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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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