材料科學(xué)與工程基礎(chǔ)習(xí)題評(píng)講-1

習(xí)題講解第一次作業(yè)英文2.6Allowedvaluesforthequantumnumbersofelectronsareasfollows:TherelationshipsbetweennandtheshelldesignationsarenotedinTable2.1.Relativetothesubshells,l＝0correspondstoanssubshelll＝1correspondstoapsubshelll＝2correspondstoadsubshelll＝3correspondstoanfsubshellFortheKshell,thefourquantumnumbersforeachofthetwoelectronsinthe1sstate,intheorderofnlmlms

,are100(1/2)and100(-1/2).WritethefourquantumnumbersforalloftheelectronsintheLandMshells,andnotewhichcorrespondtothes,p,anddsubshellsK:s:100(1/2);100(-1/2)L:s:200(1/2);200(-1/2)p:210(1/2);210(-1/2);21-1(1/2);21-1(-1/2);211(1/2);211(-1/2)M:s:300(1/2);300(-1/2)p:310(1/2);310(-1/2);31-1(1/2);31-1(-1/2);311(1/2);311(-1/2)d:320(1/2);320(-1/2);32-1(1/2);32-1(-1/2);321(1/2);321(-1/2);32-2(1/2);32-2(-1/2);322(1/2);322(-1/2)2.7Givetheelectronconfigurationsforthefollowingions:Fe2+,Fe3+,Cu+,Ba2+,Br-,andS2-.SOLUTION

Fe2+:1s22s22p63s23p63d6Fe3+:1s22s22p63s23p63d5Cu+:1s22s22p63s23p63d10Ba2+:1s22s22p63s23p63d104s24p64d105s25p6

Br-:1s22s22p63s23p63d104s24p6S2-:1s22s22p63s23p6

2.17(a)Brieflycitethemaindifferencesbetweenionic,covalent,andmetallicbonding.(b)StatethePauliexclusionprinciple.SOLUTION(a)離子鍵：無(wú)方向性球形正、負(fù)離子堆垛取決電荷數(shù)——電荷平衡體積（離子半徑）金屬鍵：無(wú)方向性球形正離子較緊密堆垛共價(jià)鍵：有方向性、飽和性，電子云最大重疊(b)原子中的每個(gè)電子不可能有完全相同的四個(gè)量子數(shù)（或運(yùn)動(dòng)狀態(tài)）2.19Computethepercentsioniccharacteroftheinteratomicbondsforthefollowingcompounds:TiO2,ZnTe,CsCl,InSb,andMgCl2.

SOLUTION由公式：

已知：TiO2,XTi=1.5andXO=3.5ZnTe,已知：XZn=1.6andXTe=2.1，故，%IC=6.05%CsCl,已知：XCs=0.7andXCl=3.0，故：

%IC=73.4%InSb,已知：XIn=1.7andXSb=1.9，故：

%IC=1.0%MgCl2,已知：XMg=1.2andXCl=3.0故：

%IC=55.5%2.24，Onthebasisofthehydrogenbond,explaintheanomalousbehaviorofwaterwhenitfreezes.Thatis,whyistherevolumeexpansionuponsolidification?

水凍結(jié)時(shí)結(jié)晶，非球形的水分子規(guī)整排列時(shí)受氫鍵方向性和飽和性的更強(qiáng)限制，不能更緊密地堆積，故密度變小，體積增大。2-7影響離子化合物和共價(jià)化合物配位數(shù)的因素有那些？離子化合物：體積電荷共價(jià)化合物：價(jià)電子數(shù)電子云最大重疊第二次作業(yè)2.18Offeranexplanationastowhycovalentlybondedmaterialsaregenerallylessdensethanionicallyormetallicallybondedones.

共價(jià)鍵需按鍵長(zhǎng)、鍵角要求堆垛，相對(duì)離子鍵和金屬鍵較疏松2.21UsingTable2.2,determinethenumberofcovalentbondsthatarepossibleforatomsofthefollowingelements:germanium,phosphorus,selenium,andchlorine.

SOLUTIONGe:4P:3Se:2Cl:12-6按照雜化軌道理論，說(shuō)明下列的鍵合形式： （1）CO2的分子鍵合C

sp雜化

（2）甲烷CH4的分子鍵合Csp3雜化

（3）乙烯C2H4的分子鍵合Csp2雜化

（4）水H2O的分子鍵合Osp3雜化

（5）苯環(huán)的分子鍵合 Csp2雜化

（6）羰基中C、O間的原子鍵合Csp2雜化

2-10當(dāng)CN=6時(shí)，K+離子的半徑為0.133nm(a)當(dāng)CN=4時(shí)，對(duì)應(yīng)負(fù)離子半徑是多少？(b)當(dāng)CN=8時(shí)，對(duì)應(yīng)負(fù)離子半徑是多少？若(按K+半徑不變)求負(fù)離子半徑,則：CN=6R=r/0.414=0.133/0.414=0.321nmCN=4R=r/0.225=0.133/0.225=0.591nmCN=8R=r+/0.732=0.133/0.732=0.182nm第三次作業(yè)3.48Drawanorthorhombicunitcell,andwithinthatcella[121]directionanda(210)plane.3.50Hereareunitcellsfortwohypotheticalmetals:

a.Whataretheindicesforthedirectionsindicatedbythetwovectorsinsketch(a)?

bWhataretheindicesforthetwoplanesdrawninsketch(b)?（a）direction1,xyzProjections0ab/2cProjectionsintermsofa,b,andc01/21Reductiontointegers012Enclosure[012]direction2,xyzProjectionsa/2b/2-cProjectionsintermsofa,b,andc1/21/2-1Reductiontointegers11-2Enclosure[112]（b）Plane1，∞：1/2：∞；0：2：0；(020)Plane2，1/2：-1/2：1；2：-2：1；(221)3.51*Withinacubicunitcell,sketchthefollowingdirections:ab（c）[012]（d）[133]（e）[111]（f）[122]（g）[123]（h）[103]3.53Determinetheindicesforthedirectionsshowninthefollowingcubicunitcell:DirectionA:xyz-2/3ab/20c-2/31/20-430[430]DirectionA:DirectionB:

xyz

xyz-2/3ab/20c2/3a-b2/3c-2/31/202/3-12/3-4302-32[430][232]DirectionCDirectionD

xyzxyz1/3a-b-ca/6b/2-c1/3-1-11/61/2-11-3-313-6[133][136]3.57

DeterminetheMillerindicesfortheplanesshowninthefollowingunitcell:planeAxyza/3b/2-c/21/31/2-1/23/12/1-2/1(322)planeB(101)3.58DeterminetheMillerindicesfortheplanesshowninthefollowingunitcell:planeA以（0，1，0）為新原點(diǎn)xyz2/3a-bc/22/3-11/23/2-1/12/13/2-2/24/2(324)planeB(221)3.61*

Sketchwithinacubicunitcellthefollowingplanes:a3.61*

Sketchwithinacubicunitcellthefollowingplanes:abc

defgh3.62Sketchtheatomicpackingof(a)the(100)planefortheFCCcrystalstructure,and(b)the(111)planefortheBCCcrystalstructure(similartoFigures3.24band3.25b).(a)FCC:(100)plane(b)BCC:(111)plane3.81ThemetaliridiumhasanFCCcrystalstructure.Iftheangleofdiffractionforthe(220)setofplanesoccursat69.22°(first-orderreflection)whenmonochromaticx-radiationhavingawavelengthof0.1542nmisused,compute(a)theinterplanarspacingforthissetofplanes,and(b)theatomicradiusforaniridiumatom.SOLUTION：已知：銥FCC的（220）晶面,2θ=69.22°;λ=0.1542nm；n=1由公式：nλ=2dsinθ和dhkl=a/()故解：（a）d220=0.1542/(2sin34.61°)=0.1357（nm）（b）∵a=d220×=2d220；又FCC的2R=a/=2d220∴R=d220=0.1357（nm）即：其（220）晶面間距為0.1357nm；銥原子半徑也為0.1357nm。2-14計(jì)算（a）面心立方金屬的原子致密度；（b）面心立方化合物NaCl的離子致密度（離子半徑r(Na+)=0.097，r(Cl-)=0.181）；（C）由計(jì)算結(jié)果，可以引出什么結(jié)論？解：（a）∵面心立方金屬∴一個(gè)晶胞中有四個(gè)金屬原子，且a=4R/√2有：PF=4(4πR3/3)/a3

=16πR3(2√2)/(3)(64R3)=0.74答：面心立方金屬的原子致密度為0.74（b）解：∵面心立方化合物NaCl,

且:r(Na+)=0.097，r(Cl-)=0.181∴一個(gè)晶胞中有四個(gè)Na離子、四個(gè)Cl離子有：PF=[4(4πr3/3)+4(4πR3/3)]/(2r+2R)3=16π(0.0973+0.1813)/3(8)(0.097+0.181)3

=0.64答：面心立方化合物NaCl的離子致密度為0.64（c）結(jié)論:(1)同種原子晶體的致密度只與晶胞類型相關(guān),與原子尺寸無(wú)關(guān)(2)化合物晶體的離子致密度與離子大小相關(guān)2-18在體心立方結(jié)構(gòu)晶胞的（100）面上按比例畫出該面上的原子以及八面體和四面體間隙。

八面體四面體2-24方向?yàn)閇111]的直線通過(guò)1/2,0,1/2點(diǎn),則在此直線上的另外兩點(diǎn)的坐標(biāo)是什么?解：∵該直線過(guò)為:u=1/2v=0w=1/2u=0: