材料科學基礎第2章習題解答

材料科學基礎第二章習題解答1.第ⅢB族元素有多少個價電子？第ⅤB族有幾個？第ⅢB族元素B（5）、Al（13）、Ga（31）、In（49）、Tl（81）有3個價電子。第ⅤB族N（7）、P（15）、As（33）、Sb（51）、Bi（83）有5個價電子。

1s/2s/2p/3s/3p/4s/3d/4p/5s/4d/5p/6s/4f/5d/6s/……原子結構描述電子運動狀態(tài)的量子數(shù)(n、l、ml、ms)主量子數(shù)n

決定了電子在核外出現(xiàn)概率最大區(qū)域(“電子層”)，離核的遠近及其能量的高低.

n決定能量，n越大，電子的能量越高；n也代表電子離核的平均距離，n越大，電子離核越遠。n相同稱處于同一電子層。n是一個非零的整數(shù)值，不同的n值，對應于不同的電子層１２３４５……..

KLMN

O……..

軌道角動量量子數(shù)描述電子云的不同形狀，也與能量有關

l

n-1,取值0，1，2，3……n-1(亞層）

s,p,d,f……(這些符號均來自光譜學)

軌道角動量量子數(shù)l

軌道角動量量子數(shù)允許的值,lnl1234亞層0000s111p22d3f

s

軌道球形p

軌道啞鈴形d軌道有兩種形狀

l值相同的電子,具有確定的電子云形狀,但在磁場中可以有不同的伸展方向.磁量子數(shù)就是描述電子云在空間的伸展方向

m可取0，±1,±2……±l（|m|

l

）

m值相同的軌道互為等價(簡并)軌道磁量子數(shù)ml磁量子數(shù)的允許值,mllml

軌道數(shù)(2l+1)

0(s)1(p)2(d)3(f)

0

＋10－1

＋2＋10－1－2

＋3＋2＋10－1－2－31357描述電子繞自軸旋轉的狀態(tài)ms取值+1/2和-1/2，分別用↑和↓表示

想象中的電子自旋兩種可能的自旋方向:

正向(+1/2)和反向(-1/2)產生方向相反的磁場相反自旋的一對電子,磁場相互抵消

電子自旋自旋量子數(shù)(ms)四個量子數(shù)與各電子層可能存在的電子運動狀態(tài)數(shù)列于下表。主量子數(shù)電子層原子軌道符號原子軌道數(shù)電子運動狀態(tài)數(shù)n=1K1s12n=2L2s2p138n=3M3s3p3d13518n=4N4s4p4d4f135732nn22n2S亞層□

p亞層d亞層軌道圖：原子軌道能級的順序

鮑林近似能級圖按原子軌道能量高低的順序排列，能級相近的放在一個方框內稱為能級組，共七組。第一能級組：1s

第二能級組：2s,2p

第三能級組：3s,3p

第四能級組：4s,3d,4p

第五能級組：5s,4d,5p

第六能級組：6s,4f,5d,6p

第七能級組：7s,5f,6d,7pl

相同的能級,n

越大,E越高

n

相同的能級,l

越大,E越高

n和l

均變動時,出現(xiàn)能級交錯

4s和3d、5s和4d2.在例題2-1中，我們說明了為什么Si和Ge相似，在周期表這一列中的下一個元素是什么？它的電子構型怎樣？ThenextgroupIVBelementintheseriesisSnwithanelectronconfiguration(fromthePeriodicTable)of[1s22s22p63s23p63d104s24p64d10]5s25p2.

NotethatthevalenceelectronconfigurationforSnisalsooftheformxs2xp2withx=5.原子結構3.你預計Ca和Zn會表現(xiàn)出類似的性質嗎？為什么？

ExaminationofthePeriodicTableshowsthatCahastheelectronconfiguration[1s22s22p63s23p64s2]whileZnhasconfiguration[1s22s22p63s23p63d10]4s2Sincebothelementshavetwovalenceelectrons(4s2)weshouldexpecttheseelementstodisplaysomesimilarproperties.AlthoughZnandCahavesimilarvalenceelectronconfigurations,theyhaveotherstructuraldifferencesthatresultindifferencesinproperties.原子結構4.如果電子的能量不是量子化的，會有哪些后果？

Althoughtherearemanypossibleanswerstothisquestion,oneofthemoreimportantresultsmightbeabreakdownintheperiodicarrangementoftheelements.ThePeriodicTableowesitsexistencetothequantizationofenergy.Ifquantizationofenergydidnotexist,wewouldlosetheabilitytounderstandandpredictpropertiesbasedonvalenceelectronconfiguration.原子結構5.Cu有多少個電子、質子和中子？DATA:FromAppendixA,theatomicnumberofCuis29andtheatomicmassis63.54g/mole.SOLUTION:Cuhas29electronsand29protons,eachprotonweighingabout1g/mole.Thebalanceoftheatomicmassisfromneutrons.COMMENTS:Elementscanhavedifferentmasses,fromhavingdifferentnumbersofneutrons.Theyarecalledisotopes.原子結構6.寫出C的電子結構。C怎樣才能形成4個相等的鍵？共價鍵為高電子密度區(qū)域，因此相互排斥，預計在共價鍵合的C中4個間的鍵合幾何（鍵角）。sp3雜化：由一個ns軌道和3個np軌道混雜形成4個能量相同的sp3雜化軌道，每個軌道含有1/4s成分和3/4p成分，各個sp3雜化軌道之間的夾角為109.5°。原子結構C原子軌道雜化激發(fā)4個sp3軌道7.大多數(shù)金屬的氧化物在能量上都是比純金屬的能量“低”的。氧化的金屬在氧化初期通常增加重量。

對于“金本位制”來講，你需要這種這特性嗎？

ASSUMPTIONS:Youwantthestandard'scriticalpropertiestobeinvariantwithtimeSOLUTION:Goldisoneofthefewmetalswhosepuremetallicstateismorethermodynamicallystablethanitsoxide.Hence,golddoesnotoxidize.Ifitdid,thenitmightgainorloseweightwithtimeofexposuretoair.COMMENTS:Thisiswhygoldisfoundinnatureasnuggets(天然金塊),whereas,forexample,ironandaluminumarefoundasoxidesorsulfides.熱力學與動力學8.在-1℃時能否得到純的液態(tài)水。

Solution:Yes,ifthepressureofthesystemisraisedaboveoneatmosphere,watercanexistat-1℃.Comments:Sincemostofourdailyexperiencesoccurat(ornear)atmosphericpressure,wetendtoforgetaboutpressureasanimportantsystemvariable.Thereare,however,manyimportantengineeringprocessesthatoccurateithersubstantiallyhigherorlowerpressures.熱力學與動力學9.糖漿的流動速率可以描述為激活能約為50KJ/mol的，當溫度從10℃變化到25℃時，流動速率變化多少？

Data:R=8.314J/mol.K,K=

C+273

Assumptions:FlowrateattemperatureTisgivenby F(T)=Foexp(-Q/RT)

Solution:Theratiooftheflowratesatanytwotemperaturesis:F(T1)=Foexp(-Q/RT1)；F(T2)＝Foexp(-Q/RT2)

F(25

C)=exp(-Q/R(1/T1-1/T2))F(10

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

Atemperatureincreaseof15

Cresultsinalmostafactorofthreeincreaseintheflowrateofmolasses（糖蜜）.熱力學與動力學10．為了形成聚合物，許多等同的小分子在化學反應中被連接在一起。聚合反應是放熱的，或者說是產生熱量的，它可描述成阿累尼烏斯過程，激活能約為80KJ/mol。如果溫度增加10℃，反應速率變化多少？Data:R=8.314J/mole-K,K=

C+273Assumption: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.Aknowledgeof

Tisnotsufficient.Wemustalsoknowthetwotemperatures.Comments:Whenthetemperatureincreasesfrom10

Cto20

C,therateincreasesbyafactorof3.19.Incontrast,atemperatureincreasefrom40

Cto50

Cresultsinarateincreaseof2.59.Thisexampleillustratesthegeneralresultthatafixedchangeintemperaturehasagreaterinfluenceonthereactionrateiftheaveragetemperatureislow.熱力學與動力學11．為什么高質量的電子導線端點要用金而不是鋼、鋁或銅制成？SOLUTION:Youdonotwanttheresistanceoftheconnectiontoincreasewithtime.Oxidesaregenerallygoodelectricalinsulators（絕緣體）.Steelpointsrustandtheoxidepreventsthemfromworking.COMMENTS:Insomeelectronicdevicesaslightimpedance（電阻，阻抗）increaseduetooxideformationcancauseacircuittofailcatastrophically,destroyinganumberofcomponents.熱力學與動力學12．將Al2O3還原為2Al和（1.5O2）是一個需要越過能障的過程。由于熱力學上有利于氧化物而不是元素的分離，怎樣才能將鋁礦石還原為純鋁？Given:OxiderepresentsalowerenergystatethanpureAl.Solution:Thermodynamicsdescribesthedirectionofspontaneouschange.Thatis,ballsrolldownhillandAlwilltransformtoAl2O3ifthekineticsarefavorableandnootherfactorsareactingonthesystem.Weknow,however,thataballcanbemoveduphillifenergyissuppliedtothesystem(i.e.ifitiscarrieduphill).Furthermore,itmayremainatahigherelevationifthereareactivationbarriersassociatedwithitsreturntothelowestenergyposition.SimilarlogicappliestothereductionofAl2O3to2Al+1.5O2.Ifmansupplies(thermal)energy,themetalcanbeextractedfromitsoreandwillremaininametastablestate.However,themetalwillreturntoitsmorestableoxideatalatertimeifconditionspermit.熱力學與動力學13．對于每種一次鍵類型，回答下面的問題：

1）這種一次鍵通常包含負電性原子、正電性原子或兩種原子嗎？

2）成鍵電子是共用的還是轉移的？

3）如果成鍵電子是共用的，它們在空間上是定域的還是不定域的？一次鍵

PrimarybondtypeTypesofatomsusuallyinvolved

BondingelectronssharedortransferredIfsharingoccursisitlocalizedordelocalizedionicelectroneg.andelectropositive

transferred

covalentelectronegativesharedlocalizedmetallicelectropositiveshared

delocalized14.確定下列每種材料中最可能的一次鍵類型：O2,NaF,InP,Ge,Mg,CaF2,SiC,(CH2)n,MgO,CaO

一次鍵ElementElectronegativityNo.ofvalenceelectronsO3.446Na0.931F3.987In1.783P2.195Ge2.014Mg1.312Ca1.002Si1.904C2.554H2.201O2,

EN=0sothatthebondisnotionic.SinceOisanelectronegativeelementandtheaveragenumberofvalenceelectronsis6,wepredictacovalentbond.NaF,

EN=EN(F)-EN(Na)=3.98-0.93=3.05.This

ENcorrespondstoabondthatisapproximately90%ionic.InP,

EN=EN(P)-EN(In)=2.19-1.78=0.41.Sincethisbondisonlyabout4%ionic,wemustexaminetheaveragenumberofelectrons,NVE.SinceNVE=(3+5)/2=4,wepredictthatthebondwillbecovalent.Ge,

EN=0sothebondisnotionic.Geisneitherstronglyelectropositiveorelectronegative,butitdoeshaveNVE=4.Thuswepredictcovalentbonding.Mg,

EN=0sothebondisnotionic.MgisanelectropositiveelementwithNVE=2.Itwillhavemetallicbonds.CaF2,

EN=EN(F)-EN(Ca)=3.98-1.0=2.98.This

ENcorrespondstoabondthatisapproximately89%ionic.SiC,

EN=EN(C)-EN(Si)=2.55-1.90=0.65.Sincethiscorrespondstoabondthatisonly

10%ionic,wemustconsiderNVE.Theaveragenumberofvalenceelectronsis(4+4)/2=4sothebondispredictedtobecovalent.(CH2),

EN=2.55-2.20=0.35sothebondisnotionic.AlthoughtheaverageNVEinthiscompoundis(4+1)/2=2.5,thebondispredictedtobecovalentbecauseHisarecognizedexceptiontothetrendandisknowntofavortheformationofcovalentbonds.MgO,

EN=EN9O)-EN(Mg)=3.44-1.31=2.13whichcorrespondstoabondthatis

68%ionic.CaO,

EN=EN(O)-EN(Ca)=3.44-1.0=2.44whichcorrespondstoabondthatis

77%ionic.15.說明C中4個未成對電子怎樣形成共價鍵，即說明電子的共用及“全填滿價電層規(guī)”的應用。

DATA:Carbonformsfourequalbonds.

SKETCH:Shownismethane（甲烷）,simplyasanexample:

SOLUTION:The8dotsrepresenttheelectrons:4comefromthecentralcarbonand1fromeachofthe4hydrogen.Theelectronsarelocalizedbetweentheatomssharingtheelectrons.一次鍵16：庫倫力是怎樣隨電荷間的距離而變化的？哪種熟知的力也表現(xiàn)出這一特點？在怎樣的電荷間隔下dF(庫侖)/dx最大和最??？Solution:TheCoulombicForcevariesinverselywiththesquareoftheseparationdistancebetweenthechargecenters.Anotherfamiliarexampleofaninversesquarelawistheforceofgravity.Afunctionofthisformhasnolocalmaximaorminimabecausethederivativeofequationhasanontrivialvaluesforwhichitisequaltozeroandthemaximumvalueoccursasxapproachesinfinity.一次鍵18：為什么共價鍵只在負電性元素間形成？Assumptions:Duringbonding,atomsseektoobtainafilledvalenceelectronshell.Solution:

Electronegativeelementsgenerallyhavenearlyfilledvalenceshellsandareseekingafewadditionalelectrons.Ifelectropositiveatomsarenearby,theycantransferelectronstotheelectronegativeatomsandanionicbondisformed.If,however,onlyelectronegativeatomsarepresentthentheonlywaytheycanallacquireextraelectronsistosharethem.Thisisthedefinitionofacovalentbond.Comments:Hydrogenalsoformscovalentbonds.一次鍵19：若O與其本身形成共價鍵，Si也與其本身形成共價鍵，那么，認為SiO成共價鍵是合理的，對嗎？Data:FromAppendixB:EN(Si)=1.90andEN(O)=3.44.Assumptions:ThepercentioniccharacterofabondisgivenbythetableinAppendixA.Solution:ItisimpossibletoformionicbondsinanypureelementsincetherewillbenodifferenceinENvaluesforidenticalatoms.InthecaseofSi-O,however,thedifferenceinelectronegativityis1.54.Thiscorrespondstoabondthatisapproximately45%ionicand55%covalent.Comments:Thistypeofprimarybondisoftendescribedasamixedionic/covalentbond.一次鍵20．陶瓷或離子固體有可能成為適當?shù)膶щ婓w嗎？需要怎樣才有可能？Assumptions:Highelectricalconductivityrequiresahighdensityofmobilechargecarriers.Solution:Inionicsolidsthereareusuallyfew,ifany,freeelectrons.Chargetransportrequiresthemotionofcomparativelylargeionswhichisgenerallyamoredifficultprocessthanelectronmotion.If,however,anionicsolidhadeitherahighfreeelectrondensityorextremelymobileions,thenitwouldbeagoodelectricalconductor.

Comments:WewillfindsomeexamplesofmaterialswiththesecharacteristicsinChapter10.一次鍵21．解釋方程2-8中指數(shù)必須遵從不等式p<q這一要求的物理

意義。Data:Equation2-8states:FA+FR=0=A'/xp-B'/xq.Solution:Thisequationdescribesthecompetingforcesofattractionandrepulsioninacovalentbond.Sincetherepulsiveforceisknowntodominateatsmallseparationdistances(x

0),werequireB'/xq>>A'/xpasx

0.Thisisequivalenttorequiringthatxq<<xp(forx

0)whichissatisfiedifandonlyifq>p.一次鍵22．鍵-能曲線可以用來獲得有關下列四種材料性能中的三種信息，哪種信息不能得到？

1）鍵能

2）平衡間距

3）一次鍵類型

4）汽化溫度鍵能曲線Thedepthofthebond-energycurvedescribesthestrengthofthebond(i.e.thebondenergy)andalsoprovidesinformationaboutthevaporizationtemperaturesinceitisanindicationoftheamountofenergythatmustbesuppliedtomoveatomstoaninfiniteseparationdistance.Thevalueofxforwhichthebondenergyisaminimumcorrespondstotheequilibriumseparationdistance.Itis,however,notpossibletogaininformationabouttheprimarybondtypefromthegeneralshapeofthebondenergycurve.23．圖2-12中所示的兩種材料中，你預計哪一個將有較高的

熔點？GIVEN:TheslopeofmaterialAatzeroforceisgreaterthanthatofB.SOLUTION:Theeaseofseparatingatomsormoleculeswithheatissimilartotheeaseofseparatingatomswithforce.Hence,AhasahighermeltingtemperaturethandoesB.鍵能曲線24.給定Al的膨脹系數(shù)為25×10-6℃-1，SiC為4.3×10-6℃-1，

預計哪種材料具有較高的熔點？鍵能曲線Assumptions:Meltingtemperatureisrelatedtothedepthofthebondenergycurveandthermalexpansioncoefficientisrelatedtotheasymmetryofthecurve.Solution:Since"deep"energywellstendtobemoresymmetric,materialswithhighmeltingtempera-turestendtohavelowexpansioncoefficients.Thus,weexpectSiCtohaveahigherTmthanAl.Comments:Th