CHAPTER 1

ATOMS: THE QUANTUM WORLD

1.1 (a) Radiation may pass through a metal foil. (b) All light

(electromagnetic radiation) travels at the same speed; the slower speed

supports the particle model. (c) This observation supports the radiation

model. (d) This observation supports the particle model;

electromagnetic radiation has no mass and no charge.

1.3 microwaves < visible light < ultraviolet light < x-rays < -rays

1.5 All of these can be determined using and . For example, in

the first entry frequency is given, so:

;

and

Frequency

(2 s.f.)

Wavelength

(2 s.f.)

Energy of photon

(2 s.f.)

Event

8.7 1014 Hz 340 nm 5.8 10-19 J Suntan

5.0 1014 Hz 600 nm 3.3 10-19 J Reading

300 MHz 1 m 2 10-25 J Microwave popcorn

1.2 1017 Hz 2.5 nm 7.9 10-17 J Dental X-ray

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1.7

1.9 (a)  

(b) 

(c) 

1.11 The energy is first converted from eV to joules:

From and we can write

1.13 (a) false. The total intensity is proportional to (Stefan-Boltzmann

Law) (b) true; (c) false. Photons of radio-frequency radiation are

lower in energy than photons of ultraviolet radiation.

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1.15 (a) Use the de Broglie relationship,

(b) 

(c) The photon needs to contain enough energy to eject the electron from

the surface as well as to cause it to move at The energy

involved is the kinetic energy of the electron, which equals

But we are asked for the wavelength of the photon, which we can get from

and or

(d) 8.6 nm is in the x-ray/gamma ray region.

1.17 To answer this question, we need to convert the quantities to a consistent

set of units, in this case, SI units.

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Use the de Broglie relationship.

1.19 From the de Broglie relationship, or we can calculate

the velocity of the neutron:

1.21 Yes there are degenerate levels. The first three cases of degenerate levels

are:

1.23 Given the expression for the cubic box is

where represents the energy of a given level having n values

corresponding to x, y and z we can describe each new level by

incrementing each n by one; any levels with the same energy will be

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degenerate. The three lowest energy levels will therefore be , ,

and . From this we can arrive at expressions for the energy of each

level: . Similarly, the other two

energy levels are determined to be and .

The 211 and the 221 levels are degenerate; that is and

1.25 (a) Refer to the plot below for parts (a) thru (d); nodes are where the

wavefunction is zero:

0.00 0.333 0.500 0.667 1.00

1.42

-1.42

0.00

(x)/m-1/2

x (m)

n = 2n = 3

(b) for n = 2 there is one node at x = 0.500 m.

(c) for n = 3 there are two nodes, one at x = 0.333 and 0.667 m .

(d) the number of nodes is equal to

Refer to the plot below for parts (e) and (f):

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x (m)

(x)2/m-1/2

n = 2n = 3

0.00 0.17 0.50 0.83 1.000.25 0.75

(e) for n = 2 a particle is most likely to be found at x = 0.25 m and x = 0.75

m.

(f) for n = 3 a particle is most likely to be found at x = 0.17, 0.50 and 0.83

m.

1.27 (a) Integrate over the “left half of the box” or from 0 to ½ L:

(b) Integrate over the “left third of the box” or from 0 to 1/3 L:

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1.29 (a) The Rydberg equation gives when from which

one can calculate from the relationship

(b) Balmer series

(c) visible, blue

1.31 For hydrogen-like one-electron ions, we use the Z-dependent Rydberg

relation with the relationship to determine the transition

wavelength. For He+, Z = 2.

1.33 (a) This problem is the same as that solved in Example 1.5, but the

electron is moving between different energy levels. For movement

between energy levels separated by a difference of 1 in principal quantum

number, the expression is

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For

Then

For an electron in a 150-pm box, the expression becomes

(b) We need to remember that the equation for was originally

determined for energy separations between successive energy levels, so

the expression needs to be altered to make it general for energy levels two

units apart:

For the expression becomes

1.35 In each of these series, the principal quantum number for the lower energy

level involved is the same for each absorption line. Thus, for the Lyman

series, the lower energy level is for the Balmer series, for

Paschen series, and for the Brackett series,

1.37 (a) 1s 2p 3d

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(b) A node is a region in space where the wavefunction passes through

0. (c) The simplest -orbital has 0 nodes, the simplest -orbital has 1

nodal plane, and the simplest d-orbital has 2 nodal planes. (d) Given the

increase in number of nodes, an f-orbital would be expected to have 3

nodal planes.

1.39 The orbital will have its lobes oriented along the x axis, the orbital

will have its lobes oriented along the y axis, and the orbital will have

its lobes oriented along the z axis.

1.41 The equation derived in Illustration 1.4 can be used:

1.43 To show that three p orbitals taken together are spherically symmetric,

sum the three probability distributions (the wavefunctions squared) and

show that the magnitude of the sum is not a function of

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1.45 (a) The probability (P) of finding an electron within a sphere of radius ao

may be determined by integrating the appropriate wavefunction squared

from 0 to ao :

(b) Following the answer developed in (a) changing the integration limits

to 0 to 2 ao:

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1.47 (a) 1 orbital; (b) 5 orbitals; (c) 3 orbitals; (d) 7 orbitals

1.49 (a) 7 values: 0, 1, 2, 3, 4, 5, 6; (b) 5 values;  (c) 3

values: (d) 4 subshells: and

1.51 (a)   (b)   (c)   (d) 

1.53 (a)   (b)   (c)   

(d) 

1.55 (a) 6 electrons;  (b) 10 electrons; (c) 2 electrons; (d) 14 electrons

1.57 (a)  five; (b)  one; (c)  seven; (d)  three

1.59 (a) six; (b) two; (c)  eight; (d) two

1.61 (a) cannot exist; (b) exists; (c) cannot exist; (d) exists

1.63 (a) The total Coulomb potential energy is the sum of the individual

coulombic attractions and repulsions. There will be one attraction between

the nucleus and each electron plus a repulsive term to represent the

interaction between each pair of electrons. For lithium, there are three

protons in the nucleus and three electrons. Each attractive Coulomb

potential will be equal to

where is the charge on the electron and is the charge on the

nucleus, is the vacuum permittivity, and is the distance from the

electron to the nucleus. The total attractive potential will thus be

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The repulsive terms will have the form

where represents the distance between two electrons a and b. The total

repulsive term will thus be

This gives

(b) The first term represents the coulombic attractions between the

nucleus and each electron, and the second term represents the coulombic

repulsions between each pair of electrons.

1.65 (a) false. is considerably affected by the total number of electrons

present in the atom because the electrons in the lower energy orbitals will

“shield” the electrons in the higher energy orbitals from the nucleus. This

effect arises because the e-e repulsions tend to offset the attraction of the

electron to the nucleus. (b) true; (c)  false. The electrons are

increasingly less able to penetrate to the nucleus as increases. (d) true.

1.67 Only (d) is the configuration expected for a ground-state atom; the others

all represent excited-state configurations.

1.69 (a) This configuration is possible. (b) This configuration is not possible

because here, so must also equal 0. (c) This configuration is

not possible because the maximum value can have is

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1.71 (a) silver

(b) beryllium

(c) antimony

(d) gallium

(e) tungsten

(f) iodine

1.73 (a) tellurium; (b) vanadium; (c) carbon; (d) thorium

1.75 (a)   (b)   (c)   (d) 

1.77 (a) 5; (b) 11; (c) 5; (d) 20

1.79 (a) 3; (b) 2; (c) 3; (d) 2

1.81

Element Electron Configuration

Unpaired

electrons

Ga 1

Ge 2

As 3

Se 2

Br 1

1.83 (a)  ; (b)   (c)   (d) 

1.85 (a) oxygen (1310 kJ > selenium (941 kJ > tellurium

(870 kJ ionization energies generally decrease as one goes down

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a group. (b) gold (890 kJ > osmium (840 kJ >

tantalum (761 kJ ionization energies generally decrease as one

goes from right to left in the periodic table. (c) lead (716 kJ >

barium (502 kJ > cesium (376 kJ ; ionization energies

generally decrease as one goes from right to left in the periodic table.

1.87 The atomic radii (in pm) are

Sc 161 Fe 124

Ti 145 Co 125

V 132 Ni 125

Cr 125 Cu 128

Mn 137 Zn 133

The major trend is for decreasing radius as the nuclear charge increases,

with the exception that Cu and Zn begin to show the effects of electron-

electron repulsions and become larger as the d-subshell becomes filled.

Mn is also an exception as found for other properties; this may be

attributed to having the d-shell half-filled.

1.89 (a) Sb3+, Sb5+; (c) Tl+, Tl3+; (b) and (d) only form one positive ion each.

1.91

1.93 (a) fluorine; (b)  carbon; (c) chlorine; (d) lithium.

1.95 (a) A diagonal relationship is a similarity in chemical properties between

an element in the periodic table and one lying one period lower and one

group to the right. (b) It is caused by the similarity in size of the ions. The

lower-right element in the pair would generally be larger because it lies in

a higher period, but it also will have a higher oxidation state, which will

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cause the ion to be smaller. (c) For example, and compounds

show the diagonal relationship, as do

1.97 Only (b) Li and Mg exhibit a diagonal relationship.

1.99 The ionization energies of the -block metals are considerably lower, thus

making it easier for them to lose electrons in chemical reactions.

1.101 (a) metal; (b) nonmetal; (c) metal; (d) metalloid; (e) 

metalloid; (f) metal 

1.103 (a) 

(b) 

(c) 1.00 mol of molecules molecules, so the energy

absorbed by 1.00 mol will be

1.105 Cu = [Ar]3d104s1 and Cr = [Ar]3d54s1; In copper it is energetically

favorable for an electron to be promoted from the 4s orbital to a 3d orbital,

giving a completely filled 3d subshell. In the case of Cr, it is energetically

favorable for an electron to be promoted from the 4s orbital to a 3d orbital

to exactly ½ fill the 3d subshell.

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1.107 This trend is attributed to the inert-pair effect, which states that the -

electrons are less available for bonding in the heavier elements. Thus,

there is an increasing trend as we descend the periodic table for the

preferred oxidation number to be 2 units lower than the maximum one. As

one descends the periodic table, ionization energies tend to decrease. For

Tl, however, the values are slightly higher than those of its lighter

analogues.

1.109 (a) The relation is derived as follows: the energy of the photon entering,

must be equal to the energy to eject the electron, plus the

energy that ends up as kinetic energy, in the movement of the

electron:

But for the photon and where m is the mass

of the object and v is its velocity. corresponds to the ionization

energy, , so we arrive at the final relationship desired.

(b) 

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1.111 By the time we get to the lanthanides and actinides— the two series of f-

orbital filling elements—the energy levels become very close together and

minor changes in environment cause the different types of orbitals to

switch in energy-level ordering. For the elements mentioned, the

electronic configurations are

La Lu

Ac Lr

As can be seen, all these elements have one electron in a d-orbital, so

placement in the third column of the periodic table could be considered

appropriate for either, depending on what aspects of the chemistry of these

elements we are comparing. The choice is not without argument, and it is

discussed by W. B. Jensen (1982), 59, 634.

1.113 (a)—(c) We can use the hydrogen wavefunction found in Table 1.2.

Remember that the probability of locating an electron at a small region in

space is proportional to

Because will be equal to some fraction times the expression will

simplify further:

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Carrying out this calculation for the other points, we obtain:

x relative probability

0.1 0.82

0.2 0.66

0.3 0.54

0.4 0.43

0.5 0.34

0.6 0.27

0.7 0.21

0.8 0.16

0.9 0.12

1 0.092

1.1 0.067

1.2 0.048

1.3 0.033

1.4 0.022

1.5 0.014

1.6 0.0081

1.7 0.0041

1.8 0.0017

1.9 0.0004

2 0.0000

2.1 0.00031

2.2 0.0011

2.3 0.0023

2.4 0.0036

2.5 0.0051

2.6 0.0067

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2.7 0.0082

2.8 0.0097

2.9 0.011

3 0.012

This can be most easily carried out graphically by simply plotting the

function from 0 to 3.

3.02.52.01.51.00.50.00.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

x

rela

tive p

rob

ab

ilit

y

The node occurs when x = 2, or when This is exactly what is

obtained by setting the radial part of the equation equal to 0.

1.115 The approach to showing that this is true involves integrating the

probability function over all space. The probability function is given by

the square of the wave function, so that for the particle in the box we have

and the probability function will be given by

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Because can range from 0 to (the length of the box), we can write the

integration as

for the entire box, we write

probability of finding the particle somewhere in the box =

1.117 (a) 4.8 x 10-10 esu; (b) 14 electrons

1.119 Given that the energy needed to break a bond is 248 kJmol-1,

Substituting this value into , this energy corresponds to a

wavelength of 3.44 10-7 m or 344 nm. This is in the ultraviolet region

of the electromagnetic spectrum.

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1.121 Given that and

:

(a) for an electron to fall from the 4d to 1s level, the energy of the photon

is = = 2.04 10-18 J (the

negative sign means that energy is released or emitted);

(b) Similarly, an electron moving from the 4d to 2p level will emit a

photon of 4.09 10-19 J;

(c) same as (b); an electron moving from the 4d to 2s emits the same

amount of energy as it would if it were moving to the 2p orbital; this is

due to the fact that all orbitals having the same principal quantum number

n are degenerate (i.e. they have the same energy).

(d) In a hydrogen atom, no photon would be emitted on moving between

orbitals possessing the same n (due to degeneracy of the 4d and the 4s

orbitals in hydrogen).

(e) Since potassium has both more electrons and protons than hydrogen,

the individual orbitals within a given shell will have different energies

(arising from the attractions and repulsions of electrons with the nucleus

and other electrons in the atom). As a result we would expect to see

emission lines for all the transitions; thus potassium should show four

lines while hydrogen only shows two.

1.123 = 1064 nm = 1.064 10-6 m; The energy of a photon of this wavelength

is ;

the energy of the ejected electron is 0.137 eV (1.602 10-19 JeV-1) =

2.195 10-20 J. The difference between these two values is 1.65 10-19 J

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or 1.65 10-22 kJ per atom of thulium. Multiplication by Avogadro’s

number gives an electron affinity of 99.2 kJ per mol of thulium.

1.125 The radial distribution of a 3s orbital is given by curve (b), while curve (a)

shows the same for a 3p orbital; this can be determined by examining the

electron density near the origin (which is the nucleus); the plot with the

most electron density closest to (0, 0) arises from the s orbital.

1.127 (a) The electron configuration of atomic chlorine is ; it has

one unpaired electron. The electron configuration of a chloride ion is

; this configuration is identical to neutral argon.

(b) Assuming a one quantum level jump, an excited chlorine should have

an electron configuration of .

(c) The energy of a given level n in a non-hydrogen atom can be estimated

by . For chlorine, Zeff is

approximately equal to 6 (Fig. 1.45). For an electron to jump from the n =

3 to n = 4 quantum level the energy needed is: =

. This energy

corresponds to a wavelength of 52.0 nm (the X-ray region).

(d) This amount of energy corresponds to 2.30×103 kJmol-1 or 23.8 eV

per chlorine atom

(e) If the proportion of in a sample of chlorine atoms is reduced to

37.89% (half of its typical value), the proportion of will be increased

to 62.11%. Based on this, the average mass of a chlorine atom will be:

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(f) thru (h): refer to the table below:

Compound Chlorine

oxidation number

Name

ClO2 +4 Chlorine dioxide

NaClO +1 Sodium hypochlorite

KClO3 +5 Potassium chlorate

NaClO4 +7 Sodium perchlorate

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