Chemistry 122 - Review of General Chemistry
Some skills you should have at the beginning of Chm 122:
1) Given a molecular formula and connectivity data, draw a Lewis Structure.
2) Recognize when a single Lewis Structure does not adequately represent a
molecule and be able to then draw appropriate resonance structures and to
judge their relative contributions to the overall structure.
3) Calculate formal charge
4) Determine approximate bond angles and molecular shape from Lewis Structure.
5) Determine bonding and molecular dipoles
6) Determine the hybridization of carbons in molecules
These topics will be reviewed quickly at the beginning of the semester.
1) Lewis Structure - see Carey 1.1-1.4
A Lewis Structure conveys the following information: connectivity of atoms, type of
bonding (single, double, triple), and location of non-bonding lone pairs of electrons. Below is an
example of how to deduce the Lewis Structure of acetone (a chemical you will encounter often in
class and lab, as well in fingernail polish remover, if you do that sort of thing) from the
condensed formula.
(CH3)2CO
acetone
oxygen will make
2 covalent bonds,
while carbon will
make four
C
O
C
C
O
H
C
H
C
C
H
H
H
H
so, begin with the
simply fill the remaining
CO group; the carbon
six valences on the two
and oxygen must be
carbons with the six hydrogens
double bonded (in the
from the condensed formula
condensed formula, the
oxygen is only bonded to
the carbon). The carbon,
then, has two additional
covalent bonds; these must
be to the other two carbons if they were to hydrogen, there
would be no more available
bonds to make to the other
carbons.
O
H
C
H
C
C
H
H
H
H
count electrons:
oxygen has six
valence electrons
and is only using
two in bonding therefore, it must
have four in nonbonding lone pairs
other atoms have
the proper number
of electrons from
bonding pairs
Learn to recognize patterns in bonding in organic molecules. Most atoms involved in stable organic
molecules will have similar - or identical - types of bonding. For example, an oxygen atom participating
in two covalent bonds (either two single bonds or one double bond) will have two lone pairs. An oxygen
with only one covalent bond will almost always have three lone pairs and a formal charge of -1. Carbon
will have four covalent bonds and no lone pairs in a stable molecule. Hydrogen will never have more
than one covalent bond. Nitrogen that has three covalent bonds will have one lone pair and a formal
charge of zero. Nitrogen that has four covalent bonds will have no lone pairs and a formal charge of +1.
The halogens will generally have one covalent bond and three lone pairs.
2. Resonance structures - see Carey 1.9
Molecules with multiple bonds, such as acetone above, generally can not be accurately
represented by a single Lewis Structure. The simplest explanation for this is that the electrons in
a multiple bond can be divided, unevenly, between the two atoms involved without breaking the
connection between the atoms. Resonance structures must have the same connectivity and the
same overall molecular charge. Neither atoms nor electrons can be added or eliminated; atoms
can not be moved (the connectivity can not change) - the structures are meant to represent a
single compound and a single compound can not have varying amounts of atoms or electrons or
different connectivity of the atoms. Each resonance structure is a correct Lewis structure for that
particular connectivity of atoms. See below for the resonance structures of acetone:
O
O
H
C
C
C
H
H
H
H
H
H
H
C
H
C
C
H
H
H
Four electrons are shared by the carbon and oxygen. Without breaking the connection between
the two atoms, two of the electrons can be moved to the electronegative oxygen. (The "curved
arrow" in the first structure represents the "movement" of these electrons. During the course, many
reactions will involve the movement of electrons and the use of curved arrows will be used to keep
track of this. The + and - signs in the structure on the right represent the formal charges of the
atoms the sign is nearest.) The two structures shown are resonance structures of the compound
acetone. Neither structure alone is adequate to explain all of the properties and behaviors of the
compound. The true structure is a hybrid of the two resonance structures with a greater
contribution from the more stable structure. In this case, that is the structure on the left. All atoms
are neutral and have full octets in contrast to the structure on the right in which two atoms are
charged and one, the carbon, does not have a full octet.
3. Formal Charge - see Carey 1.6
Molecules can be neutral or electrically charged. The overall charge of any molecule is
equal to the sum of the formal charges of the atoms in the molecule. Formal charges are
generally indicated in organic structures by a single + or - sign next to the charged atom. An
atom without a sign is assumed to be neutral. It is very unusual for an organic molecule to
possess an atom with a formal charge of more than +1 or less than -1. Using the resonance
structures for acetone shown above, the formal charges for two atoms in the CO double bond are
calculated:
The formula for formal charge is:
group # (valence electrons) - number of bonds - number of unshared electrons
So, for the structure on the left (the more stable resonance structure):
Note that for main group elements,
FC (carbon) = 4 - 4 - 0 = 0
the group number is the number of
FC (oxygen) = 6 - 2 - 4 = 0
valence electrons.
For the structure on the right (the less stable structure):
FC (carbon) = 4 -3 - 0 = +1
FC (oxygen) = 6 - 1 - 6 = -1
In both structures, the sum of the formal charges is zero.
4. Molecular shape - see Carey 1.10
A useful tool for predicting molecular shape is the VSEPR theory. Part of this theory is
that electrons (same charge) will repel one another and that molecular bonds and lone pairs are
simply electrons. Therefore, molecular shape should conform to whichever conformation
maximizes the separation of the bonds and lone pairs. Despite serious flaws with the overall
theory, it does an excellent job in approximating bond angles in molecules.
For predicting molecular shape, VSEPR theory considers only how many "things" are attached to
an atom. For this purpose, lone pairs are considered the same as atoms. In the case of most
organic molecules there are three possibilities of concern:
Four substituents (sp3 hybridized atom):
H
H
C
H
H
H
109.5o
N
H
methane
Carbon having four single bonds
is called "tetrahedral' because the
molecular shape can be accurately
represented by a tetrahedron. The
four "substituents" - atoms to which
the carbon is bonded - occupy the
corners of a tetrahedron with the carbon
at the center. The bond angles for any
two substituents and the carbon are,
therefore, 109.5 degrees. This is, of
course, a theoretical value, but
experimentally most tetrahedral carbons
are found to have bond angles quite
close to the theoretical number.
H
ammonia
Counting the lone pair,
ammonia also has four
substituents. Therefore,
the HNH bond angle will be
close to 109.5 degrees.
Three substituents (sp2 hybridized atom):
Two substituents (sp hybridized atom):
O
120o
C
H
H
Formaldehyde
Carbon having two single bonds
and a double bond has only three
substituents (the double bond is
counted only once in VSEPR).
Therefore, the maximum separation
of the bonding electrons will be if
the three substituents occupy the same
plane. The bond angles must then be
120 degrees. The carbon is trigonal
planar.
H
C
N
hydrogen cyanide
Carbon having one single bond and
one triple bond (or two double bonds)
has only two substituents.
Obviously, these two substituents will
lie on opposite sides of the carbon,
giving the molecule a linear shape and a
bond angle of 180 degrees. Note that
the nitrogen has only two substituents
as well - the carbon and the lone pair.
Therefore, the carbon and the lone pair
must be linear with the nitrogen.
5. Molecular Dipoles - see Carey 1.5, 1.11
Electrons are not distributed evenly throughout molecules. Some atoms are more
electronegative than others (you should know which) and, therefore, attract electron density
toward themselves more than other atoms. This is particularly apparent when two atoms of
widely disparate electronegativities are bonded to one another. Consider the carbon-chlorine
bond. Chlorine is extremely electronegative while carbon is not. The two electrons that make up
this bond will be "pulled" closer to the chlorine than the carbon. The bond will then be
"polarized" so that the chlorine end has some negative character and the carbon end has some
positive character. Each bond not between two identical atoms will be polarized to some extent.
The sum of these bond dipoles will confer a molecular dipole. These dipoles are important in
understanding molecular reactivity.
A useful piece of information to keep in mind is that, no matter how many polarized bonds a
molecule may contain, if the molecule is completely symmetric it will not have a molecular
dipole. Consider carbon tetrachloride; four highly polar carbon-chlorine bonds are present in the
molecule, arranged in the familiar tetrahedral pattern about the carbon. The resultant from
adding the four dipole vectors is a zero dipole. The four dipoles cancel one another and the
molecule is non-polar.
6. Orbital Hybridization - see 1.15 - 1.18
A useful way of thinking about orbitals and bonding in organic chemistry is orbital
hybridization. Carbon has four valence orbitals: one s atomic orbital and three p atomic orbitals.
In the orbital hybridization model, one imagines that carbon mixes various atomic orbitals to
make bonding orbitals. There are three ways the four orbitals can be mixed and be consistent
with observed bonding. They are shown below:
C
C
The carbon that is bonded to four
different substituents is said to be
sp3 hybridized. That is, the four
orbitals involved in bonding are
equivalent orbitals formed by mixing
one s and three p orbitals. Remember
the number of orbitals mixed must
equal the number of orbitals formed.
So, an sp3 hybridized carbon, as is the
one above, will be a tetrahedral carbon.
C
The carbon that is bonded to three
substituents, one of which is a double
bond, is said to be sp2 hybridized.
There are three equal bonding orbitals
(sp2 orbitals) that are formed by the
mixing of one s orbital and two p
orbitals. This leaves one p orbital
that is unmixed. It is this p orbital
that participates in the pi bond
(the double bond is made of one
sigma bond and one pi bond).
C
The carbon that is bonded to only
two substituents is said to be sp
hybridized. One bond to each
substituent is formed with an sp
hybridized orbital formed by the
mixing of one s and one p atomic
orbital. This leaves two p atomic
orbitals unmixed. These orbitals
participate in pi bonds.
Suggested Review Problems:
To be sure you understand these concepts, you should work the following problems.
1.6, 1.7, 1.8, 1.9, 1.10, 1.16, 1.17, 1.18, 1.19, 1.23, 1.24, 1.25, 1.26, 1.29, 1.30, 1.32, 1.33, 1.34,
1.42, 1.45