E344 2013 Summer
Solution Set 1
Solution
WAu  75%  0.5 g  0.375 g
WCu  25%  0.5 g  0.125 g
N Au 
W Au
0.375 g
 NA 
 6.02  10 23 mole 1  1.15  10 21
M Au
196.97 g / mole
N Cu 
WCu
0.125 g
 NA 
 6.02  10 23 mole 1  1.18  10 21
M Cu
63.55 g / mole
2. 2.7 Give the electron configurations for the following ions: Fe 2+, Al3+, Cu+, Ba2+, Br-, and O2-.
Solution
The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6).
Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1s22s22p63s23p63d64s2. In order to
become an ion with a plus two charge, it must lose two electrons—in this case the two 4s. Thus, the electron
configuration for an Fe2+ ion is 1s22s22p63s23p63d6.
Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1s22s22p63s23p1. In order to
become an ion with a plus three charge, it must lose three electrons—in this case two 3s and the one 3p. Thus, the
electron configuration for an Al3+ ion is 1s22s22p6.
Cu+: From Table 2.2, the electron configuration for an atom of copper is 1s22s22p63s23p63d104s1. In order
to become an ion with a plus one charge, it must lose one electron—in this case the 4s. Thus, the electron
configuration for a Cu+ ion is 1s22s22p63s23p63d10.
Ba2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the
electron configuration for one of its atoms is 1s22s22p63s23p63d104s24p64d105s25p66s2. In order to become an ion
with a plus two charge, it must lose two electrons—in this case two the 6s. Thus, the electron configuration for a
Ba2+ ion is 1s22s22p63s23p63d104s24p64d105s25p6.
Br-: From Table 2.2, the electron configuration for an atom of bromine is 1s22s22p63s23p63d104s24p5. In
order to become an ion with a minus one charge, it must acquire one electron—in this case another 4p. Thus, the
electron configuration for a Br- ion is 1s22s22p63s23p63d104s24p6.
O2-: From Table 2.2, the electron configuration for an atom of oxygen is 1s22s22p4. In order to become an
ion with a minus two charge, it must acquire two electrons—in this case another two 2p. Thus, the electron
configuration for an O2- ion is 1s22s22p6.
Solution
A. Sc 1s2 2s2p6 3s2p6 3d14s2
Zn 1s2 2s2p6 3s2p6 3d104s2
B. When n=3 and l=2 (3d),
m can be 0, ±1 and ±2 (5 different values); si can be ±1/2 (2 different values). Therefore, there are 5x2=10 atoms in
the first row of transition metals.
4. 2.14 The net potential energy between two adjacent ions, E N, may be represented by the sum of Equations 2.8 and
2.9; that is,
EN = 
A
B
 n
r
r
Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure:
1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of

EN versus r is a minimum at E0.
2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing.
3. Determine the expression for E0 by substitution of r0 into Equation 2.11.
Solution
(a) Differentiation of Equation 2.11 yields
dEN
dr
=

 A 
 B 
d 
d n 
 r 
r 
=

dr
dr
A
r (1 + 1)

nB
r (n + 1)
= 0
(b) Now, solving for r (= r0)

A
nB
= (n + 1)
2
r0
r0
or

 A 1/(1 - n)
r0 =  
nB 
(c) Substitution for r0 into Equation 2.11 and solving for E (= E0)

E0 = 


= 
A
1/(1 - n)
 A 
 
nB 
A
B
+ n
r0
r0
+
B
n/(1 - n)
 A 
 
nB 
5. 2.15 For a K+–Cl– ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance
between the ions r, according to
EA  

ER 
1.436
r
5.8  106
r9
For these expressions, energies are expressed in electron volts per K+–Cl– pair, and r is the distance in nanometers.

The net energy EN is just the sum of the two expressions above.
(a) Superimpose on a single plot EN, ER, and EA versus r up to 1.0 nm.
(b) On the basis of this plot, determine (i) the equilibrium spacing r0 between the K+ and Cl– ions, and (ii)
the magnitude of the bonding energy E0 between the two ions.
(c) Mathematically determine the r0 and E0 values using the solutions to Problem 2.14 and compare these
with the graphical results from part (b).
Solution
(a) Curves of EA, ER, and EN are shown on the plot below.
(b) From this plot
r0 = 0.28 nm
E0 = – 4.6 eV
(c) From Equation 2.11 for EN
A = 1.436
B = 5.86  10-6
n=9
Thus,
 A 1/(1 - n)
r0 =  
nB 

1/(1 - 9)
1.436

 0.279 nm


(8)(5.86  10-6) 
and

E0 = 
= 


A
 A 
 
nB 
1.436

1/(1  9)
1.436



(9)(5.86  106 ) 

B
+
1/(1 - n)
+
n/(1 - n)
 A 
 
nB 
5.86  106

9 /(1  9)
1.436



(9)(5.86  106 ) 

= – 4.57 eV
Solution
(A)
(B)

6
3
 (5 )  6  7  3  4   0
dr
r
r 

dU ( r )
6
6
 3
r7
r  3 2
3
r4
For Du, as σ=0.2750 nm,
Therefore, the interatomic spacing between two Du atoms at 0 K is:
r  3 2 * 0.2750  0.3465 nm.
(C) Importantly, the U(r) curve does not change with temperature. It describes the relationship between
the interatomic distance and the net potential energy. The curve at r0 is asymmetric and “skewed to the
right”.
Assume 3 temperatures T1 T2 and T3, (T1 > T2 >T3).
The corresponding potential energies at T1 T2 and T3 should be UT1>UT2>UT3.
The average interatomic positions are marked as the filled circles in the middle of the two positions
on the U(r) curve. As shown in the figure, from T1 to T3, the average position shifts from the right to
the left. This shift is due to the asymmetry at r0. If it is symmetric at r0, the “filled dots” are going to
have a same r = r0 value.
U(r)
r
UT1
UT2
ro
UT3
7. (after Callister 2.2SS) Generate a spreadsheet that computes the percent ionic character of a bond between atoms
of two elements, once the user has input values for the elements’ electronegativities.
A. Evaluate the percent ionicity in the bonds between Carbon (C) and C, H, O, N, and F (i.e. C-C, C-H, C-O, C-N,
and C-F).
B. Define the term "dipole," and explain which of the C-X bonds you would expect to have the largest dipole
strength?
Solution
A.
XA
2.5
2.5
2.5
2.5
2.5
XB
2.5
2.1
3.5
3
4
% ionic character
0
3.921056085
22.11992169
6.058693719
43.02171753
C-C
C-H
C-O
C-N
C-F
B.
An electric dipole exists whenever there is some separation of positive and negative portions of an atom of
molecule. C-F bond has the largest dipole strength.
8. Use the electronegativity data in Fig. 2.7 of Callister to develop an argument why the boiling point of hydrogen
sulfide (H2S) of -60 oC is so much less than that of water (H2O).
Solution
XA
2.1
2.1
XB
2.5
3.5
% ionic character
3.921056085
38.73736058
H-S
H-O
H-S dipoles are much weaker than H-O dipoles. In a H2O molecule, negative charges are highly attracted tothe
oxygen atom. The positive charge is mostly located on the hydrogen atom. The strong dipole-dipole interaction
holds the H2O, resulting in a higher boiling point.
Solution
10. 18.7 How does the electron structure of an isolated atom differ from that of a solid material?
Solution
For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each
state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron
band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of
which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each
band will equal the total number of corresponding states contributed by all of the atoms in the solid.
Solution
(C)
6.0eV 
1240eV  nm
max
 max  207nm
12. C
13. D