Spectroscopic Notation
These sections will be designated by the following letters:
Group of 2:
Group of 6:
Group of 10:
Group of 14:
s
p
d
f
l=0
l=1
l=2
l=3
Why they are s, p, d, and f is unimportant to us. They are just labels of electron placement.
1
1
d
2
3
2
3
s4
5
4
5
6
6
7
7
p
f
Now what we can do is to begin to put the elements in the “seats” just like at the baseball game!
We are going to use a method of notation called spectroscopic notation or electron configuration. This is just a shorthand method of telling a reader what row, section, and number an element is on the periodic table. Here’s how it works:
Hydrogen, element #1is in the first row so we write:
1
It is in the “s” subsection so we write:
1s
It is the first place in the 1s subsection so we finish with:
1s1
Row
Section
Place
So Hydrogen has the notation:
H = 1s1
What would the next one be? Helium is next and he kind of doesn’t fit (he’s the only one). We
know there are 2 “seats” in every “s” subsection so aside from his position on the table, we sit
him next to H (no one should sit alone).
He = 1s2
Who’s next? Lithium
1
1
2
2
Lithium has 3 total electrons. The first two are just like H and He:
Li = 1s2
But the 3rd electron can’t sit with them. There isn’t room in the 1s orbital because only 2 electrons can fit (remember there are groups of 2, 6, 10, and 14). So where does that 3rd electron
go? All we have to do is look at the Periodic Table. It is obvious that electron #3 is the first
electron in the s subsection of the 2nd energy level! Thus:
Li = 1s22s1
We have now accounted for all the electrons in Li. How about Be:
Be = 1s22s2
This has now filled the 1s level with 2 electrons and the 2s level with 2 electrons. Where do we
go from here? Boron is next do we do this:
B = 1s22s3
NO!!!!!
No! We can’t do that! There are only 2 electrons that can fit in the “s” subsection. Besides,
look at where B sits on the table. He’s not in the “s” section to the left. He’s in the “p” section
to the right. He’s sitting in the 1st seat of the “p” section in the 2nd row. Consequently:
B = 1s22s3
B = 1s22s22p1
Let’s see what all of this looks like together and add in the rest of the elements from that row:
H = 1s1
He = 1s2
Li = 1s22s1
Be = 1s22s2
B = 1s22s22p1
C = 1s22s22p2
N = 1s22s22p3
O = 1s22s22p4
F = 1s22s22p5
Ne = 1s22s22p6
What happens next? The same thing except it will be at the 3rd energy level:
1
1
2
2
3
3
Looking at the elements in that row we get:
Na = 1s22s22p63s1
Mg = 1s22s22p63s2
Al = 1s22s22p63s23p1
Si = 1s22s22p63s23p2
P = 1s22s22p63s23p3
S = 1s22s22p63s23p4
Cl = 1s22s22p63s23p5
Ar = 1s22s22p63s23p6
And it’s right about now that we begin to run into some problems. The first problem is that the
third row doesn’t come through on its promises. Remember that our energy levels and so our
rows should have:
1st Energy Level:
2
2nd Energy Level:
2 and 6
3rd Energy Level:
2 and 6 and 10
4th Energy Level:
2 and 6 and 10 and 14
But if we examine the third energy level above, we don’t have a group of 10. The group of 10
doesn’t start until the 4th energy level:
4
4
The first group of 10 (the d subshell) should be in the 3rd row, not in the 4th row.
http://image.tutorvista.com/content/atomic-structure/energy-level-diagram-multi
-electron-atom.jpeg
What’s going on here? It turns out that electrons are slightly more complicated than we
thought. These electron levels tend to overlap in complicated ways:
As the atoms get bigger and there are more
electrons, the subshells tend to overlap.
The consequence of this for our purposes is surprisingly simple. Every “d” subshell is actually
one less than the row it is in.
4
4th row but the 3d
5
6
7
5th row but the 4d
6th row but the 5d
7th row but the 6d
Anything that is in the “f” subshell is actually 2 less than the row it is in:
6th row but the 4f
6
55
Cs
132.9
56
Ba
137.3
57
La
138.9
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.2
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
7
87
Fr
(223)
88
Ra
(226)
89
Ac
227.0
90
Th
232.0
91
Pa
231.0
92
U
238.0
93
Np
237.0
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
7th row but the 5f
When it’s all put together we get this. Notice the s and p levels are the same as the row. The d
levels are one less than the row and the f levels are 2 less than the row they are in.
http://www.angelo.edu/faculty/kboudrea/periodic/config_chart.gif
To put it on a more conventional Periodic Table we get:
P
orbitals
S
orbitals
D
orbitals
F
orbitals
Here is another representation of the Periodic Table in order from left to right. The valence charges are all out of
order but it does show the progression of atomic orbitals and subshells quite nicely!
Knowing this, let’s look at the spectroscopic notation of the elements in the 4th row:
4
4
K:
Ca:
Sc:
Ti:
V:
Cr:
Mn:
Fe:
Co:
Ni:
Cu:
Zn:
Ga:
Ge:
As:
Se:
Br:
Kr:
1s22s23s23p64s1
1s22s23s23p64s2
1s22s23s23p64s23d1
1s22s23s23p64s23d2
1s22s23s23p64s23d3
1s22s23s23p64s23d4
1s22s23s23p64s23d
1s22s23s23p64s23d6
1s22s23s23p64s23d7
1s22s23s23p64s23d8
1s22s23s23p64s23d9
1s22s23s23p64s23d10
1s22s23s23p64s23d104p1
1s22s23s23p64s23d104p2
1s22s23s23p64s23d104p3
1s22s23s23p64s23d104p4
1s22s23s23p64s23d104p5
1s22s23s23p64s23d104p6
And we could keep going. However, this leads us to the our next problem. This is getting
REALLY monotonous. What’s going to happen when we have to write out the notation for
something big, like Uranium? Do we really have to write out the notation for all 92 electrons?
That sounds awful.
We don’t have to if we apply what we already know about the elements. Remember that all
elements are trying to become stable. Most of these elements are going to be stable when they
reach a number of electrons that are equal to one of our Noble Gases. The reason is that Noble
Gases have stable electron configurations. Once electrons reach those configurations, they
don’t change.
Let’s see what this has to do with examples like the Alkali Metals, group 1A. If we were to
write out the spectroscopic notation of the Alkali metals we would get:
Li: 1s22s1
Na: 1s22s22p63s1
K: 1s22s22p63s23p64s1
Rb: 1s22s22p63s23p64s23d104p65s1
Cs: 1s22s22p63s23p64s23d104p65s24d105p66s1
2 2
Fr: 1s 2s 2p63s23p64s23d104p65s24d105p66s24f145d106p67s1
Do you see the trend? Every one of these elements ends with only 1 electron in the valence or
outermost level. Look again:
Li: 1s22s1
Na: 1s22s22p63s1
K: 1s22s22p63s23p64s1
Rb: 1s22s22p63s23p64s23d104p65s1
Cs: 1s22s22p63s23p64s23d104p65s24d105p66s1
Fr: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s1
Every one of these elements ends in s1. Additionally, everything before the valence shell (what
is in bold in these examples) is the electron configuration of a Noble Gas and so therefore stable. Instead of writing out all of the notation, let’s just focus on the electrons that matter, the
valence electrons; the electrons in the outermost energy level.
If we were to write out the electron configurations of the Noble Gases, here is what we’d get:
He: 1s2
Ne: 1s22s22p6
Ar: 1s22s22p63s23p6
Kr: 1s22s22p63s23p64s23d104p6
Xe: 1s22s22p63s23p64s23d104p65s24d105p6
2 2
Rn: 1s 2s 2p63s23p64s23d104p65s24d105p66s24f145d106p6
Note that these are the same as everything above until the valence electrons:
Li: 1s22s1
Na: 1s22s22p63s1
K: 1s22s22p63s23p64s1
Rb: 1s22s22p63s23p64s23d104p65s1
Cs: 1s22s22p63s23p64s23d104p65s24d105p66s1
2 2
Fr: 1s 2s 2p63s23p64s23d104p65s24d105p66s24f145d106p67s1
Instead of writing all of this out, let’s replace the inner electron configuration with that of the
Noble Gas. Those electrons aren’t going to do anything anyway because they are stable. If we
do that, these elements become a lot simpler to manage:
Li:
Na:
K:
Rb:
Cs:
Fr:
[He] 2s1
[Ne] 3s1
[Ar] 4s1
[Kr] 5s1
[Xe] 6s1
[Rn] 7s1
This also helps us focus only on the electrons that matter, the valence or outermost electrons.
If we did the same thing with the next column, the Alkaline Earth metals we would get:
Be: [He] 2s2
Mg: [Ne] 3s2
Ca: [Ar] 4s2
Sr: [Kr] 5s2
Ba: [Xe] 6s2
Ra: [Rn] 7s2
We can now see that the Alkaline Earth metals have 2 valence electrons. Continuing this idea,
it becomes obvious to us that the number of valence electrons is simply equal to the number of
the column the element is in (ignoring the f and d subshells, mind you). It is also going to help
us know the charge as we learned before:
Valence Electrons
Charge
1
2
+1 +2
3
4 5
+3 +/-4 -3
6
-2
7
-1
8
0
We are going to see this idea come up again next unit as it is tremendously important.