1. Spectroscopy for AP Chemistry
Photoelectron Spectroscopy (PES)
Provides data for ionization energy trends and applications
Mass Spectrometry
Provides atomic/molar mass data as it ionizes

2. Electron Configuration
Is there any direct evidence that this diagram is accurately showing potential energy of electrons on the atom?

3. Photoelectron Spectroscopy
Atom
Ephoton = hv
Monochromatic
Beam of X-Rays
KE = mv2 2 e-
IEelectron = Ephoton - KE

5. Photoelectron Spectrum
Each peak is relative to the others. This indicates the relative number of electrons. If the peak is twice as big, there are twice as many electrons.
Relative Intensity = 2
Relative Intensity = 1
20 MJ/mol
10 MJ/mol
0 MJ/mol

6. Photoelectron Spectrum
Valence
19.3 MJ/mol
1.36 MJ/mol
RI = 2
RI = 2
Gap is due to increased energy of the orbital (decreased amount of energy to remove the electron)
0.80 MJ/mol
RI = 1
20 MJ/mol
10 MJ/mol
0 MJ/mol

7. Photoelectron Spectrum
Boron (Z=5)
19.3 MJ/mol
1.36 MJ/mol
Analysis:
Valence has 2 values: sp
RI is 2 to 1 in valence: s2p1
Closest core has RI 2 not 6: not pxs2p1
s2s2p1 must be 1s22s22p1
RI = 2
RI = 2
0.80 MJ/mol
RI = 1
1s2
2s2
2p1
20 MJ/mol
10 MJ/mol
0 MJ/mol
Inner orbitals require the most energy
Valence orbitals require the least energy

8. Photoelectron Spectrum
Depending on the size of the table, 1s may be intentionally cut out of view because it’s too far away and makes the graph too long
Remember IE is about REMOVING electrons, which means they are removed from the OUTSIDE to the INSIDE, and NOT in reverse order of energy! For example, 4s is removed BEFORE 3d.

9. Photoelectron Spectrum
Sodium (Z=11)
3.67 MJ/mol
1s2
2s2
3s1
104 MJ/mol
6.84 MJ/mol
0.50 MJ/mol
{ }
8 MJ/mol
4 MJ/mol
0 MJ/mol

10. Online PES Resources

11. Assign sublevels to the peaks, and identify the element

12. From the AP Sample Questions…

13. From the AP Sample Questions…
D – Ionization energy is the energy required to remove an electron from the attraction of the nucleus. Since this requires less energy, there must be less nuclear charge.

14. Which peaks in the photoelectron spectrum are representative of the binding energy of p orbital electrons?
C only c. C and E
D only d. B, C and D
C – A, B, and D are s orbitals because they all have 2 relative electrons, and the p orbital comes after s orbitals because it takes less energy to remove them.

15. Peaks A, B, and C represent the binding energies of electrons in which subshells of neon?
a. 1s,2s,2p b. 2p,2s,1s
c. 1s,1s,1s d. 2s,2p,2p

16. 2. Which of the following statements best accounts for peak A being far to the left of peaks B and C:
a. the electron configuration of neon is 1s2 2s2 2p4
b. neon has 8 electrons located in its valence shell
c. core electrons of an atom experience a much higher effective nuclear charge than valence electrons
d. peaks B and C show first ionization energies of electrons in, whereas peak A shows the second ionization energy of neon electrons

17. 3. Which of the following statements best accounts for peak C being three times the height of peak B:
a. the intensity of the photoelectron signal at a given energy is a measure of the number of electrons in that energy level
b. electrons represented by peak B have approximately triple the binding energy than those represented by peak C
c. in a photoelectron spectrum, as binding energy increases the relative number of electrons decreases
d. the height of peaks in a photoelectron spectrum does not have any relation to the structure of an atom

18. Analyzing data from PES
2.0 1s 2s
Relative Number of Electrons
84.0
4.7 +
Binding Energy (MJ/mol)
Which of the following elements might this spectrum represent?
He (B) Ne
(C) N (D) Ar
Before we get into the question that accompanies this data, let me point out a couple of features of this simulated spectrum. This spectrum is idealized, so a lot of the background noise and the other transitions within the atom have been removed so that students can focus just on the peaks that are important for establishing the subshell model.
If you look at the X axis here, you'll notice that it is in descending order. And we've reversed the X axis on purpose. This is a convention that the Test Development Committee chose to use since students can imagine that the nucleus would lie at the left-hand side of the axis right at the origin. And then if they move further to the right, then they can see that the orbitals, on average, are further from the nucleus and have less energy required to remove electrons from those orbital.
The 1S orbital then would be closest to the origin since it requires the most energy to remove and is closest to the nucleus, on average. Then we've got the 2S orbital further away, and then the 2P orbital, which requires even less energy to remove than the 2S.
Now, when we look at average distance, this is not a true direct relationship because there are greater penetrating ability of the S orbital compared to the P, but you can at least imagine first energy level, second energy level, and third. And reason out the best based on the energy.
So looking at this energy diagram, which of the following elements are most likely represented by this spectrum?
So if you look at the relative number of electrons on these peaks, you can see the amount in the 1S orbital and the 2S orbital are about the same. And then the 2P orbital appears to have about three times more electrons than the 2S orbital does. So this would be indicative of neon if we were to try to sign the electron configuration based on the data that was collected here.

19. Analyzing data from PES
7.9
1s2
2s2
3s2
3p1
Relative Number of Electrons
151
12.1
1.09
0.58
Binding Energy (MJ/mol)
Given the spectrum above, identify the element and its electron configuration:
B (B) Al
(C) Si (D) Na
Looking at the spectrum, could you assign an electron configuration for this element and identify the element as well? You could present this either as a free response question or as a multiple choice question, but we'll treat it as a multiple choice question here.
So looking at these four choices, if you go through the spectrum and analyze which sublevels are present, you can assign the number of electrons based on peak height. Doing all of that, if we look at the electron arrangements and break down these primary energy levels into the sublevels that are present and if we look at peak height and reason out proportionality of heights to number of electrons, we should be able to arrive at the electron configuration of aluminum.
Now, one note here about the axes. When you run a pure sample, the peak heights -- or really it's the area under the curve -- is proportionate to the number of electrons that have that particular binding energy. So if you look at spectra online and in published books and activities, you may see them label the axes as relative number of electrons, as I've done here.
And that's because the signal intensity is proportionate, for the most part, to the number of electrons as long as you're dealing with a pure sample. The only thing that will change the relative number of electrons is the number of electrons within each sublevel of that atom. But again, this only really works for pure samples. It does not work on mixtures.

20. Real Spectra

21. Mass Spectrometry Mass spectrometry gives the mass to charge ratio
Like PES, the relative size of the peaks indicates the relative number of particles
Separates isotopes according to mass
Used to find relative abundance and atomic/molar mass of unknown samples

22. Mass Spectrometry

23. From the AP Sample Questions…
The elements I and Te have similar average atomic masses. A sample that was believed to be a mixture of I and Te was run through a mass spectrometer, resulting in the data above. All of the following statements are true. Which one would be the best basis for concluding that the sample was pure Te?

24. From the AP Sample Questions…
Te forms ions with a -2 charge, whereas I forms ions with a -1 charge.
Te is more abundant that I in the universe.
I consists of only one naturally occurring isotope with 74 neutrons, whereas Te has more than one isotope.
I has a higher first ionization energy than Te does.
C – Since there is not a peak at 127 (the most common isotope for I), there can not be I in the sample.

25. b. there are seven isotopes of atom Y
Based on the mass spectrum of atom Y, which of the following statements is false?
a. peak A and peak D come from atoms that have the same number of electrons
b. there are seven isotopes of atom Y
c. peak C comes from the most abundant isotope of atom Y
d. peak D comes from an atom with 4 more protons than the atom that gave peak B
D – protons must be the same for atoms of the same element; neutrons make the difference on the spectrum assuming this is a single element