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 | Inorganic Chemistry 2001 Forum |
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| 9-17-01 Summary Name: Jennifer T Date: 2001-09-20 00:00:44 Link to this Comment: 240 |
· "A unit cell is the smallest bit that will generate the pattern" -Dr. B
· In Escher diagrams, the translation of the unit cell does not need to be perpendicular in the Cartesian plane. However, the translation must occur along the edges of the unit cell.
· From the fish tessellation, a unit cell may be created by connecting the same point along various objects (i.e. connecting the eyes of the fish) to form a diamond-shaped unit cell.
· Different unit cells for the same pattern can relate volumetrically. For example, a unit cell may have 1 fish and 1 frog, 2 fish and 2 frogs, etc. While any such cells may be equally good choices (chosen arbitrarily), the unit cell must be identified as either primitive or centered.
· A primitive unit cell is the most basic and can be thought of having the "least common denominator" in terms of repetition of pattern: e.g.- only one fish and one frog in a unit cell.
· A centered unit cell may be bigger than a primitive unit cell. For example, one of the tiles used in the fish/frog tessellation had 2fish/2frogs - which is twice the volume of the primitive unit cell.
· A key point is that once you have identified the shape of a unit cell, this shape may be moved about on a tessellation, which forms a lattice, (imagine the transparencies) and one will always end up with a viable unit cell.
· See Escher bird tessellation: The unit cell contains 2 white birds and 2 black fish. This is due to the fact that each bird/fish in the unit cell is angled at a different direction, therefore a unit cell can not be composed of just a bird and a fish. Instead, such a cell would be called an asymmetrical cell, the smallest piece of the pattern needed to generate the entire pattern by certain symmetry operations.
· Lastly, Dr. B explained the program used in making a wallpaper. She used 'Pencil' under the 'Tools' heading to draw a squiggle in one cell, which was then translated in the direction of the unit cell. Some of the symbols used for the unit cell/translation options: p=primitive, 1= one symmetry, e.g. - p2=primitive unit cell with symmetry by rotation then translated. Also, in 'Cells', cm=center and m=mirror planes, pmm=(like the angel/devil pattern: fourfold rotation and two mirror planes)
| Summary 9/21/01 Name: Jen and Ry Date: 2001-09-22 15:33:44 Link to this Comment: 254 |
Thank you Dr. B. for the color copies! They made us all very happy!
| lecture summary by Mo2 for Wed 9/19 Name: Modjulie K Date: 2001-09-24 15:56:05 Link to this Comment: 273 |
Lecture Date: 19 Sept 01
Lecture began by a brief review of one of the Escher diagrams from
Monday. Apparently we had decided that there was a unit cell in the
"critter" diagram, yet deceiphering the assymetric unit didn't work out
with that particular unit cell. The moral of this story seems to be that the unit cell edges have to align with the symmetry inherent in the pattern in order to correctly identify the assymetric unit.
The lecture continued with an example of once again identifying the unit cell in a pattern. This example involved the angels and demons pattern, and we thought that this example was a good one because it was very clear that you could reflect a half of the angel pattern and then rotate it to create the entire pattern. This was because the unit cell was so large so it was easier to see how to identify the assymetric unit and manipulte it to form the whole pattern.
We continued with two new symmetry operations in crystalography...
1. Glide Plane: translation + reflection over the the glide plane
**note--glide plane is PARALLEL to direction of translation** Another example of this would have been helpful.
2. Screw Axis: (think spiral staircase) Rotation + translation. The spiral staircase was very clear.
Dr. B then introduced us to a few (of many) space groups. Space groups
indicate the way the crystals pack together. The space groups were:
P(nma): P = "primitive", n = glide plan perp to a, m = mirror plane, a =
glide plane perp to c
P21/c: God's Favorite Space Group--90% of crystals pack this way. 2-1
connotes the screw axis, and c the glide plane. 2-1 and c all happening in
same direction. The space group is the mono-clinic cell in the Bravais
Lattices.
C2/c: C = Centered, 2=C2 rotation axis, c = glide plane.
We then used the handout on Bravais lattices and talked about how the
lattices are defined by a, b, and c axes, and alpha, beta, and gamma
angles. The triclinic has the lowest symmetry where none of the axes are
equal and none of the angles are equal. The cube however has the highest
symmetry where all the axes are equal and all of the angles are 90 degrees.
Lastly, she gave us a litle preview of what was to come on
Friday...apparently swift recognition of lattices, geometries and solid
state crystals of ionic solids.
Drawing lessons friday...bring colored pencils, crayons etc....
| Unit Cells!!! (friday sept. 21 & monday sept. 24) Name: Unununium Date: 2001-09-27 02:04:33 Link to this Comment: 318 |
2) NaCl lattice (cubic)
CN(Na)= 6, Oh symmetry
CN(Cl)= 6, Oh symmetry
Space Group= Fm3m (face centered, mirror plane, C3, mirror plane)
Face Centered Cubic lattice (FCC)
Fun fact: unit cell contains 192 asymmetric units
3) ZnS lattice (cubic)
CN(Zn)= 4
CN(S)= 4
Space Group= F43m (*4 is written with a bar over top of it)
(face centered, improper rotation axis, C3,
mirror plane)
FCC lattice
Fun fact: 96 asymmetric units
more dense than hexagonally packed ZnS
4) CaF2 lattice (cubic)
CN(Ca)= 8
CN(F)= 4
Space Group= Fm3m
FCC lattice
F has tetrahedral geometry
5) ZnS lattice (hexagonal prism) -Wurtzite
CN(S)= 4
CN(Zn)= 4
Space Group= P63mc (6-sub-3 refers to screw axis)
tetrahedral geometry
6) NiAs lattice (hexagonal)
CN(Ni)= 6, D3h symmetry
CN(As)= 6
trigonal prismatic geometry
7) TiO2 lattice (tetragonal-- NOT cubic) -Rutile
CN(Ti)= 6, Oh symmetry
CN(o)= 3, trigonal planar symmetry
tetragonal sides: a=2b=2c
:)
Name: Einsteiniu Date: 2001-09-27 16:32:24 Link to this Comment: 321 |
| Lecture 10-01-01 Name: Molly Flan Date: 2001-10-01 19:12:42 Link to this Comment: 359 |
Characteristics of ionic solids:
- Bonding in ionic solids is electrostatic.
- CN of ionic solids can change with a change of phase.
- High melting point
- Require highly polar solvents
- Conductive when molten
Lattice Energy (U): The energy released when gaseous ions condense to make a solid. U will be exothermic for a stable material.
U= F attractive + F repulsive
U={ A [Z+ Z- e2]/ d + Be2 / dn}
| Lecture notes 10/3/01 Name: Sarah Ledo Date: 2001-10-04 12:36:05 Link to this Comment: 395 |
Uo(calc) sometimes does not equal Uo(exp) due to polarizability. We have to consider the "squishiness" of the ions. Generally, Uo(calc) = Uo(exp) for hard ions. With soft ions, cations and anions deform and make a more covalent bond. It is difficult to group all bonds into either ionic or covalent because most fall somewhere on a range between the two types.
We also began learning about Thermochemical cycles, specifically the Born-Haber cycle. The Born Haber cycle is a series of steps that add up to a final net equation which allows us to make predictions for heat of formation. If the heat of formation is a very large positive number we know that the molecule is very unstable and may never form. If the heat of formation is very large negative number we can predict that the molecule will be stable.
| Lecture notes 10/3/01 Name: Sarah Ledo Date: 2001-10-04 12:36:19 Link to this Comment: 396 |
Uo(calc) sometimes does not equal Uo(exp) due to polarizability. We have to consider the "squishiness" of the ions. Generally, Uo(calc) = Uo(exp) for hard ions. With soft ions, cations and anions deform and make a more covalent bond. It is difficult to group all bonds into either ionic or covalent because most fall somewhere on a range between the two types.
We also began learning about Thermochemical cycles, specifically the Born-Haber cycle. The Born Haber cycle is a series of steps that add up to a final net equation which allows us to make predictions for heat of formation. If the heat of formation is a very large positive number we know that the molecule is very unstable and may never form. If the heat of formation is very large negative number we can predict that the molecule will be stable.
| Wednesday 10/10/01 Name: rachel & s Date: 2001-10-04 12:51:02 Link to this Comment: 397 |
Uo(calc) sometimes does not equal Uo(exp) due to polarizability. We
have to consider the "squishiness" of the ions. Generally, Uo(calc)
= Uo(exp) for hard ions. With soft ions, the cations and anions
deform and make a more covalent bond. It is difficult to group all
bonds into either ionic or covalent because most fall somewhere on a
range between the two types.
We also began learning about Thermochemical cycles, specifically the
Born-Haber cycle. The Born Haber cycle is a series of steps that add
up to a final net equation which allows us to make predictions for
heat of formation. If the heat of formation is a very large positive
number we know that the molecule is very unstable and may never form.
If the heat of formation is very large negative number we can predict
that the molecule will be stable.
| Aluminum reactivity Name: Jen Trewee Date: 2001-10-12 11:14:03 Link to this Comment: 475 |
Both environmental chemistry and ecology will be brought into our lecture today by discussing the abundance of Aluminum ore in nature and the recycling process of Al cans.
Where does Al come from???
There is no where to find Al in nature; it does not occur in its elemental form directly as an ore to be mined for human use. Instead, aluminum is mined commercially in two different forms - both ores - 1) bauxite and 2) corundum. Both are compounds of the form Al2O3.
Corundum
v lattice of ions
v (in demo) grey cation, which represents aluminum, has octahedral geometry and C.N.=6
v red anion, or oxygen, has a slightly distorted tetrahedral geometry and C.N.=4
v corundum is the mineral base for two gemstones in particular: ruby and sapphire.
v In ruby, a low percentage of chromium replaces some of the Al+3 cations (with Cr+3). Cr is a transition metal, and therefore adds color to the compound - in this case pink-red-purple to rubies. Chromium also gives emeralds their green colour.
v In sapphire, there is a percentage of Al+3 replaced by titanium Ti+3 and iron Fe+3; this needs to be done to the right proportions to produce the intense blue colour famous to sapphires (as well as their pink and yellow, etc tones, for sapphires come in many colors)
v Application of corundum - sandpaper - because of its characteristic hardness and stability.
To help demonstrate why Al is not found in its purely elemental form, Dr. B put some Al foil in a mystery solution to react while she went through the scientific justification on the board…
Al2O3 => 2 Al(s) + 3/2 O2 (g)
How much energy is required for this reaction to take place?
Need lattice energy and Born Haber cycle to solve
v First, Al2O3 must be changes to 2 Al(s), a reaction requiring a certain heat of formation Hf.
v 2 Al(s) goes to Al(g) through the heat of sublimation - Hsub.
v Al(g) then is converted to 2 Al+3 through the Heat of Ionization Potential HIP, in which 2*(3 e-) are lost.
v Meanwhile, 3/2 O(g) adds 3*(2 e-) to form 3 02 - given a certain HBO.
v Lastly, Al2O3 is finally reached given a lattive energy U0
To calculate lattice energy:
v A = 4.1719 (found in Huheey, or other text)
v Z+ = 3+
v Z- = 2-
v d = radius of rAl + rO = 67.5 pm + 124 pm = 191 pm (use C.N. to find appropriate rAl)
v n = 7 (table for Born exponent, also in Huheey, or text)
v N = 6
v Lattice Energy = [(N A Z+ Z- e2) / d]*[1-(1/n)] = 1.39*105 [A Z+ Z- /d][1-(1/n)]
v d = 4*pi*epsilon 0 and e’=(e/(d)1/2) (e given in coulombs in text)
v Dr. B gives e in dines:
v Uo = 1.39*105 [A Z+ Z- /d][1-(1/n)], where d is in picometers = -15,600 kJ/mol
v Very reactive!!!
v From Born Haber: Hf = 2 sub + 3/2 HBO + 2 HIP + 3 HEA + Uo
= 2(150) + 3/2(493) + 2(5139) + 3(639) + -15,600 = -2365 kJ/mol
v For ionization energy, need to sum transition for each state of Al: NOT Al3=>Al0 but instead Al3=>Al2 to Al2=>Al1 to Al1=>Al0
v As shown by the heat of formation, Al does not exist in nature because it is too unstable as elemental aluminum; one would need to put in a lot of energy to make Al(s).
The reaction with the Al foil is spontaneous because aluminum is thermodynamically unstable. Al decomposes to Al2O3 since Al2O3 is relatively unreactive. This is why a layer of Al2O3 usually forms on the surface of Al, to seal it off from reacting and protect the Al underneath. The mystery solution is mercury nitrate HgNO2. It pits Al2O3 and exposes Al metal underneath, which then reacts violently to O2 air it comes in cntact with. The smoke observed is the heat of the reaction boiling off water left over from the mercury nitrate solution.
Will Al2O3 decompose to Al and AlO?
v Al2O3 => Al(s) + AlO(s) + O2(g)
§ (in forming Al(s) and O2(g), Hf =0)
§ Hrxn = Hprod - Hreact
§ forming AlO is tricky because it is an unknown material with an unknown structure; A = ?, and r+ not in book
Predict structure: choices are 1) CsCl (CN=8), 2) NaCl (CN=6), 3) ZnS (CN=4)
v radius rule, use: 71-86pm (Mg2+) => r+/r- = 57/68 . . . points to NaCl
v use ANaCl = 1.7476 (ACsCl=1.76267 not very different though)
v calculate Uo = -3435 kJ/mol (note: a negative value)
v d=(118 + 126) pm
v calculate Hf: ionization potential, HIP is different however because there is no last e- to oxidize. Hf = -4.89 kJ/mol…almost 2000 times more unstable.
v So, Al3O2 is the most stable and the reaction to Al(s)+O2(g)+AlO does not occur; AlO, with Hf~0 has no real driving force.
| Lecture summary for 10/10/01 Name: Modjulie a Date: 2001-10-25 13:01:21 Link to this Comment: 537 |
The lecture started by recalling that Dr. Mallory had mentioned in Organic Chem that the electron pairs on oxygen in R-O-H were not the same. Why is this? We began the discussion by making a general comparison between Molecular Orbital Theory (MOT) vs. VESPR/VBT. The comparison was that the molecualr orbital diagrams similar to a branch of a tree whereas the VESPR theory gives an idea of the whole picture, a picture of the whole tree.
We were asked to consider why NH3 has an angle of 107 degrees whereas PH3 has an angle of 93 degrees. Why are these angles so different for the two molecules which are relatively similar? This investigation was continued with an intimate look at H2O and its bond angle of 104 degrees. Why doesn't water have the 109 degrees that one might expect? We had previously learned that VESPR theory said the orbitals of the lone pairs were taking up more space and thus decreasing the angle, but what does that actually mean?
This was explained through a finding the linear combinations for the orbitals and getting a better picture of exactly what is happening in the molecule. The idea was that each orbital would be a combination of the p's and S orbitals but would have different compositions in terms of p and s characters.
First we determined the symmetry of the molecule, which in the case of water is C2v. Ulimately we figured we need 4 hybrid orbitals: O/A(Hydrogren A bond), O/B(Hydrogen B bond), O/a(Lone pair a), O/b(Lone pair b). We also know that O/A = O/B, but that O/a does not equal O/b.
So our approach in a problem like this is to
1)get a structure (a first guess)
2)determine the symmetry
3)orient the molecule in the coordinate system of the preferred method, in our case the zaxis is defined as parallel to the major axis of symmetry. The problem becomes very visual here and so it would be helpful to refer to the diagrams in the notes. But after the diagram is developed we can determine the coefficients of the linear combinations of the four orbitals listed above. The linear combinations with unknown coefficients at this point are listed below:
O/A= C1(Pz) + C2(Py) + C3(S)
O/B= C1(Pz) - C2(Py) + C3(S) [ the - C2 comes from the fact that this orbital is oriented in the negative y coordinate]
Orbital of the L.P. =-C4Pz + C6 S [There's no Py component]
We figured out the values of C1 and C2 from the trigonmetric relationship given by half of our experiementally known angle of 104 degrees (52 degrees), so C1 on Pz = cos 52 degrees and C2 on Py = sin 52 degrees.
Now how do we detemine C3? We ended the lecture by understanding that the orbitals of the lone pairs must be normalized and in other words must be orthogonal. The discussion of this explanation will be continued in the next lecture.
| Solid State Chemistry Name: Jennifer T Date: 2001-11-03 20:07:06 Link to this Comment: 550 |
| helping each other Name: dr. b. Date: 2001-11-07 11:08:52 Link to this Comment: 570 |
BUT, WE CAN STILL USE THIS SPACE PRODUCTIVELY!!!!
You know the kinds of MOT questions you asked me on Monday?
_Everyone_ benefited (I know because many of you told me so).
So when you uncover such a question in your studies,
ASK IT HERE!
The Forum program notifies me immediately.
I will ASAP respond to your question here.
Remember, if YOU have a question, it is almost certain someone else in the class has that question too. You'll be helping your friends to go ahead and ask it.
