2. Growth Reflection After doing poorly on the first quiz, Dr. Roberts provided me with the opportunity to metacognate. I had to supply him with a written understanding of bonding, Lewis structures, formal charge, and resonance. I went back to the book and reread those sections so that I could formulate my metacognition. Formal charges provide a method for keeping track of electrons to draw the valid Lewis structure. So in drawing a resonance form I should look for structures with low formal charges without breaking the octet rule. I submitted the paper to Dr. Roberts and we discussed the accuracy of the content. After my growth was approved, I was given the opportunity to retake the first quiz.
Chem 501 Metacognition
In question III a. of the make-up quiz, structure B was not considered important because the carbon violates the octet rule. The carbon has more than eight valence electrons which cannot happen for elements in the second row of the periodic table. Structure D was incorrect because the oxygen and carbon had adjacent negative formal charges.
Chem 501 Make-up Quiz 1
Reflection 2. Molecular geometry.
Enduring Understanding:
Molecular geometry: shape is especially important. VSEPR predicts the shape around an atom.
This reflection shows the growth I experienced in comprehending how the Valence Shell Electron Pair Repulsion theory can be used to predict the structural geometry of molecules or ions by minimizing the electronic repulsion between regions of high electron density.
- 1. Baseline Reflection
The reflection below is from the VSEPR concept covered in Chem. 501 and Chem. 506. I present this baseline evidence to show how I could not draw a correct Lewis structure because I failed to determine the correct number of lone paired electrons around the central atom. From there I was not able to predict the geometry of the structure.
The baseline evidence shown below is from Chem. 501 quiz 3 July 17 2007. Here I failed to use the correct number of valence electrons around sulfur to predict the correct geometry of SF4. Sulfur has six valence electrons. In this evidence I drew eight electrons which gave the geometry of square planar to maximize the distance between the lone pairs.
Chem. 501 quiz 3 July 17 2007
The evidence below shows the Chem. 506 background problems that were given at the beginning of the course to assess our knowledge and ability to draw the three dimensional geometries of a given structures by indicating the correct count of electrons around the atoms and the approximate size of the angles between the bonds. In question five, the two dimensional structure of SF4 shows the size f 90 degree angles between bonds. The total of six valence electrons on sulfur was not shown. The drawing only shows four electrons around sulfur. The two missing electrons form a lone pair causing a geometry of seesaw and not planar.
Chem. 506 background problem
June 2008
2. Growth Reflection.
The VSEPR theory was covered in chem 501 and chem 506. It wasn't until half way into the inorganic chemistry 506 class when we had extra practice in determining the geometry of molecules. The growth I experienced was through realizing that to start the process of drawing the geometry of a molecule I have to first count the number of valence electrons around each atom in the molecule then compare it to the sum of valence electrons of those atoms from the periodic table. This way I can determine the number of bonds and the number of lone pairs. Both bonds and lone pairs are considered electron domains. Electron domains try to stay out of one another's way because they are negatively charged, they repel one another. The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them. Thus three domains are arranged in a trigonal-planar fashion as shown in the evidence below (Chem 506 activity 22) where BF3 has an angle of 120o for a maximum separation. The correct way to draw this structure is first by counting the number of valence electrons around the atoms to see if there is a lone pair of electrons, and then draw the structure with the maximum separation between bonds. Boron in problem one below has three valence electrons and fluorine has seven valence electrons, the sum is 24 electrons [(7x3) + 3= 24]. In counting the number of electrons for BF3 it shows 24 electrons. That indicates that boron has no lone pair only three bonds with the three atoms of fluorine which gives it the trigonal planar shape. The same evidence shows how PF3 is given the shape of trigonal pyramid. Again in counting the number of electrons around all atoms, the sum is 26 [(7x3) + 5 =26]. Seven valence electrons from each of the three fluorine atoms and five valence electron from phosphorus. This indicates that there is a lone pair of electron on the phosphorus. In total there will be four electron domains, three bonds and one lone pair, which give rise to a trigonal pyramid.
Chem 506 activity 22 July 2008 
Next evidence I present is a tutorial with Mr. Tom, the TA for Chem 506 class. The growth I experienced was through realizing that bond angles are only between bonds not between a bond and a lone pair. The presence of the lone pair reduces the bond angle because lone pair takes more space. The evidence clearly shows how two electron domains around the central atom "x" make an angle of 180 degrees, thus, the molecule is linear. If there are three electron domains with three bonds then the angle is 120 degrees giving the molecule a trigonal planar shape, but if there is a lone pair, the angle would be less than 120 degrees because lone pair of electrons occupy a larger space than the electrons making a bond which gives the molecule a bent shape.
Chem 506 tutorial
After working on the problem in the first evidence, the tutorial, and many other POGILs in Chem 506 class, I was given the opportunity for a make up quiz to assess my final comprehension and ability to draw the three dimensional geometries of a given structures. The evidence below is the make-up quiz Chem 506. It shows correctly predicted geometry for PCl5. The most stable electron domain geometry for five electron domains is the trigonal bipyramid (two trigonal pyramids sharing a base). Two of the five domains point toward what is called axial position, and the remaining three positions point toward the equatorial positions. Each axial domain makes a 90 degree angle with the equatorial domain. Each equatorial domain makes a 120 degree with either one of the other two equatorial domains and a 90 degree with either axial domain. Also this evidence shows the geometry of PCl3 where a nonbonding pair (lone pair) of electrons defines an electron domain that is located on the phosphorus. The Lewis structure for PCl3 has a total of four electron domains around the central phosphorus atom (three bonding and one nonbonding), giving it a tetrahedral geometry.
Make up quiz Chem 506
Another later evidence is a tutorial I created for Chem. 536 class named Valence shell electron pair repulsion. I used this tutorial in my classroom. It expands to hybridization theory and contains exercises for further assessments.
Here is what I can impart to my students. Chemists discuss molecular shapes in terms of two theories;
A. VSEPR
The predicted geometry of the molecule is based on the number of VSEPR:
2 electron domains results in linear geometry
3 electron domains results in trigonal planner geometry
4 electron domains results in tetrahedral geometry
5 electron domains results in trigonal bipyramid geometry
B. Hybridization theory
If there are two regions of electrons density around a central atom, the atom is said to be sp hybridized, if there are three regions of electron density around the central atom, the atom is said to be sp2 hybridized, if there are four regions of electron density around the central atom, the atom is said to be sp3 hybridized.
Reflection 3. Molecules Spectroscopy Chem. 507
Enduring Understanding: Understanding light and how light affects matter are paramount to understanding spectroscopy. Boltzmann Disribution.
I chose this reflection because I wanted to document my understanding in a concept that I have no information to submit as a baseline.
Thermal energy is defined as the energy that the system has because it is at some temperature other than 0 K.
Energy of a molecule can be separated into translational, rotational, vibrations and electronic. Within each type of those energies, the energy levels are quantized.
T=0 K all molecules are in Energy level 0. At T=300 K (~ room temperature), there is a ratio between the number of molecules in energy level 0 and energy level 1. The Boltzmann distribution calculates that ratio. At T=∞, one half of the molecules are in energy level 0, and one half of the molecules are in energy level 1.
1. Baseline Reflection
I am presenting the following baseline evidence from Chem. 507 Quiz 5 Feb7/2009. Where it shows clear misunderstanding how molecules behave when exposed to thermal energy.
Question 4 suggests two systems with the same number of molecules, but one of the systems has two degenerecies in its first energy level. The question asks to choose which system is at the higher temperature. I mistakenly chose the second system. I based my answer on the following equation:
Ni/ Nj = e-∆E/TK
I kept solving the equation mathematically using different temperatures and suggesting that Ni / Nj = gi / gj
Where N is the number of molecules and g is the number of degenerecies. This shows how I could not apply the correct form of Boltzmann equation to understand the distribution of molecules in a system.
2. Later Reflection
This is an enduring understanding named The big Idea we give back as home work at the end of each lesson. Growth is presented here through the clear explanation of Boltzmann Distribution equation.
Boltzmann developed an equation that yields the ratio of the number of molecules at two different energy levels, i and j, when the molecules are in thermal equilibrium.
Ni / Nj = gi / gj [ e-(εi – εj)/kt ], where N is the number of molecules and g is the number of degeneracies.
At T=0K
e-(εi – εj)/kt = e-1/0 = e-∞ = 1/e∞ = 1/∞ = 0
Which means that all molecules are at the lowest level zero.
At T= 300K
e-(εi – εj)/kt =0.333 = 3/9
Which means three molecules will be found at energy level one.
At T = ∞
e-(εi – εj)/kt = e1/∞ = e0 = 1
Which means there are same number of molecules in level one and level zero.
The above understanding is explained in the evidence presented from Chem 507 (chemactivity 18 Thermal Energy).
According to the hypothetical molecule described in this activity, at T=0 K all the molecules are in the lowest possible energy level because the thermal energy of the twelve molecules in Model 1 is zero, which is why state 1 is not populated at that temperature.
As the temperature increases T=300 K one fourth of the molecules ((3/12= ¼) or 3/9 when presented as a ratio) are now in the excited state; heat is added, the thermal energy is gained by the molecules.
Chem 507
(chemactivity 18 Thermal Energy)
Next I present later evidence where there are two degeneracies in energy level 1. Exercises one and two were graded homework demonstrates my growth in understanding Boltzmann equation. First evidence includes the actual exercises with explanation and the second evidence is the graded answer sheet.
Clearly for exercise 1: energy level 1 has g=2 and energy level 0 has g=1.
At T=0 K all 6.02x 1023 molecules are in energy level zero
At T = 3000 K, the ratio is 0.33 = 2/3
2/5 = x/ 6.02x 1023- x, where x represents the number of molecules in energy level 1 as a result of absorbing thermal energy.
And at T= ∞, the ratio=1 or 50:50 i.e. ½ of Avogadro’s number is in level 1
Exercises 2: the energy level 0 has g=2
Energy level 1, g=1 and energy level g=9 all together 12
2 twelfths will be in g=2.
Chem 507
(chemactivity 18 Thermal Energy)
Exercises 1,2
Graded answer sheet 