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. P 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, Cd-Cd2+, 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).