solutions ch05.doc

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Chapter 5 Structure of Molecules

Answer to 5.2:

(a) Structure of 1-pentanol

H C C C C C O H

H

H

H

H

H

H

H

H

H

H

Structure of 2-pentanolH C C C C C H

H

H

H

H

H

H

H

O

H

H

H

(b) Based on a comparison with 1-butanol and 2-butanol in Table 5.1, the prediction (and

fact) is that the boiling point of the straight-chain alcohol, 1-pentanol, is likely to be

greater than that of the branched alcohol, 2-pentanol. The basic reason is that the

dispersion forces between molecules of straight-chain alcohols are greater than the

dispersion forces between branched alcohols. Therefore, more energy is required to

change from the liquid to the gas phase, reflected in higher boiling points for straight-

chain alcohols compared to branched alcohols of the same molecular weights.

(c)

OH OH

OH

OH

OH

OH

OH

HO

(R/S) (R/S)

(R/S)

8 different structural (constitutional) isomers for C5H12O alcohols;

those with "(R/S)" to the right of the structure can exist in twoenantiomeric (right/left stereoisomeric) forms

5.3. A student followed the procedure in a laboratory manual to prepare an ionic compound, a

green solid with the composition, Co(H2NCH2CH2NH2)2Cl3. She dissolved the compound

in water, warmed the solution, and found that the solution turned reddish purple. She

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evaporated the water, recrystallized the reddish-purple crystals, and analyzed them for

cobalt, chloride, and 1,2-diaminoethane, H2NCH2CH2NH2. She found that the reddish-

purple compound had the composition, Co(H2NCH2CH2NH2)2Cl3. What conclusion(s)

can you draw about the relationship(s) between the green and reddish-purple compounds?

Explain your reasoning.

Answer to 5.3: Since the molecular formulas of the green and reddish-purple compounds

are the same, the compounds must be isomers. (In Chapter 6, metal ion complexes are

introduced. The compounds here are metal ion complexes with each of the 1,2-

diaminoethane (ethylenediamine) molecules bonded at two of the six octahedral positions

around a central Co(III) ion and chlorides bonded at the other two positions. The green

compound has the chlorides bonded on opposite sides (trans) of the metal ion as far from

one another as possible in this structure. In the reddish-purple compound the two

chlorides are bonded at adjacent positions (cis).)

Section 5.2. Lewis Structures and Molecular Models of Isomers

5.4. Predict the relative values for the boiling points of 1-propanol and 2-propanol.

OHOH

1-propanol 2-propanol

Answer to 5.4: When comparing isomeric branched and straight-chain alkanes, the

branched alkanes invariably have lower boiling points than their straight-chain isomericcousins. A similar effect applies to the isomeric variants of other functional groups.

Therefore, we predict that 2-propanol boils at a lower temperature than 1-propanol. This

prediction is borne out by experiment. They boil at 82 °C and 97 °C respectively.

5.5. This is the Lewis structure for propane, C3H8:

H C C C H

H

H

H

H

H

H

(a) Use your model kit to make this model, and then draw a representation of thisstructure in three dimensions.

(b) What is the shape (geometry) around the central carbon atom?

(c) Are there any other isomers for propane, C3H8?

Answer to 5.5:



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(a)

H

C

HCH3

CH3

(b) tetrahedral

(c) There are no other isomers. At least four carbons are necessary before another order

of linkage of carbon atoms is possible for an alkane.

5.6. This is the Lewis structure of n-pentane, C5H12.

H C C C C C H

H

H

H

H

H

H

H

H

H

H

(a) How many valence electrons are represented in this structure?

(b) Including electrons in atomic cores, what is the total number of electrons implicitly

represented?

Answer to 5.6:

(a) 12(1) + 5(4) = 32 valence electrons

(b) 12(1) + 5(6) = 42 total electrons

5.7. The boiling points of the three pentane isomers are given in Table 5.2. Make molecular

models of these compounds and cover them with plastic wrap or aluminum foil to

represent the surface of each molecule. Explain why n-pentane has the highest boiling

point and neopentane has the lowest boiling point of this set of isomers. Hint: Consider

the amount of surface each molecule has to interact with its neighbors in the liquids.Answer to 5.7: Because hydrocarbons are nonpolar, the only intermolecular forces are

dispersion forces. The molecular volume of each pentane isomer must be the same

because the volume of each simply consists of five carbons and twelve hydrogens.

However, the surface area is not the same for the three isomers. The most nearly linear

isomer (n-pentane) has the greatest surface area and therefore has the most intermolecular

attractions due to dispersion forces. This in turn leads to the highest boiling point.

Neopentane has the most nearly spherical shape and has the smallest surface area of the

three pentane isomers. Therefore, it will have the lowest boiling point. (Recall fromgeometry that the sphere is the geometric solid that minimizes the surface area for any

given volume.) In general, branching leads to more compact structures and

correspondingly lower boiling points.

5.8. In Check This 5.11, you were asked to explain the boiling point data (Table 5.1) for the

four C4H10O alcohols. Draw the three ethers that also have the formula C4H10O and



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predict which one will have the lowest boiling point. Explain the reasoning for your

selection. Hint: See Problem 5.7.

Answer to 5.8: All three isomers have polar carbon-oxygen bonds, but only the circled

isomer is branched.

5.9. In Check This 5.7 you wrote Lewis structures and made models of the four isomeric

amines having the formula C3H9N. Predict which one will have the lowest boiling point

and explain the reasoning for your selection. Hint: Consider all the possible

intermolecular interactions in the liquids.Answer to 5.9: Two of the amines are linear (A and B) and two are branched to the same

extent (C and D). However, branching is not the most important factor influencing

boiling point in this comparison. Compounds A, B, and C are all capable of forming

intermolecular hydrogen bonds, whereas compound D is not. Since hydrogen bonding is

a stronger intermolecular force than dispersion, compound D (lacking hydrogen bonds)

has the lowest boiling point.

Section 5.3. Sigma Molecular Orbitals

5.10. Which electrons, core or valence, are generally responsible for chemical bonding?

Explain why.Answer to 5.10:

Valence electrons.

5.11. Explain the difference(s) between an atomic orbital and a molecular orbital.

Answer to 5.11: Atomic orbitals describe the energy and spatial distribution of electrons

with respect to a single atomic nucleus. Molecular orbitals describe the energy and


O OO

NH2NH

NH2

N

A B C D


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spatial distribution of valence electrons with respect to two (or more) atomic cores.

Molecular orbitals are formed from atomic orbitals.

5.12. Explain the difference(s) between a bonding and a nonbonding molecular orbital.

Answer to 5.12: Bonding orbitals contain the valence electrons involved in chemical

bonds. Nonbonding orbitals contain valence electrons which remain associated with a

particular atomic core and are not involved in bonds.

5.13. Why do two hydrogen atoms combine to form a hydrogen molecule (H2)? Explain using

(a) a Lewis structure for bonding.

(b) the MO model for bonding.

Answer to 5.13:

The Lewis model places the two separate s-electrons of the starting hydrogen atoms into

a single bond linking the two atoms (H• + •H —> H–H). In the special application of the octet rule for first row atoms, a complete outer shell of electrons consists of just two

electrons. The MO model places the two separate s-electrons of the starting hydrogen

atoms into a σ molecular orbital. The σ* orbital is empty and the bond order for the

molecule is 1.0.

5.14. Why don’t two helium atoms combine to form a helium molecule (He2)? Explain using

(a) a Lewis structure for bonding.

(b) the MO model for bonding.

Answer to 5.14:

Atomic helium has two electrons and already satisfies the octet rule. In the special

application of the octet rule for first row atoms, the MO model places two s-electrons of

the starting helium atoms into a σ molecular orbital. The σ* orbital also has two

electrons and the bond order for the molecule is zero. Therefore it has no tendency to

remain bonded.

5.15. Explain why it is possible for molecules to exist, if there is always a repulsive force

between positively charged nuclei?

Answer to 5.15:

Even though both repulsion of nuclei and kinetic energy of electron waves must be

accounted for, the attractive contributions of the potential energy makes the total energy

of the molecule lower than the combined energies of the separated atoms.



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5.16. Identify the number of valence electrons in each of the following:

(a) Na (b) CO2 (c) NH3 (d) Se

Answer to 5.16:

For atoms, see what column (group) the element belongs to. For compounds, any shared

or unshared electrons (as shown in a Lewis structure) count. Be careful not to double

count the shared electrons.

(a) 1 (b) 16 (c) 8 (d) 6

5.17. Consider Figure 5.4, Relative energies

for a two-nuclei-one-electron molecular

orbital system. It has been redrawn here

with curves 1-4 labeled.

(a) What information does line 1 give

you as two nuclei move towards each

other?

(b) What information does line 2 give

you as two nuclei move towards each

other?

(c) What information does line 3 give

you as two nuclei move towards each other?

(d) What information does line 4 give you as two nuclei move towards each other?

Answer to 5.17:

(a) Line 1 tells me that as two nuclei approach each other, the kinetic energy of the one-

electron molecular orbital system increases.

(b) Line 2 is the reference baseline for energy and does not change as the two nuclei

approach each other.

(c) Line 3 tells me that as two nuclei approach each other, the total energy of the one-

electron molecular orbital system falls to a minimum and then increases significantly.

(d) Line 4 tells me that as two nuclei approach each other, the potential energy of the

system decreases.



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Section 5.4. Sigma Molecular Orbitals and Molecular Geometry

5.18. In certain gas-phase reactions, methane can lose a hydrogen cation, H+, to form the

methide anion, CH3–.

(a)Write the Lewis structure for the methide anion.

(b) How would you describe the geometry of the sigma orbitals in the methide anion?

(c) How would you describe the geometry of the methide anion?

(d) Are your answers to parts (b) and (c) the same? Explain why or why not.

Answer to 5.18:

(a)

C H

H

H

(b) The sigma orbitals, three bonding and one nonbonding, are tetrahedrally arranged

around the carbon.

(c) The geometry (the arrangement of the nuclei) is a trigonal pyramid with carbon at the

apex and the hydrogens forming the base:

C

HH

H

(d) The arrangement of sigma orbitals and the geometry are not the same, because the

nonbonding electron pair is not included in the geometric description. The structure of the

methide ion is much like the structure of ammonia, Figure 5.10.

5.19. Sodium amide, NaNH2, is a white crystalline ionic compound.

(a) Write the Lewis structure for the amide anion, NH2–.

(b) How would you describe the geometry of the sigma orbitals in the amide anion?

(c) How would you describe the geometry of the amide anion?

(d) Are your answers to parts (b) and (c) the same? Explain why or why not.

Answer to 5.19:

(a)

N H

H

(b) The sigma orbitals, two bonding and two nonbonding, are tetrahedrally arranged

around the nitrogen.

(c) The geometry of the amide ion is bent:

N

HH



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(d) The arrangement of sigma orbitals and the geometry are not the same, because the

nonbonding electron pairs are not included in the geometric description. The structure of

the amide ion is much like the structure of water, Figure 5.10.

5.20. (a)How many valence electrons are there in the Lewis structure for IF5?

(b) How many total electrons, including all of the core electrons, are there in a molecule

of IF5?

(c) Write a Lewis structure for IF5.

(d) The geometry of the fluorine atoms in the IF5 molecule is square pyramidal (a

pyramid with a square base). Why is this different from the molecular shape of PF 5, in

which the geometry of the fluorine atoms is trigonal bipyramidal, as shown in Figure

5.11(b)?

Answer to 5.20:

(a) I

FF F

F F

(b) There are 1(7) + 5(7) = 42 valence electrons in the Lewis structure for IF5.

(c) There are 1(53) + 5(9) = 98 total electrons in a molecule of IF5.

(d) Note that the Lewis structure for IF5 has 42 valence electrons, but for PF5 has 40

valence electrons. This difference is caused by the fact that iodine is in Group VIIA andphosphorus is in Group VA. Therefore, while there are 12 electrons around the central

atom in IF5, there are only 10 electrons arranged around the central atom in IF5.

5.21. For each of the following molecules or ions, write a Lewis structure, including all

non bonding pairs of electrons.

(a) I3–

(b) BF4–

(c) SF4

(d) XeF4

(e) PF6–

Answer to 5.21:

(a) I3– has 28 valence electrons total (4 σ and 18 σn). The extra electron is associated

with the central iodine, which carries the formal charge (1–). The central iodine is

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surrounded by five pairs of electrons that will adopt a trigonal bipyramidal

arrangement with the two other iodines 180° apart at the opposite ends of the

structure. For clarity, only the electron pairs on the central iodine are shown in the 3-

d structure on the right.

(b) BF4– has 32 valence electrons total (8 σ and 24 σn). The extra electron is associated

with the central boron, which carries the formal charge (1–). The boron is surrounded

by four pairs of bonding electrons in a tetrahedral arrangement.

(c) SF4 has 34 valence electrons total (8 σ and 26 σn). The sulfur has five pairs of

electrons that will adopt a trigonal bipyramidal arrangement about this central atom.The nonbonding pair is located at one of the sites on the triangular base shared by the

two pyramids. The remaining sigma bonds define the geometry that looks like a

“see-saw.”

(d) XeF4 has 36 valence electrons total (8 σ and 28 σn). The xenon has six pairs of

electrons that will adopt an octahedral arrangement with the two nonbonding pairs of

electrons 180° apart at the opposite sides of the structure. The remaining bonded

substituents are in a square planar arrangement with respect to the xenon.


F

BF

F

F

F

BF

F

F

SF F

F F

F S F

F F

F S F

F F

XeF F

F

F

F F

F

F

Xe


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(e) PF6– has 48 valence electrons total (12 σ and 36 σn). The extra electron is

associated with the phosphorus, which carries the formal charge (1–). The phosphorus is

surrounded by six pairs of bonding electrons in an octahedral arrangement.

5.22. For each molecule or ion in Problem 5.21, explain why these are the observed geometries

(shapes).

(a) I3– is linear.

(b) BF4– is tetrahedral.

(c) SF4 is shaped like a “see-saw”, , with S at the fulcrum.

(d) XeF4 is square planar, with the Xe at the center of the square.

(e) PF6– is octahedral with the P in the center of the octahedron.

Answer to 5.22: See answer to 5.21.

Section 5.5. Multiple Bonds

5.23. Which second-period elements are capable of making double bonds under normal

conditions? Why are the others not included? Explain the reasoning for your choices.

Answer to 5.23:

Carbon, Nitrogen, Oxygen. Others are not included because the molecular orbitals are

larger in higher period atoms. Thus, they are more available to react and form products

with σ bonding orbitals.

5.24. Which elements never make double bonds? Explain the reasoning for your choices.Answer to 5.24:

Hydrogen and the halogens (Fluorine, Chlorine, Bromine, and Iodine). Hydrogen only

has one valence electron. The halogens have 7 valence electrons. Thus, only one

electron from another atom can form a single bond with the halogens.


P

F

F

F

F

F

F

F

F

F

F

F

P

F


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5.25. For each molecule, write a Lewis structure and determine the molecular geometry for

each carbon atom. Hint: each molecule has a multiple bond. Molecular models may be

helpful.

(a) C3H6 (propene has one carbon-carbon double bond)

(b) C3H4 (propyne has one carbon-carbon triple bond)

(c) C3H4 (allene has two carbon-carbon double bonds)

(d) C2H4O (ethanal has one carbon-oxygen double bond)

(e) C2H3N (acetonitrile has one carbon-nitrogen triple bond)

Answer to 5.25:

(a) C3H6 (propene)

(b) C3H4 (propyne)

(c) C3H4 (allene)

(d) C2H4O (ethanal)


CCCH

H

H

H

HH

tetrahedral

trigonal planar

trigonal planar

C C C

H

HH

H

tetrahedral

linear

linear

C C C

H

H

HH

trigonal planar

linear

CCO

H

H

HH

tetrahedral

trigonal planar


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(e) C2H3N (acetonitrile)

Section 5.6. Pi Molecular Orbitals

5.26. What is(are) the difference(s) between a sigma (σ ) and a pi (π ) orbital?

Answer to 5.26: Both are molecular orbitals. Sigma obitals are cylindrically symmetric

about the axis connecting two atomic cores. Pi bonds are formed from p atomic orbitals.

They are off-axis and have two halves separated by a nodal plane which includes the twoatomic cores. Sigma bonds are typically stronger than pi bonds.

5.27. Why do all molecules have sigma bonding orbitals, but not necessarily pi-bonding

orbitals?

Answer to 5.27: There is a two-electron sigma bonding orbital between every pair of

bonded atom cores in a molecule. A sigma orbital has the greatest electron density along

the bond axis, forming a bond with the lowest possible total energy for the two nuclei

system. A pi-bonding orbital is not needed if sigma bonding orbitals successfully account

for all of the valence electrons and every second period element has an octet of electrons.

5.28. If two carbon atoms are bonded by a sigma bonding orbital, why is a second bond

between those two atoms a pi orbital rather than another sigma bonding orbital?

Answer to 5.28: The pi bonding orbital has greater electron density away from the bond

axis, not directly between the bonded atom cores. The rise in electron wave interactions

would make two sigma bonds between atom cores energetically impossible.

5.29. Why are molecules with multiple bonds generally more reactive than similar compounds

without multiple bonds?Answer to 5.29: The pi bonding orbitals, having a considerable amount of electron

density away from the bond axis, are more readily available for interactions with other

molecules.


C C N

H

HH

tetrahedral

linear


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Section 5.7. Delocalized Orbitals

5.30. Consider the following molecules and ions. Circle each one that contains a π bond. For

those molecules with a π bond, place a D next to ones where the π bond is delocalized

and an L next to ones where theπ

bond is localized. Building models may be helpful.(a) H2CCH2 (d) NH3 (g) CH4

(b) H2CO (e) CO2 (h) SO2

(c) NO3- (f) CH3C(O)OH (i) H2O

Answer to 5.30:

Molecules or ions with π bonds: (a) H2CCH2 L (b) H2CO L (c) NO3- D

(e) CO2 D (or L) (f) CH3COOH D (h) SO2 D

5.31. Consider the hydrogen carbonate anion, (HO)CO2–, and the ozone molecule, O3.

(a) What is the total number of valence electrons in each species?(b) Draw the Lewis structure for each species.

(c) How many σ bonding orbitals are there around the central atom in each species?

(d) What geometry is predicted for each species?

(e) How many π bonding orbitals are there around the central atom in each species?

(f) How many localized and how many delocalized π bonding orbitals are there in each

species? Explain how you arrive at your answer.

(g) What are the bond orders for carbon-to-oxygen in (HO)CO2– and oxygen-to-oxygen

in O3? Use sketches of the delocalized orbitals, if any, to illustrate your answer. What, ifany, similarities are there between the delocalized orbitals and the bond orders in these

two species? Explain.

Answer to 5.31:

(a) HOCO2– has 24 valence electrons and O3 has 18.

(b)

C O

O

O

H

O

OO

(c) The central C atom in HOCO2–

is surrounded by three sigma bonding orbitals and thecentral O atom in O3 has two sigma bonding orbitals (and one sigma nonbonding orbital).

(d) The O atoms around the C in HOCO2– are arranged in a plane at 120° angles; the

geometry about the C atom is trigonal planar. Ozone, O3, is a bent molecule; the three

sigma orbitals are arranged roughly trigonally (but the angles are not all equal).



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(e) There is one pi bonding orbital around each central atom in these species.

(f) Each species has one delocalized pi orbital. In the HOCO2–, the pi orbital is

delocalized over the central C atom and the two O atoms not bonded to an H atom . In

ozone, the pi orbital is delocalized over all three atoms. In each case, a second Lewisstructure, identical in energy to the ones above, can be written:

C O

O

O

H

O

O O

Whenever two or more Lewis structures with the same energy can be written for a

species, neither is an accurate representation; the pi orbitals involved are delocalized.

(g) The bond orders between the C atom and the two O atoms not bonded to an H atom in

HOCO2– are each 1.5, one-and-one-half bonds. The same is true for the bonds between

the central O atom and the other O atoms in ozone. In both cases, the sigma bond

between the atoms is a full bond and the delocalized pi bond contributes one half a bond

to each bond, for a 1.5 bond. Although these species are quite different in many ways,

some aspects of their bonding (delocalization and 1.5 order bonds) are very much alike.

5.32. (a) We have said that molecules are likely to have delocalized pi orbitals, if there is

more than one energy-equivalent way to write localized pi orbitals. Write the Lewis

structure for carbon dioxide, as you did in Consider This 5.38. Would you expect the

molecule to have delocalized pi orbitals? Explain why or why not.

(b) The molecular orbital picture for carbon dioxide, Figure 5.24, shows two pi orbitals

delocalized over all three atomic cores in the molecule. Is this consistent with your

answer in part (a)? Explain why or why not.

(c) Another way to represent the localized pi orbitals in

carbon dioxide is shown in this molecular model. (The sigma

nonbonding electrons are not represented in the model.) Is

there more than one equivalent way to model these pi orbitals?

Explain why or why not and construct a model (or models) to

demonstrate your explanation.

(d) Is your answer in part (c) consistent with the delocalized pi orbital picture for carbon

dioxide? Explain why or why not. Does it explain the observed bond lengths (Check

This 5.39)?

Answer to 5.32:



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(a) The Lewis structure for CO2 is:C OO

. There does not appear to be any other way

to arrange the electrons that is equivalent to this one, so delocalized pi orbitals are not

predicted by the Lewis structure.

(b) The Lewis structure appears to have the two pairs of pi electrons localized in two-

center bonds between the carbon and an oxygen atom. This is not consistent with the

delocalized pi orbitals in Figure 5.24 where the pi electron pairs spread over all three

atoms.

(c) The model shown is one way to arrange the pi orbitals that are perpendicular to one

another. The second picture here represents another way.

In these two models, the atoms have not switched places, but the localized pi orbitals are

shown on different pairs of atoms. There are two energy-equivalent ways to arrange the

pi orbitals.



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(d) Since there are two energy-equivalent ways to arrange the pi orbitals (which are not

represented by the two-dimensional Lewis structure), delocalization of the pi electrons

occurs and lowers the total energy of the molecule. Carbon dioxide is more stable (lower

energy) than localized pi orbitals would predict. The delocalization implied by having

two equivalent ways to arrange the pi-bond paddles is the same as that shown in Figure

5.24 and the explanation for the C-O bond lengths that are shorter than for a C-O bond

with a localized pi bond, is given in the answer to Check This 5.39(b).

5.33. Consider the nitrate ion, NO3–, the nitrite anion, NO2

–, and the nitronium cation, NO+.

(a) What is the total number of valence electrons in each ion?

(b) Draw the Lewis structure for each ion.

(c) How many σ bonding orbitals are there around the central atom in each ion?

(d) What geometry is predicted for each ion?

(e) How many π bonding orbitals are there around the central atom in each ion?

(f) How many localized and how many delocalized π bonding orbitals are there in each

ion? Explain how you arrive at your answer.

(g) What is the bond order for nitrogen-to-oxygen in each ion? Use sketches of the

delocalized orbitals, if any, to illustrate your answer.

(h) What do you predict for the relative bond lengths, shortest to longest, in these ions?

Answer to 5.33:

nitrate ion nitrite ion

a. total number of valence e– 1(5) + 3(6) + 1 = 24 e– 1(5) + 2(6) + 1 = 18 e–

b. Lewis structure

N

O

OON OO

c. σ bonding orbitals 3 2

d. geometry trigonal planar bent

e. π bonding orbitals 1 1

f. localized, delocalized 0, 1 0, 1

g. bond order 11/3 11/2

(a) NO2+ has 16 valence electrons.

(b) N OO

(c) There are two sigma bonding orbitals on the central N atom.



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(d) The ion is predicted to be linear with two sigma bonding and no sigma nonbonding

orbitals on the central N atom.

(e) There are two pi bonding orbitals around the central N atom.

(f) There are no localized pi orbitals and there are two pi orbitals that are delocalized overall three atoms and perpendicular to one another. (See the solution to Problem 5.34(c).)

(g) The bonds are both double bonds. The sigma orbital between the central N atom and

each O atom contribute a full bond. Each of the two pi delocalized orbitals spread over

the three atoms contributes half a bond between the central N atom and each O atom, for

a total of one more full bond.

(h) We would predict that the bond lengths would follow the progression of bond orders,

1-1 / 3, 1-1 / 2, and 2 for nitrate, nitrite, and nitronium ion, so nitronium would have the

shortest N-O bonds and nitrate the longest.

5.34. A single Lewis structure for the nitrate ion doesn’t represent the distribution of the π bonding orbital over all four atom cores, an alternative that has been extensively used is

to draw more than one Lewis (or 3-d) structure:

N

O

OON

O

OON

O

O O

The double-headed arrows are used to indicate that the best representation is some

intermediate structure (that is impossible to write).

(a) If all three of these structures existed separately, they would have identical energies.Explain why.

(b) If the structures have identical energies, then we would expect an intermediate

structure based on them to have equal contributions from each structure. What fraction

contribution would each make to the intermediate structure? What is your reasoning?

(c) The delocalized π bonding structure, Figure 5.23(a), led us to the conclusion that all

three of the nitrogen-oxygen bonds in nitrate can be thought of as 1 1 / 3 bonds. Show how

the same conclusion can be reached here, based on equal contributions from these three

structures to the proposed intermediate structure.Answer to 5.34:

(a) The energies are equal because the structures all have exactly the same bonding, just

electrons shifted around.

(b) Since there are three identical structures, each would be expected to contribute

exactly one-third to the intermediate.



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(c) If we focus on one of the oxygen atoms, say the one at the bottom, we see that it is

singly bonded in two structures and doubly bonded in the third. Thus, the sum of its

bonding in the three structures is 4, but each contributes only a third, so its net is 4/3 or 1-

1 / 3.

5.35. When some material is connected between the terminals of a battery, there is an electrical

potential difference between the two connections to the battery. If electrons can enter the

material at the negative terminal and leave at the positive terminal to complete the

electric circuit, we call the material an electrical conductor. Metals are good electrical

conductors. How does the delocalized molecular orbital (“sea of electrons”) model of

metallic bonding help you understand the high conductivity of metals? Present your

reasoning clearly.

Answer to 5.35: The electrons in the diffuse cloud representing the delocalized electrons

in Figure 5.27 are relatively free to move about anywhere within the metal crystal lattice.

If an extra electron enters the sea of electrons from a negative electrical terminal, another

can immediately leave the electron sea at the positive terminal, in order to maintain the

electrical neutrality of the crystal. The electron that enters does not have to be the one

that leaves, which makes the flow of electrons through the crystal relatively easy and

rapid. That is, the metal is a good conductor of electricity.

5.36. You can often break a piece of wire or thin strip of metal by bending it back and forthrepeatedly. This action causes dislocations of atoms in the crystal like those you see

represented in Figure 5.26. Explain how this effect can eventually lead to breaking the

piece of metal.

Answer to 5.36: Dislocations in a metal crystal lattice weaken the lattice because the

atomic centers are no longer symmetrically surrounded by other atom centers and the

delocalized molecular orbitals are disrupted. A discontinuity in the bonding occurs at

dislocations and, if they grow by further mechanical manipulation of the metal, the

overall bonding becomes so weakened that the piece of metal can break along thediscontinuities.

Section 5.8. Representations of Molecular Geometry

5.37. The number of isomers possible for any particular molecular formula depends on the

geometry about the atoms that make up the molecule. For each of the following



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compounds, use your models to figure out how many isomers are possible if the four

bonds about the central carbon atom (shown in bold) have a square planar molecular

shape? How many isomers are possible if the four bonds about the central carbon atom

have a tetrahedral molecular shape?

(a) CH3Br, (methyl bromide, a fumigant)

(b) CH2Cl2, (dichloromethane, a solvent used in semiconductor processing)

(c) CH3CH(OH)CH2OH, (propylene glycol, a shampoo additive)

Answer to 5.37:

The 3-d shape of a molecule determines the total number of isomers that are possible.

This is one of the ways chemists of the last century figured out that carbon (and other

atoms) assumed a tetrahedral and not a square planar geometry. See for yourself!

square planar geometry tetrahedral geometry

(a) 1 isomer possible 1 isomer possible

(b) 2 isomers possible 1 isomer possible

(c) 3 isomers possible 2 (optical) isomers possible

5.38. Skeletal structures for several molecules are given. Draw the condensed representations

for each molecule. What is the molecular formula in each case.

(a) aspirin

O

O

O

(b) nicotine N

N

(c) vitamin C

OO

HO

HO

HO OH

(d) cholesterol HO

Answer to 5.38:

Recall that each vertex or line end (which does not have a particular atom indicated) is

understood to represent a carbon atom. Because carbons will always form four bonds in

stable organic molecules, all unspecified valences are understood to be hydrogen atoms.



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The number of hydrogens at any particular carbon atom you choose to examine is simply

the mathematical difference between the number four and the number of bonds already

specified at that carbon atom.

(a) C9H8O4

(b) C10H14N2

(c) C6H8O6

(d) C27H46O

5.39. Conformations are forms of a molecule that differ only in that there have been rotations

about single bonds. Identify which of the following pairs of structures are conformers

(i.e., the same molecule) and which are actually different molecules that are not

interconvertible by any combination of rotations about single bonds or motions of themolecule as a whole. Building molecular models might be helpful in some cases.

(a)

(b)

(c)

OH

H3CHCH2CH3

OH

H3CCH2CH3H

(d)

CH2CH3

H3COHH

OH

H3CHCH2CH3

(e)

Br

Br

Br

Br

(f)

Br

Br Br

Br

Answer to 5.39:

It is extremely difficult to answer questions about molecular structure without the aid of

molecular models. If you build and manipulate molecular models a lot, the two

dimensional drawings will eventually take on three dimensional significance. Even the

most experienced chemists still resort to physical models when investigating new or

unusual structures. The answers to questions such as this one can only be arrived at

confidently if you have built the models.

(a) conformers (i.e., the same compound)



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(b) conformers

(c) mirror image isomers (i.e., different compounds related as right/left isomers)

(d) same compound (simply oriented differently in space)

(e) same compound

(f) mirror image isomers

5.40. Rotations about single bonds are rapid. This allows molecules easily to assume different

conformations. (See problem 5.39.) For each of the following molecules, can you find a

conformation that allows all the carbons to lie in a single plane? Draw 3-d structures to

illustrate your answers. Hint : molecular models are extremely valuable in answering this

question.

(a) butane, CH3CH2CH2CH3

(b) 2-methylpropane, CH(CH3)3 (common name: isobutane)

(c) 1,3-butadiene, CH2CHCHCH2

(d) 1,2-propanediene, CH2CCH2 (common name: allene)

Answer to 5.40: (a) possible (b) impossible (c) possible (d) possible

5.41. For each of the following molecules, can you find a conformation (See problem 5.39.)

that allows all the carbons AND all the hydrogens to lie in a single plane? Draw 3-d

structures to illustrate your answers.

(a) butane, CH3CH2CH2CH3

(b) 2-methylpropane, CH(CH3)3

(c) 1,3-butadiene, CH2CHCHCH2

(d) 1,2-propanediene, CH2CCH2

Answer to 5.41: (a) impossible (b) impossible (c) possible (d) impossible

5.42. For each of the following molecules, can you find a conformation (See problem 5.39.)

that allows all the atomic centers to lie in a single plane? Draw the Lewis structures first

to help answer this question. Draw 3-d structures to illustrate your answers.

(a) methanol CH3OH (wood alcohol)

(b) hydrogen peroxide H2O2 (used as a disinfectant and to bleach hair)

(c) hydrogen cyanide HCN (poisonous gas)

(d) nitric acid (HO)NO2 (useful industrial acid)

(e) nitrous acid (HO)NO (nitrite salts are suspected carcinogens)

Answer to 5.42: (a) impossible (b) possible (c) possible (d) possible (e) possible



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Section 5.9. Stereoisomerism

5.43. Stereoisomers (isomers that share the same connectivity of their bonds, but differ in their

3-d shapes) can be subcategorized as optical isomers (those stereoisomers that bear a

mirror image relationship to one another) and those that are not optical isomers. Identifywhich of the following pairs of structures are identical, which are optical isomers, and

which are other types of stereoisomers. Building models could be very helpful.

(a)

Br

Br Br

Br

(f) OH OH

(b)

Br

Br Br

BrBr

Br Br

Br

(g) OH

OH

(c) (h) Cl Cl

Cl Cl

(d) OH OH (i) Cl Cl

Cl Cl

(e) OH OH (j) Cl Cl

Cl Cl

Answer to 5.43: (a) optical isomers (b) other type of stereoisomers (c) identical (d)

other type of stereoisomers (e) optical isomers (f) identical (g) optical isomers (h) optical isomers (i) other type of stereoisomers (j) optical isomers

5.44. (a) Complete this table by filling in the blanks with either “the same” or “different.”

PropertyStructural

Isomers

Stereoisomers

Optical Isomers Other

Molar mass

Boiling point

Density

Bond connectivity

(b) For one of these data columns, all the entries are “the same.” By definition, two

compounds cannot be isomers if all of their properties are the same. What is the property

that allows us to distinguish the isomers in this column?



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Answer to 5.44:

(a)

Properties Structural Isomers Stereoisomers

Optical Isomers Other Stereoisomers

Molecular Weight same same same

Boiling Point different same different

Density different same different

Connectivity of Bonds different same same

Chemical Reactivity different same different

(b) Optical rotation (measured by a polarimeter) distinguishes the “right” and “left”

handed optical isomers from one another. Whatever the extent of rotation of plane

polarized light by one isomer, the other rotates plane polarized light an equal amount, but

in the opposite direction. One rotates clockwise; the other counterclockwise.

5.45. Use your molecular models to make as many models of isomers of C5H10 as you can.

How many are there? (If it is easy for you to visualize molecular structures, like Lewis or

condensed structures, written on a piece of paper, you can do this problem with a pencil

and paper.) How does your result compare with the number of C4H8 isomers you obtained

in Investigate This 5.55? What conclusion(s) can you draw about the relationship

between number of carbons and number of isomers? Does this conclusion make sense?

Explain why or why not.Answer to 5.45:

The non-cyclic isomers are:

1-pentene trans-2-pentene

cis-2-pentene 2-methyl-1-butene

3-methyl-1-butene 2-methyl-2-butene

The cyclic isomers are:

cyclopentane methylcyclobutane

1,1-dimethylcyclopropane



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trans-1,2-dimethycyclopropane

cis-1,2-dimethycyclopropane

There are eleven C5H10 isomers compared to six C4H8 isomers (Investigate This 5.55). As

the number of carbon atoms increases, the number of possible isomers increases. This

makes sense because there are more ways for the carbon chain to branch and rings to

form as well as more possible places for multiple bonds to be located in molecules like

these.

Section 5.10. Functional Groups -- Making Life Interesting

5.46. (a) Use your molecular model kit to construct a model of cyclopentane. (See InvestigateThis 5.45.). How flexible is the ring?

(b) Remove one of the CH2 groups and reconnect the ring to make a four-carbon ring,

cyclobutane. How easy is it to do this? How flexible is the ring?

(c) Remove another CH2 group and reconnect the ring to make cyclopropane. How easy

is it to do this? How flexible is the ring? Would you expect cyclopropane to be more

reactive, less reactive, or about the same as cyclopentane? Give the reasoning for your

answer.

Answer to 5.46: This is a model building problem. You should be able to observe howmuch harder it is to get a three-membered ring be more prone to react, since its bonds

seem ready to burst apart.

5.47. Use your model kit to make a cyclopentane model.

(a) Replace one of the hydrogens with an –OH group; the model now represents

cyclopentanol. Is there more than one structure for cyclopentanol? Why or why not? If

so, make separate models of each one. Are any of them optical isomers? How can you

tell?

(b) Add a second –OH group on the carbon adjacent to the first –OH group. The new

model is 1,2-cyclopentanediol (“di” = two). Is there more than one structure for the diol?

Why or why not? If so, make separate models of each one. Are any of them optical

isomers? How can you tell?

Answer to 5.47: Probably you can’t immediately see the relationships among the parts of

molecules and need this sort of model-building exercise to help them develop an “inner



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eye” for structure. There is, of course, only one structure for cyclopentanol, but things

get more interesting for the diol, which has cis- and trans- isomers. It is much more

difficult to find that the trans-isomer is a mixture of mirror image isomers (an enatiomeric

pair). Thus there are three isomers of the diol. This is VERY hard to see without

modeling. The easiest way to see mirror image isomers is to build both, show that they

are in fact mirror images, then show that they are nonsuperimposible. Alternatively, try

using a mirror and building the model of the image you see in the mirror.

5.48. Table 5.4 catalogs a number of important functional groups. Use the table to answer the

following questions.

(a) Which functional groups contain oxygen?

(b) Which functional groups contain nitrogen?

(c) Which functional groups in the table are most likely to ionize? (You may have to

refer back to Chapter 2.) Explain your reasoning.

Answer to 5.48:

(a) alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amide

(b) amine, amide

(c) carboxylic acid, amine [In this question, the word “ionize” refers to the reaction of

electrically neutral, polar molecules with water to form ions.]

5.49. Can you discover a mathematical relationship between the number of C’s and the numberof H’s in an alkane? [Because this is a rather abstract question, we will give you an

example to illustrate what we mean. If you wished to know the number of toes in a

crowded room, you could simply count the number of feet and multiply by 5.

Mathematically, the number of toes is a function of the number of feet. Note that this

relationship is independent of the number of feet in the room (assuming no three-toed

aliens or other exceptions)!]

(a) Can the number of H’s in an alkane be described as a function of the number of C’s?

You will need to examine a number of alkanes to be certain that you have discovered a

general relationship. Your relationship should fit branched as well as straight-chain

alkanes.

(b) Can the number of H’s in an alk ene be described as a function of the number of C’s?

(c) Can the number of H’s in an alk yne be described as a function of the number of C’s?

Answer to 5.49: (a) alkanes: CnH2n+2 (b) alkenes: CnH2n (c) alkynes: CnH2n–2



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5.50. Certain terms used in the nomenclature of carbon-containing compounds sound very

much alike but do not mean the same thing at all. Clearly explain the differences between

these pairs of words and draw a specific compound illustrating each.

(a) alkane alkene

(b) alcohol aldehyde

(c) ether ester

(d) amine amide

(e) carboxyl carbonyl

Answer to 5.50: (a) An alkane is an acyclic hydrocarbon (compound consisting of only

hydrogen and carbon) in which every carbon is connected to four other atoms. These

connections are entirely by means of single bonds. Closely related are cycloalkanes

which have one or more rings in their structures. An alkene is a hydrocarbon which has

at least one double bond as part of its structure. The doubly bonded carbons are

connected to only three other atoms because one of those connections is made with four,

not two, electrons. A cycloalkene, as you might expect, has at least one ring and one

double bond.

(b) An alcohol has an oxygen atom linked by single bonds to a hydrogen atom and a

carbon atom; two pairs of nonbonding electrons complete its octet. An equivalent

description of an alcohol focuses on the carbon atom. In this definition, the key feature is

a carbon atom singly bonded to an –OH group (also known as a hydroxyl group) and

having all three of its other bonds to either carbon or hydrogen. An aldehyde has an

oxygen atom linked by a double bond to a carbon atom. As with alcohols, two pairs of

nonbonding electrons complete the octet on oxygen. The C=O double bond is known as

a carbonyl group and is found as a substructure in many functional groups. In an

aldehyde the carbonyl carbon is linked by single bonds to (1) a hydrogen and (2) a

carbon.

(c) An ether has an oxygen atom linked by single bonds to two carbon atoms; two pairs

of nonbonding electrons complete its octet. An ester is characterized by a carbonyl liked

by single bonds to (1) a carbon atom and (2) an oxygen also bound to carbon with its

other single bond (also known as an alkoxy group).

(d) An amine has a nitrogen atom linked by three single bonds to either carbon or

hydrogen atoms. An amide has a nitrogen atom linked by a single bond to a carbonyl

carbon. The other two single bonds may connect the nitrogen to either carbon or



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hydrogen atoms. The definitions of amines and amides sound surprisingly similar, yet

their physical and chemical properties are dramatically different.

(e) The carbonyl group was defined in section (b) above, A carboxyl group (or more

precisely, a carboxylic acid functional group) consists of a carbonyl singly bonded to a

hydroxyl group and also to a carbon or hydrogen.

5.51. Draw the Lewis structures and structural formulas for each of the following compounds.

Draw bond dipoles on the structural formulas. Identify the shape of each compound.

Identify if the compound is polar or nonpolar.

(a) H2CO (formaldehyde)

(b) HO(O)CCH(NH2)CH2C(O)OH (aspartic acid)

(c) H2NCH2CH2NH2 (1,2-diaminoethane)

Answer to 5.51: Although carbon is slightly more electronegative than hydrogen, the

difference is small and we will ignore the effect of the relatively small bond dipoles

which result. In drawing structures we often write partially condensed structures, as

shown for aspartic acid and 1,2-diaminoethane.

5.52. (a) In Problem 5.2, you drew structures for isomeric alcohols with formula C 5H12O.

There are six other structures that have the formula C5H12O. Draw their structures and

identify the functional group(s) in these molecules.

(b) Predict which one of the six molecules in part (a) will have the lowest boiling point.

Explain the reasoning for your choice.Answer to 5.52:

(a) ethers

(b) The most highly branched of this set of ethers is methyl ter-butyl ether (MTBE; structure

shown).


O OO

OO O

O


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Section 5.11. Molecular Recognition

5.53. In a double helix of DNA, the nitrogenous

base cytosine is held to the nitrogenous base

guanine by hydrogen bonds. (See Investigate

This 1.33.) Dashed lines in the diagram

represent hydrogen bonds.

(a) How many hydrogen bonds can form

between, thymine and adenine? Use these two structures for thymine and adenine to draw

a diagram showing the hydrogen bonding that can take place between thymine and

adenine.

N

N

O

O

attachedto deoxyribosein DNA chain

Thymine

H3C

H


Adenine

N

N

N

N

N H

H

(b) Thymine does not pair with guanine. Offer a possible explanation for this

observation.

(c) DNA has the ability to replicate (make copies of itself) accurately and reproducibly.

How can this be related to base pairing? Explain your reasoning.

Answer to 5.53: (a) This diagram shows that thymine and adenine form two hydrogenbonds.

N

N

O

O


Thymine

N

N

N

N


N

Adenine

H

HH3C

H

(b) Thymine and guanine do not have polar groups strategically arranged to facilitate

multiple hydrogen bonding. Making models of these bases will help you visualize the

possibilities of base pairing.


NN

O

NH

H


Cytosine

NN

N

N


NH

H

H

O

Guanine


30/38


(c) The specificity of base pairing is caused by hydrogen bonding. Although there are

errors in reproducing DNA, considerable repetition is built into the DNA double helix,

helping to reduce the number of errors during the replication process.



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5.54. WEB Chap 5, Sect 5.11.1-3. Work through these pages before answering the following

questions.

(a) Explain the relationship of the activity in Sect 5.11.3 to the one you did in Check

This 5.72.(b) Make a model of aspartame. A ball-and-stick model is shown rotating in Sect 5.11.1

and here is a 3-d representation of the molecule:

CC

O

NC

H

CH2

CH

C

CH

CHHC

HC

HC

O

OCH3CH2

+H3NH

–O(O)C

An important point to note about the structure of this dipeptide (aspartyl phenylalanine

methyl ester) is that the four atoms in the center of the structure, O=C–N–H (shown in

red), lie in a plane and rotation about the C–N bond is very restricted. This bond acts

much like a double bond because interactions between the pi bond in the carbonyl and the

nonbonding electrons on the nitrogen atom delocalize the electrons over the three second

period atoms and make their structure rigid.

Two possible ways to construct the six-

membered ring are shown in these

photographs. Use whichever one you wish.

(c) As sweet receptor, cut an appropriate

size hole in a piece of paper and color its edge red, blue, and green as in sweet receptor

shown in Sect 5.11.3 (or simply label the edge with charges). Show that your aspartame

model complements the receptor, as does the correct phenyalanine isomer in the web

activity.

(d) Exchange the –H and –NH3+, on your model (left-hand side of the structure above)

and try part (c) with the new isomer. Will this isomer still taste sweet? Explain why or

why not.

(e) Undo the exchange in part (d) and then exchange the –H and one of the other groups

attached to the central carbon at the right-hand side of the structure above. Try part (c)

with the new isomer. Will this isomer still taste sweet? Explain why or why not. Discuss

the similarities and differences between your observations in parts (d) and (e).

Answer to 5.54:



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(a) Both activities involve a molecular structure with four different groups bonded

tetrahedrally to a central atom. There are two possible arrangements of the four groups

(the two optical isomers). Only one of the isomers has the four groups arranged so that

three of them can interact with three complementary groups on another molecule.

(b) Two views of the model are shown here. One is stretched out to show the parts a bit

better and the second is curled around, as it will have to be to interact with the sweet

receptor:

(c) Sheets of colored paper, red, blue, and light green, represent the negative, positive,

and neutral (hydrophobic) sites on the sweet receptor. The molecular structure is shown

here with its negative carboxylate group, -C(O)O–, interacting with the positive site, the

positive ammonium-like group, -NH3+, interacting with the negative site, and the

uncharged phenyl ring, -C6H5, interacting with the neutral site. The hydrogen atom at the

carbon whose stereochemistry is critical is marked with an arrow. Note that this hydrogen

atom is pointing out toward you. This isomer fits the “sweetness triangle” and will taste

sweet.

(d) The model with the -H and -NH3+ switched is shown here interacting with the sweet

receptor. The hydrogen atom on the critical carbon is again marked and shown in the



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same position, pointing out at you, as in the picture in part (c). Now the -C(O)O– and

-NH3+ cannot properly interact with sweet receptor and this isomer will not taste sweet.

(e) Exchange at the other stereocenter in the molecule (after the first is restored to its

original form) gives a third isomer shown here. Again, the arrow identifies the hydrogen

at the critical center. The hydrogen is pointing out toward you and both of the polar

groups are interacting with their oppositely charged counterparts on the sweet receptor.

The other end of the molecule can be rotated so that the phenyl ring can interact with the

neutral site on the sweet receptor. This isomer also ought to taste sweet.

The difference between the results in part (d) and part (e) has to do with where in the

molecule the structural change is made. If it is made at the critical stereocenter (as in (d)),

then the molecule will not “fit” the receptor. There is no way for the molecule to maintain

the same geometric relationship to the receptor (shown by the position of the hydrogen

atom with respect to the receptor) and also have the appropriate interactions of the

charged groups with the receptor. When the change is made at the other stereocenter, the

structure can rotate to put the phenyl ring in the correct geometry to interact favorably

with the sweet receptor while the overall molecule maintains its geometric relationship to

the receptor.



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Section 5.13. EXTENSION -- Antibonding Orbitals: The Oxygen Story

5.55. The double bond in O2 and a triple bond between two carbons have almost the same bond

length, 121 and 120 pm, respectively. If bond lengths get shorter as bond order increases,

why is the double bond between oxygen atoms the same length as the triple bondbetween carbon atoms? Hint : Recall the trend in atomic sizes and the reason(s) for it as

you go across a period of the periodic table.

Answer to 5.55: Oxygen atoms are smaller than carbon atoms because of higher nuclear

charge acting on electrons in same shell. This same factor is present in the sigma and pi

molecular orbitals in oxygen vs. carbon. The orbital volumes are smaller in oxygen

(increased attraction can balance greater kinetic energy) and the nuclei are closer together

for the same bond order. The double bond in carbon is about 134 pm.

5.56. Fluorine (F2) is the most reactive of the halogens. According to the Lewis model, eachfluorine atom of this covalent compound has three nonbonding pairs of electrons.

(a) How many valence electrons are there in diatomic fluorine?

(b) Assume that the molecular orbital diagram shown in Figure 5.42 is also applicable to

molecular fluorine. Draw a diagram like Figure 5.42 that shows how the valence

electrons are assigned to the various molecular orbitals. (Remember that two nonbonding

sigma orbitals are not shown in the diagram.)

(c) What is the bond order of the F-to-F covalent bond that you derive from your results

in part (b)? Explain clearly.(d) How does the molecular orbital bonding model correlate with the Lewis model for

fluorine?

Answer to 5.56: (a) Molecular fluorine has 14 valence electrons. The molecular orbitals

formed from the original p-orbitals are similar to those of molecular oxygen (Fig 4.31).

(b) There are 2 electrons in a σ molecular orbital, and 4 electrons split between two π

molecular orbitals. There are additionally 4 electrons split between two π * molecular

orbitals.

(c) Since the electrons in the π and π* molecular orbitals effectively “cancel out,” themolecular orbital model predicts a bond order of 1.

(d) The Lewis model also predicts a bond order of 1.



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General Problems

5.57. (a) Write a Lewis structure for the cyanate anion, OCN–. Can you write more than one

satisfactory structure?

(b)What geometry does(do) your Lewis structure(s) predict for the cyanate anion? How

does the structure and geometry of this ion compare to the structure and geometry of the

carbon dioxide molecule?

(c) Would you expect the cyanate anion to have delocalized π electrons? Explain the

reasoning for your response.

(d) Answer these same questions for the isocyanate anion, ONC–.

(e) Would you expect the cyanate or isocyanate anion to be the more stable isomer?

Give the reasoning for your choice.

Answer to 5.57:

(a) Two Lewis structures for the cyanate ion are:

(i)C NO

(ii) C NO

In each structure, we recognize atoms that do not have their usual complement of

bonding orbitals. In structure (i), the N atom has only two bonding orbitals, but we

usually see the N atom with three bonding orbitals, as in NH3. The N atom here is like the

N atom in the amide ion, NH2–, so we can think of the N atom in structure (i) carrying the

overall negative charge of the ion. In structure (ii), the O atom has only one bonding

orbital, so it is like the O atom in the hydroxide ion, OH–, and we can think of the O

atom in structure (ii) carrying the overall negative charge of the ion.

(b) In both structures (i) and (ii), the central C atom is bonded to the O and N atoms by

sigma bonds and it has no non-bonding electron pairs. The geometry of the ion is linear

(Figure 5.21 and Table 5.3), just like the structure of carbon dioxide.

(c) Structure (i) will have two delocalized pi orbitals like carbon dioxide. These are

shown for carbon dioxide in Figure 5.24 and further represented in the solution to

Problem 5.34. The two pi orbitals in structure (ii) are more localized between the C and N

atoms.

(d) Two Lewis structures for the isocyanate ion are:

(iii)N CO

(iv)N CO



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Again, we recognize atoms in both structures that do not have their usual complement of

bonding orbitals. In structure (iii), the C atom has only two bonding orbitals, but we

usually see the C atom with four bonding orbitals, as in CH4. The C atom in structure (iii)

is like the C atom in the methylide ion, CH22–, so we can think of this C atom as carrying

a –2 charge. The N atom has four bonding orbitals, as in NH4+, so we can think of the N

atom in both structures as carrying a +1 charge. In structure (iii), the –2 and +1 charges

sum to an overall –1 charge, as they must be to give the correct overall charge on the ion.

In structure (iv), the O atom is like the one in structure (ii) and can be considered to carry

a –1 charge. The C atom in structure (iv) is like the C atom in the methide ion, CH3–, so

we can think of this C atom as carrying a –1 charge. The sum of the charges in structure

(iv) is (–1) + (+1) + (–1) = –1, as it must be to give the correct overall charge on the ion.

The sigma-bonding framework for these two structures is, again, like that in carbon

dioxide and the isocyanate ion is linear. The pi electron distributions in structures (iii)

and (iv) are like those in structures (i) and (ii), respectively, so we expect delocalized pi

orbitals in structure (iii).

(e) The cyanate ion is likely to be more stable than the isocyanate ion. Two of the atoms

in the cyanate ion (either Lewis structure) are in a bonding environment similar to that

they would experience in stable, neutral molecules. The third atom is in a bonding

environment similar to what it would experience as an ion. At least one atom in any

Lewis structure for an ion has to be in a bonding environment like that of an ion, since

the structure carries an overall charge. In the isocyanate ion Lewis structure (iii), two of

the atoms are in bonding environments like that of ions and one of these is a multiply

charged ion. In the isocyanate ion Lewis structure (iv), all three atoms are in bonding

environments like that of ions. Lewis structures with bonding environments that are most

like stable, neutral species are more stable. (In Chapter 6, we will introduce the concept

of formal charge and these same kinds of arguments can then be made in terms of formal

charge.) Lewis structure (i) for cyanate ion is further stabilized by the delocalization of

the pi orbitals. Experimentally, we observe that isocyanates usually readily isomerize to

cyanates.



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5.58. (a) At room temperature, beryllium fluoride, BeF2, is a white crystalline solid and boron

trifluoride, BF3, is a gas. What conclusions can you draw about the bonding in these

compounds? Explain.

(b) Write the Lewis structure for BF3. What geometry do you predict for BF3? What

figure in the chapter helps you explain your choice of geometry? Explain. Hint: This

compound does not obey the octet rule.

(c) Under certain conditions BeF2 molecules can be observed in the gas phase. Write the

Lewis structure for BeF2. What geometry do you predict for BeF2? What figure in the

chapter helps you explain your choice of geometry? Explain. Hint: This compound does

not obey the octet rule.

Answer to 5.58:

(a) The gaseous compound, BF3, is likely to be a molecular compound covalently

bonded, since the molecules move about independently of one another as a gas. The

solid, crystalline compound, BeF2, is likely to be ionic (or at least have a large degree of

electron transfer from the beryllium atom to the highly electronegative fluorine atoms).

Perhaps it can be characterized as, (Be2+)(F–)2.

(b) The Lewis structure is:

B

FF

F

Since there are three sigma bonding orbitals

around the central B atom, the geometry of the B and F atoms should be trigonal planaras in Figure 5.18 and Table 5.3.

(c) The Lewis structure is:FF Be

Since there are two sigma bonding orbitals around

the central Be atom, the geometry of the Be and F atoms should be linear as in Figure

5.21 and Table 5.3. (a) The gaseous compound, BF3, is likely to be a molecular

compound covalently bonded, since the molecules move about independently of one

another as a gas. The solid, crystalline compound, BeF2, is likely to be ionic (or at least

have a large degree of electron transfer from the beryllium atom to the highly

electronegative fluorine atoms). Perhaps it can be characterized as, (Be2+)(F–)2.



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(b) The Lewis structure is:

B

FF

F

Since there are three sigma bonding orbitals

around the central B atom, the geometry of the B and F atoms should be trigonal planar

as in Figure 5.18 and Table 5.3.

(c) The Lewis structure is:FF Be

Since there are two sigma bonding orbitals around

the central Be atom, the geometry of the Be and F atoms should be linear as in Figure

5.21 and Table 5.3.

5.59. Write Lewis structures for methanoic acid, HC(O)OH, and the methanoate anion,

HC(O)O-. The C-O single and double bond lengths in the acid are 134 and 120 pm,

respectively. What do you predict for the C-O bond lengths in the anion? Explain the

basis for your prediction?Answer to 5.59: There is a single Lewis structure for the acid, but two equivalent

structures for the anion:

CHO

O

HCH

O

Oand

CHO

O

Thus, the pi electrons in the anion are delocalized over three atom centers, O-C-O, so the

electrons are shared between two C-O bonds and each has a bond order of 1-1 / 2. Thus,

we would predict a bond length intermediate between a single and double C-O bond’

about 127 pm. The experimental value is 126 pm.

solutions ch05.doc

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