Chemistry 241-2008
Exam 1 Practice Questions
1. Calculate the binding energy for Ar-40, which has an exact mass of 39.96238 g/mol and for Na-23,
which has an exact mass of 22.98977 g/mol. Other useful masses: n = 1.008664 and H-1 = 1.00794
g/mol.
Ar-40: mass = [18(1.00794) + 22(1.008664)] - 39.9623 = 0.371 g/mol = 3.71 x 10-4 kg/mol
Relative Binding energy = 3.71 x 10-4 kg/mol x (2.998 x 108 m/s)2 /40 = 8.34 x 1011 J/mol
For 23Na, relative binding energy = 7.8 x 1011 J/mol
Ar-40 is more stable.
Which has the more stable nucleus?
2. What would the universe be like if there were no strong or weak nuclear forces?
3. Sketch a chart of the relative abundance of the elements in the universe.
a. Point out (and name) a feature that comes about because of the relative stability of nuclei.
b. Point out and name a feature that comes about because of the mechanism by which different
nuclei are formed.
4. a. Starting with H-1, show a series of nuclear reactions that might lead to the creation of O-16.
Use addition of 4 H-1 nuclei to get He-4. Then use addition of He-4 to get, Be-8, C-12, and finally O-16.
Need to note that Be-8 is unstable but long lived enough to allow formation of C-12.
b. Starting with Fe-56, show a series of reactions that might lead to Ni-64.
Use a series of n-additions followed by beta-decays to move up to the next element. On an exam you
are not expected to know which isotopes undergo beta-decay, just that some will.
5. Consider the isotope Fe-45. Do you expect this isotope to be stable? If not, what type of radioactive
decay might it undergo?
Use the island of stability graphic. This isotope is below the region of stability, meaning it has too few
neutrons. Therefore, it will undergo either positron emission or electron capture.
6. Sketch a plot of the relative stability of isotopes vs. atomic number for elements numbered 1-80.
Up from H, levels off at a maximum at Fe. Then decreases gently until the end. The end isomers are
more stable than H.
7. Why is the entire universe not made of iron?
8. Balance each of these nuclear equations:
214
82
0
1
Pb
e also called a beta particle
9. Describe the effect on average planetary temperature that will occur if there are increases in each of
the following: the solar constant, the albedo, and the IR transmission factor.
10. Calculate the temperature Jupiter would have if it experienced no greenhouse effect.
The key is to know which equation to use. In this case, the lack of greenhouse effect means you do not
use the IR transmission factor, f. So, use the below equation. I didn't give you Es for Jupiter! It is about
50 W/m2. Using that, T = 88 K.
11. Draw and explain the band structure of an n-type semiconductor.
12. What are the symmetry elements and point groups for: NH3 and CO2.
NH3 = C3v CO2 = Dh
13. Explain why Ca has a higher melting point than K.
Using the band structure for Ca and K (they use the same orbitals and so have the same band
structure). Ca has twice as many bonding orbitals filled with electrons, and thus has stronger bonding,
and thus a higher melting point.
14. Explain why O2 is paramagnetic.
View MO diagram of O2. It has two unpaired electrons.
15. Compare and contrast valence bond theory and molecular orbital theory.
16. Why is a fluorine atom smaller than a Li atom?
Fluorine and Lithium are both in Period 2. F has 9 protons and electron configuration of 1s22s22p6; Li
has 3 protons and configuration 1s22s1. Moving from Li to F, 6 protons are added to the nucleus. This
increases the pull on the electron cloud, making it smaller. F also has 6 more electrons, which all repel
one another. This has the effect of making the electron cloud larger. Within the same period, the
increased pull due to the greater nuclear charge is greater than the added electron-electron repulsion.
The net effect is increased pull and contraction of the electron cloud. This is also known as increased
effective nuclear charge.
17. Why is it that fluorine is more reactive than oxygen and more reactive than neon?
Hint: Look at orbital energies and orbital vacancies.