Atomic Structure / Electron Configuration / Periodicity Test Review
1. Subatomic Particle Experiments – What were the significance and contributions to Atomic Structure
A) Cathode Ray/J.J. Thompson –
B) Millikin Oil Drop –
C) Gold Foil/Earnest Rutherford
D) Moseley K,L waves-
E) Chadwick’s discovery
2. Quantitative Analysis techniques – What are the concepts that these techniques utilize?
A)Mass spectroscopy (isotopes)-
B) PES – Photon Emission Spectroscopy
C) Spectrophotometry – – Why are we able to use this technique for certain
3. Contributions to the development of the Quantum Mechanical Model of the atom –
A) Max Plank
B) Einstein – photoelectric effect
C) Neills Bohr
D) De Broglie
F ) Heisenberg/Max Born (Copenhagen Interpretation of Quantum Mechanics_
4. Light Theories –
A) Positive theory of light – Emission Spectra
B) Negative theory of light – Absorption Spectra
5. Math questions –
| Weighted Mass Calculations:
|Weighted Mass calculations from mass spectrometers – isotopes
|| Atomic Structure – Mass Spectroscopy .pdf
|| Atomic Structure – Mass Spectroscopy Key.pdf View Download
|Wavelength, Frequency, Energy Calculations:
| wavelength, frequency and Energy calculations (side 1 of this worksheet)
|| Atomic structure 2 – bohrs.pdf
|atomic structure 2 – Bohr Key p.pdf
| Grodski review video of the front side of the worksheet above.
|wavelength, frequency and Energy calculations (side 1 of this worksheet)
atomic structure – 1 – photoelectric effect.pdf
| atomic structure – 1 – photoelectric effect KEY.pdf
| Grodski lecture
(side 1 )
|Grodski review video of the worksheet above. (side 1)
questions 1 – 3 are like the questions of the atomic structure 2 worksheet. Question 4 starts the photoelectric effect.
| Photoelectric Effect Calculations:
| I review the concepts that connect up to and including the photoelectric effect.
| Question 4 of atomic structure 2
Link at the right will start right at question 4 to review the question.
| atomic structure 1 – photoelectric effect.pdf
| Photoelectric Effect –
side 2, question 5,6
|Grodski review video of the
atomic structure – 1 – photoelectric effect.pdf (side 2)
| Question 1 of the de Broglie worksheet is a photoelectric effect
*there will be no de Broglie calculation but you do need to know his contributions
slide 85 – 90 in Atomic structure 2 presentation
| Atomic structure 3 – de Broglie.pdf
| atomic structure 3 – de Broglie Key.pdf
| Bohr Calculations-
| SIDE 2 of this worksheet
|| Atomic structure 2 – bohrs.pdf
|atomic structure 2 – Bohr Key p.pdf
| Grodski Lecture –
Reviews these concepts leading to and including the calculation using the Balmer equation for the Bohr model. Video also reviews side 2 of the atomic structure 2 worksheet.
| Question 2 –
|| Atomic structure 3 – de Broglie.pdf
|atomic structure 2 – Bohr Key p.pdf
*Important terms – Aufbau Principle, Pauli Exclusion Principle, Hund’s Rule
READ 285 – 311 – BLUE BOOK which the 3 last periodicity forms where based on:
AP Periodicity and Electron Configurations Form 1 – graded and returned – Link to view form
AP Periodicity and Electron Configurations Form 2 – graded and returned – Link to view form
AP Periodicity and Electron Configurations Form 3 – graded and returned – Link to view form
Terms and concepts:
ground state, excited state, valence electrons, core (kernel electrons), paramagnestism, diamagnetism, Ionization Energy, Electron Affinity, Photon Emission Spectroscopy (PES)
Electron configuration worksheets:
Main Presentation for this unit:
Read NOTES from week 10 !
*Connections – When we write electron configurations we are really writing the energy levels of each different electron defines by the 4 unique quantum numbers (3 of which are solutions to the Schrodinger equation).
Pauli Exclusion Principle – Electrons must have a set of 4 unique quantum numbers.
The implication of this work was to fully understand why the electron “shells” or principle energy levels held even number of electrons. The principle also extended to explain why electrons cannot occupy the same quantum state and thus must “stack” in the atom. This “stacking” repeats and results in chemicals having different chemical properties based on their valence electrons – outermost electrons.
Kernel = Core electrons = Most stable electrons
Valence electrons = Unstable electrons – used in chemical bonding
Electrons fill a groundstate (stable lowest energy) atom by following the Aufbau Principle. If they gain energy as in the case of photon of specific energy they will move away from the nucleus and into an excited state – (unstable high energy) state. This state is identified by an electron configuration that is not following the Aufbau Principle and thus not filling a lower energy orbitals (closer to the nucleus) first before filling higher energy orbitals (farther away from the nucleus). The excited state configuration represent the electron arrangement of an atom just before it emits the photons in a bright line spectrum
Using orbital diagrams (boxes for the shapes of orbitals) we can also identify excited state configurations:
Remember that electrons in the same type of orbital will occupy empty orbitals first before pairing up – HUNDS RULE.
*Notice the electrons in the same box (orbital) have the opposite electron spins while the electrons in the same sublevel have parallel spins (going in the same direction). These unpaired electrons lead to paramagnetism (the attraction to a magnetic field).
Important Vocabulary that you will need in this unit:
Z = #of protons synonyms: Z = nuclear charge and Z = Atomic Number (Thanks Moseley!)
Zeff = effective nuclear charge – (the nuclear charge that the electron feels as a result of electron – electron interactions ( screening or electron – electron repulsions).
n = principle energy level, the larger the n the larger the number of core electrons
and larger the orbitals. n defines the proximity of electrons to the
nucleus. The farther that an electron is from the nucleus the lower the
coulombic attractions that the electron feels and thus is less stable
than electrons closer to the nucleus.
Armed with Z, Zeff, and n you can explain almost everything in periodicity and electron
****Since we are continuously evaluating the energy levels of electrons that are bound in a atom or ion in this unit Ionization energy values are very helpful in determining stability of an electron.
Ionization Energy = the energy needed to remove an electron (Einstein’s Binding Energy).
Electrons with higher IE are more stable (takes more energy to remove!)
Electrons with lower IE are less stable (takes less energy to remove!)
Example for Na (sodium): IE1 + Na —-> Na+ + e-
first Ionization Energy
Ionization Energy is often described as the First Ionization Energy (1st IE) or the Second Ionization Energy (2nd IE) and so on…
IE2 + Na+ —-> Na+2 + e-
Second Ionization Energy
So the 2nd IE is the energy needed to remove the second electron. Would it require the same amount as the 1st IE? No it would require much more because Na+ the second electron would be removed from a filled principle energy level!! These are core electrons that are more stable. Do not lose site that IE is a measure of electron stability.
Stable Electrons = High coulombic attraction to nucleus = lower energy orbitals = High IE
Lower energy Lower n, Higher Zeff closer to the nucleus
Successive IE values have verified the existence of Valence electrons!!! Look at the diagram below.
Notice when a successive IE “JUMPS THROUGH THE ROOF”.
Na (atom) : 1s22s22p63s1 Na+1 (atom): 1s22s22p6
Removing valence electron (less stable) Removing a core electron (more stable)
3s electron has higher n 2p electron has lower n
3s electron has lower Zeff 2p electron has higher Zeff
Z = 11 Z = 11
IE1 = 500 kJ/mol —— 9 x increase——-> IE2 = 4560 kJ/mole
removing valence e– “jumps through the roof” removing core e–
Thus Na has 1 valence electron
Ionization Energy vs. Electron affinity
| Ionization Energy (IE)
||Electron Affinity (EA)
| Energy needed to remove an electron
|| Energy released or absorbed when electron is added
| measures stability of current electrons
|| measures stability of added electron
| Creates positive ions (cations)
|| Creates negative ions (anions)
| ∆H = positive (endothermic)
|| ∆H = negative (exothermic) or positive (endothermic)
| .50 kJ/mol + Na —–> Na+ + e–
|| F + e– ——> F– + 328 kJ/mol
| 1681 kJ/mol + F —–> F+ + e-
|| 53 kJ/mole + Na + e– ——> Na–
| Larger the IE the more stable the e–
|| Larger the negative EA the more stable the added e–
| Used for all atoms – Clear Trend
|| Used primarily for nonmetals but Trend is not clear/ many exceptions
Given the following EA for the Halogens – group 17
Fluorine (F) -328 kJ/mol
Chlorine (Cl) -349 kJ/mol
Bromine (Br) -324 kj/mol
Iodine (I) -295 kJ/mol
The EA “generally decreases” down a group because the increased shielding that occurs with more orbitals of electrons (core) between the outermost electrons (valence) when n increases is offset by the larger Z that occurs as you move down a group. As you move down a group n is the biggest factor why the outermost electron become less stable and held more loosely. That is why valence electrons are less stable than core electrons.
With Fluorine we would expect it to have the highest EA of the group since its valence electrons are in the smallest n (n= 2) and should release the greatest amount of energy (show more stability) as it grabs one electrons BUT IT DOES NOT. Because Fluorines electrons exist in a very small space with n= 2 the extra electron will be destabilized a bit by the electron – electron repulsions that will occur in this small space. The Zeff for this electron that is added will not be as high as we would expect because of the crowded small space for the electron in the second principle energy level. The rest of the group, follows the expected trend because their valence electrons in exist in larger and larger orbitals as the n increases resulting in lower Zeff due to the increased distance from the nucleus.
EA like IE also has EA2 and these values are almost always VERY positive as it will take energy to add an electron to an already negative particle (unless the Z is large enough to offset). This never the case for small values of n.
We learned that divergence of the 3d orbital is responsible for the properties of the transitional metals which are the elements in the d – block. The d 0rbital unlike the s and the p holds a maximum of 10 electrons and combined with the outermost s orbital that is very close in energy with the outermost d orbital provides a “super sublevel” where there 12 electrons reside in what becomes sort of a valence shell for these elements.
1. High Conductivity of Electricity – High number of mobile electrons (low IE) in metallic bonding
2. The Largest Paramagnetism – Largest number of degenerate orbitals that could contain the
largest number of unpaired paralleled spin electrons.
3. Multiple Oxidation States – Many choices for stability of electrons based on minimizing
electron – electron repulsions given the 6 orbitals (s and d) that
electrons can move to and from. Transitional elements cannot
achieve noble gas configurations because they cannot lose or gain
the high number electrons that this would require.
Fe would have to LOSE 8 electrons OR Gain 10 electrons to achieve
Kr or Xe electron configurations. Fe has too high of a Z to lose 8
electrons and its Z is not high enough to gain 8 electrons.
4. Valence Electrons from Multiple Principle Energy Levels (n) –
Electrons are lost by metals because of relatively low IE but
electrons lost by d – block metals are from the “super sublevel”
of (n) s and (n- 1) d electrons that are very similar in energy.
5. Form Colored Solutions – Crystal Field Splitting!!!!! Remember!!!!
Because they can have high oxidation states due the large number of electrons in their “super sublevel” they can draw electrons pairs from other molecules (ligands) to form stable complexes that cause the degenerate d orbital to split into 2 energy levels. This splitting of the d orbital based on the electrons being drawn into the d orbitals of the d – block metal because of the large coulombic attraction of d block elements that have high Z and large oxidation states creates an opportunity for these elements to absorb photons of visible light (negative theory of light) resulting in the complexes that transmit a photons of light that is not absorbed.
Mn+7 in KMnO4 = purple (in the oxidation titration lab) ——> Mn+2 (Clear)
Cu oxidized by nitric acid (in the % by mass of copper in Brass lab) = blue green solution and we used a spectrophotometer to measure how much light is missing (absorbed):
*Remember Cu solutions are blue-green because they make complexes in water:
Cu+2 + 6H2O —–> [Cu(6H2O)6]+2
The d orbitals that will interact directly with the incoming ligand (electrons from oxygen in water) will Destabilize that d orbital because of electron – electron repulsions and thus that orbital will contain electrons with lower Zeff.
The d orbitals that do not directly interact with the incoming ligand are not as destabilized and thus a GAP is created and the degenerate d orbitals are split into 2 levels by a gap small enough in energy that photons of visible light can match!
This is about Energy ” Lecture.
Divergence Lecture if you cannot get enough:
Read War and Peace to0 !!!!