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Ch02 - 재료과학 9판 솔루션
재료과학과공학
Soongsil University
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CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2 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.
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2 Chromium has four naturally-occurring isotopes: 4% of 50 Cr, with an atomic weight of 49 amu, 83% of 52 Cr, with an atomic weight of 51 amu, 9% of 53 Cr, with an atomic weight of 52 amu, and 2% of 54 Cr, with an atomic weight of 53 amu. On the basis of these data, confirm that the average atomic weight of Cr is 51 amu.
Solution The average atomic weight of silicon ( A Cr) is computed by adding fraction-of-
occurrence/atomic weight products for the three isotopes. Thus
+ = AfAfAfAfA 5454535352525050Cr CrCrCrCrCrCrCrCr
(0)(49 amu) (0)(51 amu) (0)(52 amu) (0)(53 amu) 51 amu
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2 (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 4 each electron is characterized by four quantum numbers.
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2 Relative to electrons and electron states, what does each of the four quantum numbers specify?
Solution The 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 moment on each electron.
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2 Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, O 2 - , Fe3+ and S 2 -.
Solution The electron configurations for the ions are determined using Table 2 (and Figure 2).
Fe2+: From Table 2, the electron configuration for an atom of iron is 1 s 22 s 22 p 63 s 23 p 63 d 64 s 2. In order to become an ion with a plus two charge, it must lose two electrons 4 in this case the two 4 s. Thus, the electron configuration for an Fe2+ ion is 1 s 22 s 22 p 63 s 23 p 63 d 6. Al3+: From Table 2, the electron configuration for an atom of aluminum is 1 s 22 s 22 p 63 s 23 p 1. In order to become an ion with a plus three charge, it must lose three electrons 4 in this case two 3 s and the one 3 p. Thus, the electron configuration for an Al3+ ion is 1 s 22 s 22 p 6. Cu+: From Table 2, the electron configuration for an atom of copper is 1 s 22 s 22 p 63 s 23 p 63 d 104 s 1. In order to become an ion with a plus one charge, it must lose one electron 4 in this case the 4 s. Thus, the electron configuration for a Cu+ ion is 1 s 22 s 22 p 63 s 23 p 63 d 10. Ba2+: The atomic number for barium is 56 (Figure 2), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1 s 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 become an ion with a plus two charge, it must lose two electrons 4 in this case two the 6 s. Thus, the electron configuration for a Ba2+ ion is 1 s 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, the electron configuration for an atom of bromine is 1 s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 5. In order to become an ion with a minus one charge, it must acquire one electron 4 in this case another 4 p. Thus, the electron configuration for a Br- ion is 1 s 22 s 22 p 63 s 23 p 63 d 104 s 24 p 6. O 2 - : From Table 2, the electron configuration for an atom of oxygen is 1 s 22 s 22 p 4. In order to become an ion with a minus two charge, it must acquire two electrons 4 in this case another two 2 p. Thus, the electron configuration for an O 2 - ion is 1 s 22 s 22 p 6. Fe3+: From Table 2, the electron configuration for an atom of iron is 1 s 22 s 22 p 63 s 23 p 63 d 64 s 2. In order to become an ion with a plus three charge, it must lose three electrons 4 in this case the two 4 s and one 3 d. Thus, the electron configuration for an Fe3+ ion is 1 s 22 s 22 p 63 s 23 p 63 d 5. S 2 - : From Table 2, the electron configuration for an atom of sulphur is 1 s 22 s 22 p 63 s 23 p 4. In order to become an ion with a minus two charge, it must acquire two electrons 4 in this case another two 3 p. Thus, the electron configuration for an S 2 - ion is 1 s 22 s 22 p 63 s 23 p 6.
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2 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). 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 To what group in the periodic table would an element with atomic number 114 belong?
Solution
From the periodic table (Figure 2) the element having atomic number 114 would belong to group IVA. According to Figure 2, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the right-most column of group VIII. Moving four columns to the right puts element 114 under Pb and in group IVA.
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2 Without consulting Figure 2 or Table 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) 1 s 22 s 22 p 63 s 23 p 63 d 74 s 2 (b) 1 s 22 s 22 p 63 s 23 p 6 (c) 1 s 22 s 22 p 5 (d) 1 s 22 s 22 p 63 s 2 (e) 1 s 22 s 22 p 63 s 23 p 63 d 24 s 2 (f) 1 s 22 s 22 p 63 s 23 p 64 s 1
Solution
(a) The 1 s 22 s 22 p 63 s 23 p 63 d 74 s 2 electron configuration is that of a transition metal because of an incomplete d subshell. (b) The 1 s 22 s 22 p 63 s 23 p 6 electron configuration is that of an inert gas because of filled 3 s and 3 p subshells. (c) The 1 s 22 s 22 p 5 electron configuration is that of a halogen because it is one electron deficient from having a filled L shell. (d) The 1 s 22 s 22 p 63 s 2 electron configuration is that of an alkaline earth metal because of two s electrons. (e) The 1 s 22 s 22 p 63 s 23 p 63 d 24 s 2 electron configuration is that of a transition metal because of an incomplete d subshell. (f) The 1 s 22 s 22 p 63 s 23 p 64 s 1 electron configuration is that of an alkali metal because of a single s electron.
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Bonding Forces and Energies
2 Calculate the force of attraction between a K+ and an O 2 ion the centers of which are separated by a distance of 1 nm.
Solution
The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation 2, which is just
FA = dEdrA =
d − Ar
dr =
A
r 2
The constant A in this expression is defined in Equation 2. Since the valences of the K+ and O 2 ions ( Z 1 and Z 2 ) are +1 and 2, respectively, Z 1 = 1 and Z 2 = 2, then
19 2 12 9 2
(1)(2 ) 1 10
(4)( ) (8 10 F / m) (1 10 m)
( C )
1 10 10 N
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2 The net potential energy between two adjacent ions, EN , may be represented by the sum of Equations 2 and 2; that is, N n
E A B
r r Calculate the bonding energy E 0 in terms of the parameters A , B , and n using the following procedure:
- 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 E 0.
- Solve for r in terms of A , B , and n , which yields r 0 , the equilibrium interionic spacing.
- Determine the expression for E 0 by substitution of r 0 into Equation 2.
Solution (a) Differentiation of Equation 2 yields
dEN dr =
d − Ar dr +
d rBn dr
= A
r (1 + 1)
− nB
r ( n + 1)
= 0
(b) Now, solving for r (= r 0 )
A r 02 =
nB r 0 ( n + 1)
or
r 0 = nBA
1/(1 − n )
(c) Substitution for r 0 into Equation 2 and solving for E (= E 0 )
E 0 = − rA 0
+ B
r 0 n
= − A
A
nB
1/(1 − n )+
B
A
nB
n /(1 − n )
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Thus, r 0 = A nB
1/(1 − n )
= 1.
(8)(5 × 10 − 6 )
1/(1 − 9) =0 nm
and
E 0 = − A
A
nB
1/(1 − n ) +
B
A
nB
n /(1 − n )
= − 1.
1.
(9)(5 × 10 − 6 )
1/(1 − 9) +
5 × 10 − 6
1.
(9)(5 × 10 − 6 )
9/(1 − 9)
= 4 eV
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2 Consider a hypothetical X+ 3 Y- ion pair for which the equilibrium interionic spacing and bonding energy values are 0 nm and 6 eV, respectively. If it is known that n in Equation 2 has a value of 10, using the results of Problem 2, determine explicit expressions for attractive and repulsive energies EA and ER of Equations 2 and 2.
Solution
This problem gives us, for a hypothetical X+ 3 Y- ion pair, values for r 0 (0 nm), E 0 (6. eV), and n (10), and asks that we determine explicit expressions for attractive and repulsive energies of Equations 2 and 2. In essence, it is necessary to compute the values of A and B in these equations. Expressions for r 0 and E 0 in terms of n , A , and B were determined in Problem 2, which are as follows:
r 0 = nBA
1/(1 − n )
E 0 = − A
A
nB
1/(1 − n )+
B
A
nB
n /(1 − n )
Thus, we have two simultaneous equations with two unknowns (viz. A and B ). Upon substitution of values for r 0 and E 0 in terms of n , these equations take the forms
1/(1 10) 1/ 0 nm = 10 AB = 10 AB ö ö ö ö ÷ ÷ø ø ÷ ÷ø ø
and
−6 eV= − A A 10 B
1/(1 − 10)+
B
A
10 B
10/(1 − 10)
= − A
A
10 B
−1/9+
B
A
10 B
−10/
We now want to solve these two equations simultaneously for values of A and B. From the first of these two equations, solving for A /10 B leads to
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Of course these expressions are valid for r and E in units of nanometers and electron volts, respectively.
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2 The net potential energy EN between two adjacent ions is sometimes represented by the expression
EN = − C r + D exp −
r
ρ
(2)
in which r is the interionic separation and C , D , and are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy E 0 in terms of the equilibrium interionic separation r 0 and the constants D and using the following procedure:
- Differentiate EN with respect to r and set the resulting expression equal to zero.
- Solve for C in terms of D , , and r 0.
- Determine the expression for E 0 by substitution for C in Equation 2. (b) Derive another expression for E 0 in terms of r 0 , C , and using a procedure analogous to the one outlined in part (a).
Solution
(a) Differentiating Equation 2 with respect to r yields
dE dr =
d − Cr
dr −
d D exp−ρ r
dr / 2
C Der r
At r = r 0 , dE/dr = 0, and ( / ) 0 02
C De r r
(2)
Solving for C and substitution into Equation 2 yields an expression for E 0 as
E 0 = De −( r 0 /ρ) 1 − r ρ 0
(b) Now solving for D from Equation 2 above yields ( / ) 0 02
D C e r r
Ch02 - 재료과학 9판 솔루션
Course: 재료과학과공학
University: Soongsil University
