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Size of the Atom

1 A = 1 x 10-10 m

Atomic Numbers, Mass Numbers, Isotopes

Atomic Weights

1 amu = 1.66054 x10-24 g 1 g = 6.02214 x1023 amu

Based on the measured and calculated mass of 1 Hydrogen atom

Derived from the mass of 1 Hydrogen atom – how many it would take to add up to 1 gram

Atomic Weights and Isotopes

Isotopes are just versions of atoms with the same number of protons – some heavier than others, but they are the same element.

Atomic Weight = Σ[(isotope mass) x (fractional isotope abundance)]

6[Enter Presentation Title in Header and Footer]

Periodic Table

Molecules and Molecular Compounds

Chemical Formulas: CO2 H2O C3H8

Molecular Compounds: Contain more than one type of atom – most contain non-metals

Molecular Formulas: Show the actual numbers of atomsExample: Ethane = C2H6

Empirical Formulas: Show the lowest whole number ratio of atoms – the relative amounts

Example: Ethane = CH3

Subscripts tell you how many atoms of each element are in the molecule. No subscript = 1

Structural Formulas

Structural Formula: Caffeine

Space Filling Model: Crystal Meth

Ions and Ionic Compounds

Ions and Ionic Compounds

Ions and Ionic Compounds

Common Ion Names

Memorize!

Naming Ionic and Molecular Compounds

Ionic? Metal and Non-metal

1. Cation first: If it is type II, use the correct name

2. Anion second3. Replace end of

Anion with “ide” 4. If Anion is

polyatomic, use its name

Molecular? Non-metals only

1. Least electronegative atom goes first

2. Change the ending to ide

3. Use prefixes (see page 68)

Organic Compounds and Isomers

How do we know?

hν+

-

-

hν  

Can you complete these sentences?The Bohr Model of the atom shows us…However, it doesn’t…

Be ready to share your thoughts

Tuesday 9.23.14

1. First by yourself, then with your team, use the Rydberg equation to calculate the Energy of an electron moving from n = 3 to n = 1 OR from n = 2 to n =1. Just choose one.

2. Then, using the two equations below, calculate the wavelength of of this electron. You have to be able to manipulate these equations – what do they have in common?

Equation 1 E = hv Where h is Planck’s constant = 6.626 x 10-34 Jand where v represents frequency of UV light

Equation 2 c = λvWhere again v is frequency of UV light, and λ

represents the wavelength, and c is the speed of light at 3.00 x 108 m/s

Hint: first solve for v, frequency of UV light

Bohr’s Equation: Experimental Evidence

Then, look at the emission spectra of hydrogen. Does this calculated value match any of the emission spectra peaks?

Bohr’s Equation: Experimental Evidence

1. First by yourself, then with your team, use the Rydberg equation to calculate the ΔE of an electron moving from n = 4 to n= 3

2. Then, using the two equations below, calculate the wavelength of of this electron. Show your work using dimensional analysis. The

Equation 1 E = hv Where h is Planck’s constant = 6.626 x 10-34 Jsand where v represents frequency of UV light

Equation 2 c = λvWhere again v is frequency of UV light, and λ

represents the wavelength, and c is the speed of light at 3.00 x 108 m/s

Hint: first solve for v, frequency of UV light

Warm Up: 7 mins, You might present!

Then, look at the emission spectra of hydrogen. Does this calculated value match any of the emission spectra peaks? What does this mean?

Bohr’s Equation: Experimental Evidence

- Homework: Watch Video and Do “From Shells to subshells” Assignment

- http://www.bozemanscience.com/ap-chem-004-coulombs-law

Plan for This Week and Next

Can you complete these sentences?The Bohr Model of the atom shows us…However, it doesn’t…

Be ready to share your thoughts

Tuesday 9.23.14

Atomic Orbitals

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d4f

5f

1s2s 2p3s 3p 3d4s 4p 4d 4f5s 5p 5d 5f6s 6p 6d7s 7p 8s

Further refinements to these models have occurred with new experimental results

blankenau
Jamie - not sure how you want to use this slide

3d

4d

5d

6d

4f

5f

1s1s

4s

5s

6s

7s

3s

2s

4p

5p

6p

7p

3p

2pChromium

52.00

24

Cr

63.55

CuCopper

29

[Ar]4s13d5 [Ar]4s13d10

But not all elements ‘follow the rules’

Image source: Dayah, Michael. “Dynamic Periodic Table.” Accessed Sept. 5, 2013. http://ptable.com/#Property/Ionization

Image source: http://chemistry.beloit.edu/stars/images/IEexpand.gif

Ionization Energy

Element IE1 IE2 IE3 IE4 IE5 IE6 IE7

Na 495 4,560

Mg 735 1,445 7,730

Al 580 1,815 2,740 11,600

Si 780 1,575 3,220 4,350 16,100

P 1,060 1,890 2,905 4,950 6,270 21,200

S 1,005 2,260 3,375 4,565 6,950 8,490 27,000

Cl 1,255 2,295 3,850 5,160 6,560 9,360 11,000

Ar 1,527 2,665 3,945 5,770 7,230 8,780 12,000

LO 1.5 - The student is able to explain the distribution of electrons in an atom or ion based upon data.

LO 1.6 - The student is able to analyze data relating to electron energies for patterns or relationships.

Ionization Energy: What Patterns Are Here?

𝑬=𝒉𝝂

11+

--

-

-

- -

-

-

--

-hν

Radiation Type ν E Aspects Probed

Microwaves 109 – 1011 Hz 10-7 – 10-4 MJ/mol Molecular rotations

Infrared (IR) 1011 – 1014 Hz 10-4 – 10-1 MJ/mol Molecular vibrations

Visible (ROYGBV) 4x1014 – 7.5x1014 Hz 0.2 - 0.3 MJ/mol Valence electron transitions in atoms and molecules

Ultraviolet (UV) 1014 – 1016 Hz 0.3 – 100 MJ/mol Valence electron transitions in atoms and molecules

X-ray 1016 – 1019 Hz 102 – 105 MJ/mol Core electron transitions in atoms

IE1 = 495 kJ/molIE1 = 0.495 MJ/mol

How do we probe further into the atom?

blankenau
Jamie - not sure this build is correct

11+

--

-

-

- -

-

-

--

-hν

Radiation Type ν E Aspects Probed

X-ray 1016 – 1019 Hz 102 – 105 MJ/mol Core electron transitions in atoms

Removing Core Electrons

E = 103.3 MJ/molE = 1.033 x 108 J/mol

v = E/h = (1.033 x 108 J/mol)/(6.626x10-34Js)

v = (1.559 x 1041 mol-1 s-1)x( 1 mol /6.022 x 1023 e-

vmin = 2.59 x 1017 Hz

11+

--

-

-

- -

-

-

--

-hν

Any frequency of light that is sufficient to remove electrons from the 1st shell can remove electrons from any of the other shells.

Removing Core Electrons

E1st = 103.3 MJ/mol

E2nd = 3-6 MJ/mol

PES Instrument

Image Source: SPECS GmbH, http://www.specs.de/cms/front_content.php?idart=267

3+ 3+ 3+

3+ 3+ 3+

3+ 3+ 3+

3+ 3+

3+ 3+

3+ 3+ 3+

3+ 3+ 3+3+ 3+ 3+3+ 3+

3+ 3+ 3+3+3+ 3+

3+ 3+ 3+3+3+ 3+

3+ 3+ 3+3+ 3+ 3+3+3+

3+3+ 3+

3+3+3+3+3+3+

X-ray or UV Source

Kinetic Energy AnalyzerKinetic Energy Analyzer

Binding Energy (MJ/mol)6.26 0.52

Kinetic Energy Analyzer

Negative Voltage Hemisphere

Slightly Less Negative Voltage

Hemisphere

1Volt=1  Joule1  Coulomb

1   e−=1.602 x10− 19Coulombs1  eV=1.602 x10−19 Joules

1mole of  eV=96 485 J10.364 eV=1MJ /mol

Negative Voltage Hemisphere

Positive Voltage Hemisphere

If Kinetic energy is too high…

Negative Voltage Hemisphere

Slightly Less Negative Voltage

Hemisphere

Negative Voltage Hemisphere

Positive Voltage Hemisphere

If voltage is too high…

Negative Voltage Hemisphere

Slightly Less Negative Voltage

Hemisphere

3+ 3+ 3+

3+ 3+ 3+

3+ 3+ 3+

3+ 3+

3+ 3+

3+ 3+ 3+

3+ 5+ 3+5+ 3+ 3+5+ 5+

3+ 5+ 3+5+5+ 5+

3+ 5+ 3+5+5+ 5+

3+ 3+ 3+3+ 5+ 3+5+5+

3+5+ 5+

3+3+5+3+3+5+

X-ray or UV Source

Kinetic Energy Analyzer

Binding Energy (MJ/mol)

Binding Energy (MJ/mol)19.3 0.80 1.36

Boron

6.26 0.52

Li

Kinetic Energy Analyzer

What does PES data show that emission spectra do not?

Monday 9.29.14

Chapter 2 Problem Set: Q & A The Evolving Atom: Experimental

Evidence Lecture: Bohr’s Equation and the Hydrogen Atom Lecture: Aufbau Diagrams and Ionization Energy Activity: Shells Class Activity Lecture: PES = Evidence for Subshells (Orbitals) Activity: Shells to Subshells Activity Discussion: Practice Problems• Lab: Paramagnetism and Electron Configurations

(More Evidence)

Plan for the Week

Analyzing Data from PES Experiments

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)N(C)Ne(D)Ar

Rela

tive

Num

ber o

f Ele

ctro

ns

Analyzing data from PES

Binding Energy (MJ/mol)

100 10 1

151 1.09

1s2

2p6

Rela

tive

Num

ber o

f Ele

ctro

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

Real Spectrum

11+

--

-

-

- -

-

-

- -

-hν

Auger Transitions

4

3.5

3

2.5

2

1.5

1

.5

0

Inte

nsity

(x10

5 cou

nts/

s)Real Spectrum

6

5

4

3

2

1

0

Inte

nsity

(x10

3 cou

nts/

s)Copper vs. Chromium

Jamie Benigna
You never get entirely pure samples. This Chromium has some carbon in it, so the C KLL and C 1s peaks appear. I blanked them out with a rectangle to minimize confusion.

Inte

nsity

(x10

5 cou

nts/

s)

Binding Energy (MJ/mol)

4

3.5

3

2.5

2

1.5

1

0.5

100 90 80 70 60 50 40 30 20 10 0

Mixtures of Elements

PES Sample Questions

Sample Question #1Which element could be represented by the complete PES spectrum below?

(A) Li (B) B (C) N (D) Ne

0.1110100

Binding Energy (MJ/mol)

Re

lati

ve

Nu

mb

er

of

Ele

c-

tro

ns

Sample Question #2

Which of the following best explains the relative positioning and intensity of the 2s peaks in the following spectra?

(A) Be has a greater nuclear charge than Li and more electrons in the 2s orbital(B) Be electrons experience greater electron-electron repulsions than Li electrons(C) Li has a greater pull from the nucleus on the 2s electrons, so they are harder to remove(D) Li has greater electron shielding by the 1s orbital, so the 2s electrons are easier to remove

Binding Energy (MJ/mol)14 12 10 8 6 4 2 0

Inte

nsi

ty

Be

Binding Energy (MJ/mol)14 12 10 8 6 4 2 0

Li

Inte

nsi

ty

MJ/mol

Sulfur S

1.00

16.5

2.05 22.7 239

Sample Question #3Given the photoelectron spectra above for phosphorus, P, and sulfur, S, which of the following best explains why the 2p peak for S is further to the left than the 2p peak for P, but the 3p peak for S is further to the right than the 3p peak for P?

(A) S has a greater effective nuclear charge than P, and the 3p sublevel in S has greater electron repulsions than in P.(B) S has a greater effective nuclear charge than P, and the 3p sublevel is more heavily shielded in S than in P.(C) S has a greater number of electrons than P, so the third energy level is further from the nucleus in S than in P.(D) S has a greater number of electrons than P, so the Coulombic attraction between the electron cloud and the nucleus is greater in S than in P.

MJ/mol

Phosphorus P

1.06

13.5

1.95 18.7 208

Binding Energy

Sample Question #4Looking at the complete spectra for Na and K below, which of the following would best explain the relative positioning of the 3s electrons?

Binding Energy (MJ/mol)

Inte

nsity

(c/s

)

130 105 90 75 60 45 30 15 0

Binding Energy (MJ/mol)

Inte

nsity

(c/s

)

400 350 300 250 200 150 100 50 0

Na

K

Jamie Benigna
I at first want teachers to consider the question without choices. In the script, I indicate that at the scale, it is very difficult to determine what is happening since so many peaks overlap at the righthand side. The next slide zooms in and compares the 3s sublevel for Na and K directly.
blankenau
Jamie - not sure if I have set this up properly

Sample Question #4aLooking at the spectra for Na and K below, which of the following would best explain the difference in binding energy for the 3s electrons?

Binding Energy (MJ/mol)

Inte

nsity

(c/s

)

4 3.5 3 2.5 2 1.5 1 0.5 0

Na-

3s

K-3s

(A) K has a greater nuclear charge than Na(B) K has more electron-electron repulsions than Na(C) Na has one valence electron in the 3s sublevel(D) Na has less electron shielding than K

Jamie Benigna
Updated the references in the outline. Note that with the addition of the title slide and biographical info, all slides have been bumped back by 2 from what is written in the outline.

Sample Question #4bLooking at the spectra for Na and K below, which of the following would best explain the difference in signal intensity for the 3s electrons?

(A) K has a greater nuclear charge than Na(B) K has more electron-electron repulsions than Na(C) Na has one valence electron in the 3s sublevel(D) Na has less electron shielding than K

Binding Energy (MJ/mol)

Inte

nsity

(c/s

)

4 3.5 3 2.5 2 1.5 1 0.5 0

Na-

3s

K-3s

Sample Question #5Given the photoelectron spectrum below, which of the following best explains the relative positioning of the peaks on the horizontal axis?

(A) O has more valence electrons than Ti or C, so more energy is required to remove them

(B) O has more electron-electron repulsions in the 2p sublevel than Ti and C(C) Ti atoms are present in a greater quantity than O can C in the mixture.(D) Ti has a greater nuclear charge, but the 2p sublevel experiences greater

shielding than the 1s sublevel.

Inte

nsity

(cou

nts/

s)

Image source: http://www.rsc.org/ej/JM/2010/b925409a/b925409a-f2.gif

Sample Question #6Given the photoelectron spectrum of scandium below, which of the following best explains why Scandium commonly makes a 3+ ion as opposed to a 2+ ion?

(A) Removing 3 electrons releases more energy than removing 2 electrons.

(B) Scandium is in Group 3, and atoms only lose the number of electrons that will result in a noble gas electron configuration

(C) The amount of energy required to remove an electron from the 3d sublevel is close to that for the 4s sublevel, but significantly more energy is needed to remove electrons from the 3p sublevel.

(D) Removing 2 electrons alleviates the spin-pairing repulsions in the 4s sublevel, so it is not as energetically favorable as emptying the 4s sublevel completely.

Binding Energy (MJ/mol)

Inte

nsity

(c/s

)

500 400 300 50 40 30 10 9 8 7 6 5 4 3 2 1 0

0.770.63

Jamie Benigna
I changed the language of choice C significantly on this slide. Not sure if this has to be updated elsewhere.

Binding Energy (MJ/mol)100 10 1

Inte

nsity

Hint: for additional formative assessments, use spectra from previous multiple choice questions

Example Formative AssessmentSketch the photoelectron spectrum of magnesium below, then draw the spectrum for aluminum – either on top of Mg or below it.

Quick Check – Can You Now Translate Between These Representations of Mg?

1s

2s

3s

4s

2p

3p

1s2 2s2 2p6 3s2

Binding Energy (MJ/mol)100 10 1

Inte

nsi

ty

Mg 12+

--

-

-

- -

-

-

- -

-

-

image source: http://ericsaltchemistry.blogspot.com/2010/10/jj-thomsons-experiments-with-cathode.html

http://84d1f3.medialib.glogster.com/media/f9/f9a5f2402eb205269b648b14072d9fb3a2f556367849d7feb5cfa4a8e2b3fd29/yooouu.gif

--- -

---

-

- +

+

++ +

+

++

Thomson

++++ +++

+- -

-

-

-

--

-

Rutherford

++

+++++

+

Bohr

Using Data to Makes Conclusions About Atomic Structure

Element IE1 IE2 IE3 IE4 IE5 IE6 IE7

Na 495 4560

Mg 735 1445 7730

Al 580 1815 2740 11,600

Si 780 1575 3220 4350 16,100

P 1060 1890 2905 4950 6270 21,200

S 1005 2260 3375 4565 6950 8490 27,000

Cl 1255 2295 3850 5160 6560 9360 11,000

Ar 1527 2665 3945 5770 7230 8780 12,000

Binding Energy (MJ/mol)100 10 1

151 1.09

1s2

2p6Re

lativ

e N

umbe

r of E

lect

rons

12.1

7.9

0.58

2s2 3s2 3p1

PES – Data that Shells are Divided into Subshells

63[Enter Presentation Title in Header and Footer]

Plan for Today

Questions on Packet?Lab Notebooks?Some Fun With StoichiometryWhat’s a Mole? Video(s) HW

64[Enter Presentation Title in Header and Footer]

Mole Challenge

If you lined up a mole of your shoes end to end, toe to heel, what distance would this cover? What distance is this similar to?

How many semi-trucks would it take to haul a mole of Voodoo Doughnut boxes? (Hint: you have to figure out the volume of a doughnut box, and the volume of a semi-truck container = 75.3 m3) If you lined the containers up end to end, what distance would it cover? Hint: Each container is 12.192 m long)

SP 3.2• The student can refine scientific questions

SP 3.3• The student can evaluate scientific questions

SP 6.3• The student can articulate the reasons that scientific

explanations are refined or replaced.

Applicable Science PracticesFrom the AP Chemistry Curriculum Framework:

1.5 – The student is able to explain the distribution of electrons in an atom or ion based upon data.

1.6 – The student is able to analyze data relating to electron energies for patterns and relationships.

1.7 – The student is able to describe the electronic structure of the atom, using PES data, ionization energy data, and/or Coulomb’s law to construct explanations of how the energies of electrons within shells in atoms vary.

1.8 – The student is able to explain the distribution of electrons using Coulomb’s law to analyze measured energies.

1.12 – The student is able to explain why a given set of data suggests, or does not suggest, the need to refine the atomic model from a classical shell model with the quantum mechanical model.

1.13 – Given information about a particular model of the atom, the student is able to determine if the model is consistent with specified evidence.

1.14 – The student can justify the selection of a particular type of spectroscopy to measure properties associated with vibrational or electronic motions of molecules.

Applicable Learning ObjectivesFrom the AP Chemistry Curriculum Framework:

Arizona simulated photoelectron spectrahttp://www.chem.arizona.edu/chemt/Flash/photoelectron.html

Guided inquiry activities on PES• John Gelder (Oklahoma State University) • Moog and Farrell, Chemistry: A Guided Inquiry• POGIL

Books on PES technical specs• Van der Heide, Paul. X-Ray Photoelectron Spectroscopy: An

Introduction to Principles and Practices. New Jersey: John Wiley & Sons, Inc, 2012.

• Ellis, Andrew M., Miklos Feher, and Timothy Wright. Electronic and Photoelectron Spectroscopy: Fundamentals and Case Studies. New York: Cambridge University Press, 2005.

Supporting Resources (cont.)

Supporting Resources (cont.)

Image Source: Shen Laboratory, Stanford University and SLAC National Accelerator Laboratoryhttp://arpes.stanford.edu/facilities_ssrl.html

Image source: Inna M Vishikhttp://www.stanford.edu/~ivishik/inna_vishik_files/Page452.htm

blankenau
Jamie - not sure how you wanted to use these images
Jamie Benigna
I will use this as a lead-in to the next slide

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