
Electron Configurations
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 electronelectron 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 lowerintensity, lessfrazzled 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 upwardpointing
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, 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 second because of
electronelectron 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
3p^{4}, which means everything
up to that sublevel is also full, so its electron configuration
is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}.
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] 3s^{2}3p^{4}.
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 m_{l} can
have a value between l and l,
and to get the m_{l} 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 m_{l} =
1 and m_{s} =
1/2, since the electron is the second to be placed in the orbital
and therefore must have a negative spin.
So, when determining m_{l},
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 m_{s},
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, m_{l} =
1, and m_{s} =
1/2?
Explanation
First, think about the electron configuration: n =
5 and l = 1, so it must be a 5p electron.
The m_{s} 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 5p^{4}; so it must be
tellurium.
Example
Complete the following table:
Element  Valence electron configuration  Valence orbital notation  Set of quantum numbers 

[Ar] 3d^{6}  
5, 1, 0, +1/2  
4p^{5}  
6, 0, 0, 1/2 
Answer: element  Valence electron configuration  Valence orbital notation  Set of quantum numbers (n, l, ml, ms) 

K  [Ar] 4s^{1}  4, 0, 0, +1/2  
Fe  [Ar] 4s23d^{6}  3, 2, 2, 1/2  
N  1s^{2}2s^{2}2p^{3}  2, 1, 1, +1/2  
Sn  [Kr] 5s^{2}4d^{10}5p^{2}  5, 1, 0, +1/2  
Br  [Ar] 4s^{2}3d^{10}4p^{5}  4, 1, 0, 1/2  
Ba  [Xe] 6s^{2}  6, 0, 0, 1/2 
