材料科学基础课后习题答案2.docx

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材料科学基础课后习题答案2.docx

材料科学基础课后习题答案2

SolutionsforChapter2

1.Find:

NumberofvalenceelectronsinGroupIIIBandGroup

VBelements.

Data:

PeriodicTableinAppendixA.

Solution:

Bydefinitionand/orbyexaminationofAppendix

B,GroupIIIBelementscontain3valenceelectronswhileGroupVBelementscontain5valenceelectrons.

2.Find:

Theelectronconfigurationoftheelementthatcomes

nextintheseriesSi,Ge,?

Data:

PeriodicTableinAppendixA.

Solution:

ExaminationofthePeriodicTableshowsthatSi

andGearebothGroupIVBelements.AsshowninExampleProblem2.2-1,haveavalenceelectronconfigurationoftheformxs2xp2whenx=3forSiandx=4forGe.ThenextgroupIVBelementintheseriesisSnwithanelectronconfiguration（fromthePeriodicTable）of[1s22s22p63s23p63d104s24p64d10]5s25p2.NotethatthevalenceelectronconfigurationforSnisalsooftheformxs2xp2withx=5.

Comments:

Thesimilarityoftheirvalenceelectron

configurationssuggeststhatSnshoulddisplaypropertiessimilartothoseofSiandGe.Thisistrueoveralimitedtemperaturerangebutthereareotherfactors（tobediscussedinthenextchapter）thatexplainwhySnalsohassomepropertiesthatdifferfromthoseoftheothertwoelements.

3.Find:

ShouldCaandZnexhibitsimilarproperties?

Data:

PeriodicTableinAppendixA.

Solution:

ExaminationofthePeriodicTableshowsthatCa

hastheelectronconfiguration[1s22s22p63s23p64s2]whileZnhasconfiguration[1s22s22p63s23p63d10]4s2.Sincebothelementshavetwovalenceelectrons（4s2）weshouldexpecttheseelementstodisplaysomesimilarproperties.

Comments:

AlthoughZnandCahavesimilarvalenceelectron

configurations,theyhaveotherstructuraldifferencesthatresultindifferencesinproperties.

4.Find:

Suggestsomeconsequencesofelectronenergiesnotbeingquantized.

Solution:

Althoughtherearemanypossibleanswerstothis

question,oneofthemoreimportantresultsmightbeabreakdownintheperiodicarrangementoftheelements.ThePeriodicTableowesitsexistencetothequantizationofenergy.Ifquantizationofenergydidnotexist,wewouldlosetheabilitytounderstandandpredictpropertiesbasedonvalenceelectronconfiguration.

5.FIND:

Howmanyelectrons,protons,andneutronsareinCu?

DATA:

FromAppendixA,theatomicnumberofCuis29andtheatomicmassis63.54g/mole.

SOLUTION:

Cuhas29electronsand29protons,eachprotonweighingabout1g/mole.Thebalanceoftheatomicmassisfromneutrons.

COMMENTS:

Elementscanhavedifferentmasses,fromhavingdifferentnumbersofneutrons.Theyarecalledisotopes.

6.FIND:

WhatistheelectronicstructureofC?

DATA:

FromAppendixA,theatomicnumberofCis

SOLUTION:

Carboniscapableof4covalentbondsofequalstrength.Justthinkofsomehydrocarbonsthatyouknow,suchasmethane,CH4.FourHbondtoacentralC.ThebondingofCmightbeexpectedtobe1s22s22p2,againfromAppendixA.Whatoccursinpracticeiscalledhybridization.Thefourelectronsinthe2sand2plevelshybridize,givingfourelectronsofequalbondstrengthcapability.Thebondsareasfarapartaspossibleinspace（tetrahedralbondangles）.

7.FIND:

Describethedesirableenvironmentalstabilityofa"goldstandard".

ASSUMPTIONS:

Youwantthestandard'scriticalpropertiestobeinvariantwithtime

SOLUTION:

Goldisoneofthefewmetalswhosepuremetallicstateismorethermodynamicallytablethanitsoxide.Hence,golddoesnotoxidize.Ifitdid,thenitmightgainorloseweightwithtimeofexposuretoair.

COMMENTS:

Thisiswhygoldisfoundinnatureasnuggets,whereas,forexample,ironandaluminumarefoundasoxidesorsulfides.

8.Find:

Canpurewaterexistat-1oC?

Solution:

Yes,ifthepressureofthesystemisraised

aboveoneatmosphere,watercanexistat-1oC.

Comments:

Sincemostofourdailyexperiencesoccurat（or

near）atmosphericpressure,wetendtoforgetaboutpressureasanimportantsystemvariable.Thereare,however,manyimportantengineeringprocessesthatoccurateithersubstantiallyhigherorlowerpressures.

9.Find:

Changeinflowratewhenmolassesisheatedfrom10oC

to25oC.

Given:

Activationenergy,Q,forArrheniusProcessis

50kJ/mole.

Data:

R=8.314J/mol-K,K=oC+273

Assumptions:

FlowrateattemperatureTisgivenby

F（T）=Foexp（-Q/RT）

Solution:

Theratiooftheflowratesatanytwo

temperaturesis:

F（T1）=Foexp（-Q/RT1）=exp（-Q/R（1/T1-1/T2））

F（T2）Foexp（-Q/RT2）

F（25C）=exp（-50,000J/mol/8.134J/mol-K）（1/298K-1/283K））=2.91

F（10oC）

Atemperatureincreaseof15oCresultsinalmostafactorofthreeincreaseintheflowrateofmolasses.

10.

Find:

Changeinpolymerizationratewhentemperature

increasesby10oC.

Given:

Activationenergy,Q,forArrheniusProcessis

80kJ/mole.

Data:

R=8.314J/mole-K,K=oC+273

Assumption:

PolymerizationrateattemperatureTisgiven

byP（T）=Poexp（-Q/RT）

Solution:

Theratioofthepolymerizationratesatanytwo

temperaturesis:

P（T1）=Poexp（-Q/RT1）=exp（-Q/R（1/T1-1/T2））

P（T2）Poexp（-Q/RT2）

P（T1）=exp[-Q/R（（T2-T1）/T1T2））]=exp[-Q/R（T/T1T2）]

P（T2）

Thisformoftheexpressionshowsthattheproblemcannotbesolvedwiththeinformationgiven.AknowledgeofTisnotsufficient.Wemustalsoknowthetwotemperatures.

Comments:

Whenthetemperatureincreasesfrom10oCto

20oC,therateincreasesbyafactorof3.19.Incontrast,atemperatureincreasefrom40oCto50oCresultsinarateincreaseof2.59.Thisexampleillustratesthegeneralresultthatafixedchangeintemperaturehasagreaterinfluenceonthereactionrateiftheaveragetemperatureislow.

11.

FIND:

Whyarehighqualityelectroniccableendsorcontactpointsgold-coated?

SOLUTION:

Youdonotwanttheresistanceoftheconnectiontoincreasewithtime.Oxidesaregenerallygoodelectricalinsulators.Steelpointsrustandtheoxidepreventsthemfromworking.Carpoints,forexample,needtobechangedfrequently.

COMMENTS:

Insomeelectronicdevicesaslightimpedanceincreaseduetooxideformationcancauseacircuittofailcatastrophically,destroyinganumberofcomponents.

12.

Find:

Explanationforman'sabilitytoconvertaluminum

oxidetoaluminum.

Given:

OxiderepresentsalowerenergystatethanpureAl.

Solution:

Thermodynamicsdescribesthedirectionof

spontaneouschange.Thatis,ballsrolldownhillandAlwilltransformtoAl2O3ifthekineticsarefavorableandnootherfactorsareactingonthesystem.Weknow,however,thataballcanbemoveduphillifenergyissuppliedtothesystem（i.e.ifitiscarrieduphill）.Furthermore,itmayremainatahigherelevationifthereareactiva-tionbarriersassociatedwithitsreturntothelowestenergyposition.SimilarlogicappliestothereductionofAl2O3to2Al+1.5O2.Ifmansupplies（thermal）energy,themetalcanbeextractedfromitsoreandwillremaininametastablestate.However,themetalwillreturntoitsmorestableoxideatalatertimeifconditionspermit.

13.

Find:

Characteristicsofprimarybonds.

Solution:

Providedintheformofatable.

+----------------------------------------------------------+

||involved|transferred|delocalized|

+----------+-----------------+--------------+--------------|

|ionic|electroneg.and|transferred|---------|

||electropositive|||

+----------+-----------------+--------------+--------------|

||（usuallywNVE>3）||sharing|

+----------+-----------------+--------------+--------------|

||（usuallywNVE3）||sharing|

+----------------------------------------------------------+

14.

Find:

Primarybondtypeineachofaseriesofcompounds.

Data:

Electronegativitiesandnumbersofvalenceelectrons

foreachelementcanbeobtainedfromthePeriodicTableinAppendixA.

Element|Electronegativity|No.ofvalenceelectrons

--------+-------------------+-------------------------

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:

Percentioniccharacterofabondisa

functionofthedifferenceintheelectro-negativitiesoftheelementsinvolved（SeeAppendixAforconversiontable）.Inametal,theaveragenumberofvalenceelectronsisgenerally3.Inacovalentsolidtheaveragenumberofvalenceelectronsisgenerally>3.

Solution:

•O2,EN=0sothatthebondisnotionic.SinceOisan

electronegativeelementandtheaveragenumberofvalenceelectronsis6,wepredictacovalentbond.

•NaF,EN=EN（F）-EN（Na）=3.98-0.93=3.05.ThisEN

correspondstoabondthatisapproximately90%ionic.

•InP,EN=EN（P）-EN（In）=2.19-1.78=0.41.Sincethis

bondisonlyabout4%ionic,wemustexaminetheaveragenumberofelectrons,NVE.SinceNVE=（3+5）/2=4,wepredictthatthebondwillbecovalent.

•Ge,EN=0sothebondisnotionic.Geisneither

stronglyelectropositiveorelectronegative,butitdoeshaveNVE=4.Thuswepredictcovalentbonding.

•Mg,EN=0sothebondisnotionic.Mgisanelectro-

positiveelementwithNVE=2.Itwillhavemetallicbonds.

•CaF2,EN=EN（F）-EN（Ca）=3.98-1.0=2.98.ThisEN

correspondstoabondthatisappproximately89%ionic.

•SiC,EN=EN（C）-EN（Si）=2.55-1.90=0.65.Sincethis

correspondstoabondthatisonly10%ionic,wemustconsiderNVE.Theaveragenumberofvalenceelectronsis（4+4）/2=4sothebondispredictedtobe

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