TA practice exam 2 key - CH301 Fall 2010 Practice Exam 2TA...

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Unformatted text preview: CH301 Fall 2010 Practice Exam 2TA Style Answer Key 1. Which of the following bonds is most polar? (Hint: you only need knowledge of the electronegativity trend, you dont need to know actual values.) 1. Br-Cl 2. F-I 3. I-Cl 4. F-Cl 5. Br-F 6. I-Br Fluorine and iodine are farthest apart in the halogens. So without even knowing the exact values, you can infer that thay have the greatest difference in electronegativity. 2. If a molecule has 3 identical atoms bonded to its central atom, how many non-bonding pairs of electrons must be on the central atom in order to be non-polar? 1. 3 2. 2 3. 1 4. 0 5. Two of these answers would make the molecule non-polar. The molecule described in the problem is an AB 3 U X , in Dr. Laudes system. The only value of X that would result in a non-polar molecule is 0. 3 If a molecule has 2 non-bonding electron pairs on its central atom, how many identical atoms must be bonded to the central atom in order to be polar? 1. 4 2. 3 3. 2 4. 4 or 3 5. 4 or 2 6. 3 or 2 7. 4, 3, or 2 The molecule described in the problem is an AB X U 2 , in Dr. Laudes system. The values of X that would result in a polar molecule are 3 and 2. 4. In the correct Lewis structure for XeF 4 , what bond angle(s) exist(s)? 1. 180 2. 90 3. 120 4. 180 and 90 5. 180, 90 and 120 Xenon tetrafluoride has the octahedral electronic geometry and square planar molecular geometry. 5. Valence Bond Theory is required to explain which of the following? 1. The degeneracy of non-bonding electrons found in molecule such as F 2 . 2. The bond angles observed around central atoms. 3. The degeneracy of bonds found in molecules such as CH 4 . 4. The bond angles observed around terminal atoms. VB Theory was developed to explain observations such as the degeneracy of the bonds found in molecules like methane. Based on the non-degenerate energies of the atomic orbitals of carbon, one would simply predict that the bonds they form are also different in energy. But this was not the experimental observation. VB Theory reconciles the observation with our predictive framework. 6. What is the electronic geometry of BrF 3 ? 1. linear 2. trigonal planar 3. tetrahedral 4. trigonal bipyramidal 5. octohedral Bromine trifluroide has Br as the central atom with 3 bonded atoms and 2 non-bonding pairs, for a total of 5 regions of electron density, giving it trigonal bipyramidal electronic geometry. 7. What is the molecular geometry of XeOF 4 ? 1. square pyramidal 2. octohedral 3. trigonal bipyramidal 4. square planar XeOF 4 has Xe as the central atom with 5 bonded atoms and 1 non-bonding pair, giving it square pyramidal molecular geometry. 8. How many (sigma) and (pi) bonds are in the molecule adenine?...
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TA practice exam 2 key - CH301 Fall 2010 Practice Exam 2TA...

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