Concept explainers
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 σ and π bonds are in aspartame?
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Chapter 9 Solutions
Chemistry: Cengage Technology Edition
- Chloromethane has the Lewis structure _______________________________ The carbon atom is sharing 4 electron pairs. In each shared pair the carbon atom “owns” 1 electron. The number of electrons that “belong” to carbon is ___. Carbon, being a Group ___ element would have 4 , outer shell electrons in the unbonded, neutral state. Therefore, the carbon atom in chloromethane has a formal charge of zero.arrow_forwardThe study of carbon-containing compounds and their properties is called organic chemistry. Besides carbon atoms, organic compounds also can contain hydrogen, oxygen, and nitrogen atoms (as well as other types of atoms). A common trait of simple organic compounds is to have Lewis structures where all atoms have a formal charge of zero. Consider the following incomplete Lewis structure for an organic compound called histidine (an amino acid), which is one of the building blocks of proteins found in our bodies: Draw a complete Lewis structure for histidine in which all atoms have a formal charge of zero.arrow_forwardVitamin 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_forward
- A complete Lewis structure must show all nonzero formal charges. Complete each of thefollowing Lewis structures by adding any missing formal charges.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_forwardA common trait of simple organic compounds is to have Lewis structures where all atoms have a formal charge of zero. Consider the following incomplete Lewis structure for an organic compound called methyl cyanoacrylate, the main ingredient in Super Glue. Draw a complete Lewis structure for methyl cyanoacrylate in which all atoms have a formal charge of zero.arrow_forward
- The cyclohexane carboxylate anion has a Lewis structure Pushing a pair of unshared electrons away from the negatively charged oxygen atom and, at the same time, pushing a pair of pi electrons toward the other oxygen will generate a second resonance structure. Thus,arrow_forwardShow all possible resonance structures for each of the following molecules or ions: (a) nitrate ion, NO3 (b) nitric acid, HNO3 (c) dinitrogen monoxide (nitrous oxide, laughing gas), N2O (where the bonding is in the order NNO)arrow_forwardConsider the following compounds: CO2, SO2, KrF2, SO3, NF3, IF3, CF4, SF4, XeF4, PF5, TF5, and SCl6. These 12 compounds are all examples of different molecular structures. Draw the Lewis structures for each and predict the molecular structures. Predict the bond angles and the polarity of each. (A polar molecule has a net dipole moment, while a nonpolar molecule does not.) See Exercises 25 and 26 for the molecular structures based on the trigonal bipyramid and the octahedral geometries.arrow_forward
- NOx and SOx are industrial air pollutants of grave concern. Draw the correct Lewis diagrams including formal charges on the atoms and resonance structures as appropriate for the following examples of these species: (Hint: don't use any Lewis structures with formal changes exceeding +1 or -1) Nitrous oxide, N2O, and disulfur monoxide, S2O The nitrogen trioxide radical, NO3- and sulfur trioxide, SO3; Use VSERP to assign the shape to these 4 molecules. Describe the polarity of each molecule. Which one is more polar? Justify your answer in terms of dipole moments and geometry.arrow_forwardDiazomethane (CH2N2) is an important reagent for the methylation of some organic molecules. Complete Parts 1 and 2 below about this unique reagent. Draw the Lewis structure of diazomethane (CH2N2) that contains a formal charge on carbon and nitrogen. Be sure to include all lone pairs of electrons and formal charges.arrow_forwardCarbon disulfide is a colorless, volatile, highly flammable liquid with a very unpleasant smell that is used as a solvent in some laboratory applications. The formula for carbon disulfide is CS2. Formal charges can be used to decide whether its connectivity is more likely to be C–S–S or S–C–S. Part1: Add any nonzero formal charges to the atoms as applicable for the Lewis structure shown with connectivity C–S–S. All valence electrons have been included, and this structure follows the octet rule. If formal charges are equal to zero, they should not be included. Part2: Add any nonzero formal charges to the atoms as applicable for the Lewis structure shown with connectivity C–S–S. All valence electrons have been included. If formal charges are equal to zero, they should not be included. (This structure is a resonance form of the structure in Part 1.) part 3: Add any nonzero formal charges to the atoms as applicable for the Lewis structure shown with connectivity S–C–S. All valence…arrow_forward
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