Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Textbook Question
Chapter 10, Problem 10.43QE
For each of the following molecules, complete the Lewis structure and use the VSEPR model to determine the bond angles around each central atom. Note that the drawings are only skeleton structures and may depict the angles incorrectly.
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Chemistry: Principles and Practice
Ch. 10 - Prob. 10.1QECh. 10 - Prob. 10.2QECh. 10 - Prob. 10.3QECh. 10 - Prob. 10.4QECh. 10 - Prob. 10.5QECh. 10 - Prob. 10.6QECh. 10 - Prob. 10.7QECh. 10 - Prob. 10.8QECh. 10 - Prob. 10.9QECh. 10 - Prob. 10.10QE
Ch. 10 - Which atomic orbitals overlap to form the bonds in...Ch. 10 - Prob. 10.12QECh. 10 - Identify the hybrid orbitals used by boron in BCl3...Ch. 10 - Identify the hybrid orbitals used by antimony in...Ch. 10 - Prob. 10.15QECh. 10 - Prob. 10.16QECh. 10 - Prob. 10.17QECh. 10 - Prob. 10.18QECh. 10 - Prob. 10.19QECh. 10 - Prob. 10.20QECh. 10 - Compare and contrast the molecular orbital and...Ch. 10 - Describe the bonding in molecular orbital terms...Ch. 10 - Prob. 10.23QECh. 10 - Prob. 10.24QECh. 10 - Prob. 10.25QECh. 10 - Prob. 10.26QECh. 10 - Prob. 10.27QECh. 10 - Prob. 10.28QECh. 10 - Prob. 10.29QECh. 10 - Prob. 10.30QECh. 10 - Prob. 10.31QECh. 10 - Prob. 10.32QECh. 10 - Prob. 10.33QECh. 10 - Prob. 10.34QECh. 10 - Prob. 10.35QECh. 10 - Prob. 10.36QECh. 10 - Prob. 10.37QECh. 10 - Prob. 10.38QECh. 10 - Prob. 10.39QECh. 10 - Use the VSEPR model to predict the bond angles...Ch. 10 - Prob. 10.41QECh. 10 - Prob. 10.42QECh. 10 - For each of the following molecules, complete the...Ch. 10 - Prob. 10.44QECh. 10 - Prob. 10.45QECh. 10 - Prob. 10.46QECh. 10 - Indicate which molecules are polar and which are...Ch. 10 - Prob. 10.48QECh. 10 - Indicate which of the following molecules are...Ch. 10 - Prob. 10.50QECh. 10 - Prob. 10.51QECh. 10 - Prob. 10.52QECh. 10 - Prob. 10.53QECh. 10 - Prob. 10.54QECh. 10 - Prob. 10.55QECh. 10 - Prob. 10.56QECh. 10 - Prob. 10.57QECh. 10 - Prob. 10.58QECh. 10 - Prob. 10.59QECh. 10 - Prob. 10.60QECh. 10 - Prob. 10.61QECh. 10 - Prob. 10.62QECh. 10 - Prob. 10.63QECh. 10 - Prob. 10.64QECh. 10 - Prob. 10.65QECh. 10 - Prob. 10.66QECh. 10 - Prob. 10.67QECh. 10 - Prob. 10.68QECh. 10 - Prob. 10.69QECh. 10 - Prob. 10.70QECh. 10 - Prob. 10.71QECh. 10 - Prob. 10.72QECh. 10 - Identify the orbitals on each of the atoms that...Ch. 10 - Prob. 10.74QECh. 10 - Prob. 10.75QECh. 10 - How many sigma bonds and how many pi bonds are...Ch. 10 - Give the hybridization of each central atom in the...Ch. 10 - Prob. 10.78QECh. 10 - Prob. 10.79QECh. 10 - Prob. 10.80QECh. 10 - Prob. 10.81QECh. 10 - Predict the hybridization at each central atom in...Ch. 10 - Prob. 10.83QECh. 10 - Tetrafluoroethylene, C2F4, is used to produce...Ch. 10 - Prob. 10.85QECh. 10 - Prob. 10.86QECh. 10 - Prob. 10.87QECh. 10 - Prob. 10.88QECh. 10 - Prob. 10.89QECh. 10 - Prob. 10.90QECh. 10 - Prob. 10.91QECh. 10 - Prob. 10.92QECh. 10 - Prob. 10.93QECh. 10 - Prob. 10.94QECh. 10 - Prob. 10.95QECh. 10 - Prob. 10.96QECh. 10 - Prob. 10.97QECh. 10 - Prob. 10.98QECh. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - Prob. 10.102QECh. 10 - Prob. 10.103QECh. 10 - Prob. 10.104QECh. 10 - Prob. 10.105QECh. 10 - Following are the structures of three isomers of...Ch. 10 - The ions ClF2 and ClF2+ have both been observed....Ch. 10 - Aspirin, or acetylsalicylic acid, has the formula...Ch. 10 - Aspartame is a compound that is 200 times sweeter...Ch. 10 - Prob. 10.110QECh. 10 - Prob. 10.111QECh. 10 - Calcium cyanamide, CaNCN, is used both to kill...Ch. 10 - Histidine is an essential amino acid that the body...Ch. 10 - Formamide, HC(O)NH2, is prepared at high pressures...Ch. 10 - Prob. 10.115QECh. 10 - Prob. 10.116QECh. 10 - Prob. 10.117QECh. 10 - Prob. 10.118QECh. 10 - Prob. 10.119QECh. 10 - Prob. 10.120QECh. 10 - Prob. 10.121QECh. 10 - Prob. 10.122QECh. 10 - Prob. 10.123QECh. 10 - Prob. 10.124QECh. 10 - Two compounds have the formula S2F2. Disulfur...Ch. 10 - Prob. 10.126QECh. 10 - Prob. 10.127QE
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- Vitamin B6 is an organic compound whose deficiency in the human body can cause apathy, irritability, and an increased susceptibility to infections. Below is an incomplete Lewis structure for vitamin B6. Complete the Lewis structure and answer the following questions. Hint: Vitamin B6 can be classified as an organic compound (a compound based on carbon atoms). The majority of Lewis structures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero. a. How many bonds and bonds exist in vitamin B6? b. Give approximate values for the bond angles marked a through g in the structure. c. How many carbon atoms are sp2 hybridized? d. How many carbon, oxygen, and nitrogen atoms are sp3 hybridized? e. Does vitamin B6 exhibit delocalized bonding? Explain.arrow_forwardUse the VSEPR model to predict the bond angles around each central atom in the following Lewis structures (benzene rings are frequently pictured as hexagons, without the letter for the carbon atom at each vertex). Note that the drawings do not necessarily depict the bond angles correctly.arrow_forwardMethylcyanoacrylate is the active ingredient in super glues. Its Lewis structure is (a) Give values for the three bond angles indicated. (b) Indicate the most polar bond in the molecule. (c) Circle the shortest carbon-oxygen bond. (d) Circle the shortest carbon-carbon bond.arrow_forward
- Aspartame is an artificial sweetener marketed under the name Nutra-Sweet. A partial Lewis structure for aspartame is shown below. Aspartame can be classified as an organic compound (a compound based on carbon atoms). The majority of Lewis structures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure above to give each atom a formal charge of zero when drawing the Lewis structure. Also note that the six-sided ring is shorthand notation for a benzene ring (C6H5). Benzene is discussed in Section 4-7. Complete the Lewis structure for aspartame. How many C and N atoms exhibit sp1 hybridization? How many C and O atoms exhibit sp3 hybridization? How many and bonds are in aspartame?arrow_forwardTwo different molecules have the formula C2H6O. One of the molecules has the oxygen atom bonded to both carbon atoms. The other molecule has the oxygen atom bonded to only one carbon atom while both carbon atoms are bonded to each other. Write Lewis structures for both of these compounds.arrow_forwardFormamide, HC(O)NH2, is prepared at high pressures from carbon monoxide and ammonia, and serves as an industrial solvent (the parentheses around the O indicate that it is bonded only to the carbon atom and that the carbon atom is also bonded to the H and the N atoms). Two resonance forms (one with formal charges) can be written for formamide. Write both resonance structures, and predict the bond angles about the carbon and nitrogen atoms for each resonance form. Are they the same? Describe how the experimental determination of the HNH bond angle could be used to indicate which resonance form is more important.arrow_forward
- A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.arrow_forwardThe molecular ion S3N3 has the cyclic structure All SN bonds are equivalent. (a) Give six equivalent resonance hybrid Lewis diagrams for this molecular ion. (b) Compute the formal charges on all atoms in the molecular ion in each of the six Lewis diagrams. (c) Determine the charge on each atom in the polyatomic ion, assuming that the true distribution of electrons is the average of the six Lewis diagrams arrived at in parts (a) and (b). (d) An advanced calculation suggests that the actual charge resident on each N atom is 0.375 and on each S atom is +0.041 . Show that this result is consistent with the overall +1 charge on the molecular ion.arrow_forward
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