e. MATERIALES Y MÉTODOS
1. Qué cargo desempeña en la Comercializadora Zerimar
5.1 (a) London forces, dipole-dipole, hydrogen bonding; (b) London forces; (c) London forces, dipole-dipole, hydrogen bonding; (d) London forces, dipole-dipole.
5.3 Only (b) CFLC1, (c) CH,C1,, and (d) CHCL, will have dipole- dipole interactions. The molecules CH4 and CC14 do not have dipole moments.
5.5 (a) NaCl (801°C vs. - 1 14.8°C) because it is an ionic compound as opposed to a molecular compound; (b) butanol (-90°C vs. -116°C) due to hydrogen bonding in butanol that is not possible in diethyl ether; (c) triiodomethane because it will have much stronger London dispersion forces ( -82.2°C for trifluoromeihane vs. 219°C for triiodomethane); (d) Methanol (-94 °C vs. - 169 °C) because o f the hydrogen bonding in
methanol but not in ethylene.
5.7 (a) PF3 and PBr, are both trigonal pyramidal and should have similar intermolecular forces, but PBr3 has the greater number o f electrons and should have the higher boiling point. The boiling point o f PF3 is
-101.5°C and that of PBr, is 173.2 °C; (b) S02 is bent and has a dipole moment whereas CO, is linear and will be nonpolar. S02 should have the higher boiling point. SO, boils at -10°C, whereas C 02
sublimes at -~78°C. (c) BF, and BCF are both trigonal planar, so the choice o f higher boiling point depends on the difference in total number of electrons. BCF should have the higher boiling point
(12.5°C vs. -99.9°C).
5.9 The interaction energies can be ordered based on the relationship the energy has to the distance separating the interacting species. Thus ion-ion interactions are the strongest and are directly proportional to the distance separating the two interacting species. Ion-dipole energies are inversely proportional to d 2, whereas dipole-dipole for constrained molecules (i.e., solid state) is inversely proportional to d\ Dipole-dipole interactions where the molecules are free to rotate become comparable to induced dipole-induced dipole interactions, which are both inversely related to d(\ The order thus derived is: (b) dipole-induccd dipole = (c) dipole-dipole in the gas phase < (e) dipole-dipole in the solid phase < (a) ion-dipole < (d) ion-ion.
5.11 Only molecules with H attached to the electronegative atoms F, N, and O can hydrogen bond. Additionally, there must be lone pairs available for the FTs to bond to. This is true only o f (d) UNO:.
5.13 IT because the dipoles are aligned with oppositely charged ends closest to each other thereby maximizing dipole-dipole attractions.
5.15 The ionic radius o f A r is 53 pm and that o f Be- is 27 pm. The ratio of energies will be given by
E p GC - 1* 1//
d 2
-p A l
( ~ \ z \ ^
v
cf~
J
(27 + 100)'J
5.19
5.21
The electric dipole moment o f Ihe water molecule (u) will cancel:
The attraction o f the A l 1' ion will be greater than that o f the Be2 ion.
(a) Xenon is larger, with more electrons, giving rise to larger London forces that increase the melting point, (b) Hydrogen bonding in water causes the molecules to be held together more tightly than in diethyl ether, (c) Both molecules have the same molar mass, but pentane is a linear molecule compared to dimethylpropane, which is a compact, spherical molecule. The compactness o f the dimethylpropane gives it a lower surface area. That means that the intermolecular attractive forces, which are o f the same type (London forces) for both molecules, will have a larger effect for pentane.
(a) As intermolecular forces increase, boiling point will increase because more thermal energy is required to separate molecules apart; (b) viscosity will increase with the increase o f intermolecular forces increase because stronger intermolecular forces wili make molecules less mobile (or do not let molecules move past one another easily); (c) surface tension also increases as the increase of intermolecular forces because stronger intermolecular forces intend to pull molecules together and inward. Even though the A L + ion has a larger radius, its charge is higher than that o fB e 'T making the attraction greater over all.
23 (a) c/s-Dichloroethene is polar, whereas mms-dichlorocthene, whose individual bond dipole moments cancel, is nonpolar. Therefore, cis- dichloroethene has the greater intermolecular forces and the greater surface tension, (b) Surface tension of liquids decreases with increasing temperature as a result o f thermal motion as temperature rises. Increased thermal motion allows the molecules to more easily break away from each other, which manifests itself as decreased surface tension.
25 At 50°C all three compounds are liquids. C6H6 (nonpolar) < C6H,SH (polar, but no hydrogen bonding) < C0H5OH (polar and with hydrogen bonding). The viscosity will show the same ordering as the boiling points, which are 80°C for C6H6, 169°C for C6H5SH, 182° for ChH5OH.
27 C fl4, -162°C; CH3CH3, -88.5°C; (C U})2CHCH2C H ^ 28°C;
CI h(CH2)3CH3,36°C; CH3OH, 64.5°C; CH3CH2OH, 78.3 °C;
ClhCllOHCf-L, 82.5°C; C5H9OH (cyclic, but not aromatic), I40°C: C6H5CH3OII (aromatic ring), 205°C; OHCH2CHOHCH2OH, 290°C
29 (a) hydrogen bonding; (b) London dispersion forces increase
r — diameter = — (0.15 mm) = 7.5 x 10 3 m
2 2 V1000 mmJ
(9.81 m • s' 1 )(9.97 x 10J kg-m 'J)(7.5 x 10"’ m) Remember that 1 N = l kg • rrf1 • s"“
For ethanol: d - 0.79 g • cm" h = ( 1kg 1000 g j ( 10(1 cm31 nr ; 2(22.8 xlO"3 N • m-1) (9.81 m -s"')(7.9x I0; k g • m“ ')(7.5 x 10"5 m) | = 7.9 x 10" kg • m ■' = 0.078 m or 78 mm
Water will rise to a higher level than ethanol. There are two opposing effects to consider. While the greater density o f water, as compared to ethanol, acts against it rising as high, it has a much higher surface tension.
5.33 Glucose will be held in the solid by London forces, dipole-dipole
interactions, and hydrogen bonds; benzophenone will be held in the solid by dipole-dipole interactions and London forces; methane will be held together by London forces only. London forces are strongest in
benzophenone, but glucose can experience hydrogen bonding, which is a strong interaction and dominates intermolccular forces. Methane has few electrons so experiences only weak London forces. We would expect the melting points to increase in the order Cl L (m.p. = -182°C)
< benzophenone (m.p. = 48°C) < glucose (m.p. = 148 - 155°C).
5.35 (a) network; (b) ionic; (c) molecular; (d) molecular; (e) network
5.37 Substance A: ionic; substance B: metallic; substance C: molecular solid
5.39 (a) At center: 1 center x 1 atom • center-1 = 1 atom; at 8 corners,
8 comers x -j- atom • corner 1 = 1 atom; total = 2 atoms; (b) There are eight nearest neighbors, hence a coordination number o f 8; (c) The direction along which atoms touch each other is the body diagonal o f the unit cell. This body diagonal will be composed o f four times the radius o f
the atom. In terms of the unit cell edge length a, the body diagonal will be \/3 a. The unit cell edge length will, therefore, be given by
f~ 4 V 4 -(124 pm) 4/- ~ v 3 ci or ct —■ —=■ = --- p=---_
S 73
86 pm
5.4! (a) a = length o f side for a unit cell; for an fee unit cell, a = 78 r or 272 r = 404 pm.
V = a' = (404 pm x 10~l: m • pm"1)' = 6.59 x 10 nr = 6.59x10 ' e nr .
Because for a fee unit cell there are 4 atoms per unit cell, we have , . , 1 mol A1 atoms mass(e) = 4 A! atoms x --- --- : 6.022 x 10“' atoms • mol 26.98 g__ mol A! atoms = 1.79 x 10 "" g . 1.79x10- g c! = --- -— = 2.72 g • cm ‘ 6.59 x 10-J cm1 4r 4 x 235 pm (b) a = ~j= = >/3 x/3 = 543 pm V = (543 x 1 012 m)- = 1.60x 10“:s m1 = 1.60 x 10 22 enr
There are 2 atoms per bee unit cell;
. _ 1 mol K atoms 39.1 Og
mass(g) = 2 K atoms x --- --- r x ---
6.022 x 1 ()“' atoms • mol mol K atoms
= 1 . 3 0 x 1 0 " " «
1.30 x 10 - g 1.60 x I0""2 cm’
= 0.813 g • cm
5.43 a - length o f unit cell edge
_ mass o f unit cell -
(a)
(1 unit cell)
K = a3 = •