NAME________SOLUTIONS KEY__________ Chem 51, third midterm, fall, 2004
A periodic table of the elements is included as an insert. You may use the back side for
scratch work but enter all work to be graded in the space provided with each question.
1) (50 points) Consider the atomic species B+4 in the 4p excited state with ml = 0.
This is a one-electron species with Z = 5, n = 4, l = 1, and ml = 0.
a) Calculate values with units of the following properties:
i) electronic energy
E = -RZ2/n2 = -(13.6 eV)(52/42) = -21.3 eV x 96485 C/mol = -2050 kJ/
ii) ionization energy
IE = -E = 21.3 eV (Koopman/ theorem)
iii) average location of the electron
<r> = a0n2/Z = (0.529 Å)(42/5) = 1.69 Å
iv) the orbital angular momentum in units of h/2
L = [l(l+1)]0.5(h/2π) = 20.5 h/2π
b) Will the electron ever be found at the nucleus? Briefly explain why.
No. the non-zero value of orbital angular momentum leads to centrifugal force
that prevents the electron from reaching the nucleus. There is also a planar node
for a p orbital that runs through the nucleus.
c) Suppose we have a large number of the excited boron ions in the 4p state. If
the light emitted by these ions is analyzed by a monochromator, how many
different wavelengths of light will be detected?
6 transitions corresponding to the following changes in n: 41, 43, 31, 32,
21, and 42.
d) Describe the symmetry of the electron distribution of the 4p electron.
The charge distribution will be non-spherical with a planar node (the xy plane)
perpendicular to the z axis. The charge distribution will be symmetric around the
z axis and will have equal values above and below the xy plane.
e) Will the de Broglie wavelength of the electron increase or decrease as the ion
decays to the ground electronic state? (Hint: <KE> = -E).
As n changes from 4 to 1, the total energy decreases (becomes more negative) and
the average kinetic energy increases. Hence the average speed and momentum of
the electron increase.  = h/p so the de Broglie wavelength decreases.
2) (30 points) Consider the binary ionic compound MX. Identify M and X that will yield
the largest (most negative) value of E at 0 K for the reaction M(g) + X(g) = MX(g).
Present an argument for your choice.
Three factors contribute to the overall energy change: 1) the positive ionization energy of
the metal M , 2) the negative electron affinity of the non-metal X, and 3) the Coulombic
attraction between the two ions once formed. Factors (2) and (3) argue that fluorine is
the optimal choice for X. Ionization energies increase as one moves up a column and
across a row. The electron affinity is -IE so fluorine would expect to have an optimal, i.e.
large negative EA. Furthermore, the dependence of size on n and Ze leads to a small
fluoride anion and a large Coulombic attraction since the Coulombic potential energy
depends inversely in the M-X bond length. The selection of M is less clear. The
requirement of a small IE argues for Cs. However, Cs is a large cation. Focusing on a
small cation with a resulting large Coulombic attraction would lead to Li as the optimal
choice. Qualitative reasoning does not lead to an unambiguous choice. When
experimental data are used, LiF emerges as the winner.
Electronegativity, EN, and electronegativity differences can be used as an indicator of the
extent of ionic bonding. However, all questions relating to the strength of the bond and
the stability of the molecule rest on the value of the energy change for its formation. As
discussed above, the crucial components for this energy change are the IE of M, the EA
of X, and the Coulombic attraction of the two ions.
3) (30 points) Iodine reacts with fluorine to yield several compounds with the molecular
formula IFn.
a) Which compounds are formed? Draw the electron-dot structure of each.
b) Predict the three-dimensional structure of each.
IF, linear (Cv)
IF3, trigonal bipyramid family. The F's and I form a T. (C2v)
IF5, octahedral family. The lone pair is located on the axis of a distorted
octahedron. (C4v)
IF7. 7 equivalent groups of charge are distributed around the iodine atom.
consider a pentagonal bipyramid. 5 F's form a pentagon with the I in the center.
Draw an axis through the I and perpendicular to the pentagon. Locate the final
two F's on this axis, one above and the other below the pentagon. (D5h) Another
possible structure observed in some cases is the capped octahedron. Construct an
octahedron with 6 fluorines at the 6 corners and add the seventh fluorine atom to
one of the faces of the octahedron. (C3v)
c) Only one compound with the formula FIn is formed. Explain why.
Only one FI as fluorine does not d orbitals in its valence shell.
4) (40 points) Several isomers with the molecular formula C4H8O are possible.
a) Draw electron-dot structures of any two isomers.
Many isomers are possible. Here are some of the choices. 4 structural isomers
are shown. In each case, there are two lone pairs on the oxygen.
H H
H H O
H
H
H
H
O
H
H
H HH H
H H H
H
H
H
O
H
O
H
H
H
H
H
H H
H
H
H
H
b) Select one of the two isomers from your answer to (a) and provide the
following:
This answer is for the structure in the upper-left hand corner. The molecule is
butanal. The carbons are numbered 1, 2, 3, and 4 starting from the right. Carbon
1 is doubly bonded to the oxygen.
i) estimates of bond angles
O-C1-C2, 120; O-C1-H, 109.5; C1-C2-C3,109.5; C1-C2-H, 109.5; C2C3-C4,109.5; C3-C4-H, 109.5
ii) hybridization of all heavy (non-H) atoms
O, sp2; C1, sp2; C2, sp3; C3, sp3; C4 sp3
iii) its planarity (Do all the heavy atoms lie in the same plane?)
The sp3 hybridization on carbons 2, 3, and 4 has the consequence that the
heavy atoms are not co-planar for all conformers.
iii') revised question: What type of isomerism is illustrated by your
pair of isomers?
The 4 isomers shown above are all structural isomers. The species
shown in the upper righthand corner is E-2-buten-1-ol and could be
converted into the Z geometrical isomer by rotating the double bond
by 180 degrees.
iv) description of the bonds formed between the heavy atoms
The (O, C1) double bond has two parts:a sigma-type bond represented by
the overlap of a C sp2 orbital with an O sp2 hybridized orbital and a pitype bond represented by the overlap of C and O 2pz unhybridized
orbitals.
The (C1,C2) single bond is a sigma type bond represented by the overlap
of a C sp2 hybridized orbital with a C sp3 hybridized orbital.
The (C2,C3), and (C3,C4) single bonds are sigma type bonds represented
by the overlap of C sp3 hybridized orbitals.