Spectroscopy for AP Chemistry

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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

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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. Electron Configuration. - PowerPoint PPT Presentation

Transcript of Spectroscopy for AP Chemistry

Page 1: Spectroscopy for AP Chemistry

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

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Electron ConfigurationIs there any direct evidence that this diagram is accurately showing potential energy of electrons on the atom?

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Photoelectron Spectroscopy

Atom

e-

Ephoton = hv

KE = mv2

2IEelectron = Ephoton - KE

MonochromaticBeam of X-Rays

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Photoelectron Spectrum

20 MJ/mol 10 MJ/mol 0 MJ/mol

Relative Intensity = 1

Relative Intensity = 2Each 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.

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Photoelectron Spectrum

20 MJ/mol 10 MJ/mol 0 MJ/mol

19.3 MJ/mol 1.36 MJ/mol

0.80 MJ/mol

Valence

RI = 2 RI = 2

RI = 1Gap is due to increased energy of the orbital (decreased amount of energy to remove the electron)

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Photoelectron Spectrum

20 MJ/mol 10 MJ/mol 0 MJ/mol

19.3 MJ/mol 1.36 MJ/mol

0.80 MJ/mol

RI = 2 RI = 2

RI = 1

Analysis:1)Valence has 2 values: sp2)RI is 2 to 1 in valence: s2p1 3)Closest core has RI 2 not 6: not pxs2p1

4)s2s2p1 must be 1s22s22p1

Boron (Z=5)

1s2 2s2 2p1

Inner orbitals require the most energy

Valence orbitals require the least energy

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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.

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Photoelectron Spectrum

104 MJ/mol 6.84 MJ/mol0.50 MJ/mol

8 MJ/mol 4 MJ/mol 0 MJ/mol

Sodium (Z=11)2p6

2s2

3s1

{ }

3.67 MJ/mol

1s2

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http://www.chem.arizona.edu/chemt/Flash/photoelectron.html

https://www.youtube.com/watch?v=NRIqXeY1R_I

https://www.youtube.com/watch?v=vANbxozsRSA

Online PES Resources

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Assign sublevels to the peaks, and identify the element

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From the AP Sample Questions…

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From the AP Sample Questions…

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Which peaks in the photoelectron spectrum are representative of the binding energy of p orbital electrons?a.C only c. C and Eb.D only d. B, C and D

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

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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 shellc. core electrons of an atom experience a much higher effective nuclear charge than

valence electronsd. peaks B and C show first ionization energies of electrons in, whereas peak A shows the

second ionization energy of neon electrons

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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 levelb. electrons represented by peak B have approximately triple the binding energy than those represented by peak Cc. in a photoelectron spectrum, as binding energy increases the relative number of electrons decreasesd. the height of peaks in a photoelectron spectrum does not have any relation to the structure of an atom

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Binding Energy (MJ/mol) 90 80 70 60 50 40 30 20 10 0

84.0 4.7

2.0

1s

2s

2p

Which of the following elements might this spectrum represent?

(A) He (B) Ne(C) N (D) Ar

Rela

tive

Num

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f El

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Analyzing data from PES

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Binding Energy (MJ/mol)100 10 1

151 1.09

1s2

2p6

Rela

tive

Num

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f El

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ns

12.1

7.9

0.58

Given the spectrum above, identify the element and its electron configuration:

2s2

3s2

3p1

(A) B (B) Al (C) Si (D) Na

Analyzing data from PES

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Real Spectra

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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

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Mass Spectrometry

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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?

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From the AP Sample Questions…

a. Te forms ions with a -2 charge, whereas I forms ions with a -1 charge.

b. Te is more abundant that I in the universe.c. I consists of only one naturally occurring isotope with 74 neutrons,

whereas Te has more than one isotope.d. I has a higher first ionization energy than Te does.

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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 Yc. peak C comes from the most abundant isotope of atom Yd. peak D comes from an atom with 4 more protons than the atom that gave peak B

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