FUNDAMENTALS OF MATERIALS SCIENCE AND ENGINEERING 4TH EDITION SOLUTIONS

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CHAPTER 2ATOMIC STRUCTURE AND INTERATOMIC BONDINGPROBLEM SOLUTIONSFundapsychological ConceptsElectrons in Atoms2.1 Cite the difference in between atomic mass and atomic weight.

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SolutionAtomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of theatomic masses of an atom"s normally arising isotopes.

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2.2 Chromium has actually 4 naturally-emerging isotopes: 4.34% of 50 Cr, via an atomic weight of 49.amu, 83.79% of 52 Cr, through an atomic weight of 51.9405 amu, 9.50% of 53 Cr, through an atomic weight of 52.amu, and also 2.37% of 54 Cr, through an atomic weight of 53.9389 amu. On the basis of these information, confirm that theaverage atomic weight of Cr is 51.9963 amu.

SolutionThe average atomic weight of silsymbol (A Cr) is computed by including fraction-of-occurrence/atomic

weight assets for the 3 isotopes. Thus

A Cr = f 50 CrA 50 Cr + f 52 CrA 52 Cr+ f 53 CrA 53 Cr+ f 54 CrA 54 Cr

= (0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 am

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2.4 (a) Cite two important quantum-mechanical principles associated with the Bohr design of the atom.(b) Cite two essential additional refinements that resulted from the wave-mechanical atomic version.

Solution(a) Two essential quantum-mechanical concepts connected through the Bohr model of the atom are (1) thatelectrons are particles moving in discrete orbitals, and (2) electron power is quantized right into shells.(b) Two essential refinements resulting from the wave-mechanical atomic model are (1) that electronposition is described in regards to a probcapability distribution, and (2) electron energy is quantized right into both shells andsubshells--each electron is characterized by four quantum numbers.

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2.5 Relative to electrons and electron claims, what does each of the four quantum numbers specify?

SolutionThe n quantum number designates the electron shell.The l quantum number designates the electron subshell.The ml quantum number designates the number of electron states in each electron subshell.The ms quantum number designates the spin minute on each electron.

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2.7 Give the electron configurations for the complying with ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2-. SolutionThe electron configurations for the ions are established making use of Table 2.2 (and also Figure 2.6).

Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1s 22 s 22 p 63 s 23 p 63 d 64 s 2. In orderto become an ion via a plus two charge, it must shed 2 electrons—in this instance the 2 4s. Thus, the electronconfiguration for an Fe2+ ion is 1s 22 s 22 p 63 s 23 p 63 d 6.Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1s 22 s 22 p 63 s 23 p 1. In orderto come to be an ion via a plus three charge, it should lose 3 electrons—in this case 2 3s and the one 3p. Hence,the electron configuration for an Al3+ ion is 1s 22 s 22 p 6.Cu+: From Table 2.2, the electron configuration for an atom of copper is 1s 22 s 22 p 63 s 23 p 63 d 104 s 1. Inorder to become an ion with a plus one charge, it should shed one electron—in this case the 4s. Thus, the electronconfiguration for a Cu+ ion is 1s 22 s 22 p 63 s 23 p 63 d 10.Ba2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a change facet theelectron configuration for among its atoms is 1s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 64 d 105 s 25 p 66 s 2. In order to end up being an ionthrough a plus two charge, it should lose 2 electrons—in this situation two the 6s. Therefore, the electron configuration for aBa2+ ion is 1s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 64 d 105 s 25 p 6.Br-: From Table 2.2, the electron configuration for an atom of bromine is 1s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 5. Inorder to become an ion through a minus one charge, it have to obtain one electron—in this case one more 4p. Thus, theelectron configuration for a Br- ion is 1s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 6.O2-: From Table 2.2, the electron configuration for an atom of oxygen is 1s 22 s 22 p 4. In order to becomean ion through a minus 2 charge, it need to gain 2 electrons—in this instance an additional two 2p. Hence, the electronconfiguration for an O2- ion is 1s 22 s 22 p 6.

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2.8 Sodium chloride (NaCl) exhibits primarily ionic bonding. The Na+ and also Cl- ions have electronframeworks that are the same to which two inert gases?

SolutionThe Na+ ion is just a sodium atom that has lost one electron; therefore, it has actually an electron configurationthe very same as neon (Figure 2.6).The Cl- ion is a chlorine atom that has gained one additional electron; therefore, it has actually an electronconfiguration the very same as argon.

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2.10 To what team in the regular table would an aspect via atomic number 114 belong? SolutionFrom the routine table (Figure 2.6) the element having atomic number 114 would belong to group IVA.According to Figure 2.6, Ds, having actually an atomic variety of 110 lies listed below Pt in the routine table and in the right-the majority of column of team VIII. Moving 4 columns to the appropriate puts element 114 under Pb and also in team IVA.

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2.11 Without consulting Figure 2.6 or Table 2.2, determine whether each of the electron configurationsoffered listed below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a shift metal. Justify yourchoices.(a) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2(b) 1s 2 2s 2 2p 6 3s 2 3p 6(c) 1s 2 2s 2 2p 5(d) 1s 2 2s 2 2p 6 3s 2(e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2(f) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1Solution(a) The 1s 22 s 22 p 63 s 23 p 63 d 74 s 2 electron configuration is that of a transition metal bereason of anincomplete d subshell.(b) The 1s 22 s 22 p 63 s 23 p 6 electron configuration is that of an inert gas because of filled 3s and 3p subshells.(c) The 1s 22 s 22 p 5 electron configuration is that of a halogen bereason it is one electron deficient fromhaving actually a filled L shell.(d) The 1s 22 s 22 p 63 s 2 electron configuration is that of an alkaline earth steel bereason of 2 s electrons.(e) The 1s 22 s 22 p 63 s 23 p 63 d 24 s 2 electron configuration is that of a transition steel because of anincomplete d subshell.(f) The 1s 22 s 22 p 63 s 23 p 64 s 1 electron configuration is that of an alkali metal bereason of a single s electron.

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Bonding Forces and also Energies

2.13 Calculate the pressure of attractivity between a K+ and also an O2- ion the centers of which are separated bya distance of 1.5 nm.

SolutionThe attrenergetic pressure in between 2 ions FA is simply the derivative through respect to the interatomic separation

of the attractive power expression, Equation 2.8, which is just

FA =

dEA dr

=

d−A r

⎛⎛ ⎛⎛⎛ ⎛

dr

= A

r 2

The consistent A in this expression is defined in footnote 3. Because the valences of the K+ and O 2 - ions (Z 1 and also Z 2 )are +1 and also -2, respectively, Z 1 = 1 and Z 2 = 2, then

FA = (Z 1 e) (Z 2 e)

4πε 0 r 2

= (1)(2 )(1.602 × 10

−19 C) 2

(4)(π) (8.85 × 10−12 F/m) (1.5 × 10−9 m) 2= 2.05  10 -10 N

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2.14 The net potential energy in between two surrounding ions, EN, might be represented by the sum of Equations2.8 and also 2.9; that is,

EN = −A r

+ B

rnCalculate the bonding energy E 0 in regards to the parameters A, B, and also n using the complying with procedure:

Differentiate EN via respect to r, and then set the resulting expression equal to zero, since the curve ofEN versus r is a minimum at E 0.Solve for r in regards to A, B, and also n, which yields r 0 , the equilibrium interionic spacing.Determine the expression for E 0 by substitution of r 0 right into Equation 2.11.

Solution(a) Differentiation of Equation 2.11 yields

dEN dr

=

d−A r

⎛⎝⎜⎞⎠⎟

dr

+

d B rn

⎛⎝⎜⎞⎠⎟

dr

= A r(1 + 1)

− nB r(n + 1)

= 0

(b) Now, resolving for r (= r 0 )

Ar 02

= nB r 0 (n + 1)

or

r 0 = ⎜⎝⎛ nB A⎟⎞⎠

1/(1 - n)

(c) Substitution for r 0 right into Equation 2.11 and also addressing for E (= E 0 )

E 0 = −Ar 0+ Br 0 n

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2.15 For a K+–Cl– ion pair, attractive and also repulsive energies EA and also ER, respectively, depfinish on the

distance in between the ions r, according to

EA= −1. r

ER=5.8 × 10

r 9

For these expressions, energies are expressed in electron volts per K+–Cl– pair, and r is the distance innanometers. The net power EN is simply the sum of the 2 expressions over.(a) Superimpose on a single plot EN, ER, and also EA versus r approximately 1.0 nm.(b) On the basis of this plot, identify (i) the equilibrium spacing r 0 in between the K+ and Cl– ions, and (ii)the magnitude of the bonding energy E 0 in between the 2 ions.(c) Mathematically identify the r 0 and E 0 values utilizing the options to Problem 2.14, and comparethese with the graphical results from part (b).

Solution(a) Curves of EA, ER, and EN are presented on the plot listed below.

(b) From this plot r 0 = 0.28 nm E 0 = – 4.6 eV

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(c) From Equation 2.11 for EN A = 1. B = 5.86  10 - n = 9Hence,

r 0 = ⎜⎝⎛ nB A⎟⎞⎠

1/(1 - n)

=(8)(5.86 × 101.436 -6) ⎡ ⎣⎢

⎤⎦⎥

1/(1 - 9) = 0.279 nm

and

E 0 = − AAnB⎛⎝⎜⎞⎠⎟1/(1 - n) +BAnB⎛⎝⎜⎞⎠⎟

n/(1 - n)

= − 1. 1.

(9)(5.86 × 10 −6)

⎡⎣

⎢⎢ ⎤ ⎦

⎥⎥

1/(1 − 9) +

5.86 × 10−1.

(9)(5.86 × 10−6)

⎡⎣

⎢⎢ ⎤ ⎦

⎥⎥

9 /(1 − 9)

= – 4.57 eV

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A10 B= (0.35 nm)-

In addition, from the over equation the A is equal to

A = 10B(0.35 nm)-

When the over two expressions for A/10B and A are substituted right into the above expression for E 0 (- 6.13 eV), the

following results

−6.13 eV = = − A A10 B

⎛⎝⎜

⎞⎠⎟

−1/ 9+

BA10 B

⎛⎝⎜

⎞⎠⎟

−10 / 9

= − 10 B(0.35 nm)

-<(0.35 nm)-9>−1/ 9

+ B

<(0.35 nm)-9>−10 / 9

= − 10 B(0.35 nm)

-

0.35 nm+ B(0.35 nm) 10

Or

−6.13 eV = = − 10 B(0.35 nm) 10+ B(0.35 nm) 10= − 9 B(0.35 nm) 10

Solving for B from this equation yields

B = 1.88 × 10-5 eV - nm 10

In addition, the value of A is established from among the previous equations, as follows:

A = 10B(0.35 nm)-9 = (10)(1.88 × 10-5 eV - nm 10 )(0.35 nm)-

= 2.39 eV - nm

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Thus, Equations 2.8 and also 2.9 become

EA = −2.

r

ER = 1.88 × 10

r 10

Of course these expressions are valid for r and E in systems of nanometers and also electron volts, respectively.

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