Selected Problems and Solutions for Chapters 5-9

Key points/concepts第 7-9 章1. Lattice of crystal structure: translation symmetrya lattice point = a structure motif -- unit cell2. Crystal systems (7), Bravais Lattice (14)3. Symmetry operations (point & translation) Crystallographicpoint groups(32), space groups (230), miller index of crystalplane, d-spacing etc.4. X-ray diffraction, Laue equation, Bragg’s Law, reciprocallattice, Ewald sphere, structural factor, system absence,general process of x-ray crystal structure determination.5. Close-packing of spheres (ccp/A1,hcp/A3,bcp/A2) in metalsand ionic compounds, coordination of cations.6. Crystal structures of some typical ionic compounds.

Example: p224, 7.2-- concept of lattice• A structure motif (+ occupying space) = a lattice point• Each lattice point has identical surroundings.• A lattice fulfills translation symmetry.• Differences between a real crystal structure and itslattice.Key point is to find the structure motif (basis) that fulfillstranslation symmetry!

p.227, 7.26 SnF 4Sn: (0,0,0), (1/2,1/2,1/2)F: (0,1/2,0), (1/2,0,0)(0,0,0.237), (0,0,-0.237)1) body-centred tetragonal2 lattice point within a unit cell.The black dots (2 Sn) and red balls(4 F) are defined by the coordinatesgiven!Other 4 F atoms can be obtainedby translation operation (2 LP).2) Each Sn atom is located in adistorted octahedral hole.R(Sn-F) 1 =0.237x 793 = 187.9pm; R(Sn-F) 2 = 0.5x404=202 pmKey: Figure out the atoms within a LP!

Simple cubic crystal – example: CsClp.286 9.13• Only one lattice point within a unit cell.• Each lattice point contains 2 atoms,Cl (0,0,0), Cs(1/2,1/2,1/2);• The structural factor isFhklfClTherefore, all possible diffractions are observable without systemabsence! However,2i1ffiCsee2πi(hx kyπi(hkl)iilzi)If h+k+l =2n,FhklfClfCsStrongest diffractionIf h+k+l =2n+1,FhklfClfCsWeakest diffraction

Face-centered cubic crystal – general case p227, 7.21• Lattice points (LPs): (0,0,0), (1/2,1/2,0), (0,1/2,1/2), (1/2,0,1/2)• Suppose each lattice point contains n atoms, {(x j ,y j ,z j )} (j=1,…,n)• Each unit cell contains N=4n atoms, e.g., an atom A(x i ,y i ,z i ) in oneLP has other three equivalent A atoms within the same unit cell!• Then the structure factor isSum up over all atomswithin a unit cell!F hkl [ 1ejFrom translation symmetry of fcc!• Thus, when h,k,l are neither all even nor all odd,i12i(hx kysystem absence!• Furthermore, when h,k,l are all even or all odd,F hkl 4FhklNfiediffraction observable!ilz)h kh lk l2πi( ) 2πi( ) 2πi( )ni( hx e e ] 2 22 22 22πfje1nj1fjF hkl 0e2πi(hx kyjjlzj)iijkyjlzNow sum up over allatoms within a LP!j)

Face-centered cubic crystal –Special case: NaCl p227, 7.23• Unit cell contains 4 lattice points, or 4NaCl• Each lattice point (LP) corresponds to a NaCl.• Put Cl at (0,0,0), then a neighboring Na at (1/2,0,0).• Only when h,k,l are all even or all odd can diffractions be observed!Fhklπi(hxjky j lz j ) 42 f e 4[j1jCase 1: if h = 2n (note we also have l=2n and k=2n)F hkl2Strong diffraction!Case 2: if h = 2n+1 (note we also have l=2n+1, k=2n+1)F hkl 4 [ f ClfNa 4 [ f ClfNa]]fCleWeak diffraction!fNaπhiNow sum up over all atoms within a LP!]

Face-centered cubic crystal –Special case: ZnS (sphalerite)• Unit cell contains 4 lattice points, or 4ZnS• Each lattice point corresponds to a ZnS.S(0,0,0), Zn(1/4,1/4,1/4) (different elements!)• When h,k,l are all even or all odd, diffractions observable,22πi(hxj kylz)πi(hkl) / 2fS fZnej jFhkl 4fje 4[j1] (111),(200),(220),(311),(222),(400),(331),(420),(422),…Case 1: if h+k+l = 4n, e.g., (220),(400),(440)…F hkl 4 [ f SfZn]Strongest diffraction!Case 2: if h+k+l = 4n+2, e.g., (200),(222),(420),(442)…..F hkl 4 [ f SfZn]Weakest diffraction!

Face-centred cubic crystal --Special case: Diamond O h7 - Fd3m• Lattice points: (0,0,0)+, (1/2,1/2,0)+, (0,1/2,1/2)+, (1/2,0,1/2)+• Each LP contains two C atoms (i.e., structure motif =2C)C1-- (0,0,0), C2-- (1/4,1/4,1/4) (the same element)• The other six C atoms within a unit cell can be derived as(1/2,1/2,0), (3/4,3/4,1/4) ; (0,1/2,1/2), (1/4,3/4,3/4) ;(1/2,0,1/2), (3/4,1/4,3/4)Derive the systemabsence of diamond!• Such an arrangement of C atoms produces new translationsymmetry elements, i.e., screw axes and d glide planes, which inturn introduce special system absence of diffractions (in additionto the system absence from normal FCC lattice !!!!!!!

Face-centred cubic crystal --Special case: Diamond O h7 - Fd3mDerive the systemabsence of diamond!• Such an arrangement of C atoms produces new translationsymmetry elements, i.e., screw axes and d glide planes, which inturn introduce special system absence of diffractions (in additionto the system absence from normal FCC lattice !!!!!!!4 1d glideplane1/8,3/8,5/8,7/8Sideviewtopview• Let’s derive the structural factor of diamond tounravel its system absence.

Face-centered cubic crystal –Special case: DiamondFhklffCC8i1[1 e[1 efCeh k2i( )2 2h k2i( )2 22i(hx kyieeNow sum up over all atoms within a unit cell!ilzih l2i( )2 2h l2i( )2 2k l2i( )2 2k l2i( )2 2(Note: two carbon atoms within a lattice point: (0,0,0), (1/4,1/4,1/4))System absence:h kh lk l2i( ) 2i( ) 2i( )2 22 22 2a)[1 e e e ] 0 ori(hkl) / 2b) (1 e ) 0i.e., a) h,k,l are neither all even nor all odd! & b) h+k+l = 4n+2 Observable diffractions: (111), (220),(311),(400),(331),(422) &If h+k+l=4n, F hkl = 8f C , (220),(400)… strongest diffraction!)Now sum up over all atoms within a LP!ee] 2j1](1 ee2i(hxkyi(hkl) / 2j)jlzj)

Body-center crystal – p227, 7.22General case: each lattice point contains n atoms• The total number of atoms within a unit cell is 2n;• For jth atom in a structure motif (a lattice point): (x j ,y j ,z j )• Its body-center equivalent is: (0.5+x j , 0.5+y j , 0.5+z j )Fhkl2ni1n [1 ej1fi{ feji(hkl)2πi(hx kyei2i(hx]nj1jilzkyfjjie)lzj)2i(hxjSum up over all atomswithin a unit cell!fkyjjelz12i[h(x2j)j1) k( y2j1) l(z2From translation-symmetrySum up over all atoms in a LP!j)]}While h+k+l =2n+1,eπi(hkl)e( 2n1) πi 1Fhkl0Systemabsence

Body-center crystal – special case (p.227, 7.24)• Each lattice point (LP) contains 2 atoms, A (0,0,0), B(x,y,z);• Thus in another LP, A (1/2,1/2,1/2), B (x+1/2,y+1/2,z+1/2)• The structural factor isFhkl4i1fie2πi(hx kyiilzi) [1 e[1 ei(hkl)i(hkl)] ] [f2j1AfjfeB2i(hxejkylz2i(hxkylz)jj)]Translation-symmetry term!While h+k+l =2n+1,eπi(hkl)e( 2n1) πi 1Sum up over all atoms in a LP!Fhkl 0System absence

Example:diffraction data indexing cell parameter!• 钨 为 立 方 晶 系 , 其 粉 末 衍 射 线 指 标 为 : 110,200,211,220, 310,222,321,400,…. 属 何 种 点 阵 类 型 ? 若 X 射 线 =154.4 pm,220 衍射 角 43.6°, 试 计 算 晶 胞 参 数 。 p.226, 7.18• Answer: h+k+l =2n diffractions observed! =2n+1 system absence!or (h 2 +k 2 +l 2 ): 2:4:6:8:10:12:14:16….. bcc lattice!Bragg law: 2d hkl sin = d hkl= /(2sin)since 220 = 43.6°cell parameter:adhklh2k2l2 h2k2l2/ 2sinhkl 154.4 8 / 2sin(43.6)316.6 pmThe answer given in p. 317 is wrong!Atomic radius of W atom: (bcc)4R 3a R 3a/ 4 316.6x3 / 4 137.1ppm

P227, 7.27 Bragg’s law, d-spacing• Number of S 8 in a unit cell of orthorhombic crystal:nS8~N V/ M0 6.022 1023S8~N0( abc) / MS8(10481292 2455) 1030 2.07 /(32 8)16For orthohombic crystal, 1/d hkl =[(h/a) 2 + (k/b) 2 + (l/c) 2 ] 1/2 p.200According to Bragg’s Law, we havesin = /2d hkl = [(h/a) 2 + (k/b) 2 + (l/c) 2 ] 1/2 ×/2= [(2/1048) 2 + (2/1292) 2 + (4/2455) 2 ] 1/2 ×154.18/2= [(1/1048) 2 + (1/1292) 2 + (2/2455) 2 ] 1/2 ×154.18 = 0.2273 = 13.1º(Cu K 1 x-ray, = 154.18 pm)Note: the value of is not given in question 7.27! !!!

p.257, 8.8 indexing of diffraction data!Ta metal’s x-ray diffraction data, (sin 2 ) is known.1) Indexing:sin 2 1 :sin 2 2 :sin 2 3 :sin 2 4 :sin 2 5 :sin 2 6 :sin 2 7 :sin 2 8 :sin 2 9 ….= 1:2:3:4:5:6:7:8:9:… = 2:4:6:8:10:12:14:16:18:…body-centered cubic lattice, h+k+l =2n+1 system absence!Observed hkl: (110)(200)(211)(220)(310)(222)(321)(400)(330)…2) Cell parameter: d hkl = a/(h 2 +k 2 +l 2 ) 1/2 a = d hkl (h 2 +k 2 +l 2 ) 1/2According to Bragg’s Law, we have d hkl = /2sin a = (h 2 +k 2 +l 2 ) 1/2 /2sinChoosing the (330) diffraction with sin 2 0.97826 (=154.1 pm), wehave a = (h 2 +k 2 +l 2 ) 1/2 /2sin= 154.1 x (18) 1/2 /(2x0.98907) = 330.5 pm

p.259, 8.211) Statistically, the probability for a Cuatom to appear at a lattice point isequal to the percentage (x) of Cudoping. Thus the AuCu alloybelongs to face-centered cubic latticesystem. Each unit cell contains 4lattice points and each of them is astatistic atom (Au 1-x Cu x ).2) The ordered phase belongs to simple tetragonal lattice; each unitcell/lattice point contains AuCu. Au(0,0,0), Cu (1/2,1/2,1/2);3) Cell parameter for the ordered phase:a = b = a/2 = 272.23 pm, c = a = 385 pmThe first observed diffraction is (001)d 001 = 1/(h/a) 2 +(k/a) 2 +(l/c) 2 ) 1/2 = c = aAccording to Bragg’s Law, we havesin = /2d = /2a= 0.2 = 11.5

p.259, 8.21The randomly doping phase belongsto fcc lattice. Hence its first observablediffraction is (111).d 111 = a/(h 2 +k 2 +l 2 ) 1/2 = a/3 1/2According to Bragg’s Law,sin 111 = /2d 111= 3 1/2 /2a = 0.3464 = 22.3Note: in some cases, random/multipleorientations/rotations of structural units (clusters,molecules, or groups) exist within a single crystal, whichposes difficulty in the structural determination based on X-ray diffraction data!

Example:• 银 为 立 方 晶 系 , 用 CuK 射 线 (=154.18 pm) 作 粉 末 衍 射 , 在 hkl类 型 衍 射 中 ,hkl 奇 偶 混 合 的 系 统 消 光 。 衍 射 线 经 指 标 化 后 ,选 取 333 衍 射 线 , = 78.64º, 试 计 算 晶 胞 参 数 。 已 知 Ag 的 密度 为 10.507 g·cm -3 , 相 对 原 子 质 量 为 107.87, 问 晶 胞 中 有 几 个Ag 原 子 ; 试 写 出 Ag 原 子 的 分 数 坐 标 。Answer: when h,k,l are neither all odd nor all even, system absence! Cubic F-centred.Bragg law: 2d hkl sin = d hkl= /2sincell parameter:aatoms in a unit cell:dhkl 154.18nAgh k l k l408.58pmThus each atom corresponds to one lattice point, atomiccoordinates: (0,0,0) (1/2,1/2,0) (1/2,0,1/2) (0,1/2,1/2)222λh27 / 2 sin( 78.64 ) ~ N V/ M0 6.022 1023Ag~ N a032(408.58102 / M10Ag)32/ 2 sin θ10.507 /107.87 4

Example:• 金 属 Mg 是 由 Mg 原 子 按 A3 型 堆 积 而 成 , 已 知 Mg 的 原 子 半 径 是160 pm, 求 晶 胞 参 数 。Answer: Hexagonal close-packing mode a = b = 2R = 320 pm,The height of a tetrahedron with an edgelength of 2R is2R2Rh2 R 6 /3c2h4R6 / 3522.6 pmh

Example: p.286, 9.9• KF crystal– cubic system, Mo K =70.8 pm, diffraction sin 2 :0.0132, 0.0256, 0.0391, 0.0514, 0.0644, 0.0769, 0.102, 0.115,0.127, 0.139, …..., 1) plz derive its lattice type and cellparameter; 2) Suppose the F - adopts the simple cubic packingwith cubic interstices being occupied by K + ; R K =133 pm andR F = 136 pm. plz derive the cell parameter.Answer:1) indexing: sin 2 = 1:2:3:4:5:6:8:9:10:11… simple cubicBragg law: 2d hkl sin = d hkl= /2sin2aadhkl70.8h2 k2 l2 h k11 /(2 0.139) 314 pm22l2/ 2sin3a2 2 2 2 2 2Or / 4a sin /( h k l ) 取 四 条 高 指 标 衍 射 线 求 该 式 的平 均 值 得 2 2 / 4a 0.012716 a 313. 9pm

2) For simple cubic packing of anions, cations occupy the cubicinterstices. As such, both types of ions directly contacts with eachother along the diagonals of a cube, i.e.,2(RF RK) 3a a 2( RF RK) / 3 310. 6pm2a3a

Example: p.286, 9.14• Cell parameters of NaCl-type KBr, LiBr, KF, and LiF are 658,550, 534, 402 pm, respectively. Please derive the ionic radii ofK, Li, F, and Br.Answer:1. Br anion is the largest and Li cation is the smallest! For LiBr, theanions adopt ccp structure with Li cations occupying the octahedralholes. Thus,For KBr, the K cation is large, thus, the anions may not closely contactwith each other. Instead, the cations and anions closely contact witheach other. Thus,Similarly, for KF,2aLiBr 4RBr RBr 2aLiBr/ 4 194.5pmRBr RK aKBr/ 2 RK134.5pmRF RK aKF/ 2 RF132.5pmFor LiF, the ultimate case is Li-F closely contact with each other,RF RLi aLiF/ 2 RLi 68.5pm

Example: p.288, 9.32• The cell parameters of monoclinic crystal of biphenyl: a = 824 pm,b= 573 pm, c= 951pm, = 94.5; = 1.16 gcm -3 .Answer: 1) Number of molecules within a cell,~ ~n N V/ M N [b a c cos( 90 )]/ Mbp20 (6.022 1023bpa-glideac2 1b0) 1.16 (573824 951cos 4.510bp30) /154.22) When h is odd, h0l system absence a-glide (a/2) perpendicular to b-axis.When k is odd, 0k0 system absence 2 1 screw axis parallel to b-axis. (P.215) c-centered monoclinic or P2 1 /c.3) The molecule is D 2d symmetric. Eachmolecule corresponds to a lattice point, i.e.,centering at (0,0,0), (1/2,1/2,0).

• From this example, we know that some specialtranslation symmetry, elements e.g., screw axes orglide planes, do introduce specific system absencein x-ray diffraction. So you need to keep in mindthe rules of system absence related to thesetranslation symmetry elements (see Table 7-6,p.214-215)!

Types ofreflectionhkl0kl(or h0l,hk0)00l(or h00,or 0k0systematic absence and symmetry (p214-215)Conditions for extinction Cause of extinction Centering andsymmetryelementsh+k+l = oddh+kh+lk+l= odd= odd= oddh,k,l not all even and not allodd-h+k+l not multiples of 3klk+l=odd=odd=oddk+l not multiples of 4l =oddl not multiples of 3l not multiples of 4l not multiples of 6I-centredC-centredB-centredA-centredFace-centredR-centredTranslation in (100) b/2c/2(100) glide (b+c)/2planesTranslation c/2Along c/3(001) c/4(b+c)/4Screw axis c/6ICBAFR(hexagonal)b-glide a (b,c)c-glidea (b,c)n-glidea (b,c)d-glidea (b,c)2 1 , 4 2 , 6 33 1 , 3 2 , 6 2 , 6 44 1 , 4 36 1 , 6 5

Example: p.288, 9.30SiP 2 O 7 .Answer: 1) Each Si(4+) is surroundedby 6 Oxygen anions. s(Si-O 1 ) = +4/6 = +2/3 s(P-O 1 ) = 2 - (4/6) = +4/3 s(P-O 2 )= 5- (4/3)x3 = +1 Z(O 2 ) = 2x(+1) = 2 P 2 O 74-in SiP 2 O 7 solid is stable!2) s(P-O 1 ) > s(P-O 2 ) R(P-O 1 ) < R(P-O 2 )3) Free P 2 O 74-anion is unstable!s(P-O 2 ) = +5/4 Z(O 2 ) = 2x(+5/4) = 2.5 > 2!Z s iiiZii

面 心 立 方 点 阵 的 倒 易 点 阵 为 体 心 立 方 点 阵 , 反 之 亦 然 !• 若 面 心 立 方 点 阵 单 胞 的 边 长 为 a, 其 三 个 正 交 的 单 位 矢 量 为注 意 到 面 心 立 方 点 阵 的 素 原 胞 为 棱 方 单 胞 , 其 基 矢 为 :aa( xˆ2yˆ);aa( yˆ2zˆ);aa( xˆ212 3 z 其 体 积 为 : V c a1 (a2a3) a则 其 倒 易 点 阵 的 基 矢 为 :bzˆ);b3xˆ ,ŷ,ẑ此 为 体 心 立 方 点 阵 素 单 胞 的 三 个 基 矢 , 该 体 心 立 方 点 阵 单 胞 的 边长 为 2/a./ 4ˆ)xˆ);ba 1 2 a3a1 a31a1 a 1( ˆ ˆ( ˆ ˆ21 x y 2 y z 3 ( xˆ zˆ yVcaVcaVcaˆ)故 单 胞 参 数 为 a 的 面 心 立 方 点 阵 , 其 倒 易 点阵 为 体 心 立 方 点 阵 ( 单 胞 参 数 为 2/a)!

• 已 知 ZnO 晶 体 具 有 六 方 ZnS 结 构 ( 如 下 图 1), 试 画 出 其 符 合 化 学配 比 的 (10-10)、(0001) 和 (000-1) 晶 面 的 原 子 排 列 的 平 面 示 意 图 (通 常 包 含 次 表 面 层 原 子 , 用 球 棍 模 型 表 示 ), 并 说 明 其 表 面 原 子 的配 位 数 。 (0001) 和 (000-1) 晶 面 和 c 轴 垂 直 这 两 个 表 面1均 可 用 下 图 2 表 示 .2(0001):Zn s CN=3O sub CN = 4(000-1) :O s CN=3Zn sub CN=4 (10-10) 晶 面 实 际 上 就 是 (100) 面 , 与 b、c 两 轴 平 行 ( 图 3)。Zn s /O s : CN=3, Zn sub /O sub : CN =4

第 5-6 章Key points/concepts1. Hybrid orbital theory and VSEPR2. HMO, HMO treatment of -conjugated systems andgraphical method to predefine coefficients of HMO of -conjugated systems.3. Symmetry rules for molecular reactions.4. Chemical bonds in electron-deficient molecules (boranesand carboranes etc), styx method & Wade’ (N+1) rules5. Chemical bonding in coordination complexes, crystal-fieldtheory, ligand-field theory; 18e rule and its application intransition-metal cluster compounds.

VSEPR and Hybrid orbital theory p.149, 5.8, 5.121) SCl 3+ : S + (5Ve), each Cl provides 1e to form a S-Cl bond;thus SCl 3+has a lone pair on S. Pyramidal structure like PCl 3 . Ssp 3 hybridization. S(4+)2) ICl 4- : I - (8Ve), each Cl provides 1e to form a I-Cl bond; thusICl 4-has two lone-pairs on I with a square planar structure. Iatom adopts sp 3 d 2 hybridization. I(3+).3) ICl 3 : I (7 Ve), each Cl provides 1e to form a I-Cl bond; thus ICl 3has two lone-pairs on I with a T-shaped structure. I atom adoptssp 3 d hybridization. I(3+), non-polar4) SO 3 : S (6 ve), sp 2 hybridization, trigonal planar; three S-Osigma bond, and a P 465) SO 32-: S (6 ve), sp 3 hybridization, Pyramidal structure like PCl 3 ;three S-O sigma bond.6) CO 32-: C(4 ve), sp 2 hybridization, trigonal planar; a P 46

MO theory p.149, 5.12Suppose the bonding MO of AB isψABφ*caaφφbadτcbφbφ*bφadτ0ψ*ABψABdτc2aφ*aφadτc2bφ*bφbdτc2ac2b1c2aψAB90%,c2b0.95φa10%0.32φbca 0.95;cb 0.32

Possible structures of XeO n F m (n,m=1,2,3)? P.150, 5.19• Any stable XeO n F m (n,m=0,1,2,3) compound should have evennumber of valence electrons. So m =even.• If n=0, a) m=2, XeF 2 , Xe has three lone pairs, linear structure ofD h ; b) m=4, XeF 4 , Xe has two lone pairs, planar square D 4h ; c)m=6, XeF 6 , Xe has one lone pair, thus being C 3 ; d) m=8, XeF 8 ,tetragonal antiprism structure of D 4d symmetry.• If m=0, a) n=1, XeO, Xe has 3 lone pairs, C v ; b) n=2, XeO 2 , Xehas 2 lone pairs, H 2 O-like structure of C 2v ; c) n=3, XeO 3 , Xe hasone lone pair, C 3v ; c) n=4, XeO 4 , tetrahedral structure, T d ;• If n=1, a) m=2, XeOF 2 , Xe has two lone pairs, C 2v symmetry; b)m=4, XeOF 4 , Xe has one lone pair, C 4v ; c) m=6, XeOF 6 , Xe hasno lone pair, C 3 .• If n=2, a) m=2, XeO 2 F 2 , Xe has one lone pair, C 2v symmetry; b)m=4, XeO 2 F 4 , Xe has no lone pair, D 4h ;• If n=3, a) m=2, XeO 3 F 2 , Xe has no lone pair, D 3h .

P.150, 5.201) 第 一 种 解 释 ( 杂 化 轨 道 理 论 ): 中 心 元 素 为 主 族 元 素 ( 如 P) 时 ,中 心 元 素 和 配 位 原 子 间 通 常 形 成 共 价 键 , 中 心 原 子 成 键 时 采 取sp 3 d(z 2 ) 杂 化 形 式 , 轴 向 杂 化 轨 道 包 含 了 大 量 的 nd(z 2 ) 的 组 分 ,而 赤 道 平 面 轨 道 则 只 包 含 ns 和 np 轨 道 组 分 , 故 轴 向 键 比 赤 道 面内 的 键 来 得 长 。 第 二 种 解 释 (VSEPR): 轴 向 键 和 赤 道 面 键 的夹 角 为 直 角 , 比 赤 道 面 键 角 (120 度 ) 小 , 因 此 轴 向 键 电 子 对 受到 的 排 斥 大 于 赤 道 面 键 电 子 对 , 轴 向 键 比 赤 道 面 键 来 得 长 。2) 第 一 种 解 释 (VSEPR) : 当 中 心 原 子 为 过 渡 金 属 原 子 时 , 过 渡金 属 原 子 上 d 轨 道 的 电 子 对 会 与 配 体 的 电 子 对 互 相 排 斥 。a)[Ni(CN) 5 ] 3- 为 四 角 锥 构 型 , 中 心 离 子 Ni 2+ 的 3d(x 2 -y 2 ) 轨 道 ( 赤道 面 内 ) 上 没 有 电 子 占 据 , 其 它 3d 原 子 轨 道 满 占 据 , 因 此 , 赤道 平 面 内 配 体 受 到 的 排 斥 比 轴 向 配 体 受 到 的 排 斥 小 , 所 以 轴 向配 位 键 长 于 赤 道 面 键 。b) [Cu (Cl) 5 ] 3- 为 三 角 双 锥 , 中 心 原 子3d(z 2 ) 轨 道 为 单 占 据 , 其 它 3d 原 子 轨 道 满 占 据 , 因 此 , 轴 向 配体 受 到 的 排 斥 比 赤 道 面 内 配 体 受 到 的 排 斥 小 , 所 以 轴 向 配 位 键短 于 赤 道 面 键 。同 样 地 , 可 以 解 释 当 [Cu (Cl) 5 ] 3- 为 四 角 锥 时 , 轴 向 键 比 赤 道 面内 键 长 !

P.150, 5.202) 第 二 种 解 释 ( 杂 化 轨 道 理 论 ) :a)[Ni(CN) 5 ] 3- 为 四 角 锥 构 型 时 ,中 心 离 子 Ni 2+ 的 3d(x 2 -y 2 ) 轨 道 ( 赤 道 面 内 ) 上 没 有 电 子 占 据 , 其 它3d 原 子 轨 道 满 占 据 , 中 心 离 子 采 取 dsp 3 杂 化 , 因 此 , 赤 道 平 面 内的 杂 化 轨 道 有 较 多 3d(x 2 -y 2 ) 成 分 , 轴 向 键 没 有 3d(x 2 -y 2 ) 成 分 , 因 此轴 向 配 位 键 长 于 赤 道 面 键 。b) [Cu (Cl) 5 ] 3- 为 三 角 双 锥 , 中 心 原 子3d(z 2 ) 轨 道 为 单 占 据 , 其 它 3d 原 子 轨 道 满 占 据 , 中 心 原 子 取 sp 3 d(4dz 2 ) 杂 化 , 但 轴 向 配 体 受 到 的 内 层 3d 电 子 的 排 斥 比 赤 道 面 内 配 体受 到 的 排 斥 小 , 所 以 轴 向 配 位 键 短 于 赤 道 面 键 。

From 2e, 8e, 18e rules to electron countingrules for metal cluster compounds and styxmethod for electron-deficient molecules.• 2e, 8e, 18e rules govern the stability of such molecules/groupsAX n and accounts for their geometries in combination with theVSEPR or hybrid orbital theory.• Yet, there are exceptions of these rules, e.g., PCl 5 , and PtCl 42-,whose structures can be well-understood with use of the VSEPRand hybrid orbital theory (or more precisely the generalized octetrule).• The electron counting rule for transition-metal cluster compoundsis actually a natural extension of the 18e rule, which states thatwhen the valence electrons provided by the ligands of a TMcenter are not enough to fulfill the 18e rule, valence electronsfrom neighboring TM center(s) should be involved to maintainthe TM center to have 18 VE. 1b (18n g)2

• For electron-deficient molecules/ions, the (styx) method is also anatural extension of the 8e/2e rule. In case that the total valenceelectrons of a molecule are inadequate to maintain the essentialnumber of 2c-2e bonds, formation of one or more 3c-2e (or nc-ne)bonds can make all component atoms fulfilling the 8e/2e rule.BHBB BBBBBHs t y xHx = m-st = n-sy = (2s-m)/2p sets of styxStyx methodp isomers

General rule for stability of molecules/ions• The molecules/ions should have a large HOMO-LUMO gap toenforce kinetic stability!• The 2e/8e/18e rules work well because molecules adopting suchelectron configurations always have a large HOMO-LUMO gap.• Following this general rule, the Wade’s (n+1) rule accounts for thestability of closo-boranes/carboranes, whose skeletal bondingelectrons are actually delocalized over the whole molecule.• Similarly, the Huckel (4n+2) ruleaccounts for the stability of cyclic -conjugated systems that havedelocalized electrons.

The chemical bonding in B n H n2-(n=5,6,7)? P.184, 6.4According to Wade’s (n+1) rule,• In B 5 H 52-, there are 5 B-H -bonds, one radial bonding skeletalMO and 5 tangential bonding skeletal MOs.• In B 6 H 62-, there are 6 B-H -bonds, one radial bonding skeletalMO and 6 tangential bonding skeletal MOs.• In B 7 H 72-, there are 7 B-H -bonds, one radial bonding skeletalMO and 7 tangential bonding skeletal MOs.It is known that [Co(NH 3 )Cl 2 ] has two isomers. P.184, 6.9• If the complex has a planar hexagonal structure, there should be 3isomers depending on the relative position of two Cl ligands.• If the complex has a trigonal prismatic structure, there should bethree isomers.• If the complex has a trigonal antiprismatic structure (or octahedralstructure), there should be only two isomers. Therefore thecomplex has this structure.

Examples p.185, 6.17, 6.24,• 18-e rule, TM clusterb=(18n-g)/2 b= number of M-M bonds.Or b = (18n + 8m –g)/2• Ni(CO) 4 :(m = non-metal in-cluster atom).1) Ni (10ve) + 4(CO) (8ve) = 18ve; sp 3 hybridization. Tetrahedralstructure, T d -symmetry.2) Ni (3d 10 ), 3d orbitals split into (3d z2 , 3d x2-y2 ) 4 (3d xy ,3d yz ,3d xz ) 63) no d-d excitation can be observed in such a case.• Fe 5 C(CO) 15 : g= 5x8 + 4 + 15x2 = 74; b= (18n-g)/2 = 8Square pyramid with 8 Fe-Fe bonds. (in-cluster C atom?!)• Ru 6 C(CO) 16 : g= 6x8 + 4 + 16x2 = 84; b= (18n-g)/2 = 12Octahedron with 12 Ru-Ru bonds!

Examples p.185, 6.17, 6.24,• [Rh 6 C(CO) 15 ] 2- : g= 6x9 + 4 + 15x2 +2 = 90; b= (18n-g)/2 = 9trigonal prism with 9 Rh-Rh bonds!• [Ni 8 C(CO) 16 ] 4- : g =8x10 + 4 +16x2+ 4 = 120, b = (18n-g)/2 = 12cube-shaped with 12 Ni-Ni bonds.• [Ni 8 C(CO) 16 ] 2- : g = 8x10 + 4 +16x2 + 2 = 118, b = (18n-g)/2 =13square antiprism. (actually synthesized! )

Examples p.185, 6.26,• Fe 6 (CO) 18 : g= 6x8 + 18x2 = 84; b= (18n-g)/2 = 12Octahedron with 12 Fe-Fe bonds.• [Fe 4 RhC(CO) 14 ] : g= 4x8+9+4 +14x2+1 = 74; b= (18n-g)/2 = 8square pyramid with 8 M-M bonds!• [Co 6 N(CO) 15 ] : g= 6x9 + 5 + 15x2 +1 = 90; b= (18n-g)/2 = 9trigonal prism with 9 Rh-Rh bonds!• Ni 8 (CO) 8 (PPh 3 ) 12 : g = 8x10 + 8x2+12x2 = 120, b = (18n-g)/2 =12cube-shaped with 12 Ni-Ni bonds. (wrong ligand in text book!)• Ni 8 (PPh) 6 (CO) 8 : g = 8x10 + 6x4 + 8x2 = 120, b = (18n-g)/2 =12cube-shaped with 12 Ni-Ni bonds. Each PPh ligand is m 4 -ligated!

Examples p.185, 6.25,• Fe 2 (m 2 -CO) 3 (CO) 6 :1) g= 2x8 + 9x2 = 34; b= (18n-g)/2 = 1There exists one Fe-Fe bond.2) The CO(m 2 ) forms two Fe-C bonds with the Fe atoms, adoptingC=O double bond. On the contrary, a terminal CO forms adonative Fe-CO -bond and, meanwhile, get backdonation fromFe(3d) arising from the 3d(Fe)2*(CO) bonding, whichresults in a partially weakened CO triple bond. Thus the C-Obond in a terminal CO is strong than that in a bridging CO(m 2 ).3) Why are there three m 2 -CO ligands? To fulfill the 18e rule, allvalence AOs of a Fe center should be involved, demanding each Fecenter being 6-fold coordinated (excluding Fe-Fe bond). Thus forFe 2 (CO) 9 , three m 2 -CO ligands are required to keep both Fe centersbeing 6-fold coordinated.

Example: Re 2 (CO) 4 (h 5 -C 5 H 5 ) 2Re-Re bond order =? Why are there two m 2 -CO ligands?1) g= 2x7 + 4x2 + 2x5 = 32; b= (18n-g)/2 = 2Re-Re bond order is 2.2) For a TM complex fulfilling the 18e rule, all thevalence AOs of a TM center should be involved inthe chemical bondings. As such the coordinationnumber of a Re center should be 6 (excluding theRe-Re bond). As a C 5 H 5 ligand makes use of 3valence AOs of Re, there should be 3 CO ligandscoordinated to a Re center. Thus, two m 2 -COligands are required to keep both Re center being6-fold coordinated!

根 据 Hückel 近 似 , 写 出 下 列 分 子 的 电 子 分 子 轨 道 久 期 行 列 式 :Answer: Generally, these three molecules all have a 44bond, the MO of which can be expressed as 4 cAccordingly to HMO, the seqular determinant of moleculea isαβ00EαβββEα0ββEα0ββE0i1p.150,5.21Key point: In the framework of HMO, the resonance integral exists only when two centers are directly connected!ii

The seqular determinant of molecule b isαEβ0ββ0αβEαβE0β0β0βαEThe seqular determinant of molecule c isα Eβ00β0αβEαβE0β000βαE

C 5 H 5 and its anion:symmetric MOs, boundary conditionAsymmetric MOs:)])(/cos())(/cos(A[),/cos(E)])(/cos())(/cos(A[),/cos(EA,E/m)/)sin(/sin(coscos)cos()cos()cos(cossssskkss1542333154232251115254542545252220,1,2)(m520225223222φφπφφπφψπβαφφπφφπφψπβαφψβαπθθθθθθθθθp.150, 5.24

C 5 H 5 radical and its anion:asymmetric MOs, boundary condition2cosθ sin( 2θ sin 3θ sin 2θ 2 sin( 5θ/ 2 )cos( θ / 2 ) 0 θ 2mπ/ 5 (m 1,2)EψEψas1as1as2as2) sin( α 2βcos( 2π/ 5), A[sin( 2π/ 5)(φ α 2βcos( 4π/ 5), A[sin( 4π/ 5)(φ2θ22) sin( θ ) Three symmetric and two asymmetric MOs! s1 is the lowest occupied MO (LOMO) with E s1 = 2 E as1 = E s2 , E as2 = E s3 (doubly degenerate MOs)φφ44) sin( 4π/ 5)(φ) sin( 2π/ 5)(φ11φφ55)])]

C 5 H 5 and its anion:1) For neutral C 5 H 5 , 5e in total.E total = 2( + 2) 3[+2cos(2/5)]= 5 4 6cos(2/5)For the localized system: E loc = 2x2( + ) 5 + 4 E deloc = E total – E loc = 6cos(2/5) 1.852) For C 5 H 5 anion, 6e in total.The s1,s2 and as1 MOs are fully occupied.E total = 6 + 4[1+2cos(2/5)]For the localized system: E loc = 6 + 4 E deloc = E total – E loc = 8cos(2/5) 2.47

HMO---More complex systems (p.138, )The seqular determinant of this molecule: isα Ecββ α E β 0 0β β α E β0 0 β α Eβ01β c2β c3(α E ) c4β For symmetric MOsc1c2 cos( θ/ 2 );c3 cos( 3θ0 2cosθc3 c1 c2c4/ 2 );3θθ 5θθc4 2cosθc3 c1 c2 2cosθcos- 2cos cos - cos2 2 2 25θθ 3θ 2cosθ[cos- cos ] cos ( boundary condition! )2 2 2For asymmetric MO( E)/ 2cosIt is the antibonding -MO of a C=C bond!cos=-1

Linear carbon chain with 2n carbon atoms: HOMO and LUMOsinθsin 2θsin 3θsin( 2n1)θ sin 2nθmπE α 2βcosθ,where θ 2n1(m 1,2,3,...,2n)2n2 kmπψ msin φk2n12n1(m 1,2,3,...,2n)k12ne , n occupied -MOs; so HOMO (m=n), LUMO (m=n+1).ψHOMO22n12nk1sinknπφ2n1k,EHOMOα2βcosnπ2n1ψLUMO22n12nk1sink( n 1)π2n1φk,ELUMOα2βcos( n 1)π2n1

Linear carbon chain with 2n carbon atoms: HOMOψHOMOsinθsin 2θsin 3θsin( 2n1)θ sin 2nθ22n12nkknπnπsin φk, EHOMO α 2βcos1 2n12n1c 2n2 π n( 2n1)π nπnπ2 sin sin[] sin( nπ )n2n12n12n 1If 2n=4l,nπnπc2n sin( nπ ) sin c12n12n1Similarly, c2n 1 c2,....,cn1 cn HOMO is asymmetric!If 2n=4l+2,nπnπc2n sin( nπ ) sin c12n12n1Similarly, cn c ,....,cn c HOMO is symmetric!2 12 1n

Linear carbon chain with 2n carbon atoms: LUMOψLUMOsinθsin 2θsin 3θsin( 2n1)θ sin 2nθ22n12nk( n 1)πsin φ2n1,Ek LUMOk1 ( n 1)πα 2βcos2n1c ( 2n2 2n ) π n( 2n1)π nπnπ2 sin sin[] sin( nπ )n2n12n12n 1nπnπ( n 1)πIf 2n=4l, c2n sin( nπ ) sin sin c12n12n12n1Similarly,If 2n=4l+2,c2nSimilarly,cc,....,cc LUMO is symmetric!2n12 n1nnπnπ( n 1)π sin( nπ ) sin sin c2n12n12n1c2n 1 c2,....,cn1 cn LUMO is asymmetric!1

Linear carbon chain with 2n carbon atoms: HOMO and LUMOIf 2n=4l,HOMO: cLUMO: cIf 2n=4l+2,sinθsin 2θsin 3θsin( 2n1)θ sin 2nθ casymmetric!2n1 2n12 n1nc,c,c cc,....,c,....,csymmetric!2n1 2n12 n1nc cHOMO:c2nc1,cc,....,c2n12 n1cnsymmetric!LUMO:c c,....,c c2n12 n1n asymmetric!Simplified motif of symm. & asymm. -MOs:Symmetric!p -AO of terminalcarbon atomAsymmetric withrespect to Cs operation!

Linear carbon chain with (2n+1) p -orbitals.Ψsomosinθ,sin 2θ, sin3θ, sin 4θ, .....sin(2n1)θE α 2βcosθψmN e =2n+1 n doubly occupied MOs (m=1,..,n), and a singlyoccupied MO (SOMO, m= n+1). SOMO =/2, E SOMO =, non-bonding!22n 222n 22n1 k1kmπsin φ2n 2kπ2n 1ckφkwhere ck sink12If 2n+1 = 4l +1 (e.g., 5),4 1π ( lc sin 1;c2 1 sin)πn22θ mπ/(2n 2)1If 2n+1 = 4l +3 (e.g., 7),k1(m 1,2,3,...,2n 1)(m 1,2,3,...,2n 1)symmetric4 31π ( lc sin 1;c2 1)πnsin 1asymmetric22

linear system, C5H7 anion. P.150, 5.24sin,sin2, sin3, sin4, sin5Boundary : sin6Emm 2cos262n1 k1kmsin k60 m/6m/6(m 1,2,3,...,5)(m 1,2,3,...,5)(m 1,2,3,...,5)6e, three doubly occupied MOs (m=1-3)E total 2(E1 E2 E3) 6 5.464For localized system,E local 22( ) 2 6 4Delocalization energyElocalEtotal Elocal 6 4[cos( / 6) cos(2/ 6) cos( / 2)]1.464

Cyclic system with 2n carbon atoms: HOMO and LUMOSymmetric MOs2n12n1cos θ cos θ2 2θ 2 sin( nθ)sin 02 θ mπ/ n( m 012 , , ,...,n1)E sm α 2βcos( mπ/ n ) ( m 012 , , ,...,n 1)nA1k0ψsm ( φ1 k φ2nk)cos[( 2k1)mπ/ 2n ] ( m 01 , , 2,...,n )

Cyclic system with 2n carbon atoms: HOMO and LUMOAsymmetric Mos, boundary condition:2n12n1 sin θ sin θ2 2θ 2 sin( nθ)cos 02 θ mπ/ n( m 12 , ,...,n )E as α 2βcos( mπ/ n ) ( m 1,2,...,n )nA1k0ψasm ( φ1 k φ2nk)sin[( 2k1)mπ/ 2n ] ( m 1,2,...,n )So the -conjugated molecule has 1 LOMO(s), 1 HUMO(as) and (n-1) doubly degenerate MOs. HOMO --doubly degenerated!For 2n = 4l +2, the two degenerate HOMOs are fully occupied!E HOMO α 2 β cos[( n 1)π / 2n]

Cyclic system with 2n carbon atoms: HOMO and LUMOFor 2n = 4l , the two degenerate HOMOs are singly occupied!E HOMO α 2βcos[ nπ/ 2n] αe.g., cyclobutadiene, the HOMOs are singly occupied!E HOMOE LOMO α α 2βThus,E total = 2( + 2) 2 = 4( +)For localized system: E loc = 4( +) E deloc = E total – E loc = 0No delocalization stability! Antiaromatic!

Cyclic system with 2n+1 carbon atoms: HOMO and LUMOSymmetric MOsBoundary: cos nθ cos( n 1)θ( 2n1)θ 2 sin sin2 θ 2mπ/( 2n1)θ 02( m 012 , , ,...,n )E sm α 2βcos[ 2mπ/( 2n1)]( m 01 , , 2,...,n )nψsm A φn1 ( φnk1 φnk1)cos[ 2kmπ/( 2n1)]k1

Cyclic system with 2n+1 carbon atoms: HOMO and LUMOAsymmetric MOssin nθ sin( n 1)θ( 2n1)θ 2 sin cos2 θ 2mπ/( 2n1)θ 02( m 12 , ,...,n )E asm α 2βcos[ 2mπ/( 2n1)]( m 1,2,...,n )asmAnk1( nk1nk1)cos[2km/(2n1)]• So the -conjugated molecule has 1 LOMO, and n doublydegenerateMOs (one is symmetric, one is asymmetric).• HOMO – doubly-degenerate!

Cyclic system with 2n+1 carbon atoms: HOMO and LUMO• HOMO –doubly-degenerate!a) For 2n+1 = 4l +1, the doubly degenerate HOMOs hold 3electrons! 2cos[ n/(2n1)]E HOMOThus, such type of cyclic pi-conjugation systems tends to acceptone more electron to have a stable closed-shell electronicconfiguration, which also fulfills the (4N+2) Huckel rule ofaromaticity!b) For 2n+1 = 4l +3, the doubly degenerate HOMOs hold 1electrons! 2cos[( n 1) /(2n1)]E HOMOThus, such type of cyclic pi-conjugation systems tends to lose oneelectron to have a stable closed-shell electronic configuration,which also fulfills the (4N+2) Huckel rule of aromaticity!

Cyclic system with 2n+1 carbon atoms: HOMO and LUMOe.g., C 3 H 3 , MOs! p.150Symmetric MOs, 5.23cos 2 cos 0,2/ 3; E s 1 2, E ssin 2 sin / 3;s1s2as1 A( )1A[ sin(2/ 3) sin(2/ 3)]1621 A[ cos(2/ 3)1( Asymmetric MOs,1( )1122)3233(A23 1/3E as 13)cos(2/ 3)] 2

e.g., C 3 H 3 , the HOMO is singly occupied!E HOMO α βp.1505.23E1 2Thus, E total = 2( + 2) + ( -) = 3( +)For localized system: E loc = 2( +) = 3 +2 E deloc = E total – E loc = For C 3 H 3 cation, E total = 2( + 2)For localized system: E loc = 2( +)E deloc = E total – E loc = 2Enhanced delocalization fulfilling the Hückel rule!

Cyclic system with 2n+1 carbon atoms: HOMO and LUMOe.g., C 7 H 7 cation,Delocalized systemE total = 2x(2) + 4(2cos(2/7)) 6 4(12cos(2/7))For localized system: E loc = 3x2( +) Delocalization energy:E deloc = E total – E loc = 8cos(2/7)2 = 2.99

O 3 - cyclic or linear geometry?Suppose either geometry has a P 34conjugation system.For cyclic form, the three -MOs,E1 2E2 E3 E totalcyc22E1 2E 4 2For linear form, the three -MOs,E1 2E totallineartotalE linear E22E2 , E3 2E1 2E 4 2 2totalcyclic 2( 2 -1) 0So the linear form is more stable than the cyclic form.Note: such a treatment is not too reasonable as thereare acturally 6 pi-electrons in the cyclic form!

H 3+- cyclic or linear geometry?• Let’s consider the MOs consisting of three H1s AO byusing the HMO theory!• Similar to the O 3 case, we have the following MOenergies for the cyclic and linear forms of the H 3 system.For the cyclic form,E totalcyc 2E1 2 4E1 2For the linear form, E 2E totallinear 2E1 2 2 21So H 3+prefers the cyclic form!totalE linear EE2 E3 E2 , E3 2totalcyclic 2(2 - 2) 0How about H 3 and H 3- ?

Which isomer of C 4 H 6 is more stable, 1,3-butadiene (BD) ortrimethylenemethane(TMM)?1) For BD (see textbook),Em E25BDtotal total2cos, where4k1kmsin k5 2E1 2E24P 4 m5 4(m 1,2,3,.4)4.472(m 1,2,3,4)2) For TMM, the seqular determinant is00E0 E000 E E x2 ( x2 3) 0 x x0010x0100x13,0,0111x0xor ( E E) / x

Which isomer of C 4 H 6 is more stable, 1,3-butadiene (BD) ortrimethylenemethane(TMM)?2) For TMM, E00xExE21210 E( x20 3) 00 E0 E4P 4x0010x0100x13,0,0111x0xor 3, E E ,E 3( xE TMMtotalBDEtotal2 3) 02x 3 3, E E ,E 32x 343,0,04 ( E E) / x 2E1 E2 E3 4 2 3 4 3. 4644 TMM 4( P4) Etotal( P4) BD is more stable than TMM!

C 2 H 4 + Br 2 CH 2 Br-CH 2 Br is not an elementary rxn! P.150, 5.29LUMO (Br 2 ) – u * HOMO(Br 2 ) – g *Symmetry incompatible!Overlap=0eSymmetry compatible! Wrongdirection of electron transfer!HOMO(C 2 H 4 ) – uLUMO(C 2 H 4 ) – g *• LUMO (Br 2 ) and HOMO(C 2 H 4 ) are symmetry-incompatible!• The interaction between HOMO (Br 2 ) and LUMO(C 2 H 4 ), thoughbeing symmetry-compatible, leads to electron transfer incompatiblewith the relative electronegativity of the reactants!• The reaction can not be an elementary rxn.

p.150, 5.32Key points:1. Frontier molecular orbital theory.2. HOMO of hydroquinone interacts with LUMO of quinone!• Both molecules have similar 8-center -conjugations, i.e.8c-10e in molecule a and 8c-8e in molecule b.• Thus, the LUMO of molecule b resembles the HOMO ofmolecule a, i.e., both are symmetry-compatible!• Hence, the LUMO of molecule b can interact effectivelywith the HOMO of molecule a, leading to electrontransferfrom a to b.(You may figure out the diagrams of these MOs with use ofthe HMO theory)Let’s consider a simpler case in the next page!

The reaction of [Et4N]I and tetracyanoethylene (TCNE) forms[Et4N] 2 [TCNE] 2 , which possesses [TCNE]22 with a 2.827(3) Åintradimer CC bond distance (CrystEngComm, 2001, 47,1). Plzanalyze the intradimer bonding.‣ The LUMO of neutral TCNE is the *MO of the C=C moiety.‣ This MO becomes the SOMO of TCNEmonoanion!‣ So the SOMOs of two [TCNE] aresymmetry-compatible and caneffectively interact with each other!

mPlease derive the -MOs of cyclopentadiene and explain why thereaction of two cyclopentadienes gives rise to [4+2] cycloadditionproduct, but not the [2+2] or [4+4] cycloaddition product.Answer: 1) The -MOs of cyclopentadiene are similar to those of 1,3-butadiene; the MO coefficients are given in the diagram. ThusThe boundary condition issin 5θ 0 θ mπ/ 5 (m 1,2,3,4)The four -MOs are141 k 1k1, (1/ 2) sin( k/5) , E 1. 618m m m 242 k 2k1, (1/ 2) sin(2k/5) , E 0. 618343 k 3k1, (1/ 2) sin(3k/5) , E 0. 618444 k 1k1, (1/ 2) sin(4k/5) , E 1. 618HOMOLUMO

2)Symmetryallowed[4+2]HOMO LUMOA) For [4+2] cycloaddition of two cyclopentadienes, the 1,4-site ofmolecule a interacts with the 1,2-site of molecule b. The 1,4-site ofHOMO (a) is compatible with the 1,2-site of LUMO(b); meanwhile,the 1,4-site of LUMO(a) is compatible with the 1,2-site ofHOMO(b). So the [4+2] cycloaddition is symmetry-allowed.B) For [4+4] cycloaddition, the 1,4-site of molecule a should interactwith the 1,4-site of molecule b. However, the 1,4-site of HOMO (a) isnot compatible with the 1,4-site of LUMO(b); likewise, the 1,4-site ofLUMO(a) is not compatible with the 1,4-site of HOMO(b). So the[4+4] cycloaddition is symmetry-forbidden. Similarly, the [2+2]cycloaddition is symmetry-forbidden and could not occur!

More exercises of chapter 5-9 can be found at:http://ctc.xmu.edu.cn/jiegou/wlkch/Chapter5/exercise.htmhttp://ctc.xmu.edu.cn/jiegou/wlkch/Chapter6/exercise.htmhttp://ctc.xmu.edu.cn/jiegou/wlkch/Chapter7/exercise.htmhttp://ctc.xmu.edu.cn/jiegou/wlkch/Chapter8/exercise.htmhttp://ctc.xmu.edu.cn/jiegou/wlkch/Chapter9/exercise.htm

止 于 至 善 !这 不 只 是 做 事 的 态 度 ,还 是 成 功 者 的 必 要 素 质 ,更 是 科 学 精 神 !---- 谨 以 此 校 训 与 各 位 学 生 共 勉 !