Solution Manual for Materials Science and Engineering: An Introduction, 10th Edition

Name: Solution Manual for Materials Science and Engineering: An Introduction, 10th Edition
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Solution Manual for Materials Science and Engineering: An Introduction, 10th Edition ensures effective exam preparation with structured textbook notes.

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CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2.1 Cite the difference between atomic mass and atomic weight.
Solution
Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the
atomic masses of an atom's naturally occurring isotopes.

2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460
amu, 83.79% of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an atomic weight of 52.9407 amu,
and 2.37% of 54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average
atomic weight of Cr is 51.9963 amu.
Solution
The average atomic weight of chromium is computed by adding fraction-of-occurrence/atomic
weight products for the three isotopes. Thus

2.3 Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu;
5.26% of 176Hf, with an atomic weight of 175.941 amu; 18.60% of 177Hf, with an atomic weight of 176.943 amu;
27.28% of 178Hf, with an atomic weight of 177.944 amu; 13.62% of 179Hf, with an atomic weight of 178.946 amu;.
and 35.08% of 180Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf.
Solution
The average atomic weight of halfnium is computed by adding fraction-of-occurrence—atomic
weight products for the six isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the
percent of occurrence divided by 100.) Thus
Including data provided in the problem statement we solve for as
= 178.485 amu

2.4 Bromium has two naturally occurring isotopes: 79Br, with an atomic weight of 78.918 amu, and 81Br,
with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-
occurrences of these two isotopes.
Solution
The average atomic weight of indium is computed by adding fraction-of-occurrence—atomic weight
products for the two isotopes—i.e., using Equation 2.2, or
Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or
which means that
Substituting into this expression the one noted above for , and incorporating the atomic weight values provided
in the problem statement yields
Solving this expression for yields . Furthermore, because

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2.5 (a) How many grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there
in a pound-mole of a substance?
Solution
(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the
amu/atom, g/mol, and atom/mol relationships is all that is necessary, as
= 1.66  10-24 g/amu
(b) Since there are 453.6 g/lbm,
= 2.73  1026 atoms/lb-mol

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2.6 (a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.
(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.
Solution
(a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that
electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells.
(b) Two important refinements resulting from the wave-mechanical atomic model are (1) that electron
position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and
subshells--each electron is characterized by four quantum numbers.

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2.8 Allowed values for the quantum numbers of electrons are as follows:
n = 1, 2, 3, . . .
l = 0, 1, 2, 3, . . . , n –1
ml = 0, ±1, ±2, ±3, . . . , ±l
The relationships between n and the shell designations are noted in Table 2.1. Relative to the subshells,
l = 0 corresponds to an s subshell
l = 1 corresponds to a p subshell
l = 2 corresponds to a d subshell
l = 3 corresponds to an f subshell
For the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of nlmlms, are
100(1
2 ) and 100(- 1
2 ). Write the four quantum numbers for all of the electrons in the L and M shells, and note
which correspond to the s, p, and d subshells.
Solution
For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml
values are 0 and ±1; and possible ms values are Therefore, for the s states, the quantum numbers are
and . For the p states, the quantum numbers are , , , , , and
.
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0,
±1, and ±2; and possible ms values are Therefore, for the s states, the quantum numbers are ,
, for the p states they are , , , , , and ; for the d
states they are , , , , , , , , ,
and .

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2.9 Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2-.
Solution
The electron configurations for the ions are determined using Table 2.2 (and Figure 2.8).
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 2.8), 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.

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2.10 Sodium chloride (NaCl) exhibits predominantly ionic bonding. The Na+ and Cl- ions have electron
structures that are identical to which two inert gases?
Solution
The Na+ ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the
same as neon (Figure 2.8).
The Cl- ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron
configuration the same as argon.

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2.13 Without consulting Figure 2.8 or Table 2.2, determine whether each of the electron configurations
given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your
choices.
(a) 1s22s22p63s23p63d74s2
(b) 1s22s22p63s23p6
(c) 1s22s22p5
(d) 1s22s22p63s2
(e) 1s22s22p63s23p63d24s2
(f) 1s22s22p63s23p64s1
Solution
(a) The 1s22s22p63s23p63d74s2 electron configuration is that of a transition metal because of an incomplete
d subshell.
(b) The 1s22s22p63s23p6 electron configuration is that of an inert gas because of filled 3s and 3p subshells.
(c) The 1s22s22p5 electron configuration is that of a halogen because it is one electron deficient from
having a filled L shell.
(d) The 1s22s22p63s2 electron configuration is that of an alkaline earth metal because of two s electrons.
(e) The 1s22s22p63s23p63d24s2 electron configuration is that of a transition metal because of an incomplete
d subshell.
(f) The 1s22s22p63s23p64s1 electron configuration is that of an alkali metal because of a single s electron.

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