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Answer to book chapter 3 questions ...

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Chapter 3: Solutions to Practice Problems Section 3.1

1)

2)

a) O (the most electronegative atom) has a −1 formal charge; N and C have no formal charge. b) N has a + 1 formal charge; C and O both have −1 formal charges. c) N has a −2 formal charge; O has a + 1 formal charge. d) N has a −1 formal charge; both C and O have no formal charge. e) C has a −1 formal charge; both N and O have no formal charge. The best Lewis structure has negative formal charges on the most electronegative atom; therefore a) is the best Lewis structure.

3)

In the best Lewis structure of BrO3—, the Br has a no formal charge and the single bonded O has a formal charge of −1. All other formal charges have been minimized by the formation of double bonds. a) Incorrect; formal charges not as above. b) Incorrect; the Br is missing a nonbonding pair of electrons. c) Not the best structure as the −1 formal charge has been placed on the Br, not on the O. d)

−

e) Incorrect; the O which bears the −1 formal charge has been given 9 electrons, in violation of the octet rule. The other two O’s should bear −1 formal charges. The correct answer is d).

Chapter 3 - 1

Chapter 3 - 2

5)

Lewis Structure is:

O

Cl

O

There are 4 non-bonding electrons (2 pairs) on chlorine. 6)

IF5 has one non-bonding pair of electrons on the central atom, I. SF4 has one non-bonding pair of electrons on the central atom, S.

7)

a) Correct; resonance structures differ only in the locations of the electrons. b) Incorrect; resonance structures do not differ in the locations of the nuclei. c) Correct; delocalization of electrons tends to stabilize a molecule or an ion. d) Incorrect; the only structure that actually exists is the resonance hybrid.

8)

a) Incorrect; the formal charge on the O in the second structure is 0. b) Incorrect; the formal charge on the end N atom in the first structure is 0. c) Incorrect; the formal charge on the central N atom is +1 in all structures. d) Incorrect; the formal charge on the O atom in the third structure is +1. e) Correct; the end N atom in the second structure has a formal charge of −1. The correct answer is e).

9)

Resonance structures can be drawn for all of them except for the last one.

O

N

O

H 2C CH CH 2

O O Cl O O

O

O

O

H 3C CH 2 NH

Chapter 3 - 3

10)

11)

O

O

O

3

Br

O

O

O

Br

O

O 3 more resonance structures; average Br - O bond order = 7/4

Chapter 3 - 4

O

12)

O

As

There are 3 other resonance structures for this molecule; 4 in total. O

O Average As – O bond order in AsO43― is 1.25.

13)

O

S

O

The average bond order is 2.

O

Chapter 3 - 5

15)

Molecular formula of ATP is C10H12N5 O13P34–. Each oxygen with two bonds has two non-bonding pairs for a total of 18 non-bonding pairs. Each negatively charged oxygen has three non-bonding pairs for a total of 12 non-bonding pairs. Each nitrogen has one non-bonding pair for a total of 5 non-bonding pairs. Hydrogen and phosphorus have no non-bonding electron pairs. Grand total of non-bonding pairs is 35. The indicated double bonded oxygen atoms are not completely neutral; because oxygen is more electronegative than phosphorus, each of those oxygen atoms will bear a partial negative charge.

Chapter 3 - 6

Section 3.2

1)

a)

b)

Cl

F F 120o F

90o S

F

Al Cl

Cl

F 120o

AX3, Bond Angles are Shape is triangular or trigonal planar

AX5; Shape is trigonal bipyramidal Cl

d)

c)

O H

H

P

H AX3E, Bond angles < Shape is triangular pyramidal

Cl

e)

Cl

Cl

109.5o

Cl AX4, Bond angles are 109.5o Shape is tetrahedral f)

Cl Cl

Si

Cl Sb

Cl

Cl

Cl

Cl

Cl Cl

AX4, Bond angles are 109.5o Shape is tetrahedral

g)

AX 6, Bond angles are 90o Shape is octahedral h)

S

O

F

Br

F

O O F AX3E, Bond angles < 109.5o Shape is triangular pyramidal

AX3E2, Bond angles are 90o Shape is T-shaped

Chapter 3 - 7

Chapter 3 - 8

Chapter 3 - 9

3)

i) Nonpolar molecules containing only nonpolar bonds: Cl2 ii) Nonpolar molecules containing polar bonds: CF4, BeF2, XeCl4, TeCl6, AlF3 iii) Polar molecules: SF4, OF2, BrF5, IF, PF3, HCF3

4)

AsF5 is AX5; trigonal bipyramidal, SiF5― is AX5; trigonal bipyramidal The correct answer is a).

5)

a)

CH2F2 is tetrahedral, but not symmetrical, so polar; PCl5 is trigonal bipyramidal, nonpolar

b) BeF2 is linear, nonpolar; ClF3 is T-shaped, polar c) SF6 is octahedral, nonpolar; ClF3 is T-shaped, polar d) CH2F2 is tetrahedral, but not symmetrical, so polar; ClF3 is T-shaped, polar e)

PCl5 is trigonal bipyramidal, nonpolar ; SF 6 is octahedral, nonpolar

The correct answer is e). 6)

“Main Group” means Groups 13 to 18. If the atom “A” has 3 unpaired electrons, it must have the configuration s2 p3, which is Group 15. “A” contributes 5 electrons to the total count. If there is one nonbonding pair of electrons on “A” in AFn, there are 3 electrons remaining to form bonds. The formula of AFn is AF3. This gives an AX3E geometry, which is triangular pyramidal. The correct answer is e).

Chapter 3 - 10

7)

a) CH2F2 is tetrahedral, but is not symmetrical. This molecule is polar. b) XeO3 is trigonal pyramidal. This molecule is polar. c)

BF3 is triangular planar. This molecule is nonpolar.

d) CH3OH is tetrahedral, but is not symmetrical. This molecule is polar. e) SF4 is a see-saw. The shape is irregular so this molecule is polar. 8)

a)

ClF3 is T-shaped, polar; ClF5 is a square pyramid, polar.

b) CO is linear, polar; CO2 is linear, nonpolar. c) NF3 is trigonal pyramidal, polar; ClF3 is T-shaped, polar. d) HCN is linear, polar; H2C=O is trigonal planar, polar.

9)

e)

SCl4 is see-saw, polar; ClF5 is a square pyramid, polar.

a)

H2O is bent; polar

b) F2O is bent; polar c) HOF is bent; polar d) FBeCl is linear, but not symmetrical; polar e) f)

HCl is linear, but the one bond is polar, therefore the molecule is polar. O3 is bent and there is uneven charge distribution; polar

g) BF3 is triangular planar; nonpolar h) SO2 is bent like O3; polar i) CH4 is tetrahedral; nonpolar j)

CS2 is linear; nonpolar

10) XeF3― has 12 or 6 pairs of valence shell electrons. The geometry of this molecule is AX3E3. The arrangement of the electron pairs is octahedral, but there are 3 non-bonding pairs. The correct answer is e). 11) SF4 is a see-saw,

, AsO43― is tetrahedral, PH3 is trigonal pyramidal

The correct answer is b).

Chapter 3 - 11

12)

O

O H

H

C O

a)

C O

Correct; there are two equivalent resonance structures as shown above.

b) Correct. c) Correct. All statements are correct. 13) SF42– is square planar in shape; AX4E2 14) BrF3 is T-shaped; BeCl2 is linear; The correct answer is a).

BF4― is tetrahedral

15) BF3 is triangular planar; nonpolar SnCl2 is bent; polar OCl2 is bent; polar 16) a)

CO2 is linear

b) HCN is linear c) H2S is bent d) NF3 is trigonal pyramidal e)

H2O is bent

The molecules in a) and b) are linear. 17) NO3― is trigonal planar in shape with an average bond order of 1.33 (See 1m) Answer d) is correct.

Chapter 3 - 12

Section 3.3 1)

c) is impossible; the

.

2)

The geometry of the ICl5 molecule is square pyramidal. (AX5E) There are 6 electron pairs, therefore the hybridization is sp3d2. This molecule is polar. Answer a) is incorrect.

3)

a) The N in NH4+ is sp3 hybridized. b) The B in BF4− is sp3 hybridized. c) The Xe in XeF4 is sp3d2 hybridized. d) The O in H2O is sp3 hybridized. e) The C in CH4 is sp3 hybridized. The hybridization is different in molecule c).

4)

a) Triangular pyramid is AX3E; 4 electron pairs (sp3 hybrid) but only 3 bonding pairs. Therefore, 5 electrons must come from the central atom, which must be in Group 15. Examples include NH3, PF3, AsF3, SbH3, BiCl3. b) Octahedral shape is AX6; 6 electron pairs (sp3d2 hybrid). All 6 electron pairs are bonding; one electron comes from each terminal atom. The remaining 6 electrons come from the −1 charge and the valence electrons of the central atom, of which there are 5. The central atom must be in Group 15 . Examples include PCl6−, AsF6−, SbH6−, BiF6−. c) Tetrahedral shape is AX4; 4 electron pairs (sp3 hybrid). All 4 electron pairs are bonding; one electron comes from each terminal atom. The . There must be 5 valence electrons on the central atom, which is in Group 15. Examples include NH4+, PF4+, AsCl4+, SbH4+, BiCl4+. d) T-shaped molecule is AX3E2; 5 electron pairs (this must be an sp3d hybrid). Only 3 electron pairs are bonding; one electron comes from each of the terminal atoms. This leaves 7 electrons around the central atom, so the central atom must be in Group 17. Examples include ClF3, BrCl3, IF3

Chapter 3 - 13

5)

H

a)

N:

N O

b)

H

. O: Geometry of electron pairs is tetrahedral. Hybridization is sp3.

O

C

C: Geometry of electron pairs is linear. Hybridization is sp.

N

c) H

H C

H

C

H

H

H

F

F

d) F

C

F

C F

6)

S: Geometry of electron pairs is tetrahedral. Hybridization is . C: Geometry of electron pairs is tetrahedral. Hybridization is sp3.

S

C: Geometry of electron pairs is tetrahedral. Hybridization is sp3. In this structure, each F atom has 3 non bonding pairs of electrons.

C F

F

F

sp2 hybridization results in a triangular planar arrangement of electron pairs. (AX3) a) CO2 is linear; sp hybridized. b) OCS is linear; sp hybridized. c) K2CO is the CO32− ion with K atoms bonded to the negatively charged O atoms. The CO32− molecule is triangular planar;

2

hybridized.

d) The H2C=O molecule is triangular planar; bridized. e) In C2H4, there is a double bond between the two C atoms. The arrangement of electron pairs around each C is triangular planar;

bridized.

f) The C2H2 molecule has a triple bond between the two C atoms. The arrangement of electron pairs around each C is linear; sp hybridized. g) The HCCl3 molecule is tetrahedral; sp3 hybridized. The molecules in c), d) and e) contain a C atom which is sp2 hybridized. Chapter 3 - 14

7)

a) Incorrect; a double bond consists of a σ bond and a π bond. b) Incorrect; a triple bond consists of a σ bond and two π bonds. c) Correct; rotation about the σ bond axis cannot occur when a π bond is present. d) Incorrect; a π bond consists of only one pair of electrons. e) Incorrect; a triple bond consists of a σ bond and two π bonds. Only answer c) is correct.

8)

H – C ≡ N: is the structure of HCN. The hybridization of the C atom is sp, the CN bond is a triple bond and the molecule is linear. The correct answer is c).

9)

a) CO2 is a linear molecule and therefore planar. b) CH3NH2 is tetrahedral about the C atom and triangular pyramidal about the N atom. This molecule is not planar. c) CH3+ is triangular planar. d) HCOO― is triangular planar about the C atom. Molecules a), c) and d) have all their atoms in the same plane, or are planar molecules.

10) A σ bond is a single bond, formed by the overlap of two atomic or hybrid orbitals between the atoms. A π bond is formed by sideways overlap of pure p orbitals, both above and below the plane of the σ bond. Due to the sideways overlap, the atoms involved cannot rotate.

Section 3.4 1)

The following chart lists the limitations of VB Theory and how they are addressed by MO Theory. Valence Bond Theory (Atomic Orbitals)

Molecular Orbital Theory (Atomic Orbitals)

Bonding electrons are confined to the

Bonding electrons are delocalized and are

area between the atoms; does not

associated with the entire molecule.

explain resonance structures. No prediction of paramagnetism.

Accounts for unpaired electrons and properties resulting from these unpaired electrons.

Does not describe the energies of the

There are two types of MOs and the lowest

electrons.

energy ones are always filled first. Chapter 3 - 15

2)

A node is an area where there is a zero probability of finding an electron. If there are no electrons shielding the nuclei from each other, repulsion between the nuclei occurs. This does not lead to bond formation, so reduces the stability of the molecule.

3)

A paramagnetic molecule is attracted to a magnetic field due to its unpaired electrons. A diamagnetic molecule with paired electrons is not attracted to a magnetic field.

4)

a) O22― : Bond order = 1, diamagnetic b) O2― : Bond order = 1.5, paramagnetic c) O2 : Bond order = 2, paramagnetic d) O2+ : Bond order = 2.5, paramagnetic e) O22+ : Bond order = 3, diamagnetic f) Ne2 : Bond order = 0, diamagnetic g) C2 : Bond order = 2, paramagnetic The most stable molecules have the highest bond order. From most stable to least stable: e), d), c) = g), b), a), f)

5)

The π2py and the π2pz are degenerate because they are both formed by sideways overlap of py or pz atomic orbitals. They are higher in energy than the σ2px which is formed by head on overlap of px atomic orbitals.

6)

Metals are substances which will conduct electricity. Insulators are substances which do not conduct electricity. Semiconductors are substances which have properties in between those of metals and insulators. When atoms of a semiconductor aggregate together a band of molecular orbitals is formed. The bottom half is the valence band and the top half is the conduction band. In order for a substance to conduct electricity, there must be electrons in the conduction band. If the two bands are adjacent or overlapping, a small increase in temperature will cause electrons to move from the valence band to the band. This is the case with metals. In the case of semiconductors or insulators, there is an energy gap between the two bands, which is known as the band gap energy. The band gap energy is much larger for an insulator than for a semiconductor. rature can result in the movement of electrons from . The band gap energy of an insulator is so large, that an increase in temperature will not result in this movement of electrons.

Chapter 3 - 16

Section 3.5 1)

a)

Correct; the greater the surface area, the greater the dispersion forces.

b) Incorrect; the s c)

Incorrect; dipole-dipole interactions are attractive forces between permanent dipoles of polar molecules.

d) Correct; although this attraction can also occur within a molecule.

2)

These three molecules, although they have the

, have different shapes

due to the different bonding sequences of their atoms. Due to the different bonding sequences, different shapes result and therefore, different surface areas of the three molecules. Boiling point depends on the strength of van der Waals interactions, and the smaller the surface area, the lower will be the boiling point. Melting point is more affected by the symmetry of the molecule, as symmetrical molecules can pack together more tightly, resulting in a much higher melting point than molecules which do not pack together as tightly, due to the lack of symmetry.

3)

The second molecule will have a higher boiling point as it has greater surface area.

The first molecule will have a higher boiling point as it has both dipole-dipole interactions and hydrogen bonding interactions compared to the second molecule which has hydrogen bonding only.

Chapter 3 - 17

The first molecule will have the higher boiling point due to hydrogen bonding.

The first molecule will have the higher boiling point due to its larger surface area.

The second molecule will have the higher boiling point due to hydrogen bonding, which is a stronger interaction than the dipole-dipole interaction present in the first molecule.

The second molecule exhibits a dipole-dipole interaction which is stronger than the van der Waals forces in the first molecule. The second molecule has the higher boiling point.

The second molecule will have hydrogen bonding interactions, resulting in a higher boiling point.

The first molecule has ion-ion interactions, which will result in a higher boiling point than the hydrogen bonding interactions in the second molecule.

Chapter 3 - 18

The second molecule will exhibit dipole-dipole interactions resulting in a higher boiling point than the first molecule.

The first molecule shows both hydrogen bonding and dipole-dipole interactions, resulting in a higher boiling point than the second molecule which has only dipole-dipole interactions.

Chapter 3 - 19...