材料科学基础第2章习题解答

材料科学基础第二章习题解答1.第ⅢB族元素有多少个价电子？第ⅤB族有几个？第ⅢB族元素B（5）、Al（13）、Ga（31）、In（49）、Tl（81）有3个价电子。第ⅤB族N（7）、P（15）、As（33）、Sb（51）、Bi（83）有5个价电子。

1s/2s/2p/3s/3p/4s/3d/4p/5s/4d/5p/6s/4f/5d/6s/……原子结构描述电子运动状态的量子数(n、l、ml、ms)主量子数n

决定了电子在核外出现概率最大区域(“电子层”)，离核的远近及其能量的高低.

n决定能量，n越大，电子的能量越高；n也代表电子离核的平均距离，n越大，电子离核越远。n相同称处于同一电子层。n是一个非零的整数值，不同的n值，对应于不同的电子层１２３４５……..

KLMN

O……..

轨道角动量量子数描述电子云的不同形状，也与能量有关

l

n-1,取值0，1，2，3……n-1(亚层）

s,p,d,f……(这些符号均来自光谱学)

轨道角动量量子数l

轨道角动量量子数允许的值,lnl1234亚层0000s111p22d3f

s

轨道球形p

轨道哑铃形d轨道有两种形状

l值相同的电子,具有确定的电子云形状,但在磁场中可以有不同的伸展方向.磁量子数就是描述电子云在空间的伸展方向

m可取0，±1,±2……±l（|m|

l

）

m值相同的轨道互为等价(简并)轨道磁量子数ml磁量子数的允许值,mllml

轨道数(2l+1)

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

0

＋10－1

＋2＋10－1－2

＋3＋2＋10－1－2－31357描述电子绕自轴旋转的状态ms取值+1/2和-1/2，分别用↑和↓表示

想象中的电子自旋两种可能的自旋方向:

正向(+1/2)和反向(-1/2)产生方向相反的磁场相反自旋的一对电子,磁场相互抵消

电子自旋自旋量子数(ms)四个量子数与各电子层可能存在的电子运动状态数列于下表。主量子数电子层原子轨道符号原子轨道数电子运动状态数n=1K1s12n=2L2s2p138n=3M3s3p3d13518n=4N4s4p4d4f135732nn22n2S亚层□

p亚层d亚层轨道图：原子轨道能级的顺序

鲍林近似能级图按原子轨道能量高低的顺序排列，能级相近的放在一个方框内称为能级组，共七组。第一能级组：1s

第二能级组：2s,2p

第三能级组：3s,3p

第四能级组：4s,3d,4p

第五能级组：5s,4d,5p

第六能级组：6s,4f,5d,6p

第七能级组：7s,5f,6d,7pl

相同的能级,n

越大,E越高

n

相同的能级,l

越大,E越高

n和l

均变动时,出现能级交错

4s和3d、5s和4d2.在例题2-1中，我们说明了为什么Si和Ge相似，在周期表这一列中的下一个元素是什么？它的电子构型怎样？ThenextgroupIVBelementintheseriesisSnwithanelectronconfiguration(fromthePeriodicTable)of[1s22s22p63s23p63d104s24p64d10]5s25p2.

NotethatthevalenceelectronconfigurationforSnisalsooftheformxs2xp2withx=5.原子结构3.你预计Ca和Zn会表现出类似的性质吗？为什么？

ExaminationofthePeriodicTableshowsthatCahastheelectronconfiguration[1s22s22p63s23p64s2]whileZnhasconfiguration[1s22s22p63s23p63d10]4s2Sincebothelementshavetwovalenceelectrons(4s2)weshouldexpecttheseelementstodisplaysomesimilarproperties.AlthoughZnandCahavesimilarvalenceelectronconfigurations,theyhaveotherstructuraldifferencesthatresultindifferencesinproperties.原子结构4.如果电子的能量不是量子化的，会有哪些后果？

Althoughtherearemanypossibleanswerstothisquestion,oneofthemoreimportantresultsmightbeabreakdownintheperiodicarrangementoftheelements.ThePeriodicTableowesitsexistencetothequantizationofenergy.Ifquantizationofenergydidnotexist,wewouldlosetheabilitytounderstandandpredictpropertiesbasedonvalenceelectronconfiguration.原子结构5.Cu有多少个电子、质子和中子？DATA:FromAppendixA,theatomicnumberofCuis29andtheatomicmassis63.54g/mole.SOLUTION:Cuhas29electronsand29protons,eachprotonweighingabout1g/mole.Thebalanceoftheatomicmassisfromneutrons.COMMENTS:Elementscanhavedifferentmasses,fromhavingdifferentnumbersofneutrons.Theyarecalledisotopes.原子结构6.写出C的电子结构。C怎样才能形成4个相等的键？共价键为高电子密度区域，因此相互排斥，预计在共价键合的C中4个间的键合几何（键角）。sp3杂化：由一个ns轨道和3个np轨道混杂形成4个能量相同的sp3杂化轨道，每个轨道含有1/4s成分和3/4p成分，各个sp3杂化轨道之间的夹角为109.5°。原子结构C原子轨道杂化激发4个sp3轨道7.大多数金属的氧化物在能量上都是比纯金属的能量“低”的。氧化的金属在氧化初期通常增加重量。

对于“金本位制”来讲，你需要这种这特性吗？

ASSUMPTIONS:Youwantthestandard'scriticalpropertiestobeinvariantwithtimeSOLUTION:Goldisoneofthefewmetalswhosepuremetallicstateismorethermodynamicallystablethanitsoxide.Hence,golddoesnotoxidize.Ifitdid,thenitmightgainorloseweightwithtimeofexposuretoair.COMMENTS:Thisiswhygoldisfoundinnatureasnuggets(天然金块),whereas,forexample,ironandaluminumarefoundasoxidesorsulfides.热力学与动力学8.在-1℃时能否得到纯的液态水。

Solution:Yes,ifthepressureofthesystemisraisedaboveoneatmosphere,watercanexistat-1℃.Comments:Sincemostofourdailyexperiencesoccurat(ornear)atmosphericpressure,wetendtoforgetaboutpressureasanimportantsystemvariable.Thereare,however,manyimportantengineeringprocessesthatoccurateithersubstantiallyhigherorlowerpressures.热力学与动力学9.糖浆的流动速率可以描述为激活能约为50KJ/mol的，当温度从10℃变化到25℃时，流动速率变化多少？

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）.热力学与动力学10．为了形成聚合物，许多等同的小分子在化学反应中被连接在一起。聚合反应是放热的，或者说是产生热量的，它可描述成阿累尼乌斯过程，激活能约为80KJ/mol。如果温度增加10℃，反应速率变化多少？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.热力学与动力学11．为什么高质量的电子导线端点要用金而不是钢、铝或铜制成？SOLUTION:Youdonotwanttheresistanceoftheconnectiontoincreasewithtime.Oxidesaregenerallygoodelectricalinsulators（绝缘体）.Steelpointsrustandtheoxidepreventsthemfromworking.COMMENTS:Insomeelectronicdevicesaslightimpedance（电阻，阻抗）increaseduetooxideformationcancauseacircuittofailcatastrophically,destroyinganumberofcomponents.热力学与动力学12．将Al2O3还原为2Al和（1.5O2）是一个需要越过能障的过程。由于热力学上有利于氧化物而不是元素的分离，怎样才能将铝矿石还原为纯铝？Given:OxiderepresentsalowerenergystatethanpureAl.Solution:Thermodynamicsdescribesthedirectionofspontaneouschange.Thatis,ballsrolldownhillandAlwilltransformtoAl2O3ifthekineticsarefavorableandnootherfactorsareactingonthesystem.Weknow,however,thataballcanbemoveduphillifenergyissuppliedtothesystem(i.e.ifitiscarrieduphill).Furthermore,itmayremainatahigherelevationifthereareactivationbarriersassociatedwithitsreturntothelowestenergyposition.SimilarlogicappliestothereductionofAl2O3to2Al+1.5O2.Ifmansupplies(thermal)energy,themetalcanbeextractedfromitsoreandwillremaininametastablestate.However,themetalwillreturntoitsmorestableoxideatalatertimeifconditionspermit.热力学与动力学13．对于每种一次键类型，回答下面的问题：

1）这种一次键通常包含负电性原子、正电性原子或两种原子吗？

2）成键电子是共用的还是转移的？

3）如果成键电子是共用的，它们在空间上是定域的还是不定域的？一次键

PrimarybondtypeTypesofatomsusuallyinvolved

BondingelectronssharedortransferredIfsharingoccursisitlocalizedordelocalizedionicelectroneg.andelectropositive

transferred

covalentelectronegativesharedlocalizedmetallicelectropositiveshared

delocalized14.确定下列每种材料中最可能的一次键类型：O2,NaF,InP,Ge,Mg,CaF2,SiC,(CH2)n,MgO,CaO

一次键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.说明C中4个未成对电子怎样形成共价键，即说明电子的共用及“全填满价电层规”的应用。

DATA:Carbonformsfourequalbonds.

SKETCH:Shownismethane（甲烷）,simplyasanexample:

SOLUTION:The8dotsrepresenttheelectrons:4comefromthecentralcarbonand1fromeachofthe4hydrogen.Theelectronsarelocalizedbetweentheatomssharingtheelectrons.一次键16：库伦力是怎样随电荷间的距离而变化的？哪种熟知的力也表现出这一特点？在怎样的电荷间隔下dF(库仑)/dx最大和最小？Solution:TheCoulombicForcevariesinverselywiththesquareoftheseparationdistancebetweenthechargecenters.Anotherfamiliarexampleofaninversesquarelawistheforceofgravity.Afunctionofthisformhasnolocalmaximaorminimabecausethederivativeofequationhasanontrivialvaluesforwhichitisequaltozeroandthemaximumvalueoccursasxapproachesinfinity.一次键18：为什么共价键只在负电性元素间形成？Assumptions:Duringbonding,atomsseektoobtainafilledvalenceelectronshell.Solution:

Electronegativeelementsgenerallyhavenearlyfilledvalenceshellsandareseekingafewadditionalelectrons.Ifelectropositiveatomsarenearby,theycantransferelectronstotheelectronegativeatomsandanionicbondisformed.If,however,onlyelectronegativeatomsarepresentthentheonlywaytheycanallacquireextraelectronsistosharethem.Thisisthedefinitionofacovalentbond.Comments:Hydrogenalsoformscovalentbonds.一次键19：若O与其本身形成共价键，Si也与其本身形成共价键，那么，认为SiO成共价键是合理的，对吗？Data:FromAppendixB:EN(Si)=1.90andEN(O)=3.44.Assumptions:ThepercentioniccharacterofabondisgivenbythetableinAppendixA.Solution:ItisimpossibletoformionicbondsinanypureelementsincetherewillbenodifferenceinENvaluesforidenticalatoms.InthecaseofSi-O,however,thedifferenceinelectronegativityis1.54.Thiscorrespondstoabondthatisapproximately45%ionicand55%covalent.Comments:Thistypeofprimarybondisoftendescribedasamixedionic/covalentbond.一次键20．陶瓷或离子固体有可能成为适当的导电体吗？需要怎样才有可能？Assumptions:Highelectricalconductivityrequiresahighdensityofmobilechargecarriers.Solution:Inionicsolidsthereareusuallyfew,ifany,freeelectrons.Chargetransportrequiresthemotionofcomparativelylargeionswhichisgenerallyamoredifficultprocessthanelectronmotion.If,however,anionicsolidhadeitherahighfreeelectrondensityorextremelymobileions,thenitwouldbeagoodelectricalconductor.

Comments:WewillfindsomeexamplesofmaterialswiththesecharacteristicsinChapter10.一次键21．解释方程2-8中指数必须遵从不等式p<q这一要求的物理

意义。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.一次键22．键-能曲线可以用来获得有关下列四种材料性能中的三种信息，哪种信息不能得到？

1）键能

2）平衡间距

3）一次键类型

4）汽化温度键能曲线Thedepthofthebond-energycurvedescribesthestrengthofthebond(i.e.thebondenergy)andalsoprovidesinformationaboutthevaporizationtemperaturesinceitisanindicationoftheamountofenergythatmustbesuppliedtomoveatomstoaninfiniteseparationdistance.Thevalueofxforwhichthebondenergyisaminimumcorrespondstotheequilibriumseparationdistance.Itis,however,notpossibletogaininformationabouttheprimarybondtypefromthegeneralshapeofthebondenergycurve.23．图2-12中所示的两种材料中，你预计哪一个将有较高的

熔点？GIVEN:TheslopeofmaterialAatzeroforceisgreaterthanthatofB.SOLUTION:Theeaseofseparatingatomsormoleculeswithheatissimilartotheeaseofseparatingatomswithforce.Hence,AhasahighermeltingtemperaturethandoesB.键能曲线24.给定Al的膨胀系数为25×10-6℃-1，SiC为4.3×10-6℃-1，

预计哪种材料具有较高的熔点？键能曲线Assumptions:Meltingtemperatureisrelatedtothedepthofthebondenergycurveandthermalexpansioncoefficientisrelatedtotheasymmetryofthecurve.Solution:Since"deep"energywellstendtobemoresymmetric,materialswithhighmeltingtempera-turestendtohavelowexpansioncoefficients.Thus,weexpectSiCtohaveahigherTmthanAl.Comments:Th