NAME:_________KEY______________Chem. 51, Fourth Midterm Exam.
A periodic table of the elements is included as an insert with this examination. You may
use the back for scratch work but enter all work to be graded in the space provided with
each question. Show your work in order to receive credit in problems involving
computation. h = 6.626 x 10-34 J-s, c = 2.9979 x 108 m/s, NA = 6.0221 x 1023.
1) (40 points) In a study designed to better understand molecules involved in vision, your
instructor has employed microwave spectroscopy to determine the three-dimensional
structure of 2,4-hexadienal.
a) Provide the following information for the isomer shown below: the
hybridization of each heavy (i.e. not H) atom and estimates of the bond angles.
H
H
H
H
H
O
H
H
H
The methyl carbon on the left has sp3 hybridization. Therefore the bond
angles centered on this carbon will be close to the tetrahedral angle of
109.5. All other heavy atoms including the oxygen are sp2 hybridized.
Consequently, the bond angles involving the backbone atoms, e.g. /_CCC .
will be close to 120
b) Provide a complete description of the bonding between carbon and oxygen.
The carbon, oxygen double bond consists of a -type single bond represented by
an overlap of a C sp2 hybridized orbital with an O sp2 hybridized orbital and a type single bond represented by an overlap of a C pz unhybridized orbital with a
O pz unhybridized orbital.
c) Draw the structure of one structural isomer of the species given in (a).
There are many correct answers to this question. Correct answers will have zero
formal charge on all atoms, tetravalent carbon, divalent oxygen, and monovalent
hydrogen. Here is an acceptable solution.
H
O
H
H
H
H
H
H
H
d) Draw the structure of one geometrical isomer of the species given in (a).
The stereochemistry (geometrical arrangement) of the two carbon-carbon double
bonds in the starting structure is E or trans. Acceptable solutions involve
changing either or both to a Z or cis arrangement. The structure below is for
Z,E-2,4-hexadienal. Rotation about a single bond which is facile corresponds to
generating a conformer and not an isolatable isomer.
H
H
H
H
H
H
H
H
O
e) Provide the basis for the experimentally observed planarity of the heavy atoms
in 2,4-hexadienal.
Two factors contribute to the planarity. The local geometry about each sp2
hybridized carbon is planar and this carbon and the 3 atoms bonded to it lie in a
plane. Five such planes exist in the molecule. These 5 planes are co-planar as a
result of the overlap of the unhybridized pz orbitals on the 6 atoms that are sp2
hybridized. This overlap yields the pi molecular orbitals.
2) (60 points) This problem deals with the alkali metal potassium, atomic number 19.
a) The first ionization energy of potassium is 418.8 kJ/mol (4.3407 eV).
Calculate the effective nuclear charge seen by the ionized electron.
IE = -E = (13.6 eV)Ze2/n2 so Ze =[(4.3407 eV)(42)/(13.6 eV)]0.5 = 2.26. The
configuration of K is [Ar]4s so n of the most readily ionized electron, the sole
valence electron is 4. If the core were perfectly shielding, Ze would be 19-18 =1.
The actual value reflects the ability of the 4s electron that has no orbital angular
momentum to see the full nuclear charge part of the time.
b) Ionization can be caused by the absorption of a photon. Calculate the
wavelength of light that would barely ionize ground-state potassium.
We require the energy per atom which is (418800 J/mole)/(6.022 x 123
molecule/mole) = 6.964 x 10-19 J. This energy change of the potassium valence
electron that is ionized is the energy of the ionizing photon, E.
From Einstein's model, E = h = hc/. Solving for , one obtains 286 nm.
c) Calculate the average distance from the nucleus of the electron with the 418.8
kJ ionization energy.
<r> = a0n2/Ze = (0.529 Å)(42)/(2.26) = 3.74 Å
d) In an article on the high pressures in the centers of planets, S. Scandolo and R.
Jeanloz note that potassium becomes a transition metal under these conditions.
Succinctly discuss the plausibility of their statement.
It was shown in class that a convincing argument could be made for two
electronic configurations for K: [Ar} (4s) and [Ar] (3d). The two states are
close in energy. Empirical data summarized in the periodic table show that
under conditions close to standard conditions the first is the ground state. The
high-pressure data indicate that under high-pressure and therefore high-density
conditions, the electron distribution is perturbed and the second case becomes
the ground state. The [Ar] (3d) configuration with an unfilled d subshelll is the
signature of a transition metal. N.B. Zn is not a transition metal! Also note that
high pressure does not imply temperature. One can construct a diamond anvil
pressure that can apply pressures in excess of several GPa and at the same time
maintain the sample at ambient temperature.
e) The ninth and tenth ionization energies of potassium are 16964 kJ and 48610
kJ, respectively. Explain why one number is large and the other is very large.
We shall use the valence electron of unionized K as the benchmark. n is 4 and Ze
is close to 1. In contrast, n for the 9'th electron to be ionized is 3. Since the
orbital energy has an inverse, quadratic dependence on n, the change in n alone
argues for a large increase in the IE. Furthermore, with 8 electrons already
removed, the ninth electron with a configuration of [Ne] (3s) will see a Ze of
approximately 19-10 = 9. The electron as a result is held more tightly. After this
electron is removed, the next in line for ionization has an n of 2. We are now in
the regime of small values of n where changes in n yield very large energy
changes. The change in Ze is not very dramatic. The major change in IE on
going from the 9th. to the 10th lies mostly in the value. of n. Magic incantations
such as the favorite mantra "closed shell, closed shell, closed shell" may score
points on an AP exam but are not explanations based on fundamental principles.
3) (35 points) Draw a correct Lewis electron dot (Valence Bond) structure for the
following two species. Make sure that formal charge is optimized. If the molecule
exhibits resonance, draw the Lewis structures of all resonance forms. Also, briefly
describe the three-dimensional structure of the species. For each species, the central
atom is given in bold face.
a) XeOF4 (Xe, Z = 54)
b) PO3- (P, Z = 15)
The electron dot structures are given below. Note on the case of XeOF4
that a double, not a single Xe,O bond is formed. This preferred structure
has zero formal charge on all atoms. In contrast, a structure with a Xe, O
single bond and three lone pairs on the oxygen has a formal charge of -1
on the oxygen and +1 on the xenon. The xenon compound has 6 groups of
charge which would be expected to arrange themselves at the 6 corners of
an octahedron. The heavy atoms attached to the xenon would form a
square pyramid.
Three resonance structures that minimize formal charge are possible for
the phosphite anion. Three groups of charge are present and a perfect
trigonal planar structure is expected with a P-O-P bond angle of exactly
120 and a P,O bond length intermediate between that of P,O single and
double bonds.
4) (15 points) An atomic orbital is displayed below. Provide the following information as
well as the basis for your answers: the value of the quantum number l, the value of the
quantum number |ml|, and the probability of finding an electron in the orbital close to the
nucleus.
We see three nodal planes in the orbital so l = 3. (Note the change in the
sign of the lobes as one moves from top to bottom, then left to right, and
finally front to rear.) Two of these nodal planes contain the z axis so |ml|
= 2.
The three nodal planes intersect at the origin so the value of the
wavefunction and therefore its square, the probability, will be exactly zero
on these planes and at the nucleus. The question, however, asked about
the probability close to the nucleus. A second principle can be applied to
address this question. A moderately high value of l results in significant
orbital angular momentum and hence a high centrifugal force that will
prevent a electron represented by this orbital from getting close to the
nucleus. Therefore, the probability of finding the electron rapidly
decreases as one approached the nucleus. Quantitatively, the probability
is proportional to r6 [6 = 2(l) = 2(3)].