I. OLYMPIC HA HC VIT NAM: K THI CHN HC SINH GII QUC GIA NM 2003 (BNG A) 1. Nhm clorua khi ho tan vo mt s dung mi hoc khi bay hi nhit khng qu cao th tn ti dng dime (Al2Cl6). nhit cao (7000C) dime b phn li thnh monome (AlCl3). Vit cng thc cu to Lewis ca phn t dime v monome; Cho bit kiu lai ho ca nguyn t nhm, kiu lin kt trong mi phn t ; M t cu trc hnh hc ca cc phn t . 2. Phn t HF v phn t H2O c momen lng cc, phn t khi gn bng nhau (HF 1,91 Debye, H2O 1,84 Debye, MHF 20, M H 2 O 18); nhng nhit nng chy ca hidroflorua l 830C thp hn nhiu so vi nhit nng chy ca nc l 00C, hy gii thch v sao? BI GII: 1. * Vit cng thc cu to Lewis ca phn t dime v monome. Nhm c 2 s phi tr c trng l 4 v 6. Ph hp vi quy tc bt t, cu to Lewis ca phn t dime v monome:

Monome

Cl

Al Cl

ClCl; dime

Cl Al Al Cl

Cl Cl

Cl

* Kiu lai ho ca nguyn t nhm : Trong AlCl3 l sp2 v Al c 3 cp electron ho tr; Trong Al2Cl6 l sp3 v Al c 4 cp electron ho tr . Lin kt trong mi phn t: AlCl3 c 3 lin kt cng ho tr c cc gia nguyn t Al vi 3 nguyn t Cl. Al2Cl6: Mi nguyn t Al to 3 lin kt cng ho tr vi 3 nguyn t Cl v 1 lin kt cho nhn vi 1 nguyn t Cl (Al: nguyn t nhn; Cl nguyn t cho). Trong 6 nguyn t Cl c 2 nguyn t Cl c 2 lin kt, 1 lin kt cng ho tr thng thng v lin kt cho nhn. Cl * Cu trc hnh hc: 1200 1200 Phn t AlCl3: nguyn t Al lai ho kiu sp2 (tam gic phng) Al nn phn t c cu trc tam gic phng, u, nguyn t Al tm cn 3 nguyn t Cl 3 nh ca tam gic. Cl Cl1200

Phn t Al2Cl6: cu trc 2 t din ghp vi nhau. Mi nguyn t Al l tm ca mt t din, mi nguyn t Cl l nh ca t din. C 2 nguyn t Cl l nh chung ca 2 t din. Al O Cl 2. * Phn t

O O O O O O

H-F

M = 20 Jt ; = 1,91 Debye

H-O-H

M = 18 = 1,84 Debye

c th to lin kt hidro HF

c th to lin kt hidro HO

* Nhit nng chy ca cc cht rn vi cc mng li phn t (nt li l cc phn t) ph thuc vo cc yu t: - Khi lng phn t cng ln th nhit nng chy cng cao. - Lc ht gia cc phn t cng mnh th nhit nng chy cng cao. Lc ht gia cc phn t gm: lc lin kt hidro, lc lin kt Van der Waals (lc nh hng, lc khuch tn). *Nhn xt: HF v H2O c momen lng cc xp x nhau, phn t khi gn bng nhau v u c lin kt hidro kh bn, ng l hai cht rn phi c nhit nng chy xp x nhau, HF c nhit nng chy phi cao hn ca nc (v HF momen lng cc ln hn, phn t khi ln hn, lin kt hidro bn hn). Tuy nhin, thc t cho thy Tnc (H2O) = 00C > Tnc(HF) = 830C. * Gii thch: Mi phn t H-F ch to c 2 lin kt hidro vi 2 phn t HF khc hai bn H-FH-FH-F. Trong HF rn cc phn t H-F lin kt vi nhau nh lin kt hidro to thnh chui mt chiu, gia cc chui lin kt vi nhau bng lc Van der Waals yu. V vy khi un nng n nhit khng cao lm th lc Van der Waals gia cc chui b ph v, ng thi mi phn lin kt hidro cng b ph v nn xy ra hin tng nng chy. Mi phn t H-O-H c th to c 4 lin kt hidro vi 4 phn t H2O khc nm 4 nh ca t din. Trong nc mi phn t H2O lin kt vi 4 phn t H2O khc to thnh mng li khng gian 3 chiu. Mun lm nng chy nc cn phi ph v mng li khng gian 3 chiu vi s lng lin kt hidro nhiu hn so vi HF rn do i hi nhit cao hn K THI CHN HC SINH GII QUC GIA NM 2003 (BNG A)

Kim loi A phn ng vi phi kim B to hp cht C mu vng cam. Cho 0,1 mol hp cht C phn ng vi CO2 (d) to thnh hp cht D v 2,4 gam B. Ha tan hon ton D vo nc, dung dch D phn ng ht 100 ml dung dch HCl 1 M gii phng 1,12 l kh CO2 (ktc). Hy xc nh A, B, C, D v vit cc phng trnh phn ng xy ra. Bit hp cht C cha 45,07 % B theo khi lng; hp cht D khng b phn tch khi nng chy. BI GII:nHCl = 0,1 mol ; nCO2 = 0,05 mol+

nH 0, 1 2 = = 0, 05 1 n CO2 suy ra hp cht D l mui cacbonat kim loi. Hp cht D khng b phn tch khi nng chy, vy D l cacbonat kim loi kim. 2 H+ + CO32- = H2O + CO2 C + CO2 = D + B C l peroxit hay superoxit, B l oxi. t cng thc ho hc ca C l AxOy . 3, 2.100 Lng oxi trong 0,1 mol C (AxOy ) l 16 x 0,05 + 2,4 = 3,2 (g); mC = = 7,1 gam 45, 07 Mc = 7,1 : 0,1 = 71 (g/mol). mA trong C = 7,1 - 3,2 = 3,9 (g). 3, 9 3, 2 : x:y= MA = 39 (g). Vy A l K ; B l O2 ; C l KO2 ; D l K2CO3 M A 16Dung dch D phn ng ht 0,1 mol HCl gii phng kh CO2

K + O2 KO2 4 KO2 + 2 CO2 2 K2CO3 + 3O2 K2CO3 + 2 HCl 2 KCl + H2O + CO2 K THI CHN HC SINH GII QUC GIA NM 2004 (BNG A) Cc phng trnh phn ng: Vit phng trnh ho hc cho mi trng hp sau: a) Cho kh amoniac (d) tc dng vi CuSO4.5H2 O. b) Trong mi trng baz, H2O2 oxi ho Mn2+ thnh MnO2. Trong mi trng axit, H2O2 kh MnO4- thnh Mn2+. BI GII: a) C th vit CuSO4.5H2O dng [Cu(H2O)4] SO4.H2O. Do khi phn ng xy ra, NH3 s th cc phn t H2O cu ni: [Cu(H2O)4] SO4.H2O + 4 NH 3 b) H2O2 + 2e Mn2+ + 4 OH 2 e Mn2+ + H2O2 + 2 OH c) 2 5 MnO4 + 8 H3O+ + 5 e H2O2-

[Cu(NH3)4] SO4.H2O + 4 H2O 2 OH MnO2 + 2 H2O MnO2 + 2 H2O Mn2+ + 12 H2O O2 2 Mn + 2 H3O2+ +

S kh S oxi ho S kh S oxi ho 14 H2O

+ 2 H2O 2 e 6 H3O+

2MnO4 + 5 H2O2 +

+ 5 O2 +

K THI CHN HC SINH GII QUC GIA NM 2005 (BNG A) Nhng hai tm km, mi tm c khi lng 10 gam vo hai dung dch mui kim loi ho tr hai. Sau mt thi gian xc nh, ly hai tm km ra khi dung dch, ra sch, lm kh ri cn li. Kt qa cho thy mt tm c khi lng 9,5235 gam, tm kia c khi lng 17,091 gam. Cho bit: Mt trong hai dung dch mui kim loi ho tr hai l mui st (II); lng km tham gia phn ng hai dung dch l nh nhau. 1. Gii thch hin tng xy ra mi dung dch. 2. Cho bit kim loi no tham gia vo thnh phn dung dch mui th hai. BI GII: 1. Khi nhng tm km vo dung dch mui Fe(II): Zn2+ + Fe (1) Zn + Fe2+ V: MFe < MZn nn khi lng tm km gim i. Khi nhng tm km vo dung dch mui th hai X2+ Zn + X2+ Zn2+ + X (2) V: MZn < MX nn khi lng tm km tng ln. 2. Gi x l s mol Zn phn ng, theo (1) ta c: (10 65,38 x) + 55,85 x = 9,5235 x = 0,05 (mol) V lng Zn tham gia phn ng 2 trng hp l nh nhau, theo (2) ta c: (10 65,38 0,05) + MX 0,05 = 17,091 MX = 207,2.

Vy X2+ l Pb2+, X l Pb Zn + Pb2+

Zn2+ + Pb

K THI CHN HC SINH GII QUC GIA NM 2005 (BNG A) Hon thnh cc phng trnh phn ng sau y: 1. NaCl + H2SO4 c, nng 2. NaBr + H2SO4 c, nng 3. NaClO + PbS 4. FeSO4 + H2SO4 + HNO2 5. KMnO4 + H2SO4 + HNO2 6. NaNO2 + H2SO4 long BI GII: 1. NaCl + H2SO4 (c, nng) hoc 2. 2 NaCl + H2SO4 (c, nng) 2 NaBr + 2 H2SO4 (c, nng) 2 HBr + H2SO4 (c, nng)

HCl

+ NaHSO4

2 HCl + Na2SO4 2 NaHSO4 + 2 HBr SO2 + 2 H2O + Br2

3. 4. 5. 6.

2 NaBr + 3 H2SO4 (c, nng) 2 NaHSO4 + SO2 + 2 H2O + Br2 4 NaClO + PbS 4 NaCl + PbSO4 2 FeSO4 + H2SO4 + 2 HNO2 Fe2(SO4)3 + 2 NO + 2 H2O 2 KMnO4 + 3 H2SO4 + 5 HNO2 K2SO4 + 2 MnSO4 + 5 HNO3 + 3 H2O 3 NaNO2 + H2SO4 (long) Na2SO4 + NaNO3 + 2 NO + H2O

K THI CHN HC SINH GII QUC GIA NM 2005 (BNG A) t chy kim loi magie trong khng kh. Cho sn phm thu c tc dng vi mt lng d dung dch axit clohidric, un nng ri c dung dch n cn kh. Nung nng sn phm mi ny v lm ngng t nhng cht bay hi sinh ra trong qa trnh nung. Hy vit cc phng trnh phn ng xy ra trong th nghim trn v cho bit c nhng cht g trong sn phm ngng t c. BI GII: Cc phn ng: 2 Mg + O2 2 MgO 3 Mg + N2 Mg3N2 MgCl2 + to o

MgO + 2 HCl Mg3N2 + 8 HCl MgCl2.6 H2O NH4Cl II. t

H2O 2 HCl + 5 H2O NH4Cl

3 MgCl2 + 2 NH4Cl MgO NH3 + + HCl

Sn phm c ngng t: NH4Cl ; H2O ; HCl. OLYMPIC HA HC QUC T:

OLYMPIC HA HC QUC T LN TH 28: S polime ha lp th ca hydrocacbon cha no thng c coi l mt trong nhng phn ng quan trng trong ha hc hu c cng nghip. Mui ca cc cation ln khng thng hng mang in tch ln tn cng vo mt electron phn b dc theo cc lin kt ca phn t olefin thng c dng lm cht xc tc trong cc qa trnh ny. Anion cloroaluminat (AlCl4-) thng c dng do c n tch m c bt nh cao. Nhu cu pht trin cht xc tc mi thuc loi ny thc y cc nh ha hc kho st tng tc trong h thng A B, trong A = Te (kt tinh) v B = TeCl4 + 4AlCl3. Hp phn th hai B - c coi nh l mt cht tng t Te(IV) cloroaluminat Te[AlCl4]4, nhng li khng th c lp c hp cht ny di dng hp cht ring bit. Ngwofi ta thy rng tng tc ca cc hp phn A v B c th dn n s to thnh ba hp cht mi I, II v III trong cc h thng lc u c cha theo th t 77,8; 87,5 v +1,7% theo s mol ca hp phn A. Ngi ta cng nhn thy trong trng hp cc hp cht II v III khng c s to thnh sn phm ph km theo, nhng s hnh thnh hp cht I c km theo s gii phng 1 mol TeCl4 d bay hi ng vi 2 mol I. C hai hp cht u c mu tm hng v c hai phn li thnh 3 ion m kho st tnh dn in trong NaAlCl4 cho thy: T php o nghim lnh tng NaAlCl4 nng chy cho php tnh c khi lng mol phn t ca I v II ln lt bng 112643g/mol v 86748g/mol. Ph hng ngoi IR ca c hai hp cht ch quan st c c mt di c th c gn cho kiu rung ca lin kt tOo bi mt nguyn t Te. Di ny nm tjai 133cm-1 v v vy c nng lng thp n ni khng nghi ng g na lin kt ny l mt loi Te-Te. S liu ph cng hng t ht nhn 27Al NMR ca cc phc I v II cho thy ch c mt loi nhm ph tr t din trong mi hp cht. Tuy nhin di ha hc quan st c ca nhm vi cc hp cht I v II khc nhua, nh th cho thy cc nguyn t Al trong cc mi trng khc nhau. 1. Hy xc nh t l nguyn t ti thiu ca Te:Al:Cl vi cc phc I, II v III. 2. Vit cng thc phn t ca cc hp cht I v II. 3. Vit cng thc cc anion v cation trong hp cht I v II. 4. Vit cng thc ha lp th ca cc cation v anion trong cu to ca I v II. Gi thit rng cc cation trong I v II l nhng v d ca h thng thm v c. 5. I hay II c bn nhit cao hn. Bit rng AlCl3 l mt cht cc k d bay hi. 6. Nu mt trong cc hp cht I v II c th chuyn thnh cht kia khi un nng th hy vit phng trnh phn ng tng ng. BI GII: 1. C th dng s lu v hm lng Te(kt tinh) xc nh cc t l Te:Al:Cl, nh vy: 77,8% Te (kt tinh) tng ng vi: 7Te (kt tinh) + 2TeCl4 + 8AlCl3 v t l nguyn t b nht cho thnh phn m khng tch lng d TeCl4 l Te:Al:Cl = 9:8:32, trong hm lng Al v Cl l chn v c th chia ht cho 4. tng lc hm lng Te vt qa bi s ca 4 l 1. Tr mt mol TeCl4 t t l thu c ri chia cho 2 ta c 4Te+4Al+14Cl v t l l Te:Al:Cl = 2:2:7, t l ny c th c kim tra li bng cch so snh vi khi lng phn t cho. 87,5% Te(kt tinh) ng vi: 7Te (kt tinh) + TeCl4 + 4AlCl3 = 8Te + 4Al + 16Cl v t l l: Te:Al:Cl = 2:1:4 91,7% Te(kt tinh) ng vi: 11Te (kt tinh) + TeCl4 + 4AlCl3 = 12Te + 4Al + 16Cl v t l l Te:Al:Cl = 3:1:4 2. Cng thc phn t c th c suy t cc khi lng phn t. C hai u ng vi gp i cng thc n gin nht: Vi cht I: Te4Al4Cl14. Vi cht II: Te4Al2Cl8.

3. Thnh phn ca cc ion c trong cu to ca I v II c th c xc nh bng cch da trn c im l c hai cht I v II u l nhng cht in ly to bi 3 ion v tt c nhng nguyn t telu phi tng ng theo ph IR v ch c lin kt vi nhau, cng nh l cc nguyn t Al trong c hai hp cht l tng t v c phi tr t din. t nht trong mt trng hp, c th l anion AlCl4- ph hp vi cht II nn v th c cng thc l [Te4]2+[AlCl4]2-. Do mu tng t nn cc cation trong c hai trng hp l tng t, cu to I phi cha cation [Te4]2+ v anion [Al2Cl7]-, iu ny ph hp vi cc s liu ph NMR, gn cho cc nguyn t Al trong cht I c hnh hc t din khc so vi hnh hc t din ca cht II. Anion [Al2Cl7]Hp cht I: Cation [Te4]2+ Hp cht II: Cation [Te4]2+ Anion [AlCl4]4. Hnh hc ca cc anion: AlCl4- c cu to t din nCl Cl Cl Al Cl

Al2Cl7- gm hai t din c chung nh mt nguyn t cloCl Cl Cl Al Cl Al Cl Cl Cl

Hnh hc ca cation: [Te4]2+ c cu trc vung phng do c tnh thm. Cu hnh vung phng l u i v cation c tnh thm, c ngha l ng phng v c bn lin kt bng nhau cho cc cnh ca vng thm tng ng:Te2+

Te Te

Te

5. bn nhit ca cht II phi cao hn cht I; c hai u l cc hp cht ion vi im nng chy cao, nhng hp cht I c th chuyn thnh II bng phn ng tch AlCl3, l mt cht rn d tch ra khi bay hi v c th tch ra mt cch tng i d dng nh un nng. 6. Te4[Al2Cl7]2 = Te4[AlCl4]2 + AlCl3. OLYMPIC HA HC QUC T LN TH 28: Gii hn d l mt trong nhng tham s c bn tng vic phn tch nh lng mt lng rt nh (vt) ca cc nguyn t. Gii hn d c m t l khi lng nguyn t b nht c th c xc nh bng mt phng php cho trc vi mt mc chnh xc cho trc. V d, xt hai phng php c dng xc nh vi lng ca bitmut. Nm 1927, Bergh, nh ha hc ngi c, ngh rng bitmut c th c kt ta di dng mui khng tan: 8-hydroxiqiunonlin tetraiodobitmutat [C9H6(OH)NH][BiI4] c M = 862.7g/mol. 1) a) Vit cng thc cu to cation v anion ca mui ny. b) Mc oxy ha ca nguyn t Bi trong hp cht ny l bao nhiu? 2) Xc nh khi lng nguyn t b nht ca bitmut (theo mg) c th c xc nh mt cch ng tin cy bng phng php Bergh, nu khi lng b nht ca cht kt ta c th o c mt cch ng tin cy l 50,0mg.

Mt phng php khc do R.Belcher v cng s t Birmingham pht trin xc nh hm lng vt ca bitmut c gi l phng php a bi. Theo phng php ny, ngi ta tin hnh mt chui cc phn ng ri tin hnh chun sn phm cui c m t chi tit di y. Bc 1: Thm 50mg kali hexathioxianatocromat (III), K3[Cr(SCN)6] vo mt lng nh khong 2mL dung dch lnh c axit ho c cha mt lng vt Bi3+ dn n kt ta nh lng ca bitmut. 3) Vit v cn bng phng trnh phn ng di dng ion thu gn. Bc 2: Lc ly kt ta, ra bng nc lnh v x l vi 5mL dung dch natri hydrocacbonat 10%. Vic x l ny gip chuyn kt ta ban u thnh mt kt ta khc oxobitmut cacbonat (BiO)2CO3 vi s gii phng ion hexathioxianatocromat (III) vo dung dch. 4) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 3: Nc lc c axit ha nh v chuyn sang phu tch, sau thm 0,5mL dung dch iot bo ho trong clorofom v lc mnh hn hp. Iot tin hnh phn ng oxi ha - kh vi ligand ca ion phc v to thnh sn phm c c xianogen iodua [ICN] v ion sunfat. 5) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 4: Sau 5 pht, thm 4mL dung dch H2SO4 2M vo hn hp. S axit ha ny dn n phn ng oxy ha - kh vi s to thnh iot phn t. 6) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 5: Tch nh lng iot bng 4 phn clorofom. Lp dung dch nc c chuyn vo mt bnh cha, thm vo 1mL nc brom v lc hn hp trong 5 pht. Brom ly d c th phn ng c vi hydro xianua to thnh xianogen bromua BrCN v vi iot to thnh IO3-. 7) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 6: loi lng d brom phn t, thm 3mL dung dch axit fomic 90% vo hn hp. 8) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 7: Thm lng d kali iodua (khong 1,5g) vo dung dch hi axit ny. Iot c th phn ng vi BrCN theo cch tng t nh vi ICN to thnh iot phn t. 9) Vit v cn bng phng trnh phn ng ny di dng ion thu gn. Bc 8: Chun dung dch thu c bng dung dch Na2S2O3 0,00200M tiu chun. Nh vy kt qa thu c s dng tnh hm lng bitmut trong mu em phn tch. 10) a) Vi mi mol bitmut c bao nhiu mol thiosunfat trong mu ban u? b) Hm lng b nht ca bitmut c th xc nh c bng phng php Belcher l bao nhiu? Gi thit rng s xc nh ny l ng tin cy nu dng khng di 1mL dung dch Na2S2O3 0,00200M tiu chun. 11) Phng php phn tch a bi ny ca Belcher nhy hn phng php phn tch trng lng ca Bergh bao nhiu ln? BI GII: 1) a) Cng thc cu to ca cation v anion: Cation:

N H OH

Anion:

I I Bi I I

hay:I Bi I I I

Mc oxy ha ca bitmut: +3 2) m = 12,1mg 3) Bi3+ + [Cr(SCN)6]3+ = Bi[Cr(SCN)6] 4) 2Bi[Cr(SCN)6] + 6HCO3- = (BiO)2CO3 + 2[Cr(SCN)6]3- + 3H2O + 5CO2. 2Bi[Cr(SCN)6] + 5HCO3- + 5OH- = (BiO)2CO3 + 2[Cr(SCN)6]3- + 3H2O v.v (C th cn c nhng dng khc) 5) [Cr(SCN)6]3- + 24I2 + 24H2O = Cr3+ + 6SO42- + 6ICN + 42I- + 48H+ 6) ICN + I- + H+ = I2 + HCN 7) a) 3Br2 + I- + 3H2O = IO3- + 6Br- + 6H+. b) Br2 + HCN = BrCN + Br- + H+. 8) Br2 + HCOOH = 2Br- + CO2 + 2H+. 9) a) IO3- + 5I- + 6H+ = 3I2 + 3H2O b) BrCN + 2I- + H+ = I2 + HCN + Br10) a) Chun iot bng thiosunfat theo phn ng: I2 + 2S2O32- = S4O62- + 2IGi thit rng dung dch ban u cha 1mol Bi. Trong phn ng (5) mi mol Bi dn n s hnh thnh 42 mol ion iodua ( thun tin, chia tt c cc h s ca phn ng cho 2), trong 6 mol ion Ic tiu th phn ng (6). Nh vy 36 mol iodua phn ng theo phng trnh 7a to 36mol IO3m vi phn ng 9a cho 108 mol I2 nn cn 216 mol thiosunfat chun . Mt khc mt mol Bi3+ to ra 6 mol HCN theo cc phn ng (5) v (6). Brom oxy ha HCN trong phn ng 7b cho 6 mol BrCN, m theo phn ng 9b to 6 mol iot nn cn thm 12 mol thiosunfat na. Nh vy, tng s mol thiosunfat cn thit l 228mol. b) ta c th tnh c kt qa l: 1,83.10-3mg = 1,83g 11) 6600 OLYMPIC HA HC QUC T LN TH 31: Ligand (L) c th to phc c vi nhiu kim loi chuyn tip. L c tng hp do s un nng hn hp gm bipiridin, axit axetic bng v hydro peoxit n 70-80oC trong 3 gi. Sn phm cui L kt tinh di dng hnh kim nhuyn v c khi lng phn t bng 188. Mt phn ng tng t vi piridin l:

b)

[O]N N O

Phc ca L vi Fe v Cr c cng thc l FeLm(ClO4)n.3H2O (A) v CrLxCly(ClO4)z.H2O (B). Thnh phn phn tch nguyn t v tnh cht l hc ca chng c cho trong bng 1 v 2. Quan h gia mu v bc sng cho trong bng 3: Bng 1: Thnh phn phn tch nguyn t: Phc Phn tch nguyn t (% khi lng) A Fe: 5,740; C: 37,030; H: 3,090; Cl: 10,940; N: 8,640 Cr: 8,440; C: 38,930; H: 2,920; Cl: 17,250; N: 9,080 B Dng cc s liu sau: S hiu nguyn t: Cr = 24; Fe = 26 Khi lng nguyn t: H = 1; C = 12; O = 16; Cl = 35,45; Cr = 52; Fe = 55,8 Bng 2: Tnh cht vt l: Phc Mu Momen t (B.M) A 6,13 Vng B Khng o Tm Bng 3: Quan h gia bc sng v mu: Bc sng (nm) v mu hp th Mu b sung (mu b) 400 (tm) Lc vng 450 (xanh) Vng 490 (lc xanh lam) Vng cam 500 (lc) 570 (lc vng) Tm 580 (vng) Xanh lam 600( Vng cam) Lc xanh lam 650 () Lc 1. Vit cng thc phn t ca L. 2. Nu L l mt ligand cng cua hai rng, vit cu tjao ca bipiridin dng. Vit cu to ca L 3. Ligand L c in tch tng cng l bao nhiu? 4. Vit cu to khi mt phn t L lin kt vi mt ion kim loi M 5. T cc s lu ghi trong bng 1 hy xc nh cng thc thc nghm ca A. Ga tr ca m v n trong FeLm(ClO4)n.3H2O l bao nhiu?. Vit tn y ca A theo quy tn\c IUPAC. Khi A ho tan trong nc th t l gia cation v anion l bao nhiu? 6. S oxy ha ca Fe trong A l bao nhiu? C bao nhiu electron d c mt trong ion Fe ca phc? Vit cu hnh spin cao v spin thp c th tn ti c i vi phc ny. Cu hnh no, cao hay thp l cu hnh ng? C chng c no l tt nht minh ho kt lun chn? 7. T bng 3, c lng bc sng (nm) ca A. 8. Phn tch chi tit B cho thy n c cha ion Cr3+. Hy tnh momen t ch vi spin ca hp cht ny. 9. Hp cht B l loi cht in phn 1:1. Hy xc nh cng thc thc nghim ca B v cc ga tr ca x, y, z trong CrLxCly(ClO4)z.H2O.

BI GII: 1. Bit L c tng hp t bipiridin v trong phn ng bipiridin b oxy ha n gin thnh bipiridin oxit.Khi lng phn t ca bipiridin l 156 (ca C10H8N2) trong khi khi lng phn t ca L l 188. Khc bit 32 do hai nguyn t oxy. V vy cng thc phn t ca L l C10H8N2O2. 2. Cu to ca bipiridin:

N

N

Cu to c th c ca L:

N O O

N

N O O

N

N O O

N

3. in tch tng cng ca ligand L: khng. 4. Cu to:

N O M

N O

5. Cng thc thc nghm ca A l: FeC30H30Cl3N6O21 m = 3; n = 3 Cng thc ca phc: [FeL3](ClO4)3.3H2O T l cation v anion l: 1:3 6. S oxy ha ca Fe: +3 S electron d trong ion Fe ca phc: 5 Cu hnh spin cao: (t2g)3(eg)2 Cu hnh spin thp: (t2g)5(eg)0 Cu hnh ng: spin cao do c momen t tnh c da vo s e c thn l 5,92B.M cn spin thp l 1,73B.M. So snh vi ga tr momen t o c trong bng 2 ta rt ra c kt lun trn. 7. 450nm 8. 3,87B.M 9. CrC20H18N4Cl3O9 x=2

y=2 z=1 OLYMPIC HA HC QUC T LN TH 32:

Phc vung phng cis-diaminodicloroplatin (II) l mt dc phm quan trng iu tr ung th. 1. Vit cc ng phn cis v trans ca phc. Mt s ion cng c cng thc nguyn Pt(NH3)2Cl2. 2. Vit tt c cng thc c th c ca ion trn nhng phi tha mn cc iu kin sau: - C cng thc nguyn Pt(NH3)2Cl2. - Anion v cation phi c vit r v tt c phi c cu trc vung phng. - Anion v cation phi th hin c s tn ti ca mi phc platin (II) ring bit ca mi hp cht. 3. Lp 5d ca platin c bao nhiu electron? S tch mc nng lng trong gin nng lng obitan d ca phc vung phng lin quan n phc bt din do lien kt kim loi ligand: Nu cc ligand nm trn trc z bin mt m lin kt kim loi ligand vi cc ligand nm trn trc x v y tr nn mnh hn. 4. Trong s 5 obitan 5d ca platin, trong phc Pt vung phng th obitan no c mc nng lng cao nht?BI GII:

1. Cng thc cu to cc dng ng phn ca phn t cis-diaminodicloroplatin (II): (1 im) H3N Cl Cl NH3 Pt Pt Cl NH3 Cl NH3 trans cis 2. [Pt(NH3)4][PtCl4]. [Pt(NH3)3Cl][Pt(NH3)Cl3] [Pt(NH3)3Cl]2[PtCl4]

[Pt(NH3)4][Pt(NH3)Cl3]2 3. 8 4. 5d x 2 y 2 . Trong phc t din 4 ligand u nm trn ng phn gic ca hai trc x v y.

Nu c y electron th mt electron s cao hn.OLYMPIC HA HC QUC T LN TH 33: Phim en trng cha lp ph bc bromua trn nn l xenluloz axetat. 1. Vit phn ng quang ha xy ra khi chiu nh sng vo lp AgBr ph trn phim. 2. Trong qu trnh ny th lng AgBr khng c chiu sng s b ra bng cch cho to phc bi dung dch natri thiosunfat. Vit phng trnh phn ng. 3. Ta c th thu hi bc t dung dch nc thi bng cch thm ion xianua vo, tip theo l km. Vit cc phn ng xy ra. BI GII: h 1. Phn ng: 2 AgBr (r ) 2 Ag (r ) + Br2 / 2 Br 2. AgBr(r) + 2Na2S2O3 Na3[Ag(S2O3)2] + NaBr 3. [Ag(S2O3)2]3- + 2CN- [Ag(CN)2]- + 2S2O3-

2[Ag(CN)2]- + Zn [Zn(CN)4]2- + 2Ag+ OLYMPIC HA HC QUC T LN TH 36: Trong qa kh c mt s cng trnh v vic iu ch cc hp cht ca canxi ha tr 1. Mc d bn cht ca cc hp cht y vn cha c bit nhng chng to ra c mt s quan tm rt ln i vi cc nh ha tinh th. C th chuyn CaCl2 thnh CaCl bng: canxi, hydro, cacbon. 1. Vit cc phn ng xy ra. Cho CaCl2 phn ng vi Ca theo t l 1:1 ta ch nhn c mt cht mu xm khng ng nht. Nhn di knh hin vi ta thy c mt phn mu bc v tinh th khng mu. 2. Tinh th khng mu v phn c nh kim l ca cht no? Cho CaCl2 phn ng vi hydro nguyn t ngi ta thu c sn phm mu trng cha 52,36% Ca v 46,32% Cl v khi lng. 3. Xy dng cng thc thc nghim cho hp cht trn. Khi CaCl2 tc dng vi cacbon nguyn t ta thu c tinh th mu . T l s lng nguyn t xc nh c bng php phn tch nguyn t l n(Ca):n(Cl) = 1,5:1. Thu phn cht ny thi thu c sn phm nh thy phn Mg2C3. 4. a) Vit hai ng phn (mch khng vng) ca cht sinh ra khi thy phn. b) Vit phng trnh phn ng xy ra (gi s khng sinh ra CaCl) Khng phn ng no trong cc phn ng trn cho php ta xc nh c cu trc ca CaCl. C th gi thit rng CaCl c mng tinh th tng t nh cu trc tinh th n gin. Cho bit t l bn knh cation ca cc hp cht MX c cho bng: S phi tr ca M Hnh dng ca cc T l bn knh: Kiu cu trc Nng lng mng nhm X bao quanh rM/rX li ca CaCl LHo 3 Tam gic 0,155-0,225 BN -663,8 kJ.mol-1 4 T din 0,225-0,414 ZnS -704,8 kJ.mol-1 6 Bt din 0,414-0,732 NaCl -751,9 kJ.mol-1 8 Lp phng 0,732-1,000 CsCl -758,4 kJ.mol-1 5. a) CaCl c kiu cu trc tinh th no?. Bit r(Ca2+) 120pm; r(Cl-) 167pm Mun bit CaCl c bn nhit ng hay khng ta khng ch s dng LHo m cn s dng Hs. b) Tnh Hs ca CaCl vi cc s liu cho xc nh CaCl c bn nhit ng hay khng ta phi tnh Ho ca qa trnh 2CaCl Ca + CaCl2 (b qua S). 6. S dng tnh ton cho bit phn ng trn c xy ra hay khng? BI GII: 1. CaCl2 + Ca = 2CaCl. 2CaCl2 + H2 = 2CaCl + 2HCl 4CaCl2 + C = 4CaCl + CCl4. 2. Tiu phn c nh kim mu bc: Ca Tinh th khng mu: CaCl2. 3. n(Ca) : n(Cl): n(H) = 1 : 1 : 1 Cng thc thc nghim: CaClH Lu rng phn ng gia CaCl2 v hydro khng th no dn n CaCl m thay vo l hydrua CaClH. Cu trc ca hp cht ny c xc nh php phn tch ph tia X nhng phng php ny khng phi l mt phng php tt tm thy cc nguyn t nh nh hydro. Chnh v s bin mt ca hydro trn ph tia X m trong mt thi gian di CaClH b tng lm l CaCl.

4. a) Cc cu trc ng phn:C H C C H H C C CH3

b) Cng thc thc nghim: Ca3C3Cl2. Lu rng: Nu t l s nguyn t n(Ca):n(Cl) = 1,5:1 (hay tt hn l 3:2 v c th vit li l CaCl2.2Ca2+ = Ca3Cl44+) v sn phm kh phi cha anion C34- nn phi cn hai cation Ca2+ trung ho in nn chnh v vy nn cng thc Ca3C3Cl2 c chp nhn. 5. a) T l bn knh l r(Ca2+)/r(Cl-) = 0,719. Vy kiu tinh th s l kiu NaCl. b) Tnh nhit sinh ca phn ng hnh thnh CaCl da vo chu trnh Born Haber: LHo Ca+(k) + I1 Cl-(r) CaCl(r)

ECl

Hs Ca(k) Hthng hoa Cl(k) 0,5Hphn ly

Ca(r) + 0,5Cl2(k) Thay s vo ta tnh c: Hs = -231,9kJ/mol 6. 2CaCl Ca + CaCl2 c Hs = -332,2kJ/mol. Vy phn ng trn khng xy ra. OLYMPIC HA HC QUC T LN TH 37: S tng hp v xc nh tnh cht ca vng kch thc nano l mt lnh vc ang rt pht trin. Phng php Bruff Schiffrin tng hp vng kch thc nano (AuNP) da vo s bn nhit ng v bn khng kh ca AuNP ta c th iu ch AuNP dng a phn tn c kch thc nm trong khong t 1,5 v 5,2nm. Phng php ny c ni dung nh sau: Dung dch HAuCl4 c trn ln vi toluen trong dung mi tetra n octylamoni bromua. Dung dch thu c em trn ln vi dodecanthiol v c un nng vi NaBH4. Nhng ht AuNP mu ti c sinh ra v nm pha hu c. Sau 24 gi th dung mi toluen s c cho bay hi v khi rn thu c c ra vi hn hp dung mi etanol v hexan tch thiol. Cc ht AuNP c th c phn lp v ho tan tr li trong dung mi hu c thng thng m khng b phn hy hay t hp thun nghch (irrvesible aggregation). 1. Phng php ny cn s tip cn t trn xung hay t di ln? a) Tip cn t trn xung lm cho ht nano c kch thc nh nht. b) Tip cn t di ln c th bin cc phn t v nguyn t ring l thnh cu trc nano. 2. Trimetyl n octylamin bromua cn c th c s dng nh l mt cht chuyn pha. N c th chuyn AuCl4- t pha nc sang pha hu c. Tnh cht no gip cho n c th lm c iu ny?

a) Mt pha ca phn t mang in dng, u cn li mang in m. b) Mt pha c tnh a nc, mt pha c tnh k nc. c) Mt pha mang tnh axit, mt pha mang tnh baz. 3. Vai tr ca NaBH4 trong qa trnh iu ch ny l g? a) Tc nhn kh ha. b) Tc nhn oxy ha. c) Tc nhn trung tnh. d) Tc nhn to phc. 4. Nu ng knh trung bnh ca ht vng kch thc nano l 3nm th s nguyn t vng s l bao nhiu trong mi phn nano? (bn knh nguyn t ca vng l 0,144nm). Ch ra bng tnh ton a) 102 b) 103 c) 104 d) 105 5. Bng tnh ton hy ch ra phn trm s nguyn t vng trn b mt l bao nhiu v cho bit cu tr li no ng. a) 20 30% b) 40 50% c) 60 70% d) 80 90% BI GII: 1: b) 2: b) 3: a) 4: b) 4 3 V AuNP = . .rAuNP 3 4 3 V Au = . .rAu 3

r3 V N = AuNP = AuNP = 1000 = 10 3 r3 V Au Au 5: b) 4/3..rAuNP3 = 4/3..rAu3.NAu. Din tch b mt ca mt tiu phn nano cu vng: S = 4..rAuNP2 S = 4..rAu2.NAu2/3 NS SAuNP/.rAu2 = 4NAu2/3. P = NS/NAu = 4NAu1/3. NAu = 1000 P = 40% III. BI TO CHUN B CHO OLYMPIC HA HC QUC T: OLYMPIC HA HC QUC T LN TH 30: Mt c nghin cu sinh nhn mt l hng gm cc halognua kim loi kim nhng cc bnh u mt nhn tr mt bnh cha kali bromua. Phng th nghim ni c lm vic khng h c bt k loi ph k no v vy c dng ct trao i ion nhn bit cc mu halogenua kim loi kim mt nhn . Loi nha c chn l loi nha polystiren mng li kiu axit mnh, cha cc nhm axit sunfonic (-SO3H)

3

nn ch cc proton c th trao i. C phn tch c su mu kim loi kim (v c KBr kim chng phng php) theo cch sau: C cn 5,00 0,01g mi mu, ri ho tan vi nc ct trong ng ong 100mL. Cho 40mL mi dung dch qua ct; dung dch ra c thu vo ng ong c th tch 250mL, ra ct hai ln vi nc ct; dung dch ra ny c thm nc c 250mL. Trc khi mu k tip c cho vo ct, c ti to proton cho nha trong ct bng cch ra vi lng cn thit HCl 1M ri vi nc ct. C chun cc mu 50mL ca mi dung dch ra, lm ba ln vi dung dch NaOH (nng l thuyt 3,26.10-2M) dng cht ch th l phenolphtalein thu c cc kt qa sau: Mu th nghim Th tch chun trung bnh A 21,15 0,1mL B 29,30 0,1mL C 7,40 0,1mL D 21,20 0,1mL E 10,30 0,1mL F 29,15 0,1mL KBr 10,25 0,1mL phn tch cc kt qa ny ta c th gi thit rng: Mi mu th t >99% tinh khit. Mi bnh u y cht, khng b nhim nc v khng kh Khng c trng hp hai bnh cha cng mt kim loi kim halogenua; l ha cht ch gm florua, clorua, bromua v iodua, khng c hp cht ca atatin. a) Hy cho bit l do v sao phi tin hnh cc th tc nu trn? Vit phng trnh phn ng ho hc ca bt k phn ng no xy ra. b) Mu th no c th chc chn c nhn bit t s phn tch ny? Mu th no c th gii hn kt qa ch cn hai hoc ba kh nng? c) Dng cc dng c c trong phng th nghim: knh thy tinh, giy qy, dung dch natri pesunfat (Na2S2O8) trong mi trng axit v mt l cha dung dch h tinh bt c c th nhn bit c c su mu th. Khng cn bit kt qa th nghim ca c vi cc ha cht nu trn, hy gii thch lm th no vi cc vt liu trn l nhn bit c tt c cc mu th cha nhn bit c cu b). d) Tnh cht no ca kim loi kim halogenua ngn cn khng th nhn bit r rng mt s mu nhn bit bng k thut trao i ion dng y? Liu mt hiu ng nh th c phi l tr ngi ng k trong mt n lc tng t nhn bit mt s halogenua ca kim loi kim th MX2 chng hn? BI GII: a) Hin nhin, cn xc nh khi lng mol phn t ca mi mu th Mr(MX) bng cch trao i M+ vi H+ nh ct trao i ion v bng cch chun xc nh lng H+. Cc phn ng gm: M+ + [RSO3H] H+ + [RSO3-M+] H+ + OH- H2O b) Phn tch kt qa: S mol M+ trong 5g = s mol OH- .(250/50).(100/40) = th tch chun .0,326.5.2,5 M(r)(MX) = khi lng mu th (5g)/s mol M+ trong 5g Thu c cc kt qa sau:

Mu th nghim Mr(g/mol) D on cht MX c th l: A 58,01 NaCl(58,44); KF(58,10) B 41,88 LiCl(42,39); NaF(41,99) C 165,81 KI(166,00); RbBr(165,37); CsCl(168,36) D 57,88 NaCl(58,44); KF (58,10) E 119,13 KBr(119,00); RbCl(120,92) F 42,09 LiCl(42,39); NaF(41,99) KBr 119,71 KBr(119,00); RbCl(120,92) Cc kt qa trn cho thy, k thut th nghim hin nhin cha chnh xc c th xc nh r rng tng mui. V d: nu mi mu u khc nhau, nu bit c mt mu chc chn l KBr th khi y mu E chc chn phi l RbCl. Khi lng phn t ca RbCl ln hn KBr vy m mu E li c Mr hi nh hn da trn kt qa chun . Cn thy rng khi lng mol phn t qa gn nhau ca hai (c khi ba halogenua) ca kim loi kim loi tr kh nng xc nh A, B, C, D, F. c) Knh thy tinh: Cc mui liti ht m mnh, do li mt lng nh B v F trn cc knh thy tinh k cn s cho php xc nh liti clorua: n s ht m v chy ra (nho) trong mt thi gian ngn (tr khi khng kh trong phng th nghim qa kh). Giy qy: Ion florua l mt ion lin hp ca axit yu HF nn dung dch mui florua c tnh kim. Do giy qy s xc nh c dung dch long no ca B hoc F l NaF v dung dch long no ca A hoc D l KF. Pesunfat axit ha + h tinh bt: Pesunfat oxy ha I- thnh I2, to phc mu xanh thm vi h tinh bt. Nh vy, nu C l KI s cho mu xanh thm ca phc vi tinh bt. Pesunfat cng oxy ha Clhoc Br- nhng khng c ch th c trng vi h tinh bt nn tc nhn ny khng gip phn bit RbBr v CsCl (Tuy nhin, c th xc nh C vi cc phng php nu nn C l KI) Khng c php th tm NaCl: bng phng php loi tr, cht no trong hai cht A v D khng lm xanh qy tm l NaCl. d) iu ngn cn vic xc nh r rng cc halogenua khc nhau ca cc kim loi kim MX bng s trao i ion chnh l khi lng mol phn t gn trng nhau ca cc halogenua y. Nguyn nhn l v cc hp cht ny c hai tiu phn u c ha tr I, ng thi cc kim loi kim c s hiu nguyn t ch hn cc nguyn t halogen ng trc hai n v: do vy, bt i mt lp y ca kim loi (chng hn K Na) v thm mt lp y vo halogen (nh F Cl) s cho hp cht c khi lng phn t gn nh ging ht hp cht ban u. Cc halogenua kim loi kim th ni chung khng c tr ngi ny, nu cng xt nh trn: bt i mt lp y ca M (nh Ca Mg) v thm mt lp y vo halogen (nh F Cl) cho ra cc cht c khi lng phn t hon ton khc nhau (78,08 v 95,21 g.mol-1 theo th t cho CaF2 v MgCl2). OLYMPIC HA HC QUC T LN TH 30: S kho st cc phn ng to phc ca cc ion kim loi chuyn tip. Mn+ + mL- MLm(n-m)+ thng tr nn phc tp do s ccnh tranh ng thi ca cc qa trnh cn bng khc: v d nh ligand L- thng l baz lin hp ca mt axit yu nn nng ca n trong dung dch ty thuc nhiu vo pH. Trong trng hp ny, thng phi vit li hng s to phc ca cc kim loi chuyn tip m. n ML(m m ) + m = n M n + L bng cch thay [L-] bng L CT(L), trong CT(L) l nng tng cng ca L tt c cc dng trong dung dch (nh djang HL hay L- hoc MLi(n-i)+) v L l t l ca dng thch hp L- so vi L tng

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cng. Phng php ny thng c dng, v d nh trong phn tch bng php o chun dng EDTA, v EDTA (H4Y) l mt axit bn chc yu ch c kh nng cho phn ng to phc dng hon ton mt proton Y4-. Khi y: K a1 K a 2 K a 3 K a 4 Y 4 = 4 3 2 H + + K a1 H + + K a1 K a 2 H + + K a1 K a 2 K a 3 H + + K a1 K a 2 K a 3 K a 4 Vi Ka,i l hng s ion ha th i ca EDTA (vi tr s ln lt bng 1,02.10-2;2,14.10-3; 6,92.10-7 v 5,50.10-11). a) Xc nh cc tr s ca (Y4-) pH ln lt bng 2, 6, 10. Tnh nng d ca anion Y4- hon ton mt proton trong 500mL dung dch c cha 3,252g EDTA tr s pH nu trn. b) Hng s to phc KY cho s to phc ca Mn+ vi Y4-, c ga tr 6,3.1021 (Hg2+); 2,1.1014 (Fe2+) v 5,0.1010 (Ca2+). Ion kim loi no s tp phc vi EDTA c hu sut hn 99,9% trong dung dch c cha 5,00.10-3 M EDTA tng cng, c m vi pH l: i) 2 ii) 6 iii) 10 c) Ion thu ngn (II) Hg2+ c i lc mnh vi clorua:

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Hg2+ + 4Cl- HgCl42Vi hng s to phc l Cl = 3,98.1015. Vi dung dch clorua 0,5M tng cng v 5,00.10-3M dung dch EDTA tng cng, hy xc nh t l thu ngn tn ti di dng ion t do, di dng anion tetraclorua v di dng phc EDTA ti pH bng: i) 2 ii) 6 iii) 10 (Nn ga thit rng nng kim loi tng cng rt nh hn 0,05M) e) Mt loi hn hng ch cha thu ngn, natri v canxi. Cho 5,218g mu ny tc dng vi mt tc nhn oxy ha thch hp ri pha thnh 500mL. Ly 25mL dung dch ny; m ti pH = 2,6; em chun vi dung dch 0,0122M dung dch MgY2-: tr s chun trung bnh l 44,19 mol.L-1. Khi ly 10mL mu dung dch, m ti pH = 9,5 c tr s chun 57,43mol.L-1. Hy xc nh phn trm khi lng ca thu ngn, canxi, natri trong hn hng. BI GII: a) Tnh c cc tr s sau: pH [Y4-](M) (Y4-) 2 3,712.10-14 8,26.10-16 -6 6 2,249.10 5,01.10-8 10 0,3548 7,90.10-3 Cc nng ca anion c xc nh nh CT(EDTA) = 0,02226M Mr(C10H16N2O8) = 292,25g.mol-1 b) Ta cn xc nh t l [MY2-]/[M2+], t l ny (nh xc nh hng s to phc) c tr s KY[Y4-]. Ta c th xc nh [Y4-] t x bit cu a. Tnh c cc tr s sau: pH [Y4-] [HgY2-]/[Hg2+] [FeY2-]/[Fe2+] [CaY2-]/[Ca2+] -16 6 2 1,856.10 1,17.10 0,039 9,28.10-6 6 1,1245.10-8 7,08.1013 2,36.106 562 -3 19 11 10 1,774.10 1,12.10 3,73.10 8,87.107 C th thy rng, ti pH = 2, ch Hg2+ to phc vi lng ng k; ti pH = 6, c Hg2+ v Fe2+ u to phc, trong khi ti pH = 10, c ba ion kim loi u to phc vi hiu qa cao.

c) Do HCl l axit mnh, cn bng gia Hg2+ v Cl- phi khng ph thuc pH; ta c th tnh c t s [HgCl42-]/[Hg2+] bng 2,488.1014 vi [Cl-] = 0,5M. Dng cc tr s [HgY2-]/[Hg2+] c c t cu b) ta tnh c: pH [HgY2-]/[Hg2+] %(Hg2+) %(HgY2-) %(HgCl42-) 2 1,17.106 4.10-13 5.10-7 >99,9 13 -13 6 7,08.10 3.10 22,2 78,8 10 1,12.1019 9.10-18 >99,9 2,2.10-3 d) Cc kt qa thu c t cu b) ch ra rng, ti pH = 2, s to phc EDTA ca Ca2+ l b qua c: ta c th gi thit iu y vn ng ti pH = 2,6; v nh th EDTA ch phn ng vi Hg2+ ti pH thp nh vy. Ti pH = 10, s to phc ca c Hg2+ cng nh Ca2+ l ng k, v ti pH = 9,5 s chun cho bit lng tng cng ca Hg2+ v Ca2+. Ta phi gi thit rng EDTA khng phn ng ng k vi Na+.Ti pH = 2,6, s mol Y4- = 5,391.10-4M S mol ny bng s mol (Hg2+) trong 25mL; nn s mol Hg2+ trong 500mL = 1,078.10-4 mol. V khi lng nguyn t ca Hg l 200,59g.mol-1, cho php xc nh khi lng thu ngn trong mu th l 2,163g. Ti pH = 9,5; s mol Y4- = 5,391.10-4 mol. S mol ny bng tng s mol (Hg2+ + Ca2+) trong 10mL, nn tng s mol (Hg2+ + Ca2+) trong 50mL = 3,503.10-2 mol, v v vy s mol Ca2+ trong 500mL = 2,425.10-2 mol tng ng vi 0,972g canxi trong mu th. Suy ra khi lng natri trong mu th (gi s khng ln tp cht khc) l 2,083g. Nh vy hm lng cc cht trong mu th l: Hg (41,45%); Na (39,32%), Ca (18,63%). OLYMPIC HA HC QUC T LN TH 30: S bc x kh nh knh (Greenhouse gas) l mt mi quan tm hng u v mi trung. S gia tng nng kh nh knh (CO2) trong kh quyn c ghi nhn vi thp nin qua. a) Ta s c lng s phn b cc kh nh knh trong kh quyn tri t bng cch gi thit rng nhng kh ny (CO2, hi H2O..) c khoanh vng trong tng c cao 10 11km (Cch m t nh th khng chnh xc, nhng c mc ch minh ho mt s khi nim). nh hnng ca tng kh - nh knh nh trn s ra sao i vi nhit kh quyn ti cao 5km? b) nh hng ca tng kh nh knh ln nhit kh quyn ti 15km. Gii thch mt cch nh tnh hiu ng ph ny ca tng nh knh (greenhouse layer). c) S bin i nhit ti phn thp hn ca kh quyn s nh hng ra sao n cn bng ca CO2 v ca H2O gia pha kh v pha ho tan trong nc? Liu s dch chuyn cn bng ca cc tiu phn ny, t n, c nh hng g n nhit ca phn thp hn ca kh quyn? d) Nay xt n tng cao hn tng nh knh l tng ozon. Ozon c to thnh cng nh b phn hy u do cc qa trnh quang ha (photochemical). ngh mt c ch gii thch s to thnh v phn hy ozon trong mt kh quyn cha oxy tinh khit. e) Entanpy to thnh ca O v O3 l Hof(O) = 249kJ.mol-1 v Hof(O3) = 143kJ.mol-1. Xc nh phtn vi di sng ln nht c kh nng quang phn oxy v ozon, theo th t. f) Trong vng cc, s tan r ca ozon ti tng bnh lu quan st c trong ma xun. Ngi ta ngh rng iu kin dn n s tan r tai ho ny (c bit di tn l thng ozon) ty mt s yu t, trong c cn bng sau: HCl(k) + PSC = HCl*PSC (1) ClONO2(k) + PSC = ClONO2*PSC (2) ClO(k) + ClO(k) = ClOOCl (3) PSC (vit tt ca Polar Stratospheric Cloud) ch my bnh lu ti cc vng cc to thnh t s ngng t hi nc v cc cht bay hi khc ti tng rt cao ny. S to thnh PSC thng xy ra

ti tng bnh lu Nam Cc trong ma ng v u ma xun, nhng t hn ti cc Bc v nhit khng qa thp. c bit rng s tan r ozon cng nghim trng khi cc cn bng trn di theo chiu thun. Cn nh rng s to thnh lin kt thng l to nhit, vy nhit s nh hng th no n cc cn bng trn? g) S to thnh ozon trong tng bnh lu ti cc Bc l mt hin tng c ghi nhn gn y: thot u, ngi ta hy vng cc Bc c min nhim vi s to thnh l thng ozon, nhng tht ra khng phi vy. Cn c trn cc thng tin c trn y, ngh xem l gii no c th ng tin cy c gii thch cho s pht trin l thng ozon ti cc Bc. i) Mc CFC trong tng bnh lu ti Bc bn cu gn y tng ln bng mc CFC m tng bnh lu ca Nam cc t trong thp nin trc. ii) S trn xung lin tc ca cc kh nh knh vo phn thp hn ca kh quyn cng lm gim lin tc nht tng bnh lu ti cc Bc. iii) S gia tng nng hi nc trong tng bnh lu ti cc Bc lm cho s to thnh PSC d dng hn trc y. iv) S gia tng lng phng x hng ngoi n tng bnh lu ti cc Bc gy ra s gia tng quang phn ca ozon ti cc Bc. BI GII: a) Cc kh nh knh lm nng h tng kh quyn, v mt s photon hng ngoi (IR) c ngun gc t b mt tri t l ra thot khi kh quyn li b hp th v bn tr li; s bn tr li v b mt tri t tng t nh khi bn i, lm cho h tng kh quyn nng ln. b) Cc kh nh knh lm lnh thng tng kh quyn: v c t photon hng ngoi n c cao ny t b mt, s hp th hng ngoi ny xy ra t cao 15km t hn l ra phi c. Hp th hng ngoi t dn n nhit gim. c) Cn bng xy ra l: V H2O(l) H2O(h). C hai cn bng ny u di theo chiu thun khi nhit tng: vy nng hi nc v CO2 s tng theo T. Do c hai u l kh nh knh, nn dn n hiu ng nh knh dng. d) Trong kh quyn oxy tinh khit, s hnh thnh ozon bng qa trnh quang ha dn n s quang phn oxy: O2 + h O + O (i) Sau : O + O2 O3 (ii) S quang phn cng ph hy ozon: O3 + h O2 + O (iii) (Mt qa trnh phn hy ozon khc c th l: O3 + O O2 + O2 (iv)) e) T entanpy to thnh cho sn (Hof(O2) = 0kJ.mol-1), tnh c entanpy ca phn ng (i) v (iii): Hoi = 498kJ/mol = 8,27.10-19J/phn t Hoiii = 106kJ/mol = 1,76.10-19J/phn t Vi E = h = hc/ ta c: 1 = 2,40.10-7m = 240nm 2 = 1,129.10-6m = 1129nm CO2(aq) CO2(k)

y l cc photon c di sng di nht c nng lng quang phn O2 v O3 theo th t (thc t, s quang phn c hiu qa ozon cn di sng ngn hn ga tr ngh, v nhng l do khng xt n trong phm vi cu hi ny). f) C ba cn bng ny khi xt theo chiu thun u c s to thm mt lin kt, nn nhit cao, cn bng s di qua tri v di qua phi khi nhit h thp. g) Ba gi thuyt c th b bc b nh sau: i. Khi nim rng mc CFC Bc bn cu t hn Nam Bn cu l iu khng tng: mc cng nghip ha mnh nht, v phng thch CFC nhiu nht l Bc bn cu (Thc vy, nng CFC h tng kh quyn l tng t nhau trn ton a cu: chng c trn rt u) iii. Trong khi s gia tng hi nc lm tng kh nng to PSC, khng c chng c g v s chuyn dch hi nc n tng Bnh cc Bc (c chng ch l chuyn n h tng kh quyn). iv. nh hng ca s gia tng nng cc kh nh knh (v chng vn ang gia tng!) l lm gim lng hng ngoi n c tng bnh lu. Ngoi ra d cc photon gn - hng - ngoi c nng lng cao hn mnh lin kt O2 O, chng thc s khng quang phn ozon. Gi thuyt ii) l cu tr li c ngha nht cc kh nh knh s lm nng h tng kh quyn v lm lnh thng tng kh quyn (Tuy nhin, iu khng c ngha l c ch ny l nguyn nhn ng thin nhin lun phc tp hn ta mong i). OLYMPIC HA HC QUC T LN TH 31: Tng hp mt hp cht ca crom. S phn tch nguyn t cho thy rng thnh phn c Cr (27,1%); C (25,2%), H(4,25%) theo khi lng, cn li l oxy. a) Tm cng thc thc nghim ca hp cht ny. b) Nu cng thc thc nghim gm mt phn t nc, ligand kia l g? Mc oxy ha ca Cr l bao nhiu? c) Kho st t tnh cho thy hp cht ny l nghch t, phi gii thch t tnh ca hp cht ny nh th no? V th cu to ph hp ca cht ny. BI GII: a) Cng thc thc nghim CrC4H8O5. b) T cng thc thc nghim CrC4H8O5, hp cht l [Cr(CH3COO)2(H2O)]. Nh vy, ligand l cc nhm axetat. Do nhm (CH3COO-) c in tch 1 nn mc oxy ha ca Cr l +2. c) Ion Cr2+ l h d4, ngha l h c 4e thuc obitan d. S phn b 4 electron phi thuc loi spin nng lng cao do ligand yu. Ch yu t ny cho thy [Cr(CH3COO)2(H2O)] c tnh thun t. Tuy nhin t cc kt qa thc nghim, hp cht ny li c tnh nghch t l do hp cht ny dng nh hp c cu to nh sau:

CH3 C OO OH2 O C H3C Cr C

CH3

O O Cr H2O

O O O C CH3

Trong cu to ny, hai nguyn t Cr to lin kt bn, bao gm mt sigma, hai pi v mt delta, vi bc lin kt tng cng l 4. S hnh thnh lin kt bn i hi tt c cc electron thuc obitan d u phi cp i. V vy da theo tnh cht t, hp cht dng nh hp l nghch t. OLYMPIC HA HC QUC T LN TH 32: Clorat v peclorat c s dng trong s ch to dim qut, pho v cht n. Bc th nht trong vic sn xut kali clorat l s in phn dung dch nc ca kali clorua. a) Vit cc phng trnh phn ng xy ra ti hai in cc. Cl2 c to thnh anot v OH- c to thnh catot b) Clo to thnh phn ng vi ion hydroxit to thnh clorat. Vit phng trnh phn ng. c) Hy tnh khi lng kali clorua v in lng (theo Ah, ampe gi) cn thu c 100g KClO3. BI GII: a) Catot: 2H2O(l) + 2e = 2OH-(aq) + H2(k) Anot: 2Cl-(aq) = Cl2(aq) + 2e b) Phn ng: 3Cl2(k) + 6OH-(aq) = ClO3-(aq) + 5Cl-(aq) + 3H2O(l) c) mKCl = 60,83g Q = 131Ah OLYMPIC HA HC QUC T LN TH 32: Mi trng d nhim ch lun l mt iu ng lo ngi. Trong c th con ngi, mc c hi ca ch c th c gim bt vi liu php chelat bng cch s dng cc ligand c tim nng hnh thnh cc phc Pb2+ bn c th c thn bi tit. Ligand EDTA4- c dng cho mc ch ny nh s hnh thnh phc [Pb(EDTA)]2- rt bn vng (hng s bn vng, K(Pb) = 1018M-1). Ligand c cung cp bng cch tim truyn dung dch Na2[Ca(EDTA)], mui natri ca phc canxi tng i km bn (K(Ca) = 1010,7M-1). S trao i ca canxi vi ch ch yu din ra trong mch mu. a) Mc ch c trong mu ca mt bnh nhn l 83g/dL. Hy tnh nng mol ca ch trong mu bnh nhn ny. b) Trong mt th nghim lm mu, ngi ta iu ch mt dung dch cha Ca(NO3)2.4H2O v Na2[Ca(EDTA)] c nng mi cht theo th t bng 2,5mM v 1,0mM. Thm Pb(NO3)2 rn vo t c nng ch tng ng vi nng ch trong mu bnh nhn nu trn. Hy tnh tr s gn ng ca t l [Pb(EDTA)]2-/Pb2+ trong dung dch thu c ti cn bng. Khng xt tnh cht axit baz ca cc tiu phn c lin quan v s thay i th tch dung dch coi nh khng ng k.

c) S bi tit ca phc [Pb(EDTA)]2- qua thn hin nhin l mt qa trnh bc nht theo nng [Pb(EDTA)]2- trong mu. Sau 2 gi, nng ca phc [Pb(EDTA)]2- trong mu ca hu ht cc bnh nhn thng gim 60%. Hy tnh chu k bn hy sinh hc ca phc [Pb(EDTA)]2BI GII: a) [Pb2+] = 83/(207,2.0,10) = 4,00M b) Xt phn ng: [Ca(EDTA)]2-(aq) + Pb2+(aq) [Pb(EDTA)]2-(aq) + Ca2+(aq) Vi hng s cn bng K = K(Pb)/K(Ca) = 107,3. Do kh nng to phc mnh v ion Ca2+ c d so vi tng lng ligand EDTA4-, hin nhin tt c ligand s ni kt trong phc ch hoc phc canxi. V [Pb(EDTA)2-] P) hay tg3eg1 (P) hay tg3eg2 (P) hay tg4eg2 (P) hay tg5eg2 (