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Now let’s discuss how to determine the electron configuration for an atom—in other words, how electrons are arranged in an atom. The first and most important rule to remember when attempting to determine how electrons will be arranged in the atom is Hund’s rule, which states that the most stable arrangement of electrons is that which allows the maximum number of unpaired electrons. This arrangement minimizes electron-electron repulsions. Here’s an analogy. In large families with several children, it is a luxury for each child to have his or her own room. There is far less fussing and fighting if siblings are not forced to share living quarters: the entire household experiences a lower-intensity, less-frazzled energy state. Likewise, electrons will go into available orbitals singly before beginning to pair up. All the single–occupant electrons of orbitals have parallel spins, are designated with an upward-pointing arrow, and have a magnetic 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, nNumber of sublevelsNames of sublevels

11s

22s, p

33s, p, d

44s, 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.Sublevelspdf

Number of orbitals1357

Maximum number of electrons261014

Quantum number, l0123

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 second because of electron-electron interference. It takes less energy for an electron to be placed in 4s than in 3d, so 4s fills before 3d. Notice that the middle of the periodic table contains a square of 10 columns: these are the elements in which the d orbitals are being filled (these elements are called the transition metals). Now look at the two rows of 14 elements at the bottom of the table. In these rare earth elements, the f orbitals are being filled. 　　One final note about electron configurations. You can use the periodic table to quickly determine the valence electron configuration of each element. The valence electrons are the outermost electrons in an atom—the ones that are involved in bonding. 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 art above. 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 that particular row has, which will be very helpful in determining Lewis dot structures. 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 3p4, which means everything up to that sublevel is also full, so its electron configuration is 1s22s22p63s23p4. You can check your answer—the neutral sulfur atom has 16 protons, and 16 electrons. Add up the number of electrons in your answer: 2 + 2 + 6 + 2 + 4 = 16.　　Another way of expressing this and other electron configurations is to use the symbol for the noble gas preceding the element in question, which assumes its electron configuration, and add on the additional orbitals. So sulfur, our example above, can be written [Ne] 3s23p4.　　Orbital Notation　　Orbital notation is basically just another way of expressing the electron configuration of an atom. It is very 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 into their own orbitals before electrons 8, 9, and 10 entered, forcing pairings in the 2p sublevel; the same thing happens in the 3p level.　　Now we can determine the set of quantum numbers. First, n = 3, since the valence electron (the outermost electron) is a 3p 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 3p sublevel alone. It looks like this:

Simply number the blanks with a zero assigned to the center blank, with negative numbers to the left and positive 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 the middle, negative numbers to the left, and positive numbers to the right. 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 5p 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 5p4; so it must be tellurium. 　　Example　　Complete the following table:ElementValence electron configurationValence orbital notationSet of quantum numbers

[Ar] 3d6

5, 1, 0, +1/2

4p5

6, 0, 0, -1/2

Answer: elementValence electron configurationValence orbital notationSet of quantum numbers (n, l, ml, ms)

K[Ar] 4s1

4, 0, 0, +1/2

Fe[Ar] 4s23d6

3, 2, -2, -1/2

N1s22s22p3

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