Please turn in on separate sheets, stapled together
Problem Set 1
(Due beginning of class, Friday, July 1, 2011)
1. a. Based on periodic trends, would you expect Cl
to be a better
By “better”, we mean “stronger”, and certainly Cl
is the stronger oxidant. Chlorine is
nearer the top of the halogens column in the periodic table, and would thus more readily
accept and stabilize an electron than iodine would.
b. Based on periodic trends, would you expect K or Li to be a better
Again, by “better”, we mean “stronger”, and potassium is the stronger reductant.
Potassium is nearer the bottom of the alkali metals column in the periodic table, and
would thus more readily donate an electron than lithium would.
c. Based on periodic trends, would you expect B or C to have a higher
By “ionization potential”, we mean “ionization energy”, and carbon should have a higher
one than B, if only because C will have a higher Z
than B due to its larger nuclear
charge and same number of core electrons.
2. Rationalize the following values for successive ionization potentials of
IP(n) corresponds to the energy required to remove the nth electron.
IP(n) (MJ mol
Definitely a jump in ionization energy from n = 3 to n = 4, as might be expected for
boron, which has an electron configuration of (1s)
. By removing the three
electrons in the incomplete second shell, boron becomes isoelectronic with helium, and
thus gains the stability of having only complete electronic shells. Two subtle points: the
first electron removed is from the 2p orbital, being unpaired, and thus is a lower energy
cost to remove. The last electron removed (the last 1s electron) will be a very high energy
cost to remove because it is attracted by all 5 protons in the nucleus with no screening by
any other electrons.
3. Explain the decrease in
first ionization energy
(equivalent to the ionization potential)
for oxygen versus nitrogen in the table of electronic properties found in Resource Section
2 of Atkins (p. 785).
The electron configuration of oxygen is (1s)