Chem2090-PS6Key - P.4d; Lev-x 5r if 6; 111;? 61.18. In an...

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Unformatted text preview: P.4d; Lev-x 5r if 6; 111;? 61.18. 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. Y3+ < Sr2+ < Rb+ < Br" < Sez' e. $.28. The electron affinity of fluorine is —328.0 kJ/rnol (Figure 9—10) and the first and second ionization energies of Mg (Table 10.4) are 737.7 kJ/mol and 1451 kJ/mo], respectively. q Mg(g)—) Mg+(‘g)+ e“ Il =737.7 kJ/rnol Mg+ (g)—> Mg2+ (g)+ e' I2 = 1451 kJ/mol 2F(g)+26‘—->2F(g) 23.4.: 2(-—328.0)kJ/mol Mg(g)+2 F(g)-mg“ (n+me 7, 34. First we write the electron configuration of the element, 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. (a) Ni 3d,"‘45z —_; Ni2+ [Ar] 30!8 Two unpaired e' (b) Cu [AI] 3d'04s‘ ——> cu2+ [Ar] 3d9 One unpaired e' (c) Cr [Af] 3d543‘ cr3+ [Ar] 3613 Three unpgaired e' 7.40. (a) Size increases down a group and from right to left in a period. Ba is closest to the lower left corner of the periodic table and thus has the largest size. (b) Ionization energy decreases down a group and from right to left in a period. Although Pb is closest to the bottom of its group, Sr is farthest left in its period (and only one period above Pb). Sr should have the lowest first ionization energy. _ Electron affinity becomes more negative from left to right in a period and from bottom to top in a group. Cl is closest to the upper right in the periodic table and has the most negative (smallest) electron affinity. (d) The number of unpaired electrons can be determined from the orbital diagram for each species. Atomic nitrogen ([He]2s22p3), by virtue of having three half—filled p— orbitals, possesses the greatest number of unpaired electrons. H :0: H I il l .0. H—CIZ—C—CII~H ' H ' H H to O- I. ll .0 (0.4. (a) §=c=§ (b) (c) :gl—C—{J-l: (a) :E—N=9 [0.10. (a) 36—61—65 3 (has 20 valence electrons, whereas the molecule C102 has 19 valence electrons. This is a proper Lewis structure for the chlorite ion, although the brackets and the minus charge are missing. A plausible Lewis structure for the molecule (310213 (3261—6 : on 00 0. (b) Here, 53—9 = is improperly written as a covalent Lewis structure. CaO is actually .. 2- 2+ _ . an ionic compound. [ca] '9 ' is a more plausible Lewis structure for C130. (0916. (a) [:6—H1'[Ca]1+[:{):—H]' .§_ (b) H l -- - H—I‘lJ—H [: Br2] H (c) now—{3:1 :1' {oa]2+[:(:):—c:::1:]' :0. 22. We calculate formal charge 2 # valence electrons — # lone-pair e' — 'A # bond-pair e" (a) Hus—gags | H H—c'li—fi—H H §=C=§i (b) es=s=ss =s—r=@ s=fi—@= r—fi=@ E=fi-§= (‘1) 000 O O _of N=5—2—3=0 of O=6-—4—2=0 of N=5-—4—2=—1 of 0=6—2—3=+1 of C=4—0—4=0 f s=6—4—2=0 of C=4-4—2=-—2 ofcentral S=6—0—4 oftermin S=6——4— of N=5—6—1=—2 of -O=6—4—2=0 of F=7—2—3=2+ of N=5-4—2=-1 of O=6-—6—I=—1 f F=7i—2-3=2+_ of N=5—2—3=0 f O=6~4—2=0 f F=7—6—1=0 f 0=6—6—1=-—1 of N=5—2—3=0“ f F=7—4—~2=l+ ofS=6-—6—l=—l ofO=6—2—3=+1 ofCl=7—6—1=O ofS=6—2—3=+1 ofO=6—6fi-l=—1 ofCl=7—6—l=0 ofS=6—6—1=-—1 ofO=6—6——1=-—l ofCl=7—4—2=+l most plausible most plausible 0 should have the negative formal charge most plausible 26 electrons present; Cl should not have a more positive f.c. than S l O. 30. In C302 there are (3 x 4) +(2 x 6) = 24 valence electrons or 12 valence electron pairs. A plausible Lewis structure follows: 6=C=C=C=6 [6-, 36. (a) 3 F2 cannot possess a dipole moment, since both of the atoms in the diatomic molecule are the same. This means that there is no electronegativity difference between atoms, and hence no polarity in the F—WF bond. (b) 5=I<Ih§= Each nitrogen-to-oxygen bond in this molecule is polarized toward , lecule is polar (c) Although each B-F bond is polarized toward F in this trigonal __ “I: i planar AX3 molecule these bond dipoles cancel. The molecule :F-—B——F' : is nonpolar. . u (d) H—fir: The H—Br bond is polar toward Br, and this molecule is polar as Well. (e) The H——-C bonds are not polar, but the C——Cl bonds are, toward Cl. H The molecular shape is tetrahedral (Ax,) and thus these two (2 —— Cl :éi_('3__c'], dipoles do not cancel each other; the molecule is polar. - " “ - H (1) Although each Si—F bond is polarized toward F, in this tetrahedral : i5 : AX 4 molecule these bond dipoles oppose and cancel each other. = 1';_Sli__i5 : As a result, the molecule is nonpolar. L ‘r -o I .. t (g) 62C=§ In this linear molecule, the two bonds fi-om carbon both are pUlHJlLUU away from carbon. But the C=O bond is more polar than the C=S bond, and hence the molecule is polar. (a) In CO2 , there are 4 +(2 x 6) = 16 valence electrons, or 8 electron pairs. 6.20:6 (b) In 0C1”, there are 6+ 7 +1 =14 valence electrons, or 7 electron pairs, 9" :O—él: (c) In C0322 there are 4 + (3 x 6.) + 2 = 24 Valence electrons, or 12 electron pairs (see below). (d) In OH‘, there are 6 + 1+1 = 8 valence electrons, or 4 electron pairs. .. 2- 2- <—-- ‘9’ «——~ CFC—*9? ©“C*€.5= 1.3%: I to. 48. '4 (a) lo, 62. (b) (a) (b) (C) (d) (e) C2H5 has a total of (5 x l) + (2 x 4) = 13 valence electrons or Iii III 6 electron pairs and a lone electron on C. A plausible Lewis H—C—C - structure is given to the right. Ill III H02 has a total of (2 x 6) +1 = 13 valence electrons or 6 electron pairs (2 bonding and 4 non-bonding) and a lone Hw—5—6 - electron on O. A plausible Lewis structure is given to the right. ClO has a total of 7 + 6 = 13 valence electrons or 6 electron _ -- '- pairs (1 bonding and 5 non-bonding) and a lone electron on either C1 or O. A plausible Lewis structure is given to the right. In OSF2 , there are a total of 6 + 6 +(2 x 7) = 26 valence electrons, or 13 electron pairs. S is the central atom. A plausible Lewis structure is shown to the right. This molecule is of the type AX3E1 It has a tetrahedral electron-group geometry and a trigonal pyramidal shape. In OZSFZ, there are a total of (2 x 6) + 6 + (2 x 7) = 32 valence electrons, or 16 electron pairs. S is the central atom. One plausible Lewis structure is shown to the right. The molecule is of the AX4 type It has a tetrahedral electron-group geometry and a tetrahedral shape. The structure with all single bonds is preferred because it avoids an expanded octet. In SF; there are 6 + (5 x 7) + 1 = 42 valence electrons, or 21 electron pairs. A plausible Lewis structure is shown to the right. The ion is of the AXSE type. It has an octahedral electron-group geometry and a square pyramidal molecular shape. In C104- , there are a total of 1+ 7 +(4 x 6) = 32 valence electrons, or 16 electron pairs. Cl is the central atom. A plausible Lewis structure is shown to the right. "lhis ion is of the AX4 type. It has a tetrahedral electron-group geometry and a tetrahedral shape. In C10; the total number of valence electrons is l + (3 x 6) + 7 = 26 valence electrons, or 13 electron pairs. A plausible Lewis structure is shown to the right. The molecule is of the type AX3 E. It has‘a tetrahedral electron-group geometry and a trigonal pyramidal molecular shape. - or ('0. 76. lo. 82. First we need to draw the Lewis structure of each of the compounds cited, so that we can determine the order, and hence the relative length, of each O-to-O bond. ' (a) In H202, there are (2 x l) + (2 x 6) = 14 valence electrons or 7 electron pairs. A plausible Lewis structure is H—é — é — H (b) In 02 , the total number of valence electrons is (2 x 6 =)12 valence electrons, or 6 electron pairs. A plausible Lewis structure is E = E (c) In 03, the total number of valence electrons is (3 x 6) = 18 valence electrons, or 9 electron pairs.'A plausible Lewis structure is §=§ ~91 <--+ I 0 — O=Q (Two most stable resonance contributors) Thus, 02 should have the shortest O-to-O bond, because the O atoms are joined via a double bond. The'Single O — 0 bond in H202 should be longest. ' The reaction in terms of Lewis structures is §=6 -—_6J + §=fiu ——> éxN —§ - + §=§ The net result is the breakage of an 0 — 0 bond (142 kJ/mol) and the formation of an N—O bond (222 kJ/mol). AH =142 kJ/rnol - 222 kJ/rnol = ~80. icJ/filol ...
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This note was uploaded on 10/25/2009 for the course CHEM 2090 at Cornell University (Engineering School).

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Chem2090-PS6Key - P.4d; Lev-x 5r if 6; 111;? 61.18. In an...

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