MSE160 Custom Textbook Solutions

5/20/2018 MSE160 Custom Textbook Solutions

1/541


2/541

Chapter 8

153

Practice Exercises

8.1 = (588 nm)91 10 m

1 nm

= 5.88 107 m

=c

=8

7

3.00 10 m /s

5.88 10 m

= 5.10 1014s1= 5.10 1014Hz

8.2 = (10.9 m)6

1 10 m

1 m

= 1.09 105m

=c

=

8

5

2.998 10 m / s

1.09 10 m

= 2.75 1013s1= 2.75 1013Hz

8.3 =8 1

6 1

c 2.998 10 m s=

92.3 10 s

= 3.25 m

8.4 12 21 1 1= 109,678 cm

4 6

= 109,678 cm1 (0.06250.02778)

1

= 3.808 103cm1

= 2.63 104cm = 2.63 m

8.5 12 2

1 1 1= 109,678 cm

2 3

= 109,678 cm1 (0.25000.1111)

1

= 1.5233 104cm1

= 6.565 105cm = 656.5 nm, which is red.

8.6 (a) n = 4, l= 2

(b) n = 5, l= 3

(c) n = 7, l= 0

8.7 When n= 2, = 0, 1. Thus we haves, and p subshells.

When n= 5, = 0, 1, 2, 3, 4. Thus we haves,p, d,f, andgsubshells.

The number of subshells spans the values: 0,1,2,3,---, n1

Thus,

Shell 1: 1 subshell

Shell 2: 2 subshells



Shell 5: 5 subshellsShell 6: 6 subshells

8.8 (a) Mg: 1s22s22p63s2

(b) Ge: 1s22s22p63s23p63d104s24p2

(c) Cd: 1s22s22p63s23p63d104s24p64d105s2

(d) Gd: 1s22s22p63s23p63d104s24p64d104f75s25p65d16s2


3/541

Chapter 8

154

8.9 The electron configuration of an element follows the periodic table. The electrons are filled in the order of

the periodic table and the energy levels are determined by the row the element is in and the subshell is

given by the column, the first two columns are the s-block, the last six columns are thep-block, the d-block

has ten columns, and thef-block has 14 columns.

8.10 (a) O: 1s22s22p4

S: 1s22s22p63s23p4

Se: 1s22s22p63s23p63d104s24p4

(b) P: 1s22s22p63s23p3

N: 1s22s22p3

Sb: 1s22s22p63s23p63d104s24p64d105s25p3

The elements have the same number of electrons in the valence shell, and the only differences

between the valence shells are the energy levels.

8.11 (a) Na:

(b) S

(c) Fe

1s 2s 2p 3s 3p 4s 3d

8.12 (a) Mg

0 unpaired electrons

(b) Ge

1s 2s 2p 3s 3p 3d 4s 4p 2 unpaired electrons

(c) Cd

1s 2s 2p 3s 3p 3d 4s 4p

5s 4d 0 unpaired electrons

(d) Gd

1s 2s 2p 3s 3p 3d 4p

5s 4d 5p 6s 5d 4f 8 unpaired electrons

8.13 Yes, Ti, Cr, Fe, Ni, and the elements in their groups have even numbers of electrons and are paramagnetic.

Additionally, oxygen has eight electrons, but it is paramagnetic since it has two unpaired electrons in the 2p

orbitals.

8.14 (a) P: [Ne]3s23p3


4/541

Chapter 8

155

[Ne]

3s 3p (3 unpaired electrons)

(b) Sn: [Kr]4d105s25p2

4d 5s 5p

[Kr]

(2 unpaired electrons)

8.15 Based on the definition of valence, there are no examples where more than 8 electrons would occupy the

valence shell. For representative elements the valence shell is defined as the occupied shell with the highest

value of n. In the ground state atom, onlysandpelectrons fit this definition. The transition elements have

outer electron configurations: (n-1)dnnsmso the valence shell is the nssubshell.

8.16 (a) Se: 4s24p4 (b) Sn: 5s25p2 (c) I: 5s25p5

8.17 (a) Sn (b) Ga (c) Cr (d) S2

8.18 (a) P (b) Fe3+ (c) Fe (d) Cl

8.19 (a) Be (b) C

8.20 (a) C2+ (b) Mg2+

Review Questions

8.1 Light is a form of energy that results from small oscillations in the electrical and magnetic properties of

particles.

8.2 In general, frequency describes the number of times an event occurs in a finite time period. The frequency

of light is the number of times a wave crest passes a specific point in space in a given time interval. The

symbol for frequency is the Greek letter nu, , and has the units of inverse seconds, s1.

8.3 Wavelength is the distance between consecutive maxima of a wave. The symbol is the Greek letterlambda, .

8.4 See Figure 8.2.

8.5 The amplitude affects the brightness of light. The color of light is affected by the wavelength or frequency.

The energy of the light is affected by the frequency or wavelength.

8.6 gamma rays < X rays < ultraviolet < visible < infrared < microwaves < TV waves

8.7 By the visible spectrum, we mean that narrow portion of the electromagnetic spectrum to which our eyes

are sensitive. These are the wavelengths from about 400 nm to 750 nm.

8.8 violet < blue < green < yellow < orange < red

8.9 = c, where c is a constant equal to the speed of light, is the wavelength and is the frequency.

8.10 E = h, where E is the energy, h is Plancks constant and is the frequency.

8.11 A photon is one unit of electromagnetic radiation whose energy is the product of h and .

8.12 Since = c/, we can substitute into equation 8.2 to get E = hc/.


5/541

Chapter 8

156

8.13 (a) infrared

(b) visible light

(c) X-rays

(d) ultraviolet light

8.14 The quantum is the lowest possible packet of energy, that of a single photon.

8.15 An atomic spectrum consists of a series of discrete (selected, definite and reproducible) frequencies (and

therefore of discrete energies) that are emitted by atoms that have been excited. The particular values for

the emission frequencies are characteristic of the element at hand. In contrast, a continuous spectrum, such

as that emitted by the sun or another hot, glowing object, contains all frequencies and, therefore, photons of

all energies.

8.16 An electron in an atom can have only certain specific values for energy. Aside from these discrete

energies, other energies are not allowed. When an excited atom loses energy, not just any arbitrary amount

can be lost, only specific amounts of energy can be lost. In other words, the energy of an electron is

quantized.

8.17 Bohr proposed a model similar in design to a solar system. The nucleus is at the center and the electrons

orbit the nucleus in specific orbits that are a constant fixed distance from the nucleus.

8.18 When an electron falls from an orbit of higher energy (larger radius) to an orbit of lower energy (smallerradius), the energy that is released appears as a photon with the appropriate frequency. The energy of the

photon is the same as the difference in energy between the two orbits.

8.19 The lowest energy state of an atom is termed the ground state.

8.20 Bohr's model was a success because it accounted for the spectrum of the hydrogen atom, but it failed to

account for spectra of more complex atoms.

8.21 Very small particles have properties that are reminiscent of both a particle and a wave. Massive particles

also have this but the wavelike properties are too small to be observed.

8.22 Diffraction is a phenomenon caused by the constructive or the destructive interference of two or more

waves. The fact that electrons and other subatomic particles exhibit diffraction supports the theory thatmatter is correctly considered to have wave nature.

8.23 To determine whether a beam was behaving as a wave or as a stream of particles, a diffraction experiment

would have to be done.

8.24 Wave/particle duality is that light and matter have both wave-like properties and particle-like properties.

8.25 In a traveling wave, the positions of the peaks and nodes change with time. In a standing wave, the peaks

and nodes remain in the same positions.

8.26 The collapsing atom paradox comes from classical mechanics and asks, "Why doesn't the electron fall into

the nucleus?"

8.27 Quantum mechanics resolves the collapsing atom paradox by allowing only certain energy levels for the

electron. The electron located in the nucleus does not give one of the allowed energies.

8.28 Wave mechanics and quantum mechanics.

8.29 This is the orbital of the electron.

8.30 First, we are interested in the energies of orbitals, because it is the energies of the various orbitals that

determine which orbitals are occupied by the electrons of the atom. Secondly, we are interested in the


6/541

Chapter 8

157

shapes and orientations of the various orbitals, because this is important in determining how atoms form

bonds in chemical compounds

8.31 n= 1, 2, 3, 4, 5,

8.32 (a) n= 1 (b) n= 3

8.33 Every shell contains the possibility that = 0.

8.34 (a) 1 (b) 3 (c) 5 (d) 7

8.35 Yes, if the value for for this electron is 3 or larger.

8.36 The only impossible set of legitimate quantum numbers would be an electron having the exact same values

for the four quantum numbers. Recall the Pauli Exclusion Principle.

8.37 The electron behaves like a magnet, because the revolving charge (spin) of the electron creates a magnetic

field. Recall that the electron does not actually revolve but behaves in a fashion reminiscent of an electro-

magnet.

8.38 Atoms with unpaired electrons are termed paramagnetic.

8.39 No two electrons in the same atom can have exactly the same set of values for all of the four quantum

numbers. This limits the allowed number of electrons per orbital to two, since with other quantum numbers

being necessarily the same, two electrons in the same orbital must at least have different values of ms.

8.40 ms= +1/2 or1/2

8.41 The distribution of electrons among the orbitals of an atom.

8.42 The orbitals within a given shell are arranged in the following order of increasing energy: s


7/541

Chapter 8

158

8.50 As explained by the Heisenberg Uncertainly Principle, the position and momentum of an electron cannot be

known with 100% certainty, therefore it is impossible to know where an electron is, but we can determine

the probability of finding an electron in a given space. An electron is visualized as being within a cloud

around the nucleus. This electron cloud defines a volume in space where the probability of finding an

electron is high.

8.51 (a) See Figures 8.22a and 8.23 (b) See Figures 8.24a, 8.25 and 8.26.

8.52 As nbecomes larger, the orbital becomes larger.

8.53 The threeporbitals of a givenpsubshell are oriented at right angles (90) to one another.

8.54 A nodal plane is a plane in which there is zero probability of finding an electron.

8.55 A radial node is a spherical shaped node on which the electron density is zero.

8.56 Aporbital has 1 nodal plane; a dorbital has 2 nodal planes.

8.57 See Figure 8.27.

8.58 The effective nuclear charge is the net nuclear charge that an electron actually experiences. It is different

from the formal nuclear charge because of the varying imperfect ways in which one electron is shielded

from the nuclear charge by the other electrons that are present. The effective nuclear charge remains nearly

constant from top to bottom in any one group of the periodic table, and it increases from left to right in any

one row of the periodic table.

8.59 The larger atoms are found in the lower left corner of the periodic table; the smaller atoms are found in the

upper right corner of the periodic table.

8.60 The size changes within a transition series are more gradual because, whereas the "outer" electrons are in

anssubshell, the electrons that are added from one element to another enter an inner ( n1)dsubshell.

8.61 Ionization energy is the energy that is needed in order to remove an electron from a gaseous atom or ion.

These are positive values because the force of attraction between an electron and the nucleus of either anatom or a positive ion (cation) must be overcome in order to remove the electron.

8.62 (a) O(g)O+(g) + e

(b) O2+(g)O3+(g) + e

8.63 Ionization energy increases from left to right in a row of the periodic table because the effective nuclear

charge increases from left to right. The latter trend occurs because of the consequences of the increasingly

imperfect shielding of electrons by other electrons within the same level. The ionization energy decreases

down a periodic table group because the electrons reside farther from the nucleus with each successive

quantum level that is occupied. The farther the electrons are from the nucleus, the less tightly they are held

by the nucleus.

8.64 Removing a second electron involves pulling it away from a greater positive charge because of the positivecharge created by the removal of the first electron. Hence, more energy must be spent to ionize the second

electron than the first.

8.65 The fifth electron must be removed from a different, lower nshell. Also, an electron must be removed

from an ion having a 4+ rather than only a 3+ charge.

8.66 The last valence electron of aluminum begins the occupation of the 3pset of orbitals. This electron is

therefore well shielded from the nuclear charge by the 3selectrons.


8/541

Chapter 8

159

8.67 The last valence electron of sulfur is placed in an orbital with another electron, and the two electrons are

required to be spin-paired by the Pauli exclusion principle. This destabilizes the last electron, making it

easier to ionize than is the case for the last electron of phosphorus.

8.68 Electron affinity is the enthalpy change associated with the addition of an electron to a gaseous atom:

X(g) + eX(g)

8.69 S(g) + eS(g), first electron affinity

S(g) + eS2(g), second electron affinity

The first electron affinity is exothermic, while the second electron affinity should be endothermic because

work must be done to force the electron into the negative Sion.

8.70 The value for fluorine is low because of the especially small size of the atom, which causes addition of an

electron to be relatively unfavorable. The comparison between chlorine and bromine follows the normal

trend on descent of a group, the electron affinity decreasing with ionization energy.

8.71 The second electron affinity is always unfavorable (endothermic) because it requires that a second electron

be forced onto an ion that is already negative.

8.72 The electron affinity becomes more negative (exothermic) as effective nuclear charge increases. Fluorine

therefore has the more exothermic electron affinity because it has the larger effective nuclear charge.

Review Problems

The number used for the speed of light, c, depends on the number of significant figures. For one to three significant

figures, the value for c is 3.00 108m/s, for four significant figures, the value for c is 2.998 10 8m/s.

8.738

9

c 3.00 10 m/s= =

436 10 m

= 6.88 1014s1= 6.88 1014Hz

8.748

9

c 3.00 10 m/s= =

295 10 m

= 1.01 1015s1= 1.01 1015Hz

8.758

6

c 3.00 10 m/s= =

6.85 10 m

= 4.38 1013s1= 4.38 1013Hz

8.768

6

c 3.00 10 m/s= =

0.48 10 m

= 6.25 1014s1= 6.25 1014Hz

8.77 295 nm = 295 109m8

9

c 3.00 10 m/s= =

295 10 m

= 1.02 1015s1= 1.02 1015Hz

8.788

9

c 2.99792458 10 m/s= =

632.99139822 10 m

= 4.73670145 1014s1= 4.73670145 1014Hz

The speed of light is only given to 9 significant figures in the text.

8.79 101.1 MHz = 101.1 106Hz = 101.1 106s18

6 1

c 2.998 10 m/s= =

101.1 10 s

= 2.965 m


9/541

Chapter 8

160

8.80 5.09 1014Hz = 5.09 1014s1

8

7

14 1 9

c 3.00 10 m/s 1 nm= = = 5.89 10 m = 589 nm

5.09 10 s 1 10 m

8.81

8

1

c 3.00 10 m/s= =

60 s

= 5.0 10

6m = 5.0 103km

8.82 1.50 1018Hz = 1.50 1018s18

18 -1

c 3.00 10 m/s= =

1.50 10 s

= 2.00 1010m = 2.00 101nm = 2.00 102pm

8.83 E = h= 6.63 1034J s (4.0 1014s1) = 2.7 1019J

19 23J 2.7 10 J 6.02 10 photons=

mol 1 photon 1 mol

= 1.6 105J mol1

8.84 E = h= hc/, and 563 nm = 563 109m

34 8 ms

9

6.63 10 J s 3.00 10hcE = =

563 10 m

= 3.53 1019J

8.85 (a) violet (see Figure 8.7)

(b) = c/= 2.998 108m s1/ 410.3 109m = 7.307 1014s1

(c) E = h= (6.626 1034J s) (7.307 1014s1) = 4.842 1019J

8.86 (a) yellow (See Figure 8.7)

(b)

8

14 1

9

m3.00 10c s 5.09 10 s589 10 m

(c) E = h= (6.63 1034Js) (5.09 1014s1) = 3.37 1019J

8.87 1 12 2

1 1 1= 109,678 cm = 109,678 cm 0.1111 0.02778

3 6

= 9.140 103cm1

= 1.094 104cm = 1094 nm

We would not expect to see the light since it is not in the visible region.

8.88 1 12 2

1 1 1= 109,678 cm = 109,678 cm 0.250 0.0400

2 5

= 2.303 104cm1

= 434.2 nm, this will be purple and will just be visible

8.89 12 2

1 1 1= 109,678 cm

4 10

= 5.758 103cm1

= 1.737 106m, this is in the infrared region

8.90 1 12 2

1 1 1= 109,678 cm = 109,678 cm 1 0.0625

1 4

= 102,823 cm1

= 9.73 106cm = 97.3 nm, which is in the ultraviolet region.


10/541

Chapter 8

161

E =h c

=

34 8 118

9

6.626 x 10 J s 3.00 x 10 m s2.04 x 10 J

97.3 x 10 m

8.91 (a) p (b) f

8.92 (a) 3 (b) 2

8.93 (a) n= 3, = 0 (b) n= 5, = 2

8.94 (a) n=4, =1 (b) n= 6, = 3

8.95 0, 1, 2, 3, 4, 5

8.96 n= 8

8.97 (a) m= 1, 0, or1 (b) m

= 3, 2, 1, 0,1,2, or3

8.98 5,4,3,2,1, 0, 1, 2, 3, 4, 5

8.99 When m=4 the minimum value of is 4 and the minimum value of nis 5.

8.100 If l= 5 we should have 2l+ 1 orbitals which gives eleven orbitals. There are eleven values for m:5,4,

3,2,1, 0, 1, 2, 3, 4, and 5.

8.101 n m ms

2 1 1 +1/2

2 1 1 1/2

2 1 0 +1/2

2 1 0 1/2

2 1 +1 +1/2

2 1 +1 1/2

8.102 n m ms

3 2 2 +1/2

3 2 2 1/2

3 2 1 +1/2

3 2 1 1/2

3 2 0 +1/2

3 2 0 1/2

3 2 1 +1/2

3 2 1 1/2

3 2 2 +1/2

3 2 2 1/2

8.103 21 electrons have = 1, 20 electrons have = 2

8.104 12 electrons have = 0, 12 electrons have m= 1

8.105 (a) S 1s22s22p63s23p4

(b) K 1s22s22p63s23p64s1

(c) Ti 1s22s22p63s23p63d 24s2


11/541

Chapter 8

162

(d) Sn 1s22s22p63s23p63d 104s24p64d 105s25p2

8.106 (a) As 1s22s22p63s23p64s23d 104p3

(b) Cl 1s22s22p63s23p5

(c) Ni 1s22s22p63s23p64s23d8

(d) Si 1s22s22p63s23p2

8.107 (a) Mn is [Ar]4s23d5, five unpaired electrons, paramagnetic

(b) As is [Ar] 3d104s24p3, three unpaired electrons, paramagnetic

(c) S is [Ne]3s23p4, two unpaired electrons, paramagnetic

(d) Sr is [Kr]5s2, zero unpaired electrons, not paramagnetic

(e) Ar is 1s22s22p63s23p6, zero unpaired electrons, not paramagnetic

8.108 (a) Ba is [Xe]6s2, zero unpaired electrons: diamagnetic

(b) Se is [Ar]4s23d104p4, two unpaired electrons: paramagnetic

(c) Zn is [Ar]4s23d10, zero unpaired electrons: diamagnetic

(d) Si is [Ne]3s23p2, two unpaired electrons: paramagnetic

8.109 (a) Mg is 1s22s22p63s2, zero unpaired electrons

(b) P is 1s

2

2s

2

2p

6

3s

2

3p

3

, three unpaired electrons(c) V is 1s22s22p63s23p63d34s2, three unpaired electrons

8.110 (a) Cs is [Xe]6s1, 1 unpaired electron

(b) S is [Ne]3s23p4, 2 unpaired electrons

(c) Ni is [Ar]4s23d8, 2 unpaired electrons

8.111 (a) Ni [Ar]3d84s2

(b) Cs [Xe]6s1

(c) Ge [Ar] 3d104s24p2

(d) Br [Ar] 3d104s24p5

(e) Bi [Xe] 4f145d106s26p3

8.112 (a) Al [Ne]3s23p1

(b) Se [Ar]4s23d104p4 or [Ar]3d104s24p4

(c) Ba [Xe]6s2

(d) Sb [Kr]5s24d105p3 or [Kr]4d105s25p3

(e) Gd [Xe]6s24f75d1 or [Xe]4f75d16s2

8.113 (a) Mg

(b) Ti

1s 2s 2p 3s 3p 4s 3d

8.114 a) As:

1s 2s 2p 3s 3p 4s 3d 4p (b) Ni:

8.115 (a) Ni


12/541

Chapter 8

163

4s 3d 4p

[Ar]

(b) Cs

(c) Ge

4s 3d 4p

[Ar]

(d) Br

4s 3d 4p

[Ar]

8.116 (a) Al

3s 3p

[Ne]

(b) Se

4s 3d 4p

[Ar]

(c) Ba

(d) Sb

5s 4d 5p

[Kr]

8.117 The value corresponds to the row in which the element resides:

(a) 5 (b) 4 (c) 4 (d) 6

8.118 The value corresponds to the row in which the element resides:

(a) 3 (b) 4 (c) 6 (d) 5

8.119 (a) Na 3s1 (b) Al 3s23p1 (c) Ge 4s24p2 (d) P 3s23p3

8.120 (a) Mg 3s2 (b) Br 4s24p5 (c) Ga 4s24p1 (d) Pb 6s26p2

8.121 (a) Na

3s (b) Al

3s 3p (c) Ge

4s 4p (d) P

3s 3p

8.122 (a) Mg


13/541

Chapter 8

164

3s 3p (b) Br

4s 4p (c) Ga

4s 4p (d) Pb

6s 6p

8.123 (a) There are 10 core electrons in Na so the valence electron would see 1110 or +1 as the effective

nuclear charge.

(b) There are 10 core electrons in S so the valence electrons would see 1610 or +6 as the effective

nuclear charge.

(c) There are 28 core electrons in Br so the valence electrons would see 3528 or +7 as the effective

nuclear charge.

8.124 (a) 2 (b) 4 (c) 7

8.125 (a) Mg (b) Bi

8.126 (a) Al (b) Tl

8.127 Sb is a bit larger than Sn, according to Figure 8.29. (Based upon trends, Sn would be predicted to be larger,

but this is one area in which an exception to the trend exists. See Figure 8.29.)

8.128 Since these atoms and ions all have the same number of electrons, the size should be inversely related to

the positive charge:

Mg

2+

< Na

+

< Ne < F

< O

2

< N

3

8.129 Cations are generally smaller than the corresponding atom, and anions are generally larger than the

corresponding atom:

(a) Na (b) Co2+ (c) Cl

8.130 Cations are generally smaller than the corresponding atom, and anions are generally larger than the

corresponding atom:

(a) S2 (b) Al (c) Au+

8.131 (a) N (b) S (c) Cl

8.132 (a) Li (b) F (c) F

8.133 (a) Br (b) As

8.134 (a) S (b) N

8.135 The element with the largest difference between the second and third ionization potential would be the

element with two valence electrons. The third ionization would remove an electron from the core, which is

much higher in energy than removing electrons from valence levels. Mg has the valence structure:


14/541

Chapter 8

165

8.136 Si

Additional Exercises

8.137 (speed = wavelength frequency) v =

v

(a) longest wavelength =m

s

1

330

20 s

= 16. m

shortest wavelength =m

s

1

330

20,000 s

= 0.016 m

(b) longest wavelength =ms

11500

20 s

= 75 m

shortest wavelength =m

s

1

1500

20,000 s

= 0.075 m

8.138 We proceed by calculating the energy of a single photon:

34 8 ms 23

3

6.626 10 J s 3.00 10hcE = = = 6.63 10 J

3.00 10 m

It requires 4.184 J to increase the temperature of 1.00 g of water by 1 Celsius degree. So,

photons = 4.184 J 6.63 1023J/photon = 6.32 1022photons

8.139 12 2 9

2 1

1 1 1 1 1 nm 1 m= 109,678 cm

410.3 nm 100 cmn n 1 10 m

1

5 2 2

4 1

1 2

2

1 1 1 109,678 cm

4.103 10 cm 2 x

2.437 10 cm 1 1 0.222

4109678 cm x

1 1 0.222

4x

x 6

8.140 This corresponds to the special case in the Rydberg equation for which n1= 1 and n2= .

1 12 2

1 1 1= 109,678 cm = 109,678 cm 1 0

1

= 109,678 cm1

= 9.12 106cm = 91.2 nm.

8.141 A transition from high energy to low energy may result in light emission. The transition from 4p3d,

5f4d, and 4d2pare the only possibilities.


15/541

Chapter 8

166

8.142 (a) Start by calculating

1 12 2

1 1 1= 109,678 cm = 109,678 cm 1.000 0.04000

1 5

= 1.053 105cm1

= 9.498 106cm = 94.98 nm, which is in the ultraviolet region of the visible spectrum.

(b) 1 12 21 1 1= 109,678 cm = 109,678 cm 0.2500 0.06252 4

= 2.056 104cm1

= 4.863 105cm = 486.3 nm, which is in the visible region of the spectrum.

(c) 1 12 2

1 1 1= 109,678 cm = 109,678 cm 0.0625 0.0278

4 6

= 3.808 103cm1

= 2.626 104cm = 2626 nm, which is in the infrared region of the spectrum.

8.143 (a) Thesshell is not completely filled.

(b) Using the shorthand notation of [Kr], we imply that the 4s, 3dand 4pshells are filled but the rest

of the configuration negates this assumption.

(c) There is nothing wrong.

(d) Thesshell should be completely filled. If moving an electron from thessubshell would have

filled the dsubshell, then we could have this case but the dsubshell is still two electrons short offilling.

8.144 (a) This diagram violates the aufbau principle. Specifically, thes-orbital should be filled before

filling the higher energyp-orbitals.

(b) This diagram violates the aufbau principle. Specifically, the lower energys-orbital should be

filled completely before filling thep-orbitals.

(c) This diagram violates the aufbau principle. Specifically, the lower energys-orbital should be

filled completely before filling thep-orbitals.

(d) This diagram violates the Pauli Exclusion Principle since 2 electrons have the same set of quantum

numbers. This electron distribution is impossible.

8.145 14

8.146 The 4selectrons are lost; n= 4, = 0, m= 0, ms= 1/2

8.147 (a) This corresponds to the special case in the Rydberg equation for which n1= 1 and n2= .

For a single atom, we have:

1 1 12 2

1 1 1= 109,678 cm = 109,678 cm 1 0 109,678 cm

1

= 9.12 106cm = 91.2 nm.

Converting to energy, we have:

E = hc/

E =

34 8 ms

9

6.626 10 J s 3.00 10hc=

91.2 10 m

= 2.18 1018J

(b) Conversion to kJ/mol gives us:


16/541

Chapter 8

167

kJ/mol =18 232.18 10 J 1 kJ 6.02 10 photons

photon 1000 J mole

= 1.31 103kJ/mol

8.148

2s 2p

2s 2p

+ 2 e-

O2-

O

Thus the oxide ion, O2, has the electron configuration of [Ne].

Since the oxide ion has a closed shell electron configuration, the addition of a third electron is extremely

difficult.

8.149 We simply reverse the electron affinities of the corresponding ions.

(a) F(g)F(g) + e, H = 328 kJ/mol

(b) O(g)O(g) + e, H = 141 kJ/mol

(c) O2(g)O(g) + e, H =844 kJ/mol

The last of these is exothermic, meaning that loss of an electron from the oxide ion is favorable from the

standpoint of enthalpy.

8.150 In problem 8.149 parts b and c we can determine that

O(g) + 2eO2(g) H = +703 kJ

From table 8.2 we can see that O(g)O+(g) + eH = +1314 kJ/mol.

It takes more energy to ionize oxygen than to create O2(g)

Multi-Concept Problems

8.151 E =1

2mv2

2 Ev

m first determine E for a single particle

E = (2080 103J/mol)23

1 mole

6.022 10 particles

= 3.454 1018J/particle

v =18

31

2(3.454 10 J)

9.109 10 kg

= 2.754 106m/sec

8.152 (a) We must first calculate the energy in joules of a mole of photons.

E = 34 8

9

6.63 10 J s 3.00 10 m/shc=

600 10 m

= 3.32 1019J/photon


17/541

Chapter 8

168

(3.32 1019J/photon)(6.02 1023photons/mol) = 2.00 105J/mol

Next, we calculate the heat transfer problem as in Chapter 7:

Heat = (specific heat)(mass)(change in temperature)

2.00 105J = 4.18 J g1C1 X g 5.0 C

X = 9.57 103g

(b) E = 6.63 1019J/photon

E = 3.99 105J/mol

X = 1.91 104g

8.153 To solve this problem use E = hc/, where is our unknown quantity.

=

34 8 ms 23 photons

mol3 Jmol

6.63 10 J s 3.00 10hc = 6.02 10

E 328 10

= 3.65 107m = 365 nm

8.154 From Table 8.2 we find that the first ionization energy for sodium is 496 kJ mol-1. We need to determine of

a wavelength of 23.7 nm would have enough energy to remove the 3s 1 electron. If this wavelength

corresponds to an energy greater than 496 kJ mol-1, then the excess energy would be observed as the

electrons kinetic energy.

E = h= hc/

E = [6.63 x 10-34J s x 3.00 x 108 m s-1/23.7 x 10-9 m] x 6.02 x 1023atoms mol-1= 5.05 x 106 J

or 5050 kJ

The maximum kinetic energy of the ionized electron would be the difference between the energy of the

photon and the ionization energy.

KE = 5050kJ496 kJ = 4554 kJ

8.155 The ionization energies of B are given in Table 8.2. All of the following values are in kJ mol-1.

1st IE = 800 2ndIE = 2426 3rdIE = 3659 4thIE = 25,020 5thIE = 32,820

To determine the kinetic energy of the ionized electron we need to first determine the energy of the

ionizing photon. The kinetic energy of each ejected electron will be the difference between this energy and

the energy required to ionize the electron.

Ephoton= [6.63 x 10-34J s x 3.00 x 108m s-1/23.7 x 10-9m] x 6.02 x 1023atoms mol-1= 5.05 x 106J

Or 5050 kJ

IE KE

1st 5050800 kJ = 4250 kJ

2nd

50502426 kJ = 2624 kJ3rd 50503659 kJ = 1391 kJ

The photons energy is not great enough to remove electrons beyond these. The 4thionization

energy is 25,020 kJ or about 5 times the energy of the ionizing photon.


18/541

Chapter 9

169

Practice Exercises

9.1 There is one electron missing, and it should go into the 5sorbital, and the 5porbital should be empty.

1s12s22p63s23p63d104s24p64d105s2

9.2 Cr: [Ar]3d54s1

(a) Cr2+: [Ar]3d4 The 4selectron and one 3delectron are lost.

(b) Cr3+: [Ar]3d3 The 4selectron and two 3delectrons are lost.

(c) Cr6+: [Ar] The 4selectron and all of the 3delectrons are lost.

9.3 S2: [Ne] 3s23p6

Cl: [Ne] 3s23p6

The electron configurations are identical.

9.4

CaI Ca2+ I2+I

9.5

MgO Mg2+ O2-

+

9.6

aldedyde

amine

acid

ketone

alcohol


19/541

Chapter 9

170

9.7 (a) CH3NHCH3will produce a basic solution

(b) HCOOH will produce an acidic solution

(c)

9.8 = q r

q = 0.167 e19

1.602 10 C

1 e

= 2.675 1020C

r = 154.6 pm = 154.6 1012 m

q = (2.675 1020C) (154.6 1012 m) = 4.136 1030C m

in debey units:

q = 4.136 1030C m30

1 D

3.34 10 C m

= 1.24 D

9.9 q =r

= 9.00 D303.34 10 C m

1 D

= 3.006 1029C m

r = 236 pm = 236 1012m

q =29

12

3.006 10 C m

236 10 m

= 1.27 1019C

In electron charges

q = 1.27 1019C19

1 e

1.602 10 C

= 0.795 e

On the sodium the charge is +0.795 eand on the chlorine, the charge is0.795 e.

This would be 79.5% positive charge on the Na and 79.5% negative charge on the Cl.

9.10 The bond is polar and the Cl carries the negative charge.

9.11 (a) Br (b) Cl (c) Cl

9.12

P OO HH

O

O

Number of valence electrons:

O 6 each and 4 O atoms total 24 electrons

P 5 electrons

H 1 each and 2 H atoms total 2 electrons

Negative charge 1 electron

Total 32 valence electrons


20/541

Chapter 9

171

9.13 SO2

S OO

NO3

N OO

O

HClO3

H3AsO4

As OO HH

O

O

H

9.14 SO2has 18 valence electrons

SeO42has 32 valence electrons

NO+has 10 valence electrons

9.15

F O F

H H

H

N

H +

O S O

O ON

O

_

F FCl

F

O

O

Cl

O

H O


21/541

Chapter 9

172

9.16 The negative sign should be on the oxygen, so two of the oxygen atoms should have a single bond and

three lone pairs and the sulfur should have one double bond, two single bonds, and a lone pair.

9.17

(a) (b)

N N O

-2

+1 +1

S C N

_

-100

9.18 In each of these problems, we try to minimize the formal charges in order to determine the preferred Lewis

structure. This frequently means violating the octet rule by expanding the octet. Of course, this can onlybe done for atoms beyond the second period as the atoms in the first and second periods will never expand

the octet.

(a) SO2

O S O

(b) HClO3

(c) H3PO4

P

O

H O

HO

HO

9.19

There is no difference between the coordinate covalent bond and the other covalent bonds.


22/541

Chapter 9

173

9.20

OH H+ OH H+

coordinate covalent bond

9.21

There are four resonance structures.

P

O

O

O

O PO

O

O

O

P

O

O

O

O P O

O

O

O3- 3- 3- 3-

9.22

O

CO O

H

O

CO O

H

9.23

BrO

O

O BrO

O

O Br

O

OO

_ _ _

Review Questions

9.1 A stable compound forms from a collection of atoms when bonding results in a net lowering of the energy.

The process of bonding the atoms together must release energy to make the bonded compound more stable

than the original collection.

9.2 The ionic bond is the attraction between positive and negative ions in an ionic compound. It is largely an

electrostatic attraction, and it gives rise to the lattice energy of the ionic compound. The lattice energy is

the net gain in stability when the gaseous ions are brought together to form the crystalline ionic compound.

9.3 Ionic bonds tend to form upon combining an element having a high EA with an element having a low IE.

9.4 The lattice energy is the energy necessary to separate a mole of an ionic solid into its constituent ions in the

gas phase. It is also the energy that is released on forming the ionic solid from the gaseous ions. Asdiscussed in the answer to review questions 9.1 and 9.2, it is the lattice energy that is primarily responsible

for the stability of ionic compounds.

9.5 Magnesium ([Ne]3s2) can achieve the electron configuration of the nearest noble gas (Ne) by losing only

two electrons: Mg2+ 1s22s22p6. Magnesium will not form the Mg3+ion because an extremely high amount

of energy would be required to break into the 2s22p6core to remove an electron.

9.6 When chlorine gains one electron to form Cl, it has filled an orbital and achieved a noble gas

configuration. To make the Cl2ion, an electron would have to be placed in the next higher shell. The


23/541

Chapter 9

174

amount of energy required for this to occur is extremely high and makes the creation of a Cl 2ion

energetically unfavorable.

9.7 Many of the transition metals in Period 4 have an 4s2outer-shell electron configuration. Since these

characteristically are the first electrons to be lost when a transition metal atom is ionized, it is common that

a 2+ ion should be formed.

9.8 The largest difference in ionization energies would be between the third and the fourth successive

ionization energies because of the extremely high amount of energy that would be required to break into the

2s22p6core to remove an electron.

9.9 (a) Al2O3

(b) BeO

(c) NaCl

9.10 The valence electrons are primarily responsible for chemical bonds.

9.11 (a) correct

(b) incorrect

(c) correct

(d) incorrect

9.12 Ionic bonding does not occur between two nonmetal elements because more energy must be provided in the

form of IE and EA than can be recovered from the lattice energy.

9.13 As two hydrogen atoms approach each other in forming the H2molecule, the electron density of the two

atoms shifts to the region between the two nuclei.

9.14 The energy drops to some optimum or minimum value when the nuclei have become separated by the

distance called the bond distance. The electron spins become paired.

9.15 Bond formation is always exothermic.

9.16 The bond distance in a covalent bond is determined by a balance (compromise) between the separate

attractions of the nuclei for the electron density that is found between them, and the repulsions between thelike-charged nuclei and those between the like-charged electrons. These attractions and repulsions oppose

one another, and a bond distance is achieved that maximizes the attraction while minimizing the repulsions.

9.17 The octet rule is the expectation that atoms tend to lose or gain electrons until they achieve a noble gas-type

electron configuration, namely eight electrons in the valence, or outer-most, shell. It is the stability of the

closed-shell electron configuration of a noble gas that accounts for this.

9.18 (a) one (b) four (c) two (d) three (e) one

9.19 The valence shell of a period 2 element can hold only eight electrons since only two types of subshells,s

andp, are available in this period. The valence shell of elements in row three can hold as many as eighteen

electrons. This results because there are three types of subshells,s,pand d, that can hold electrons. Thes

subshells hold 2 electrons,psubshells hold 6 electrons and dsubshells hold 10 electrons. The total is,therefore, 18 electrons.

9.20 (a) single bond: a covalent bond formed by the sharing of one pair of electrons.

(b) double bond: a covalent bond formed by the sharing of two pairs of electrons.

(c) triple bond: a covalent bond formed by the sharing of three pairs of electrons.


24/541

Chapter 9

175

9.21

CH N

9.22 Since the outer shell (or valence shell) of hydrogen can hold only two electrons, hydrogen is not said toobey the octet rule. It does, however, readily satisfy its requirement for a closed shell electron

configuration through the formation of one covalent bond.

9.23 Methane

Ethane

Propane

9.24

The is an example of an alkane, carbon-hydrogen compounds with single bonds. The general formula for

alkanes is CnH2n+2so we should have 14 hydrogen atoms. The formula is C6H14, and is named n-hexane.


25/541

Chapter 9

176

9.25 There are three isomers of pentane.

9.26 A carbonyl group is shown below.

This group is found in organic acids such as benzoic acid (shown below), ketones, aldehydes, amides, and

esters.

9.27 acid

ketone

alcohol

aldehyde

aminehydrocarbon

9.28 CH3CH2COOH(aq) + H2O CH3CH2COO-(aq) + H3O

+(aq)

The acid is propanoic acid, a weak acid


26/541

Chapter 9

177

The anion formed is the propanoate ions:

9.29 CH3NHCH2CH3(aq) + H2O CH3NH2CH2CH3+(aq) + OH-(aq)

9.30 CH3CH2COOH(aq) + CH3NHCH2CH3(aq) CH3CH2COO

-

(aq) + CH3NH2CH2CH3+

(aq)

9.31 Ethylene acetylene

9.32 A polar covalent bond is one in which the electrons of the bond are not shared equally by the atoms of the

bond, and this causes one end of the linkage to carry a partial negative charge while the other end carries a

corresponding partial positive charge. In other words, there is a dipole in a polar bond.

9.33 = q r

A dipole moment is the product of the amount of charge on one end of a polar bond (which behaves as a

dipole) and the distance between the two partial charges that compose the dipole. It is also normally taken

to be the product of the charge in the dipole of a polar bond and the internuclear distance in the polar bond.

1 debye = 3.34 1030coulomb meter

9.34 Electronegativity is the attraction that an atom has for the electrons in chemical bonds to that atom. Pauling

based his scale of electronegativity on the greater bond energy polar bonds have than would be expected if

the opposite ends of the bonds were electrically neutral.

9.35 Fluorine has the largest electronegativity, whereas oxygen has the second largest electronegativity.

9.36 (b) and (d) are the only ones with an electronegativity difference greater than 1.7.

9.37 Elements having low electronegativities are metals. Elements with low ionization potentials and low

electron affinities tend to have low electronegativities.

9.38 This has to do with the ease of oxidation. Aluminum is easier to oxidize than iron.


27/541

Chapter 9

178

9.39 The most reactive metals are in the left-most groups of the periodic table, namely the metals of Groups IA

and IIA. The least reactive metals are found in the second and third rows of the transition elements.

9.40 The lower the electronegativity, the more reactive is the metal.

9.41 calcium > iron > silver > iridium

9.42 (a) NR

(b) 2NaI(aq) + Cl2(g)2NaCl(aq) + I2(s)

(c) 2KCl(aq) + F2(aq)2KF(aq) + Cl2(g)

(d) CaBr2(aq) + Cl2(g)CaCl2(aq) + Br2(l)

(e) 2AlBr3(aq) + 3F2(g)2AlF3(aq) + 3Br2(l)

(f) NR

9.43 (a) F2 (b) P4 (c) Br2

(d) S8 (e) Cl2 (f) S8

9.44 (a) four electrons or two bonding pairs

(b) six electrons, all in bonding pairs

(c) two electrons for each hydrogen

9.45 We expect a minimum of ten electrons, in bonding pairs between As and each of five Cl atoms.

9.46 N is in the second period, it can only have an octet of electrons, so it cannot form five bonds. Whereas, As

is in the fourth row, it can have an expanded octet and form five bonds; therefore, it can form AsCl 3 and

AsCl5.

9.47 Bond length is the distance between the nuclei of two atoms that are linked by a covalent bond.

Bond energy is the energy that is needed to break a chemical bond; conversely, the energy that is released

when a chemical bond is formed.

9.48 The number of electron pairs that comprise a covalent linkage (bond) between two atoms is called the bond

order. As bond order increases, so does the strength of the bond. Therefore, as bond order increases, bond

energy increases and bond length decreases, in keeping with increased bond strength.

9.49 The HCl bond energy is defined to be the energy required to break the bond to give atoms of H and Cl,

not the ions H+and Cl. In other words, when the bond is broken between the H and the Cl atoms, one of

the two electrons of the bond must go to each of the atoms. When ions are obtained, however, both

electrons of the bond are given to the chlorine atom and the lattice energy would be the more appropriate

term to use.

9.50 The formal charge on an atom is its apparent charge in a particular Lewis diagram. For any particular atom

in a Lewis diagram, the formal charge is calculated by subtracting the number of bonds to the atom, and the

number of unshared electrons on the atom, from the number of valence electrons that the unbonded atom

normally has in the ground state.

9.51 It is generally held that the Lewis structure having the smallest formal charges is most favored. This isbased on the notion that either an accumulation of too much charge, or too much charge separation, is

destabilizing.

9.52 The formal charges on the atoms in HCl are 0. The actual charges are H: +0.17eand Cl:0.17e. The

formal charges arise from the bookkeeping in Lewis structures, and are not the same as actual charges.

9.53 A coordinate covalent bond is one in which both electrons of the bond are contributed by (or donated by)

only one of the atoms that are linked by the bond.


28/541

Chapter 9

179

9.54 Once formed, a coordinate covalent bond is no different than any other covalent bond.

9.55

9.56 Lewis structures are often inadequate, without the concept of resonance, to describe the distribution of

electrons in certain molecules and ions. Sometimes more than one Lewis structure is needed to describe a

molecule. These structures differ only in the placement of electrons. Resonance more fully describes the

distribution of electrons in a molecule.

9.57 A resonance hybrid is the true structure of a molecule or polyatomic ion, whereas the various resonance

structures that are used to depict the hybrid do not individually have any reality. The hybrid is a mix, or

average, of the various resonance structures that compose it.

9.58

The resonance structures of benzene are more stable than the ring containing three carbon-carbon double

bonds because six bonds with a bond order of 1.5 is more stable than three single bond and three double

bonds together.

9.59

Review Problems

9.60 Na(g)Na+(g) + e IE = 496 kJ/mol

Cl(g) + eCl(g) EA =348 kJ/mol

Na(g) + Cl(g)

Na

+

(g) + Cl

(g) H = 148 kJ/mol

Na(g)Na+(g) + e IE = 496 kJ/mol

Na+(g)Na2+(g) + e IE = 4563 kJ/mol

2Cl(g) + 2e2Cl(g) EA = 2(348 kJ/mol)

Na(g) + 2Cl(g)Na2+(g) + 2Cl(g) H = 4.36 10

3kJ/mol

In order for NaCl2to be more stable than NaCl, the lattice energy should be almost 30 times larger.

4.36 103 kJ/148 kJ = 29.5

B

Cl

Cl

Cl

O H

H

B

Cl

Cl

Cl

O H

H

C

C

C

C

C

C

H

H

H

H

H

H

C

C

C

C

CH

H

H

H

H

H

C

CC

C

C

CH

H

H

H

H

C

CC

C

C

CH

H

H

H

H

C

CC

C

C

CH

H

H

H

H

C

CC

C

C

CH

H

H

H

H

C

CC

C

C

CH

H

H

H

H

H

H

H

H

H

H

H

H

H H H H H


29/541

Chapter 9

180

9.61

Mg(s) + 1/ 2O2(g)

Mg(g) + O (g)

Mg+(g) + O (g)

Mg2+(g) + O(g)

Mg2+(g) + O-(g)

Mg2+(g) + O2-(g)

MgO(s)

+150 kJ

+737 kJ

+1450 kJ

-141 kJ

+844 kJ

-602 kJ

-3642 kJ

The lattice energy for NaCl is 787 kJ. The large difference in lattice energy is a result of charge (divalent

versus univalent) and size of ions.

9.62 Magnesium loses two electrons:

MgMg2+ + 2e

[Ne]3s2 [Ne]

Bromine gains an electron:

Br + eBr

[Ar]3d104s24p5 [Kr]

To keep the overall change of the formula unit neutral, two Brions combine with one Mg2+ion to form

MgBr2:

Mg2++ 2BrMgBr2

9.63 Nitrogen (1s22s22p3) gains three electrons to achieve the electron configuration of the next noble gas, neon:

N31s22s22p6

Lithium (1s22s1) loses an electron to achieve the electron configuration of the closest noble gas helium:

Li1+1s2

To keep the overall change of the formula unit neutral, three Li+ions combine with one N3ion to form

Li3N:

3Li++ N3Li3N

9.64 Pb2+: [Xe] 4f145d106s2

Pb4+: [Xe]4f145d10

9.65 Bi3+: [Xe]6s24f145d10

Bi5+: [Xe]4f145d10


30/541

Chapter 9

181

9.66 Mn3+: [Ar]3d4 4 unpaired electrons

9.67 Co3+: [Ar]3d6 4 unpaired electrons

9.68 (a) (b) (c)

Si

Sb Ba

(d) (e)

Al S

9.69 (a) (b) (c)

K Ge

As

(d) (e)

Br

Se

9.70 (a)

Ca BrBr Ca2+ Br2+

(b)

O O OAl Al 2 Al+ + O32-

(c)

K S 2 K+ + SK 2-

9.71 (a)

Mg S Mg2+ + S2-

(b)

MgCl Mg2+ Cl2+Cl

(c)

NMg Mg 3 Mg2+ + N23-

N Mg

9.72 = q r

= 0.16 D303.34 10 C m

1 D

= q (115 pm)12

1 m

10 pm

5.34 1031Cm = q (115 1012m)


31/541

Chapter 9

182

q = 4.64 1021C

q = 4.64 1021C

19

1 e

1.60 10 C

= 0.029 e

The charge on the oxygen is0.029 eand the charge on the nitrogen is +0.029 e. The nitrogen atom is

positive.

9.73 = q r

= 1.42 D303.34 10 C m

1 D

= q (176 pm)12

1 m

10 pm

4.74 1030Cm = q (176 1012m)

q = 2.69 1020C

q = 2.69 1020C

-19

1 e

1.60 10 C

= 0.17 e

The charge on the fluorine is0.17 eand the charge on the bromine is +0.17 e. The bromine atom is

positive.

9.74 = q r

= 1.83 D303.34 10 C m

1 D

= q (91.7 pm)12

1 m

10 pm

6.11 1030Cm = q (91.7 1012m)

q = 6.66 1020C

q = 6.69 1020C

-19

1 e

1.60 10 C

= 0.42 e

The charge on the fluorine is0.42 eand the charge on the hydrogen is +0.42 e. The fluorine atom is

negative.

9.75 = q r

= 7.88 D30

3.34 10 C m1 D

= q (0.255 nm)91 m

10 nm

2.63 1029Cm = q (0.255 109m)

q = 1.03 1019C

q = 1.03 1019C

-19

1 e

1.60 10 C

= 0.64 e

The charge on the fluorine is0.64 eand the charge on the cesium is +0.64 e. The cesium atom is

positive.

9.76 J /molecule = (242.6 103J/mol)

23

1 mole

6.022 10 molecules

= 4.029 1019J/molecule

9.77 We can use Hesss Law, Tables 8.2 and 8.3 to answer this question.

(1) H(g) H+(g) + e- IE = 1312 kJ mol-1

(2) Cl(g) + e- Cl-(g) EA = -348 kJ mol-1

(3) HCl(g) H(g) + Cl(g) BE = 431 kJ mol-1


32/541

Chapter 9

183

If we reverse all three chemical equations and add them together the results are:

H+(g) + Cl-(g) HCl(g)

The energy released from this reaction is: H = -1312 kJ mol-1+ 348 kJ mol-1431 kJ mol-1= -1395 kJ mol-1

9.78 E = h=hc hc

E

E = (348 103J/mol)23

1 mol

6.022 10 molecules


=

34 8 1

19

6.626 10 J sec 3.00 10 m s

5.78 10 J

= 3.44 107m = 344 nm

Ultraviolet region

9.79 E = h=hc hc

E

E = (242.6 103J/mol)23

1 mol

6.022 10 molecules


=

34 8 1

19

6.626 10 J sec 3.00 10 m s

4.029 10 J

= 4.934 107m = 493.4 nm

9.80 (a)

Br Br + Br Br

(b)

2 H O+ H O H

(c)

3 H N+ H N H

H

9.81 (a)

Cl N Cl

Cl


33/541

Chapter 9

184

(b)

Cl C Cl

Cl

Cl

(c)

Cl S Cl

(d)

Br Cl

9.82 (a) We predict the formula H2Se because selenium, being in Group VIA, needs only two additional

electrons (one each from two hydrogen atoms) in order to complete its octet.

(b) Arsenic, being in Group VA, needs three electrons from hydrogen atoms in order to complete its

octet, and we predict the formula H3As.

(c) Silicon is in Group IVB, and it needs four electrons (and hence four hydrogen atoms) to complete

its octet: SiH4.

9.83 (a) Each chlorine atom needs one further electron in order to achieve an octet, and the phosphorus

atom requires three electrons from an appropriate number of chlorine atoms. We conclude that a

phosphorus atom is bonded to three chlorine atoms and that each chlorine atom is bonded only

once to the phosphorus atom: PCl3.

(b) Since carbon needs four additional electrons, the formula must be CF4. In this arrangement, each

fluorine atom acquires the one additional electron that is needed to reach its octet.

(c) Each halogen atom needs only one additional electron from the other: ICl.

9.84 Here we choose the atom with the smaller electronegativity:

(a) S (b) Si (c) Br (d) C

9.85 (a) I (b) I (c) F (d) N

9.86 Here we choose the linkage that has the greatest difference in electronegativities between the atoms of the

bond: NS.

9.87 The least polar bond of the four is PI because it is the bond that has the smallest difference in

electronegativities between the linked atoms.

9.88 (a) (b)

Cl As Cl

Cl

Cl

+

O Cl O_


34/541

Chapter 9

185

(c) (d)

O N OH

F Xe F

9.89 (a) (b)

F

Te

F

F

F

F

F

F

F

F

Cl

(c) (d)

P

F

F F

F

F

F

_

Xe

F

F F

F

9.90 (a) (b)

Cl Si Cl

Cl

Cl

F P F

F

(c) (d)

H P H

H

Cl S Cl

9.91 (a) (b)

O I O

O

H

O C O

O

H H

(c) (d)


35/541

Chapter 9

186

O C O

O

H

_

Cl P Cl

Cl

Cl+

9.92 (a) (b)

S C S

9.93 (a) (b)

O Se O

O O Se O

9.94 (a) (b)

H As H

H

O Cl OH

(c) (d)

O Se O

O

H H

O As O

O

O

H

H

H

9.95 (a) (b)

N O+

N O

_O

(c) (d)

Sb

Cl

Cl Cl

Cl

Cl

Cl

_

O I O

O

_

9.96 (a) (b)

HCH

O

ClSCl

O


36/541

Chapter 9

187

9.97 (a) (b)

Cl Ge Cl

Cl

Cl

O C O

O 2-

(c) (d)

O P O

O

O3-

O O

2-

9.98 (a) (b)

O Cl OH

0 +1 -10

OSO

O1-

1-

2+

0

(c)

OSO1-

1+

0

9.99 (a) (b)

ONCl

O

1+

0

1-0

F N O

F

F

1-1+

0

00

(c)

F S F

O

O

0

1-

0

1-

2+


37/541

Chapter 9

188

9.100

O Cl O

O

O

H

1-

1-

1-

0

0

3+

O Cl O

O

O

H

9.101

ClSCl

O 1-

1+

0 0

ClSCl

O

0 0

0

0

9.102 The formal charges on all of the atoms of the left structure are zero, therefore, the potential energy of this

molecule is lower and it is more stable. Group 2 atoms normally have only two bonds.

9.103 The one on the right is better. The one on the left gives one chlorine atom a formal charge of +1, which is

not a likely situation for such a highly electronegative element.

9.104

H O H H H O H

H

+ +

9.105

F B F F F B F

F

F F_

_

9.106 The average bond order is 4/3

O C O

O 2-

O C O

O 2-

O C O

O 2-

9.107

N N

O

O O

O

N N

O

O O

O

N N

O

O O

O

N N

O

O O

O

The average bond order is 3/2.


38/541

Chapter 9

189

9.108 The Lewis structure for NO3is given on page 395, and that for NO2

is given below.

Resonance causes the average number of bonds in each NO linkage of NO3to be 1.33. Resonance

causes the average number of electron pair bonds in each linkage of NO2to be 1.5. We conclude that the

NO bond in NO2should be shorter than that in NO3

.

9.109 The Lewis structures of the four must be compared. Carbon monoxide has a triple bond, :OC:; the formate

ion (see text) has an average CO bond order of 1.5; CO2has double bonds; and CO32has a bond order of

1.33 (from review problem 9.106)

The order of increasing CO bond length is: CO < CO2< HCO2< CO3

2

9.110

O C O O C O

These are not preferred structures, because in each Lewis diagram, one oxygen bears a formal charge of +1

whereas the other bears a formal charge of1. The structure with the formal charges of zero has a lower

potential energy and is more stable.

9.111

O Cl

O

O

1-

0 0

0

O Cl

O

O

1-

0 0

O 0

0

O Cl

O

O O Cl

O

O O Cl

O

O

_ _ _

O Cl

O

O

O

O

Cl O

O

O OCl

O

O

O

O

ClO

O

O

_ _ _ _

The average bond order in ClO3

is 1.67 and the average bond order in ClO4

is 1.75. Since ClO4

has thelarger bond order, it will have the shorter bond length.


39/541

Chapter 9

190

Additional Exercises

9.112

Ca(s) + C l2(g)

Ca(s) + 2Cl(g)

Ca(g) + 2Cl(g)

Ca2+(g) + 2Cl(g)

Ca2+(g) + 2 C l-(g)

CaCl2(s)

+242.6 kJ

+192 kJ

+1146 kJ

-348 kJ x 2

-795 kJ

-2269

Ca+(g) + 2Cl(g)

+589.5 kJ

9.113

Electron affinity for Br =331.4 kJ/mol


40/541

Chapter 9

191

9.114

Cl Sn Cl

Cl

Cl

9.115 (a) PCl3 (b) GeF4 (c) SiCl4 (d) SrO

9.116 Eighteen electrons: 3s2, 3p6, 3d10

9.117 (a) Carbon nearly always makes four bonds and this structure has only 3. Additionally, the formal

charge on the carbon atom is 1. Also, there are too many electrons. The structure should have 24

but there are 26 shown.

(b) The formal charge on the oxygen atom bonded to the hydrogen is +1 which is unlikely since it is

the most electronegative element in the molecule.

(c) Carbon never makes more than four bonds.

9.118 The two structures are not resonance structures because the atoms do not have the same connectivity. The

first structure with the phosphorous in the middle is a resonance structure of POCl 3. The other structure is

not because the atoms are in a different order, and have a different connectivity. The latter structure is not

reasonable one because it has five bonds to oxygen, so the first structure is the more likely structure for

POCl3.

9.119 Given: Better (lower energy) structure:

H N N N

0

1+

1+

2-

H N N N

0 1+

-10

9.120

O

As

O

O O H

H

H

0

00

0

00

1+1-

O

As

O

O O H

H

H

9.121 Of the four resonance structures, the one in the upper left is the best possible structure. Each of the

terminal nitrogen atoms has a1 formal charge and the central atom has a formal charge of +1. The

structure in the upper right is also a good structure except that the formal charge on the rightmost nitrogen

atom is2 and the central nitrogen is +1 while the leftmost nitrogen has a formal charge of zero. The

bottom two resonance structures are both poor because each violates the octet rule. The left bottom

structure has too many electrons around the central nitrogen atom while the right bottom structure has too

few electrons on the central atom.


41/541

Chapter 9

192

9.122 The average bond orders are: SO3(1.00), SO2(1.50), SO32(1.33), and SO4

2(1.50). Therefore the SO

bond lengths should vary in the following way: SO3> SO32> SO2SO4

2.

9.123

Cl S S Cl

9.124 Because this is an acid it is probable that the hydrogen atom is bonded to the most electronegative atom.

For this molecule, that is O. So, the structures HOCN and HONC are the most likely.

9.125 = q r

8.59 D303.34 10 C m

1 D

= q 217 pm12

1 m

10 pm

q = 1.32 1019C

The charge of an electron is 1.60 1019C.

The amount of charge is19

19

1.32 10 C

1.60 10 C

Therefore, the amount of charge on the K is +0.826 eand the amount of charge on the F is0.826 e.

9.126 The formula for the hydrocarbon is C3H8and the proper name for this compound is propane.

Multi-Concept Problems

9.127

9.128


42/541

Chapter 9

193

9.129

Moles of acid

3 5

1 mol12.5 g x C H COOH x 0.145 mol

86.09 g

Moles of Base

4 9 2

1 mol17.4 g C H NH x 0.239 mol

73.14 g

Since the reaction involves one mole of acid reacting with one mole of base, the base is in excess.

Grams of unreacted base = (0.239 mol0.145 mol) x 73.14 g mol-1= 6.78 g

9.130 Let q be the amount of energy released in the formation of 1 mol of H2molecules from H atoms: 435

kJ/mol, the single bond energy for hydrogen.

q = specific heat mass T

mass = q (specific heat T)

g H2O =

3

Jg C

(435 10 J)

4.184 100 C 25 C

= 1.4 103g

9.131 E = energy required to atomize H2(bond energy) + IEH+ EAH

E = 435 kJ/mol + 1312 kJ/mol + (73 kJ/mol)

E = 1674 kJ/mol23

1000 J 1 mol

1 kJ 6.022 10 molecule


E = h E =hc

2.78 1018J/molecule =34 8(6.626 10 Js)(3.00 10 m/s)

7.15 108m = 71.5 nm

(These are very high energy photons, in the X-ray region of the electromagnetic spectrum.)


43/541

Chapter 9

194

9.132 First we need to determine the empirical formula of the compound using the combustion analysis data.

Since the data is in mg it is just as convenient to work in mmoles, 1/1000 of a mole, rather than convert

masses to grams. Molar masses remains numerically the same, that is mg mmol -1rather than g mol-1.

mmol C = 22

2

1 mmol CO 1 mmol C37.54 mg CO x x 0.853 mmol

44.01 mg mmol CO

mmol H = 22

2

1 mmol H O 2 mmol H7.684 mg H O x x 0.853 mmol

18.01 mg mmol H O

To determine the moles of oxygen on the compound we need to subtract the mass of C and H from the

sample mass of 38.40 mg.

mg of C =12.01 mg C

0.812 mmol x = 9.75 mgmmol

mg of H =1.008 mg H

0.853 mmol H x = 0.860 mgmmol

mg of O = 38.40 mg(9.75 mg C + 0.860 mg H) = 27.82 mg

mmol of O =1 mmol O

27.79 mg O x = 1.74 mmol16.00 g

empirical formula: C0.853H0.853O1.74

Divide by the smallest mmol value to obtain whole numbers.

CHO2.14 This is most likely CHO2taking into consideration experimental error in the combustion data.

The molar mass is 90 and the empirical mass is 45 so there are two empirical units per molecule.

The molecular formula is C2H2O4

The titration data can be used to determine how many acid protons are in the compound.

Moles of base used =0.0625 mmol

14.28 mL x = 0.8925 mmolmL

Mole of acid =1 mmol

40.2 mg x = 0.447 mmol90 mg

number of acid protonsmmoles of acid x = mmol of base

mmol of acid

number of acid protons 0.8925 mmol= = 2

mmol of acid 0.447 mmol

Thus, we know that both protons in the compound are acid protons.


44/541

Chapter 9

195

The Lewis structure for the compound, oxalic acid, is shown below.


45/541

Chapter 10

196

Practice Exercises

10.1 SeF6should have an octahedral shape (Figure 10.4) because it has six electron pairs around the central

atom.

10.2 SbCl5should have a trigonal bipyramidal shape (Figure 10.4) because, like PCl

5, it has five electron pairs

around the central atom.

10.3 HArF should have a linear shape (Figure 10.7) because although it has five electron pairs around the central

Ar atom, only two are being used for bonding.

10.4 IBr2should have a linear shape (Figure 10.7) because although it has five electron pairs around the central

I atom, only two are being used for bonding.

10.5 RnF4should have a square planar shape (Figure 10.10) because although it has six electron pairs around the

central Xe atom, only four are being used for bonding.

10.6 In SO32, there are three bond pairs and one lone pair of electrons at the sulfur atom, and as shown in Figure

10.5, this ion has a trigonal pyramidal shape.

In PbCl4, there are four bonding domains (4 single bonds) around the lead atom, and as shown in

Figure10.4, this molecule has a tetrahedral shape.

In XeO4, there are four bond pairs of electrons around the Xe atom, and as shown in Figure 10.5, this

molecule is tetrahedral.

In OF2, there are two bond pairs and two lone pairs of electrons around the oxygen atom, and as shown in

Figure 10.5, this molecule is bent.

10.7 SF4is distorted tetrahedral and has one lone pair of electrons on the sulfur, therefore it is polar.

10.8 (a) TeF6is octahedral, and it is not polar.

(b) SeO2is bent, and it is polar.(c) BrCl is polar because there is a difference in electronegativity between Br and Cl.

(d) AsH3, like NH3, is pyramidal, and it is polar.

(e) CF2Cl2is polar, because there is a difference in electronegativity between F and Cl.

10.9 The HCl bond is formed by the overlap of the halffilled 1satomic orbital of a H atom with the halffilled

3pvalence orbital of a Cl atom:

Cl atom in HCl (x= H electron):

x

3s 3p

The overlap that gives rise to the HCl bond is that of a 1sorbital of H with a 3porbital of Cl:

10.10 The halffilled 1satomic orbital of each H atom overlaps with a halffilled 3patomic orbital of the P atom,

to give three PH bonds. This should give a bond angle of 90.

P atom in PH3(x= H electron):


46/541

Chapter 10

197

x

3s 3p

xx

The orbital overlap that forms the PH bond combines a 1sorbital of hydrogen with a 3porbital of

phosphorus (note: only half of each p orbital is shown):

z

y

x

10.11 BF3usessp2hybridized orbitals since it has only three bonding electron pairs and no lone pairs of

electrons.

Thesp2hybrid orbitals on the B, x= F electron

x 2psp2

xx

10.12 BeF2usessphybridized orbitals since it has only two bonding electron pairs and no lone pairs of electrons.

Thesphybrid orbitals on the Be; x= F electron

2psp

xx

10.13 sp3

sp3

xx xx

10.14 Since there are five bonding pairs of electrons on the central phosphorous atom, we choosesp3d

hybridization for the P atom. Each of phosphorouss fivesp3dhybrid orbitals overlaps with a 3patomic

orbital of a chlorine atom to form a total of five PCl single bonds. Four of the 3datomic orbitals of P

remain unhybridized.

10.15 VSEPR theory predicts that AsCl5will be trigonal bipyramidal. Since there are five bonding pairs of

electrons on the central arsenic atom, we choosesp3dhybridization for the As atom as a trigonal bipyramid.

Each of arsenic's fivesp3dhybrid orbitals overlaps with a 3patomic orbital of a chlorine atom to form a

total of five AsCl single bonds. Four of the 4datomic orbitals of As remain unhybridized.

10.16 sp3d2

10.17 (a) PCl3has four pair of electrons around the central atom, P, so the hybridization is sp3

(b) BrCl3has five pair of electrons around the central atom, Br, so the hybridization is sp3d


47/541

Chapter 10

198

10.18 NH3issp3hybridized. Three of the electron pairs are use for bonding with the three hydrogens. The fourth

pair of electrons is a lone pair of electrons. This pair of electrons is used for the formation of the bond

between the nitrogen of NH3and the hydrogen ion, H+.

10.19 Since there are six bonding pairs of electrons on the central phosphorous atom, we choosesp3d2

hybridization for the P atom. Each of phosphoruss sixsp3d2hybrid orbitals overlaps with a 3patomic

orbital of a chlorine atom to form a total of six PCl single bonds. Three of the 3datomic orbitals of P

remain unhybridized.

P atom in PCl6(x= Cl electron):

xx xx x 3d

sp3d2

x

The ion is octahedral because six atoms and no lone pairs surround the central atom.

10.20 Atom 1has three electron domains:sp2 Atom 2 has four electron domains:sp3

Atom 3 has three electron domains:sp2

There are 10 bonds and 2 bonds in the molecule.

10.21 Atom 1has two electron domains:sp Atom 2 has three electron domains:sp2

Atom 3 has four electron domains:sp3

There are 9 bonds and 3 bonds in the molecule.

10.22 CN

has 10 valence electons and the MO diagram is similar to that of C and N. The bond order of the ion is3 and this does agree with the Lewis structure.

2s

2s

2pz

2px

2px

2pz

2py

2spy

10.23 NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 10.1 for O2, except that

one fewer electron is employed at the highest energy level


48/541

Chapter 10

199

2s

2s

2pz

2px

2px

2pz

2py

2spy

8 bonding e 3 antibonding e 5Bond Order = =

2 2

The bond order is calculated to be 5/2.

Review Questions

10.1 (a) See Figure 10.4. The angles are 120.

(b) See Figures 10.4 and 10.5 The bond angles are 109.5.

(c) See Figures 10.4 and 10.10. The bond angles are 90.

10.2 (a) See Figure 10.4. The bond angle is 180.

(b) See Figures 10.4 and 10.7. The bond angles are 120 between the equatorial bonds and 180

between the two axial bonds.

10.3 VSEPR Theory is based on the principle that adjacent electron pairs repel one another, and that these

destabilizing repulsions are reduced to a minimum when electron pairs stay as far apart as possible.

10.4 An electron domain is a region in space where electrons can be found.

10.5 In HCHO, there are three bonding domains around the C; one bonding domain around each H, and onebonding domain around O and two nonbonding domains around O.

10.6 (a) 90bond angles

(b) 120 bond angle between equatorial bonds and 180 bond angle between axial bonds

(c) 90 bond angles between equatorial bonds and 90 bond angles between equatorial bonds and

axial bond

10.7 (a) Planar triangular, otherwise known as trigonal planar

(b) Octahedral

(c) Tetrahedral

(d) Trigonal bipyramidal


49/541

Chapter 10

200

10.8 Polar molecules attract one another, and that influences the physical and chemical properties of substances.

10.9 A bond's dipole moment is depicted with an arrow having a + sign on one end, where the "barb" of the

arrow is taken to represent the location of the opposing negative charge of the dipole:

10.10 A molecule will be polar if the bonds are polar and these polar bonds are arranged in such a geometry thatthey do not cancel one another.

10.11 A molecule having polar bonds will be nonpolar only if the bond dipoles are arranged so as to cancel one

another's effect.

10.12

S

O O

The individual bond dipoles do not cancel one another.

10.13 Both VB and MO theory have wave mechanics as their theoretical basis. In each theory, bonds areconsidered to arise from the overlap of orbitals. Valence bond theory uses the overlap of atomic orbitals

while molecular orbital theory uses the molecular orbitals to describe the bonding.

10.14 Lewis structures do not explain how atoms share electrons, nor do they explain why molecules adopt

particular shapes. Also, electrons are known to be delocalized, a fact which only MO theory addresses

from the beginning. Also, odd-electron systems can be more effectively discussed with MO theory.

10.15 VB theory views atoms coming together with their orbitals already containing specific electrons. The

bonds that are formed according to VB theory do so by the overlap of orbitals on neighboring atoms, and

this is accompanied by the pairing (sharing) of the electrons that are contained in the orbitals.

MO theory, on the other hand, considers a molecule to be a collection of positive nuclei surrounded by a set

of molecular orbitals, which, by definition, belong to the molecule as a whole, rather than to any specificatom. The electrons of the molecule are distributed among the molecular orbitals according to the same

rules that govern the filling of atomic orbitals.

10.16 Orbital overlap occurs when orbitals from different atoms share the same space. This overlap provides a

more stable region for the electrons, which find themselves under the stabilizing influence of the positive

charge of two nuclei.

10.17 The greater the orbital overlap the greater the bond energy.

10.18 This is the same as the HF molecule, shown in Figure 10.16. A halffilled valenceporbital of Br overlaps

with the halffilled 1sorbital of the hydrogen atom.

10.19 Hybrid orbitals provide better overlap than do atomic orbitals, and this results in stronger bonds.

10.20 These are shown in Figures 10.21.


50/541

Chapter 10

201

10.21 sp3d sp3d2

10.22 Elements in period 2 do not have a dsubshell in the valence level.

10.23 Lewis structures are something like shorthand representations of VB descriptions of molecules.

10.24 The VSEPR model gives the shape of the electron domains around the central atom; this information is

used to predict the hybridization of the atom.

10.25 90

10.26 sp3C atom

sp3N atom

sp3O atom

There are zero, one and two lone pairs, respectively, on these atoms whensp3hybridization is utilized to

form bonds.

10.27 (a)

H

H H

H

c

(b)

H H

H

N

(c)

H

H

O

10.28 This angle would have had to be 90, the angle between one atomicporbital and another.


51/541

Chapter 10

202

10.29 See Figure 10.25. See Figure 10.33

Valence shell for boron Valence shell for carbon

2s 2p 2s 2p

sp2hybridized boron atom sp2hybridized carbon atom

sp2 2p sp2 2p

10.30 sp3dtrigonal bipyramid sp3d2octahedral

10.31 The ammonium ion has a tetrahedral geometry, with bond angles of 109.5.

H

NH

H

H

+

10.32 The geometry of the boron BCl3is trigonal planar; after the addition of the water molecule, it becomes

tetrahedral. The hybridization changes fromsp2tosp3. The geometry of the oxygen in the water molecule

changes from bent to trigonal pyramidal after coordination to the boron atom. Its hybridization does not

change, but remains atsp3.

10.33 bondThe electron density is concentrated along an imaginary straight line joining the nuclei of the

bonded atoms.

bondThe electron density lies above and below an imaginary straight line joining the bonded nuclei.

10.34 The characteristic sidetoside overlap ofpatomic orbitals that characterizes bonds is destroyed upon

rotation about the bond axis. This is not the case for a bond, because regardless of rotation, a bond is

still effective at overlap.

10.35 See Figure 10.33.

10.36 See Figure 10.35.

10.37 Two resonance hybrids should be drawn.

sp2

hybrid orbitals are used for each carbon atom. Consequently, each bond angle is 120. See Figure 9.42.

10.38 If an electron is forced to occupy the higher energy MO, the molecule loses stability and the bond is made

weaker than if an electron, or a pair of electrons, occupies the lower energy (bonding) MO.


52/541

Chapter 10

203

10.39 The overlap will create a bonding MO.

10.40 There are two regions of overall between the two 3dorbitals so this forms a orbital.

10.41 In the hypothetical molecule He2, both the bonding and the antibonding MO are doubly occupied, and the

net bond order is zero, as shown in Figure 10.39. H2has a bond order of 1

10.42 As shown in Table 9.1, the highest energy electrons in dioxygen occupy the doubly degenerate

antibonding level:

x y

* *

2p 2p

Since their spins are unpaired, the molecule is paramagnetic.

10.43 As shown in Table 10.1, the bond order of Li2is 1.0. The bond order of Be2would be zero. Yes, Be2+

could exist since the bond order would be 1/2.

10.44 (a) 2.5 (b) 1.5 (c) 1.5

10.45 As bond order increases, bond energy (and strength) increases.

10.46 See Figure 10.40.

10.47 A delocalized MO is one that extends over more than two nuclei. Benzene is planar and each carbon atomhas a half-filled p orbital perpendicular to the plane of the molecule. These orbitals overlap to form a pi

bonding system throughout the molecule.

10.48 Delocalization increases stability.

10.49 Delocalization energy is a term used in MO theory to mean essentially the same thing as the term resonance

energy, which derives from VB theory. They both represent the additional stability associated with a

spreading out of electron density.

10.50 For conductors, metals, the valence band and the conduction band are the same so electrons can easily

move throughout the metal.

For insulators, non-metals, the valence band is filled and the gap between the valence band and conductionband is very large so electrons cannot move into the conduction band.

For semiconductors, the valence band is filled but the separation between the valence band and the

conduction band is small so electrons can easily be promoted into the conduction.

10.51 The band containing the valence electrons of the element is called the valence band.

The conduction band is the empty or partial filled band existing throughout the element.


53/541

Chapter 10

204

10.52 Increasing the temperature of a semiconductor excites more electrons from the valence band into the

conduction band.

10.53 The 2sband is completely filled and much lower in energy than the conduction band. It is not possible to

excite electrons from the 2sband into the conduction band. Calciums valence band is formed from the 4s

electrons and therefore, it is that band that conducts electrons.

10.54 p- type transistors are those formed by doping silicon with elements containing three valence electrons, the

3A family of elements. The doping creates a site with a positive hole. It is the movement of the hole

throughout the semiconductor that results in current flow.

An n-type transistor if formed by doping silicon with elements containing five valence electrons, the 5A

family of elements. Doping results in an extra electron at the doping site. This electron can move into the

conduction band and cause current flow.

10.55 Any element in the 3A family such as boron or gallium would form a p-type germanium transistor.

Any element in the 5A family such as phosphorus or arsenic would form an n-type germanium transistor.

10.56 A solar cell operates on the principle that light falling on the cell disturbs the equilibrium of the p-n

junction causing a flow of current.

LEDs operate on the opposite principle of a solar cell. That is, a current flows through the p-n junction

causing an emission of light.

10.57 Of the nonmetals, only the noble gases exist in nature as isolated atoms.

10.58 Period 2 elements form strong bonds because the atom size is small which enables atoms to approach

each other very closely. Period 3 elements (and all other elements) are too large to accommodate the close

proximity required to form strong bonds. Because the period 3 and higher elements are large and form

only weak bonds, it is much more effective to form only bonds than to form a bond and a bond.

10.59 Because the halogens need only a single electron to complete their valence shell, these elements do not

require the formation of bonds. The formation of a single bond creates an extremely stable diatomicmolecule.

10.60 An allotrope is a different structure or physical form of an element. An isotope is a form of an atom that

differs from other atoms of the same element in the number of neutrons in the nucleus. An example is H

and D, two of the isotopes of hydrogen.

10.61 The allotropes of oxygen are dioxygen, O2, and ozone, O3.


54/541

Chapter 10

205

10.62 The two unpaired electrons are located in the antibonding molecular orbitals.

2s

2s*

2py, 2pz

2px

2py*, 2pz*

2px*

The overall bond order is (8Be4ABe)/2 = 2.

10.63

O O O

The molecule is nonlinear and usessp2hybrid orbitals. There is a lone pair of electrons on the central

oxygen, consequently, ozone is indeed a polar molecule. Additionally, the formal charges suggest that O3is

polar.

10.64 The presence of ozone in the upper atmosphere shields the earth from harmful ultraviolet

radiation.

10.65 Diamond is a network covalent solid in which each carbon atom is the corner of a tetrahedron. Each carbon

atom is bonded to four other carbon atoms usingsp3hybrid orbitals.

10.66 Graphene is a monolayer of graphite and is formed by the overlap of neighboring 2porbitals to form a -

type bonding throughout the layer.

Carbon forms sheets of sp2sp2sigma () bonds.

10.67 The weak bonds that exist between the sheets ofsp2bonded carbon are easily broken. The lubricating

properties of graphite are a result of the weak attraction between parallel sheets.

10.68 C60

, or buckminsterfullerene, resembles a soccer ball. It consists of hexagonal and pentagonal arrangements

of carbon atoms. This arrangement naturally assumes a spherical geometry. The molecule is named in

honor of R. Buckminster Fuller, an architect. Other roughly spherical arrangements of carbon atoms are

also possible. While not perfectly spherical, they resemble buckminsterfullerene and are given the general

name of fullerenes.

10.69 Carbon nanotubes have the same molecular structure as graphite, but instead of forming sheets, the layer of

carbon curves and forms a tube.

10.70 S8


55/541

Chapter 10

206

10.71 White phosphorous, P4, is a tetrahedron. See figure 10.53

.

10.72 The bond angle in P4is 60. Using only p orbitals, a bond angle of 90is expected. Because the observed

bond angle is much less, the molecule is extremely reactive.

10.73 Red phosphorus consists of tetrahedra joined as in Figure 10.54. While it too is used in explosives, it is

much less reactive than white phosphorus.

10.74 Black phosphorus, like graphite, exists as parallel sheets of atoms.

10.75 Silicon has a molecular structure identical to diamond, i.e., a network covalent solid arranged in a

tetrahedral manner. Silicon, unlike carbon, does not form multiple bonds. Consequently, there is no

graphite-like allotrope of silicon.

Review Problems

10.76 (a) Bent (central N atom has two single bonds and two lone pairs)

(b) Planar triangular (central C atom has three bonding domainsa double bond and two single

bonds)

(c) T-shaped (central I atom has three single bonds and two lone pairs)

(d) Linear (central Br atom has two single bonds and three lone pairs)

(e) Planar triangular (central Ga atom has three single bonds and no lone pairs)

10.77 (a) trigonal pyramidal (b) tetrahedral

(c) tetrahedral (d) nonlinear (bent)

(e) linear

10.78 (a) Bent (central F atom has four electron domains; two are lone pair)

(b) Trigonal bipyramidal (central As atom has five electron domains)

(c) Trigonal pyramidal (central As atom has four electron domains; one is a lone pair)

(d) Trigonal pyramidal (central Sb atom has four electron domains; one is a lone lair)

(e) Bent (central Se atom has four electron domains; two are lone pair)

10.79 (a) distorted tetrahedral (b) octahedral

(c) nonlinear (d) tetrahedral

(e) tetrahedral

10.80 (a) Tetrahedral (central I atom has four electron domains)

(b) Square planar (central I atom has six electron domains; two are lone pair)

(c) Octahedral (central Te atom has six electron domains)

(d) Tetrahedral (central Si atom has four electron domains)

(e) Linear (central I atom has five electron domains; three are lone pair)

10.81 (a) linear (b) square planar

(c) Tshaped (d) trigonal pyramidal

(e) planar triangular


56/541

Chapter 10

207

10.82 BrF4+ The central Br atom has five electron domains, one is a lone pair

10.83 PF3

10.84 180 There are two electron domains around each carbon atom so the molecule is linear

10.85 all angles 120

10.86 (a) 109.5 (b) 109.5

(c) 120 (d) 180

(e) 109.5

10.87 (a) 109.5 (b) 109.5

(c) 180 (d) 120

Documents

MSE160 Custom Textbook Solutions