|
| |
Shortcuts
Shortcuts for Electron Configurations
Next, let's use the periodic table to provide a short cut method of getting the last
part of the electron configuration for nearly any element.
We will be able to determine both the outermost electron configurations and also
the electron configuration of the last-occupied sublevel.
| Let's take calcium, element number 20, as our first example. It is in the
fourth period and in the s block, so its electron configuration will end with 4s. Calcium
is in the second column of the s block, so it has two electrons in that sublevel and the
outermost electron configuration is 4s2.
(Its last-occupied sublevel configuration is also 4s2.) |
|
| Iron, element number 26, is also in the fourth period but it is in the d
block. You need to remember that 3d gets electrons after 4s. Because iron is in the sixth spot of the d block, the last
electron in the configuration is 3d6. The s electrons are also important to the
chemistry of the transition metals so they are usually included as outermost electrons.
Therefore the electron configuration ends with 4s23d6 (or
alternately 3d64s2). The
last-occupied sublevel configuration is just the 3d6
(because that is the place where the last electron is
located). |
Fourth
Period |
3d1 |
3d2 |
3d3 |
3d4 |
3d5 |
26
Fe
4s23d6 |
|
|
| For elements in the p block, count down the periods to get the energy
level of the last electrons and count over from the edge of the p block to determine how
many electrons are in the last sublevel. For example, nitrogen is in the second period and
is the third element over in the p block, so its electron configuration ends with 2p3.
If you are figuring the electron configuration for the purpose of determining the number
of valence electrons, be sure to include
electrons in the s sublevel. For nitrogen that would be 2s22p3 for a
total of 5 valence electrons. What would the
last-occupied sublevel configuration be? Right - 2p3! |
|
| Another example is iodine, element number 53. It is in the fifth period
and it is the fifth element over in the p block, so its electron configuration ends with
5p5. Again, if our concern is valence electrons we would include the s
sublevel, 5s25p5. |
Fifth
Period |
5p1 |
5p2 |
5p3 |
5p4 |
53
I
5s25p5 |
|
|
Practice
For practice, try figuring out the last orbital electron configuration for the elements
shown below (they are also listed in example 4 in your workbook). Use the periodic table
in workbook example 1 for reference. Then use the periodic table in workbook example 2 to
check your answers. That table includes the s electrons for the p block and d block
elements. You may include them or not as you wish.
Shortcuts and Complete Configurations
This short-cut method can also be used to determine the complete
electron configurations. Use the location of the element on the periodic table to figure
out the last entry in the configuration, as you have just done. Use the shape of the
periodic table to remind you of the order in which the sublevels fill. Do this by reading
the blocks in each period in order. First 1s. Then 2s, 2p. Then 3s, 3p. Then 4s, 3d, 4p.
Next 5s, 4d, 5p, and so on. Keep adding electrons to that pattern until you have reached
what you know to be the end of the electron configuration for that element based on its
location on the periodic table.
| 1s |
|
1s |
| 2s |
|
2p |
| 3s |
3p |
| 4s |
3d |
4p |
| 5s |
4d |
5p |
| 6s |
* |
5d |
6p |
| 7s |
§ |
6d |
7p |
| |
*
§ |
4f |
|
| 5f |
|
