Show your work in all questions involving computation or derivation to receive credit. A periodic table of the elements is printed on one insert. Figure 1, which contains information required to answer Problem 1, is on a second insert. You may use the backs of the inserts for scratchwork but enter ALL work to be graded in the space provided with each question. This exam has 6 questions and is worth 150 points.
1) (20 points) This question applies to orbital A which is displayed at the top of Figure 1. The coordinate system for the five orbitals displayed on Figure 1 is defined by the diagram in the upper left-hand corner.
a) Determine the values of the quantum numbers l and the absolute value of ml of an atomic orbital represented by orbital A. Provide the basis for your answer.
Orbital A contains 3 angular nodes which are planes so l = 3. Only one of the three, the xy plane, does not contain the z axis so the absolute value of ml = 2.
b) Determine in units of h/2p the value of the orbital angular momentum and the absolute value of the z-component of the orbital angular momentum of an electron represented by orbital A.
L = [l(l+1]0.5 h/2p = [3(4)]0.5 h/2p = [12]0.5 h/2p
The absolute value of Lz = the absolute value of ml times h/2p = 2 h/2p
c) Orbital A is related by a symmetry operation to another orbital displayed on Figure 1. Which orbital? What is the symmetry operation? Which quantum number is related to the symmetry operation?
Orbitals A and E are related by a rotation about the z-axis of 90º/|ml| = 90º/2 = 45º.
d) Give the name or symbol of an element with a valence electron in an unfilled subshell that could be represented by orbital A. Would this element be paramagnetic?
An element with an f electron (l =3) with an unfilled f shell is required. Any of the actinides, Th through No, or any of the lanthanides, Ce through Yb, would be suitable. The elements Lr and Lu have a closed f subshell and would not be suitable. The elements Ac and La are not actinides and lanthanides, respectively. The suitable elements will have at least one unpaired f electron and hence will be paramagnetic.
e) Could an electron described by orbital A ever be found at the nucleus? Briefly explain.
No! The high value of l indicates a high value of the orbital angular momentum and a resulting large centrifugal force that will force the electron from the nucleus. At short r, the centrifugal force is always stronger than the Coulombic force.
2) (20 points) Compounds of chromium in the +6 oxidation state are used in the electroplating industry. The EPA claims that workers in an electroplating plant must be protected from the vapors of the chromium compounds which if inhaled would cause lung cancer. The foreman of the plant insists that the government's charge is unwarranted since compounds of metals such as sodium chloride are known to be very non-volatile (i.e. have a very low vapor pressure). Who is correct? Discuss.
The series of ionizations required to produce a +6 cation requires a prohibitively high input of energy that cannot be compensated by stabilizing terms such as Coulombic attraction. Ionic bonding is not possible for chromium(VI) compounds. The compounds do exist and in fact are quite stable. The bonding is therefore covalent. Although the intramolecular forces in a typical compound such as chromium(VI) oxide are quite strong, the intermolecular forces are weak. Hence, these compounds will be quite volatile and the EPA is justified that special measures such as very efficient ventilation be taken in the handling of chromium(VI) compounds.
3) (25 points) The nitrogen +6 cation is a one-electron species. A N+6 cation is excited to a 1000p excited state.
a) Calculate the ionization energy of this excited atomic species.
IE = E(n = infinity) - E(n = 1000) = 0 - -R(Z2/n2) = R(Z2/n2) = (13.6 eV)(10002/72) = 0.00066 eV.
Multiply the result by 96485 C/mole to convert the answer to J/mole. The result is a very low 64 J/mole. A 160 GHz photon will effect the ionization.
b) Suppose that several N+6 cations were excited to the 1000p state. Propose experimental results that could be used to demonstrate that they were in fact excited to this state.
A nitrogen +6 cation in the highly excited and hence metastable 1000p state can revert to the ground 1s electronic state by an emission of a single photon. The ion can also drop down by steps. There is no restriction on the change in the n quantum number. Hence there is a large number of pathways connecting the 1000p and 1s states. A large population of exited ions will decay via all of these pathways and a broad spectrum of photons with wavelengths ranging from the X-ray to the microwave will be emitted. The value of n in the excited state can be determined from the wavelength of the photon with the highest energy.
c) Calculate the rms radius of this excited atomic species. Could this excited atomic species be "seen" with a conventional microscope whose resolution is 1 (m?
rrms = a0(n2/Z) = (0.529 Å)(10002/7) = 76,000 Å = 7.6 mm. This result is indeed within the range of a convention optical microscope so one might be able to design an experiment that would allow one to "see" an atom, albeit one in an highly excited state. Such an atom is in effect a macroscopic object and therefore classical mechanics could be used to describe its dynamics.
4) (25 points) Draw acceptable electron-dot structure(s) for the chlorite anion, ClO2-. What consequences does your answer have for the bond angle and bond lengths in the anion?
An acceptable electron-dot structure should not have formal charge on the chlorine atom. There should be a formal charge of -1 on one of the oxygen atoms such the species is a -1 anion. There are 2 acceptable structures which are equivalent resonance structures. The two Cl-O bonds will have equal bond lengths that will be intermediate in value between a single and double bond. Counting the lone pairs, there are 4 groups of charge around the chlorine so a bent structure is predicted.
5) (30 points) Use your knowledge of the periodic table and propose a compound that would have as high an ionic character as possible. Support your choice with a succinct argument. Be sure to consider all the factors!
To optimize ionic character, one requires a high electron affinity, a low ionization energy, and a large Coulombic force between the cation and anion. Optimization of the Coulombic term requires small ions and high charges on the ions. The high charges usually requires counterproductive high ionization so we shall have to settle for +1 and -1 ions.
Fluorine is the obvious choice for the non-metal in our ionic salt. Fluorine has a very large although not quite the largest electron affinity BUT has a very small ionic radius. In fact, the fluoride anion is the anion with the smallest ionic radius. The choice of the cation is not as straightforward. One of the group Ia elements would be a straightforward choice. Focusing on just the ionization energy would direct one's attention to francium or if one preferred to avoid radioactive elements to cesium. However, the francium and cesium cations are quite large. The Coulombic argument would favor the tiny lithium cation. Alas, lithium has the highest ionization energy of the group Ia elements. The optimum lies somewhere between lithium fluoride and cesium fluoride.
6) (30 points) Several isomers with the molecular formula C2H5N are possible.
a) Draw an acceptable electron-dot structure of one of these isomers.
b) Predict the bond angles of your isomer and give the hybridization of each of the heavy (non-H) atoms.
Four isomers- I , II, III, and IV -are possible.
Hybridization of the heavy atoms:
isomer I: both carbons are sp2 hybridized and the nitrogen is sp3 hybridized.
isomer II: the methyl (i.e. CH3) carbon is sp3 hybridized and the nitrogen and the middle carbon are sp2 hybridized.
isomer III: the methyl carbon is sp3 hybridized and the other carbon and the nitrogen in the middle are sp2
hybirdized.
isomer IV: all three heavy atoms are sp3 hybridized.
c) Describe the bonds formed between the heavy atoms. Use the s and p formalism in your description.
isomer I: A double bond is formed between the two carbon atoms that consists of a s-type overlap of two sp2 hybridized orbitals on the two carbons and a p-type overlap of two unhybridized orbitals on the two carbons. A single bond is formed between the nitrogen and the middle carbon that consists of a s-type overlap of sp3 hybridized orbitals on the carbon and nitrogen.
isomer II: A double bond is formed between the middle carbon atom and the nitrogen that consists of a s-type overlap of two sp2 hybridized orbitals on the two atoms and a p-type overlap of two unhybridized orbitals on the two atoms. A single bond is formed between the two carbon atoms that consists of a s-type overlap of sp3 hybridized orbitals on the two.
isomer III: A double bond is formed between the nitrogen atom in the middle and a carbon atom that consists of a s-type overlap of two sp2 hybridized orbitals on the two atoms and a p-type overlap of two unhybridized orbitals on the two. The carbon-nitrogen single bond consists of a s-type overlap of a sp3 hybridized orbital on the carbon atom with a sp2 hybridized orbital on the nitrogen atom.
isomer IV: A single bond is formed between each pair of heavy atoms that consists of a s-type overlap of two sp3 hybridized orbitals on the two atoms.
d) What chemistry (i.e. types of reactions) would be possible for your isomer as a consequence of its electronic structure?
isomer I: The lone pair on the nitrogen indicates that the isomer will function as a weak base. The double bond indicates that the isomer will readily react with electron-greedy reagents called electrophiles.
isomers II and III: The lone pair on the nitrogen indicates that the isomer will function as a weak base. The double bond indicates that the isomer will readily react with electron-greedy reagents called electrophiles.
iosmer IV: The lone pair on the nitrogen indicates that the isomer will function as a weak base. The sp3 hybridization on the heavy atoms suggests a preference for a tetrahedral bond angle of 109.5º. However, geometry dictates a bond angle of 60º for the cyclic structure. Geometry and chemistry are in conflict and the ring is highly strained as a result. The compound is therefore highly reactive and will readily react with any species that will break the ring and relieve the strain.
16 November 1999
