diffusion, effusion, root mean square speed, average kinetic energy
Graham's Law
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)
2) dipole-dipole interactions
3) London Dispersion forces (75% completed in today's class)
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 25:
EXAM 3 is coming soon to the lecture theatre near you! For a sneak preview
click here
.
Ch. 10 continued - LOTS of calculations similar to the ones you will find
on pp. 363 - 373 in BLB. Keep paying attention to units!
Make sure that your units are consistently applied in calculations
in Ch. 10!!!
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 23:
Homework 4 was handed back.
More about hybridizations:
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 - we have practiced determining hybridizations of ALL central
atoms in glycine, aspirin, carbon dioxide, etc... 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
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.
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!!!
April 18:
Homework 4 was collected, Quiz 6 was administered! Ch. 9.3 - polarity of polyatomic
molecules - it can be determined if you know both:
(i) polarities of individual bonds, and
(ii) the SHAPE of the molecule.
Note: highly symmetrical molecules tend to be NONPOLAR.
Note: an introduction of lone pair(s) often introduces POLARITY!!!
Ch. 9.4-9.5: Introduction to the Valence Bond Theory.
New terms: hybridization, hybrid orbitals.
Hybridization prepares atomic orbitals, AO's, for the bond formation,
by providing unpaired electrons needed. 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 sp3
d2
hybrids! Both Valence Bond Theory and VSEPR theory often
give us the same predictions of molecular shapes. Compare Tables 9.1 and 9.4 and see the similarities!April 16:
Ch. 9.1-9.2: examples, examples, examples.
Differential repulsions: lone pairs exert more repulsions than
bonding pairs, multiple bonds more than single bond. All this leads to
distortions of ideal shapes (see p. 310 in BLB).April 11:
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.
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.
April 9:
Exam 2 was handed back. Ch. 8.
7 More about resonance - always calculate
formal charges to decide which Lewis structure is the best, and hence
contributes the most to the resonance hybrid.
Ch. 8.8 Exceptions to the octet rule - yes, you can have those,
and sometimes breaking the octet is better than adhering to it!
(i) electron deficient species: BeCl2, BF3
- How do these species deal with less than 8 electrons on the central
atom?
(ii) octet expansion: for species containing larger atoms as
central atoms (n = or > 3), eg.: S, P, Sb, etc... 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
