Evaluation of Students in General Chemistry Who Di

Evaluation of Students in General Chemistry Who Did and Did Not Utilize Spartan ProJulie B. EalyPenn State UniversityLehigh Valley - Berks : Evaluation of Students in General Chemistry Who Did and Did Not Utilize Spartan Pro Julie B. Ealy Penn State University Lehigh Valley - Berks American Chemical Society Boston, Massachusetts August 2002

Molecular Modeling Lab Content : Molecular Modeling Lab Content Ealy H2, HF, F2: Lewis dot, bond length, electrostatic potential surface, dipole CH4, CH3Br: Bond length, bond angle CH4, NH3, H2O: Lewis dot, bond angle, orbital notation HF, HCl, HBr, HI: Bond length, electrostatic potential surface, dipole Organic molecules: identify by functional group, hybridization

Molecular Modeling Lab Content : Molecular Modeling Lab Content Moog & Farrell Bond order and bond strength Bond length and bond strength Resonance: benzene Bond order and bond energy Intermolecular forces: alkane, alcohol, ketone

Questions to be answered : Questions to be answered What will be the nature of incorrect responses written by lab students for long response questions related to molecular structure? Would students who completed molecular modeling exercises in lab perform at a significantly higher level on a posttest than on a pretest related to molecular structure?

Questions to be answered : Questions to be answered 3. Would nonlab students who had only completed molecular structure and concepts in lecture perform at a significantly higher level on a posttest than on a pretest related to molecular structure? 4. Would the lab students who completed molecular modeling exercises in lab perform at a significantly higher level than the nonlab students on questions related to molecular structure and concepts on the semester exam?

The Students and the Procedure : The Students and the Procedure All 23 students were enrolled in first semester chemistry 11 students were enrolled in lab as well as lecture – hereafter referred to as the lab students 12 students were only enrolled in lecture – hereafter referred to as the nonlab students All students do not have to take lab Some students will take lab at a later time Average grade on the semester exam: Lab Nonlab 69 72 t = -.45  = .67

Incorrect responses of students who utilized molecular modeling in lab : Incorrect responses of students who utilized molecular modeling in lab What contributes to a greater bond length? Greater electronegativity – with lower electronegativity the electrons are shared more evenly Lone pairs on one atom push atoms attached to a central atom further away

Incorrect responses of students who utilized molecular modeling in lab : Incorrect responses of students who utilized molecular modeling in lab What contributes to a greater bond length? Lone pair electrons cause a bond angle to be smaller which, in turn, causes the bond length to be longer Bond length between H and F and H and Cl should be the same because each molecule has a bond order of 1 A longer bond length shows a lower electron density

Incorrect responses on the lab test : Incorrect responses on the lab test Stronger – Intramolecular bonds or Intermolecular forces? Intermolecular forces are stronger because they occur between particles, not inside. Between particles is where the bonds are taking place Intermolecular forces are stronger because London dispersion and dipole-dipole forces are present

Incorrect responses on the lab test : Incorrect responses on the lab test Bond length combined with intermolecular and intramolecular: The reason the bond length is longer is because lone pair electron forces in a molecule would be greater than intramolecular forces based on the presence of a dipole moment which increases the strength of intermolecular forces.

Pretest and Posttest QuestionsIncorrect responses were written from previous students’ incorrect responses on a lab test after completion of a unit on molecular modeling. : Pretest and Posttest Questions Incorrect responses were written from previous students’ incorrect responses on a lab test after completion of a unit on molecular modeling. 1. On Structure 15, blue represents a more positive area and red represents a more negative area. In other words, blue corresponds to low electron density and red to high electron density. Structure 15 represents the electron density surface of fluorine, F2. A ball and spoke model of fluorine is represented by Structure 10. Which of the following statements is correct regarding Structures 15 and 10?

Pretest and Posttest QuestionsF2 – electrostatic surface and ball and spoke : Pretest and Posttest Questions F2 – electrostatic surface and ball and spoke A. Electron density is around the nucleus of the fluorine atoms. B. The blue area on the electron density surface (Structure 15) represents the strongest forces. C. Electronegative concentration is high around the fluorine atoms. D. The most negative area on the fluorine atoms is because of the lone pairs of electrons around fluorine.

Pretest and Posttest QuestionsBall and spoke – CH4 and CH3Br : Pretest and Posttest Questions Ball and spoke – CH4 and CH3Br 2. On Structure 6 the bond length between the central atom (carbon – black) and the atom at the top (hydrogen – white) is shorter than the bond length in a similar position on Structure 5 (bromine – red). The difference can best be explained by the following

Pretest and Posttest QuestionsBall and spoke – CH4 and CH3Br : Pretest and Posttest Questions Ball and spoke – CH4 and CH3Br A. The Br atom elongates the bond because of its bonding pair – bonding pair repulsion. B. Bromine has more energy levels than hydrogen and its additional electron density causes a longer C-Br bond length. C. Lone pairs of electrons exhibit repulsion against other bonding pairs to cause the bond to lengthen. D. A higher bond energy between carbon and bromine contributes to a longer bond length.

Pretest and Posttest QuestionsBall and spoke: BeCl2, HgBr2, and XeF2  : Pretest and Posttest Questions Ball and spoke: BeCl2, HgBr2, and XeF2 3. Structures 25, 26, and 27 are all linear molecules. They represent the following molecules:, respectively. BeCl2, HgBr2, and XeF2. An acceptable explanation that explains that Structure 27 is XeF2 is

Pretest and Posttest Questions Ball and spoke: BeCl2, HgBr2, and XeF2  : Pretest and Posttest Questions Ball and spoke: BeCl2, HgBr2, and XeF2 A. XeF2 would have the shortest bond length because F2 wants very much to have an octet of electrons around it. B. The number of energy levels in Xe causes it to be the largest central atom and the small size of the fluorine atom enables it to approach closer to a central atom than Cl or Br could approach. C. Xe has a high number of protons and would repel its many electrons very far away which would create a large electron density around it causing much repulsion with the fluorine atoms. D. Xe is medium sized with lots of electrons and because of its low electronegativity its pull on electrons are not that great.

Pretest and Posttest QuestionsBall and wire – alkane, alcohol, ketone : Pretest and Posttest Questions Ball and wire – alkane, alcohol, ketone 4. In order from 37-39, the strongest intermolecular (between molecules) force of attraction is A. Induced dipole, hydrogen bonding, dipole-dipole B. Hydrogen bonding, dipole-dipole, induced dipole C. Hydrogen bonding, induced dipole, dipole-dipole D. Induced dipole, dipole-dipole, hydrogen bonding

Pretest and Posttest QuestionsBall and wire – alkane, alcohol, ketone : Pretest and Posttest Questions Ball and wire – alkane, alcohol, ketone 5. The molecular geometry on structures 37-39 at locations A, B, and C is A. Trigonal planar, tetrahedral, bent B. Trigonal planar, bent, tetrahedral C. Bent, trigonal planar, tetrahedral D. Tetrahedral, trigonal planar, bent

Pretest and Posttest QuestionsBall and spoke: C2H4 (19), C2H6 (23) Ball and wire: C6H6 (21), C2H2 (24) : Pretest and Posttest Questions Ball and spoke: C2H4 (19), C2H6 (23) Ball and wire: C6H6 (21), C2H2 (24) 6. Rank structures 19, 21, 23, and 24 in terms of the bond length between the two carbon atoms (the dark ones) from shortest to longest C-C bond length A. 19, 23, 21, 24 B. 24, 21, 23, 19 C. 24, 19, 21, 23 D. 19, 23, 24, 21

Results of the Pretest and the Posttest  of number correct : Results of the Pretest and the Posttest  of number correct  Lab Pretest Posttest  1.5 3.6 t = -3.9  = .00042*   Nonlab Pretest Posttest  1.6 2.1 t = -0.91  = .19

Results of the Pretest and the Posttest  of number correct : Results of the Pretest and the Posttest  of number correct Pretest Lab Nonlab  1.5 1.6 t = -.070  = .53   Posttest Lab Nonlab  3.6 2.1 t = 2.86  = .0047*

Molecular Structure and Concept  Semester Exam Questions  % of Correct Responses LAB NONLAB : Molecular Structure and Concept Semester Exam Questions % of Correct Responses LAB NONLAB 4. Consider the following information: 64 25 Molecule Bond Energy Chlorine 243 Carbon dioxide 804 Nitrogen 945

Semester Exam Cl2, CO2, N2 : Semester Exam Cl2, CO2, N2 The best explanation for the difference in bond energy is the difference in the a) polarity of the molecules b) type of bonds in the molecules c) electronegativity of the atoms that make up each molecule d) intermolecular forces between molecules

Semester Exam : Semester Exam Consider the following table of information for questions 10-12. Molecule Ionization Bond Bond Boiling Electron Energy Energy Length Point Affinity Chlorine 1255.7 243 199 -34.0 -349 Bromine 1142.7 224 228 59.5 -324 Iodine 1008.7 151 266 185.2 -295

Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 10. The increase in bond length 55 42 from chlorine to iodine is best explained by the following explanation regarding iodine atoms. They have the a) greatest electron density b) lowest electronegativity c) more lone pairs of electrons d) strongest intramolecular forces

Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 11. Chlorine has the lowest boiling point because of the 55 58 a) short bond length in a chlorine molecule b) strong bond energy in chlorine c) weak intermolecular forces d) energy needed to remove an electron from a chlorine atom

Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 12. A strong relationship exists 64 50 between bond energy and a) ionization energy b) bond length c) boiling point d) electron affinity

 Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 30. All of the following pairs of 64 50 molecules and molecular geometries are correctly matched EXCEPT  Compound Molecular Geometry a) HBr linear b) CBr4 tetrahedral c) AsH3 pyramidal d) BeBr2 angular e) CH4 tetrahedral

  Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 43. Which response includes 64 8 all of the following substances in which dispersion forces are the most significant factors in determining boiling points?

Semester Exam : Semester Exam I. Cl2 II. HF III. Ne IV. KNO3 V. CCl4   a) I, III, and V b) I, II, and III c) II and IV d) II and V e) III, IV, and V

  Semester Exam  % of Correct Responses LAB NONLAB : Semester Exam % of Correct Responses LAB NONLAB 52. Which response represents the 55 50 intermolecular forces for C3H8 and C3H7Cl, respectively?   a) London dispersion; London dispersion b) London dispersion; Dipole-dipole c) Dipole-dipole; London dispersion d) Dipole-dipole; Dipole-dipole

Semester Exam Results : Semester Exam Results LAB NONLAB  4.2 2.9 t = 1.9  = .035*   Mean based on the number correct for each group Questions 4, 10-12, and 52 were written by Julie B. Ealy Questions 30 and 43 are from the Whitten, Davis, and Peck testbank.

Conclusions : Conclusions Students’ incorrect answers to long response questions are varied and indicate misconceptions that need to be addressed. Lab students did significantly better on a posttest than on a pretest. Nonlab students did not do significantly better on a posttest than on a pretest. 4. Lab students did significantly better than the nonlab students on the semester exam.