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AK9 - 18 84(E(a This is an excited state silicon atom(3p...

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Unformatted text preview: 18. 84. (E) (a) This is an excited state silicon atom (3p electrons should remain unpaired with same Spin i (b) This is an excited state phosphorus atom; the three 3p orbital electrons should have the same spin (violates Hund’s rule). (c) This is a ground state sulfur atom. (d) This is an excited state sulfur atom. The two unpaired electrons should have the same qi “3»?3 n 86. (E) (3) Te: {anidmss'lsp‘l . (d) *l’t: [Xe]6s34r"‘sd3 (b) CSZ[X61631 (e) Os: [Xc]6524fl45c16 (c) semantiladmifi . . (1) Cr: [Ar]4513d5 *Note that this is the expected electron configuration of Pt based on its position in the periodic table. Experiment reveals that the true ground state configuration is in fact [Xe]osI 4fM SdL". lfltimately, it is experiment and not the position of the element in the periodic table that has the final say on the true ground state electronic configuration. 88. (E) (a) arsenic; (b) sulfur; (c) scandium; (d) rtitheniiim*; (e) not an element *Note that this is the expected electron configuration ofRu based on its position in the periodic table. Experiment reveals that the true ground state electron configuration is in fact [to] 4d” 5s‘. 14. (M) The hydrogen ion contains no electrons. only a nucleus. It is exceedingly tiny, much smaller than any other atom or electron—containing ion. Both H and H’ have a nuclear charge of l but Hm has two electrons to H‘s one, and thus is larger. Both He and H" contain two electrons, but He has a nuclear charge of 2+, while H’ has one of only 1+. The smaller nuclear charge of H‘ is less effective at attracting electrons than the more positive nuclear charge of He. The only comparison left is between H and He; we expect He to be smaller since atomic size decreases from left to right across a’ period. Thus, the order by increasing size: H+ < He < H < H' . (E) In an isoelectronic series, all of the species have the same number and types of electrons. The size is determined by the nuclear charge. Those Species with the largest (positive) nuclear charge are the smallest. Those with smaller nuclear charges are larger in size. Thus, the more positively charged an ion is in an isoelectronic series, the smaller it will be. ‘i < Sr} < Rbl < Bi" < Se} 24. (E) First we convert the mass of Na given to an amount in moles of Na. Then we compute the energy needed to ionize this much Na. lg 1molNa x495'8kJ=0.0216kJ>< x 1000 J : 1000 mg 22.99 g Na 1 mol Na lkl "2l.6.l Energy = 1.00 mg Na >< t i l l l i 28. (M) The electron affinity of fluorine is —328.0 kJ/rnol (Figure 9—1 1) and the first and second ionization energies of Mg (Table 9.4) are 737.7 kJ/mol and 1451 kJ/mol. respectively. Mgtgle Melt silt 6' 1: 737.7 it] 711101 Mg'its)—> Ms“’(s)—- 0' V 73 21451 kJ711~1i11 2 Fts)+2 6‘ -> 2 F’ts) 2E.A.i2(—328.0) 1o 11101 Mgtgl+ 2 Fts) ; Mei—(Q5773 F’tg) ALE—e +1533 141111101 Endothermic. 2_9_. (E) The electron is being removed from a species with a neon electron configuration. But in the case of Nah . the electron is being removed from a species that is lett with a 2 7- charge. while in f the case ofNei the electron is being removed from a species with a 1+ charge. The more highly; charged the resulting species. the more difficult it is to produce it by removingI an electron. 1' 36. (E) First we write the electron configuration ot‘the clement, then that of the ion. In each case, the number of unpaired electrons written beside the configuration agrees with the data given in the statement of the problem. (20M [Ar] 3d“4.€ #» N1“ [Ar] 3L7“ Two unpaired 6' (b)Cu [Arj 377 "45‘ 77—» C11“ [Ar] 3d“ One unpaired e’ (c) Cr [Ar] 3d545‘ —w-> Cr?” [Ar] 3t!" Three unpaired e' M_ electron configuration ot‘the iron atom. [ Fe] :l sitrl3c1’”—ls3 Then the two ions result from the loss of both 45 electrons. followed by the loss ot‘a 3d electron in the case of FeJT. 38. (E) The electron configuration of each iron ion can be determined by starting with the l [Few : lArl3d" four unpaired electrons f l7 Fell ”j :‘ lAt] 37!; ll ve unpaired electrons 44. (M) (a) The boilint:r point increases by 52 C’C from Cll4 to Sil-lJ , and by 22 °C from SiH4 l t . i . e l to GeHi. One expects another increase in boiling pornt trom GetLl to SnHy . E probably by about 15 DC. Thus. the predicted boiling,J point of SnH4 is r75 C . i (The actual boilingr point of SnH4 is 752 ”C). (b) The boiling point decreases by 39 0C from H 1Te to HISe. and by 20 0C from HlSe l to HIS . it probably will decrease by about. ‘10 ”C to reach H30. Therefore. one i predicts a value for the boiling:r point of H30 ot’ approximately —7l OC. Ot‘course, the actual boiling point of water is 100’ C. The prediction is seriously in error (~ 170 DC) because we have neglected the hydrogen bondingI between water molecules. a topic that is discussed in Chapter 13. F “1619513105 l unpaired e" N [He] 35237)} 3 unpaired e" ‘ 5; [NC] 3-5737?“ t.) unpaired e' M9" [He] 1‘73?“ 0 unpaired e' E . r "y 1"! l _ . __ V"), 4' '1 l - Scl- [Ne] gas 3;), t) unpaired e ll [Ar] 3“, lunpaired c 46. (E) (a) 6. Tl‘s electron configuration [Xe] filfHSd1U 6.916;)! has one p electron in its outermost shell. (1)) 8. Z = 70 identifies the element as Yb. an f —bloek of the periodic table, or an inner transition element. (c) 5. Ni has the electron configuration [Ar] 452 3d“ . It also is a d —bloek element. (d) 1. An 52 outer electron configuration, with the underlying configuration of a noble gas, is characteristic of elements of group 2(2A), the alkaline earth elements. Since the noble gas core is [Ar-ll the element must be Ca. (e) 2. The element in the fifth period and Group 15 is Sb, a metalloid. (f) 4 and 6. The element in the fourth period and Group 16 is Se, a nonmetal. (Note. 13 is a nonmetal with electron confiUuration 1322322 l (havin one electron in the shell _ . _ o P g P of highest n). 6 rs also a possrble answer. 60. (E) (a) A; Element “A" (K) the alkali metals. (b) B; Element “B” (As) is a nowmetal. It is easier for As to gain an electron to form As' than for K to form K. (e) A; Element “A“ {K} has the larger atomic radius In general, size decreases going across a period. (d) B; Element “"8 (As) has the greatest electron aliinity. Non—metals have a greater tendency to add electrons compared with metals. has an electron configuration consistent with group 1A, which is . PS 9 Ex 2: A) As n increases (down the periodic table), EA increases (becomes less negative). Electron affinity is directly proportional to Zeff (charge) and inversely proportional to n (distance from the nucleus). n increases faster than Zefi because although shielding by electrons gets worse as n increases (greater Zefi), the electrons are farther from the nucleus . The combination of these two opposing trends results in an increasing EA as n increases. B) For n = 2 and n = 3, the electrons are still fairly close to the nucleus. Since electrons in the n = 3 shell shield worse than electrons in the n = 2 shell, Zefi increases faster than n. C) As you move from left to right, EA decreases (becomes more negative). As more electrons are added, they screen Zeff less and less effectively. D) The EAs in group 15 increase instead of decrease. Addition of pairing energy when an extra electron is added makes EA less favorable: 1- e' 4—4—34 ————> 4L 4: —4— II' P l” E) Additional electrons have to go in the (n+1) orbitals, which are far from the nucleus and have relatively well screened Zefi. 4— (mos + e' 4¥4Pt~l+ *——> 4L—-—’ll--’H- n r“ (a we): (gust) 15 =1310‘57mol(§g“ ‘ “'5 ‘S/mol ...
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