BI TP V CU TO NGUYN TCu 1: Cho phn t v cac ion sau: [PtCl6] 2 ; [Ni(CN)4]2 ; NH4+; SF6 ; NF3 . Nu trang thai lai ho a cua cac nguyn ttrung tm va cu truc hi nh ho c cua cac phn ttrn.GiiCht Lai ho Cu trc hnh hc [PtCl6] 2 [Ni(CN)4]2 NH4+ SF6 NF3 d2sp3dsp2sp3sp3d2sp3bt dinvung phngt dinbt dinchpCu 2: Da trn cu to phn t, hy gii thch:a. Ti sao NO2 c khuynh hng ime ho to thnh N2O4?b. Ti sao BCl3 c th kt hp vi NH3 to ra NH3BCl3?c. Ti sao AlCl3 tn ti dng ime Al2Cl6?GiiNguyntNtrongNO2chabn c 7e lp ngoi cngv cn 1e c thnnn c khuynh hng gp chung vi nguyn t N trong phn t th 2:ONO.ONO.+ONOONONguyn t N trong phn t NH3 cn cp e t do to lin kt phi tr vi nguyn t B trong phn t BH3:HN HH:HB HH+HN HHHB HHNguyn t Clo ca mi phn t AlCl3 u cn 3 cp e t do c kh nng to lin kt phi tr lm bn ho nguyn t Al (trc c 6e)AlClClClAlClClClCu 31. 137Ce tham gia phn ng trong l phn ng ht nhn, c chu k bn hy 30,2 nm. 137Ce l mt trong nhng ng v b pht tn mnh nhiu vng ca Chu u sau mt v tai nn ht nhn. Sau bao lu lng cht c ny cn 1% k t lc tai nn xy ra.2. Da theo thuyt MO, hy gii thch t tnh ca phn t F2 v ion CO+.GiiVy sau 200,46 nm th lng cht c trn cn 1% k t lc tai nn xy ra.2. Cu hnh electron ca phn t F2:(lks 2 )2(*s 2 )2(lkp 2z )2(lkp 2x = lkp 2y)4(*p 2x = *p 2y)4Phn t F2 khng c electron c thn nn nghch t.

Cu hnh electron ca ion CO+:(lks 2 )2(*s 2 )2(lkp 2x = lkp 2y)4(lkp 2z )1Ion CO+ c electron c thn nn thun t. Cu 4Vit cu hnh electron ca cc ion v phn t : O2, O2,O22, O2+theo thuyt MO. Tnh s lin kt, so snh di lin kt trong cc phn t v cho bit cht no thun t, cht no nghch t.Gii Cu hnh electron :(8-4)/2 = 2 O2 :2s2 *2s2 z2 x2 y2 *x1 *y1 (8-5)/2 = 1,5 O2 : 2s2 *2s2 z2 x2 y2 *x2 *y1(8-6)/2 = 1 O22: 2s2 *2s2 z2 x2y2 *x2 *y2(8-3)/2 = 2,5 O2+ : 2s2 *2s2 z2 x2 y2 *x1S lin kt gim, di lin kt tng theo th tO22>O2>O2> O2+O2, O2 v O2+ u c electron c thn => c tnh thun t. O22khng c electron c thn => c tnh nghch tCu 51. Cho bit trong s cc t hp s lng t sau y, t hp no l khng hp l:a. n=3,l=2,m=2,s= +12b. n=3,l=3,m=2,s= -12c. n=4,l=2,m=-3, s= +121. p dng cng thc:K = NNlgK303 , 2tNNlgt303 , 2NNlnt1o o o M k = NNlg693 , 0T 303 , 2tT693 , 0o (0,5)72 , 200693 , 02 . 2 , 30 . 303 , 2100 lg .693 , 02 , 30 . 303 , 2100NNlg693 , 02 , 30 . 303 , 2too (nm)d. n=4,l=3,m=-3, s= +12e. n=4,l=2,m=-2,s= -12 2.Cho bit cu trc phn t ca cc phn t v ion sau: NH4+ ; PCl5 ; SF6; XeF4. Hy cho bit : hnh dng phn t v v n gin phn tGii: 1.(b) do vi phm: l =0(n-1)(c) do vi phm: m = -l l2. NH4+: t din u PCl5: lng chp tam gic

HNHHH + ClPClClClClSF6: bt din u XeF4: vung phng

FSFF FFFFXeFFFCu 6 1. Cho phn t v cac ion sau: [PtCl6] 2-; [Ni(CN)4]2 ; NH4+; SF6 ; NF3

. Nu trang thai lai ho a cua cac nguyn ttrung tm va cu truc hi nh ho c cua cac phn ttrn2. Cho 2 b s lng t sau y ng electron c mc nng lng cao nht ca2 nguyn t 2 nguyn t X, Yn l mlmsX 3 2 1 -1/2Y 3 3 2 +1/2a. Nguyn t no khng tn ti ?b. Hy xc nh v tr ca nguyn t cn li trong Bng tun hon. .Gii1.Lai ho Cu trc hnh hc [PtCl6] 2- [Ni(CN)4]2- NH4+ SF6 NF3 sp3d2sp2dsp3sp3d2sp3bt dinvung phngt dinbt dinchp2. a. Nguyn t Y khng c tht ,b s lng t khng c tht v vi n = 3 th l ti a l 2. b.Nguyn tX c : n=3, l=2=> 3dml=1, ms = -1/2

=> electron l 3d7 2 1 0 -1-2- Cu hnh ca electron nguyn t X: 1s2 2s2 2p6 3s2 3p6 3d7 4s2 - V tr X : Z = 27 => th 27 4 lp electron => chu k 4 C 9 electron ha tr, nguyn t d => nhm VIIIA Cu 7 a) Cho bit trng thi laiha v dng hnh hc ca cc phn t : BrF3 , CBr4, PCl5, NH3 .b) Nguyn t (X) c nhiu dng th hnh, c m in nh hn oxi v ch to hp cht cng ho tr vi halogen. Electron cui cng in vo phn lp tha mn iu kin tng 4 s lng t bng 5,5 ;n + l = 4. Vit cu hnh electron v gi tn (X).Giia)Cng thc phn t Trang thai lai ho a cua nguyn ttrung tmDng hi nh ho cBrF3sp3d Ch TCBr4sp3T din uPCl5sp3d Lng thp tam gicNH3sp3Hnh thp y tam gicb) Xc nh cc trng hp:n = 1 2 3l =m =00,10, 1, 2ss, ps, p, dChn n = 3, l = 1, m = 1, ms = + Cu hnh electron ca (X): 1s2 2s2 2p6 3s2 3p3 X l P (photpho)Cu 81) Hy gii thch ti sao kh F2 mt mt electron thnh 2F+ th bn lin kt tng; kh N2 chuyn thnh 2N+ th bn lin kt gim.2) Hp cht A c to thnh t cc ion u c cu hnh electron 1s2 2s2 2p6 3s2 3p6. Trong mt phn t A c tng s ht proton, ntron v electron bng 164. Xc nh cng thc phn t ca A. Bit rng A tc dng c vi mt nguyn t n cht c trong thnh phn A theo t l mol 1 : 1 to thnh chtB. Vit cng thc cu to ca A v B.Gii1) F2 :[KK]2 *2 2 4 * * 4S S Z x Y x y ( ) ( )Bc ni: (8 6) : 2 = 12F+ : [KK]2 *2 2 4 *2 *1S S Z x y x y ( ) Bc ni:(8 5) : 2 = 1,5V bc lin kt trong 2F+ ln hn trong F2 nn 2F+c bn lin kt ln hn.N2 : [KK]2 *2 4 2s s x y z ( ) Bc ni: (8 2) : 2 = 32N+: [KK]2 *2 4 1s s x y z ( ) Bc ni(7 2) : 2 = 2,5 V bc lin kt trong 2N+nh hn trong N2 nn lin kt trong 2N+ km bn hn.2) V cu hnh electron ca cc ion l : 1s2 2s2 2p6 3s2 3p6.Do mi ion u c 18e Gi s trong phn t (A) c x ion :s electron = 18x s proton = 18xV t N trong phn t (A) l N, ta c : 18x + 18x + N = 164 36x + N = 164N = 164 36x (1)Vi ng v bn, ta c : N1 1, 52Z Nn N1 1, 5218x (2)Gii (1) v (2) ta c : 03 , 3 6 , 2 x Chn x = 3 N = 56* TH1: A gm 2 cation M+ v 1 ion X2 Cng thc A l M2XTa c : ZX =16 233 . 18 l lu hunh (S)ZM =19 133 . 18 + l kali (K)Cng thc A lK2S * TH2 : (A) gm 1 cation M2+ v anion XCng thc A l MX2Ta c : ZX =17 1354 l clo (Cl)H2OOHOHZM =20 2354 +l canxi (Ca)Cng thc A lCaCl2 V A tc dng vi nguyn t n cht c trong thnh phn ca A to thnh B, nn A l K2S.K2S + S K2S2Cng thc cu to ca A v B K S KK S S K (0,25)Cu 9: a) Vit cng thc cu to ca Ca3(PO4)2 v Fe3O4 . b) Lin kt to thnh gia Al v Clo l lin kt g ? gii thch . c) Trong t nhin tn ti hp cht Al2Cl6 . Gii thch s hnh thnh lin kt trong phn t ca hp cht ny .Giib) Lin kt cng ha tr do 3,16 1,61 = 1,55 < 1,7 c) Phn t Al2Cl6 c hnh thnh do s nh hp AlCl3, Lin kt gia hai phn t AlCl3 hnh thnh do vic to lin kt phi tr gia nguyn t Cl ca phn t AlCl3 ny vi nguyn t Al ca phn t AlCl3 khc Cu 10S dng thuyt Gillespie, hy xc nh cu trc ca cc phn t v ion sau , ng thi cho bit kiu lai ha ca cc AO ha tr ca nguyn t trung tm v so snh gc lin kt gia chng . a) CF4 v NF3b) NO2 , NO2+, NO2- c) OF2, OCl2d) NH3 ; PH3 ; AsH3; SbH3.Giia) CF4 bn mt u ; NF3 chp tam gic. C v N u c lai ha sp3 , gc FCF = 109028 > gc FNF.b) NO2 gp khc , NO2+ thng , NO2- gp khc. N trong NO2 v NO2- c lai ha sp2. N trong NO2+ c lai ha sp. NO2+ gc ONO = 1800 > gc ONO NO2-c) OF2 v OCl2 u c dng gp khc v Oxi u c lai ha sp3. Gc FOF < gc ClOCl v m in ca F > m in ca Cl.d) C 4 phn t u c cu trc chp tam gic. nguyn t trung tm u c lai ha sp3. Gc HXH gim dn t NH3 n SbH3 trong nhm , v m in ca nguyn t trung tm gim dn theo dy N-P-As-Sb.Cu 1 1 1) Bit rng trong c th sinh vt ang sng, 1 gam cacbon ca c th c 15,3 phn r phng x 14C trong 1 giy. Trong mt mu c vt, ngi ta thy 1 gam cacbon c 10,5 phn r phng x 14C trong 1 giy. Hy tnh tui ca mu c vt . Bit chu k bn hy ca ng v phng x 14C l 5730 nm.2)Vng c mng li lp phng tm din, mng n v c cnh a=4,070.10-10m. Hy tnh bn knh ca nguyn t vng v t khi ca vng. Bit nguyn t khi ca vng l 197,0 amu.3) i vi phn t H2O thc nghim o c momen lng cc l 1,85D, di lin kt O-H l 95,7pm, gc lin kt HOH l 104,5o.H2OOHOH a) Hy tnh momen lng cc ca lin kt O-H theo n v D v C.m. b) Tnh ion ca lin kt O-H trong phn t H2O.Gii1)0 0121215, 3N Nlnlnln kt10, 5NNt t 5730x 3112ln 2 ln 2ln 2 kt ;2)Xt mng n v ca Au hnh lp phng, c 8 nguyn t Au 8 nh v 6 nguyn t Au 6 mt. ng cho d ca hnh vung trn mi mt lp phng bng 4 ln bn knh nguyn t Au: d=4rmt khc: d2=2a2 d=a2=2x4,070.10-10m r=14x2x4,070.10-10=1,439.10-10mMi mng n v c 4 nguyn t Au ( mi nh c 18nguyn t x 8 nh + mi mt c 12nguyn t x 6 mt) nn t khi Au l khoi lng4nguyentAuThetchomang323 8 34x197, 019, 4g.cm6, 023x10 x(4, 070x10 cm) 3)a) Da vo php cng vect theo hnh v sau:Ta c: H O 30 2OH o1, 85D1, 511D 5, 031.10 C.m.2cos 2cos52, 25 b) Tnh OH theo l thuyt:OH(l thuyt)=e.R=1,602.10-19C x 95,7.10-12m=15,331.10-30C.m ion ca lin kt OH:H2OOHOH

30305, 031.10 C.m32, 82%15, 331.10 C.mCu 12 1) Cho cac phn tsau: POF3 ; BF3 ; SiHCl3 ; O3. Nu trang thai lai ho a cua cac nguyn ttrung tm va vecu truc hi nh hoc cua cac phn ttrn. 2) Kim loi ng c cu trc mng tinh th kiu lp phng tm din. - Cho bit s nguyn t ng cha trong t bo s ng ny.- Tnh di cnh lp phng a (nm) ca mng tinh th, bit rng nguyn t ng c bn knh bng 0,128 nm.Gii 1) Cng thc phn t Trang thai lai ho a cua nguyn t trung tmCu truc hi nh hocPOF3sp3POFFFBF3sp2BFF FHSiCl3sp3SiHClClClO3sp2OOO. .2) - S nguyn t Cu = 881 + 621= 4 - Xt ng cho ca 1 mt, ta c di ng cho = 4Cur = a2 a = 24Cur = 2128 , 0 4 = 0,362 nmCu 13 1. Trong s cc phn t v ion: CH2Br2, F - , CH2O, Ca2+, H3As, (C2 H5 )2O , phn t v ion no c th to lin kt hiro vi phn t nc? Hy gii thch v vit s m t s hnh thnh lin kt . 2. Nng lng lin kt ca N-N bng 163 kJ.mol1, ca N Nbng 945 kJ.mol1. T 4 nguyn t N c th to ra 1 phn t N4 t din u hoc 2 phn t N2 thng thng. Trng hp no thun li hn? Hy gii thch.Gii1. Cc vi ht CH2Br2, Ca2+, H3As khng c nguyn t m in mnh nn khng th to lin kt hiro vi phn t nc.Cc vi ht F - , CH2O, (C2 H5 )2O c nguyn t m in mnh nn c th to lin kt hiro vi phn t nc: 2. a) Xt du ca nhit phn ng H =i Ei -j Ej i jTrong i, j l lin kt th i, th j cht tham gia, cht to thnh tng ng ca phn ng c xt; Ei ; Ej l nng lng ca lin kt th i, th j .b) Xt c th vi nit : * Phn ng 4 N N4(1)C H1 = 4 EN-EN4 = 0,0- 5 163 ; vy H1 =- 815 kJ . * Phn ng 4 N 2 N2(2)C H2 = 4 EN- 2 EN2 = 0,0- 2 945 ; vy H2 =- 1890 kJ .Ta thy H2 < H1. Vy phn ng 4 N 2 N2 xy ra thun li hn phn ng4 N N4.Cu 1 4:Cho cac phn t sau: PH3 ; AsH3 ; POF3 ; POCl3 ; BF3 ; SiHCl3 ; NF3 ; O3. a. Nu tra ng thai lai ho a cua cac nguyn ttrung tm vavecu truc hi nh hoc cua cac phn ttrn. b. So sa nh goc lin kt H X H gia hai phn t PH3 va AsH3. Giai thich. c. Trong hai phn tNF3 va BF3, phn t nao co momen l ng cc ln hn khng?Giia)Cng thc phn t Trang thai lai hoa cu a nguyn t trung tmCu truc hinh ho cPH3sp3PHHHAsH3sp3AsHHHPOF3sp3POFFFPOCl3sp3POClClCl. . .HCHHOH. . .F OHHC2H5OC2H5HOHOBF3sp2BFF FHSiCl3sp3SiHClClClNF3sp3NFFF. .O3sp2OOO. .b) Gc HPH > HAsH v m in ca nguyn t trung tm P ln hn ca As nn lc y mnh hn. c)NFFF. .BFF FC cu trc bt i xngnn c momen lng cc ln hn khng.C cu trc i xng nn c momen lng cc bng 0 khng.Cu 15 Nguyn t (X) c nhiu dng hnh th, c m in nh hn oxi v ch to hp cht cng ho tr vi halogen. Electron cui cng in vo phn lp tha mn iu kin tng 4 s lng t bng 5,5 ;n + l = 4.a. Vit cu hnh electron v gi tn (X). b. (X) to vi hir nhiu hp cht cng ho tr, c cng thc chung l XaHb. Dy hp cht ny tng t dy ng ng ca ankan. Vit cng thc cu to 4 cht ng ng u tin.Gii:a) Xc nh cc trng hp:n = 1 2 3l =m =00,10, 1, 2Ss, p s, p, dChn n = 3, l = 1, m = 1, ms = + Cu hnh electron ca (X): 1s2 2s2 2p6 3s2 3p3 X l P (photpho)b)4 cht u tin l: PH3, P2H4, P3H5, P4H6Cng thc cu to: PH H HH HP PH HCu 1 5 A, B l hai nguyn t k tip nhau trong cng mt chu k ca bng tun hon trong B c tng s lng t (n+l) ln hn tng s lng t (n+l) ca A l 1. Tng s i s ca b 4 s lng t ca e cui cng ca cation Aa+ l 3,51. Xc nh b 4 s lng t ca e cui cng trn A, B.2. Vit cu hnh e v xc nh tn A, B.Gii1/V hai nguyn t k tip nhau trong cng 1 chu k => hai nguyn t c cng s lp e (cng n)M tng (n+l) ca B ln hn tng (n+l) ca A l 1=> Cu hnh e lp ngoi cng caA, B l:A: ns2, B: np1 Mt khc A c 2 e lp ngoi cng cation A c dng A2+ (A thuc phn chnh nn d nhng 2e lp ngoi cng)Vy tng s i s ca 4 s lng t cui cng ca A2+ l:(n-1)+1+1-1/2=3,5 n=3Vy 4 s lng t ca :A: n=3; l=0; ml=0; ms=-1/2B: n=3; l=1; ml=-1;ms=+1/2b)Cu hnh e:A: 1s22s22p63s2(Mg)B:1s22s22p63s23p1(Al) Cu 16 H H P PH PP H HH HH P HPP HH Tnh nng lng mng li ion ca CaCl2, bit rng :-0298,sH ca tinh th CaCl2 = -795 kJ/mol- Nhit thng hoa 0aH ca Ca(r) Ca(k) = 192 kJ/mol- Nng lng ion ha : Ca(k) Ca2+ (k) + 2e ; I1+ I2 = 1745 kJ/mol- Nng lng lin kt (Elk)Cl-Cl trong Cl2= 243 kJ/mol- i lc electron (E) ca Cl(k) = -364 kJ/molGiiC th thit lp chu trnh Born-Haber tnh ton theo nh lut Hess:Ca(r)+Cl2(k) CaCl2 (r)

0298, SH

0aH Cl ClEH

1 2( )CaI I + Ca(k)+2Cl(k) Ca2+ (k)+2Cl-(k) 2ClENu s dng phng php ni trn ta c :Ca(r)+Cl2(k) CaCl2 (r)0298, SH Ca(k) Ca(r)0aH 2Cl(k) Cl2(k)Cl ClECa2+ (k)+ 2e Ca(k)1 2( )CaI I +2[Cl-(k) Cl(k)+e]2ClE Cng cc phng trnh ta c :Ca2+ (k)+2Cl-(k) CaCl2 (r)Nng lng ca qu trnh ny l nng lng mng li ion ca CaCl2 H = 0298, SH 0aH Cl ClE1 2( )CaI I + 2ClE H = (-795) 192 243 1745 -2(-364) = -2247kJ/mol Cu 17Hai nguyn t A, B c electron cui cng ng vi 4 s lng t sau:A( n =2, l = 1 , m = -1 , s = -1/2 ); B ( n = 3, l = 1 , m = 0 , s = -1/2 )a) Vit cu hnh electron v xc nh v tr A , B trong bng h thng tun hon. Gi tnb) Vit cng thc cu to cc hp cht trong cng thc phn t c 3 nguyn t A, B v hidro. Cho bit loi lin kt trong cc hp cht ny.Giia)Nguyn t A: n = 2 ( lp 2); l =1 ( phn lp p ) ; m= -1 ( obitan px ) s= -1/2Cu hnh electron ca A : 1s22s22p4 Tng t : B c phn lp ngoi cng l 3p5Cu hnh electron ca A : 1s22s22p63s23p5 V tr trong HTTH :A: th 8 , chu k 2 , nhm VIA => A l O ( oxi )B: th 17 , chu k 3 , nhm VIIA => B l Cl ( clo )b) H-Cl-O ; lin kt : H-O ; O-Cl l lin kt cng ha tr c ccHClO2 :

H-O-ClO : 2 lin ktcng ha tr c cc v 1 lin kt cho- nhnHClO3 : H-O-ClO : 2 lin ktcng ha tr c cc v 2 lin kt cho- nhnO O HClO4 : H-O-ClO : 2 lin ktcng ha tr c cc v 3 lin kt cho- nhn OCu 18 1. Tinh th NaCl c cu trc lp phng tm mt. Tnh bn knh ca ion Na+ v khi lng ring ca NaCl tinh th, bit rng cnh a ca mng n v c s bng 5,58 0A, bn knh ca ion Cl- l 1,810Av khi lng mol ca Na v Cl ln lt l 22,99g mol-1 v 35,45g mol-1.2. Tnh c kht C ca loi cu trc ny.Gii1.a = (r++r-)2 r+ =098 , 0 81 , 1258 , 52A ra

tnh khi lng ring ca NaCl tinh th, ta cn tnh s phn t NaCl trong mng c s. Trong mt mng cha 4 phn t NaCl. Khi lng mt mol NaCl l: 22,99 + 35,45 = 58,44 gam mol-1 Khi lng mt phn t NaCl l m v khi lng 4 phn t NaCl l 4m:

2310 . 023 , 64 . 44 , 584 m Khi lng ca NaCl l: 34amd

gam d 23 , 2) 10 . 58 , 5 ( 10 . 023 , 644 , 58 . 43 8 23 cm-3 2. Cu trc lp phng tm mt, s nguyn t trong mng c s l 4 Th tch 4 nguyn t trong mng: 334. 4 r Va3 = 324

,_

r Vy: 74 , 02431633

,_

rrC Hay 74% Cu 19Kim loi ng c cu trc mng tinh th kiu lp phng tm din. Hy:a. V cu trc mng t bo c s v cho bit s nguyn t ng cha trong t bo s ng ny.b. Tnh di cnh lp phng a (nanomet) ca mng tinh th, bit rng nguyn t ng c bn knh bng 0,128 nm.c. Xc nh khong cch gn nht gia 2 nguyn t ng trong mng.d. Tnh khi lng ring d ca ng theo n v g/cm3Giia. Mng t bo c s ca ng: Theo hnh v s nguyn t ng l:- 8 nh ca hnh lp phngS nguyn t Cu = 881= 1 - 6 mt ca hnh lp phngS nguyn t Cu = 621 = 3 Tng s nguyn t Cu = (1 + 3) = 4b. Tnh di cnh lp phng: Xt mt ABCD ta c ng cho AC = 22 = 4Cur a = 24Cur = 2128 , 0 4 = 0,362 nma = 0,362 nm c. Tnh khong cch ngn nht gia 2 nguyn t Cu: Theo hnh v khong cch ngn nht gia 2 nguyn t ng l an AMTa c:AM = 222a AC= 22 362 , 0 = 0,2559 nm d. Khi lng ring ca ng: (0,5 )Ta c:D =43 a NM = 4) 10 62 , 3 ( 10 023 , 6643 8 238,9 g/cm3CU I :(5,0 im)1/ AlCl3 khi ha tan vo mt s dung mi hoc khi bay hi nhit khng qu cao th tn ti dng ime (Al2Cl6).a. Vit cng thc cu to ca phn t ime v monome ,cho bit kiu lai ha ca nguyn t Al, kiu lin kt trong mi phn t ?b. M t cu trc hnh hc ca phn t .2/ Hy dng k hiu lng t biu din cc trng hp s lng electron trong mt obitan nguyn t. 3/ Mt phn tRY3 , c tng s ht c bn bng 196, trong s ht mang in nhiu hn s ht khng mang in l 60 , sht mang din ca R t hns ht mang in ca Y l 76. Xc nh k hiu ho hc RY3 . Vit cu hnh electron ca nguyn t R v Y, xc nh v tr R v Y trong bng tun hon.CU I:( 5IM)Cl 1/(2,75) Cl Cl Cl |Al AlAl ClCl ClClCl

CTCT ( 0,5) (0,25)a. Kiu lai ha ca nguyn t Al: (0,5 x 2)- Trong AlCl3 l sp2( v Al c 3 cp electron ho tr)- Trong Al2Cl6 l sp3(v Al c 4 cp electron ho tr) Kiu lin kt ca mi phn t :(0,5) - AlCl3 : c 3 lin kt cng ha tri 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 l nguyn t nhn , Cl l nguyn t cho).Trong 6 nguyn t Cl c 2 nguyn t Cl c 2 loi lin kt : 1lk cng ho tr v 1 lk cho nhn.b. Cu trc hnh hc (0,5) - Phn t AlCl3 c cu trc tam gic phng u.- Phn t Al2Cl6 c cu trc 2 t din ghp vi nhau,mi nguyn t Al l tm ca mt t din ,mi nguyn t HCl l nh ca t din , c 2 nguyn t Cl l nh chung ca t din 2/ (0,25 x 3) 3/(1,5) - t h, gii h(0,5)- RY3 l AlCl3 (0,25) -Cu hnh electron(0,25) v v tr(0,5)Cu I: (4 im) 1) Cation X+ v anion Y- c cu hnh electron ging cu hnh electron ca kh him Ar. Vit cu hnh electron ca X, Y. Cho bit v tr ca X, Y trong bng h thng tun hon cc nguyn t ha hc.1) 1 im - Cu hnh e ca X : 1s22s22p63s1 - Cu hnh e ca Y : 1s22s22p5- X nm th 11, chu k 3, phn nhm chnh nhm I- Y nm th 9, chu k 2, phn nhm chnh nhm VIICu I ( 5,0 im): 1. ( 0,75 im). Cho khi lng ring ca Li l 0,53g/cm3; ca Ag l 1,05g/cm3. Hy so snh dn in ca 2 kim loi trn v gii thch bng tnh ton. 2. ( 3,5 im). C 3 bnh in phn mc ni tip, bnh 1 cha dung dch CuCl2, bnh 2 cha dung dch Na2SO4, bnh 3 cha dung dch AgNO3. Cho ngun in mt chiu i qua 3 bnh. a. Vit phng trnh phn ng xy ra 3 bnh.b. Khi cc m bnh 1 thu c 3,2g kim loi th cc m ca cc bnh cn li thu c bao nhiu? cc dng cc bnh thot ra bao nhiu lt kh ( ktc)? 3. ( 0,75 im ). Ch cho bm kh CO2; dung dch NaOH khng r nng ; hai cc thu tinh c khc . Hy iu ch dung dch Na2CO3khng c ln NaOH hay NaHCO3 m khng dng thm phng tin no khc. - So nguyen t Li co trong 1 cm3: 202310 . 456710 . 02 , 6 . 53 , 0 nguyen t- So nguyen t Ag co trong 1cm3:202310 . 58510810 . 02 , 6 . 5 , 10nguyen tV nguyen t Ag co trong 1cm3 cua Ag ln hn Li nen Ag dan ien tot hnPhng trnh ien phan xay ra cac bnh :* Bnh 1: catot: Cu2+ + 2e = Cu anot: 2Cl 2e = Cl2 Phng trnh ien phan: 2 2Cl Cu CuCl + p (1)* Bnh 2: catot: 2H2O + 2e = H2 + 2OH anot: H2O 2e = 2H+ + 21O2 Phng trnh ien phan: 2 2 2O21H O H + p(2)* Bnh 3: caot: Ag+ + 1e = Ag anot: H2O 2e = 2H+ + 21O2 Phng trnh ien phan: 3 2 2 3HNO 2 O21Ag 2 O H AgNO 2 + + +pmol 05 , 0642 , 3nCu Theo nh luat Faraday, ta co:9650 It2 . 96500ItnCu Do cac bnh mac noi tiep nen It(1) = It(2) = It(3) Vay cc am cua bnh (2), so mol H2 tao thanh:mol 05 , 096500 2965096500 2Itn2Hg 1 , 0 2 . 05 , 0 m2H cc am bnh (3):g 8 , 10 108 . 1 , 0 mmol 1 , 09650096501 . 96500ItnAgAg Theo cac phng trnh ien phan ta co: cc dng bnh (1) ta co:lit 12 , 1 V mol 05 , 0 n n2 2Cl Cu Cl cc dng bnh (2) ta co:lit 56 , 0 V mol 025 , 0 n21n2 2 2O H O cc dng bnh (3) ta co:lit 56 , 0 V mol 025 , 0 n41n2 2O Ag O Cho vao hai coc thuy tinh co khac o mot lng ddNaOH nh nhau, anh so coc (1) va (2) Bm kh CO2 vao coc (1) cho en dCO2 + NaOHNaHCO3 Lay dd thu c coc (1) o vao coc (2) ta c Na2CO3NaHCO3 + NaOH Na2CO3 + H2O2)Cho b bn s lng t ca electron cui cng ca nguyn t cc nguyn t A,X,Z nh sau :A : n = 3, l = 1, m = -1, s = - 1/2X : n = 2, l = 1, m = -1, s = -1/2Z :n = 2, l =1, m = 0, s = +1/2 a) Xc nh A,X,Zb)Cho bit trng thi lai ha v cu trc hnh hc ca cc phn t v ion sau:AX2,2 23 4, AX AX , ZA2

3) Cho bit s bin i trng thi lai ha ca nguyn t nhm trong phn ng sau v cu to hnh hc ca AlCl3,4AlCl:AlCl3 + Cl- 4AlCl 4)a) Nguyn t R thuc nhm no ? Phn nhm no trong bng h thng tun hon? Bit s oxi ha ca R trong hp cht oxit cao nht l mO, trong hp cht vi hidro l mH v : 1 mO -1 mH1= 6 b) Xac inh nguyn tR, bit trong h p cht vi hidro co%H = 2,74% vkh i lng.Vi t cng thc phn toxit cao nh t cua R va hp cht cua R vi hidro.c) Hoa n thanh cac phng trinh pha n ng sau y : NaRO + SO2 + H2O HRO + I2 + H2O FeR3 + SO2 + H2O KRO3 + HIR languyn t trn (cu b).2)(1 im)a) 0,5 b) 0,5 ima) Nguyn t A : n = 3, l = 1, m = -1, s = - 3p4 A l SNguyn t X : n = 2, l = 1, m = -1, s = -1/2 2p4 X l ONguyn t Z :n = 2, l =1, m = 0, s = +1/2 2p2 Z l Cb) Phn t,ion Trng thi lai ha ca nguyn t trung tmCu trc hnh hcCS2sp ng thngSO2 sp2Gc 23SO sp3Chp y tam gic u

24SO sp3T din u3) (0,5 im) Trc pha n ng trang thai lai ho a cua Al la: sp2 Sau pha n ng trang thai lai ho a cua Al la: sp3 Cu tao hi nh ho c : * AlCl3: tam giac ph ng * 4AlCl: t din u 4) (1,5 im) a) 0,25 imb) 0,25 imc) 1 ima) Ta co :1 mO -1 mH1= 6 1 mO+1 mH1= 8

1 mO= 71 mH1 = 1b) Hp cht vi hidro cua R co cng thc : RH 97, 2635, 51 2, 74RHm Rm . R languyn tclo. Cng thc phn toxit cao nht cua R laCl2O7 vah p cht cua R vi hidro laHCl.c) NaClO + SO2 + H2O NaHSO4 + HCl5HClO + I2 + H2O 2HIO3 + 5HCl 2FeCl3 + SO2 + 2H2O 2FeCl2 + H2SO4 + 2HCl KClO3 + 6HI 3I2 + KCl + 3H2OCu 1(4,5 )1/. Th tch trng trong km kim loi chim khong bao nhiu phn trm? Tnh khi lng ring ca nguyn tkm bit KLNT ca Zn l 65,38 vc ; bn knh ca nguyn t Zn l 1,37.10-8 cm.So snh KLR ca nguyn t Zn vi KLR ca km kim loi ( d = 7,113 g/cm3 ).2/.Vit CTCTca cccht sau: Pb3O4; Fe3O4; Al2(SO4)3; Na2S2O3; NH4NO3; K2S2O8 ; H4P2O7 ; Al2C6

3/. Cn bng cc phng trnh phn ng sau y:a) CuFeS2+ Fe2(SO4)3 + H2O + O2 CuSO4 +FeSO4 +H2SO4b) Zn+NO3- +OH - ZnO22-+NH3 +H2Oc) Cr2S3 + Mn(NO3)2 + K2CO3K2CrO4+ K2SO4+K2MnO4+ NO + CO2Cu1 (4,5)1/. Th tch trng trong km kim loi chim khong 25%0,25) Khi lng ring ca nguyn tkm dZn = 10g/cm3(0,5)KLR ca nguyn t Znln hn KLR ca km kim loi v khong cch gia cc nguyn t Zn trong Zn kim loi rt gn nhau.(0,25)2/. CTCT ca 8 cht * 0,25(2,0)3/. Cn bng cc phng trnh phn ng 0,5 *3 (1,5)a) CuFeS2+x Fe2(SO4)3 + H2O + yO2 CuSO4 +FeSO4 +H2SO4Theo nh lut bo ton electron : x+ 2 y = 8. Do c v s nghim.b) 4 Zn+NO3- +7 OH - 4 ZnO22-+NH3 + 2 H2Oc) Cr2S3+15Mn(NO3)2 +20 K2CO3 2K2CrO4 +3K2SO4+15K2MnO4 +30NO +20 CO2Cu2(3,5 )Hp cht M to bi X+ v Y3- c 2 ion u c 5 nguyn t ca 2 nguyn t to nn.A l mt nguyn t trong X+ c ha tr m l a. B l mt nguyn t trong Y3- Trong cc hp cht A,B u c th c ha tr dng cao nht l a+2.Phn t lng ca hp cht M l 149 vc trong MY3- > 5MX+Tm cng thc phn t ca hp cht M.Cu1 (3,5)* A,B thuc nhm a+2* A c ha tr m a A c ha tr vi H l a (0,25)a = 8-(a+2) a = 3 . A c ha tr m l -3 (0,25) Aphi kt hp vi H. Vy X+ l AH4+ (0,5)* B c ha tr dng l +5. (0,25) B phi kt hp vi O. Vy Y3- l BO43- (0,5)CTPT(AH4)3BO4 M = 3A + 12 + B + 64 = 149 3A + B = 73(1)(0,5)* MY3- > 5 B +64 > 5A + 20 (2) (0,5)MX+(1) ,(2) A3NFCu trc phn t NH3 v NF3 u c cu trc thp, nguyn t N u trng thi lai ho sp3 v cn mt cp electron cha chia. Momen lng cc ca cc lin kt N Hv F H xp x nhau. Tuy nhin v momen lng cc ca 3 lin kt N H trong phn t NH3 nh hng cng chiu vi momen lng cc gy ra bi cp electron cha chia.Trong NF3 momen lng cc tng hp ca 3 lin kt N F nh hng theo chiu ngc li / C6H5Cl v C6H4Cl2 (1,5 )Phn t C6H4Cl2 tn ti di 3 dng (o, m, p)

C6H5Cl m C6H4Cl2 p C6H4Cl2< C6H5Cl < 0 C6H4Cl2b. p C6H4FCl=C6H5F-C6H5Cl=1,7 1,58 = 0,12 DCu IV (4im) :1- C cu hnh electron :1s22s22p63s23p63d54s1.(1)a) Dng k hiu lng t biu din cu hnh electron (1)b) (1) l cu hnh electron ca nguyn t hay ion ? Ti sao ?c) Cho bit tnh cht ha hc c trng ca ion hay nguyn t ng vi cu hnhelectron (1) .Cho v d .2- Kim loi A phn ngvi phi kim B to hp cht C c mu vng cam. Cho 0,1mol hp cht C phn ngvi CO2 (d) to thnh hp cht D v 2,4 gam B. Ha tan hon ton D vo nc,dung dch D phn nght 100ml dung dch HCl 1M gii phng 1,12 lit kh CO2 (kc) . 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. Cu IV (4im) 1-(2im) a- Vit lng t (0,5)b- Cu hnh electron nguyn t Crom (0,5)c- Tnh cht ho hc c trng l tnh kh v tnh lng tnh .Vd (1)2-(2im) Dung dch D phn ng vi HCl gii phng kh CO2. Suy ra hp cht D l mui cacbonat ca kim loi .D khng b phn tch khi nng chy , vy D l cacbonat ca kim loi kim . C + CO2 D + B .Vy C l peoxit hay superoxit, B l oxi (0,5) t cng thc ca C l AxOy. T d kin tnh c MA= 39 Vy A l K , B l O2, C l KO2, D l K2CO3(0,75)Cc phn ng: K + O2KO24 KO2 + 2 CO2 2 K2CO3 + 3 O2Cu I (5 im): 1)Cho b bn s lng t ca electron cht cng trn nguyn t ca cc nguyn t A, X, Z nh sau:A : n = 3 , l = 1 , m = -1 , s = -1/2X : n = 2 , l = 1 , m = -1 , s = -1/2Z : n = 2 , l = 1 , m = 0 ,s = +1/2a) Xc nh A, X, Z. b) Cho bit trng thi lai ha v cu trc hnh hc ca cc phn t v ion sau: ZA2 , AX2 , AX32-, AX42-.1) 3,5 ima) 1,5 . Nguyn t A: n = 3, l = 1, m = -1, s = -1/2 3p4. A l S. Nguyn t X: n = 2, l = 1, m = -1, s = -1/2 3p4. X l O.Nguyn t Z: n = 2, l = 1, m = 0, s = +1/2 2p2. Z l C. b) 2 Phn t, ionTrng thi lai ha ca nguyn t trung tmCu trc hnh hcCS2 sp ng thngSO2 sp2GcSO32-sp3Chp y tam gic uSO42-sp3T din uBi I/2: (2,0 im)a) Trong bng tun hon c mt ghi 546 , 63s 4 d 329X1 10a1) Hy cho bit ngha ca ch v cc s trong .a2) Xc nh v tr ca X trong bng tun hon.a3) Hon thnh phng trnh phn ng theo s sau:O X2XOX XCl2Cho 0,2 mol XO ( cu trn) tan trong dung dch H2SO4 20% un nng, sau lm ngui dung dch n 100C. Tnh khi lng tinh th XSO4.5H2O tch ra khi dung dch, bit rng tan ca XSO4 100C l 17,4gam/100gam nc X: k hiu nguyn t. 3d10 4s1: c 10 e obitan 3d v 1 e obitan 4s. 29: s th t nguyn t. 63,546: nguyn t khi trung bnh ca X.X l Cu : Chu k 4. Nhm IB. 29.Cc phn ng:( )( )( )( ) O H 4 NO 2 CuCl 3 HNO 2 HCl 6 Cu 3 4)O H Cu H CuO 3O21O Cu CuO 2 2CuO 2 O21O Cu 12 2 322t22 2tt2 2000+ + + ++ ++ +3NH CO, Al, bang H thay the (Co

Cu 3: 5 im1) Mt cht thi phng x c chu k bn hy l 200 nm, c cha trong thng kn v chn di t. Hi trong thi gian bao lu tc phn r gim t 6,5.1012 nguyn t/pht xung cn 3,00.10-3 nguyn t/pht.2)Phn ng x phng ha este etyl axetat bng dung dch NaOH 10oC c hng s tc bng 2,38 mol-1lit ph-1. Tnh thi gian cn x phng ha 50% etyl axetat 100C khi trn 1 lt dung dch etyl axetat 0,05M vi: a) 1 lt dung dch NaOH 0,05Mb) 1 lt dung dch NaOH 0,10 M 3)a) Tnh h s nhit ca tc phn ng bit 393oK phn ng kt thc sau 18 pht, 453oK phn ng kt thc sau 1,5s. b)H s nhit ca 1 phn ng bng 2. Cho bit 0oC, phn ng kt thc sau 1024 ngy, vy 300oC, phn ng kt thc sau bao nhiu lu?( ) 2( ) 1( ) 3 ( ) +3HNO HCl hhCu 3 : (5 im)1) 1 imHng s phng x :K=1/ 20, 693t =0, 693200 = 347.10-5/nmMt khc :K = 2, 303tlgoNN347.10-5/nm = 2, 303tlg1236.50.103, 00.10

t = 1,02.104nm2) 2 im :Phn ng x phng ha etyl axetat l phn ng bc hai:CH3COOC2H5+ OH-

otCH3COO-+C2H5OHa) Nng u 2 tc cht bng nhau :T biu thc : 1[ ] A =kt+ 1[ ]oA ta rt ra:t = 1k [ ] ([ ] )o oxA A x Vi k = 2,38, [A]o = 0,025M, x = 0,0125M, ta c :t =0, 01252, 38.0, 025(0, 025 0, 0125) = 16,8 ph = 16 ph 48s b) Nng 2 tc cht khng bng nhau :t=[ ] [ ] 1 1. ln .([ ] [ ] ) [ ] [ ]o oo o o oB A xk A B A B x Vi [A]o = 0,025 M , [B]o = 0,05 M, x = 0,0125 M ta c :t=1 0, 05 0, 025 0, 0125ln .2, 38(0, 025 0, 05) 0, 025 0, 05 0, 0125 = 6 ph 49s3) 2 im a)Theo Vant Hoff :2 12 101t tvvTa c : 2 1453 39372010 10t t Nn6720 2, 99 3 b)Khi nhit tng t 0oC n 300oC, theo Vant Hoff, s ln tc tng l230.V thi gian hon tt phn ng t l nghch vi tc phn ng nn ta c : 03030011024 2v C CC v 0 00Thi gian 300 Thi gian 300Thi gian 0Thi gian 3000C = 3010242 ngy 301024 24 60 600, 082s s CU 11 . Hai nguyn t A v B cng thuc chu k 3 . Hy vit cu hnh electron ca chng , bit rng nng lng Ion ha (I) ca chng ln lt c cc gi tr sau (tnh theo kJ/mol)(A) :I1 I2 I3I4I5 I6 577 1816 274411576 1482918375(B) :I1 I2 I3I4I5 I6 1012190329104956627822230CU 11 .(A) : Sau I3 c bc nhy t ngt v nng lng ion ha , vy nguyn t A c 3 electron ha tr . n thuc chu k 3 , nhm IIIA ( l Al) .c.h.e : 1s22s22p63s23p1 .Tng t (B) : l P c c.h.e : 1s22s22p63s23p3 .