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 Eh and  c . For example, in
the first entry frequency is given, so:
 
c


2.998×108 m  s-1
= 3.4×10-7 m = 340 nm ;
14 -1
8.7×10 s
and E  h   6.626×10-34 J  s  8.7×10-14 s-1  = 5.8×10-19 J
Frequency
Wavelength
Energy of photon
(2 s.f.)
(2 s.f.)
(2 s.f.)
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
1m
2  10-25 J
Microwave popcorn
1.2  1017 Hz
2.5 nm
7.9  10-17 J
Dental X-ray
SM-50
Event
1.7
Wien’s law states that T max  constant  2.88  103 K  m.
If T /K  1540C  273C  1813 K, then max 
2.88  103 K  m
1813 K
max  1.59  106 m, or 1590 nm
1.9
(a) From c  n and E  hn, we can write
E  hc 1
 (6.626 08  1034 J  s) (2.997 92  108 ) (589  10 9 m) 1
 3.37  1019 J
 5.00  103 g Na 
23
1
E 
 (6.022  10 atoms  mol )
1
 22.99 g  mol Na 
(b)
 (3.37  1019 J  atom 1 )
 44.1 J
(c) E  (6.022  1023 atoms  mol1 )(3.37  1019 J  atom1 )
 2.03  105 J or 203 kJ
1.11
The energy is first converted from eV to joules:
E  (140.511  103 eV) (1.6022  1019 J  eV1 )  2.2513  1014 J
From E  hv and c  v we can write
hc
E
(6.626 08  1034 J  s) (2.997 92  108 m  s 1 )

2.2513  1014 J

 8.8237  1012 m or 8.8237 pm
1.13
(a) false. The total intensity is proportional to T 4 . (Stefan-Boltzmann
Law) (b) true;
(c) false. Photons of radio-frequency radiation are
lower in energy than photons of ultraviolet radiation.
SM-51
1.15
(a) Use the de Broglie relationship,   hp 1  h(mv)1 .
me  (9.109 39  1028 g) (1 kg/1000 g)  9.109 39  1031 kg
(3.6  103 km  s 1 ) (1000 m  km1 )  3.6  106 m  s 1
  h(mv)1

6.626 08  1034 J  s
(9.109 39  1031 kg) (3.6  106 m  s 1 )
 2.0  1010 m
(b) E  h
 (6.626 08  1034 J  s) (2.50  1016 s 1 )
 1.66  1017 J
(c) The photon needs to contain enough energy to eject the electron from
the surface as well as to cause it to move at 3.6  103 km  s 1 . The energy
involved is the kinetic energy of the electron, which equals
1
2
mv 2 .
1
mv 2
2
1
J  (9.109 39  1031 kg) (3.6  106 m  s 1 ) 2
2
J  5.9  1018 J
Ephoton  1.66  1017 J 
 1.66  1017
 1.66  1017
 2.25  1017 J
But we are asked for the wavelength of the photon, which we can get from
E  hv and c  v or E  hc 1 .
2.25 1017 kg  m 2  s 2  (6.62608 10 34 kg  m 2  s 1 )
 (2.99792 108 m  s 1 ) 1
  8.8 109 m
 8.8 nm
(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.
SM-52
(5.15 ounce) (28.3 g  ounce 1 ) (1 kg/1000 g)  0.146 kg
 92 mi   1 h   1 km   1000 m 
1


  41 m  s


 h   3600 s   0.6214 mi   1 km 
Use the de Broglie relationship.
  hp 1  h(mv) 1
 h(mv) 1

6.626 08  1034 J  s
(0.146 kg) (0.041 km  s 1 )
6.626 08  1034 kg  m 2  s 1

(0.146 kg) (41 m  s 1 )
 1.1  1034 m
1.19
From the de Broglie relationship, p  h 1 or h  mv , we can calculate
the velocity of the neutron:
v
h
m
(6.626 08  1034 kg  m 2  s 1 )

(remember that 1 J  1 kg  m 2  s 2 )
27
12
(1.674 93  10 kg) (100  10 m)
 3.96  103 m  s 1
1.21
Yes there are degenerate levels. The first three cases of degenerate levels
are:
n1 = 1, n2 = 2 is degenerate with n1 = 2,n2 = 1
n1  1, n2  3 is degenerate with n1  3, n2  1
n1  2, n2  3 is degenerate with n1  3, n2  2
1.23
 n2  n2  n2 
h2  x
y
z

Given the expression for the cubic box is E


xyz
2
8m 
L


where Exyz represents the energy of a given level having n values
corresponding to x, y and z we can describe each new level by
SM-53
incrementing each n by one; any levels with the same energy will be
degenerate. The three lowest energy levels will therefore be E111 , E211 ,
and E221 . From this we can arrive at expressions for the energy of each


h2 12 12 12
3h2

 
level: E111 
. Similarly, the other two
8 m 
 8 mL2
L2
energy levels are determined to be E211 
3 h2
9 h2
and E221 
.
2
2
4 mL
8 mL
The 211 and the 221 levels are degenerate; that is E211 E121 E112 and
E
1.25
221
 E
212
 E
122
(a) Refer to the plot below for parts (a) thru (d); nodes are where the
wavefunction is zero:
1.42
n=2
n=3
(x)/m -1/2
0.00
-1.42
0.00
0.333
0.500
0.667
1.00
x (m)
(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 n  1
Refer to the plot below for parts (e) and (f):
SM-54
n=2
n=3
(x)2/m-1/2
0.00
0.17 0.25
0.50
0.75 0.83
1.00
x (m)
(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:

L
2
0
2 L2 
n x 
sin

 dx

L 0
L 
2
2 
2  1
n x
n x x  2 
 
 cos
 sin
  
L  2n
L
L
20 


given n is an integer:
L

2  L / 2   1

0 
L  2   2
(b) Integrate over the “left third of the box” or from 0 to 1/3 L:

L
2
0
2 L
n x 
   2  sin
 dx
0
L 
L 
2
2
2  1
n x
n x
 
 cos
 sin

L  2n
L
L

1
n
n 1

cos
 sin

L  n 
3
3 3
x 3 
 
20 

L
SM-55
if n is a multiple of 3, the first term in this sum is zero and the
probability of finding an electron in the left third of the box is
1/ 3. Also, as n becomes large the probability of finding the
electron in the left third of the box approaches 1/3.
1.29
(a) The Rydberg equation gives  when   3.29  1015 s1 , from which
one can calculate  from the relationship c  .
 1
1 
   2  2 
 n1 n2 
and c    2.99792  108 m  s 1
 1
1 
c   2  2 
 n1 n2 
1 1 
2.99792  108 m  s 1  (3.29  1015 s 1 )  
 4 16 
  4.86  10-7 m  486 nm
(b) Balmer series
(c) visible, blue
1.31
For hydrogen-like one-electron ions, we use the Z-dependent Rydberg
relation with the relationship c   to determine the transition
wavelength. For He+, Z = 2.
 1
1
1
1

 2 2 3.29  1015 s1  2  2   9.87  1015 Hz
2
2
1
2 
 n1 n2 
  Z 2 

1.33
c


 

2.99792  108 m  s1
 3.04  10-8 m  30.4 nm
15 1
9.87  10 s
(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
SM-56
(n  1)2 h2 n 2 h 2 (2n  1)h 2
E  En 1  En 


8mL2
8mL2
8mL2
5h2
For n  2 and n  1  3, E 
8mL2
Then 3,2
hc 8mhcL2 8mcL2



E
5h
5h2
For an electron in a 150-pm box, the expression becomes
3,2 
8(9.109 39  1031 kg) (2.997 92  108 m  s 1 ) (150  1012 m) 2
5(6.626 08  1034 J  s)
 1.5  108 m
 1.48  10 8 m
(b) We need to remember that the equation for E 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:
(n  2) 2 h 2 n 2 h 2 (n 2  4n  4  n 2 )h 2 (4n  4)h 2
E  En  2  En 



8mL2
8mL2
8mL2
8mL2
hc
hc(8mL2 )
8mcL2

 2
 2
E h [4n  4] h [4n  4]
For n  2, the expression becomes
λ

8mcL2
8mcL2

h[( 4  2)  4]
12h
8(9.10939  10 31 kg) (2.99792  10 8 m  s 1 ) (150  10 12 m) 2
12(6.62608  10 34 kg  m 2  s 1 )
 6.18  10 9 m
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 n  1; for the Balmer series, n  2; for
Paschen series, n  3; and for the Brackett series, n  4.
SM-57
1.37
(a) 1s
2p
3d
(b) A node is a region in space where the wavefunction  passes through
0.
(c) The simplest s -orbital has 0 nodes, the simplest p -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 px orbital will have its lobes oriented along the x axis, the p y orbital
will have its lobes oriented along the y axis, and the pz orbital will have
its lobes oriented along the z axis.
1.41
The equation derived in Illustration 1.4 can be used:
e
2(0.55 a0 ) / a0
 2 (r  0.55a0 , ,  )
 a03

 2 (0, ,  )
 1 

3 
  a0 
1.43
 0.33
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  or  .
px  R(r )C sin  cos 
p y  R (r )C sin  sin 
pz  R(r )C cos 
1
 3 2
where C  

 4 
Squaring the three wavefunctions and summing them:
SM-58
R(r ) 2 C 2 sin 2  cos 2   R(r ) 2 C 2 sin 2  sin 2   R(r ) 2 C 2 cos 2 
 R(r ) 2 C 2  sin 2  cos 2   sin 2  sin 2   cos 2  

 R(r ) 2 C 2 sin 2   cos 2   sin 2    cos 2 

Using the identity cos2 x  sin 2 x  1 this becomes
R(r ) 2 C 2  sin 2   cos 2    R(r ) 2 C 2
With one electron in each p orbital, the electron distribution is
not a fuction of  or  and is, therefore, spherically symmetric.
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 :
 2r 
4 ao
P  2  r 2 exp   dr
ao 0
 ao 
This integral is easier to evaluate if we allow the following
change of variables:
z
2r
ao
a 
 z  2 when r  ao , z  0 when r  0, and dr   o  dz
 2
2
1 2
1
P   z 2 exp( z )dz   ( z 2  2 z  2) exp( z )
0
2
2
0
1
   (4  4  2) exp(2)   2 
2
 0.323 or 32.3%
(b) Following the answer developed in (a) changing the integration limits
to 0 to 2 ao:
a
z  4 when r  2ao , z  0 when r  0, and dr   o
 2
4
1 4
1
P   z 2 exp( z )dz   ( z 2  2 z  2) exp( z )
0
2
2
0
1
   26 exp(4)   4 
2
 0.761 or 76.1%
SM-59

 dz

1.47
(a) 1 orbital;
(b) 5 orbitals;
(c) 3 orbitals;
1.49
(a) 7 values: 0, 1, 2, 3, 4, 5, 6;
(d) 7 orbitals
(b) 5 values; 2,  1, 0, 1, 2;
(c) 3
values: 1, 0, 1; (d) 4 subshells: 4s, 4 p, 4d , and 4 f
1.51
(a) n  6; l  1;
1.53
(a)  1, 0,  1;
(b) n  3; l  2;
(c) n  2; l  1;
(b)  2,  1, 0,  1,  2;
(d) n  5; l  3
(c)  1, 0,  1;
(d)  3,2,  1, 0,  1,  2,  3.
1.55
(a) 6 electrons;
(b) 10 electrons;
1.57
(a) 5d , five;
1.59
(a) six;
1.61
(a) cannot exist;
1.63
(a) The total Coulomb potential energy V ( r ) is the sum of the individual
(b) 1s, one;
(b) two;
(c) 6 f , seven;
(c) eight;
(b) exists;
(c) 2 electrons;
(d) 14 electrons
(d) 2 p, three
(d) two
(c) cannot exist; (d) exists
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
(e)(3e) 3e 2

4 0 r
4 0 r
where e is the charge on the electron and 3e is the charge on the
nucleus,  0 is the vacuum permittivity, and r is the distance from the
electron to the nucleus. The total attractive potential will thus be
SM-60
 3e2   3e2   3e2   3e2   1 1 1 




   
 4 0 r1   4 0 r2   4 0 r3   4 0   r1 r2 r3 
The repulsive terms will have the form
(e)(e)
e2

4 0 rab
4 0 rab
where rab represents the distance between two electrons a and b. The total
repulsive term will thus be
e2
e2
e2
e2



4 0 r12 4 0 r13 4 0 r23 4 0
 1
1 1 
   
 r12 r13 r23 
This gives
 3e2  1 1 1 
e2  1
1
1 
V (r )  



 


 
 4 0  r2 r2 r3  4 0  r12 r13 r23 
(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. Zeff 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 l increases.
1.67
(d) true.
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 l  0 here, so m1 must also equal 0.
(c) This configuration is
not possible because the maximum value l can have is
n  1; n  4, so lmax  3.
SM-61
1.71
(a) silver
[Kr]4d 10 5s1
(b) beryllium
[He]2s 2
(c) antimony
[Kr]4d 10 5s 2 5 p3
(d) gallium
[Ar]3d 10 4s 2 4 p1
(e) tungsten
[Xe]4f 14 5d 4 6s 2
(f) iodine
[Kr]4d 10 5s 2 5 p5
1.73
(a) tellurium;
1.75
(a) 4 p;
1.77
(a) 5;
(b) 11;
1.79
(a) 3;
(b) 2;
(b) vanadium;
(b) 4 s;
(c) 6 s;
(c) 5;
(c) 3;
(c) carbon;
(d) thorium
(d) 6s
(d) 20
(d) 2
1.81
Unpaired
Element
Electron Configuration
electrons
Ga
[Ar] 3d 10 4s 2 4 p1
1
Ge
[Ar] 3d 10 4s 2 4 p 2
2
As
[Ar] 3d 10 4s 2 4 p 3
3
Se
[Ar] 3d 10 4s 2 4 p 4
2
Br
[Ar] 3d 10 4s 2 4 p5
1
(c) (n  1)d 5 ns 2 ;
(d) (n  1)d 10 ns1
1.83
(a) ns1 ;
1.85
(a) oxygen (1310 kJ  mol1 ) > selenium (941 kJ  mol1 ) > tellurium
(b) ns 2 np1 ;
(870 kJ  mol1 ); ionization energies generally decrease as one goes down
SM-62
a group.
(b) gold (890 kJ  mol1 ) > osmium (840 kJ  mol1 ) >
tantalum (761 kJ  mol1 ); ionization energies generally decrease as one
goes from right to left in the periodic table.
(c) lead (716 kJ  mol1 ) >
barium (502 kJ  mol1 ) > cesium (376 kJ  mol1 ) ; 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 electronelectron 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
P3  S2  Cl
1.93
(a) fluorine;
1.95
(a) A diagonal relationship is a similarity in chemical properties between
(b) carbon;
(c) chlorine;
(d) lithium.
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
SM-63
cause the ion to be smaller. (c) For example, Al3 and Ge 4  compounds
show the diagonal relationship, as do Li  and Mg 2 .
1.97
Only (b) Li and Mg exhibit a diagonal relationship.
1.99
The ionization energies of the s -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)

c
 3600 cm 1
  c(3600 cm 1 )
  (2.997 92  108 m  s 1 ) (3600 cm 1 )
  (2.997 92  1010 cm  s 1 ) (3600 cm 1 )
  1.1  1014 s 1
(b) From E  h : E  (6.626 08  1034 J  s)(1.079  1014 s 1 )
 7.2  1020 J.
(c) 1.00 mol of molecules  6.022  1023 molecules, so the energy
absorbed by 1.00 mol will be
(7.151  1020 J  molecule1 )(6.022  1023 molecules  mol1 )
 4.3  104 J  mol1 or 43 kJ  mol1 .
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.
SM-64
1.107 This trend is attributed to the inert-pair effect, which states that the s 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,
Etotal , must be equal to the energy to eject the electron, Eejection , plus the
energy that ends up as kinetic energy, Ekinetic, in the movement of the
electron:
Etotal  Eejection  Ekinetic
 1
But Etotal for the photon  h and Ekinetic    mv 2 where m is the mass
 2
of the object and v is its velocity. Eejection corresponds to the ionization
energy, I , so we arrive at the final relationship desired.
(b)
Etotal  h  hc 1
 (6.62608  10 34 J  s) (2.99792  10 8 m  s 1 ) (58.4  10 9 m)1
 3.401  10 18 J
 Eejection  Ekinetic
1
  1
Ekinetic   mv 2     (9.10939  10 28 g) (2450 km  s1 )2
2
  2
 1
   (9.10939  10 31 kg) (2.450  10 6 m  s1 )2
 2
 2.734  10 18 kg  m 2  s2  2.734  10 18 J
3.401  10 18 J  E ejection  2.734  10 18 J
E ejection  6.67  10 19 J
SM-65
1.111 By the time we get to the lanthanides and actinides— the two series of forbital 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
[Xe] 6s 2 5d 1
Lu
[Xe] 6s 2 4f 14 5d 1
Ac
[Rn] 7s 2 6d 1
Lr
[Rn] 7 s 2 5 f 14 6d 1
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), J . Chem. Ed . 59, 634.
1.113 (a)—(c) We can use the hydrogen 2s wavefunction found in Table 1.2.
Remember that the probability of locating an electron at a small region in
space is proportional to  2 , not  .
1
 2s
 2s
2
1  1 2 
r
 
2
3  
4  2 a0  
a0
 1 
r

2
3 

a0
 32 a0  
  2 a0
e

r
2
  ar
 e

0
2
 1 
r  a
2




 e
3
a0 
 2 s 2 (r , ,  )  32 a0  

0
 2 s 2 (0, ,  )
 1  2 a
2 e

3 
 32 a0 
r
0
0
2

r  a
2   e
a0 

4
r
0
Because r will be equal to some fraction x times a0 , the expression will
simplify further:
SM-66
2
xa0

xa0   a
2


 e
a0 
 2 s 2 ( r , ,  ) 
(2  x) 2 e  x


4
4
 2 s 2 (0, ,  )
0
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
SM-67
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
(2  x) 2 e x
function f ( x) 
from 0 to 3.
4
0.80
0.70
relative probability
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x
The node occurs when x = 2, or when r  2a0 . 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
1
 2 2
 n x 
    sin 

 L
 L 
and the probability function will be given by
2
 n x 
 2    sin 2 

 L
 L 
SM-68
Because x can range from 0 to L (the length of the box), we can write the
integration as
x

0
2
x 2 
 n x 
dx     sin 2 
 dx
0  L 
 L 
for the entire box, we write
probability of finding the particle somewhere in the box =

L
0
 2  2  n x 
  sin 
 dx
 L
 L 
2 x
 n x 
probability     sin 2 
 dx
0
 L
 L 

L
2
0
2 L
n x 
   2  sin
 dx
L 0
L 
2
2
2  1
n x
n x
 
 cos
 sin

L  2n
L
L

L
x 
 
2  0 
2  1
L 
 cos n  sin n    0 


L  2n
2 
if n is an integer, sin n will always be zero and
probability 
2 L
1
L  2 
1.117 (a) 4.8 x 10-10 esu; (b) 14 electrons
1.119 Given that the energy needed to break a C – C bond is 248 kJmol-1,
348 kJ
1 mol C–C bonds
1000 J


= 5.78  10 -19 J
23
mol C–C bonds
6.022  10 C–C bonds
1 kJ
Substituting this value into   hc E , 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.
SM-69
h
2.18  10 18 J
 
1.121 Given that E  
and
n
n2
n2
E
 E
final
 E
initial
:
(a) for an electron to fall from the 4d to 1s level, the energy of the photon
 1
1 
is E  E1 E4 = 2.18  10 18 J 
= - 2.04  10-18 J (the


2
2
1
4 
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 E 
hc


6.626  1034 J  sec2.998  108 m
1.064  10 6 m
 1.867  10 19 J ;
the energy of the ejected electron is 0.137 eV  (1.602  10-19 JeV-1) =
2.195  10-20 J. The difference between these two values is 1.65  10-19 J
SM-70
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 [Ne]3s 2 3p5 ; it has
one unpaired electron. The electron configuration of a chloride ion
is [Ne]3s2 3p6 ; this configuration is identical to neutral argon.
(b) Assuming a one quantum level jump, an excited chlorine should have
an electron configuration of [Ne]3s2 3p 4 4s1 .
(c) The energy of a given level n in a non-hydrogen atom can be estimated


2
Z eff
2.18  10 18 J
2
Z eff
h
 
by E  
. For chlorine, Zeff is
n
n2
n2
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: E  E  E =
4
3

 1
1
 62 2.18  1018 J 
   3.82  1018 J . This energy
 4 2 32 
corresponds to a wavelength of 52.0 nm (the X-ray region).
(d) This amount of energy corresponds to 2.30×103 kJmol-1 or 23.8 eV
per chlorine atom
(e) If the proportion of
37
Cl in a sample of chlorine atoms is reduced to
37.89% (half of its typical value), the proportion of
35
Cl will be increased
to 62.11%. Based on this, the average mass of a chlorine atom will be:
 37.89% 
 62.11% 
Clave mass  
6.139  10 23 g  
5.807  10 23 g

 100 
 100 
Clave mass  5.933  10 23 g/atom  35.73 g/mol
SM-71
(f) thru (h): refer to the table below:
Compound
Chlorine
Name
oxidation number
ClO2
+4
Chlorine dioxide
NaClO
+1
Sodium hypochlorite
KClO3
+5
Potassium chlorate
NaClO4
+7
Sodium perchlorate
SM-72