What if sp3 hybrid orbitals were higher in energy than the p orbitals in the free atom? How would this affect our model of bonding?
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Textbook Solutions for Chemical Principles
Question
What if the p2p orbitals were lower in energy than the s2p orbitals? What would you expect the B2 molecular orbital energy level diagram to look like (without considering p-s mixing)? Compare your expected diagram to Figs. 14.37 and 14.38 and state the differences for each.
Solution
The first step in solving 14 problem number 2 trying to solve the problem we have to refer to the textbook question: What if the p2p orbitals were lower in energy than the s2p orbitals? What would you expect the B2 molecular orbital energy level diagram to look like (without considering p-s mixing)? Compare your expected diagram to Figs. 14.37 and 14.38 and state the differences for each.
From the textbook chapter Orbitals you will find a few key concepts needed to solve this.
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full solution
What if the p2p orbitals were lower in energy than the s2p
Chapter 14 textbook questions
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Chapter 14: Problem 14 Chemical Principles 8
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Chapter 14: Problem 14 Chemical Principles 8
What if the p2p orbitals were lower in energy than the s2p orbitals? What would you expect the B2 molecular orbital energy level diagram to look like (without considering p-s mixing)? Compare your expected diagram to Figs. 14.37 and 14.38 and state the differences for each.
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Chapter 14: Problem 14 Chemical Principles 8
What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of bonding orbitals and antibonding orbitals which is lower in energy? Explain.
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Chapter 14: Problem 14 Chemical Principles 8
Explain the difference between the s and p MOs for homonuclear diatomic molecules. How are bonding orbitals and antibonding orbitals different? Why are there two p MOs and one s MO? Why are the p MOs degenerate?
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Chapter 14: Problem 14 Chemical Principles 8
Compare Figs. 14.38 and 14.40. Why are they different? Because B2 is known to be paramagnetic, the p2p and s2p MOs must be switched from our first prediction. What is the rationale for this? Why might one expect the s2p to be lower in energy than the p2p? Why cant we use diatomic oxygen to help us decide whether the s2p or p2p is lower in energy?
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Chapter 14: Problem 14 Chemical Principles 8
Which of the following would you expect to be more favorable energetically? Explain. a. An H2 molecule in which enough energy is added to excite one electron from a bonding to an antibonding MO b. Two separate H atoms
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Chapter 14: Problem 14 Chemical Principles 8
Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as s or p
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Chapter 14: Problem 14 Chemical Principles 8
Which is the more correct statement: The methane molecule (CH4) is a tetrahedral molecule because it is sp3 hybridized or The methane molecule (CH4) is sp3 hybridized because it is a tetrahedral molecule? What, if anything, is the difference between these two statements?
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Chapter 14: Problem 14 Chemical Principles 8
Compare and contrast the MO model with the LE model. When is each useful?
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Chapter 14: Problem 14 Chemical Principles 8
What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?
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Chapter 14: Problem 14 Chemical Principles 8
Why do we hybridize atomic orbitals to explain the bonding in covalent compounds? What type of bonds form from hybrid orbitals, s or p? Explain.
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Chapter 14: Problem 14 Chemical Principles 8
What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigonal planar arrangement of electron pairs? A linear arrangement of electron pairs? How many unhybridized p atomic orbitals are present when a central atom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? What are the unhybridized p atomic orbitals used for?
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Chapter 14: Problem 14 Chemical Principles 8
Why are d orbitals sometimes used to form hybrid orbitals? Which period of elements does not use d orbitals for hybridization? If necessary, which d orbitals (3d, 4d, 5d, or 6d) would sulfur use to form hybrid orbitals requiring d atomic orbitals? Answer the same question for arsenic and for iodine.
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Chapter 14: Problem 14 Chemical Principles 8
The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double bond and a triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?
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Chapter 14: Problem 14 Chemical Principles 8
Give the expected hybridization for the molecular structures illustrated below. a. d. b. c. e.
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Chapter 14: Problem 14 Chemical Principles 8
Use the localized electron model to describe the bonding in CCl4.
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Chapter 14: Problem 14 Chemical Principles 8
Use the LE model to describe the bonding in H2O.
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Chapter 14: Problem 14 Chemical Principles 8
Use the LE model to describe the bonding in H2CO and C2H2. Carbon is the central atom in H2CO, and C2H2 exists as HCCH.
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Chapter 14: Problem 14 Chemical Principles 8
The space-filling models of ethane and ethanol are shown below. H O C Ethane (C2H6) Ethanol (C2H5OH) Use the LE model to describe the bonding in ethane and ethanol.
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Chapter 14: Problem 14 Chemical Principles 8
The space-filling models of hydrogen cyanide and phosgene are shown below. Cl N H O C Hydrogen cyanide (HCN) Phosgene (COCl2)Use the LE model to describe the bonding in hydrogen cyanide and phosgene.
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Chapter 14: Problem 14 Chemical Principles 8
Give the expected hybridization of the central atom for the molecules or ions in Exercises 63, 64, and 66 from Chapter 13.
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Chapter 14: Problem 14 Chemical Principles 8
Give the expected hybridization of the central atom for the molecules in Exercises 99 and 100 from Chapter 13.
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Chapter 14: Problem 14 Chemical Principles 8
Urea, a compound formed in the liver, is one of the ways humans excrete nitrogen. The Lewis structure for urea is A AA H A O H H H O O NO OC N Using hybrid orbitals for carbon, nitrogen, and oxygen, determine which orbitals overlap to form the various bonds in urea
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Chapter 14: Problem 14 Chemical Principles 8
Indigo is the dye used in coloring blue jeans. The term navy blue is derived from the use of indigo to dye British naval uniforms in the eighteenth century. The structure of the indigo molecule is C N H H O C C C C C C H H H C O C H H C C C C C C H H H C N a. How many s bonds and p bonds exist in the molecule? b. What hybrid orbitals are used by the carbon atoms in the indigo molecule?
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Chapter 14: Problem 14 Chemical Principles 8
For each of the following molecules, write a Lewis structure, predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. a. CF4 g. AsF5 b. NF3 h. KrF2 c. OF2 i. KrF4 d. BF3 j. SeF6 e. BeH2 k. IF5 f. TeF4 l. IF3
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Chapter 14: Problem 14 Chemical Principles 8
Predict the hybrid orbitals used by the sulfur atom(s) in each of the following. Also predict the molecular structure, including bond angle(s). a. SO2 b. SO3 c. S2O3 2 S O O O S A A O O 2d. S2O8 2 O O O O O O O O S S A A O O A A O O O 2 e. SO3 22 h. SF4 f. SO4 22 i. SF6 g. SF2 j. F3SOSF
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Chapter 14: Problem 14 Chemical Principles 8
Why must all six atoms in C2H4 be in the same plane?
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Chapter 14: Problem 14 Chemical Principles 8
The allene molecule has the following Lewis structure: G D G D P P H H H H C C C Must all four hydrogen atoms lie in the same plane? If not, what is the spatial relationship among them? Why?
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Chapter 14: Problem 14 Chemical Principles 8
Biacetyl and acetoin are added to margarine to make it taste more like butter. CH3 CH3 O O OC C O O O O O A B B B CH3 CH CH3 OH C O Biacetyl Acetoin Complete the Lewis structures, predict values for all COCOO bond angles, and give the hybridization of all the carbon atoms in these two compounds. Must the four carbons and two oxygens in biacetyl lie in the same plane? How many s bonds and how many p bonds are there in biacetyl and acetoin?
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Chapter 14: Problem 14 Chemical Principles 8
Many important compounds in the chemical industry are derivatives of ethylene (C2H4). Two of them are acrylonitrile and methyl methacrylate. H H H H C C C C C CH3 CH3 O O H C N O O G G G S J DG D D D a P P b c d e f Acrylonitrile Methyl methacrylate Complete the Lewis structures for these molecules, showing all lone pairs. Give approximate values for bond angles a through f, and give the hybridization of all carbon atoms. In acrylonitrile and methyl methacrylate indicate which atoms in each molecule must lie in the same plane. How many s bonds and how many p bonds are there in acrylonitrile and methyl methacrylate?
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Chapter 14: Problem 14 Chemical Principles 8
One of the first drugs to be approved for use in treatment of acquired immune deficiency syndrome (AIDS) was azidothymidine (AZT). Complete the Lewis structure of AZT. A A A A A A B B O O O C C C C N N N N N C C C O C H HH H H H H C H H H OO CH3 a. How many carbon atoms use sp3 hybridization? b. How many carbon atoms use sp2 hybridization? c. Which atom is sp hybridized? d. How many s bonds are in the molecule? e. How many p bonds are in the molecule? f. What is the NONON bond angle in the azide (ON3) group? g. What is the HOOOC bond angle in the side group attached to the five-membered ring? h. What is the hybridization of the oxygen atom in the OCH2OH group?
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Chapter 14: Problem 14 Chemical Principles 8
Hot and spicy foods contain molecules that stimulate pain-detecting nerve endings. Two such molecules are piperine and capsaicin: G D CHP (CH2)3 CH2 CH2 CH3 CH3 O O CHCH PCH C NO H H H O H O N H3CO O H C C H H CH CH CH HO H H B O a g h j i k l b c d f HH H HH H H H H H e G D G G G G D D B J A Piperine Capsaicin Piperine is the active compound in white and black pepper, and capsaicin is the active compound in chili peppers. The ring structures in piperine and capsaicin are shorthand notation. Each point where lines meet represents a carbon atom.a. Complete the Lewis structures for piperine and capsaicin, showing all lone pairs of electrons. b. How many carbon atoms are sp, sp2, and sp3 hybridized in each molecule? c. Which hybrid orbitals are used by the nitrogen atoms in each molecule? d. Give approximate values for the bond angles marked a through l in the above structures.
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Chapter 14: Problem 14 Chemical Principles 8
Two molecules used in the polymer industry are azodicarbonamide and methyl cyanoacrylate. Their structures are O O O O N N N NH2 N H2C CH3 C C C C C H H a b e f h d g c Azodicarbonamide Methyl cyanoacrylate Azodicarbonamide is used in forming polystyrene. When added to the molten plastic, it decomposes to nitrogen, carbon monoxide, and ammonia gases, which are captured as bubbles in the molten polymer. Methyl cyanoacrylate is the main ingredient in super glue. As the glue sets, methyl cyanoacrylate polymerizes across the carboncarbon double bond. (See Chapter 21.) a. Complete the Lewis structures showing all lone pairs of electrons. b. Which hybrid orbitals are used by the carbon atoms in each molecule and the nitrogen atoms in azodicarbonamide? c. How many p bonds are present in each molecule? d. Give approximate values for the bond angles marked a through h in the above structures.
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Chapter 14: Problem 14 Chemical Principles 8
The antibiotic thiarubin-A was discovered by studying the feeding habits of wild chimpanzees in Tanzania. The structure for thiarubin-A is H3C C C C C CCCC CH CH2 CC S S H H a. Complete the Lewis structures showing all lone pairs of electrons. b. Indicate the hybrid orbitals used by the carbon and sulfur atoms in thiarubin-A. c. How many s and p bonds are present in this molecule?
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Chapter 14: Problem 14 Chemical Principles 8
The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide (CO2), and carbon suboxide (C3O2). The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).
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Chapter 14: Problem 14 Chemical Principles 8
Consider the following molecular orbitals formed from the combination of two hydrogen 1s orbitals: a. Which is the bonding molecular orbital, and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?
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Chapter 14: Problem 14 Chemical Principles 8
Sketch the molecular orbital and label its type (s or p; bonding or antibonding) that would be formed when the following atomic orbitals overlap. Explain your labels. a. + + b. + + c. + + d. +
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Chapter 14: Problem 14 Chemical Principles 8
What are the relationships among bond order, bond energy, and bond length? Which of these can be measured? Distinguish between the terms paramagnetic and diamagnetic. What type of experiment can be done to determine if a material is paramagnetic?
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Chapter 14: Problem 14 Chemical Principles 8
What modification to the molecular orbital model was made from the experimental evidence that B2 is paramagnetic?
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Chapter 14: Problem 14 Chemical Principles 8
A Lewis structure obeying the octet rule can be drawn for O2 as follows: P O O Use the molecular orbital energy-level diagram for O2 to show that the above Lewis structure corresponds to an exicted state.
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Chapter 14: Problem 14 Chemical Principles 8
How does molecular orbital theory explain the following observations? a. H2 is stable, whereas He2 is unstable. b. B2 and O2 are paramagnetic, whereas C2, N2, and F2 are diamagnetic. c. N2 has a very large bond energy associated with it. d. NO1 is more stable than NO2.
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Chapter 14: Problem 14 Chemical Principles 8
Why does the molecular orbital model do a better job in explaining the bonding in NO2 and NO than the hybrid orbital model?
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Chapter 14: Problem 14 Chemical Principles 8
Show how a hydrogen 1s atomic orbital and a fluorine 2p atomic orbital overlap to form bonding and antibonding MOs in the hydrogen fluoride molecule. Are they s or p MOs?
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Chapter 14: Problem 14 Chemical Principles 8
Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. H2 1, H2, H2 2, H2 22 b. N2 22, O2 22, F2 22 c. Be2, B2, Ne2
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Chapter 14: Problem 14 Chemical Principles 8
Which charge(s) for the N2 molecule would give a bond order of 2.5?
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Chapter 14: Problem 14 Chemical Principles 8
In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed? a. H2 b. B2 c. C2 22 d. OF
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Chapter 14: Problem 14 Chemical Principles 8
In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. CN or NO O2 21 or N2 21
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Chapter 14: Problem 14 Chemical Principles 8
Using the molecular orbital model to describe the bonding in F2 1, F2, and F2 2, predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?
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Chapter 14: Problem 14 Chemical Principles 8
The transport of O2 in the blood is carried out by hemoglobin. Carbon monoxide can interfere with oxygen transport because hemoglobin has a stronger affinity for CO than for O2. If CO is present, normal uptake of O2 is prevented, depriving the body of needed oxygen. Using the molecular orbital model, write the electron configurations for CO and for O2. From your configurations, give two property differences between CO and O2.
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Chapter 14: Problem 14 Chemical Principles 8
Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. a. CN1 b. CN c. CN2
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Chapter 14: Problem 14 Chemical Principles 8
As compared with CO and O2, CS and S2 are very unstable molecules. Give an explanation based on the relative abilities of the sulfur and oxygen atoms to form p bonds.
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Chapter 14: Problem 14 Chemical Principles 8
Consider the following electron configuration: (s3s)2(s3s*)2(s3p)2(p3p)4(p3p*)4 Give four species that, in theory, would have this electron configuration.
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Chapter 14: Problem 14 Chemical Principles 8
What type of molecular orbital would result from the combination of two dxz atomic orbitals shown below? Assume the x axis is the internuclear axis. z x z x + +
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Chapter 14: Problem 14 Chemical Principles 8
Using an MO energy-level diagram, would you expect F2 to have a lower or higher first ionization energy than atomic fluorine? Why?
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Chapter 14: Problem 14 Chemical Principles 8
Use Figs. 14.45 and 14.46 to answer the following questions. a. Would the bonding MO in HF place greater electron density near the H or the F atom? Why? b. Would the bonding MO have greater fluorine 2p character, greater hydrogen 1s character, or an equal contribution from both? Why? c. Answer parts a and b for the antibonding MO in HF.
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Chapter 14: Problem 14 Chemical Principles 8
The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 pm and 424.7 kJ/mol, respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of a pz orbital from oxygen and the 1s orbital of hydrogen (the OOH bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1s character? c. Can the 2px orbital of oxygen form MOs with the 1s orbital of hydrogen? Explain. d. Knowing that only the 2p orbitals of oxygen interact significantly with the 1s orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. Energy 2pz 2px 2py 2s AO of hydrogen Molecular orbitals 1s AOs of oxygen Increasing stability e. Estimate the bond order for OH. f. Predict whether the bond order of OH1 is greater than, less than, or the same as that of OH. Explain.
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Chapter 14: Problem 14 Chemical Principles 8
What is delocalized p bonding, and what does it explain? Explain the delocalized p bonding system in C6H6 (benzene) and SO2.
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Chapter 14: Problem 14 Chemical Principles 8
Describe the bonding in the O3 molecule and the NO2 2 ion using the LE model. How would the molecular orbital model describe the p bonding in these two species?
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Chapter 14: Problem 14 Chemical Principles 8
Describe the bonding in the CO3 22 ion using the LE model. How would the molecular orbital model describe the p bonding in this species?
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Chapter 14: Problem 14 Chemical Principles 8
The space-filling model for benzoic acid is shown below. Benzoic acid (C6H5CO2H) H O C Describe the bonding in benzoic acid using the LE model combined with the molecular orbital model.
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Chapter 14: Problem 14 Chemical Principles 8
The space-filling model for benzoic acid is shown below. Benzoic acid (C6H5CO2H) H O C Describe the bonding in benzoic acid using the LE model combined with the molecular orbital model.
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Chapter 14: Problem 14 Chemical Principles 8
The infrared spectrum of 1H79Br shows the y 5 0 to y 5 1 transition at 2650. cm21. Determine the vibrational force constant for the HBr molecule. The atomic mass of 1H is 1.0078 u, and the atomic mass of 79Br is 78.918 u
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Chapter 14: Problem 14 Chemical Principles 8
If the force constant of 14N16O is 1550. N m21, determine the wave number of a line in the infrared spectrum of NO. The atomic mass of 14N is 14.003 u, and the atomic mass of 16O is 15.995 u.
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Chapter 14: Problem 14 Chemical Principles 8
The microwave spectrum of 12C16O shows that the transition from J 5 0 to J 5 1 requires electromagnetic radiation with a wavelength of 2.60 3 1023 m. a. Calculate the bond length of CO. See Exercise 60 for the atomic mass of 16O. b. Calculate the frequency of radiation absorbed in a rotational transition from the second to the third excited state of CO
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Chapter 14: Problem 14 Chemical Principles 8
Draw the idealized NMR spectra for the following compounds. a. CH2 CO H H H H H CH3 O Assume that the five hydrogen atoms in the benzene ring are equivalent with no spinspin coupling. b. CH3 C O C CH3 CH3 CH3 c. CH2CH3 H H H CH3CH2 CH2CH3
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Chapter 14: Problem 14 Chemical Principles 8
Consider the following idealized NMR multiplets. i. ii. iii. iv. Which is the correct multiplet for the underlined group in the following molecules?a. CH3 2 OCH CH2 CH3 b. CH3 CO CH2 O CH3 c. (CH3)3C C O CH2 O CH2 CH3 d. H2C CH CH2 C O CH2 CH3
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Chapter 14: Problem 14 Chemical Principles 8
The NMR spectra on the right are for the organic compounds C6H12 and C4H10O. Deduce the structures for these compounds. See Exercise 74 for a discussion of the bonding in organic compounds. The structure of C6H12 has one double bond, with the rest being single bonds, and the structure of C4H10O has only single bonds. Note that the TMS reference has been omitted in each spectrum. C6H12 C4H10O
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Chapter 14: Problem 14 Chemical Principles 8
Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. XeO3 c. XeOF4 e. XeO3F2 b. XeO4 d. XeOF2
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Chapter 14: Problem 14 Chemical Principles 8
Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule. a. BH3 H H B* H b. N2F2 FiN*iN*iF c. C4H6 H C C H H C H C H * * * * H
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Chapter 14: Problem 14 Chemical Principles 8
Two structures can be drawn for cyanuric acid: CO N N H N C O H H C OO O O G G C O N N N C O H H H C O G B M D a. Are these two structures the same molecule? Why or why not? b. Give the hybridization of the carbon and nitrogen atoms in each structure. c. Use bond energies (Table 13.6) to predict which form is more stable; that is, which contains the strongest bonds?
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Chapter 14: Problem 14 Chemical Principles 8
Using bond energies from Table 13.6, estimate the barrier to rotation around a carboncarbon double bond. To do this, consider what must happen to go from C C to H H Cl Cl G G D D P CC Cl H H Cl G G D D P in terms of making and breaking chemical bonds; that is, what happens to the p bond?
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Chapter 14: Problem 14 Chemical Principles 8
Show how a dxz atomic orbital and a pz atomic orbital combine to form a bonding molecular orbital. Assume the x axis is the internuclear axis. Is a s or a p molecular orbital formed? Explain.
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Chapter 14: Problem 14 Chemical Principles 8
Describe the bonding in the first excited state of N2 (the one closest in energy to the ground state) using the MO model. What differences do you expect in the properties of the molecule in the ground state and in the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.
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Chapter 14: Problem 14 Chemical Principles 8
Consider three molecules: A, B, and C. Molecule A has a hybridization of sp3. Molecule B has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule C consists of two s bonds and two p bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.
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Chapter 14: Problem 14 Chemical Principles 8
Complete the following resonance structures for POCl3: A A OO B (A) (B) Cl O Cl ClP A Cl OO O Cl ClP a. Would you predict the same molecular structure from each resonance structure? b. What is the hybridization of P in each structure? c. What orbitals can the P atom use to form the p bond in structure B? d. Which resonance structure would be favored on the basis of formal charges?
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Chapter 14: Problem 14 Chemical Principles 8
Using molecular orbital theory, explain why the removal of one electron in O2 strengthens bonding, while the removal of one electron in N2 weakens bonding.
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Chapter 14: Problem 14 Chemical Principles 8
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.] C N C C C O O H H H CC C H C H O H H H H b c e f d a g a. How many s bonds and p 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 p bonding? Explain.
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Chapter 14: Problem 14 Chemical Principles 8
Consider the following computer-generated model of caffeine: H O N C Complete a Lewis structure for caffeine in which all atoms have a formal charge of zero (as is typical with most organic compounds). How many C and N atoms are sp2 hybridized? How many C and N atoms are sp3 hybridized? sp hybridized? How many s and p bonds are there?
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Chapter 14: Problem 14 Chemical Principles 8
Aspartame is an artificial sweetener marketed under the name Nutra-Sweet. A partial Lewis structure for aspartame is shown below: C C OH O O C O H2N CH NH OCH3 CHCH2 CH2 Note that the six-sided ring is shorthand notation for a benzene ring (OC6H5). Benzene is discussed in Section 14.5. Complete the Lewis structure for aspartame. How many C and N atoms exhibit sp2 hybridization? How many C and O atoms exhibit sp3 hybridization? How many s and p bonds are in aspartame? Aspartame is an organic compound, and the Lewis structure follows the guidelines outlined in Exercise 74.
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Chapter 14: Problem 14 Chemical Principles 8
The N2O molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement is correct, NNO or NON? Explain your answer. b. On the basis of your answer in part a, write the Lewis structure of N2O (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in N N q OO be described in terms of orbitals?
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Chapter 14: Problem 14 Chemical Principles 8
Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: NCl3(g) 88n NCl2(g) 1 Cl(g) DH 5 375 kJ/mol ONCl(g) 88n NO(g) 1 Cl(g) DH 5 158 kJ/mol Rationalize the difference in the values of DH for these reactions, even though each reaction appears to involve the breaking of only one NOCl bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)
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Chapter 14: Problem 14 Chemical Principles 8
Draw the Lewis structures for SeO2, PCl3, NNO, COS, and PF3. Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit sp3 hybridization by the central atom? Which of the compounds have a linear molecular structure?
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Chapter 14: Problem 14 Chemical Principles 8
Draw the Lewis structures for TeCl4, ICl5, PCl5, KrCl4, and XeCl2. Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit d2sp3 hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?
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Chapter 14: Problem 14 Chemical Principles 8
A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. FClO3 is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for F3ClO, F2ClO2 1, and F3ClO2. What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?
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Chapter 14: Problem 14 Chemical Principles 8
Pelargondin is the molecule responsible for the red color of the geranium flower. It also contributes to the color of ripe strawberries and raspberries. The structure of pelargondin is: HO HC HC C C C OH CH CH OH OH H C C C C H H C O + C C C 1 4 2 3 How many s and p bonds exist in pelargondin? What is the hybridization of the carbon atoms marked 14?
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Chapter 14: Problem 14 Chemical Principles 8
Complete a Lewis structure for the compound shown below, then answer the following questions. How many carbon atoms are sp2 hybridized? How many CON bonds are formed by the overlap of an sp3 hybridized carbon with an sp3 hybridized nitrogen? How many lone pairs of electrons are in the Lewis structure of your molecule? How many p bonds are present? N N C C O O C C C C N H H H H N C C H H H H H H
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Chapter 14: Problem 14 Chemical Principles 8
Which of the following statements concerning SO2 is(are) true? a. The central sulfur atom is sp2 hybridized. b. One of the sulfuroxygen bonds is longer than the other(s). c. The bond angles about the central sulfur atom are about 120 degrees. d. There are two s bonds in SO2. e. There are no resonance structures for SO2.
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Chapter 14: Problem 14 Chemical Principles 8
Consider the molecular orbital electron configurations for N2, N2 1, and N2 2. For each compound or ion, fill in the table below with the correct number of electrons in each molecular orbital. MO N2 N2 1 N2 2 s2p* ________ ________ ________ p2p* ________ ________ ________ s2p ________ ________ ________ p2p ________ ________ ________ s2s* ________ ________ ________ s2s ________ ________ _____
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Chapter 14: Problem 14 Chemical Principles 8
Place the species B2 1, B2, and B2 2 in order of increasing bond length and increasing bond energy
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Chapter 14: Problem 14 Chemical Principles 8
Given that the ionization energy of F2 2 is 290. kJ/mol, do the following. a. Calculate the bond energy of F2 2. You will need to look up the bond energy of F2 and ionization energy of F2. b. Explain the difference in bond energy between F2 2 and F2 using MO theory.
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Chapter 14: Problem 14 Chemical Principles 8
Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some He2 molecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why He2 molecules can come to exist and why they dissociate.
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Chapter 14: Problem 14 Chemical Principles 8
a. A flask containing gaseous N2 is irradiated with 25-nm light. Using the following information, indicate what species can form in this flask during irradiation. N2(g) 88n 2N(g) DH 5 941 kJ/mol N2(g) 88n N2 1(g) 1 e2 DH 5 1501 kJ/mol N(g) 88n N1(g) 1 e2 DH 5 1402 kJ/mol b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of N2 (1501 kJ/mol) is greater than the first ionization energy of atomic nitrogen (1402 kJ/mol).
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Chapter 14: Problem 14 Chemical Principles 8
Use the MO model to determine which of the following has the smallest ionization energy: N2, O2, N2 22, N2 2, O2 1. Explain your answer.
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Chapter 14: Problem 14 Chemical Principles 8
Cholesterol (C27H46O) has the following structure: CH3 CH3 CH3 H3C CH3 H H H HO In such shorthand structures, each point where lines meet represents a carbon atom, and most H atoms are not shown. Draw the complete structure, showing all carbon and hydrogen atoms. (There will be four bonds to each carbon atom.) Indicate which carbon atoms use sp2 or sp3 hybrid orbitals. Are all carbon atoms in the same plane, as implied by the structure?
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Chapter 14: Problem 14 Chemical Principles 8
Arrange the following from lowest to highest ionization energy: O, O2, O2 2, O2 1. Explain your answer.
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Chapter 14: Problem 14 Chemical Principles 8
Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that CO is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: MOCqO a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule, Figs. 14.45 and 14.46.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.
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Chapter 14: Problem 14 Chemical Principles 8
Use the MO model to explain the bonding in BeH2. When constructing the MO energy-level diagram, assume that the Be 1s electrons are not involved in bond formation
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Chapter 14: Problem 14 Chemical Principles 8
In Exercise 71 in Chapter 13, the Lewis structures for benzene (C6H6) were drawn. Using one of the Lewis structures, estimate DHf for C6H6(g) using bond energies and given the standard enthalpy of formation of C(g) is 717 kJ/mol. The experimental DHf value for C6H6(g) is 83 kJ/mol. Explain the discrepancy between the experimental value and the calculated DHf value for C6H6(g).
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Chapter 14: Problem 14 Chemical Principles 8
For each of the following chemical formulas, an NMR spectrum is described, including relative overall areas (intensities) for the various signals given in parentheses. Draw the structure of a compound having the specific formula that would give the described NMR spectrum. (Hint: All of these formulas represent organic compounds. Lewis structures for organic compounds typically have all atoms in the compound with a formal charge of zero. This is the case in this problem.) a. C2H3Cl3; NMR has one singlet signal. b. C3H6Cl2; NMR has a triplet (4) and a quintet (2) signal. c. C3H6O2; NMR has a singlet (1), a quartet (2), and a triplet (3) signal. d. C5H10O; NMR has a heptet (1), a singlet (3), and a doublet (6) signal. e. C3H6O; NMR has a triplet (3), a quintet (2), and a triplet (1) signal.
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Chapter 14: Problem 14 Chemical Principles 8
Structural isomers are compounds that have the same chemical formula but the atoms are bonded together differently giving different compounds. Consider the two structural isomers having the formula C2H6O. Draw the two isomers, and describe what should be seen in the NMR spectrum for each isomer (types of signals and relative overall intensities). (Hint: Both of the isomers are organic compounds. Lewis structures for organic compounds typically have all atoms in the compound with a formal charge of zero. This is the case in this problem.)
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Chapter 14: Problem 14 Chemical Principles 8
The sp2 hybrid atomic orbitals have the following general form: F1 5 "1/3Fs 1 2AFpx F2 5 "1/3Fs 2 AFpx 1 BFpy F3 5 "1/3Fs 2 AFpx 2 BFpy where Fs, Fpx, and Fpy represent orthonormal (normalized and orthogonalized) atomic orbitals. Calculate the values of A and B.
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