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Chapter 2

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University of Toronto St. George

Materials Science & Engineering

MSE101H1

Scott Ramsay

Spring

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CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
2.3 Allowed values for the quantum numbers of electrons are as follows:
n = 1, 2, 3, . . .
l = 0, 1, 2, 3, . . . , n –1
m = 0, ±1, ±2, ±3, . . . , ±l
l
m = 1
s 2
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 nlmsm , are
100( ) and 100( 1). Write the four quantum numbers for all of the electrons in the L and M shells, and note which
2 2
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 m
l
values are 0 and ±1; and possible m values are 1. Therefore, for the s states, the quantum numbers are
s 2
1 1 1 1 1 1
200 (2) and 200 2) . For the p states, the quantum numbers are 210 (2) , 210 (2 ), 211( 2 , 211 ( 2),
1 1
21( 1)( ), and 21( 1( ) .
2 2
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible m values are 0,
l
±1, and ±2; and possible m values are 1. Therefore, for the s states, the quantum numbers are 300 ( ),
s 2 2
1 1 1 1 1 1 1
300 ( ), for the p states they are10 ( ), 310 ( ), 311 ( ), 311 ( ), 31(1)( ) , and 31(1)( ); for the d
2 2 2 2 2 2 2
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 1 1 1 1 1 1 1 1 1
states they are320 (2) ,320 ( 2), 321 (2) , 321 2 ), 32(1)(2) , 32(1) 2 ), 322( 2 , 322 (2 ), 32( 2( 2 ,
1
and 32( 2)( ).
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.4 Give the electron configurations for the following ions: P , P , and Ni .
Solution
The electron configurations for the ions are determined using Table 2.2.
5+ 2 2 6 2 3
P : From Table 2.2, the electron configuration for at atom of phosphorus is 1s 2s 2p 3s 3p . In order to
become an ion with a plus five charge, it must lose five electrons—in this case the 3s and 3p electrons. Thus, the
5+ 2 2 6
electron configuration for a ion is 1s 2s 2p .
3- 2 2 6 2 3
P : From Table 2.2, the electron configuration for at atom of phosphorus is 1s 2s 2p 3s 3p . In order to
become an ion with a minus three charge, it must acquire three electrons, which in this case will be added to and fill
3- 2 2 6 2 6
the 3p subshell. Thus, the electron configuration forion is 1s 2s 2p 3s 3p .
2+ 2 2 6 2 6 8 2
Ni : From Table 2.2, the electron configuration for at atom of nickel is 1s 2s 2p 3s 3p 3d 4s . In order
to become an ion with a plus two charge, it must lose two electrons—in this case the 4s electrons. Thus, the electron
2+ 2 2 6 2 6 8
configuration for an Niion is 1s 2s 2p 3s 3p 3d .
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.6 Without consulting Figure 2.6 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.
2 2 6 2 5
(a) 1s 2s 2p 3s 3p
(b) 1s 2s 2p 3s 3p 3d 4s 4p2 6
2 2 6 2 6 10 2 6 5 2
(c) 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s
Solution
2 2 6 2 5
(a) The 1s 2s 2p 3s 3p electron configuration is that of a halogen because it is one electron deficient
from having a filled p subshell.
2 2 6 2 6 10 2 6
(b) The 1s 2s 2p 3s 3p 3d 4s 4p electron configuration is that of an inert gas because of filled 4s and
4p subshells.
(c) The 1s 2s 2p 3s 3p 3d 6 10 4s 4p 4d 5s electron configuration is that of a transition metal because of
an incomplete d subshell.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Bonding Forces and Energies
2+ 2–
2.8 Calculate the force of attraction between a Ca and an O ion the centers of which are separated by
a distance of 1.25 nm.
Solution
The attractive force between two ions A is just the derivative with respect to the interatomic separation of
the attractive energy expression, Equation 2.8, which is just
A
dE d
F = A = r = A
A dr dr r2
2+ 2-
The constant A in this expression is defined in footnote 3. Since the valences of theand O ions (Z1and Z 2
are both 2, then
(Z e)(Z e)
F A = 1 2
4 0 2
19 2
= (2)(2)(1.6 10 C)
(4)()(8.85 1012 F/m)(1.25 10 9 m) 2
= 5.89 10-10 N
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.9 The net potential energy between two adjacent ions, EN, may be represented by the sum of Equations
2.8 and 2.9; that is,
A B
EN = + n (2.11)
r r
Calculate the bonding energy E 0n terms of the parameters A, B, and n using the following procedure:
1. Differentiate N with respect to r, and then set the resulting expression equal to zero, since the curve of
ENversus r is a minimum at E 0
2. Solve for r in terms of A, B, and n, which yie0ds r , the equilibrium interionic spacing.
3. Determine the expression for 0 by substitution o0 r into Equation 2.11.
Solution
(1) Differentiation of Equation 2.11 yields
A B
d d
dE N r rn
dr = dr dr
= A nB = 0
r(1 + 1) r(n + 1)
(2) Now, solving for r (= 0 )
A nB
=
r2 r (n + 1)

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