Document 6594095
Transcription
Document 6594095
Chemistry l04A Exam 2-1013012014 Valence Atomic Orbital Energies tU" (15 points) MO Theory, UV-Vis Spectra (or partially Electrons can be promoted from full (or partially full) orbitals to vacant in the occuls vacant) orbitals by photons of light. The lowest energy absorption usually visible or ultraviotei (uv) regio-n of the spectrum, and results in the promotion of an (LUMO) electron from the highLst o..rli.a MO @OMO) to the lowest unoccupied MO of the molecule or-ion. Using MO theory, explain the following observation: F2 is virtually colorless, while Clz is yellow; Br2 is deep red, and 12 is purple. fo" aW htl'o6orr+ U) , l'+0rlo 7 4, ",nA Lu440' fi, tr'- 'b2o -V* &\/ LV"V<- W/4"- LLLlt4o A T) 4t : 9cr'- ovbto * IL ,t --rd It nbs W,,nu ^br*bg bb^<---*d,t*,'t Cy: abro'V* 0*; - r^*L Cen'il. Ko"t/u-lu* *;" .-. .!_ 4d r pozoL o,l--l"f zL-?,'\ 'ghu& s*,o,/,( )",,r/- ilrl*t,*' r>z&'"+ UI A,@l)u -/>--.- / 1\ f\-'l? a >v : l*q) boq/)- 5'latgPwealvn) e- r^zrl/r+ l^.p t2]. (15 points) MO Theory, Electron -'/ I ?Oor- a/%W \erru,lrL ftaao-cu* g^e n1 L* ?-' -n- r'-i,l t- Affinity rl -?_ oz (0.40 ev) Explain why the electron affinity of cz is much larger (3.54 eV) than that of orN2 (-2.2 eV). 4o ftD Pi^'3''*' q"4) /,-\ \ t-eA"_,fut/ ;:--''- r r,-_q, I ? 'lb 7b'L @@ uc fr) -4..4b.. ,-T ly -1t" fuupts.cl+ aalX e cenn";'4", lt4 e, f+t* d*{ {,,ilu f^ unlret, 2>, 0- f,,/t h'04 v( l- W qrfii b,r/,1 4,{.o> (u# ) s'tryA L\a,o / q ^W7b A- grnru,/'[,- AA, giri^rtr 13] (30 points) Valence Bond Theory Experimentally, we know the axial and equatorial positions in PHs are non-equivalent, with bond tengths such that raxiar ) requatoriat: Valence bond theory treatment of such trigonal bipyrimidal molecules requires sp3d orbital hybridization. Determine the wavefunctions for the 5 hybridizedorbitals use the following in-plane coordinate system, and explain why rrgrl ) fequaroriar. 5 + cr % - c6tu +4"!; ?.r= r,t) -Lr?1 -Q(r 4c6J{ f,, = L\i + C7 fr + cq olr* ?., :. c1 frvn 6u,,tof47. % ^ evtu { nurqil4''5,. L( E6 u',/'/e-ra c,o* )* V)a,ur-l,,*t A I +c{= /or*' 'b r* +f-a; '(0.-, -i-fu *b"e; ?q, C, +i,-?g -,boo \ ?o, Ofurf "lo8, ct c,t + gcy = o 7 6rbiW ! Ar-* + 5c? = 2ci ;5*: ft)r*- fr,- z =+ ,t=b B) \2 ' l+C= .I*r;_ C; .b = )-otx - ,fo ' 6 fu = *, 'op. '?, lg tE= I i> C.r =O. An'oX +;rn f / ,o/.A gSZ, s, 677" ? br/+; toZ &^M*1 bon&>" 0 brq*. g .U t/UlAt Cl4^rrv+'r-' a- dko*r- t*ff Vvld6e- i^ eEuM.,/ b4 *w -t'r-^, ( d''4, "kn L oUrn*+--^ )i,r nv',r^ bDruLg Wr+ loryr--), lq @A points) Molecular Orbital Theory 1. (a). Start with a C2 cluster and sketch its MO diagram. Fill in the electrons, and find the LUMO and HOMO. Sketch the shape of the LUMO and HOMO. What is the CL4C ,,8, /r\ Ir--^ ,\ ttvfI u '/ \ \ /t l---"---j -'loilol ett. ''/'-T -t/ [',R. 2ro ''4L'ry-' e*q lzt -'\1,qil 'z -.. .+lrfi'5 \1b " tq \*o,h,, lz ,1 G c-- c @ L.uw" 69 L-c ^7 €9 @ e,o-3-s/ f;*;TW an4,'-ba"/, T (b). Now let's take a look at a CO molecule. Similarly, sketch its MO diagram. Fill in the electrons, and find the LUMO and HOMO. Sketch the shape of the LUMO and HOMO. What is the bond order of the CO molecule? Sketch the photoelectron spectrum for CO. What would you look for in the PE spectrum to distinguish CO from Cz? L c.o O 7@"' -; lz' -1N,/:<-\\ t l. t\ \ t, '-'lL ^_ ,4+Y 1*-- -11,*'v-- " ,-'\: -15 X<-r/ '7u=^ lf ', '1 )- 46'€e-/ ' lo W4to: Lo fEt '.u''*,Q h""v- Af orll;*'oar,l. , b* 0'@ oo >U,l/t : I zP q -o c @ @ UW r1*rtv/ *&3, ) (rrr"l- ba"ah'g ) (9 kAL, ..^ (c). Which species (C2 or CO) is more stable? Why? D br- wLo'rt/ gWe o\t*- *" 'U7 borr.& arok^. Orw cenW 4s, a/Xtac{-{4'"f C'Lo v/o'4\il ,1o99e99e9 A- hn'gh.-- No'{//) *rr(- bwu Lt4,t/z-t, ,a.akiV tV"r-.1-;,rv. Lwk: CO cat, bc-- {r*& q 79d,*.rX, A C",rr, rrt t ) i+ lU Wor^t-,' . (d)" Consider the following reactions, Hr*C=C+HC=CH BHr+CO+H3B-CO Which MO is used for Cz and CO? fr. Q'. LU,//-() / o/c-,rla,,y> tldl 4 Hu fta-U<: lcz>taff o'*'^l'"6' ,LDV: WLuw -/(- e ff2"ue I L-a C fu- CC', fro,qo W I Ce"@ jh @..o@ 14 vo{z'. Co Lo<,trts vouloL Xi"'<- L.z bol,*, du'oh 6{6,n>r? vna,,4rrlr, ul^er^, art qu*4V'n, Le*iz tl (,) l4'ru12+*r\- oo C- 6 WW @ @ c ' (e). C+ can be considered as the combination of two units of Cz. Let's consider two possible geometries: linear Ca (butadiene form) and square planar Ca (cyclobutadiene form). Sketch a partial MO diagram for each configuration considering only frontier MOs and draw the orbital interactions. Is there any difference between the two configurations in the bonds formed? What if we combined two Cz2' units (isoelectronic to CzHz)? How would the bonding change? Aar|h Llne,,<'. \ brt* on ?X*&l ( Naw/tu,tto, W,ilz//P./@, cuta/tu,a), 5r, y(ort.---, (aL"'c-=2 D I f= ? /22 v -. I l,fuwt tzr W, , ,1/ ,o- '"'1''# -.1r 4H '-__W_'Z 4t ,'.1\//-- -r3--;'J e->sltq ./rg*.1.-ra2 o 1/No,wt ffo,llo .z- ban l< o No,r,tol /t4o,(//2 ^4-b''A/ *Bf** I ) i O -La'"t LU,IID ;Lu-nct ) I @e@ I I I f* C- : PaD l)/,,t/ttiy,** l)w,-'. O@oO Q=Q--C= @-6L@ {G\Lu4,L(,, t",i 9-=P v -o L7 " dW&il/), Aoq @ra.@ @42@ ioilud'brbo{4 ,ray z-ba,rl ,; (( ve{fl dz4'eaz/'t''7 21'1 er,L|e-r"<-?(r^ frl^i Jhz1e- -z-btil "L{ot Cl,i. t)"Jln".^ 1a.,.- gcfuo- /n' wiT yogil"}* {4/!u tur, cvr Aq9,r/*.? ru,di'bonfu'_V olrlqt'tttt.) ,ttUt #"* , r,*^!oV