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Lecture

Problem Sheet 1 Solution.pdf

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Department
Engineering
Course Code
ENG 2002
Professor
Hany E.Z.Farag

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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. 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. 50 2 . 2 Chromium has four naturally-occurring isotopes: 4.34% of Cr, with an atomic weight of 49.9460 amu, 83.79% of Cr, 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 silicon (ACr is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes. Thus A = f A + f A  f A  f A Cr 50C r 50C r 52C r 52C r 53Cr 53C r 54Cr 54Cr  (0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 amu 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. 2.3 (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 mol 1 g/mol  # g/amu =    6.022  10 23 atoms 1 amu/atom  = 1.66  10 -24g/amu (b) Since there are 453.6 g/lb m , 1 lb- mol = (453.6 g/lb )m6.022  10 23 atoms/g- mol) = 2.73  10 26 atoms/lb-mol 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. 4 . 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--each electron is characterized by four quantum numbers. 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. 5 . 2 Relative to electrons and electron states, what does each of the four quantum numbers specify? Solution e h T n quantum number designates the electron shell. e h T l quantum number designates the electron subshell. e h T m quantum number designates the number of electron states in each electron subshell. l e h T m suantum number designates the spin moment on each electron. 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 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 nlms are 100( ) and 100(  1). Write the four quantum numbers for all of the electrons in the L and M shells, and note 2 2 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 poslible m values are 0 and ±1; and possible m values are  1 . Therefore, for the s states, the quantum numbers are200( ) s 2 2 1 1 1 1 1 1 and 200( 2) . For the p states, the quantum numbers are 210 (2) ,210 ( 2) ,211 (2) , 211( 2) ,21( 1)(2), and 21( 1)(1 ). 2 For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible m values are l 1 1 0, ±1, and ±2; and possible m s values are  2. Therefore, for the s states, the quantum numbers are 300 (2) , 1 1 1 1 1 1 1 300 ( ), for the p states they ar310 ( ) ,310 ( ), 311 ( ) ,311 ( ), 31(1) ( ), and 31(1)( ); for the d 2 2 2 2 2 2 2 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. 2.7 Give the electron configurations for the following ions: Fe 2+, Al , Cu , Ba , Br , and O .2- Solution The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6). 2+ 2 2 6 2 6 6 2 e F : From Table 2.2, the electron configuration for an atom of iron 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 two 4s. Thus, the electron configuration for an Fe2+ ion is 1s 2s 2p 3s 3p 3d .6 l A 3+ : From Table 2.2, the electron configuration for an atom of aluminum is 1s 2s 2p 3s 3p . 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 3+ 2 2 6 electron configuration for an Al ion is 1s 2s 2p . u C +: From Table 2.2, the electron configuration for an atom of copper is 1s 2s 2p 3s 3p 3d 4s . In order to become an ion with a plus one charge, it must lose one electron—in this case the 4s. Thus, the electron + 2 2 6 2 6 10 configuration for a Cu ion is 1s 2s 2p 3s 3p 3d . a B 2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 6s . In order to become an ion with a plus two charge, it must lose two electrons—in this case two the 6 s. Thus, the electron configuration for a Ba 2+ion is 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p . 2 6 r B -: From Table 2.2, the electron configuration for an atom of bromine is 1s 2s 2p 3s 3p 3d 4s 4p . In2 5 order to become an ion with a minus one charge, it must acquire one electron—in this case another 4p. Thus, the - 2 2 6 2 6 10 2 6 electron configuration for a Br ion is 1s 2s 2p 3s 3p 3d 4s 4p . O 2: From Table 2.2, the electron configuration for an atom of oxygen is 1s 2s 2p . 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 2- 2 2 6 configuration for an O ion is 1s 2s 2p . 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. + - 8 . 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 + a N e h T ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon (Figure 2.6). - l C e h T ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon. 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. The Periodic Table 9 . 2 With regard to electron configuration, what do all the elements in Group VIIA of the periodic table have in common? Solution Each of the elements in Group VIIA has five p electrons. 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. 2.10 To what group in the periodic table would an element with atomic number 114 belong? Solution From the periodic table (Figure 2.6) the element having atomic number 114 would belong to group IVA. According to Figure 2.6, 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. 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. 2.11 Without consulting Figure 2.6 or T able 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 6 7 2 (a ) 1s 2s 2p 3s 3p 3d 4s (b ) 1s 2s 2p 3s 3p2 6 2 2 5 (c) 1s 2s 2p (d ) 1s 2s 2p 3s6 2 2 2 6 2 6 2 2 (e) 1s 2s 2p 3s 3p 3d 4s (f) 1s 2s 2p 3s 3p 4s 6 1 Solution (a) The 1 s 22s 2p 3s 3p 3d 4s electron configuration is that of a transition metal because of an incomplete d subshell. (b) The 1s 2s 2p 3s 3p electron configuration is that of an inert gas because of filled 3s and 3p subshells. 2 2 5 (c) The1 s 2s 2p electron configuration is that of a halogen because it is one electron deficient from having a filled L shell. 2 2 6 2 (d) The 1s 2s 2p 3s electron configuration is that of an alkaline earth metal because of two s electrons. (e) The 1 s 2s 2p 3s 3p 3d 4s electron configuration is that of a transition metal because of an incomplete d subshell. (f) The 1s 2s 2p 3s 3p 4s electron configuration is that of an alkali metal because of a single s electron. 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. 2.12 (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 4f subshell is being filled for the rare earth series of elements. (b) The 5f subshell is being filled for the actinide series of elements. 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- 2.13 Calculate the force of attraction between a K and an O ion the centers of which are separated by a distance of 1.5 nm. Solution The attractive force between two ions F is just the derivative with respect to the interatomic separation of A the attractive energy expression, Equation 2.8, which is just  A  d  F = dE A =  r = A A dr dr 2 r The constant A in this expression is defined in footnote 3. Since the valences of the K and ions (Z and Z ) are 1 2 +1 and -2, respectively, Z = 1 and Z = 2, then 1 2 (Z1e)(Z2e) FA = 2 4 0 (1)(2 ).602  10 19 C )2 = (4)()(8.85  10 12 F/m)(1.5  10 9 m) 2 = 2.05  10-10N 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. 2.14 The net potential energy between two adjacent ions, EN, may be represented by the sum of Equations 2.8 and 2.9; that is, E =  A  B N r rn Calculate the bonding energy E 0in 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 yiel0s r , the equilibrium interionic spacing. 3. Determine the expression for 0 by substitution of0r into Equation 2.11. Solution (a) Differentiation of Equation 2.11 yields  A  B  d  d   dE N  r  r  dr = dr  d r A nB = (1 + 1)  (n + 1)=0 r r (b) Now, solving for r (= r ) 0 A = nB r2 r (n + 1) 0 0 or 1/(1 - n) r = A  0 nB  (c) Substitution for0r into Equation 2.11 and solving for E (=0E ) A B E0=  + n r0 r0 A B =  1/(1 - n) n/(1 - n) A  A  nB  nB  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. + – 2.15 For a K –Cl ion pair, attractive and repulsive energies E A and E ,Rrespectively, depend on the distance between the ions r, accordi
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