Chem 130 Third Exam Key

Transcription

1 Nme Chem 130 Third Exm Key On the following pges re eight questions tht consider topics rnging from precipittion solubility, cid bse, nd oxidtion reduction rections to metl lignd complexes nd coordintion compounds. Red ech question crefully nd consider how you will pproch it before you put pen or pencil to pper. If you re unsure how to nswer one question, then move on to nother question; working on new question my suggest n pproch to the one tht is more troublesome. If question requires written response, be sure tht you nswer in complete sentences nd tht you directly nd clerly ddress the question. Question 1 /15 Question /17 Question 3 /15 Question 5 /15 Question 6 /8 Question 7 /15 Question 4 /15 Totl /100 Potentilly useful equtions nd constnts: c = ln KE = hn W E = hn = hc/l 1 = λ 1 nm n1 n 1 FC B = V - N V Q $Q d - AVEE = xie, + yie / + zie 1 x + y + z (vlence shell electrons only) δ = V - N - B æ EN ç è EN + EN b ö ø c = m/s h = Js N A = mol 1 Other potentilly useful informtion is vilble s seprte hndouts.

2 Problem 1. For ech pir of cids, circle the one tht is the stronger cid nd, in no more thn 3 sentences ech, explin the reson(s) for your choice. For the compounds in (b) nd in (c), the cidic hydrogen is ttched to oxygen. For ll three pirs, the stronger cid is the compound with the wekest X H bond, where X is the element to which the cidic hydrogen is bound. () NH 4 + vs. PH 4 + : In this cse we re compring N H bond to P H bond nd the length of the bond is the importnt considertion. Becuse P hs lrger rdius thn N, the P H bond is longer nd weker, mking PH 4 + the stronger of the two cids. (b) HOBr vs. HOI: In this cse this cidic hydrogen is bound to n oxygen for both molecules. Here the importnt considertion is the electronegtive element to which oxygen is bound nd its effect on the strength of the O H bond s it pulls electron density wy from this bond, wekening the bond. The more electronegtive Br hs greter effect thn the less electronegtive I, mking HOBr the stronger of the two cids. (c) H CO 3 vs. HCO 3 : In this cse the cidic hydrogen is bound to n oxygen tht is, in turn, bound to crbon. There re no differences in the electronegtivities of the elements in the two compounds, nor re there ny differences in oxidtion sttes tht might explin difference in cidity. Here, the importnt considertion is chrge; it is more difficult to remove positive chrge ( proton, H + ) from n nion thn it is from neutrl molecule; thus, H CO 3 is the stronger cid. Problem. The following informtion is known for rections between the elements A, B, C, nd D, nd their +1 ctions, A +, B +, C +, nd D + ; note tht NR mens there is no rection: () C + B + C + + B (b) C + + A NR (c) A + + D NR (d) D + B + D + + B Predict the products of the following two rections, writing NR if there is no rection. (e) D + + C C + + D (f) A + + B NR In the spce below, present convincing rgument tht your responses re correct by rrnging the ction/element pirs into rectivity series ordered from the strongest oxidizing gent to the wekest oxidizing gent nd explining how you rrived t this order; in ddition, explin why your rectivity series supports your predictions. Limit the written portion of your response to 3 6 sentences. strongest oxidizing gent B + /B D + /D A + /A C + /C wekest oxidizing gent From rections () nd (d) we know tht B + is stronger oxidizing gent thn D + nd C +, which plces B + /B bove D + /D nd C + /C. From rection (b) we know tht C + is weker oxidizing gent thn A +, which plces C + /C below A + /A. From rection (c) we know tht D + is stronger oxidizing gent thn A +, which plces D + /D bove A + /A, giving the finl order shown on the left. The predictions bove for the rections recognize tht stronger oxidizing gent (upper left) will rect with stronger reducing gent (lower right).

3 Problem 3. Suppose you hve solution tht might contin one or more of the following ctions: Pb + Co + Fe 3+ B + NH 4 + To identify the ions in the solution you dd HCl to the solution nd observe precipitte. After centrifuging, you remove the superntnt solution from the precipitte, dd H SO 4 to this solution, nd observe precipitte. After centrifuging, you remove the superntnt from this second precipitte, dd NOH to this solution, nd observe tht no precipitte forms. In the tble below, plce ech of the five ctions in the cell tht mtches your prediction tht it must be present, tht it must be bsent, or tht you do not hve sufficient informtion to determine if it is present or if it is bsent. ction must be present ction must be bsent insufficient informtion Pb + nd B + Co + nd Fe 3+ NH 4 + In the spce below nd in short prgrph of 3 5 sentences, present convincing rgument tht your ssignments for ll five ctions re correct. Note: to receive full credit your explntion here must support your predictions. The solution to this problem requires three things: solubility rules, recognizing tht positive test for n ion requires tht you hve n opportunity to see it precipitte or not precipitte, nd recognizing tht once n ion is precipitted the superntnt no longer contins tht ion. With these in mind, we note tht dding HCl produces precipitte of PbCl (s) s this is the only ction here tht forms precipitte with Cl. Adding H SO 4 to the superntnt produces precipitte of BSO 4 (s) s this is the only remining ction tht forms precipitte with SO 4. Adding NOH to the superntnt does not produce precipitte, which mens neither Co + nor Fe 3+ re present becuse they would form Co(OH) (s) nd Fe(OH) 3 (s). Finlly, none of the regents cn rect with NH 4 +, so we hve insufficient informtion to determine if it is or is not present. Problem 4. The drug Nipride, N [Fe(CN) 5 NO], is metl-lignd complex tht is used s source of moleculr NO during surgery where it helps moderte blood pressure. Although the compound releses moleculr NO into the blood strem, it is present in the metl-lignd complex s the ction NO +. When we first described metl-lignd complex we chrcterized the metl in terms of its coordintion vlency nd its ordinry vlency. Using Nipride s n exmple, in -4 sentences, clerly explin the difference between these two types of vlencies nd report their vlues for the metllignd complex in Nipride. The ordinry vlency is the metl s oxidtion stte. In this cse, the two N + ions tell us tht the metl-lignd complex crries chrge of. Ech of the five cynide lignds hs chrge of 1 (CN ) nd the nitrosyl lignd hs chrge of +1 (NO + ); thus, the lignds combine for chrge of 4, which mens the iron is present with chrge of +. The coordintion vlency is the number of electron pirs donted to the metl to form metl-lignd bonds. Ech of the six lignds dontes n electron pir, so the coordintion number is 6. The iron in this metl-lignd complex is expected to be low-spin due to the presence of the cyno nd the nitrosyl lignds. How mny unpired electrons re in iron s d-orbils? Explin how you rrived t your response in -3 sentences. Iron(II) hs vlence shell electron configurtion of 3d6. For low spin complex, the three lower energy d-orbitls fill completely before electrons enter the two higher energy d-orbitls. With six electrons, this mens tht ech of the three lower energy d-orbitls hs two pired electrons nd there re not unpired electrons.

4 Problem 5. Octhedrl metl-lignd complexes of chromium (III) re quite colorful. For exmple, the following is list of severl such metl-lignd complexes nd their colors. Cr(cc) 3 : red bsorbs wvelength of pproximtely 55 nm [Cr(H O) 6 ] 3+ : violet bsorbs wvelength of pproximtely 575 nm [CrCl (H O) 4 ] + : green bsorbs wvelength of pproximtely 665 nm [Cr(ure) 6 ] 3+ : green bsorbs wvelength of pproximtely 665 nm [Cr(NH 3 ) 6 ] 3+ : yellow bsorbs wvelength of pproximtely 415 nm Cr(CH 3 CO ) 3 (H O) 3 : blue-violet bsorbs wvelength of pproximtely 590 nm where the lignds re cetylcetone (written here in shorthnd nottion s cc), wter, chloride, ure, mmoni, nd cette (which is CH 3 CO ). Arrnge the lignds in order of incresing vlues for the octhedrl field splitting, D o. Explin how you rrived t your ordering in prgrph of 4 6 sentences. Note: to receive full credit your explntion here must support your predictions. The color we see is the opposite of the color of light bsorbed. Using color wheel, therefore, gives the wvelengths bsorbed by the compounds tht re listed bove. Becuse there is n inverse reltionship between energy nd wvelength nd direct reltionship between energy nd D o, shorter wvelength mens lrger D o. This is sufficient to put the compounds in order of smllest D o to lrgest D o, but the problem sks us to order the lignds, not the compounds. The lrgest D o belongs to [Cr(NH 3 ) 6 ] 3+, so NH 3 hs the lrgest D o. The next lrgest D o is for Cr(cc) 3, so cc hs the second lrgest D o. The ion [Cr(H O) 6 ] 3+ comes next, so H O hs the third lrgest D o. The compound Cr(CH 3 CO ) 3 (H O) 3 comes next nd, s we know tht H O fvors higher D o, then cette must hve smller D o. Finlly, both [Cr(ure) 6 ] 3+ nd CrCl (H O) 4 ] + hve the sme color, but Cl must hve the smller D o given tht the compound lso includes H O, which fvors higher D o. The order, therefore, is, from smllest-to-lrgest D o : Cl < ure < cette < H O < cc < NH 3. Problem 6. Provide the chemicl formul for the coordintion compound tht hs the following nme: hexmminecoblt(iii) pentchlorocuprte(ii). The formul is [Co(NH 3 ) 6 ][CuCl 5 ]. As check, note tht the ction hs chrge of +3 nd the nion hs chrge of 3. Provide the nme for the coordintion compound tht hs the following formul: (NH 4 ) 4 [Fe(ox) 3 ] where ox is short for the oxlte nion, C O 4. The nme is mmonium trioxltoferrte(ii). Note tht the number of mmoniums is implied by the formul of the nion, tht the nme of the oxlte nion is chnged to oxlto, nd tht the metl s nme tkes on suffix of te nd gives the oxidtion stte. Iron is nmed s ferrte (for ferrum), but ironte is ccepted here s well.

5 Problem 7. Consider the hypotheticl octhedrl metl-lignd complex MABC D, which consists of the lignds A, B, C, nd D. Drw ll possible geometric isomers for this metl-lignd complex. Do not consider opticl isomers. Be creful to drw only once ech unique geometric isomer. There re more cells vilble to you in the tble below then there re unique isomers. You my use the spce t the bottom of the pge or on the bck of this pge to consider possibilities, but only those structures plced in the tble will be evluted. Be sure your structures re cler nd esy to understnd. If mking chnge to structure mkes it too messy, then plce lrge X through the structure nd redrw it in new cell. To drw ll of the isomers it helps to note tht the two Cs cn only be cis or trns to ech other, tht the two Ds cn only be cis or trns to ech other, nd the A nd B cn only be cis or trns to ech other. This leds us to six unique isomers tht pper below nd tht re represented s well using the shorthnd nottion geometry for A & B/geometry for C & C/geometry for D & D Note tht the C/C/C rrngement ppers twice becuse A cn be either trns to C or trns to D (s noted by the symbols AC T nd AD T ).