hsg cau tao chat

7/29/2019 HSG Cau Tao Chat

1/36

PHN CU TAO CHTTrng THPT chuyn Thi Bnh

A- MT S VN CHUNG V LI THUYT

I. cc c trng v cu to phn t:

Mt phn t hnh thnh c v tn ti bn nh kt qu ca tng tc gia cc ht nhn v electron dnn mt nng lng h cc tiu (nng lng ny ca phn t phi thp hn nng lng ca h ban u).

Trong phn t c s phn b v tr tng i gia cc ht nhn nguyn t nn c c hnh dng khnggian ca phn t vi di lin kt v gc xc nh.1) Nng lng lin ktNng lng lin kt gia hai nguyn t A v B l nng lng cn thit va ph v lin kt hay

nng lng to ra khi hai nguyn t A v B trng thi c bn kt hp vi nhau. Tuy nhin nng lnglin kt l su ca cc tiu nng lng trn c cong th nng

Th d: phn ng H2 2 H cn nng lng bng 436 kJ.mol-1. Phn t H2 bn vng nn khi cho hainguyn t H kt hp vi nhau: 2 H H2 to ra mt nng lng bng 436 kJ.mol-1. Nh vy hai gitr nng lng bng nhau v gi tr v ngc nhau v du. Quy c rng nng lng lin kt c dudng bin lun rng lin kt cng bn th nng lng lin kt cng ln??? nn EH-H= 436 kJ.mol-1.Trong phn t c nhiu lin kt th nng lng lin kt c tnh trung bnh.

2) di lin kt di ca mt lin kt trong phn t l khong cch trung bnh gia hai ht nhn nguyn t to ralin kt khi phn t trng thi nng lng thp nht. di lin kt thng c k hiu l d.

Phng php ph vi sng hay phng php nhiu x electron thng c dng xc nh dilin kt. Tr s di lin kt trong khong t 0,74 (phn t H 2) n 4,47 (phn t CS2); thngthng trong khong 1,0 2,0 i vi lin kt gia hai nguyn t ca cc nguyn t chu k 2, 3, 4.

di ca mt lin kt no thng gn ng l mt hng s trong cc phn t khc nhau. Chnghn lin kt n C-C trong hu ht cc phn t hirocacbon khng lin hp vo khong 1,53-1,54.Trong C6H6 (benzen) di lin kt gia hai nguyn t C cnh nhau bng 1,40 . Tr s ny nm trongkhong di mt lin kt C-C l 1,54 v di mt lin kt i C=C l 1,34. di lin kt cngnh, lin kt cng bn.

Bn knh lin kt: T cc s liu c th thy rng di lin kt d AB xp x bng 1/2(dAA + dBB) vidAA, dBB l di lin kt A-A, B-B tng ng. Chng hn, coi A l Cl, B l Cl; bit dCl-Cl = 1,99A,vy dC-Cl = 1/2(dC-C + dCl-Cl) = 1/2(1,54 + 1,99) = 1,765. Tr s thc nghim cho bit d C-Cl = 1,766. Do ngi ta coi 1/2dAA l bn knh lin kt hay bn knh cng ho tr rA ca nguyn t .3) Gc lin kt :

Gc lin kt l gc to bi hai na ng thng xut pht t mt ht nhn nguyn t i qua hai htnhn ca hai nguyn t lin kt vi nguyn t .

Cc trng hp in hnh v gc lin kt theo nh ngha trn l:- Phn t thng, gc lin kt bng 180o (2); chng hn C2H2, CO2,- Phn t c gc, gc lin kt khc 180o, chng hn BF3 hay C2H4 c gc 120o, H2O c gc 104,5o,

- Phn t t din, gc lin kt bng 109

o

28, chng hn CH4,Trong mt s trng hp, ngi ta ch n gc c to ra t 4 nguyn t hay 2 mt phng, l gcnh din hay gc xon (hay gc vn). Di y l hnh nh mt s phn t cho thy chng c kch thcring.

1


2/36

4) Cc dng lin kt ho hcXt mt cch i cng, lin kt ho hc c bn dng:- Lin kt cng ho tr (hay lin kt nguyn t)- Lin kt ion (hay lin kt in ho tr)- Lin kt kim loi- Lin kt hiro, tng tc Van de Van; gi chung l tng tc yu.

Thc t khng c ranh gii r rt gia cc dng lin kt . Tuy nhin, thun li khi xem xt, ngi

ta vn cp ring tng dng , hai dng u thng c cp n nhiu hn.II. Quy tc bt t (Octet):

T s phn tch kt qu thc nghim v cu to ho hc ca cc phn t, nm 1916 nh ho hcCxen (Kossel) v Liuyx (Lewis) a ra nhn xt m ngy nay gi l quy tc bt t (hay quy tcoctet): Khi to lin kt ho hc, cc nguyn t c xu hng t ti cu hnh lp ngoi cng bn vngca nguyn t kh tr vi 8e.

Cn lu l quy tc ch p dng c cho mt s gii hn cc nguyn t, ch yu l cc nguynt chu k 2. Quy tc bt t (octet) th hin trong tng dng lin kt c th. Thng thng trong linkt ion, sau khi cho nhn electron lp v ngoi cng c s electron nh cc nguyn t kh him.Thc t quy lut y ch ng cho a s cc trng hp nguyn t nhm A. (Hc vin ly th d v

nhng trng hp khng tun theo quy tc bt t).III. Thuyt liuytx (Lewis) (nm 1916):

1. Ni dung ca thuyt:Trong phn t c to ra t nguyn t cc nguyn t phi kim, lin kt ho hc gia hai nguyn t

c thc hin bi cp (i) e dng chung, nh m mi nguyn t u c c cu hnh lp ngoicng bn vng ca nguyn t kh tr vi 8e.

Electron ca mi nguyn t c th tham gia c lin kt l e ho tr. i e to lin kt phi c spini song.

V d: Phn t Cl2 c lin kt gia hai nguyn t Cl c thc hin nh cp e gp chung.cp e ny l cp e lin kt, c k hiu hay , cc e cn li c gi l e khng lin kt.a) Phn loi lin kt cng ho trCn c vo v tr cp e dng chung so vi ht nhn nguyn t tham gia lin kt, ngi ta chia lin

kt cng ho tr thnh hai loi:- Lin kt cng ho tr khng phn cc (hay khng c cc): i e dng chung gia khong cch

hai ht nhn nguyn t. l lin kt trong cc phn t n cht nh Cl2, Br2... (trng hp hiu m in 0,4).

- Lin kt cng ho tr c cc (hay phn cc): i e dng chung lch v pha nguyn t ca nguynt c tnh phi kim mnh hn (hay c m in ln hn). l lin kt ho hc trong cc phn thp cht nh H2O, NH3, CH4,... (hiu m in trong khong 0,40 1,70).

b) Tnh nh hng khng gian ca lin kt cng ho tr

Lin kt cng ho tr c tnh nh hng khng gian. Trong lin kt cng ho tr, cp e dng chung(hay cp e lin kt) c phn b khong khng gian gia hai ht nhn tham gia lin kt.Lin kt cng ho tr c tnh cht bo ho. Chng hn trong hp cht gia Cl vi H, ch c 1 nguyn

t H lin kt vi 1 nguyn t Cl to thnh HCl; khng th c nhiu hn mt nguyn t H lin kt vimt nguyn t Cl. Do vy s nguyn t lin kt vi mt nguyn t cho trc b hn ch bi ho trca nguyn t .

Bi tp: Cho cc nguyn t H, F, Cl, Br, I.1) Hy vit CTPT ca cc cht c to ra t cc nguyn t cho.2) Trong s cc cht nu cht no c lin kt khng c cc, c cc? Hy ch r v tr ca cp

electron lin kt trong mi cht.2. Cng thc cu to Liuytx (Lewis):

Biu din lin kt v cu to phn t kh trc quan1. Cng thc:Mi du chm biu th mt electron. Hai chm hay mt vch ch mt cp electron trong nguyn t

hay phn t. Cc electron ny l cc electron ho tr. Cng thc ho hc ch r th t lin kt gia cc2


3/36

nguyn t v cc k hiu ch s phn b electron ho tr c gi l cng thc Lewis (do Lewis xng). Thng thng cc cp electron lin kt vit bng du vch, electron khng lin kt biu din

bng chm. Cng thc Lewis khng ch dng cho cc hp cht c lin kt cng ho tr m dng ccho cc hp cht c lin kt ion.

Bi tp:Vit cng thc Lewis cho cc phn t : a) nit, b) nc, c) Canxi clorua.2. Cch vit cng thc Lewis:

a) Cc khi nim cn dng:+ Nguyn t trung tm v phi t: Trong mt cng thc ho hc, c nguyn t trung tm l nguyn

t cn nhiu e nht to c cu hnh tm electron (octet) lp ngoi cng ca n (hay nguyn tc s oxi ho cao nht); cc nguyn t khc v c cp electron khng lin kt ca nguyn t trung tmc gi l phi t. V d: trong phn t NH 3, nguyn t trung tm l N, phi t gm 3H v 1 cp ekhng lin kt ca N ( v ho tr). Trong phn t HCN, nguyn t trung tm l C, phi t gm 1H v1N ( y khng c cp e khng lin kt v ho tr).

+ Li ca nguyn t: Phn li ca mt nguyn t (khi nguyn t ny l thnh phn ca mt cngthc ho hc c xt) gm ht nhn v cc electron cc lp bn trong. V d: Xt li nguyn t cacc nguyn t trong HCN ta c: li nguyn t N gm ht nhn v hai e phn lp 1s 2; li nguyn t Cgm ht nhn v 2 electron phn lp 1s2; li nguyn t H ch gm ht nhn, thc t H thng ccoi l trng hp ngoi l.

+ in tch:- in tch li nguyn t: l s n v in tch ca nguyn t khi ta b cc electron lp ho tr i

nn l mt s nguyn dng, c tr s bng s e ho tr vn c ca nguyn t .- in tch hnh thc ca mt nguyn t = (in tch ca li nguyn t - tng s e ring ca nguyn

t tng s e to lin kt c nguyn t tham gia/2).V d: Xc nh in tch hnh thc ca N trong NH3, NH4+

- Trong NH3: T cu to Lewis, ta thy:in tch li ca N l 5S e khng lin kt ca N l 2Tng s e to lin kt c N tham gia l 6 (hay c 3 lin kt)

Vy in tch hnh thc ca N = 5 2 6/2 = 0

- Trong NH4+

: Xt tng t nh trn, ch N khng cn e khng lin kt v N tham gia 4 lin ktvi 4 H.Vy in tch hnh thc ca N = 5 0 8/2 = +1y chnh l in tch ca c nhm NH4+.

b) Cc bc vit cu to Lewis: HCNBc 1: Vit cng thc cu to s b ca cht da vo ho tr ca cc nguyn t v gi thit rng

ch c lin kt n c hnh thnh. Nu cha bit th t lin kt gia cc nguyn t, hy dng githit vit th t .

y ta c: H : C : N (a) hay H : N : C (b)Bc 2: gi n1 l tng s e ho tr ca cc nguyn t.- Thng thng da vo cu hnh e ca cc nguyn t

H: 1s2

1eC: 1s2 2s2 2p2 4eN: 1s2 2s2 2p3 5e

Vy n1 = (1 + 4 + 5) e = 10 eCh : Nu cng thc l:

+ Ion m: 1 n v in tch m do c cng thm 1e vo tng trn.+ Ion dng: 1 n v in tch dng do tr i 1e t tng trn.

HCN l phn t trung ho nn khng p dng phn ny.Bc 3: Tm cng thc Lewis (gn ng)- gi n2 l tng s e to lin kt trong cng thc a ra bc 1. S e cn li khng tham gia lin

kt n3 = n1 n2- S e cn ly to bt t cho nguyn t m in nht trong cng thc ban u bng n 4.Khi p dng ba bc trn cho HCN. n2 = 4e, vy n3 = n1 n2 = 6e.

3


4/36

Trong (a), N m in hn C nn phi to bt t cho N. Trong cng thc ban u N mi c 2e, ncn 6e na mi thnh 8 e. Nh vy n4 = 6e.Bc 4: Tm cng thc Lewis ng- Tm s e cn li, k hiu n5 = n3 n4

+ Nu n5 = 0: tnh in tch hnh thc mi nguyn t trong cng thc va vit bc 3.+ Nu n5 0: chnh l s e cn dng to bt t cho nguyn t trung tm.

Ch : Vic ny ch c thc hin khi nguyn t trung tm l nguyn t ca nguyn t thuc chuk 3 tr i.

Sau tnh li in tch hnh thc cho mi nguyn t trong cng thc va vit.

p dng: Vi HCN c n3 = n4 = 6e nn n5 = 0.

Do tnh in tch hnh thc cho cc nguyn t trong phn t H : C : :..

..N (c)

H: 1 - 1 = 0C: 4 2 = 2

N: 5 6 1 = -2- Sau khi thc hin nh trn, nu nguyn t trung tm l nguyn t ca nguyn t chu k 2 cha t

c bt t, ta phi chuyn mt hay mt s cp e khng lin kt ( nguyn t m in hn) thnh cpe lin kt, sao cho c c bt t i vi nguyn t trung tm .

T (c) ta thy nguyn t trung tm C cn thiu 4e mi c c bt t. Vy ta phi chuyn 4e (2 cp

e) khng lin kt ca N thnh 4e (2 cp) lin kt:H : C : :

..

..N H :

..

..C N :

(f)Tnh li in tch hnh thc ca cc nguyn t trong (f):

H: 1 1 = 0C: 4 4 = 0

N: 5 2 3 = 0Kt lun: (f) l cng thc Lewis cn tm cho HCN. (Hc vin t xc nh loi b cng thc (b).Bi tp p dng: Tm cng thc Lewis ca PCl3.

c) Xc nh cng thc Lewis ca CO32-.

+ Cng thc gi nh l: (a)+ Khi tnh s e ho tr, cn ch y l mt anion c in tch 2:n1 = (6 x 3 + 4 x 1 + 2) e = 24 eT (a) c n2 = 6e n3 = n1 n2 = (24 6) e = 18e+ S e cn to bt t cho 3 O l n 4 = 6e. 3 = 18eVy n5 = n3 n4 = 0.+ Tnh in tch hnh thc cho tng nguyn t trong : (b)

C: 4 3 = +1O: 6 6 1 = -1;

Nh vy nguyn t cacbon cha t bt t.

+ T (b) c 3 kh nng chuyn 2 electron t mt trong ba nguyn t O cho nguyn t C C c 8electron v thu c 3 cng thc cu to tng ng nhau vi 1 lin kt i C = O v mt in tchhnh thc cho mi nguyn t oxi:

(I) (II)(III)

Tnh in tch hnh thc trn mi nguyn t.C: 4 4 = 0

O: 6 6 1 = -1 (c 2 O)O: 6 4 2 = 0Vy 3 cng thc Lewis u ng cho CO32-.

4

O : C : O..O

: O : : C : O :..

O..: :

.. ..

..

O : C : O..O

: :

: :

..

..

..

..

..

: O : C : :O..O..: :

.. ......

: O : C :O :

O:

.... ....

:


5/36

+ Thc nghim cho bit ion CO32- c cu to phng, 3 nguyn t O 3 nh ca tam gic u, gcOCO 120o, 3 lin kt C-O c di u bng nhau, l 131 pm.

gii thch kt qu ny, ngi ta gi thit rng c s cng hng gia 3 cu to (I), (II), (III)vi nhau. Ba cng thc Lewis tng ng nhau; chng c gi l cc cng thc cng hng caCO32-.

Gi thuyt v s cng hng cc cu to c th p dng cho cc trng hp ion v phn t khc nhNO3-, SO42-, PO43-, C6H6,...

+ p dng gi thuyt ny ta tnh c in tch trn mi nguyn t oxi trong CO32- theo cng thc:in tch hnh thc trn mi nguyn t xc nh = in tch ton nhm/ S cu to cng hng.

C th: in tch hnh thc trn mi nguyn t oxi = - 2/3Kt qu ny cho thy: in tch trong ion CO32- khng c nh trn mt nguyn t oxi no, in tch

c phn b cho cc nguyn t oxi trong nhm. Ngi ta ni: c s gii to in tch.Thc cht s gii to in tch ny l gii to electron. Bng cch tm cu to Lewis cho cng thc

ho hc vi quan nim gii to electron gp phn khng nh vai tr ng ch ca cng thc cu toLewis trong nghin cu v ging dy ho hc.

+ Bc ca mt lin kt c xc nh bng tng s cc cp electron to ra lin kt .T ta c: bc bng 1 i vi mt lin kt n C - C;

bc bng 2 i vi mt lin kt i C = C;bc bng 3 i vi mt lin kt ba C C;

+ Khi c cu to cng hng th:Bc ca mt lin kt bng tng cc bc lin kt thuc v hai nguyn t ang xt trong cc cu tocng hng chia cho s cu to cng hng.

p dng cho CO32-: Chn bt c mt lin kt no trong ba lin kt gia C vi O, ta u thy tng bcca lin kt l: 2 + 1 + 1 = 4.

Ta c 3 cu to cng hng ((I), (II), (III)).Vy bc ca lin kt gia C vi O trong CO32- bng:

4/3 = 13

1(*)

Kt qu ny cho thy lin kt gia C vi O trong CO32- c bc trung gian gia lin kt i C = O(bng 2) vi lin kt n C - O (bng 1). Tr s v di lin kt ph hp vi kt qu :

Lin kt C = O trong H2C = O bng 121 pm.Lin kt C - O trong H3C OH bng 143 pm. Nh trn bit, thc nghim xc nh c di

lin kt gia C vi O trong CO32- bng 131 pm.Bi tp: Hy vit cu to Lewis cho NO3- (ch trnh by r cc bc, cc cu to cng hng, xc

nh bc lin kt gia N vi C).

5


6/36

IV. Thuyt sc y gia cc cp electron:

1. M hnh s y gia cc cp electron v ho tr:Mi lin kt cng ho tr gia hai nguyn t c to thnh nh cp electron lin kt hay cp

electron dng chung. i e lin kt phn b trong khong khng gian gia hai ht nhn nguyn t tora lin kt . Lin kt cng ho tr l lin kt c nh hng khng gian lm cho phn t c hnhdng nht nh c trng cho phn t v cho cht cho.Nhiu nguyn t sau khi gp chung e to lin kt cn c cc e khng lin kt. Chng hn trong N

NH3 ngoi 3 cp electron lin kt vi 3 nguyn t H, cn c 1 cp e khng lin kt. Cc cp

electron d lin kt hay khng lin kt ny s y nhau do cng tch in m.Trong phn t AXn, A l nguyn t trung tm, X l phi t; n l s phi t X c trong AXn. Nu Acn c m cp e khng lin kt, mi cp c k hiu l E, ta c k hiu AX nEm. M hnh VSEPR xts phn b khng gian gia A vi X, vi E. Coi nguyn t trung tm A c dng cu. Tm ca hnhcu l ht nhn nguyn t A v cc electron phi ho tr bn trong (li), v qu cu l cc e lp ngoicng (e ho tr). Mi cp e ho tr chim mt khong khng gian no ca qu cu.Nh vy, mt mc nht nh, hnh dng ca phn t ph thuc vo khong khng gian chim

bi cc e ho tr ca nguyn t trung tm A. Hnh dng phn t ph thuc ch yu vo s phn b cccp e hay cc m my e ho tr ca nguyn t A.

2. Ni dung ca thuyt sc y gia cc cp e ho tr (VSEPR)Vo nhng nm 1940, N. Sidgwick, H. Powell a ra thuyt sc y gia cc cp electron ho tr v

sau c cc nh bc hc khc, trong c R. Gillespie, b sung v hon chnh.+ Cu hnh cc lin kt ca nguyn t (hay ion) ph thuc vo tng s cp electron ho tr lin kt

hay khng lin kt ca nguyn t.+ Cc obitan c cc cp e ho tr c phn b u nhau v cch nhau xa nht c lc y nh

nht gia chng.C s khng tng ng gia cp e lin kt v cp e khng lin kt. i e lin kt chu lc ht

ng thi ca hai ht nhn nguyn t A v X to ra lin kt nn chuyn ng ch yu vngkhng gian gia hai ht nhn. Trong khi , cp e khng lin kt ch chu lc ht ca ht nhn A nnc th chuyn ng ra xa hn. Kt qu l cp e khng lin kt chim khong khng gian rng hn sovi khong khng gian chim bi cp e lin kt.

+ Obitan c cp electron khng lin kt chim khng gian ln hn so vi obitan cha cp electronlin kt v th sc y gia cc cp electron lin kt gim hn so vi cp khng lin kt. Th d gclin kt trong cc phn t CH4, NH3 v H2O tng ng bng 109o28, 107o v 104,5o do cc phn t cs cp electron khng lin kt bng 0, 1 v 2.

+ Khng gian ca cp electron lin kt s gim nu m in ca cc nguyn t lin kt ln. gclin kt trong NF3 ch l 102o so vi 107o ca NH3. Tng t gc lin kt tgim trong dy: PI3 (102o),PBr3 (101,5o), PCl3 (100,3o) v PF3 (97,8o).

a) M hnh s y gia cc cp electron lin kt: AXn vi n = 2 6n=2: hai cp e c phn b trn ng thng. Phn t thng nh BeH2. Gc lin kt XAX bng

180o.n=3: ba cp e c phn b trn ba nh ca tam gic u. Phn t c hnh tam gic u, phng.

Gc XAX bng 120o

. V d: BF3, AlCl3,...n=4: bn cp e c phn b bn nh t din u, tm l A. Phn t c hnh t din u. GcXAX bng 109o28. V d: CH4, NH4+,...

n=5: c s phn b ng u 5 cp e trn mt cu. Khi k cc ng ni cc nguyn t X vi nhauta c hnh lng thp tam gic. V d minh ho: phn t PCl 5 c phn b lng chp tam gic nntrong phn t c ba loi gc lin kt: trong mt phng c gc ClPCl bng 120 o, trong khi cn c gcClPCl bng 90o v gc 180o. Theo m hnh VSEPR, nm cp e c phn b trn mt cu v ho trnh sau: Ba cp cng vi ht nhn A trong 1 mt phng, ba cp ny ba nh tam gic u tm A.Cc lin kt c to ra trong mt phng ny c gi l lin kt ngang. Hai cp e cn li c phn

b trn hai u ca on thng vung gc vi mt phng tam gic u ti A. Hai cp e ny to ra hailin kt trc. Do s phn b nn di lin kt ngang ngn (2,02 nm) hn di lin kt trc (2,14

m). Kt qu s sp xp trn a ti mt lng thp tam gic y chung l tam gic u tm A, hainh l hai cp e to ra hai lin kt trc.

n = 6: AX6. vi s phn b 6 cp e trn mt cu v ho tr. Khi k cc ng ni cc nguyn t Xvi nhau ta c hnh bt din.V d minh ho: phn t SF6. Su cp e ca phn t ny c phn b

6


7/36

trn v ho tr ca nguyn t A su nh ca mt hnh bt din u. Trong trng hp ny khng cs phn bit gia lin kt ngang vi lin kt trc v di. V d minh ho ca trng hp ny l

phn t SF6. Trong phn t ny, gc gia hai trc lin kt cnh nhau bng 90o.b) M hnh s y gia cc cp e lin kt v khng lin kt: AXnEm n + m 6

Nguyn t trung tm A va c n cp e lin kt, va c m cp e khng lin kt (k hiu l E). Dokhng c s tng ng gia cp e lin kt vi cp e khng lin kt, hai trng hp AX n vi AXnEmc cng s cp e nhng hnh dng hai phn t khng hon ton ging nhau. Xt mt s trng hpthng gp:

1) (n + m) 4

* AX2E:Ba nguyn t A, X, X khng cn nm trn cng ng thng nh trong trng hp AX 2. Lc ny

phn t c dng gp khc. V d: SnCl2, gc ClSnCl 120o. C th dng phn t SO2 lm th d vis y ca hai cp electron lin kt S = O vi 1 cp electron khng lin kt ca nguyn t S.

* AX3E:Nguyn t trung tm A c tng cng 4 cp e nhng phn t khng c dng t din u nh AX 4 m

c hnh tam gic. V d in hnh l NH3, gc HNH 107o.* AX2E2:

Nguyn t trung tm A cng c 4 cp e nhng do c 2 cp e khng lin kt nn hnh dng phn tkhc vi AX4 hay AX3E. Phn t ny c gc gn vi gc t din. V d: H2O c gc HOH 104o.

Bi tp: Dng m hnh VSEPR hy d on hnh dng phn t AXE3.2) (n + m) = 5 (nguyn t trung tm A c 5 cp e).Nu AX5 th phn t c hnh lng thp tam gic. Song s c mt ca cp e khng lin kt s lm

bin dng hnh .*AX4E:

S phn b E v 4 cp e lin kt c m t nh hnh trn. Kt qu phn t c hnh ci bp bnh. Vd: phn t SF4.

*AX3E2:Ba cp e lin kt to 3 lin kt A-X; s y tng h gia 3 cp e ny vi nhau v vi 2 cp e khng

lin kt (E), kt qu to ra phn t hnh ch T. V d: ClF 3, HClO2.3) (n + m) = 6(nguyn t trung tm A c 6 cp e).

* AX5E:Bn trong nm cp e lin kt c phn b trong mt phng; 1 cp e lin kt cn li c phn b

trong trc gn vung gc vi mt phng trn. Do 5 cp e lin kt (hay phn AX 5) to ra hnh thpvung. Trong cu to ny, lin kt ngang v lin kt trc khng tng ng hnh hc. Chng hntrong BF5, lin kt trc c di 169 pm cn lin kt ngang c di ti 177 pm.

* AX4E2:Theo m hnh VSEPR, 2 cp e khng lin kt (E) c phn b trans so vi 4 cp e lin kt; m 4

cp e lin kt ny c phn b trong mt phng to ra hnh vung phng. V d: phn t XeF4.Vic p dng m hnh VSEPR gii thch c hnh dng ca cc hp cht c lin kt bi (C 2H2,

C2H4,...); m in ca cc nguyn t l phi t cng c nh hng n hnh dng phn t. Hnhdng phn t cng gn lin vi tnh cht ho hc ca phn t.Nhc im: kh khn khng th vt l cc halogenua kim loi kim th dng MX2 ch c cu

hnh thng i vi hp cht ca Be cn i vi hp cht ca Ba u c cu to gp khc. Nguynnhn l do nh hng ca cc obitan khc trong nguyn t Bari.

V. Thuyt lin kt ho tr (VB):

1. Nhng lun im chnh ca thuyt lin kt ho tra) Lin kt ho hc c to nn bi cp electron c spin ngc chiu nhau ca hai nguyn t.

Trong c s che ph gia nhng obitan ca hai nguyn t to nn mt vng c mt intch cao dn n vic gim th nng ca h, ngha l lin kt c hnh thnh.

b) Lin kt c hnh thnh theo phng trong khng gian m kh nng che ph gia cc obitan

nguyn t l cc i (thng gi gn l nguyn l che ph cc i).c) Hai obitan nguyn t che ph nhau cng nhiu, lin kt c to nn cng bn.+ Theo quan nim ca thuyt VB, lin kt ho hc gia 2 nguyn t hnh thnh c l do s xen

ph 2 AO ho tr, mi AO c 1 e c thn. C th hnh dung r hn khi xt mt s giai on trong s

7


8/36

hnh thnh phn t H2 t 2 nguyn t H. Mc xen ph 2 AO hay 2 m my e ca 2 nguyn ttham gia lin kt c trng cho bn ca lin kt. Lin kt ho hc c to ra cng bn nu s xen

ph 2 AO cng mnh.+ Lin kt ho hc gia 2 nguyn t s c phn b theo phng no m c s xen ph ln

nht (cc i) ca 2 AO tham gia lin kt (2 my electron ca 2 AO tham gia lin kt). S nh hngho tr ca nguyn t tham gia lin kt th hin s c tr s xc nh ca gc lin kt (gc ho tr)trong phn t nhiu nguyn t.

xc nh c s lin kt CHT ta cn xt n phn b electron trn cc lng t trng thi cbn v kch thch. gii quyt vn hnh dng phn t cn dng n hnh dng obitan nguyn bn

ca nguyn t hay cc obitan lai ho ca n.a) Mt s p dng

Gc lin kt (gc ho tr) l mt trong s cc tnh cht phn t c thc nghim pht hin. Da vothuyt ho tr nh hng (VB), ta c th gii thch c kt qu .

- Phn t H2S+ Thc nghim cho bit, trong H2S c gc HSH 92o.+ Ta bit S c cu hnh e l [Ne] 3s 2 3p4. Nh vy S c 6e ho tr phn b trong 4 AO: 4px, 4py v

4pz. H c mt electron duy nht trn obitan 1s1. Theo thuyt VB, 2e c thn ca S to 2 lin kt S-H vi 2e ca 2 nguyn t H. 2 AO c 2 e c thn ny (chng hn l AO-3px, AO-3py) c trcvung gc vi nhau, tc l gc gia hai trc ca 2 AO l 90 o. V AO-s c i xng cu nn vngc s xen ph ca 1 AO-s vi 1 AO-3p s phn b trn trc ca AO-3p tng ng. iu ny c nghal thng thng gc ca phn t H2S, tc HSH bng 90o. Tuy nhin hai vng c mt e cao cnhnhau s c s y nhau. S y ny lm cho gc HSH m rng ra. Kt qu l gc bng 92o nh ktqu thc nghim.

+ Cng phn nhm chnh th 6 vi S cn c Se, Te u to hp cht vi H 2 c gc tng t nhtrn, c th l H2Se c gc HSeH 91o, H2Te c gc HTeH 90o.

- H2OThc nghim thu c gc HOH 104,5o

+ O cng nhm chnh th 6 vi S. Cc e ho tr vcc AO tng ng:

R rng y khng th cho rng O dng 2 AO-p

nguyn cht xen ph vi 2 AO-1s ca 2 H nh trongH2S c. Mun c s che ph cc i gia cc obitan1s ca H v 2py, 2pz ca O th cc nguyn t H phi tinli nguyn t O theo phng ca trc y v trc z. Nh vyth gc ho tr HOH phi l 90o. Gc thc nghim thu c104,5o ln hn gc l thuyt 90o rt nhiu.

+ Theo s liu trn, gc t din l 109o29 rt gn vi tr s 104,5o. Do ngi ta gi nh rng Otrong H2O trng thi lai ho sp3. Trong 4 AO-sp3 c 2 AO m mi AO c 2e (1 cp) c gi l cpe khng lin kt ca O, 2 AO-sp3 cn li th mi AO ch c 1e. Theo thuyt VB, to 2 lin kt O-Hcn c s xen ph 2 AO-1s ca H vi 2 AO-sp3 m mi AO mi c 1e. Vy th gc lin kt (gc hotr) HOH phi ng bng gc t din 109o29. Ngi ta cho cho rng 2 AO-sp3 c 2 cp e khng lin

kt to ra 2 vng c mt e cao, s y nhau. S y ny lm hp gc to bi 2 trc ca 2 AO-sp 3cn li, kt qu ta quan st c gc l 104,5o.Nh vy mc d cn rt nh tnh, thuyt ho tr nh hng gii thch c kt qu thc nghim

v gc lin kt ca H2O.Theo thuyt lin kt ho tr, lin kt cng ho tr c hnh dung l s ghpi electron ca hai nguyn t lin kt v lin kt s nm theo phng m cc obitan nguyn t che

ph nhau nhiu nht. Vy c th xc nh phng ca lin kt da vo ch tiu che ph cc i giacc obitan nguyn t, ngha l theo gi tr cc i ca cc obitan nguyn t. Xc nh c phngca lin kt l c th xc nh c hnh dng ca phn t. y l mt trong nhng thnh tu ln nhtca c hc lng t p dng vo l thuyt v cu to phn t.

b) Lp lun tng t nh trn c th gii thch hnh dng ca cc phn t XH3:NH3 PH3 AsH3 SbH3


9/36

Vy da vo s che ph cc i gia cc obitan nguyn t s v p, c th gii thch c hnh dngca nhiu phn t nhng cch gii thch hon ton khng th p dng cho nhng hp cht ca ccnguyn t C, Si, B, Be, Hg...

2. Khi nim lai ho ca thuyt lin kt ho tr:Mt trong cc cn c xy dng thuyt VB l kt qu gii bi ton H 2 ca Heitle -London. H2 l

trng hp n gin nht v nguyn t H ch c AO ho tr 1s. c c s l gii ph hp thcnghim cho cc phn t phc tp, bt buc phi m rng ti cc AO ho tr ns, np,... m n 2. Thuytlai ho nhm gii quyt vn .Lai ho l s t hp cc AO ho tr c s lng t l khc nhau ca cng mt nguyn t to ra cc

AO mi c cng nng lng.- iu kin cc AO ho tr tham gia lai ho c l phi c nng lng gn nhau.- S AO tham gia t hp bng tng s cc lai ho thu c.- Cc AO lai ho l cc AO suy bin, ngha l cc AO c nng lng v kch thc hon ton nh

nhau nhng khc vi nhau v s nh hng (phng) trong khng gian..- c im hnh hc ca AO lai ho l c mt u (hay mt phn) n rng cn u kia b thu hp.- S nh hng cc AO lai ho trong khng gian th hin s phn b mt electron trong khng

gian.Lai ho l khi nim thun tu ton hc. V nguyn tc, cc AO tham gia t hp l cc nghim ca

phng trnh Schrodinger, l hm ring ca ton t Hamilton ca h H, vy theo nguyn l chng chttrng thi, t hp tuyn tnh ca cc hm cng l nghim ca phng trnh Schrodinger. Lai ho cth coi l mt khi nim gi nh c dng gii thch cc kt qu thc nghim.

Xt phn t CH4 : Nguyn t C c cu hnh electron trng thi c bn 1s 2 2s2 2px1 2py1 v trngthi kch thch 1s2 2s1 2px1 2py1 2pz1... Theo cch lp lun nh trn, phn t CH 4 c hnh dng nh sau: che ph cc i gia cc obitan nguyn t 1s ca H vi cc obitan 2s v 2p ca C th ba nguyn tH tin li nguyn t C theo phng ca cc trc x, y, z cn nguyn t H th t tin li che ph viobitan 2s ca C lm vi ba lin kt C- H trn nhng gc nh nhau v bng 125 o (v chu lc y nhnhau ca ba cp electron lin kt ). Nhng trn thc t phn t CH 4 l phn t t din u, ngha lc 4 lin kt C-H u nh nhau v lm vi nhau nhng gc nh nhau bng 109o28.

Pauling cho rng trong phn t CH4, mt obitan 2s v 3 obitan 2p ca C trn ln vi nhau, hay nitheo ngn ng ton hc, chng t hp vi nhau to nn nhng t hp mi tng ng nhau v c

gi l cc obitan lai ho. Bn obitan lai ho mi, mi obitan mang 1/4 tnh cht ca obitan 2s v 3/4tnh cht ca obitan 2p ca nguyn t C hng v 4 nh ca 1 t din u. y l nh hng tt nhtcho s xen ph vi 4 obitan 1s ca 4 nguyn t H c th to thnh 4 lin kt C H bn vng v tora phn t CH4 t din.

3. Kiu lai ho v hnh dng ca phn tThc t thng xt hp cht ca cc nguyn t chu k 2. Cc AO ho tr ca mi nguyn t ca

nguyn t chu k 2 l 2s, 2px, 2py, 2pz. Ta s xt cc dng lai ho c cc AO ho tr ny tham gia.b) Lai ho sp

- Lai ho sp l lai ho trong AO 2s t hp tuyn tnh vi 1 AO 2p z (thng dng AO z) tora 2 AO lai ho sp.

- C th hnh dung qu trnh lai ho xy ra nh sau:

Lai ho sp: 1AO-s+1 AO-pZ = 2AO-sp

2AO-sp l: d1 =2

1(s+pZ)

d2 =2

1(s-pZ)

+ AO 2pz c i xng i vi trc z. AO - 2s c i xng cu. T hp tuyn tnh 2 AO to ra 2AO mi cng nm trn trc z; mi AO mi ny c phn m rng, phn b thu hp. C 2 AO lai ho spu nm trn cng mt ng thng: trc z. Do ngi ta gi lai ho sp l lai ha thng, k hiu lhi. Vy ta c 2 AO lai ha sp l h i1, hi2. Gc gia 2 trc 2 AO - sp l 180o.

9

1AOs + 1AOp


10/36

- Biu thc ca 2 AO lai ho sp l:hi1 = (2s + 2pz)/ 2hi2 = (2s - 2pz)/ 2

- Lai ho sp c dng gii thch lin kt ho hc trong phn t BeH 2, C2H2 (ankin) . Ccphn t u c dng thng.Lai ho sp2: 1AO-s+2 AO-p = 3AO-sp2

3AO-sp2 l: t1 =3

1(s+ 2 px)

t2 =61 ( 2s-px+ 3 py)

t3 =6

1( 2s-px- 3 py)

c) Lai ho sp2

- Lai ho sp2 l lai ho trong AO - 2s t hp tuyn tnh vi 2 AO - 2p (thng chn l AO 2p z,AO 2py) to ra 3 AO mi c cng nng lng. Tng t nh AO lai ho sp, AO-sp 2 cng b bindng so vi AO c bn, c phn m rng v phn b thu hp.

- Hnh dung qu trnh lai ho v s nh hng cc AO trong khng gian c m t nh sau:

Ba AO-sp2 cng nm trong mt mt phng, gc to bi hai trc ca hai AO cnh nhau l 120 o. Do

lai ho sp2

c gi l lai ha tam gic. K hiu AO-sp2

l t, km theo s th t: t1, t2, t3.Lai ho sp3: 1AO-s+3 AO-p = 4AO-sp3

4AO-sp3 l: te1 =2

1(s+px+py+pZ)

te2 =2

1(s + px - py - pZ)

te3 =2

1(s - px+ py - pZ)

te4 =

2

1(s - px - py + pZ)

s =4

1; px =

4

3cos cos ;

py =4

3cos sin ; pZ =

4

3cos

- Lai ho sp2 c p dng gii thch lin kt ho hc trong cc anken CnH2n (n2), benzen, ...Cc phn t c gc lin kt bng 120o.

d) Lai ho sp3

- Lai ho trong AO - 2s t hp tuyn tnh vi 3 AO - 2p to ra 4 AO mi c cng nng lng

c gi l lai ho sp

3

.- C th hnh dung qu trnh lai ho sp3 nh sau:

10


11/36

- Hnh dng ca mi AO lai ho sp 3 cng tng t nh hnh dng AO lai ho sp, sp 2 va xt. 4 AO-sp3 hng ra 4 nh ca t din u m tm ca t din l nguyn t (chnh xc l ht nhn nguyn t)

c cc AO lai ho. Do lai ho sp3 c gi l lai ha t din. AO-sp3 c k hiu l te. Vy ta c 4AO-sp3 l te1, te2, te3, te4.

- Lai ho sp3 c p dng gii thch lin kt ho hc trong cc ankan C nH2n+2 (n1), in hnh ltrong CH4. Ngoi ra n cng c p dng trong cc trng hp khc nhau H 2O, NH4+,... cc phnt c gc lin kt gn bng 109o28Ngoi ra cn c cc kiu lai ho sp3d2 hay sp2d vi s tham gia ca cc obitan d.S dng quan nim lai ho c th gii thch v hnh dng cc phn t kh ng.Kiu laiho

Gc gia cc obitan lai ho Dng phn t V d

spsp2

sp3

dsp2

sp3d2

180o

120o

109o2890o

90o

ng thngTam gic uT din uHnh vungBt din u

BeH2, BeCl2, ZnCl2, CO2BF3, NO3-, CO32-

CH4, CCl4, NH4+, ClO4-, SO42-,PO43-

PtCl4, CuCl42-, Ni(CN)42-

SF6, AlF63-, SiF62-

Kiu lai ho cc obitan ca nguyn t cho thy s obitan lai ho c to nn ng thi l s phitr ti a ca nguyn t . Bng di y h thng li kh nng lai ho cc obitan ca nguyn t ccnguyn t v s phi tr ti a m nguyn t c:

Nguyn t chu k Kiu lai ho v s phi tr (vit trong dungoc)

II

IIIIVV v VI

sp(2), sp2(3), sp3(4)

sp3

(4), dsp3

(5), d2

sp3

(6), sp3

d2

(6)sp3(4), dsp3(5), d2sp3(6), sp3d2(6)d2sp3(6), d2sp3(7)

Da vo bng , c th hiu c cng thc cc dy hp cht v anion sau y ca cac nguyn tthuc cng nhm trong bng tun hon:

Nhm IVA CH4 SiF62- GeF62- SnF84- PbF84-

CO32- SiO44- GeO44- SnO68- PbO68-

Nhm VIA H2SO4 H6TeO6Nhm VIIA HClO4 H5IO6Tuy nhin, gii thch hnh dng ca phn t, ngoi s lai ho, cn vn dng thm mt s gi thit

khc na. V d nh i vi nhng phn t sau y, cc nguyn t trung tm u cng mt kiu lai

ho sp3 ca cc obitan, s bin i ca gc ho tr c gii thch nh sau:Gc ho tr gim xung v vai tr ca s trong s lai ho sp3 gim xung

Gc linkt trongcc phn t

H2O (105o)NH3 (107o)CH4 (108o28)

H2S (92o)PH3 (94o)

H2Se (91o)AsH3 (92o)

H2Te (90o)SbH3 (90o)

B. Cu to cc phn t n gin

1. Phn t O2Nguyn t O c cu hnh electron: 2s2 2p4

Hai nguyn t O lin kt vi nhau bng hai cp electron chung:

11


12/36

Cng thc cu to phn t vi lin kt cp gia hai nguyn t O ph hp nng lng ca lin kt l494 kJ/mol v di ca lin kt l 1,21A v gii thch c hu ht tnh cht ca oxi tr t tnh.

Oxi trng thi kh, lng hay rn u c tnh thun t. T tnh o c cho thy s c mt ca haielectron c thn trong phn t O2. Bi vy gii thch tnh cht ngi ta buc phi gi thit thmrng lin kt cng ho tr cng c th c to nn nh 3 electron gi l lin kt ba electron, ngha l

phn t O2 c cu to:

Trong ngoi lin kt cng ho tr bnh thng c to nn bng cp electron (vch lin) cn c

hai lin kt c to nn nh ba electron (vch ri). Lin kt ba electron cn c gi l lin kt mtelectron v thc t trong ba electron ch c mt electron c dng chung gia hai nguyn t:

Tuy nhin s lin kt gia hai nguyn t O cng ch l hai. (Thuyt obitan phn t gii thch mtcch n gin tnh thun t ca O2 bng s tn ti ca hai electron c thn trn hai obitan phn t

phn lin kt)2. Phn t N2

Nguyn t N c cu hnh electron 1s2 2s2 2p3

Hai nguyn t N lin kt vi nhau bng ba cp electron chung:

Hai obitan 2px ca hai nguyn t N che ph nhau to thnh lin kt .

Cp obitan 2py v cp obitan 2pz ca hai nguyn t N che ph nhau, theo tng cp mt, to thnh hailin kt .

Nh vy, lin kt trong phn t l lin kt ba. Lin kt ba ny c nng lng rt ln (942 kJ/mol)nn phn t N2 rt bn. Nit kh tham gia phn ng nhit thng.

3. Phn t NOTrc y ngi ta cho rng phn t NO c cu to:

Ngha l trong phn t c lin kt i.Nhng thc nghim cho thy rng phn t ny c mmen lng cc rt b ( = 0,15D) v di

ca lin kt N-O l 1,44A, ngha l trung gian gia di lin kt i trong ion NO - (1,18A) v calin kt ba trong ion NO+ (1,06A). Vy bc ca lin kt trong NO khng th bng 2 m bng 2,5,ngha l phn t c cu to:

12


13/36

Trong ngoi hai lin kt cng ho tr bnh thng (c to nn nh cp electron chung) cn cmt lin kt ba electron na (ion NO+ c mt trong cc hp cht NOCl, NOClO4, ion NO- c mttrong NaNO).

4. Phn t COTrc y ngi ta cho rng phn t CO c cu to:

ngha l trong phn t c lin kt i.Nhng thc nghim cho thy rng phn t c mmen lng cc rt b ( = 0,1118D), mt phn tgn nh khng c cc, v nng lng ca lin kt rt ln (1070 kJ/mol), ln hn c nng lng calin kt ba trong N2.

Bi vy, ngy nay ngi ta cho rng nguyn nhn lm gim mnh phn cc ca CO l nguyn tO a ra mt cp electron ca mnh dng chung vi nguyn t C, ngha l to thnh mt lin ktcho - nhn ngoi hai lin kt cng ho tr bnh thng:

Nh vy, lin kt C-O l lin kt ba. Tht vy, lin kt ny c nng lng rt ln so vi cc lin ktho hc.

Qua cu to ca cc phn t xt trn y, ta thy i vi phn t ch c hai nguyn t, hnh dngca phn t lun lun l ng thng ni lin hai ht nhn nguyn t. Nhng i vi nhng phn tgm ba nguyn t hay hn na, gii thch hnh dng ca chng, cn vn dng s lai ho cc obitanca nguyn t trung tm.

5. Than chNh bit, tinh th than ch c cu to lp. Cc lp lin kt vi nhau bng lc Van de Van, lc ny

rt yu hn lc ca lin kt ho hc cho nn cc lp d tch ri nhau v than ch mm trong khi kimcng l cht cng nht.

Trong lp than ch, mi nguyn t C trng thi lai ho sp2 lin kt cng ho tr vi ba nguyn t Cbao quanh cng nm trong mt mt phng to thnh vng 6 cnh. Nhng vng ny lin kt vi nhauto thnh mt lp v tn. Sau khi to thnh nhng lin kt nh vy, mi nguyn t C cn c mtelectron c thn trn obitan 2p c trc vung gc vi mt phng ca lp.Nhng obitan khng lai ho che ph nhau to nn lin kt vi mt trong ba nguyn t C bao

quanh:

di ca lin kt C-C trong cc lp l 1,415A, hi ln hn di ca lin kt C-C trong vngbenzen (1,39A) c bi l 1,5, ngha l bi ca lin kt C-C trong lp than ch ~ 1,33. Nhngkhc vi benzen, lin kt trong lp tinh th than ch l khng nh ch trong ton lp. Bi vykhc hn vi kim cng, than ch c mu xm, c nh kim, dn in v dn nhit.

6. Phn t O3Phn t O3 c dng ng gy vi gc


14/36

Mt obitan lai ho c mt electron c thn cn hai obitan lai ho na, mi mt c mt cp electron:

Mt obitan lai ho c mt cp electron che ph vi mt obitan 2p ca nguyn t O ( bn tri cahnh v) to thnh lin kt cho nhn, mt obitan lai ho khc c electron c thn che ph vi obitan2p ca nguyn t O khc ( bn phi ca hnh v) c electron c thn to thnh lin kt cng ho tr:

hayMt obitan 2p cn li khng lai ho ca nguyn t O trung tm c electron c thn che ph vi

obitan 2p khc ca nguyn t O (bn phi) cng c electron c thn to thnh lin kt :

( n gin, trong hnh khng biu din s che ph ca cc obitan to thnh lin kt ) Vy cngthc cu to ca phn t O3 l:

Ngy nay n gin, ngi ta hay dng mt cng thc linh hot hn, trong lin kt c chiai cho c hai lin kt ( hai bn), ngha l mt lin kt khng nh ch c k hiu bng vch ri:

7. Phn t SO2Phn t SO2 c dng ng gy ging nh O3 vi gc


15/36

Nu mt cch gn ng ngi ta tha nhn rng trong phn t SO2 ch nhng obitan 3s v 3p ca Stham gia to thnh lin kt th cch m t s to thnh cc lin kt s tng t nh i vi phn t O 3,ngha l phn t SO2 c cng thc cu to:

hay cng thc linh hot hn vi mt lin kt khng nh ch:

bi ca lin kt S-O l 1,5Nhng vic rt ngn mnh di ca lin kt S-O (1,43A) trong SO 2 so vi di ca lin kt n

(1,55A) ni ln rng lin kt l lin kt i v phn t SO2 c cu to:

c ho tr bn, nguyn t S phi trng thi lai ho sp 2 v c cu hnh electron trng thi kchthch, ngha l mt electron 3p chuyn sang obitan 3d:

Mt obitan lai ho c mt cp electron t do v mi mt obitan lai ho cn li c mt electron cthn che ph vi obitan 2p ca hai nguyn t O cng c electron c thn to thnh lin kt :

Obitan 3p khng lai ho ca S c electron c thn che ph vi obitan 2p khc ca nguyn t O (gis bn tri ca hnh v) c electron c thn to thnh mt lin kt v mt obitan 3d khng lai ho

ca S c electron c thn che ph vi obitan 2p khc ca O (gi s bn phi hnh v) c electronc thn to thnh mt lin kt na.

8. Phn t NO2Phn t NO2 c dng gp khc gn ging nh O3 v SO2

S to thnh cc lin kt trong phn t c m t tng t nh i vi phn t O 3 v SO2, ngha lNO2 c cng thc cu to:

15


16/36

hay cng thc vi lin kt khng nh ch

9. Phn t SO3Phn t SO3 c dng hnh tam gic u, nguyn t S nm trng tm ca tam gic v ba nguyn t

O nm nh:

Nu mt cch gn ng ngi ta tha nhn rng trong phn t SO3 ch nhng obitan 3s v 3p ca Stham gia to thnh lin kt th cch m t s to thnh lin kt gn tng t nh i vi phn t O 3.

hay mt cng thc lin kt khng nh ch:

bi ca lin kt S-O l 1,33 v S c ho tr 4.Nhng vic rt ngn di ca lin kt S-O (1,43A) trong SO3 cng ging nh trong SO2 l lin kt

i v phn t SO3 c cu to:

c ho tr su, nguyn t S phi trng thi lai ho sp2 v c cu hnh electron trng thi kchthch, ngha l mt electron 3s v mt electron 3p chuyn sang cc obitan 3d:

16


17/36

Ba obitan lai ho ca S, mi obitan c mt electron c thn che ph vi obitan 2p electron c thnca ba nguyn t O to thnh ba lin kt cng ho tr. Ngoi ra mt obitan 3p v hai obitan 3d khnglai ho ca S, mi mt c mt electron c thn che ph vi obitan 2p cn li c electron c thn ca

ba nguyn t O to thnh ba lin kt . Nh vy trong phn t SO3, nguyn t S c ho tr su.10. Phn t NH3Phn t NH3 c dng hnh chp tam gic, nguyn t N nh v ba nguyn t H nh ca tam gic

u:

Trong phn t NH3, nguyn t N trng thi lai ho sp3:

Mt obitan lai ho c cp electron khng lin kt, cn ba obitan lai ho khc, mi obitan c mtelectron c thn che ph vi obitan 1s c electron c thn ca ba nguyn t H to thnh ba lin kt

cng ho tr:

Vy phn t NH3 c cng thc cu to:

11. Ion NH4+

Ion NH4+ c dng hnh t din u, nguyn t N nm trung tm v bn nguyn t H nm nhca t din:

Trong ion NH4+, s to thnh ba lin kt cng ho tr bi ba cp electron chung gia N v H xy ratng t nh trong phn t NH3. Ch khc y l obitan lai ho c cp electron che ph vi obitan

17


18/36

1s trng electron ca H+ to thnh lin kt cho nhn. Tuy nhin c bn lin kt N - H u ging nhauv u l lin kt cng ho tr to nn bi cp electron:

12. Phn t H2SPhn t H2S c dng gp khc ging cc phn t O3, SO2:

Trong phn t H2S, nguyn t S trng thi lai ho sp3. Hai obitan lai ho, mi obitan c mt cpelectron khng lin kt cn hai obitan lai ho cn li, mi mt c mt electron c thn che ph viobitan 1s c electron c thn ca hai nguyn t H to thnh hai lin kt cng ho tr :

13. Phn t CO2Phn t CO2 c dng ng thng, nguyn t C gia hai nguyn t O:

trong phn t CO2, nguyn t C trng thi lai ho sp v c cu hnh electron trng thi kch

thch:

Hai obitan lai ho sp, mi mt c mt electron c thn, che ph vi obitan 2p c electron c thnca hai nguyn t O hai nm to thnh hai lin kt cng ho tr:

Hai obitan khng lai ha ca C, mi mt c electron c thn che ph vi obitan 2p khc celectron c thn ca hai nguyn t O to thnh lin kt :

v phn t c cng thc cu to:

18


19/36

trong C c ho tr 4.

C- BAI TP AP DUNG

I- BAI T P CO HNG DN GIAI

Cu 1:1. So snh, c gii thch.

a. ln gc lin kt ca cc phn t:

CH4; NH3; H2O. H2O; H2S.

b. Nhit nng chy ca cc cht : NaCl; KCl; MgOc. Nhit si ca cc cht : C2H5Cl; C2H5OH; CH3COOH

2. i vi cc phn t c cng thc tng qut AXn (n 2 ), lm th no xc nh phn t phn cchay khng phn cc ?3.Hy gii thch ti sao PbI2 ( cht rn mu vng) tan d dng trong nc nng, v khi ngui li ktta di dng kim tuyn ng nh ?Cu 2 :

Xt cc phn t BF3, NF3 v IF3. Tr li cc cu hi sau :1. Vit cng thc chm electron Lewis ca cc cht trn2. Da vo thuyt lai ha obitan nguyn t hy cho bit tr ng th i lai ha ca nguyn t trung tm vdng hnh hc ca mi phn t3. Xc nh xem phn t no l phn cc v khng phn cc. Gii thch kt qu chnCu 3:Cho hai nguyn t A v B c tng s ht l 65 trong hiu s ht mang in v khng mangin l 19. Tng s ht mang in ca B nhiu hn ca A l 26.a) Xc nh A, B; vit cu hnh electron ca A, B v cho bit b 4 s lng t ng vi electron sau

cng trong nguyn t A, B.b) Xc nh v tr ca A, B trong HTTH.c) Vit cng thc Lewis ca phn t AB2, cho bit dng hnh hc ca phn t, trng thi lai ho ca

nguyn t trung tm?

d) Hy gii thch ti sao phn t AB2 c khuynh hng polime ho?CU 41. Vit cng thc cu to Lewis, nu trng thi lai ha v v dng hnh hc ca cc phn t sau:

(a) B2H6 (b) XeO3 (c) Al2Cl6Gii thch v sao c Al2Cl6 m khng c phn t B2F6?

2. Trnh by cu to ca cc ion sau: O +2

, O 22

theo thuyt MO (cu hnh electron, cng thc cuto). Nhn xt v t tnh ca mi ion trn.

3. So snh v gii thch ngn gn phn cc (momen lng cc) ca cc cht sau: NH3, NF3, BF3.4. Ha tan 2,00 gam mui CrCl3.6H20 vo nc, sau thm lng d dung dch AgNO3 v lcnhanh kt ta AgCl cn c 2,1525 gam. Cho bit mui crom ni trn tn ti di dng phc cht.

4.1. Hy xc nh cng thc ca phc cht .

4.2. Hy xc nh cu trc (trng thi lai ha, dng hnh hc) v nu t tnh ca phc chttrn.

CU 5: Cho biet trang thai lai hoa cua ngt trung tam va dang hnh hoc cua cac phan tsau :

H2O , H2S , H2Se , H2Te .- Hay sap xep theo chieu tang dan o ln goc lien ket va giai thch s sap

xep o.- Tai sao ieu kien thng H2O the long,con H2S , H2Se , H2Te the kh?- Hay sap xep theo chieu tang dan tnh kh cua cac chat tren.Giai thch.

Cu 61. X, Y l hai phi kim. Trong nguyn t X, Y c s ht mang in nhiu hn s ht khng

mang in ln lt l 14 v 16.Hp cht A c cng thc XYn, c c im:

- X chim 15,0486% v khi lng

19


20/36

- Tng s proton l 100- Tng s ntron l 106

a. Xc nh s khi v tn nguyn t X, Y. Cho bit b bn s lng t ca e cui cng trn X,Y

b. Bit X, Y to vi nhau hai hp cht l A, B. Vit cu trc hnh hc v cho bit trng thi laiho ca nguyn t trung tm ca A, B.

c. Vit cc phng trnh phn ng gia A vi P2O5 v vi H2OVit cc phng trnh phn ng gia B vi O2 v vi H2O

2. Cho bit tr s nng lng ion ho th nht I1(eV) ca cc nguyn t thuc chu k II nh sau:

Chu k II Li Be B C N O F NeI1 (eV) 5,39 9,30 8,29 11,2

614,54 13,6

117,41 21,55

Nhn xt s bin thin nng lng ion ho th nht ca cc nguyn t trn. Gii thch.Cu 7

Nguyn t C c electron cui cng ng vi 4 s lng t:n = 3, l = 1, m = 0, ms = -1/2

Hai nguyn t A, B vi ZA < ZB < ZC ( Z l in tch ht nhn ).Bit rng: - tch s ZA. ZB. ZC = 952

-t s ( ZA + ZC ) / ZB = 3.1. Vit cu hnh electron ca C, xc nh v tr ca C trong bng H thng tun hon, t suy ra

nguyn t C?2. Tnh ZA, ZB. Suy ra nguyn t A, B?3. Hp cht X to bi 3 nguyn t A, B, C c cng thc ABC. Vit cng thc cu to ca X. trng

thi lng, X c tnh dn in.Vy X c hnh thnh bng cc lin kt ha hc g?

Cu 81) Cho X, Y l 2 phi kim trong nguyn t X v Y c s ht mang in nhiu hn s ht khng mang

in ln lt l 14 v 16. bit trong hp cht XYn. X chim 15,0486 % v khI lng. Tng s proton l 100. Tng s ntron l 106a. Xc nh s khi v tn X, Y

b. Xc nh CTCT XYn v cho bit kiu lai ha ca nguyn t X dng hnh hc ca XYn.c. Vit phng trnh phn ng gia XYn vi P2O5 v vi H2O

2) a. Ti sao SiO2 l mt cht rn nhit phng nng chy 1973K trong khi CO2 li l cht kh nhit phng nng chy 217K

b. Cht dicloetilen (C2H2Cl2) c ba ng phn k hiu X,Y,Z- Cht X khng phn cc cn cht Z phn cc

- Cht X v cht Z kt hp vi Hidro cho cng sn phmX (hc Z) + H2 Cl - CH2 - CH2 Cl. Vit cng thc cu to X, Y, Z. Cht Y c momen lng cc khng ?

Cu 9:1. Silic c cu trc tinh th ging kim cng vi thng s mng a = 0,534 nm.Tnh bn knh nguyn t

cng ha tr ca Silic v khi lng ring (g.cm-3) ca n. Cho bit MSi= 28,086 g.mol-1. Kim cngc cu trc lp phng tm din, ngai ra cn c 4 nguyn t nm 4 hc t din ca mng c s.

2. C cc phn t XH32.1. Hy cho bit cu hnh hnh hc ca cc phn t PH3 v AsH3.2.2. So snh gc lin kt HXH gia hai phn t trn v gii thch.

2.3. Nhng phn t no sau y c moment lng cc ln hn 0 ? BF3, NH3, SiF4, SiHCl3, SF2, O3.Cho bit Zp = 15, ZAs = 33, ZO = 16, ZF = 9, ZCl = 17, ZB = 5, ZN = 7, ZSi = 14, ZS = 16.Cu 10

20


21/36

103o15

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 = +1/2

1 Xc nh A, X, Z.2 Cho bit trng thi lai ho v cu trc hnh hc ca cc phn t v ion sau: ZA 2, AX2, AX32-, AX42-.3 Bng thuyt lai ho gii thch s to thnh phn t ZX. Gii thch v sao ZX c moment lng cc b.Gii thch s hnh thnh lin kt trong phn t phc trung ho Fe(CO)5 bng thuyt VB.4 Gii thch v sao AX32- li c kh nng ho tan A to thnh A2X32-.

CU 11(1) Cho bit s bin i trng thi lai ho ca nguyn t Al trong phn ng sau v cu to

hnh hc ca AlCl3, AlCl

4.

AlCl3 + Cl AlCl4(2) Biu din s hnh thnh lin kt phi tr trong cc trng hp sau:(o): Sn phm tng tc gia NH3 v BF3.(b): Sn phm tng tc gia AgCl vi dung dch NH3.(3): Gii thch s khc nhau v gc lin kt trong tng cp phn t sau:(a)

S O

Cl Cl Cl Cl(b)O O

F F Cl Cl

Cu 121. Cu hnh electron ngoi cng ca nguyn t X l 5p5. T s ntron v in tch ht nhn bng

1,3962. S ntron ca X bng 3,7 ln s ntron ca nguyn t thuc nguyn t Y. Khi cho4,29 gam Y tc dng vi lng d X th thu c 18,26 gam sn phm c cng thc l XY.Hy xc nh in tch ht nhn Z ca X v Y v vit cu hnh electron ca Y tm c.

2. Hy cho bit trng thi lai ha v dng hnh hc ca hp cht XCl3.

3. Bn knh nguyn t Cobalt l 1,25. Tnh th tch ca n v ca tinh th Co nu trong 1trt t gn xem Co kt tinh dng lp phng tm mt.

Cu 13A, B l 2 nguyn t k tip nhau trong cng mt chu k ca bng tun hon trong B c tng s lngt ( 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 electroncui cng ca cation A +a l 3,5.

a)Xc nh b 4 s lng t ca electron cui cng trn A, B.b)Vit cu hnh electron v xc nh tn ca A, B.

Cu 14: 1) Cho cc cht sau y:

CO2 , SO2 , C2H5OH, CH3COOH, HIHy cho bit cht no c nhit si cao nht? Gii thch?2) Dng thuyt ni ha tr, hy cho bit c cu lp th (biu din bng hnh v) v trng thi

lai ha ca nguyn t trung tm ca cc phn t v ion sau:H2SO4 , [Ni(CN)4]2- , ICl3 , XeF4

Cu 15:a. Hy cho bit cu trc hnh hc , kiu lai ha ca cc phn t : SF 6 , XeF2 , OF2b. Da vo cu hnh electron ca uran [Rn]5f36d17s2. Hy cho bit hai hp cht X,Y ca uran vi flo ,

cho bit ti sao c c 2 hp cht ny . Hon thnh phn ng sauClF3 + A B + Cl2

c. Ti sao ozon tan nhiu trong nc , nhng oxi t tan trong nc

Cu 16:Nguyn t ca nguyn t phi kim A c electron cui cng c b 4 s lng t tha mn m + l = 0 v n+ ms = 3/2 ( quy c cc gi tr m t thp n cao )

21

111o103ov

111ov


22/36

1. Xc nh s hiu nguyn t, gi tn nguyn t A. Vit cng thc electron, cng thc cu to ca phnt A2. Kim chng s lin kt v tnh cht thun t ca A2 bng cu hnh electron ca phn t.2. Ion A3B2- v A3C2- ln lt c 42 v 32 electron2.1. Tm 2 nguyn t B v C ( s hiu nguyn t, tn, k hiu )2.2. Dung dch mui ca A3B2- v A3C2- khi tc dng vi axit clohidric cho kh D v F tng ng.- M t dng hnh hc ca phn t D v E.- Nu phng php ha hc phn bit D v E.- Kh no trong 2 kh c th kt hp vi O2 ? Ti sao?Cu 17

1. So snh bn knh ca cc ht sau: Al, Al3+, Na, Na+, Mg, Mg2+, F-, O2-.2. Trong s cc cu trc c th c sau y:

a) Ca ICl4(-):

I

Cl Cl

Cl Cl

. .

. .

I

Cl Cl. .

Cl

Cl

(a) (b)

.

.

b) Ca TeCl4:

Te

Cl

Cl

ClCl

.

.

(c)

TeCl

Cl

ClCl

(d)

. .

c) Ca ClF3:

Cl

F

F

F

..

..

()

ClF

F

F

..

(g)

. .

Cl

(e)

F

F

F

. .

. .

nhng cu trc no c kh nng tn ti u tin hn? V sao?3. Ti sao nc nh hn nc lng? (c v hnh minh ha)Cu 18.

1. Hy cho bit dng hnh hc v trng thi lai ha ca nguyn t trung tm i vi phn t H2O vH2S. So snh gc lin kt trong 2 phn t v gii thch.2. Bn knh nguyn t ca cc nguyn t chu k 3 nh sau, hy nhn xt v gii thch:

Nguynt

Na Mg Al Si P S Cl

Bn knh ( oA ) 1,86 1,60 1,43 1,17 1,10 1,04 0,99

3. Kh N2 v kh CO c mt s tnh cht vt l ging nhau nh sau:

Nng lng phn lyphn t (kJ/mol)

Khong cch gia cc

ht nhn ( oA )Nhit nng chy

(oC)

N2 945 1,10 210

CO 1076 1,13 205Da vo cu hnh MO ca phn t N2 v phn t CO gii thch s ging nhau .

4. Gii thch bn phn t v tnh kh ca cc hp cht hydrohalogenua.

22


23/36

HNG DN GIICu 1.1. a. CH4 > NH3 > H2OGii thch:

H|

C N H O

H | H H H H HH

S cp e cha tham gia lin kt cng nhiu cng y nhau, gc lin kt cng nh.

b. H2O > H2SGii thch: V m in ca O > S, m in ca nguyn t trung tm cng ln s ko my ca ie- lin kt v pha n nhiu hn lm tng ln gc lin kt.

c. So snh nhit nng chy ca cc cht:MgO > NaCl > KCl

Gii thch: bn knh ion K+ > Na+in tch ion Mg2+ > Na+ v O2- > Cl-(Nng lng phn li t l thun vi in tch ion v t l nghch vi bn knh ion)

3.So snh nhit si ca cc cht:C2H5Cl < C2H5OH < CH3COOH

Gii thch:-C2H5Cl khng c lin kt hiro-Lin kt hidro gia cc phn t axit bn hn lin kt hidro gia cc phn t ru.C2H5 O H O

H C2H5

O H OCH3 C C CH3O H O

2. Mun xc nh mt phn t c cc hay khng, trc ht cn phi bit s sp xp ca cc nguyn ttrong phn t(dng hnh hc ca phn t)

Momen lng cc (o phn cc) l mt i lng c ln v c chiu.Trong phn t, nu cc lin kt phn cc c sp xp i xng nhau, momen lng cc ccng ln v ngc chiu.Chng s trit tiu nhau v phn t khng phn cc.

Ngc li nu cc lc khng cn bng, phn t s c cc.

3. PbI2 d tan trong nc nng v qu trnh ha tan PbI2 thu nhit ln:

PbI2

Pb2+

+ 2I-

H > 0Cn khi ngui th xy ra qu trnh ngc li, ta nhit (H < 0).V qu trnh ngui t t, s mm kt tinh t, nn tinh th c to thnh d dng.

Nu lm ngui nhanh s thu c dng bt vng PbI2.

Cu 2 :

SFF

F

NF F

F

I FF

F

Lai ha sp2 Lai ha sp3 Lai ha sp3dTam gic phng Thp y tam gic Hnh ch T

23


24/36

Khng cc v momenlng cc lin kt btrit tiu

C cc v lng cc linkt khng trit tiu

C cc v lng cc linkt khng trit tiu

Cu 3a) Gi ZA, ZB ln lt l s proton trong nguyn t A, B.

Gi NA, NB ln lt l s notron trong nguyn t A, B.Vi s proton = s electron

Ta c h :

=

=

=

=+

=

=++=+++

17Z

4Z

13ZZ

21ZZ

262Z2Z

19)N(N)2Z(2Z65)N(2Z)N(2Z

B

A

AB

BA

AB

BABA

BBAA

(0,5)

ZA = 4 A l Be Cu hnh e : 1s22s2

B 4 s lng t: n = 2, l= 0, m = 0, ms =2

1

ZB = 17 B l Cl Cu hnh e : 1s22s22p63s23p5

B 4 s lng t: n = 3, l= 1, m = 0, ms =2

1

b) Ta c Z = 4 Be th 4, c 2 lp e Be chu k 2.Nguyn t s, c 2e ngoi cng phn nhm chnh nhm II.

Tng t cho Cl: th 17, chu k 3, phn nhm chnh nhm VII.

c) :Cl....

:Be:Cl....

:

Hnh dng hnh hc ca phn t: ng thngTrng thi lai ho : sp

d) Khi to thnh phn t BeCl2 th nguyn t Be cn 2 obitan trng; Cl t trng thi bn vng v cnc cc obitan cha 2 electron cha lin kt do nguyn t clo trong phn t BeCl2 ny s a ra cpelectron cha lin kt cho nguyn t Be ca phn t BeCl 2 kia to lin kt cho-nhn. Vy BeCl2 ckhuynh hng polime ho:

Cu 41.

24

Cl ClBe

........

Cl

B

e

Cl

Cl

B

e

Cl

Cl

B

e

Cl

Cl

B

e

Cl

Cl

B

eCl


25/36

Xe

OO

O

Xe lai ha sp3 phn t

dng thp tam gic

B B

HHH

H H H

B lai ha sp3, phn t B2H6 g

2 t din lch c 1 cnh ch

lin kt BHB l lin kt 3 t

ch c 2 electron, 1 electro

H v 1 electron ca B.

Al

ClCl

Al

ClCl

Cl

Cl

Al lai ha sp3, phn t Al2Cl6 gm 2 t din l

c 1 cnh chung, c 2 lin kt cho nhn

to thnh do cp e khng lin kt ca Cl v

obitan trng ca Al.Trong Al2Cl6 nguyn t Al

t -c cu trc bt t vng bn.

2. O+

2 : ()lk

s 2 (*

s

) 2)( lk

z 2

)( lkx

2 =)( lk

y 2

)( *x

1=)( *y

OO

O2

2 : ()lk

s 2 (*

s

) 2)( lk

z 2

)( lkx

2 =)( lk

y 2

)( *x

2=)( *

y 2

OO

2

O +2

c electron c thn nn thun t. O 22

khng c electron c thn nn ngch t.3.

N

HH

H

Cc vectmomen l- ng ccca cc lin kt v cpelectron khng lin kt c ngchiu nn momen l- ng ccca phn t l n nht.

N

FF

F

Cc vectmomen l- ng ccca cc li n kt v cpelectron khng lin kt ng- cchiu nn momen l- ng ccca phn tb hn NH3.

B

F

F

F

Phn t dng tam gic uCc vectmomen l- ng ccca cc lin kt trit tiu lnnhau(tng bng khng) phnt khng phn cc.

4.n(AgCl) = (2,1525:143,5) = 0,015; n(CrCl3 . 6H2O) = (2:266,5) = 7,5.10-3

n(Cl- to phc) = 3(7,5.10-3) - 0,015 = 7,5.10-3

Trong phn t phc cht t l mol Cl: Cr3+ = (7,5.10-3) : (7,5.10-3) = 1:1

Cng thc ca phc: [Cr(H2O)5Cl]2+

24Cr3+ (1s2 2s2 2p6 3s2 3p6 3d3) 24Cr3+ : [Ar] 3d3

3d3 4s 4p

Cr lai ha sp3d2

Ar

Phc thun t

A

H2O

Cl

900

900

H2OH2O

H2O

H2O

Bt din u

25

C phn t Al2Cl6 v nguyn t Al t cu trc btt vng bn.

Khng c phn t B2F6 v: phn t BF3 bn do clin kt pi khng nh ch c to thnh giaobitan trng ca B vi cp electron khng lin ktca F v kch thc ca nguyn t B b so vinguyn t F nn tng tc y gia 6 nguyn t Fln lm cho phn t B2F6 tr nn km bn.


26/36

Cu 5. -Trong cc phn t H2O , H2S, H2Se, H2Te; O, S, Se, Te (R) trng thi lai to sp3, phn t ccu to dng gc :

RH H

- V m in ca O ln nht nn cc cp e lin kt b ht v pha O mnh khong cchgia 2 cp e lin kt trong phn t H2O l nh nht nn lc y tnh in mnh nht gclin kt ln nht .Th t tng dn gc lin kt l : H2Te , H2Se, H2S, H2O .- iu kin thng nc th lng l do cc phn t nc c kh nng to lin kt H lin

phn t :... O H ... O H ...

H H- Trong cc phn t H2R , R u c s oxi ho -2, tuy nhin t O n Te bn knh R li tng ln

kh nng cho e tng t O n Te, tc l tnh kh tng theo th t H2O, H2S, H2Se, H2Te .

Cu 61. a. Gi PX,NX ln lt l s proton v ntron ca X

PY,NY ln lt l s proton v ntron ca YTa c: PX + nPY = 100 (1)NX + nNY = 106 (2)

T (1) v (2): (PX+NX) + n(PY+NY) = 206 AX+nAY = 206 (3)Mt khc: AX / (AX+nAY) = 15,0486/100 (4)

T (3), (4): AX = PX+NX = 31 (5)Trong X c: 2PX - NX = 14 (6)

T (5), (6): PX = 15; NX = 16 AX = 31X l photpho 15P c cu hnh e l : 1s22s22p63s23p3 nn e cui cng c b bn s lng t l:

n =3, l=1, m = +1, s = +1/2Thay PX = 15; NX = 16 vo (1), (2) ta c nPY = 85; nNY = 90

nn: 18PY 17NY = 0 (7)Mt khc trong Y c: 2PY NY = 16 (8)T (7), (8): PY = 17; NY = 18 AY = 35 v n = 5

Vy: Y l Clo 17Cl c cu hnh e l 1s2 2s22p63s23p5,nn e cui cng c b bn s lng t l: n = 3; l =1; m = 0, s = -1/2

* Xc nh ng mi cht 0,5 , ng mt b bn s lng t 0,25 .b. Cl

A: PCl5; B: PCl3 Cl

Cu to ca A: (0,5) Cl P- PCl5 c cu trc lng thp tam gic- Nguyn t P trng thi lai ho sp3d

Cl ClCu to ca B: ..

- PCl3 c cu trc thp tam gic P- Nguyn t P trng thi lai ho sp3

Cl Cl Clc. ng mi pt:

3 PCl5 + P2O5 = POCl3PCl5 + 4H2O = H3PO4 + 5 HCl2PCl3 + O2 = POCl3

PCl3 + 3H2O = H3PO3 + 3 HCl2. Nhn xt:

a. Nhn chung nng lng ion ho tng dn

26


27/36

Gii thch: T tri sang phi trong mt chu k, in tch ht nhn ca cc nguyn t tng dn v s engoi cng cng tng thm c in vo lp n ang xy dng d. Kt qu cc e b ht v ht nhnmnh hn lm bn knh nguyn t gim, dn n lc ht ca nhn vi e ngoi cng tng lm e cng kh

b tch ra khi nguyn t lm nng lng ion ho tngb.Be v N c nng lng ion ho cao bt thngGii thch: Be c cu hnh e: 1s22s2 c phn lp s bo ho. y l cu hnh bn nn cn cung cp

nng lng cao hn ph v cu hnh nyN c cu hnh e: 1s22s22p3 phn lp p bn bo ho, y cng l mt cu hnh bn nn cng cn cung

cp nng lng cao hn ph v cu hnh ny

Cu 71. Nguyn t C c cu hnh electron cui cng :3p5

+1 0 -1Cu hnh electron ca C:1s2 2s2 2p6 3s2 3p5

V tr ca C: STT 17, chu k 3, nhm VII A. C l Clo.2. ZC = 17 ZB . ZA = 56 ZA = 7 , A l Nit

ZA + 17 = 3ZB ZB = 8 , B l Oxi

3. CTCT X Cl - N = ONOCl trng thi lng c tnh dn in vy trong cht lng phi c cc ion NO+ v Cl-. Do trongphn t NOCl c lin kt ion v lin kt cng ha tr.

Cu 8I.1)

a. Gi Px, PY l s proton X, Ynx, ny l s ntron X, Y

Px + nPy = 100 (1)Nx + nNy = 106 (2)Px + Nx + n(PY + Ny) = 206Ax + nAy = 206 (3)

Ax Ax + nAy=> Ax = 31Trong nguyn t X : 2Px Nx = 14

Px = 15Nx = 16

Thay Px, Nx vo (1) , (2)n (Ny Py) = 5 ( 5)2Py ny = 16 (6) => Ny = 2Py - 16n(Py 16) = 5

Py =5 16n

n

+

n 1 2 3 4 5Py 21 18,8 17,67 17,25 17

Py = 17, n =5 , Ay = 35 => Y l clo

b. PCl5 : nguyn t P lai ha sp3d dng lng thp tam gic.

Cl

Cl P Cl

Cl Cl

c. P2O5 + PCl5 = 5POCl3PCl5 + H2O = H3PO4 + 5HCl

I.2)

27

=

15,0486

100 (4)

=> X l photpho


28/36

- C v Si u c bn electron ha tr tuy nhin khc vi CO2 (O = C = O) SiO2 khng phi l mtphn t n gin vi lin kt Si =O. nng lng ca 2 lin kt i Si=O km xa nng lng ca bn linkt n Si-O v vy tinh th SiO2 gm nhng t din chung nh nhau.

O

O Si O

O

a. SiO2 l tinh th nguyn t lin kt vi nhau bng lin kt cng ha tr bn

trong khi CO2 rn l tinh th phn t, lin kt vi nhau bng lc Vanderwall yu.b. X khng phn cc vy X tn ti dng trans Z phn cc.Vy Z tn ti dng Cis

HCl

HCCH

ClH

CTCT

HCl

CC

ClH

= (X)

ClCl

CC

HH

=

(X) (Z)

CTCT Y s l

ClH

CC

ClH

=

C-H C-ClX= 2,5 2,1 = 0,4 X = 0,5

Vy Y phn ccCu 91. rSi = (a. 3 )/8 = (0,534. 3 )/8= 0,118

S nguyn t Si c trong mt mng c s: 8.(1/8) + 6.(1/2) + 4 = 8Khi lng ring ca Si = 2,33 g.cm-1.

2.

2.1. P : 1s

2

2s

2

2p

6

3s

2

3p

3

; As : 1s

2

2s

2

2p

6

3s

2

3p

6

3d

10

4s

2

4p

3

P v As u c 5 electron ha tr v c 3 electron c thn trong XH 3

X

HH H

X trang thai lai hoa sp3.

2.2. XH3 hnh thp tam gic, gc HPH > gc AsH, v m in ca nguyn t trung tm P ln hnso vi As nn lc y mnh hn.

2.3.

28

hoc Z + H2


29/36

N FF

FSi

ClH

ClCl

sp3

S

FF

O

O Osp3 sp3 sp2

B

F

FF

SiF

FF

F

sp2 sp3

4 cht u tin c cu to bt i xng nn c moment lng cc > 0.

Cu 101. 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 2p4 X l ONguyn t Z: n = 2, l = 1, m = 0, s = +1/2 2p2 Z l C

2.Phn t, ion Trng thi lai ho cu

nguyn t trung tmCu trc hnh hc

CS2 sp ng thngSO2 sp2 Gc

SO 23 sp3 Chp y tam gic u

SO 24 sp3 T din u

3C: [He] 2s2 2p2

O: [He] 2s2 2p4

Cacbon dng 1 obitan 2s t hp vi 1 obitan 2p to ra 2 obitan lai ho sp hng ra hai pha khc nhau,trong c mt obitan bo ho v 1 obitan cha bo ha.Cacbon dng 1 obitan lai ho cha bo ho xen ph xichma vi 1 obitan p cha bo ho cu oxi v

dng 1 obitan p thun chng cha bo ho xen ph pi vi 1 obitan p cha bo ho cn li cu oxi. Oxidng 1 obitan p bo ho xen ph v obitan p trng ca cacbon to lin kt pi kiu p p.Cng thc cu to: :C O:+ CO c moment lng cc b v trong phn t c lin kt phi tr ngc cp electron ca nguyn toxi cho sang obitan trng cu nguyn t cacbon lm gim phn cc ca lin kt nn lm gimmoment lng cc.+ S hnh thnh lin kt trong phn t Fe(CO)5Fe (Z = 26) [Ar] 3d6 4s2 4p0

Fe* [Ar] 3d8 4s0 4p0

29


30/36

trng thi kch thch, nguyn t Fe dng 1 obitan 3d trng t hp vi 1obitan 4s v 3 obitan 4p tothnh 5 obitan lai ho dsp3 trng hng ra 5 nh ca hnh lng chp y tam gic u tm l nguynt Fe.CO dng cp electron t do cha lin kt trn nguyn t cacbon to lin kt phi tr vi cc obitan laiho trng ca st to ra phn t phc trung ho Fe(CO)5

Fe* [Ar] 3d8 4s0 4p0

:CO :CO :CO : CO :CO

4. S [Ne] 3s2 3p4

S* [Ne] 3s2 3p4

SO32- c kh nng kt hp thm 1 nguyn t S to S 2O32- v trn nguyn t S trong SO32- cn c mtcp electron t do cha lin kt c kh nng cho vo obiatn 3p trng ca nguyn t S to lin kt chonhn.

2-

S O

S

OO

2-

S O

OO

S

Cu 11

AlCl3 + Cl AlCl4(1) - Trc phn ng trng thi lai ho ca Al l: sp2

- Sau phn ng trng thi lai ho ca Al l: sp3

- Cu to hnh hcCl Cl

Al Al

Cl Cl Cl ClCl

Tam gic phng T din(2)

H F

H N+ B- F NH3 Ag+ NH3 Cl

H F* Nit cn 1 cp electron t do * Ag+ cn obital/ho tr trng

* B cn obital ho tr trng(3) Trong cc phn t, nguyn t trung tm u c trng thi lai ho sp3 v c cu to gc.

30


31/36

(a). S sai bit gc ho tr trong phn t SCl2 v OCl2 l do s khc bit v m in caoxi v lu hunh. ca nguyn t trung tm cng nh th cc cp electron lin kt b ynhiu v pha cc nguyn t lin kt, nn chng chim vng khng gian nh xung quanh nguynt trung gian. ca oxi ln hn S nn gc ho tr Cl O Cl ln hn Cl S Cl.

(b). S sai bit gc ha tr trong phn t OF2 v OCl2 cng do s khc bit v ca cc nguyn tlin kt. Nguyn t lin kt c cng ln th gc ha tr cng nh. Flor c ln hn Clor nn

gc ha tr F O F nh hn Cl O Cl.Cu 121.Cu hnh y ca nguyn t X l: 1s2 2s2 2p6 3s2 3p6 3d10 4s24p6 4d10 5s25p5

Vy ZX = 53 = s proton X3962,1=

X

X

Z

N NX = 74

AX = ZX + NX = 53 + 74 = 127Ta c :

7,3=Y

X

N

N NY = 20

Y + X XYMY MXY4,29 18,26

3926,18

127

29,426,18

29,4=

+==

Y

YY

XY

YM

MM

M

M(g/mol)

Vy : ZY + NY = 39 ZY = 19Cu hnh electron ca Y : 1s2 2s2 2p6 3s2 3p6 4s1

2. X c 7 electron lp ngoi cng, cn obitan d trng nn trong hp cht XCl 3 X lai ha sp3d,dng hnh hc l ch T

ClX

Cl

Cl 3.

1,25

1,25

5

A B

C D

A B

C D

AD = 1,25 . 4 = 5 ()

AB = 54,32

52

= ( )

Vy th tch ca mng n v ca Co : V = (3,54)3 = 44,36 ()3

Cu 13a)V 2 nguyn t k tip nhau trong cng mt chu k nn 2 nguyn t c cng s lp electron

( cng n ). M tng ( n + l ) ca B ln hn tng ( n + l ) ca A l 1 nn:Cu hnh electron lp ngoi cng ca A, B l:

A: ns2. B: np1

Mt khc A c 2e lp ngoi cng cation A c dng A2+.Vy tng i s ca 4 s lng t ca A2+ l:

(n 1 ) + 1 + 1 - 21

= 3,5Vy 4 s lng t ca :

31


32/36

A: n = 3 l = 0 m = 0 s = -2

1

B: n = 3 l = 1 m = - 1 s = +2

1

b)Cu hnh electron ca A, B:A: 1s22s22p63s2 ( Mg ).B: 1s22s22p63s23p1 ( Al ).

Cu 14.1) C2H5OH v CH3COOH to c lin kt hidr gia cc phn t nn c nhit si cao hncc cht kia.

t0 si ca CH3COOH > t0 si ca C2H5OH v cOHHCCOOHCH MM 523 > v cho lin kt hidr trong CH3COOH bn hn trong C2H5OH (trong CH3COOH,

nhm C = O ht in t lm tng in tch dng trn H ca nhm OH, nn lc ht tnh in gia Hny vi O ca phn t CH3COOH th nh mnh hn). V vy CH3COOH c nhit si cao nht.

2) H2SO4 : S trng thi lai ha sp3 , c cu t in lch

[Ni(CN)4]2-Ni2+ trng thi lai ha sp2d c cu hnh vung

2

NCNC

Ni

NCNC

ICl3 : I trng thi lai ha sp3d, c cu lng thp tam gicXeF4 : Xe trng thi lai ha sp3d2, c cu bt din u

Cu 15 :b.-T cu hnh e ca U [Rn]5f36d17s2 nguyn t U c 4e c thn , to phn t UF4 :Ngoi ra, U U6+ + 6e

[Rn]5f36d17s2 [Rn]Nn to phn t UF6

Phn ng : 2 ClF3 + 3UF4 3UF6 + Cl2

Cu 16 :1.

2.2.1

Trng hp 1: ms= +1/2 => n=1 => l=0 =>m=0Vy cu hnh electron ca nguyn t A : 1s1 => HydrTrng hp 2: ms= -1/2 => n=2 => l=0 => m=0

hoc l=1 => m= -1* Vi ms= -1/2; n=2; l=0; m=0 => Cu hnh electron l 1s22s2 : B-ri* Vi ms= -1/2; n=2; l=1; m= -1 => Cu hnh electron l 1s22s22p4 : -xyV A l phi kim nn hoc A l Hydro (H) hoc A l O-xi (O)Vi A l Hydro- CTPT : H2

- CT electron : H:H- CTCT: H - H ( 1 lin kt)- Cu hnh e ca phn t 1s2

- S lin kt : N= 2/2 =1- Khng c electron c thn nn l cht nghch t.Vi A l -xi

- CTPT: O2- CT electron:- CTCT: O = O

Gia 2 nguyn t oxy c 1 lin kt cng ha tr bnh thng v 2 lin kt c bit 3e(3lectron ny do 1 nguyn t a ra 1, nguyn t kia a ra 2 gp chung), trong ch c 1 electron c dng chung. Vy s electron chung gia 2 nguyn t l 4,v trong phn t c 2e c thn.Cu hnh electron ca phn t :

32


33/36

2.2.

2s2 2s*2 z2x2y2x*1y*1

S lin kt : N = (8-4)/2 = 2C 2 electron c thn nn O2 l cht thun t

Ion A3B2- c 42 electron.* Nu A l Hidr, ta c: 3.1 + ZB = 42 -2 ; ZB = 37Loi v khng tn ti ion RbH32-

* Vy A l oxi.Lc 3.8 +ZB = 42 - 2 ; ZB = 16 ( B l lu hunh ) Chn

Ion A3C2- :Ta c : 3.8 + ZC = 32 -2 => ZC = 6 ( C l cc-bon) ChnVy A3B2- l SO32-

A3C2- l CO32-

SO32- + 2H+ SO2 + H2OCO32- + 2H+ CO2 + H2OD l SO2 ; E l CO2- Dng hnh hc phn t :SO2 : nguyn t S trng thi lai ha sp2 nn phn t c cu to gc

CO2: nguyn t C trng thi lai ha sp nn phn t c cu to ng thng O=C=O- Phn bit SO2 v CO2Dng dung dch brm nhn ra SO2 qua hin tng mu nu ca dung dch

brm nht dnSO2 + Br2 + 2H2O = H2SO4 + 2HBr

- Kh SO2 c th kt hp vi O2 to SO3 do lu hunh trong SO2 cn cp electron tdo. CO2 khng c kh nng ny do cc-bon trong phn t khng cn electron cthn.

Cu 171.- Khi i t trisang phai trong mt chu k bn knh nguyn t gim dn nn: Na > Mg > Al.

- V cc ion Na+, Mg2+, F - , O2 u c cu hnh electron ging Ne : 1s2 2s2 2p6, nn bn knh ca chnggim xung khi in tch ht nhn tng:8O2 > 9F > 11Na+ > 12Mg2+ > 13Al3+ .- V cu hnh electron ca Al l 1s2 2s2 2p6 3s2 3p1 rt ln hn so vi O2 .- Do bn knh gim dn nh sau:

Na > Mg > Al > O 2-> F > Na+ > Mg2+ > Al3+2.a) Cu trc (a) c kh nng tn ti thc t v n m bo cho lc y gia cc cp elctron khng linkt l nh nht.

b) Cu trc (C) c kh nng tn ti trong thc t v tng tc y cu trc ny b nht.c) Cu trc c s gim nhiu lc y gia cc electron khng lin kt v cp electron lin kt.3. Do c lin kt hidro nn nc c cu trc c bit. Cc nguyn t Oxi nm tm v bn nh camt t din u. Mi nguyn t hidro lin kt chnh vi mt nguyn t oxi v lin kt hidro vi mtnguyn t oxi khc. Cu trc ny tng i xp nn c t khi nh. Khi tan thnh nc lng cu trcny b ph v nn th tch gim v do t khi tng. Kt qu l nc nh hn nc.

33


34/36

O

H

H

H

H

O

O

H

O

H

H

O

HH

H

Cau truc tdien cua tinh thenc aCu 18

1. Phn t H2O v H2S u l phn t c gc v chng thuc dng AX2E2. Trng thi lai ha ca oxi v lu hunh u l sp3. Oxi c m in ln hn lu hunh, my electron lin kt b ht mnh v pha nguyn ttrung tm s y nhau nhiu hn, lm tng gc lin kt. V vy gc lin kt trong phn t H 2Oln hn gc lin kt trong phn t H2S.2. Nhn xt: T u n cui chu k bn knh nguyn t gim dn.Gii thch: Trong chu k, s lp electron nh nhau nhng do in tch ht nhn tng dn, selectron lp ngoi cng tng dn, lm cho lc ht gia ht nhn vi lp ngoi cng mnh dndn n bn knh nguyn t gim.

3. Cu hnh MO ca phn t N2: (2s)2 (2s*)2 (x)2 = (y)2 (z)2 bc lin kt = 3Cu hnh MO ca phn t CO: (2s)2 (2s*)2 (x)2 = (y)2 (z)2 bc lin kt = 3Lin kt trong phn t N2 v CO rt ging nhau dn n mt s tnh cht vt l ging nhau.

4. Cc hp chthydrohalogenua: HF HCl HBr HI bn phn t gim t HF n HI v m in gim v bn knh nguyn t tng t F n I. Tnh kh tng t HF n HI v m in gim t F n I lm cho kh nng nhng electrontng t F1 n Cl1.

II- BAI T P TGIAI

Cu 1. Xc nh hnh dng ca cc phn t sau: COS , ClNO , COCl2, SeF4 , BF4- , BF3 , H3S+ , XeF4 ,

HCN , SbF3 , CH3+

, CH3-

, SeF6 , ICl3 , SbF5 , NO2F , O2XeF2.Cu 2. .Cho bit : Sn ( Z = 50)a/ Hy vit cu hnh v s phn b electron trong cc obitan ha tr ca Sn

b/ Hy v cu trc hnh hc ca SnCl4 .Trn c s hy gii thch ti sao SnCl4 l mt axitLewis.

c/ Vit cc phn ng ca SnCl4 vi Br-( tl mol 1:1 v 1:2 ) v vi etilen iamin .Hy s dngthuyt s y cc cp electron ha tr(VSEPR) v cu trc khng gian ca cc sn phm ca cc

phn ng .d/ Trn c s cu trc hy cho bit trng thi lai ha ca Sn(IV) trong cc tiu phn .

Cu 3. Cho cc phn t XeF2, XeF4, XeOF4, XeO2F2.a/ Vit cng thc cu to Li - uyt (Lewis) cho tng phn t.

b/ p dng quy tc y gia cc cp electron ho tr, hy d on cu trc hnh hc ca cc phnt .

c/ Hy cho bit kiu lai ho ca nguyn t trung tm trong mi phn t trn.Cu 4. a/ Xc nh trng thi lai ha v cu trc phn t ca NH3 v NF3. So snh gc lin kt HNHv FNF. Gii thch?

b/ Da vo cu trc phn t trn hy gii thch ti sao trong iu kin thng NH3 l cht khkhng mu v ha lng ti -33,350C v ha rn ti -77,750C cn NF3 l cht kh khng mu ha lng -1290C v ha rn ti -2090C.

c/ So snh tnh baz v tnh kh ca NH3 v NF3 .d/ So snh tnh bn nhit ca NH3 v NF3 .

Cu 5. Cho bit hnh dng ca NH3 v H3O+.

b/ Trn c s ca thuyt lai ha hy gii thch hnh dng ( cu trc) ca NH3 v ca H3O+. Sosnh gc to bi gia HNH v HOH.Cu 6. Cho Xe (Z=54) ; P ( Z = 15) ; F ( Z = 9) v O ( Z = 8 ).

34


35/36

a/ Vit cu hnh electron ca Xe v xc dnh v tr ca Xe trong bng HTTH.b/ Xc nh cu trc hnh hc ca cc phn t v ion sau :

XeF2 ; XeF4 ; XeO4 , [XeF4]2+ ; PF -6 v PBr+4 .Gii thch cu trc va xc nh.

Cu 7. Hy cho bit cu to Lewis, dng lai ha( nu c ) , hnh dng theo m hnh VSEPR. Momenlng cc ( = 0 hay 0 ) ca mi phn t sau: SF4, HClO2 , HClO , ICl-4 , IF7 , BrF5, HNO3 vC2H6.Cu 8. Thc nghim cho bit phn t amoniac c cu trc hnh chp trong gc lin kt HNH = 1070.Cho bit s hiu nguyn t ca N = 7 v ca H = 1.

1/ Trn c s ca thuyt lai ha hy gii thch kt qu thc nghim trn.2/ Khi NH3 kt hp proton chuyn thnh ion NH+4. Xc nh cu trc ca NH+4 v so snh gc

lin kt ca HNH trong phn t NH3 vi trong ion NH+4 . Gii thch.3/ Xt phn ng : N2(k) + 3H2(k) 2NH3(k) (1)Ti iu kin chun i vi cc cht c:

S0

,298 puK = -197,9 J/K v H0

,298 puK = -91,8 kJDa vo cu trc v c im lin kt trong cc phn t N2 , H2 v NH3 hy gii thch ti sao

phn ng (1) li ta nhit.Cu 9. Cc hp cht halogen - halogen.

Tt c cc hp cht gia cc halogen cc cng thc chung l XX n trong X c m in

ln hn v n l cc s phi tr bng 1, 3, 5 v 7.1/ Gii thch ti sao ch s n l l cc s l.2/ Clo v brom u to c vi Flo hp cht l ClF3 v BrF3 . C hai phn t ny u cu trc

hnh ch T. Trong gc lin kt FBrF = 86012,6 cn gc lin kt FClF l 87029Gii thch gc lin kt v cho bit trng thi lai ha ca clo v brom trong cc phn t .3/ ICl3 iu kin thng l cht rn mu vng gm nhng phn t ime ( ICl3)2 do 2 phn t

ICl3 cng c cu trc hnh ch T trng hp m thnh. Xc nh bn cht lin kt ca cc nguyn ttrong cc phn t .

4/ Gii thch ti sao Iot c th to vi flo hp cht IF 7 song li khng to c vi clo hp chtICl7. Trn c s ca l thuyt v sc y ca cc cp electron ha tr ( VSEPR). Hy xc nh cu trckhng gian ca phn t IF7 .

5/ Ti nhit khong 3500C, IF7 phn hy thnh IF5 v F2 . Trn c s ca thuyt VB hy xcnh trng thi lai ha ca Iot trong phn t IF5 v cu trc phn t ca IF5 .Cu 10. Hy gii thch lin kt ha hc trong cc phn t sau theo thuyt lai ha, dng m hnhVSEPR

a/ NH3; b/ H2O; c/ CH4; d/ C2H2 ; e/ C2H4 f/ C6H6.Cu 11. PF3 l cht kh khng mu, phn hy chm, c to ra khi nung PF5 nhit trn 4000C.

1/ S dng m hnh VSEPR, hy d on cu trc hnh hc ca PF3 v PF5. Trn c s chobit trng thi lai ha ca P trong cc phn t .

2/ Khi cho PF3 tc dng vi O2 to ra cht kh X duy nht. Xc nh cng thc ca kh X v cutrc khng gian ca phn t X. Hy cho bit trng thi lai ha ca photpho trong phn t v m t cclin kt to thnh trong phn t .

3/ Cho PF3 tc dng vi Cl2 thu c PCl2F3 (Y).a/ S dng m hnh VSEPR v thuyt VB, hy xc nh cu trc khng gian ca Y v trng thi

lai ha ca P trong phn t PCl2F3.b/ Trn thc t. trong khong nhit t 00C n 1400C, PCl2F3 tn ti dng hp cht ion l

[PCl+4][PF -6] . Xc nh cu trc ca cc ion trong phn t PCl 2F3 v cho bit trng thi lai ha ca Ptrong cc ion .Cu 12. Hp cht photpho petahalogenua c nhng c im cu to khc vi d on thng thng. trng thi lng, chng tn ti dng hp cht ion to nn t 1 cation in tch 1+ v 1 anion in tch1-.

1/ Photphopentaclorual cht rn mu trng, km bn nhit. ti trng thi rn c thnh phn lP2Cl10. Hy xc nh cc ion trong phn t P2Cl10. Cho bit trng thi lai ha ca photpho trong cc hpcht .

2/ Hp cht ion A c cng thc n gin l PBr2F3.

35


36/36

a/ Xc nh cc ng phn c th c ca phn t A.b/ Hy cho bit cng thc ng vi ng phn no bn nht? km bn nht? Gii thch?3/ Photphopentabromua l hp cht ion c cng thc phn t l PBr5. Xc nh cu trc ca

phn t ny v cho bit trng thi lai ha ca photpho trong phn t ny?Hy cho bit ti sao photphopentabromua khng th c cng thc phn t l P2Br10.

Cu 13. Khi cho AsF3 tc dng vi Cl2 nhit 500C - 600C sau lm lnh n -40 0C, ngi ta thuc hp cht ion X c cng thc l As2Cl4F6.

a/ Xc nh cc ng phn c th c ca X.b/ Xc nh ng phn bn nht ? ng phn km bn nht ? Gii thch ?

Cu 14. i lu hunh i florua l kh khng mu tn ti 2 dng ng phn l (-SF)2 v S(S)F2.1/ Xc nh cu trc ca 2 dng ng phn . Cho bit trng thi lai ha ca lu hunh trong

cc ng phn .2/ So snh bn ca 2 dng ng phn . Gii thch?3/ So snh nhit si v nhit nng chy ca 2 ng phn . Gii thch?3/ Khi un nng nhit >1800C, S2F2 phn hy thnh SF4 v S.a/ Vit phng trnh phn ng xy ra .

b/ Da vo m hnh VSEPR v thuyt VB, hy xc nh cu trc khng gian ca SF4. Trn c s hy cho bit trng thi lai ha ca nguyn t lu hunh trong phn t .

c/ So snh di cc lin kt ca lu hunh v flo trong phn t . Gii thch?4/ Khi cho SF4 tc dng vi BF3 v vi CsF, ngi ta thu c cc hp cht ion A v B theo s

sau : SF4(l) + BF3 [SF3][BF4] (1)(A)

SF4(l) + CsF Cs[SF5] (2)(B)

SF4 + Cl2 + CsF Cs[SClF5] (3)(E)

Xc nh cu trc ca cc ion trong phn t A v ca anion tron phn t B, E. Trong cc phnt, lu hunh trng thi lai ha g?

5/ Tin hnh in phn SF4 trong HF lng ngi ta thu c kh H 2 v SF6. Vit cc phngtrnh phn ng xy ra ti cc in cc v phn ng chung trong bnh in phn.

6/ Xc nh cu trc khng gian ca SF6. Trng thi lai ha ca lu hunh trong cc hp cht .

hsg cau tao chat

Documents