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

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CHAPTER 2ATOMIC STRUCTURE AND INTERATOMIC BONDINGPROBLEM SOLUTIONSFundamental ConceptsElectrons in Atoms2.1Cite the difference between atomic mass and atomic weight.SolutionAtomic mass is the mass of anindividual atom, whereas atomic weight is the average (weighted) of theatomic masses of an atom's naturally occurring isotopes.

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2.2Chromium has four naturally-occurring isotopes: 4.34% of50Cr, with an atomic weight of 49.9460amu, 83.79% of52Cr, with an atomic weight of 51.9405 amu, 9.50% of53Cr, with an atomic weight of 52.9407 amu,and 2.37% of54Cr, with an atomic weight of 53.9389 amu.On the basis of these data, confirm that the averageatomic weight of Cr is 51.9963 amu.SolutionTheaverage atomic weight of chromiumis computed by adding fraction-of-occurrence/atomicweight products for the three isotopes. Thus

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2.3Hafnium has six naturally occurring isotopes: 0.16% of174Hf, with an atomic weight of 173.940 amu;5.26% of176Hf, with an atomic weight of 175.941 amu; 18.60% of177Hf, with an atomic weight of 176.943 amu;27.28% of178Hf, with an atomic weight of 177.944 amu; 13.62% of179Hf, with an atomic weight of 178.946 amu;.and 35.08% of180Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf.SolutionTheaverage atomic weight of halfniumis computed by adding fraction-of-occurrence—atomicweight products for the sixisotopes—i.e., using Equation 2.2.(Remember: fraction of occurrence is equal to thepercent of occurrence divided by 100.) ThusIncluding data provided in the problem statement we solve foras= 178.485amu

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2.4Bromium has two naturally occurring isotopes:79Br, with an atomic weight of 78.918 amu, and81Br,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.SolutionThe average atomic weight of indiumis computed by adding fraction-of-occurrence—atomic weightproducts for the two isotopes—i.e., using Equation 2.2, orBecause there are just two isotopes, the sum of thefracture-of-occurrences will be 1.000; orwhich means thatSubstituting into this expression the one noted above for, and incorporating the atomic weight valuesprovidedin the problem statement yieldsSolving this expression foryields. Furthermore, because

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then

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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 thisbasis, how many atoms are therein a pound-mole of a substance?Solution(a)In order to determine the number of grams in one amu of material, appropriate manipulation of theamu/atom, g/mol, and atom/mol relationships is all that is necessary, as= 1.6610-24g/amu(b) Since there are 453.6 g/lbm,= 2.731026atoms/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) thatelectrons 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 electronposition is described in terms of a probability distribution, and (2) electron energy is quantized into both shells andsubshells--each electron is characterized by four quantum numbers.

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2.7Relative to electrons and electron states, what does each of the four quantum numbers specify?SolutionThenquantum number designates the electron shell.Thelquantum number designates the electron subshell.Themlquantum number designates the number of electron states in each electron subshell.Themsquantum number designates the spin moment on each electron.

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2.8Allowed values for the quantum numbers of electrons are as follows:n = 1, 2, 3, . . .l = 0, 1,2, 3, . . . , n–1ml= 0, ±1, ±2, ±3, . . . , ±lThe relationships between n and the shell designations are noted in Table 2.1. Relative to the subshells,l = 0 corresponds to an s subshelll = 1 corresponds to a p subshelll = 2 corresponds to a d subshelll = 3 corresponds to an f subshellFor the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of nlmlms, are100(12) and 100(-12). Write the four quantum numbers for all of the electrons in the L and M shells, and notewhich correspond to the s, p, and d subshells.SolutionFor theLstate,n= 2, and eight electron states are possible. Possiblelvalues are 0 and 1, while possiblemlvalues are 0 and ±1;and possiblemsvalues areTherefore, for thesstates, the quantum numbers areand. For thepstates, the quantum numbers are,,,,, and.For theMstate,n= 3, and 18 states are possible. Possiblelvalues are 0, 1, and 2;possiblemlvalues are 0,±1, and ±2;and possiblemsvalues areTherefore, for thesstates, the quantum numbers are,, for thepstates they are,,,,, and; for thedstates they are,,,,,,,,,and.

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2.9Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2-.SolutionThe electronconfigurations for the ions are determinedusing Table 2.2 (and Figure 2.8).Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1s22s22p63s23p63d64s2. In order tobecome an ion with a plus two charge, it must lose two electrons—in this case the two 4s.Thus, the electronconfiguration for an Fe2+ion is 1s22s22p63s23p63d6.Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1s22s22p63s23p1. In order tobecome an ion with a plus three charge, it mustlose three electrons—in this case two 3sand the one 3p. Thus, theelectron configuration for an Al3+ion is 1s22s22p6.Cu+: From Table 2.2, the electron configuration for an atom of copper is 1s22s22p63s23p63d104s1. In orderto become an ion with a plus one charge, it must lose one electron—in this case the 4s.Thus, the electronconfiguration 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 theelectron configuration for one of its atoms is 1s22s22p63s23p63d104s24p64d105s25p66s2. In order to become an ionwith a plus two charge, it must lose two electrons—in this case two the 6s.Thus, the electron configuration for aBa2+ion is 1s22s22p63s23p63d104s24p64d105s25p6.Br-: From Table 2.2, the electron configuration for an atom of bromine is 1s22s22p63s23p63d104s24p5. Inorder to become an ion with a minus one charge, it must acquire one electron—in this case another 4p.Thus, theelectronconfiguration 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 anion with a minus two charge, it must acquire two electrons—in this case another two 2p.Thus, the electronconfiguration for an O2-ion is 1s22s22p6.

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

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The Periodic Table2.11With regard to electron configuration, what do all the elements in Group VIIA of the periodic tablehave in common?SolutionEach of the elements in Group VIIA has fivepelectrons.

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2.12To what group in theperiodic table would an element with atomic number 119belong?SolutionFrom the periodic table (Figure 2.8) theelement having atomic number 119 would belong to group IA.According to Figure 2.8,Uuo, having an atomic number of118 belongs to Group 0 (or 18) of the periodic table.The next column to the right is actually the left-most column, Group IA (or 1).

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2.13Without consulting Figure 2.8or Table 2.2, determine whether each of the electron configurationsgiven below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify yourchoices.(a)1s22s22p63s23p63d74s2(b)1s22s22p63s23p6(c)1s22s22p5(d)1s22s22p63s2(e)1s22s22p63s23p63d24s2(f)1s22s22p63s23p64s1Solution(a) The 1s22s22p63s23p63d74s2electron configuration is that of a transition metal because of an incompletedsubshell.(b) The 1s22s22p63s23p6electron configuration is that of an inert gas because of filled 3sand 3psubshells.(c)The 1s22s22p5electron configuration is that of a halogen because it is one electron deficient fromhaving a filledLshell.(d) The 1s22s22p63s2electron configuration is that of an alkaline earth metal because of twoselectrons.(e) The 1s22s22p63s23p63d24s2electron configuration is that of a transition metal because of an incompletedsubshell.(f) The 1s22s22p63s23p64s1electron configuration is that of an alkali metal because of a singleselectron.

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2.14(a)What electron subshell is being filled for the rare earth series of elements on the periodic table?(b)What electron subshell is being filled for the actinide series?Solution(a) The 4fsubshell is being filled for the rare earth series of elements.(b) The 5fsubshell is being filled for the actinide series of elements.

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