Solutions for Chapter 2 1. Find: Number of valence electrons in Group IIIB and Group VB elements. Data: Periodic Table in Appendix A. Solution: By definition and/or by examination of Appendix B, Group IIIB elements contain 3 valence

electrons while Group VB elements contain 5 valence electrons. 2. Find: The electron configuration of the element that comes next in the series Si, Ge? Data: Periodic Table in Appendix A. Solution: Examination of the Periodic Table shows that Si and Ge are both Group IVB elements. As

shown in Example Problem 2.2-1, have a valence electron configuration of the form xs2xp2 when x=3 for Si and x=4 for Ge. The next group IVB element in the series is Sn with an electron configuration (from the Periodic Table) of [1s22s22p63s23p63d1O4s24p64d1O] 5s25p2. Note that the valence electron configuration for Sn is also of the form xs2xp2 with x=5.

Comments: The similarity of their valence electron configurations suggests that Sn should display properties similar to those of Si and Ge. This is true over a limited temperature range but there are other factors (to be discussed in the next chapter) that explain why Sn also has some properties that differ from those of the other two elements.

3. Find: Should Ca and Zn exhibit similar properties? Data: Periodic Table in Appendix A. Solution: Examination of the Periodic Table shows that Ca has the electron configuration

[1s22s22p63s23p64s2] while Zn has configuration [1s22s22p63s23p63d10] 4s2. Since both elements have two valence electrons (4s2) we should expect these elements to display some similar properties.

Comments: Although Zn and Ca have similar valence electron configurations, they have other structural differences that result in differences in properties.

4. Find: Suggest some consequences of electron energies not being quantized. Solution: Although there are many possible answers to this question, one of the more important results

might be a breakdown in the periodic arrangement of the elements. The Periodic Table owes its existence to the quantization of energy. If quantization of energy did not exist, we would lose the ability to understand and predict properties based on valence electron configuration.

5. Find: How many electrons, protons, and neutrons are in Cu? Data: From Appendix A, the atomic number of Cu is 29 and the atomic mass is 63.54 g/mole. Solution: Cu has 29 electrons and 29 protons, each proton weighing about 1 g/mole. The balance of the

atomic mass is from neutrons. Comment: Elements can have different masses, from having different numbers of neutrons. They are

called isotopes. 6. Find: What is the electronic structure of C? Data: From Appendix A, the atomic number of Cis Solution: Carbon is capable of 4 covalent bonds of equal strength. Just think of some hydrocarbons that

you know, such as methane, CH4. Four H bond to a central C. The bonding of C might be expected to be 1s22s22p2, again from Appendix A. What occurs in practice is called hybridization. The four electrons in the 2s and 2p levels hybridize, giving four electrons of equal bond strength capability. The bonds are as far apart as possible in space (tetrahedral bond angles).

7. Find: Describe the desirable environmental stability of a "gold standard". Assumptions: You want the standard's critical properties to be invariant with time

Solution: Gold is one of the few metals whose pure metallic state is more thermodynamically table than its oxide. Hence, gold does not oxidize. If it did, then it might gain or lose weight with time of exposure to air.

Comments: This is why gold is found in nature as nuggets, whereas, for example, iron and aluminum are found as oxides or sulfides.

8. Find: Can pure water exist at -1°C? Solution: Yes, if the pressure of the system is raised above one atmosphere, water can exist at -1°C.

Comments: Since most of our daily experiences occur at (or near) atmospheric pressure, we tend to forget about pressure as an important system variable. There are, however, many important engineering-processes that occur at either substantially higher or lower pressures.

9. Find: Change in flow rate when molasses is heated from 10°C to 25°C. Given: Activation energy, Q, for Arrhenius Process is 50 kJ/mole. Data: R= 8.314 J/mol-K, K=oC+273 Assumptions: Flow rate at temperature Tis given by F(T)= Foexp(-Q/RT) Solution: The ratio of the flow rates at any two temperatures is: F(T1) = Foexp(-Q/RT1) = exp(-Q/R (1/Tl-1/T2) ) F(T2) Foexp (-Q/RT2) F(25°C) = exp(-50,000J/mol/8.134J/mol-K) (1/298K-1/283K)) = 2.91 F(10°C) A temperature increase of 15°C results in almost a factor of three increase in the flow rate of

molasses. 10. Find: Change in polymerization rate when temperature increases by 10°C. Given: Activation energy, Q, for Arrhenius Process is 80 kJ/mole. Data: R= 8.314 J/mole-K, K= oC+273 Assumption: Polymerization rate at temperature T is given by P(T)= Poexp(-Q/RT) Solution: The ratio of the polymerization rates at any two temperatures is: P(T1) = Poexp (-Q/RT1) = exp(-Q/R(1/T1 -1/T2)) P (T2) Poexp (-Q/RT2) P(T2) Poexp (-Q/RT2) P(T1) = exp [-Q/R ( (T2-T1) /T1T2) ) J = exp [-Q/R (∆T/T1T2) ] P(T2) This form of the expression shows that the problem cannot be solved with the information

given. A knowledge of ∆T is not sufficient. We must also know the two temperatures. Comments: When the temperature increases from 10°C to 20°C, the rate increases by a factor of 3.19. In

contrast, a temperature increase from 40°C to 50°C results in a rate increase of 2.59. This example illustrates the general result that a fixed change in temperature has a greater influence on the reaction rate if the average temperature is low.

11. Find: Why are high quality electronic cable ends or contact points gold-coated? Solution: You do not want the resistance of the connection to increase with time. Oxides are generally

good electrical insulators. Steel points rust and the oxide prevents them from working. Car points, for example, need to be changed frequently.

Comments: In some electronic devices a slight impedance increase due to oxide formation can cause a circuit to fail catastrophically, destroying a number of components.

12. Find: Explanation for man's ability to convert aluminum oxide to aluminum. Given: Oxide represents a lower energy state than pure Al. Solution: Thermodynamics describes the direction of spontaneous change. That is, balls roll down hill

and Al will transform to Al203 if the kinetics are favorable and no other factors are acting on the

system. We know, however, that a ball can be moved uphill if energy is supplied to the system (i.e. if it is carried uphill). Furthermore, it may remain at a higher elevation if there are activation barriers associated with its return to the lowest energy position. Similar logic applies to the reduction of Al203 to 2Al + 1.5 O2. If man supplies (thermal) energy, the metal can be extracted from its ore and will remain in a metastable state. However, the metal will return to its more stable oxide at a later time if conditions permit.

13. Find: Characteristics of primary bonds. Solution: Provided in the form of a table.

Primary bond type

Types of atoms usually involved

Bonding electrons shared or transferred

If sharing occurs, is it localized or delocalized

Ionic Electronegative and electropositive

Transferred -------

Covalent Electronegeative (usually with NVE > 3)

Shared Localized

Metallic Electropositive (usually with NVE ≤ 3)

Shared delocalized

14. Find: Primary bond type in each of a series of compounds. Data: Electronegativities and numbers of valence electrons for each element can be obtained from

the Periodic Table in Appendix A.

Element Electronegativity No. of valence electrons O 3.44 6 Na 0.93 1 F 3.98 7 In 1.78 3 P 2.19 5 Ge 2.01 4 Mg 1.31 2 Ca 1.00 2 Si 1.90 4 C 2.55 4 H 2.20 1

Assumptions: Percent ionic character of a bond is a function of the difference in the electronegativities of the

elements involved (See Appendix A for conversion table) .In a metal, the average number of valence electrons is generally ~ 3. In a covalent solid the average number of valence electrons is generally > 3.

Solution: • 02, ∆EN=O so that the bond is not ionic. Since 0 is an electronegative element and the average number of

valence electrons is 6, we predict a covalent bond. • NaF, ∆EN = EN(F)-EN(Na) = 3.98-0.93 = 3.05. This AEN corresponds to a bond that is approximately

90% ionic. • InP, ∆EN = EN(P)-EN(In) = 2.19-1.78 = 0.41. Since this bond is only about 4% ionic, we must examine the

average number of electrons, N~. Since N~ = (3+5)/2 = 4, we predict that the bond will be covalent.

• Ge, ∆EN=O so the bond is not ionic. Ge is neither strongly electropositive or electronegative, but it does have N~=4. Thus we predict covalent bonding.

• Mg, AEN=O so the bond is not ionic. Mg is an electro- positive element with N~ =2. It will have metallic bonds.

• CaF2, ∆EN = EN(F)-EN(Ca) = 3.98-1.0 = 2.98. This ~EN corresponds to a bond that is appproximately 89% ionic.

• SiC, ∆EN = EN(C)-EN(Si) = 2.55-1.90 = 0.65. Since this corresponds to a bond that is only ~10% ionic, we must consider N~. The average number -of valence electrons is (4+4)/2 = 4 so the bond is predicted to be covalent.

• (CH2), AEN = 2.55-2.20 = 0.35 so the bond is not ionic. Although the average N~ in this compound is (4+1)/2 = 2.5, the bond is predicted to be covalent because His a recognized exception to the trend and is known to favor the formation of covalent bonds.

• MgO, AEN = EN90)-EN(Mg) = 3.44-1.31 = 2.13 which corresponds to a bond that is ~68% ionic. • CaO, AEN = EN(O)-EN(Ca) = 3.44-1.0 = 2.44 which corresponds to a bond ~77% ionic Comments: All of these predictions are in agreement with experimental observations. 15. Find: Show the octet rule in C. Data: Carbon forms four equal bonds. Sketch: Shown is methane, simply as an example:

Solution: The 8 dots represent the electrons: 4 come from the central carbon and 1 from each of the 4 hydrogen. The electrons are localized between the atoms sharing the electrons.

16. Find: Characteristics of the Coulombic Force. Data: Fa(x) is given in Equation 2.4-1 and sketched in Figure 2.4-2(a) . Solution: The Coulombic Force varies inversely with the square of the separation distance between the

charge centers. Another familiar example of an inverse square law is the force of gravity. A function of this form has no local maxima or minima because the derivative of equation 2.4-1 has a nontrivial value for which it is equal to zero and the maximum value occurs as x approaches infinity (for the sign convention adopted in Figure 2.4-2(a)).

17. Find: Definitions of ionization potential and electron affinity. Solution: Ionization potential is the energy required to remove an electron from an isolated neutral atom.

The energy released when an isolated neutral electronegative atom gains an electron is its electron affinity.

Comments: The significance of these variables is that the difference between the ionization potential and the electron affinity is the amount of work that must be done to create a pair of isolated monovalent ions. This quantity is important in the determination of the total bond energy in an ionic bond.

18. Find: Reason why covalent bonds are restricted to electro- negative elements. Assumptions: During bonding, atoms seek to obtain a filled valence electron shell. Solution: Electronegative elements generally have nearly filled valence shells and are seeking a few

additional electrons. If electropositive atoms are nearby, they can transfer electrons to the electronegative atoms and an ionic bond is formed. If, however, only electronegative atoms are present then the only way they can all acquire extra electrons is to share them. This is the definition of a covalent bond.

Comments: Hydrogen also forms covalent bonds. 19. Find: Is the Si-O bond covalent? Given: O forms covalent bonds with itself and Si also forms covalent bonds with other Si atoms. Data: From Appendix B: EN(Si) = 1.90 and EN(O) = 3.44. Assumptions: The percent ionic character of a bond is given in the table in Appendix A.

Solution: It is impossible to form ionic bonds in any pure element since there will be no difference in EN values for identical atoms. In the case of Si-O, however, the difference in electronegativity is 1.54. This corresponds to a bond that is approximately 45% ionic and 55% covalent.

Comments: This type of primary bond is often described as a mixed ionic/covalent bond. 20. Find: Can an ionic solid be a good electrical conductor? Assumptions: High electrical conductivity

requires a high density of mobile charge carriers. Solution: In ionic solids there are usually few, if any, free electrons. Charge transport requires the motion

of comparatively large ions which is generally a more difficult process than electron motion. If, however, an ionic solid had either a high free electron density or extremely mobile ions, then it would be a good electrical conductor.

Comments: We will find some examples of materials with these characteristics in Chapter 10. 21. Find: Physical significance of p<q in Equation 2.4-7. Data: Equation 2.4-7 states: FA+FR = O = A'/xpB'/xq. Solution: This equation describes the competing forces of attraction and repulsion in a covalent bond.

Since the repulsive force is known to dominate at small separation distances (x→0), we require B'/xq » A'/xpas x→0. This is equivalent to requiring that xq « xp (for x→0) which is satisfied if and only if q>p.

22. Find: Types of information available from the bond-energy curve. Solution: The depth of the bond-energy curve describes the strength of the bond (i.e. the bond energy)

and also provides information about the vaporization temperature since it is an indication of the amount of energy that must be supplied to move atoms to an infinite separation distance. The value of x for which the bond energy is a minimum corresponds to the equilibrium separation distance. It is, however, not possible to gain information about the primary bond type from the general shape of the bond energy curve.

23. Find: Identify which material shown in Fig. 2.5-1 has a higher melting temperature. Given: The slope of material A at zero force is greater than that of B. Solution: The ease of separating atoms or molecules with heat is similar to the ease of separating atoms

with force. Hence, A has a higher melting temperature than does B. 24. Find: Relationship between xth and melting temperature. Data: xth (Al) = 25x10-6 °C-1 and xth(SiC) = 4.3 x 10-6 °C-1. Assumptions: Melting temperature is related to the depth of the bond energy curve and thermal expansion

coefficient is related to the asymmetry of the curve. Solution: Since "deep" energy wells tend to be more symmetric, materials with high melting

temperatures tend to have low expansion coefficients. Thus, we expect SiC to have a higher Tm than Al.

Comments: This prediction is consistent with experiment - Tm(SiC) = 2700°C and Tm(Al) = 660°C. 25. Find: Relationship between stiffness and thermal expansion. Assumptions: Stiffness is related to the

curvature (second derivative) of the bond energy curve at its minimum and expansion coefficient is related to symmetry of the curve.

Solution: Bond energy curves that are sharply curved have large second derivatives and high stiffness. However, such curves will also be relatively symmetric so they will exhibit low coefficients of thermal expansion. In order to obtain a stiff material with a high expansion coefficient it would require a tightly curved but asymmetric shape. This combination is difficult to achieve.

26. Find: Explanation for the similarity of the moduli of oxide ceramics. Solution: Elastic modulus is one of the properties that can be determined from the bond energy curve. If

several materials have similar bond characteristics then we should expect those materials to display similar modulus values. This is the case with many oxide ceramics and other ionic solids.

27. Find: Estimate relative deflections of oxide and polymer glasses. Given: E(oxide) = 10 E(polymer) Data: Deflection is inversely proportional to modulus (stiffness). Solution: The material with the higher modulus will deflect less under the same load. Therefore, the

plastic will display 10 times as much deflection as the oxide.

28. Find: Sketch bond curves for a material with Xth < 0. Solution: Sketch

Comments: The bond energy curve is steeper to the right of Xo than to the left. This means that the midpoints of the constant energy line segments shift to the left as energy (temperature) increases.

29. Find: Calculate the atomic separation of pure Ti and Cu at 625°C. Given: The atomic separation at room temperature (25°C) for Ti is 2.94 A and the thermal expansion

coefficient is 9 x 10-6 /°C. Data: The atomic separation of Cu is 1.28 Ax 2 = 2.56 A according to Appendix C. The thermal

expansion coefficient of Cu is 17 x 10-6/°C, according to Table 2.5-1. Solution: The atomic spacing of Ti at 625°C is: 2.94 A + 600°C x (9 x 10-6/ °C) = 2.94 A + 0.0054 A = 2.9454 A. Similarly, the atomic spacing of Cu at 625°C is: 2.56 A + 600°C x (17 x 10-6/ °C) = 2.94 A + 0.0102 A = 2.57 A. Comments: There is a problem with significant digits in this problem. The atomic spacing at 25°C was

provided only to 2 decimal places. Better accuracy is required in the atomic spacings if better accuracy in the calculated results are required.

30. Find: Methods for measuring Young’s Modulus and coefficient of thermal expansion in the lab. Solution: Young’s Modulus relates the stiffness or deflection of a material to the magnitude of the force

or load causing that deflection. The modulus of a material could be measured using either of the methods sketched above, that is, by measuring the deflection of a cantilever beam or tne change in length of an axially leaded rod. Coefficient of thermal expansion is measured by observing the change in length of a sample resulting from a temperature change (see Equation 2.5-4).

31. Find: CNs for ions in CaF2 Given: r(Ca)=0.197nm, r(Ca2+) = 0.106nm, r(F) = 0.06nm, r(F-) = 0.133nm Data: r/F ranges for various CNs given in Table 2.6-1. Solution: Since this is an ionic compound use r(Ca2+)/R(F-) = 0.106nm/0.133nm = 0.797. From Table

2.6-1, this corresponds to CN (Ca2+) = 8. Since the anion:cation ratio is 2:1, the CN(F-) is given by:

CN(F-) = ½[CN(Ca2+)] = 4.

Comments: This prediction is consistent with experiment.

32. Find: CNs of cations in NiO, ZnS, and CsI. Data: Radii in Appendix B→r (Ni2+) = 0.078, r (O2-) = 0.132, r(Zn2+ = 0.083, r(S2-) = 0.165, r(I-) =

0.220 Assumptions: Ionic bonding in all three compounds. This could be checked by investigating ∆EN values. Solution: For NiO: r(Ni2+) / R(O2-) = 0.105/0.132 = 0.795, so that from Table 2.6-1 we can find CN(Ni2+) =

6. For ZnS: r(Zn2+) / R(S2-) = 6. For CsI: r(Cs+) / R(I-) = 0.165/0.220=0.750, so that from Table 2.6-1 we find CN(Cs+) = 8.

33. Find: Sketch the structure of methane and as in Fig. 2.6-5. Sketch: Methane is a covalently-bonded material.

Comments: In methane the C is in the center of the cube and the H are centered on the appropriate cube

corners. The bond angles are 109.5° and the bond lengths are all the same. In the ketone the C is located within an equilateral triangle. The O bond and the other two single bonds are planar, essentially 120° apart.

34. Find: Relationship between bond type and density. Solution: Density depends on the mass and radius of the atoms in the solid and on the efficiency with

which the atoms are packed together. The later factor is a function of bond type through its influence on coordination numbers. Metals tend to have high CNs, typically 8 or 12. Ionic solids typically have CNs ranging from 3 to 8. Covalent solids typically have CNs ranging from 2 to 4. Therefore, if all other factors are roughly equal, covalent solids will display the lowest densities and metals will have the highest densities.

Contents: For covalent solids CN = 8-NVE and NVE # 4. This combination implies that CN(covalent) # 4. 35. Find: When do you use r/R to predict CN in a covalent compound? Solution: Since in covalent compounds CN is determined by the 8-NVE rule, the r/R rule is never used to

predict CN in covalent compounds. 36. Find: Find CX in pure Ge. Given: Ge is a Group IV covalently bonded compound. Solution: In a covalent compound CN=8-NVE. For Group IV Ge, NVE = 4. Therefore, CN(Ge) = 8-4=4. Comment: This prediction is consistent with experiment. 37. Find: Radius range for CN=4. Assumption: Assume ionic bonding Solution: The appropriate figures are sketched in Table 2.6-1. The minimum r/R ratio is found using the

sketch on the left. For this geometry:

r + R = ap3/2 [anions touch cations along ½ a body diagonal]

and

R + R = ap2 [anions touch each other along face diagonals]

Dividing the first equation by the second gives:

(r+R)/2R = 3/2p2 or

r/R = (p3 / p2) – 1 = 0.225 The maximum r/R for CN=4 in the minimum value for CN=6. Using the geometry on the right:

(r + R) = a/2

and

(R + R) = ap2/2 Dividing the first equation by the second gives:

r/R = (2/p2) - 1 = 0.414 Therefore, the radius ratio range for CN=4 is 0.225 ≤ (r/R) < 0.414. 38. Find: Characteristics of an ionic bond. Given: A2B compound with r(A) = 0.12nm, r(B) = 0.15nm, r(B+) = 0.14nm Solution: A. Anions are generally larger than their neutral counterparts because the added electrons

increase electron-electron mutual repulsion and decrease the relative magnitude of the nuclear charge. Therefore, r(A-) > r(A). Using the inverse argument we predict r(B+) < r(B).

B. If the compound is ionic we nust use an nfl ratio to predict the CX of the smaller ion (in this case the anion).

r(A-)/R(B+) = 0.13/0.14= 0.929 From Table 2.6-1, this implies CN(A-} = 8. Since the anion:cation ratio is 2:1, the

coordination number for the cation is predicted to be CN(B+) =2[CN(A-) = 16. A CN of 16, however, is not possible. Therefore, the most likely values are CN(B+) = 12 and CN(A-) = 6. Recall that lower CN values are always possible but are generally not energetically favorable.

39. Find: Relative size of atoms if CN(A) = CN(B) = 12. Solution: A coordination number of 12 for all atoms/ions in a compound suggests that all of the

atoms/ions are the same size. That is r/fl1. 40. Find: Bond characteristics of Si and C. Solution: Both Si and C are Group IV elements and in covalent compounds they will each have CN=4.

Thus we should expect some similarity between C-based (organic) structures and Si—based structures.

Comments: These similarities will be investigated in Chapter 6. 41. Find: Structure of C2H6 Assumptions: This is a covalent compound Solution/Sketch:

H H

Comments: CN(C) = 4 and CN(H) = 1. 42. Find: Explanation for no solids with CN = 5, 7 or 9. Solution: It is geometrically impossible to fill three-dimensional space with the polvhedrons (solid

geometrical figures) that result from CNs of 5, 7, or 9. Comment: To get a feeling~ for this statement you might try to “fill two dimensional space with a series of

equal sized (regular) pentagons or heptagons.

43. Find: Location of tacks in a room to maximize separation distance. Solution: The “tacks” or atoms should be placed in four of the eight corners of the cube (room) such that

no two adjacent corners are occupied. The separation is ao2. 44. Find: Bond angle H-N-H in NH3 Assumptions: Covalent bonding with CN(N) = 3 and CN(H) = 1. Sketch: Solution: The bond angle is close to the tetrahedral angle of 109.5° (see Figure 2.6-4 (a) and associated

text). Comments: See Example Problem 2.6-4. 45. Find: Type of bond between (NH4)

+ and Cl-. Solution: The bonding in (NH4) essentially results in four exposed protons. These H+ ions can form

hydrogen bridges between the electronegative N and Cl ions resulting in comparatively strong secondary bonds.

46. Find: Predict which material has the higher Tm Assumptions: For compounds with secondary bonds, the important factors are the relative strength of the

bonds and the size of the molecules (since larger molecules have a greater surface area over which secondary bonds can occur).

Solutions: A. will have a higher Tm than because the latter is permanent dipole while the former is too symmetric to have strong

secondary bonds. B. will have a lower TV than because the latter is permanent dipole while the former is too symmetric to have strong

secondary bonds. C. Although their structures are very similar, C~4H3o will have a higher T. than Cit, because it is a

larger molecule with more surface for secondary bond formation.

47. Find: Suggest whether natural polymers – amides and cellulose - are moisture sensitive. Given: Amides contain the group and cellulose contains and cellulose contains – OH’s. Solution: N contains a lone pair of electrons. Hence, the N tends to be partially negative and the H

partially positive. O has two lone pairs of electrons, so O is partially negative and the H is partially positive. Hence, both groups are dipolar. Water is also a dipole. Hence, water and NH’s and OH’s are attracted to one another. Cellulose arid amides are moisture sensitive. Their properties - volume, strength, mass, etc. depend on relative humidity.

Comments: This is in part why cotton (cellulose) and wool (a polyamide) are such comfortable fibers. 48. Find: Describe what binds the molecules in a mass of molten polymer. Given: Polymers have high molecular weights. Atoms within each molecule are covalently bonded. Solution: There are several forces that can hold polymer molecules together in a melt. The most

important one is perhaps a secondary bond force. The concerted action of even van der Waals bonds along the length of a polymer can be substantial. Another force that keeps polymer molecules together in the melt are entanglement forces. The molecules are coiled on themselves and one another. It can be difficult, dine consuming and energetically demanding to untangle the molecules.

49. Find: Is the dipole in C≡N stronger than that in C-H? Data: From Appendix A, C has an electronegativity of 2.55, H 2.10, and N 3.05. Solution: N has one lone pair of electrons, so we anticipate its electronegativity to be substantial.

Appendix A confirms this. Hence, C≡N is a strong permanent dipole. C-H is not regarded as a dipole.

50. Find: Compare the bond strength of a permanent dipole to that of a primary ionic bond. Given: A primary bond involves complete transfer of whole charges. A secondary ionic bond may not

involve transfer of a whole charge. Solution: The strength of an individual secondary bond can be significant. In charged polymer solutions,

for example, whole charges can be associated with individual atoms or groups of atoms. There are, however, not as many of these charged species per unit volume as there are in an ionic solid. An example of a charged polymer solution (a polyelectrolyte) is the natural polymer guar, which is made from guar beans and is solubilized in water. Guar gum is used in synthetic, or inexpensive, ice cream. It is the reason that cheap ice cream does not really “melt’.

The concerted action of secondary bonds is generally much weaker than that of the primary bonds. Either the individual bonds are weak and many, or they are strong but few. Secondary bonding is responsible for a number of critical properties, such as lack of a melting temperature in polymers like cellulose, rayon, wool, and silk. One class of polymers, ionomers, has charged groups built into the molecules. They are not present in high density, but they impart interesting rheological and mechanical properties to the polymer. Golf ball covers are made from ionomers.

Coments: When molecules are aligned in one direction in a fiber or film, then the product can be 100 times stronger in one direction than in the other. This is one reason that fiber producers strive to align molecules along the fiber axis.

51. Find: Examples of materials with mixed bonding. Assumptions: Mixed primary-primary bonding occurs when each bond displays characteristics of two of the

primary bond types. Mixed primary-secondary bonding occurs when some of the bonds are primary bonds while other bonds in the compound are secondary bonds.

Solution: Mixed bonding at the primary-primary level can occur to ceramics that are ionic/covalent. This occurs when ∆EN is in the range of 1.2-2.2. Examples include SiO2 (∆EN=1.54), Al2O3 (∆EN = 1.83, ZnO (∆EN 1.79), and many others. Mixed bonding at the primary-secondary level occurs in graphite and in most TP polymers including all of the vinyls arid nylon.

Comments: Mixed primary-primary bonding can also occur in metallic/covalent solids. 52. Find: Discuss the effect of external fields on electrical conductivity. Solution: Both covalent and ionic solids are electrical insulators. Hence, a shift from one type of bonding

to the other will not notably change electrical conductivity in normal circumstances. If, however, the field caused bonding to become more metallic and less covalent, then the electrical conductivity might increase substantially. Such a material might make an excellent switching device.

53. Find: How might an effective dopant change electrical conductivity? Given: Most polymers are inherently insulating. Assumptions: Charge can hop from host to dopant and back again along the length of a sample. Solution: The doping will increase the electrical conductivity of the system by several orders of

magnitude. Comments: Such systems have been researched for many years and are now used commercially. 54 Find: Bond characteristics of Ni3Al Data: From Appendix B: EN(Ni) = 1.91, EN(Al) = 1.61 Solution: Since ∆EN = .030, the table in Appendix A states that this bond is ≈2% ionic. This bond does

not display any significant ionic characteristics. 55. Find: Describe the mechanism of a hair “perm”. Given: Human hair is similar to a crosslinked thermoset polymer with sulfur forming the atomic bridges

between chains. These crosslinks give hair its shape “memory”. Solution: If you want to change the shape of your hair, you must from it into the new shape and then

create additional crosslinks to retain the new shape. The additional crosslinks are formed using a sulfur-containing chemical.

56. Find: Silk, wool, hair, cotton — all natural fibers — do they melt? Given Recall secondary bonding is strong in all these polymers. Solution: A simple experiment is on order. Take a fiber of any or each of the above materials. Light a

match and slowly bring it closer and closer. PET will melt. Do the natural fibers melt? No. They burn, or degrade, before the melting temperature is reached. The strength of the secondary bonds, thanks to strong hydrogen bonding, is sufficient to raise the melting temperature above the degradation temperature.

Comments: If you could break up the secondary bonding, such as by modifying the cellulose with an acetate or nitrate, then

57. Find: Imagine what might occur when a polymerization reactor goes out of control. Given: The reduction of a double bond to a single bond, such as for ethylene, H3C = CH3, liberates a

vast amount of heat. Assumptions: The heat cannot be removed fast enough, so the temperature builds. Solutions: Kaboom! Comments: Many precautions are taken to prevent the temperature from running away. For example, in

one type of reactor, the mer (say ethylene} is in the form of tiny droplets suspended in water. Each droplet is a reaction vessel. If the rate were to become very high in one droplet, then the surrounding water would absorb the liberated heat and the kaboom is avoided. Thus, in today’s modern reactors, runaway polymerization is no longer a significant problem.

55. Find: Should you use PVC or PZ for hot water piping? Solution: The potential problem is that the hot water may degrade the properties of the polymer pipe.

Since the mer in Pe is symmetric there will be only relatively weak secondary bonds between the chains. In contrast tine highly electronegative Cl atom in the PVC mer will result in much stronger secondary bonds. Therefore, PVC pipe would be a better choice for this application.

Comments: Much of the plastic pipe available in hardware stores is fabricated from PVC. 59. Find: Are all plastics recyclable? Solution: The answer to this question depends on the definition of recyclable. It the intent is for the

polymer to be remelted and then formed into the same shape (like glass bottles and aluminum cans), then only TP polymers are recyclable. Thermosets cannot be recycled in this way since they generally char and burn, rather than melt, when reheated. However, most polymers can be reused in a different form. For example, some previously un-recyclable polymers are being reused as additives in concrete and paving applications.

60. Find: How does the stiffness of a rubber change with oxygen or sulfur additions? Solution: Both C and S act as crossiinking agents in rubber. Since increasing the crosslink density

restricts molecular motion, we should expect the stiffness of the rubber to increase as the amount of 0 or S increases.