Since all comparisons are in a row, the answers use essentially the same reasoning.

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1 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 1 The keys are done quickly. I do make mistakes. If you think you found an error, please bring it to my attention so I can correct it for others. Thank you. Problem 1 (p 2) What is the total nuclear charge and effective nuclear charge for each of the atoms below? ow does this affect the electron attracting ability of an atom? Z total core electrons = = Li a K B Al i P l Br I e e Ar Z effective = Atoms with higher Z eff tend to hold onto their valence elctrons tighter. Atoms with the same Z eff, but lower valence shells also hold tighter because the valence electrons are closer to the same Z eff e- 2e- 10e- 18e- 2e- 10e- 2e- 10e- 2e- 10e- 2e- 10e- 2e- 10e- 28e- 46e- 0e- 2e- 10e- Problem 2 (p 2) - Which atom in each pair below probably requires more energy to steal away an electron (also called ionization potential IP)? Why? Are any of the comparisons ambiguous? Why? heck your answers with the data in the I.P. table on page 7. a. vs b. vs c. vs d. vs e ince all comparisons are in a row, the answers use essentially the same reasoning. e Both and are in the same row and use the same n=2 valence shell. has a Z eff = +5 and holds onto its electrons tighter than with a Z eff = +4. Both and are in the same row and use the same n=2 valence shell. has a Z eff = +6 and holds onto its electrons tighter than with a Z eff = +5. Both and are in the same row and use the same n=2 valence shell. has a Z eff = +7 and holds onto its electrons tighter than with a Z eff = +6. Both and e are in the same row and use the same n=2 valence shell. e has a Z eff = +8 and holds onto its electrons tighter than with a Z eff = +7. Problem 3 (p 4) Write and atomic configuration for, B,,,,, e,, l, Br, I. hydrogen: 1s 1 boron: 1s 2, 2s 2, 2p 1 carbon: 1s 2, 2s 2, 2p 2 nitrogen: 1s 2, 2s 2, 2p 3 oxygen: 1s 2, 2s 2, 2p 4 fluorine: 1s 2, 2s 2, 2p 5 neon: 1s 2, 2s 2, 2p 6 sulfur: 1s 2, 2s 2, 2p 6, 3s 2, 3p 4 chlorine: 1s 2, 2s 2, 2p 6, 3s 2, 3p 5 bromine: 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 3d 10, 4s 2, 4p 5 Iodine: 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 3d 10, 4s 2, 4p 6, 4d 10, 5s 2, 5p 5

2 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 2 Problem 4 (p 5) - Explain the atomic trends in ionization potential in a row (a vs i vs l) and in a column ( vs l vs Br). These three elements (a vs i vs l) are all in the same row. The Z eff increases across a row (in the same shell) so the rightmost element should hold onto its electron the tightest and have the highest ionization potential (a = 118 > i = 189 > l = 300 all kcal/mole). These three elements ( vs l vs Br) are all in the same column. The Z eff is +7 for all of them. luorine s valence electrons are in the n=2 shell (closest to the +7 Z eff ), chlorine s in the n=3 and bromine s in n=4 (farthest from the +7 Z eff ). Being closer to the Z eff should produce a stronger attraction, thus fluorine has the highest ionization potential ( = 402 > l = 300 > Br = 273 all kcal/mole). Problem 5 (p 6) - Explain the atomic trends in atomic radii in a row (Li vs vs ) and in a column ( vs i vs Ge). Across a row Z eff increases, in the same shell and the valence electrons are held tighter and have smaller radii (Li = 167, = 67, = 42). In a column, each atom has the same Zeff, but there are additional layers, so larger radii are found ( = 67, i = 111, Ge = 125). Valence electrons have been lost in cations so there is a deficiency of electron density. Across a row, the higher positive Z eff should pull in the core electron tighter and produce a smaller radius (Li + = 90, Be +2 = 59 and B +3 = 41). Valence electrons have been added in anions so there is an excess of electron density, which repel one another (found on the right side, in a row). There is greater excess on the nitrogen (with a lower Z eff = +5), than the oxygen, (Z eff = +6), than the fluorine, (Z eff = +7). The nitrogen electron cloud expands the most and has the largest radius ( -3 = 132, -2 = 126 and -- = 119). ation and anion radii in picometers (pm) = m [100 pm = 1 angstrom] Li = 90 Be = 59 B = 41 = = 132 = 126 = a = 116 Mg = 86 Al = 68 i = P = = 170 l = 167 K +1 = 152 a = 114 3d elements -2-1 Ga = 76 Ge = As = e = 184 Br = Rb = 166 r = 132 4d elements In = 94 n = b = Te = 207 I = 206 Problem 6 (p 6) - In the tables above: a. Explain the cation distances compared to the atomic distances. Each time an electron is lost from an element the positive charge goes up and the element holds on tighter to its remaining electrons. Also, electron/electron repulsion goes down as each electron is lost. This should contract the electron cloud more and more. b. Explain the anion distances compared to the atomic distances. Each time an electron is gained to an element the negative charge goes up and there is more and more electron/electron repulsion. This should expand the electron cloud more and more. c. Explain the cation distances in a row (Li = 90A, Be = 59A, B = 41A, A = angstrom). imilar to a. Each time an electron is lost from an element the positive charge goes up and the element holds on tighter to its remaining electrons. Also, electron/electron repulsion goes down as each electron is lost. This should contract the electron cloud more and more. d. Explain the anion distances in a row ( = 132A, = 126A, = 119A, A = angstrom). n the right side, in a row, there is an excess of electrons, which repel each other. There is greater excess on the nitrogen, Z eff = +5, than the oxygen, Z eff = +6, than the fluorine, Z eff = +7. The nitrogen electron cloud expands the most and has the largest radius. -1

3 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 3 Problem 7 (p 7) lassify each bond type below as pure covalent, polar covalent or ionic according to our simplistic guidelines above. If a bond is ionic, rewrite it showing correct charges. a b c d e +2 f Br Mg Br g m pure covalent X = = 0 pure covalent X = = 0.3 ionic polar covalent polar covalent polar covalent X = = 1.8 polar covalent X = = 0.5 X = = 1.0 X = = 1.5 mp = 711 o X = = 0.5 bp = decomposes (bp indicates ionic) h i j k l a a l polar covalent ionic X = = 2.3 polar covalent X = = 1.8 pure covalent pure covalent mp = -84 o mp = 801 o X = = 1.3 X = = 0 X = = 0 bp = +20 o bp = 1413 o (bp indicates molecular) (bp indicates ionic) n a o p q r polar covalent polar covalent polar covalent X 2.6 = = polar covalent polar covalent X = = 0.5 X = = 1.0 X = = 0.5 mp = 318 o bp = 1388 o X = = 0.7 X = = 0.8 (bp indicates ionic) ionic l Br s t u v w Li l a polar covalent X = = 0.5 ionic X = = 2.0 mp = 390 o bp = 430 o (bp indicates ionic) pure covalent pure covalent X = = 0.0 X = = 0.3 ionic X = = 2.1 mp = 210 o bp = 400 o (bp indicates ionic) Problem 8 (p 12) - Predict the formula for the combination of the following pairs of ions. What kinds of melting points would be expected for these salts? K K 2 K 3 K 2 K 3 K 3 P 4 K 2 3 Ba 2 Ba( 2 ) 2 Ba( 3 ) 2 Ba Ba( 3 ) 2 Ba 3 (P 4 ) 2 Ba 3 Zn 2 Zn( 2 ) 2 Zn( 3 ) 2 Zn Zn( 3 ) 2 Zn 3 (P 4 ) 2 Zn 3 Al 3 Al( 2 ) 3 Al( 3 ) 3 Al 2 3 Al( 3 ) 3 AlP 4 Al 2 ( 3 ) 3 All of these salts have very high melting points.

4 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 4 Problem 9 (p 13) upply dipole arrows to any polar bonds above (according to our arbitrary rules). Make sure they point in the right direction. nly the polar bonds have been redrawn below. If a molecule is completely nonpolar, it is not shown Problem 10 (p 19) - The following pairs of molecules have the same formula; they are isomers. Yet, they have different melting points. Match the melting points with the correct structure and provide an explanation for the difference. int: single bonds can rotate and pi bonds tend to be rigid, fixed shapes. 3 a b. mp = -32 o mp = -126 o c. d. mp = -105 o mp = -185 o mp = -95 o mp = +6 o mp = +79 o mp = -88 o Molecules with rigid, symmetrical shapes tend to pack tighter (closer) in their crystal lattice structures. loser contact makes for stronger intermolecular forces of attraction, meaning it is harder to break down the lattice structure, meaning a higher melting point (all other things being equal). In each part the molecules have the same formula (or almost = c) and similar weak dispersion forces. The compound with the more regular shape and higher melting point is the first one in a, b and d, while the second structure has a more regular shape in part c.

5 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 5 Problem 11 (p 23) Provide an explanation for the different boiling points in each column. (µ = dipole moment, X = difference in electronegativities) 4B 5B 6B 7B 4 mp = -182 o bp = -164 o = 0.3 i 4 mp = -185 o bp = -112 o = 0.3 Ge 4 mp = -165 o bp = -88 o = 0.2 n 4 mp = -146 o bp = -52 o = mp = -78 o bp = -33 o = 1.42 D = 0.8 P 3 mp = -132 o bp = -88 o = 0.0 As 3 mp = -111 o bp = -62 o = 0.0 b 3 mp = -88 o bp = -17 o = mp = 0 o bp = +100 o = 1.80 D = mp = -82 o bp = -60 o = 1.0 D = e mp = -66 o bp = -41 o =? D = Te mp = -49 o bp = -2 o =? D = 0.1 mp = -84 o bp = +20 o = 1.86 D = 1.8 l mp = -114 o bp = -85 o = 1.0 D = 1.0 Br mp = -87 o bp = -67 o = 0.8 D = 0.8 I mp = -51 o bp = -35 o = 0.4 D = 0.5 8B e mp = -272 o bp = -269 o = A e mp = -249 o bp = -246 o = A Ar mp = -189 o bp = -185 o = A Kr mp = -157 o bp = -153 o = A Temp ( o ) e row 2 boiling points ( o ) Answer: Each dotted line represents the boiling points in the hydrides in a column in the periodic table, plus the oble gases. In the oble gases there is a continuning trend towards higher boiling point as the main atom gets larger with more and more polarizable electron clouds (stronger dispersion forces). Where bonds to hydrogen become polar, hydrogen bonding becomes important, leading to stronger attractions for neighbor molecules and higher boiling points deviate from the trends observed due to dispersion forces. Deviations are clearly seen with 3, 2 and. Melting points are often less predictable and are not plotted. 2 l P 3 row 3 i 4 e 2 e As 3 Ge 4 row 4 Br Ar 2 Te b 3 I n 4 Kr row 5 2 a b c 3 d e f = 0.08 D = 1.30 D bp = -42 o bp = -22 o = 1.69 D = 1.69 D = 1.68 D mp = -188 o mp = -141 o bp = +65 o bp = +78 o bp = +98 o mp = -98 o mp = -114 o mp = sol. = 40mg/L 2 sol. = 71g/L o 2 sol. = miscible 2 sol. = miscible 2 sol. =? h i g =? D bp = +138 o mp = -78 o 2 sol. = 22g/L =? D bp = +157 o mp = -50 o 2 sol. = 6g/L =? D bp = +195 o mp = -16 o 2 sol. 0 = 1.66 D bp = +118 o mp = -90 o 2 sol. = 73g/L tructure a, b and c are similar in size. tructure c only has dispersion forces, while structure b has dispersion forces and polar bonds, while structure c has dispersion forces, polar bonds and hydrogen bonds. Intermolecular forces of interaction increase in order of dispersion forces < polar bonds < hydrogen bonds, so c has the highest bp, followed by b and lowest is a. ompound d is like c except it has an extra methyl, thus greater dispersion forces, which is similar for e, f, g, h and i. All have increasing boiling points because of their extra dispersion forces (on top of their hydrogen bonds). Problem 12 (p 24) Provide an explanation for the different boiling points in each series. In part b, an - bond is expected to be more polar than an - bond, so why does boil lower than 2? Think of an " bond" donor site as a hand that can grab and a lone pair as a handle that can be held onto. If you only had one hand, it really doesn't make any difference how many handles are available to grab hold of and if you only have one handle, it doesn t make any difference how many hands you have. But, if you have two hands and two handles a much stronger grip is possible with twice the handholds.

6 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 6 a. and b no polarity, only weak dispersion forces ot attraction for neighboring molecules =0D bp = -161 o mp = -182 o 2 hydrogen bonding is possible, but only 2 per molecule on average, also, - is less polar than - =1.47D bp = -33 o mp = -78 o 3 hydrogen bonding is possible, 4 per molecule on average, is our most polar 'typical' bond =1.78D bp = +100 o mp = 0 o 2 2 hydrogen bonding is possible, - is our most polar 'typical' bond, but only 2 per molecule on average, so lower bp than water =1.85D bp = +20 o mp = -84 o c = 0 D bp = -89 o mp = -183 o = 1.47 D bp = -7 o mp = -93 o = 1.78 D bp = +65 o mp = -97 o = 1.85 D bp = -78 o mp = -142 o bonding is the most important factor affecting bp's in these molecules. 3 makes stronger bonds than 3 2 and the other two do not have bonds. While 3 has the most polar bond, fluorine is so electronegative that it does not share its lone pairs much and is not very polarizable. Ethane has no polarity and only has weak dispersion forces to form attractions with neighbor molecules, so has the lowest bp. d. donate bond = 1.2 D bp = +48 o mp = -83 o = 1.0 D bp = +36 o mp =?? o = 0.6 D 3 bp = +3 o mp = -117 o zero 0 bonds More bonds possible on the first structure, so stronger attractions for neighbor molecules and a higher bp. The third structure does not have any polarized protons (-) to hydrogen bond with. accept bond 1 1 1

7 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 7 e = 0.08 D bp = 0 o mp = -136 o = 1.22 D = 1.69 D bp = +97 o bp = +195 o mp = -126 o mp = -13 o bonding possibilities = bonding forms stronger connections to neighbor molecules and requires more energy to remove a molecule from the liquid (higher bp). The third structure can bond at two ends, the second structure at one end and the first structure cannot bond at all. f. 3 = 2.33 D bp = -19 o mp = -92 o = 84 o = 58 o = 26 o = 1.78 D bp = +65 o mp = -97 o 3 = 2.7 D bp = +20 o mp = -123 o = 1.69 D bp = +78 o mp = -114 o = 2.91 D bp = +56 o mp = -95 o = 1.66 D bp = +82 o mp = -89 o In each part we are comparing two similar structures, one with a polar = bond with a polar - bond and bonding possibilities. The alcohols, with bonding, have stronger attractions for the neighbor molecules, thus higher bp's. The differences are larger on the smaller moleules, where the bonding represents a larger fraction of the total interactions. Problem 13 (p 25) Provide an explanation for the different boiling points in each series. a. l Br I b. bp = +85 o bp = +130 o bp = +155 o bp = +188 o imilar molecules. The highest boiling points go with the largest dispersion forces (greatest polarizability, I > Br > l > ) bp = -0.5 o bp = +5 o bp = +78 o tructure 3 has bonding and polarity (strongest attractions), structure 2 has polar bonds and structure 1 only has dispersion forces (weakest attractions) c. bp = -47 o bp = +20 o bp = +101 o 2 bp = +210 o bp = -78 o (sublimes) tructure 1 only has weak dispersion forces, structure 2 had polar bonds (stronger), structure 3 has bonds too (stronger), structure 4 has the strongest polarity and bonds with both - bonds (highest). inally 2 has polar bonds, but its symmetry cancels out its polarity. The oxygen atoms hold so tightly to their electrons that they are less polarizable than structure 1 and sublimes at a lower temperature. d. bp = -42 o bp = -23 o bp = +82 o tructure 1 (propane) only has weak dispersion forces, and has a bent shape that prevents closer contact with neighbor molecules. tructure 2 (propyne) only has weak dispersion forces, but has a linear shape that allows closer contact with neighbor molecules, so stronger attractions. tructure 3 has the shape of propyne, and relatively strong polarity which makes stronger attractions and a higher bp.

8 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 8 e bp = +36 o bp = -89 o 3 bp = +30 o bp = +10 o 3 More carbons have greater contact surface area so stronger dispersion forces (2 < 5). traight chains can approach closer than when branches are present (branches tend to push neighbor molecules away = weaker dispersion forces). More branches weaken those attractions even more (D < < A). e. 4 3 l 2 l 2 l 3 l 4 bp = -164 o bp = -24 o bp = +40 o bp = +61 o bp = +77 o hlorine is a larger atom (than ) and has polarizable lone pairs. More chlorines = more dispersion forces = higher bp. B, and D have polarity too. A and E are not polar, but E has four chlorine atoms and larger dispersion forces dominate here (highest bp). f bp = -42 o bp = +17 o bp = +78 o Bp's follow bonding possibilities. tructure 3 has the strongest bonds because - is more polar. tructure 2 has some bonding possibilitiies so a higher bp than structure 1 which only has dispersion forces. All 3 molecues are similar in size, so comparisons are fair. g hexane, 6 14 cyclohexane, benzene, 6 6 mp = -91 o bp = +69 o mp = +6 o bp = +81 o mp = +5 o bp = +80 o All 3 structures only have weak dispersion forces and are approximately the same size. The rings are more regular in shape and probably pack closer in the solid, thus the much higher mp's (similar to one another) and are also closer in the liquid with no branches wildly swinging around to bump neighbor molecules farther away, thus the higher bp's too (again, rings similar to one another). Problem 14 (p 25) Match the given boiling points with the structures below and give a short reason for your answers. (-7 o, +31 o, +80 o, +141 o, 1420 o ) 2-butanone 2-methyl-1-butene propanoic acid Kl potassium chloride 2-methylpropene MW = 72 g/mol MW = 70 g/mol MW = 74 g/mol MW = 74.5 g/mol MW = 56 g/mol 1420 o +141 o -7 o +80 o +31 o tructure 4 has ionic bonds, by far the strongest forces to overcome and has an extremely high bp. tructure 3 has strong bonding possiblities and the second high bp. tructure 1 has polar = bond so higher than structures 2 and 5, which only have weak dispersion forces. tructure 2 is larger and has a larger surface area and more dispersion forces than structure 5, so has a higher bp.

9 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 9 Problem 15 (p 28) Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DM) that makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use DM (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the difference from the methanol picture. = cations are smaller because they have lost electrons and hold onto the remaining electrons tighter. = anions are larger because they have extra electron density that is repulsive and expands the electron clouds polar Methanol is good at solvating positive and negative charge strong bond dipole DM is very good at solvating positive charge but it is poor at solvaing n egative charge. 3 3 Problem 16 (p 29) a. exane (density = 0.65 g/ml) and water (density = 1.0 g/ml) do not mix. Which layer is on top? b. arbon tetrachloride (density = 1.59 g/ml) and water (density = 1.0 g/ml) do not mix. Which layer is on top? hexane water 3 2 = 1.8 D = 0 D Differences in polarity keep these two from mixing and water is more dense than hexane and sinks to the bottom. b water carbon tetrachloride l l l l = 0 D = 1.8 D Differences in polarity keep these two from mixing and carbon tetrachloride is more dense than water and sinks to the bottom. Problem 17 (p 29) - The melting point of al is very high ( 800 o ) and the boiling point is even higher ( 1400 o ). Does this imply strong, moderate or weak forces of attraction between the ions? onsidering your answer, is it surprising that al dissolves so easily in water? Why does this occur? onsider another chloride salt, Agl. ow does your analysis work here? What changed? Lattic structure - depends on the size and charge of the ions. Each ion is surrounded on many sides by oppositely charged ions. To introduce the disorder of a liquid (melt) or a gas (boil) requires a very large input of energy (mp indicates the amount of energy required to break down the ordered lattice structure and boiling point indicates the amount of energy required to remove an ion pair from the influence of all neighbors. Ionic bonds (ionic attractions on all sides) can only be bronke at great expense in energy.

10 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 10 owever, many water molucules working together can break down this structure by solvating both the cations and anions (solvation energy). When solvation energy > lattice energy, the salt will dissolve (al). When solvation energy < lattice energy the salt will not dissolve (Agl). al Agl lattice energy solvation energy lattice energy solvation energy 2 mall difference between two very large energy values. DG = dissolves DG = does not dissolve Problem 18 (p 29) a. Which solvent do you suspect would dissolve al better, DM or hexane? Explain your choice? b. Which solvent do you suspect would dissolve al better, methanol or benzene? Explain your choice? a b 3 very strong interactions K overall polar much weaker interactions, but still possible. 3 3 exane cannot help dissolve ions at all because it has no polarity. The salt sinks to the bottom (if more dense). 3 3 Methanol is good at solvating positive and negative charge. Benzene is not good at dissolving ions at all because it has no polarity, although its pi electrons can weakly interact with cations. The salt sinks to the bottom (if more dense). Problem 19 (p 30) The terms "hydrophilic" and "hydrophobic" are frequently used to describe structures that mix well or poorly with water, respectively. Biological molecules are often classified in a similar vein as water soluble (hydrophilic) or fat soluble (hydrophobic). The following list of well known biomolecules are often classified as fat soluble or water soluble. Examine each structure and place it in one of these two categories. Explain you reasoning.

11 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 11 a b c vitamin A ydrophobic - mostly nonpolar vitamin B2 (riboflavin) ydrophilic - lots of "" and polarity. P vitamin B6 (pyridoxine) ydrophilic - lots of "" and polarity and charge. d ydrophilic - lots of 2 e f "" and polarity ydrophilic - lots of and charge. "" and polarity. P P P ATP vitamin (ascorbic acid) vitamin D2 ydrophobic - mostly nonpolar Problem 20 (p 31) a. arbohydrates are very water soluble and fats do not mix well with water. Below, glucose is shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point out why each is classified in the manner indicated. = glucose (carbohydrate) typical saturated triglyceride (fat)

12 hem 201/Beauchamp Topic 1: Atoms,Bonding,Dipoles,IP, salvation - Key 12 b. All of the groups in glucose can be methylated. What do you think this will do to the solubility of glucose? Why? ne of these structures is soluble in carbon tetrachloride the other one is not. Which one is it and why? 3 glucose methylation reaction methylated glucose The methylation reaction converts the alcohol groups () to ether groups ( 3 ) and the ability to hydrogen bond goes down, as does water solubility. owever, the ability to dissolve in nonpolar carbon tetrachloride goes up, so B will be much more soluble in l 4. Problem 21 (p 31) Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your body to solubilize the fats, so that they can be absorbed and transported in the aqueous. The major bile salt glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good compromise structure that can mix with both the fat and aqueous. Use the rough 3D drawings below to help your reasoning, or better yet, build models to see the structures for yourself (though it s a lot of work). 3 cholesterol synthesized in many, many steps in the body glycolate (bile salt) nonpolar fats inside Glycolate has a nonpolar, hydrophobic face that can cover the inside of a fat ball and a hydorphilic face that can point outward toward the aqueous, which allows fats to be transported throughout the body to reach fat storage cells and other essential locations