Chem 111
Review for Exam 2
Remember that when I make up my multiple choice questions, I look at this sheet and write
the questions.
For molecular orbital theory, you should realize that the molecular orbital, MO, is a new orbital that
exists through the molecule, rather than being localized to a particular atom. Also the MO is formed
from a combination of atomic orbitals and can be classified as sigma, pi, etc based on whether the
MO lies along the internuclear axis or above and below it. Remember that sigma MOs are formed
from combinations of s orbitals or p orbitals that overlap along the internuclear axis. MOs are
formed from combinations of p orbitals that overlap above and below the axis. For our purposes
we have talked about two types of MOs - bonding and antibonding. Only two electrons can go in
any MO. You should be familiar with the molecular orbital electronic energy diagram. Be able to
fill in a molecular orbital energy diagram for all electrons or maybe just the valence electrons, and
from that be able to tell the bond order, B.O. = ½ (# of electrons in bonding MOs - # of electrons in
antibonding MOs). Also be able to assign para or diamagnetism and write down the molecular
electronic configuration. You should be able to do this for diatomic molecules and diatomic ions.
You should be able to name simple compounds, simple ionic compounds (including ones that
contain molecular anions and cations), binary molecular compounds, acids, hydrates and organic
compounds (methane (CH4), ethane (C2H6). In order to accomplish this, you must first decide which
type of compound it is that you are working with, since the way that you name it depends very much
on the type of compound that it is. Ionic compounds are named by taking the cation name with the
anion name. The cation name is just the name of the metal, or for transition metals: the name of the
metal followed by the charge in Roman numerals in parentheses. This is because the transition
metals can have several charges on them (what are the exceptions? (Al makes only a 3+ cation, CdCd2+, Zn-Zn2+, Ag-Ag+). Can you justify this based on the periodic table? The anion is named by
taking the stem name and adding ide or if the anion is a polyatomic anion, the anion name is simply
the polyatomic anion name. You must know the table of polyatomic ions (molecular ions-why call
it this?). When is it appropriate to use per, hypo, hydrogen, and dihydrogen prefixes on some
anions, and ite and ate endings? What determines if a compound is the ite or ate or hypo ite, etc.
You should know in addition to nitrate, sulfate, phosphate, and chlorate the ite compounds and hypo
ite and per ate for halogens like perchlorate.
Acids are compounds which when dissolved in water give up one or more protons. HCl(aq),
HBr(aq), HI(aq), H2SO4(aq), H2SO3(aq) are examples. The (aq) distinguishes the acid from the
binary molecule in the case of the 1st three. Remember acids which come from anions ending in
ide have the acid named from hydro followed by the stem name with ic and the word acid.
Oxoacids which are from anions ending in ate have the acid named from the stem name with ic
and the word acid (or something close to that). Those acids which come from anions ending in ite
have the acid named from the stem name followed by ous and the word acid following.
The binary molecular compounds are named using the Greek Prefixes, mono, di, tri, tetra, penta,
hexa, hepta, octa, nona, and deca for both the 1st and 2nd atom. Rember that the mono prefix is
usually dropped for the 1st atom. The name then is a combination of the 1st atom prefix and
name, and the prefix, stem name and ide from the second atom.
You should also be able to name simple organic molecules like alkanes an alcohols. Remember
that these names were given out in lab. (Methane, ethane, propane, butane). Alcohols are
named from the alkane name by taking off the trailing e and replacing it with ol. Hydrates are
ionic compounds with varying numbers of water molecules attached to them. The hydrate is named
then by naming the ionic compound as usual with the attachment prefix and the word
Ahydrate@ where the prefix comes from the Greek prefixes reflecting the # of water molecules.
The nomenclature learned in Chapter 2 is still very important for the new concepts learned in
Chapter 3. We learned that the term molecule is reserved for covalently bonded atoms, and
formula unit must be used for ionic compounds but can be used for covalent ones as well.
Remember a molecule is a formula unit, but a formula unit is not always a molecule. The molecular
weight or molecular mass is simply the sum of the average atomic masses of the atoms making up
the molecule and the formula mass is the sum of the masses of the atoms making up the formula
unit. In molecule of propane, how many atoms of carbon are there?
In Chp. 3 we learned about the concept of a mole. Remember a mole of material is the
same number as the # of Carbon-12 atoms needed to make 12.00 g of C-12. The number
is often called Avogadro’s number, 6.022 x 1023. Thus 1 Carbon-12 atom has a mass of
12 amu and 1 mole of carbon atoms has a mass of exactly 12.00 grams. The weights
on the periodic table can be thought of as amu/atom OR g/mole. The mole concept
can be used in the following way. If you have a dozen water molecules, H2O, you have
two dozen hydrogen atoms, so if you have a mole of water, then you have two moles
hydrogen atoms. The mole is also important because a chemical eqn like N2 + 3H2 
2NH3 means that 1 molecule of nitrogen and 3 molecules of hydrogen make 2 molecules
of ammonia, or since a mole is just a certain # of molecules, 1 mole of nitrogen and 3
moles of hydrogen make 2 moles of ammonia. Also, the molecular formula or formula
unit has much information as well. For instance NH3 tells us that if we had 2 molecules
of NH3 there are six atoms of H present, or if I had 2 moles of NH3 then there are 6 moles
of H atoms. The beauty of the mole over the molecule is that we can use a balance to
measure moles of a substance whereas we cannot measure out individual molecules. You
must be able to convert between mass, moles, and # of molecules with great ease. You
must memorize Avogadro’s number.
You must be able to balance a chemical reaction. This may mean taking a statement
from words, coming up with the formulas, putting the reactants and products on right side
of arrows, putting in phase labels and information above the arrow, and then balancing
the equation by changing only the coefficients in front of the molecules or atoms.
Remember the following molecular substances: H2, O2, N2, F2, Cl2, Br2, I2, S8, and P4.
These are the molecular forms of these substances and should be used in the chemical
equation unless other information is given. Any other elemental substances can be
assumed to be just a single atom. A chemical equation is balanced when there is an
overall atom and overall charge balance.
Often once a new substance is made in the course of a reaction, its formula is not known.
What can be determined is the % by mass of the atomic components making up that
formula unit. This can be done by combustion for organic substances or by other
methods. From the elemental mass % you should be able to determine the empirical
formula or simplest formula. You cannot get the exact molecular formula unless you have
the molecular weight. You should be able to get the molecular formula from the mass %
of the elements in the molecule and the molecular weight. Remember that since the
formula can be thought of molecules and atoms or moles of molecules and atoms, then
you must convert to moles to get the empirical formula.
A very important part of the chapter involved balancing simple reactions and then use of
the chemical STOICHIOMETRY from the balanced chemical eqn. The chemical eqn is
the only BRIDGE between calculations involving amounts of reactants and products. You
MUST be able to calculate the amounts of product formed or reactant needed. Remember
the limiting reactant or limiting reagent. The amount of product formed from the limiting
reagent is called the theoretical yield. The actual amount formed is often less than the
theoretical amount and is called the actual yield. The percent yield is then defined as:
(actual yield/theoretical yield) x 100%.
In CHP 4 we defined acids, bases, strong electrolytes, weak electrolytes, aqueous
solutions, solutes, and solvents. What are the six strong acids? What are the strong bases?
We also learned the solubility rules which allow us to determine if a given salt is soluble
in water or not. The solubility rules and the lists of the strong acids and bases (you must
memorize these) helped us to understand and allow us to classify reactions in aqueous
solutions into different categories. We talked about acid-base reactions, combustion
reactions, single displacement, double displacement, and oxidation reduction reaction. Of
these, the double displacement or metathesis reactions and oxidation reduction reactions
are the ones upon which we focused.
We learned some classifications of chemical eqns: molecular eqn, ionic eqn. and net ionic
eqn. to help get at the root of the reactions that occur in aqueous solutions. Remember
that the soluble salts and strong acids and bases are written as totally dissociated in the
ionic eqn. You should be able to write these three types of eqns. We focused on double
displacement reactions, or metathesis rxns. The three types were? Remember the gas
formation reactions involved an acid reacting with a carbonate or an acid reacting with a
sulfite and the formation of the gas drove the reaction in the forward direction as reflected
in the net ionic equation. What drove the other two types of metathesis reactions. You
should be able to determine how to synthesize a salt product based on precipitation,
neutralization, or gaseous formation (H2CO3 and H2SO3).
In order to be able to study oxidation-reduction reactions (redox rxns) we had to first be
able to assign oxidation states to all of the atoms in a chemical eqn. You should know
the rules and be able to determine the values of oxidation # for the atoms in a molecule.
In an oxidation reduction reaction one species is reduced (the oxidation # is reduced and
electron(s) are a reactant) and one species is oxidized (the oxidation # is increased and the
electrons are a product). Both oxidation and reduction must occur and so in a sense the
species being oxidized causes the reduction and is called the reducing agent. What is an
oxidizing agent. Since oxidation reduction reactions often involve ionic species in
aqueous solution they are hard to balance both charge and atoms by simple inspection.
You should be able to write down the unbalanced oxidation and/or reduction half
reactions and balance them.
In the last part of the chapter we defined units of concentration such as molarity. You
should know how to find the molarity of a solution, and be able to do dilution
calculations (MiVi = MfVf). You should also be able to determine the volume of one
reactant solution necessary to react with a given volume of another solution. What is a
titration? You should be able to perform stoichiometric calculations for solutions
including acid-base and redox titrations, and precipitation rxns.
Overall:
The setup for this exam will be similar to the last. Expect multiple choice questions and
short answers questions and longer problems. Chapter summaries in your book and this
sheet should be used. Look at quizzes and homework problems. All that you need for the
exam is a pencil or pen and a calculator. The calculator must not be used for anything
other than mathematical calculations. Problems from Chapter 10 will be of this type:
1) knowing how to construct a molecular electronic energy diagram, the bond order,
molecular electronic configuration, and if it is para or diamagnetic, sigma and pi bonds.
The problems from Chapter 3 will be of the following type:
1) Converting between mass, moles, and molecules or atoms (guaranteed)
2) Balancing chemical reactions, mass-mole relations from the molecular formula or
formula unit
3) Determining the simplest or empirical formula from mass % information, and
determining the molecular formula from empirical formula and molecular wt.
4) Calculating limiting reagent, theoretical yield, moles of reactant needed, moles of
product formed, percent yield
The problems for Chapter 4 could be of the following type:
1) using solubility rules, strong acid identities, and strong base identities to determine the
products or reactants needed to make a certain product in metathesis type reactions and
writing the molecular, ionic, and net ionic eqns. (see the quiz)
2) assigning oxidation numbers, and writing the oxidation and reduction half reactions,
identifying oxidizing and/or reducing agent.
3) using the definitions of concentration (molarity) and the balanced chemical eqn to
determine the volume of a reactant necessary for an acid-base or a redox titration, or the
concentration of a solution that causes a precipitation (solution stoichiometry).