8s 8p 8d 8f 8g . . . . . .
7s 7p 7d 7f 7g . . . .
6s 6p 6d 6f 6g . .
5s 5p 5d 5f 5g
4s 4p 4d 4f
3s 3p 3d
2s 2p
1s
Chemistry 218
Name___Crib__________________________ Problem Set #1
This problem set is due by the end of class on Friday, January 21st. Please begin answering the questions
directly on this sheet and add as many additional sheets as necessary. Where an explanation is
requested, the majority of the points depend on the explanation. For quantitative problems, all
work must be shown.
1. Consider the real element with the following successive ionization energies:
Ionization Energies (MJ @ mol!1)
I II III IV VVI VII VIII IX X
0.7865 1.5771 3.2316 4.3555 16.091 19.785 23.786 29.252 33.877 38.733
To what group does this element belong? Briefly explain your reasoning.
On the basis of the relatively large increase between IE4 and IE5, it is most
likely that the fifth ionization involves a core electron rather than a valence
electron. As a result, there are four valence electrons. The group in the periodic
table whose atoms have only one valence electron is group IV (14). Therefore the
element is a member of group 14. (The ionization energies belong to silicon).
2. Element 117, which has yet to be discovered, will be a member of the halogen family. What will be
the atomic number of the halogen immediately below element 117? Explain. Deduce the full (i.e., no
shorthand notation) ground state electron configuration of the element immediately below element 117.
We can expand the normal diagonal filling order as
shown to the right. Beginning in the n = 5 shell, g
orbitals can be populated. This means that the filling
order will be
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s
5f 6d 7p 8s 5g 6f 7d 8p
The halogens all have an ns2 np5 valence shell
configuration, so the electron configuration of the
halogen below element 117 would be:
