Now let’s discuss how to determine the electron configuration for anatom—in other words, how electrons are arranged in an atom. The firstand most important rule to remember when attempting to determine howelectrons will be arranged in the atom is
Hund’s rule,which states that the most stable arrangement of electrons is thatwhich allows the maximum number of unpaired electrons. This arrangementminimizes electron-electron repulsions. Here’s an analogy. In largefamilies with several children, it is a luxury for each child to havehis or her own room. There is far less fussing and fighting if siblingsare not forced to share living quarters: the entire house
holdexperiences a lower-intensity, less-frazzled energy state. Likewise,electrons will go into available orbitals singly before beginning topair up. All the single–occupant electrons of orbitals have parallelspins, are designated with an upward-pointing arrow, and have amagnetic spin quantum number of +1/2.
As we mentioned earlier, each principal energy level,
n, has
n sublevels. This means the first has one sublevel, the second has two, the third has three, etc. The sublevels are named
s,
p,
d, and
f.
| Energy level principal quantum number, n | Number of sublevels | Names of sublevels |
| 1 | 1 | s |
| 2 | 2 | s, p |
| 3 | 3 | s, p, d |
| 4 | 4 | s, p, d, f |
At each additional sublevel, the number of available orbitals is increased by two:
s = 1,
p = 3,
d = 5,
f = 7, and as we stated above, each orbital can hold only two electrons, which must be of opposite spin. So
s holds 2,
p holds 6 (2 electrons times the number of orbitals, which for the
p sublevel is equal to 3),
d holds 10, and
f holds 14.
| Sublevel | s | p | d | f |
| Number of orbitals | 1 | 3 | 5 | 7 |
| Maximum number of electrons | 2 | 6 | 10 | 14 |
| Quantum number, l | 0 | 1 | 2 | 3 |
We can use the periodic table to make this task easier.
Notice there are only two elements in the first
period (the first row of the periodic table); their electrons are in the first principal energy level:
n = 1. The second period (row) contains a total of eight elements, which all have two sublevels:
s and
p;
s sublevels contain two electrons when full, while
p sublevels contain six electrons when full (because
p sublevels each contain three orbitals).
The third period looks a lot like the secondbecause of electron-electron interference. It takes less energy for anelectron to be placed in 4
s than in 3
d, so 4
s fills before 3
d. Notice that the middle of the periodic table contains a square of 10 columns: these are the elements in which the
dorbitals are being filled (these elements are called the transitionmetals). Now look at the two rows of 14 elements at the bottom of thetable. In these rare earth elements, the
f orbitals are being filled.
One final note about electronconfigurations. You can use the periodic table to quickly determine thevalence electron configuration of each element. The
valence electronsare the outermost electrons in an atom—the ones that are involved inbonding. The day of the test, as soon as you get your periodic table(which comes in the test booklet), label the rows as shown in the artabove. The number at the top of each of the rows (i.e., 1A, 2A, etc.)will tell you how many valence electrons each element in thatparticular row has, which will be very helpful in determining Lewis dotstructures. More on this later.
Example
Using the periodic table, determine the electron configuration for sulfur.
Explanation
First locate sulfur in the periodic table; it is in the third period, in the
p block of elements. Count from left to right in the
p block, and you determine that sulfur’s valence electrons have an ending configuration of 3
p4, which means everything up to that sublevel is also full, so its electron configuration is 1
s22
s22
p63
s23
p4.You can check your answer—the neutral sulfur atom has 16 protons, and16 electrons. Add up the number of electrons in your answer: 2 + 2 + 6+ 2 + 4 = 16.
Another way of expressing this and otherelectron configurations is to use the symbol for the noble gaspreceding the element in question, which assumes its electronconfiguration, and add on the additional orbitals. So sulfur, ourexample above, can be written [Ne] 3
s23
p4.
Orbital Notation
Orbital notation is basically justanother way of expressing the electron configuration of an atom. It isvery useful in determining quantum numbers as well as electron pairing.The orbital notation for sulfur would be represented as follows:

Notice that electrons 5, 6, and 7 went intotheir own orbitals before electrons 8, 9, and 10 entered, forcingpairings in the 2
p sublevel; the same thing happens in the 3
p level.
Now we can determine the set of quantum numbers. First,
n = 3, since the valence electron (the outermost electron) is a 3
p electron. Next, we know that
p sublevels have an
l value of 1. We know that
ml can have a value between
l and
-l, and to get the
ml quantum number, we go back to the orbital notation for the valence electron and focus on the 3
p sublevel alone. It looks like this:

Simply number the blanks with a zeroassigned to the center blank, with negative numbers to the left andpositive to the right of the zero. The last electron was number 16 and“landed” in the first blank as a down arrow, which means its
ml = -1 and
ms = -1/2, since the electron is the second to be placed in the orbital and therefore must have a negative spin.
So, when determining
ml,just make a number line underneath the sublevel, with zero in themiddle, negative numbers to the left, and positive numbers to theright. Make as many blanks as there are orbitals for a given sublevel.For assigning
ms, the first electron placed in an orbital (the up arrow) gets the +1/2 and the second one (the down arrow) gets the -1/2.
Example
Which element has this set of quantum numbers:
n = 5,
l = 1,
ml = -1, and
ms = -1/2?
Explanation
First, think about the electron configuration:
n = 5 and
l = 1, so it must be a 5
p electron. The
ms quantum number corresponds to this orbital notation picture:

Be sure to number the blanks and realize that the -1/2 means it is a pairing electron! The element has a configuration of 5
p4; so it must be tellurium.
Example
Complete the following table:
| Element | Valence electron configuration | Valence orbital notation | Set of quantum numbers |
|
|  |
|
| [Ar] 3d6 |
|
|
|
|  |
|
|
|
| 5, 1, 0, +1/2 |
| 4p5 |
|
|
|
|
| 6, 0, 0, -1/2 |
| Answer: element | Valence electron configuration | Valence orbital notation | Set of quantum numbers (n, l, ml, ms) |
| K | [Ar] 4s1 | 无忧托福社区门户,yc:~r+xM)d
| 4, 0, 0, +1/2 |
| Fe | [Ar] 4s23d6 | 无忧托福社区门户Yl#oD!r'xG&L9d_A
| 3, 2, -2, -1/2 |
| N | 1s22s22p3 |  | 2, 1, 1, +1/2 |
| Sn | [Kr] 5s24d105p2 |  | 5, 1, 0, +1/2 |
| Br | [Ar] 4s23d104p5 |  | 4, 1, 0, -1/2 |
| Ba | [Xe] 6s2 |  | 6, 0, 0, -1/2 |