NAME:______SOLUTIONS_____Chem 1b, 2007, 1st. Midterm Exam
The insert contains important data and the fundamental constants. You may use its back
for scratch work but enter ALL work to be graded in the space provided with each
question. Show your work in questions involving calculations in order to receive credit.
1) (25 points) Sodium, atomic number 11, has a single stable isotope with a mass of
22.98997700 amu.
a) The nucleus of this isotope has a spin, s, of 3/2. What are the allowed values of
the ms quantum number?
mss-ms in integer steps so ms = -3/2,-1/2,1/2,3/2
b) Elemental sodium is paramagnetic and sodium chloride is diamagnetic.
Provide a brief explanation for this difference in magnetic properties.
The configuration of Na is [Ne]3s yielding one electron pair which yields
paramagnetism. In contrast, Na+ and Cl- , the species in NaCl, have closed-shell
configurations, e.g,. [Ne] and [Ar], respectively where the electrons are paired.
The pairing of the electrons cancels the magnetism from the spin so the species
are diamagnetic.
c) A sample of sodium is irradiated by a beam of neutrons. Write a balanced
equation for the nuclear reaction.
n +
23
Na  24Na
d) X, the product of the nuclear reaction in part (a), is an unstable isotope. Write
a balanced equation for its decay.
24
Na has an excess of neutrons and beta decay is expected with an internal decay
of one of its neutrons. 24Na  24Mg+ + e- + 
e) The unstable isotope X has a half life of 14.96 hour. Disposal of a sample of
irradiated sodium is possible when the radiation has decayed to 1.00% of the level
present immediately after neutron activation. Calculate the minimum storage
time required before safe disposal is possible.
f = N/N0 = exp(-kt) where k = ln(2)/ so t = - (1/k)ln[N/N0]
 = 14.96 h so k = ln(2)/(14.96 h) = 4.63 x 10-2 h-1.
f = 1/100 so t = -(1/(0.0463 h-1)(ln[1/100]) = 99.4 h.
2) (38 points) Atomic physicists generated a beam of Be+3 ions in an excited state.
a) They analyzed the light emitted by the ions and found that all the emitted
photons had the same wavelength. What is the principal quantum number n of
the excited ions. Briefly defend your answer.
The ion is a one-electron species and the energy depends only on n. The excited
state must be the first excited state with n =2 for which there is only one emission
pathway, n=21. For the other, higher excited states, there are more pathways.
For example, if n were 3, three wavelengths would characterize the emissions
from a population of states, e.g. the n=31, n=32, and n=21 transitions
.
b) Calculate the ionization energy of one of the excited Be+3 ions.
From Koopman’s theorem, IE = -E = -(-Z2/n2) = (13.6 eV)(42/22) = 54.4 eV.
Multiplying this result by F, one obtains 5249 kJ.mole.
c) Calculate the average distance from the nucleus of the electron in one of the
Be+3 ions.
<r>  a0(n2/Z) = (0.529 Å)(22/4) = 0.529 Å
d) Does the Pauli Exclusion Principle apply to the Be+3 ion? Explain.
No. there is only one electron and the Pauli principle only applies if there is more
than one identical, indistinguishable particle. The Pauli principle does apply to
the nucleus where there are more than one indistinguishable, identical particles.
d) Pictures of several atomic orbitals are shown below. Which orbitals might be
possible for the excited Be+3 ion? Explain.
For n = 2 there are two values of l, l = 0 and 1. The corresponding angular wave
functions for these 2s and 2p states have no and one angular nodes, respectively.
Orbital (iii) with no nodes and orbital (i) with one angular node, a plane, are
acceptable. Orbitals (ii) and (iv) with two and three nodes, respectively, are
unacceptable.
3) (25 points) The insert provides values of ionization energies for all known group 2
elements. Discuss how the vertical and horizontal trends in the data support the shell
model for the atom.
An application of Aufbau leads to the electronic configuration [IG] (ns)2 (IG stands for
one of the inert gases, group 18) for the alkaline-earth metals. The two ns electrons are
valence electrons with high n (high compared with the core) and a low Z(effective) close
to 2. The first two IE’s are expected to be low as they are. Further ionization requires
removal of core electrons with lower n and higher Z(effective) and from the Bohr
formula greatly increased IE’s. This core effect is also clearly shown by the data.
An examination of vertical trends, e.g. 1IE across the family, is also instructive. The core
electrons serve as a screen for the two valence electrons, resulting in Z(effective) close to
2, independent of the actual charge on the nucleus. As one goes down the column, n for
the two valence electrons increases from 2 to 7 and from its dependence on 1/n2, the
ionization energy is expected to decrease. A monotonic decrease of IE with n is
observed. However, radium behaves anomalously and does not follow the trend. Its
behavior can be attributed to relativistic effects. With radium, Z is quite large (Z = 88)
and the average speed of the core electrons approaches the speed of light. Special
relativity makes contributions to the IE of radium. Note that these trends and the
anomalous value for radium is also shown by the second ionization enerby.
4) (12 points) In an accurate determination of the binding energy of a core 1s electron in
sodium, a spectroscopist irradiates sodium vapor with X-ray radiation having a
wavelength of 9.890 Å (0.9890 nm) and observes that the liberated photoelectron has a
kinetic energy of 2.919 x 10-17 J. Calculate the binding energy of the 1s electron.
E = BE + KE(e-) so BE = E - KE(e-).
E = h = hc/ = (6.626 x 10-34 J-s)(2.998 x 108 m/s)/(9.980 x 10-10 m) = 2.009 x 10-16 J.
BE = (2.009 x 10-16 J) - (0.2909 x 10-16 J) = 1.717 x 10-16 J/atom.
INSERT PROVIDED WITH THE EXAMINATION
VALUES OF FUNDAMENTAL CONSTANTS
h = 6.626 x 10-34 J-s
c = 2.998 x 108 m/s
e = 1.602 x 10-19 C
NA = 6.022 x 1023 molecules/mole
me = 9.109 x 10-31 Kg
F = NAe = 96485 C/mole
1 Å = 10-8 cm
nano = 10-9
SUCCESSIVE IONIZATION ENERGIES (IE) IN ELECTRON VOLTS (eV)
Element
1IE
2IE
3IE
4IE
Be
9.322
18.211
153.897 217.719
Mg
7.646
15.035
80.144 109.266
Ca
6.113
11.871
50.913 67.27
Sr
5.695
11.030
42.89
57
Ba
5.211
10.003
37.3
?
Ra
5.279
10.147
?
?