CH301 Fall 2008 Practice Exam 2 Key 1. Which of the following statements are true? I. Bond polarity can be calculated using the differences in electronegativity values. II. Bond polarity describes to what degree electrons are shared between two atoms (i.e. Is one atom “hogging” the electrons and how much?). III. Homonuclear bonds are always polar. IV. F—Br is a more polar bond than F—I V. Li—B is less polar than O—H VI. O—O is more polar than H—H 1. I, II, and VI 2. I only 3. I and II 4. I, II,V correct 5. II, V, VI 6. II, IV, V 7. II and V Explanation: I is true – you will need to be able to do these calculations on the test because they will allow you to rank the polarities of various bonds. II is true and describes bond polarity conceptually. III is false because homonuclear bonds, bonds between two atoms of the same element (i.e. H—H or O—O) are always NON-POLAR because the difference in EN is always zero. IV is false because the EN difference between F and I is larger than the EN difference between F and Br, implying a more polar bond. We know this because EN decreases down the periodic table. V is true. You must have memorized the EN values of all elements in the second row of the periodic table. You will need them on the test. In this case we see Li—B has a difference in EN of 1.2 while O—H has an EN difference of 1.3. Remember than a larger EN difference is a more polar bond. VI is false. These are both homonuclear bonds with an EN difference of ZERO meaning both bonds are completely nonpolar. 2. Which of the following molecules is non-polar? 1. CO2 correct 2. H2O 3. CH3F 4. O3 5. NO Explanation: The only symmetrical molecules is carbon dioxide and it therefore cannot possibly be polar. You might be tempted to pick ozone (O3) because it has nonpolar bonds but if you draw the lewis structure, you’ll see that resonance puts a partial negative charge on the peripheral oxygens and that the structure is angular (see Worksheet 4 2007) – so the molecule ends up being polar! 3. Which of the following statements are correct? I. If a central atom is bonded to five other atoms, it is hypovalent and trigonal bipyramidal. II. If a central atom is bonded to three identical atoms, it must be nonpolar. III. A molecule of generalized formula AB2U2 always has an angular shape. 1. II only 2. I only 3. III only correct 4. I and II 5. I and III6. II and III 7. I, II, and III Explanation: I is false because having five bonds would mean TEN electrons on one atom which has more than the octet rule requires (namely eight) so it is HYPERvalent not hypovalent. The shape is correct. II is false because it could be like ammonia, NH3, and be trigonal pyramidal and polar. III is true – look it up. You should be able to name the shape of any molecule, generalized or specific. Here we have used A to mean the central atom, B to represent a bond to another atom, and U to represent a lone pair. 4. What bond angles exist in PCl5? 1. 109.5º, 180º 2. 90º, 60º 3. 120º 4. 30º, 109.5º 5. 90º, 120º, 180º correct Explanation: The shape is trigonal bipyramidal so the angles are 90, 120 and 180. Look it up if you don’t believe me. You should know the bond angles between every three atoms in all of our structures. 5. Which of the following statements are true regarding Valence Bond Theory? I. The number of orbitals is conserved when hybridizing. II. An sp orbital has twenty five percent more s character than an sp3 orbital. III. Valence bond theory is lame. 1. I only 2. II only 3. III only 4. I and II correct 5. I and III 6. II and III 7. I, II and III Explanation: I is true – if you hybridize three orbitals together you get three out. For instance, one s + one p + one p = three sp2. II is true – the percent character of a hybrid orbital is just what proportion of the original orbitals is that type. So for sp 1 s / 2 orbitals = 50% s character. For sp3, 1 s / 4 orbitals = 25% s character. This describes how much the orbital is like an s orbital. So you would expect the sp orbital to be more spherical and closer to the nucleus, while the sp3 orbital would be more dumb-bell like and farther from the nucleus. III is false – valence bond theory is totally awesome, duh. Are you even TRYING to learn? 6. Which of the following molecules is matched with its correct electronic geometry? 1. BH3 , trigonal planar correct 2. H2O, angular 3. NH3 , trigonal planar 4. XeF4 , tetrahedral 5. CH4, square planar Explanation: BH3 is the only correctly assigned one. Angular and square planar are both MOLECULAR geometries so they are obviously wrong without even having to draw them. Obviously… And NH3 has a lone pair so it would be tetrahedral. XeF4 has TWO lone pairs so it would be octahedral. Make sure you know the difference between electronic and molecular geometries and can assign them readily. 7. What molecular geometries can you find in this molecule? H.. | H3C—C≡C-O—C—NH2 ‘’ | H I. tetrahedral II. linear III. angular IV. trigonal bipyramidal V. trigonal planar VI. trigonal pyramidal VII. Waldo 1. I, II, III, VI correct 2. I, II, V 3. I II, VI 4. II and V 5. II, III and VI 6. VII only Explanation: From left to right, the geometries around the central atoms are tetrahedral, linear, angular, and trigonal pyramidal. If you picked VII. Waldo you are awesome at life but you fail at chemistry. 8. How many sigma (σ) and pi (π) bonds are in the Lewis structure for HCOCHCHCH3? 1. 7 σ and 1 π 2. 10 σ and 2 π correct 3. 12 σ and 0 π 4. 8 σ and 2 π 5. 7 σ and 2 π Explanation: A single bond is one sigma bond, a double bond is one sigma and one pi bond, and a triple bond is one sigma and two pi bonds. The molecule looks like this so ….just count them? :O: H H H || | | | H—C—C=C—C—H | H 9. In carbon dioxide, the carbon must hybridize its ______ atomic orbitals to make _____ to bond to the oxygens. 1. 2s and 2p; an sp orbital 2. 1s and 1p; two sp orbitals 3. 2s and 2p; two sp orbitals correct 4. 1s, 2s, and 2p; three sp orbitals 5. 1s, 2s and 2p; two sp orbitals Explanation: Bonding involves the valence electrons. For carbon, this means its 2s and 2p electrons. It would not use its 1s electrons because they are not valence. It would
View Full Document
Unlocking...