Chem 104A, Fall 2016, Midterm 1 Key

Transcription

1 hm 104A, ll 2016, Mitrm 1 Ky 1) onstruct microstt tl for p 4 configurtion. Pls numrt th ms n ml for ch lctron in ch microstt in th tl. (Us th formt ml m s. Tht is spin -½ lctron in n s oritl woul writtn s 0 - ) Provi th trm symols rprsnting th stts construct from th tl. Mk sur to inclu th spin-orit coupling quntum numr, J. Thn trmin which of ths stts is th groun stt. (15 pts) ML\M microstts totl. Trm ymols: 1 D2, 3 P2, 3 P1, 3 P0, 1 0 Groun tt: 3 P2 1 pt for corrct ML vlus 1 pt for corrct M vlus 2 pts for corrct numr of microstts 3 pts for corrct microstt numrtions 5 pts for corrct trm symols (-1 for ch wrong trm symol) 3 pts for corrct groun stt

2 2) Th following sris of compouns hv incrsing ipol momnts: N3 = 0.23 D, P3 = 1.03 D, As3 = 2.57 D. Provi t lst two rgumnts to xplin this trn. (10 pts) Th iffrnc in lctrongtivity twn th two toms incrss with incrsing ipol momnt. Th on istncs lso incrs with incrsing ipol momnt u lrgr tomic rii of th cntrl toms. This llows for strongr ipol momnt. As you go own th sris thr is lss s-p mixing. This rsults in on ngls closr to 90. Th mor cut ons woul thn rsult in highr ipol momnt. 5 points for ch rgumnt. 3) Assign th point group of th molculs low. Thn using ltr s ruls trmin th ffctiv nuclr chrg for 3 lctron in th following mtls Mn 0, Ni 0, n o 2+. (12 pts) Mn N Ni o o o o 2v D2h T Mn 0 : 25 (0.35*4 + 18) = 5.6 Ni 0 : 28 (0.35*7 + 18) = 7.55 o 2+ : 27 (0.35*6 + 18) = points for ch point group 2 points for ch Zff

3 4) onsir th clips 6-coorint molcul hxhyri tungstn, 6, shown low. Ech of th six hyri ligns r lll -f. c f z x y ) List ll of th symmtry lmnts in this molcul. (6 pts) E, 3, 3 2, 32, σh, 3, 3 5, 3σv 1/2 point pr symmtry lmnt, -1/2 point pr incorrct lmnt ) ht is its point group? (2 pts) D3h c) Provi symmtry lmnt tht intrchngs only th givn st(s) of hyri ligns. If this is not possil, writ not possil. (4 pts). / n / σv. / n /f n c/ 2 ) Apply th following symmtry oprtions on th structur ov: 3, 2(y), 3. Drw th rsulting structur with th hyris lll, illustrting how thy hv trnsform ftr ll of th symmtry oprtions hv n ppli. ht is th nt symmtry oprtion? (5pts) c 3 c 2 (y) f 3 c f f c f σ (yz) 1 point for ch structur, 2 points for corrct nt symmtry oprtion

4 5) Illustrt grphiclly th convntionl shps n phss of th following oritls. Thn rw its corrsponing ril istriution function long th x xis. (10 pts) 3 x 2-y2, 5px, 2s, 4z2, 3xy 1 point for ch oritl igrm n ch ril istriution function.

5 6) Elmntl sulfur cn us s n lctro mtril for lithium-sulfur ttris. At mort potntils, it forms molculs with th chmicl formul 8, chrctriz y n -- ngl of Drw th Lwis structur of this molcul. (3 pts). Qulittivly, wht is th p-chrctr of th - ons rltiv to tht of th lon pirs? (4 pts) Th p chrctr of th - ons is grtr thn tht of th lon pirs. c. Th molcul cn oxiiz y two lctrons to form An itionl - on is form in this oxition. Drw th most likly structur of this molcul. (3 pts). In lithium-sulfur ttris, 8 cn ruc y 2 lctrons to form Li28. Drw th structur of this molcul. (3 pts)

6 7) Drw th Lwis structur for thiosulfuric ci (223). ht is its molculr gomtry n point group? To which toms r th hyrogn toms oun? Explin. (10 pts) Th molculr gomtry is ttrhrl. 3 pts Th point group is s. 3pts Th hyrogns r oun to oxygn toms cus is mor lctrongtiv thn. This rsults in forming mor stl conjugt s whn th ngtiv chrg is plc on th inst of th. 4 pts 8) A molcul will hv ipol momnt if th summtion of ll of th iniviul on momnt vctors is nonzro. Th prsnc of crtin symmtry lmnts rquir tht molcul s ipol momnt qul to zro cus for th nt cnclltion of th iniviul on ipols. Ths symmtry lmnts r: n invrsion cntr, i; horizontl mirror pln, σ h ; two or mor istinct n xs. Th ipol momnt of 5 is 0. Th ipol momnt of I 5 is 2.18 D.. Us VEPR to trmin th molculr gomtry for ch molcul. Thn trmin thir point groups. (8 pts) I Trigonl ipyrmil D3h qur pyrmil 4v 1 point for ch structur, 1 point for ch gomtry, 2 points for ch point group. Explin how ths xprimntl t r consistnt with VEPR concpts (n th symmtry lmnts ch molcul posssss)? (5 pts) 5 contins σh n two istinct symmtry xs incluint 3 n 2. Tht is why its ipol momnt is zro. I5, howvr, hs non of th ncssry symmtry lmnts. Thus, it hs ipol momnt.