2011 -Cau Tao Chat

7/31/2019 2011 -Cau Tao Chat

1/38

TT LUYN THI & BI DNG KIN THC NGY MI 18A/88INH VN T - TP. HI DNG

HOAHOC.ORG NG XUN QUNH09798.17.8.85 09367.17.8.85 - netthubuon Website: www.hoahoc.org - [email protected]

PHN CU TO CHT

Trng THPT chuyn Thi Bnh

A- MT S VN CHUNG V L 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 dn n

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 khng gian ca

phn t vi di lin kt v gc xc nh.

1) Nng lng lin kt

Nng 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 lng linkt 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 hai

nguyn t H kt hp vi nhau: 2 H H2 to ra mt nng lng bng 436 kJ.mol-1. Nh vy hai gi tr

nng lng bng nhau v gi tr v ngc nhau v du. Quy c rng nng lng lin kt c du dng

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 ra lin 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 di lin

kt. Tr s di lin kt trong khong t 0,74 (phn t H2) n 4,47 (phn t CS2); thng thng

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. Chng hn

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 trong khong di

mt lin kt C-C l 1,54 v di mt lin kt i C=C l 1,34. di lin kt cng nh, lin kt cng

bn.Bn knh lin kt: T cc s liu c th thy rng di lin kt dAB xp x bng 1/2(dAA + dBB) vi dAA,

dBBl 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 dC-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 ht nhn

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 104,5

o,

7/31/2019 2011 -Cau Tao Chat

2/38

TI LIU LUYN THI & BI DNG KIN THC NM 2011


- Phn t t din, gc lin kt bng 109o28, 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 gc nh

din hay gc xon (hay gc vn). Di y l hnh nh mt s phn t cho thy chng c kch thc ring.

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 tavn cp ring tng dng , hai dng u thng c cp n nhiuhn.

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 hc Cxen(Kossel) v Liuyx (Lewis) a ra nhn xt m ngy nay gi l quy tc bt t (hay quy tc octet): Khi tolin kt ho hc, cc nguyn t c xu hng t ti cu hnh lp ngoi cng bn vng ca nguyn t khtr vi 8e.

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

7/31/2019 2011 -Cau Tao Chat

3/38



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 nguyn t ctnh phi kim mnh hn (hay c m in ln hn). l lin kt ho hc trong cc phn t hp 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 trLin 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 vi mtnguyn t Cl. Do vy s nguyn t lin kt vi mt nguyn t cho trc b hn ch bi ho tr ca nguynt .

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 cc nguynt v cc k hiu ch s phn b electron ho tr c gi l cng thc Lewis (do Lewis xng). Thngthng cc cp electron lin kt vit bng du vch, electron khng lin kt biu din bng chm. Cngthc Lewis khng ch dng cho cc hp cht c lin kt cng ho tr m dng c cho cc hp cht c linkt 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 t c soxi ho cao nht); cc nguyn t khc v c cp electron khng lin kt ca nguyn t trung tm c gil phi t. V d: trong phn t NH3, nguyn t trung tm l N, phi t gm 3H v 1 cp e khng lin ktca N ( v ho tr). Trong phn t HCN, nguyn t trung tm l C, phi t gm 1H v 1N ( y khngc 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 cng thcho hc c xt) gm ht nhn v cc electron cc lp bn trong. V d: Xt li nguyn t ca cc

nguyn t trong HCN ta c: li nguyn t N gm ht nhn v hai e phn lp 1s

2

; li nguyn t C gmht nhn v 2 electron phn lp 1s2; li nguyn t H ch gm ht nhn, thc t H thng c coi ltrng hp ngoi l.

7/31/2019 2011 -Cau Tao Chat

4/38



+ 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 kt vi 4H.

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: HCN

Bc 1: Vit cng thc cu to s b ca cht da vo ho tr ca cc nguyn t v gi thit rng ch clin kt n c hnh thnh. Nu cha bit th t lin kt gia cc nguyn t, hy dng gi thit vitth 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 4e

N: 1s2 2s2 2p3 5eVy n1 = (1 + 4 + 5) e = 10 e

Ch : 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 n4.

Khi p dng ba bc trn cho HCN. n2 = 4e, vy n3 = n1 n2 = 6e.Trong (a), N m in hn C nn phi to bt t cho N. Trong cng thc ban u N mi c 2e, n cn 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 chu k 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.

7/31/2019 2011 -Cau Tao Chat

5/38



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 cp elin 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 tch2:n1 = (6 x 3 + 4 x 1 + 2) e = 24 e

T (a) c n2 = 6e n3 = n1 n2 = (24 6) e = 18e

+ S e cn to bt t cho 3 O l n4 = 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 8

electron v thu c 3 cng thc cu to tng ng nhau vi 1 lin kt i C = O v mt in tch hnhthc cho mi nguyn t oxi:

(I) (II)(III)

Tnh in tch hnh thc trn mi nguyn t.C: 4 4 = 0O: 6 6 1 = -1 (c 2 O)O: 6 4 2 = 0

Vy 3 cng thc Lewis u ng cho CO32-.

+ Thc nghim cho bit ion CO32- c cu to phng, 3 nguyn t O 3 nh ca tam gic u, gc OCO

120o, 3 lin kt C-O c di u bng nhau, l 131 pm.

O : C : O..O

: O : : C : O :..

O..: :

.. ..

..

O : C : O..O

: :

: :

..

..

..

..

..

: O : C : :O..

O..: :

.. ..

....

: O : C :O :

O:

.... ....

:

7/31/2019 2011 -Cau Tao Chat

6/38



gii thch kt qu ny, ngi ta gi thit rng c s cng hng gia 3 cu to (I), (II), (III) vinhau. Ba cng thc Lewis tng ng nhau; chng c gi l cc cng thc cnghng ca CO3

2-.Gi thuyt v s cng hng cc cu to c th p dng cho cc trng hp ion v phn t khc nh

NO3-, SO4

2-, 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 CO3

2- 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 hohc vi quan nim gii to electron gp phn khng nh vai tr ng ch ca cng thc cu to Lewistrong 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 li

n 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 to cng

hng chia cho s cu to cng hng.p dng cho CO3

2-: Chn bt c mt lin kt no trong ba lin kt gia C vi O, ta u thy tng bc calin 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 CO3

2- 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 (bng2) 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. Nhtrn 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).

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 to ra lin kt. Lin kt cng ho tr l lin kt c nh hng khng gian lm cho phn t c hnh dng nht nh ctrng 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 dlin 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 A cnc m cp e khng lin kt, mi cp c k hiu l E, ta c k hiu AXnEm. M hnh VSEPR xt s phn

b khng gian gia A vi X, vi E. Coi nguyn t trung tm A c dng cu. Tm ca hnh cu l ht nhnnguyn t A v cc electron phi ho tr bn trong (li), v qu cu l cc e lp ngoi cng (e ho tr). Micp e ho tr chim mt khong khng gian no ca qu cu.

7/31/2019 2011 -Cau Tao Chat

7/38



Nh vy, mt mc nht nh, hnh dng ca phn t ph thuc vo khong khng gian chim bicc e ho tr ca nguyn t trung tm A. Hnh dng phn t ph thuc ch yu vo s phn b cc cp ehay 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 haykhng 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 nhtgia chng.

C s khng tng ng gia cp e lin kt v cp e khng lin kt. i e lin kt chu lc ht ngthi ca hai ht nhn nguyn t A v X to ra lin kt nn chuyn ng ch yu vng khng gian

gia hai ht nhn. Trong khi , cp e khng lin kt ch chu lc ht ca ht nhn A nn c th chuynng ra xa hn. Kt qu l cp e khng lin kt chim khong khng gian rng hn so vi khong khng

gian chim bi cp e lin kt.

+ Obitan c cp electron khng li

n kt chim khng gian ln hn so vi obitan cha cp electron li

nkt v th sc y gia cc cp electron lin kt gim hn so vi cp khng lin kt. Th d gc lin kttrong cc phn t CH4, NH3 v H2O tng ng bng 109

o28, 107o v 104,5o do cc phn t c s cpelectron 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. gc linkt trong NF3 ch l 102

o so vi 107o ca NH3. Tng t gc lin kt tgim trong dy: PI3 (102o), PBr3

(101,5o), PCl3 (100,3o) v PF3 (97,8

o).

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 h nh tam gic u, phng. GcXAX 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. Gc XAXbng 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 nhau ta c

hnh lng thp tam gic. V d minh ho: phn t PCl5 c phn b lng chp tam gic nn trong phn tc ba loi gc lin kt: trong mt phng c gc ClPCl bng 120o, trong khi cn c gc ClPCl bng 90o vgc 180o. Theo m hnh VSEPR, nm cp e c phn b trn mt cu v ho tr nh sau: Ba cp cngvi 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 onthng vung gc vi mt phng tam gic u ti A. Hai cp e ny to ra hai lin 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 timt lng thp tam gic y chung l tam gic u tm A, hai nh 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 X vinhau ta c hnh bt din.V d minh ho: phn t SF6. Su cp e ca phn t ny c phn b trn v hotr ca nguyn t A su nh ca mt hnh bt din u. Trong trng hp ny khng c s phn bitgia 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

7/31/2019 2011 -Cau Tao Chat

8/38



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

1) (n + m) 4* AX2E:

Ba nguyn t A, X, X khng cn nm trn cng ng thng nh trong trng hp AX2. Lc ny phn

t c dng gp khc. V d: SnCl2, gc ClSnCl 120o. C th dng phn t SO2 lm th d vi s y cahai 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 t khc

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 bindng 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. V d:

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 linkt (E), kt qu to ra phn t hnh ch T. V d: ClF3, 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 trongtrc gn vung gc vi mt phng trn. Do 5 cp e lin kt (hay phn AX5) to ra hnh thp vung.Trong cu to ny, lin kt ngang v lin kt trc khng tng ng hnh hc. Chng hn trong BF5, linkt 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 (C2H2,

C2H4,...); m in ca cc nguyn t l phi t cng c nh hng n hnh dng phn t. Hnh dngphn 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 hnhthng i vi hp cht ca Be cn i vi hp cht ca Ba u c cu to gp khc. Nguyn nhn l donh 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 in tch cao dnn vic gim th nng ca h, ngha l lin kt c hnh thnh.

7/31/2019 2011 -Cau Tao Chat

9/38



b) Lin kt c hnh thnh theo phng trong khng gian m kh nng che ph gia cc obitannguyn 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 hnh thnhphn t H2 t 2 nguyn t H. Mc xen ph 2 AO hay 2 m my e ca 2 nguyn t tham 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 hng ho trca nguyn t tham gia lin kt th hin s c tr s xc nh ca gc lin kt (gc ho tr) trong phn tnhiu nguyn t.

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

othuyt 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] 3s2 3p4. Nh vy S c 6e ho tr phn b trong 4 AO: 4px, 4py v 4pz. Hc mt electron duy nht trn obitan 1s1. Theo thuyt VB, 2e c thn ca S to 2 lin kt S-H vi 2eca 2 nguyn t H. 2 AO c 2 e c thn ny (chng hn l AO-3px, AO-3py) c trc vung gc vinhau, tc l gc gia hai trc ca 2 AO l 90o. V AO-s c i xng cu nn vng c s xen ph ca 1AO-s vi 1 AO-3p s phn b trn trc ca AO-3p tng ng. iu ny c ngha l thng thng gc ca

phn t H2S, tc HSH bng 90o. Tuy nhin hai vng c mt e cao cnh nhau s c s y nhau. S

y ny lm cho gc HSH m rng ra. Kt qu l gc bng 92onh kt qu thc nghim.+ Cng phn nhm chnh th 6 vi S cn c Se, Te u to hp cht vi H2c gc tng t nh trn,

c th l H2Se c gc HSeH 91o, H2Te c gc HTeH 90

o.- H2O

Thc nghim thu c gc HOH 104,5o+ O cng nhm chnh th 6 vi S. Cc e ho tr v cc 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 trong H2S c. Mun c s che

ph cc i gia cc obitan 1s ca H v 2py, 2pz ca O th ccnguyn t H phi tin li nguyn t O theo phng ca trc y v trcz. Nh vy th gc ho tr HOH phi l 90o. Gc thc nghim thuc 104,5o 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 sp

3. Trong 4 AO-sp3 c 2 AO m mi AO c 2e (1 cp) c gi l cp ekhng lin kt ca O, 2 AO-sp3 cn li th mi AO ch c 1e. Theo thuyt VB, to 2 lin kt O-H cnc s xen ph 2 AO-1s ca H vi 2 AO-sp3 m mi AO mi c 1e. Vy th gc lin kt (gc ho tr) 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-sp3

cn li, kt qu taquan st c gc l 104,5o.

7/31/2019 2011 -Cau Tao Chat

10/38



Nh vy mc d cn rt nh tnh, thuyt ho tr nh hng gii thch c kt qu thc nghim vgc lin kt ca H2O.Theo thuyt lin kt ho tr, lin kt cng ho tr c hnh dung l s ghp ielectron ca hai nguyn t lin kt v lin kt s nm theo phng m cc obitan nguyn t che ph nhaunhiu nht. Vy c th xc nh phng ca lin kt da vo ch tiu che ph cc i gia cc obitannguyn t, ngha l theo gi tr cc i ca cc obitan nguyn t. Xc nh c phng ca lin kt l cth xc nh c hnh dng ca phn t. y l mt trong nhng thnh tu ln nht ca 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

7/31/2019 2011 -Cau Tao Chat

11/38



b) Lai ho sp- Lai ho sp l lai ho trong AO 2s t hp tuyn tnh vi 1 AO 2pz(thng dng AO z) to ra 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 2pzc i xng i vi trc z. AO - 2s c i xng cu. T hp tuyn tnh 2 AO to ra 2 AOmi cng nm trn trc z; mi AO mi ny c phn m rng, phn b thu hp. C 2 AO lai ho sp u

nm trn cng mt ng thng: trc z. Do ngi ta gi lai ho sp l lai ha thng, k hiu l hi. Vy tac 2 AO lai ha sp l hi1, hi2. Gc gia 2 trc 2 AO - sp l 180

o.- Biu thc ca 2 AO lai ho sp l:

hi1 = (2s + 2pz)/ 2

hi2 = (2s - 2pz)/ 2 - Lai ho sp c dng gii thch lin kt ho hc trong phn t BeH2, C2H2 (ankin) . Cc phn

t u c dng thng.Lai ho sp

2: 1AO-s+2 AO-p = 3AO-sp2

3AO-sp2 l: t1 =3

1

(s+ 2 px)

t2 =6

1( 2 s-px+ 3 py)

t3 =6

1( 2 s-px- 3 py)

c) Lai ho sp2- Lai ho sp2l lai ho trong AO - 2s t hp tuyn tnh vi 2 AO - 2p (thng chn l AO 2pz, AO

2py) to ra 3 AO mi c cng nng lng. Tng t nh AO lai ho sp, AO-sp2 cng b bin dng 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:

1AOs + 1AOp

7/31/2019 2011 -Cau Tao Chat

12/38



Ba AO-sp2 cng nm trong mt mt phng, gc to bi hai trc ca hai AO cnh nhau l 120o. Do laiho sp2c 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 =

43

cos cos ; py =43

cos sin ; pZ =43

cos

- Lai ho sp2c p dng gii thch lin kt ho hc trong cc anken CnH2n (n2), benzen, ... Ccphn 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 nnglng c

gi l lai ho sp3.

- C th hnh dung qu trnh lai ho sp3

nh sau:

- Hnh dng ca mi AO lai ho sp3 cng tng t nh hnh dng AO lai ho sp, sp2 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 ccAO lai ho. Do lai ho sp3c gi l lai ha t din. AO-sp3c k hiu l te. Vy ta c 4 AO-sp

3 lte1, te2, te3, te4.

- Lai ho sp3c p dng gii thch lin kt ho hc trong cc ankan CnH2n+2 (n1), in hnh ltrong CH4. Ngoi ra n cng c p dng trong cc trng hp khc nhau H2O, NH4

+,... cc phn t cgc 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 lai ho Gc gia cc obitan lai ho Dng phn t V dsp

sp2sp3dsp2sp3d2

180o

120o109o2890o90o

ng thng

Tam gic uT din uHnh vungBt din u

BeH2, BeCl2, ZnCl2, CO2

BF3, NO3-, CO32-CH4, CCl4, NH4

+, ClO4-, SO4

2-, PO43-

PtCl4, CuCl42-, Ni(CN)4

2-SF6, AlF6

3-, SiF62-

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

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

7/31/2019 2011 -Cau Tao Chat

13/38



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

Nhm IVA CH4 SiF62- GeF6

2- SnF84- PbF8

4-CO32- SiO4

4- GeO44- SnO6

8- PbO68-

Nhm VIA H2SO4 H6TeO6Nhm VIIA HClO4 H5IO6

Tuy 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 hosp3 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 lin kt trong cc 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 O2:Nguyn t O c cu hnh electron: 2s2 2p4

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

Cng thc cu to phn t vi lin kt cp gia hai nguyn t O ph hp nng lng ca lin kt l 494kJ/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 phnt O2 c cu to:

Trong ngoi lin kt cng ho tr bnh thng c to nn bng cp electron (vch lin) cn c hailin kt c to nn nh ba electron (vch ri). Lin kt ba electron cn c gi l lin kt mt electronv 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 p.t gii thch mt cch ngin 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 .

7/31/2019 2011 -Cau Tao Chat

14/38



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

lin kt .

Nh vy, lin kt trong phn t l lin kt ba. Lin kt ba ny c nng lng rt ln (942 kJ/mol) nnphn 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 calin kt N-O l 1,44A, ngha l trung gian gia di lin kt i trong ion NO- (1,18A) v ca lin kt batrong ion NO+ (1,06A). Vy bc ca lin kt trong NO khng th bng 2 m bng 2,5, ngha l phn t ccu to:

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

- c mt trongNaNO).

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 t gnnh khng c cc, v nng lng ca lin kt rt ln (1070 kJ/mol), ln hn c nng lng ca lin kt

ba trong N2.Bi vy, ngy nay ngi ta cho rng nguyn nhn lm gim mnh phn cc ca CO l nguyn t O

a ra mt cp electron ca mnh dng chung vi nguyn t C, ngha l to thnh mt lin kt cho -nhn ngoi hai lin kt cng ho tr bnh thng:

Nh vy, lin kt C-O l l.kt ba. Tht vy, lin kt ny c nng lng rt ln so vi cc l.kt ho hc.Qua cu to ca cc phn t xt trn y, ta thy i vi phn t ch c hai nguyn t, hnh dng ca

phn t lun lun l ng thng ni lin hai ht nhn nguyn t. Nhng i vi nhng phn t gm banguyn t hay hn na, gii thch hnh dng ca chng, cn vn dng s lai ho cc obitan ca nguynt 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 kim cng lcht cng nht.

7/31/2019 2011 -Cau Tao Chat

15/38



Trong lp than ch, mi nguyn t C trng thi lai ho sp2 lin kt cng ho tr vi ba nguyn t C baoquanh cng nm trong mt mt phng to thnh vng 6 cnh. Nhng vng ny lin kt vi nhau to thnhmt lp v tn. Sau khi to thnh nhng lin ktnh vy, mi nguyn t C cn c mt electron c thntrn 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 vng benzen(1,39A) c bi l 1,5, ngha l bi ca lin kt C-C trong lp than ch ~ 1,33. Nhng khc vi

benzen, lin kt trong lp tinh th than ch l khng nh ch trong ton lp. Bi vy khc hn vi kimcng, 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

7/31/2019 2011 -Cau Tao Chat

16/38



Mt 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 cng thccu to ca phn t O3 l:

Ngy nay n gin, ngi ta hay dng mt cng thc linh hot hn, trong lin kt c chia icho 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

7/31/2019 2011 -Cau Tao Chat

17/38



Nhng vic rt ngn mnh di ca lin kt S-O (1,43A) trong SO2 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 h

nh electron trng thi kch

thch, 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 c thn

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 (gi s

bn tri ca hnh v) c electron c thn to thnh mt lin kt v mt obitan 3d khng lai ho ca Sc electron c thn che ph vi obitan 2p khc ca O (gi s bn phi hnh v) c electron c 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 O3 v SO2, ngha lNO2 c cng thc cu to:

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:

7/31/2019 2011 -Cau Tao Chat

18/38



Nu mt cch gn ng ngi ta tha nhn rng trong phn t SO3 ch nhng obitan 3s v 3p ca S tham

gia to thnh lin kt th cch m t s to thnh lin kt gn tng t nh i vi phn t O3.

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:

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

7/31/2019 2011 -Cau Tao Chat

19/38



ca S, mi mt c mt electron c thn che ph vi obitan 2p cn li c electron c thn ca ba nguynt 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 ng.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 mt electronc 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, ng.t N nm trung tm v bn ng.t H nm nh ca t din:

Trong ion NH4+, s to thnh ba lin kt cng ho tr bi ba cp electron chung gia N v H xy ra

tng t nh trong phn t NH3. Ch khc y l obitan lai ho c cp electron che ph vi obitan 1strng electron ca H+ to thnh lin kt cho nhn. Tuy nhin c bn lin kt N - H u ging nhau v ul lin kt cng ho tr to nn bi cp electron:

7/31/2019 2011 -Cau Tao Chat

20/38



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 cp

electron khng lin kt cn hai obitan lai ho cn li, mi mt c mt electron c thn che ph vi obitan1s 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 thn ca

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 c electron c

thn ca hai nguyn t O to thnh lin kt :

v phn t c cng thc cu to:

trong C c ho tr 4.

7/31/2019 2011 -Cau Tao Chat

21/38



C- BI TP P DNG

I-BI TP C HNG DN GII

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 cc haykhng phn cc ?3.Hy gii thch ti sao PbI2 ( cht rn mu vng) tan d dng trong nc nng, v khi ngui li kt tadi 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 v dnghnh 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 mang inl 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 cngtrong 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 AB2c 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 , O2

2 theo thuyt MO (cu hnh electron, cng thc cu to).

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 lc nhanh ktta 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 cht trn.

CU 5: Cho bit trng thi lai ho ca ngt trung tm v dng hnh hc ca cc phn t sau :H2O , H2S , H2Se , H2Te .

- Hy sp xp theo chiu tng dn ln gc lin kt v gii thch s sp xp .- Ti sao iu kin thng H2O th lng,cn H2S , H2Se , H2Te th kh?- Hy sp xp theo chiu tng dn tnh kh ca cc cht trn.Gii 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.

7/31/2019 2011 -Cau Tao Chat

22/38



Hp cht A c cng thc XYn, c c im:- X chim 15,0486% v khi lng- 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, Yb. Bit X, Y to vi nhau hai hp cht l A, B. Vit cu trc hnh hc v cho bit trng thi lai ho

ca nguyn t trung tm ca A, B.c. Vit cc phng trnh phn ng gia A vi P2O5 v vi H2O

Vit cc phng trnh phn ng gia B vi O2 v vi H2O2. 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 11,26 14,54 13,61 17,41 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 nguynt 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 cng c 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 XH3

2.1. Hy cho bit cu hnh hnh hc ca cc phn t PH3 v AsH3.

7/31/2019 2011 -Cau Tao Chat

23/38



103o15

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: 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 = -1/2

Z: n = 2, l = 1, m = 0, s = +1/21 Xc nh A, X, Z.2 Cho bit trng thi lai ho v cu trc hnh hc ca cc phn t v ion sau: ZA2, AX2, AX3

2-, AX42-.

3 Bng thuyt lai ho gii thch s to thnh phn t ZX. Gii thch v sao ZX c moment lng cc b. Giithch s hnh thnh lin kt trong phn t phc trung ho Fe(CO)5 bng thuyt VB.4 Gii thch v sao AX3

2- 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, AlCl4

.

AlCl3 + Cl AlCl 4 (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 OCl Cl Cl Cl

(b)

O OF 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 cho 4,29gam Y tc dng vi lng d X th thu c 18,26 gam sn phm c cng thc l XY. Hy xcnh 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 1 trt 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 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 electron cui

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, HI

Hy 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

111o103ov

111ov

7/31/2019 2011 -Cau Tao Chat

24/38



Cu 15:a. Hy cho bit cu trc hnh hc , kiu lai ha ca cc phn t : SF6 , 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 sau

ClF3 + A B + Cl2c. 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 )1.Xc nh s hiu nguyn t, gi tn nguyn t A. Vit cng thc electron, cng thc cu to ca phn tA2. Kim chng s lin kt v tnh cht thun t ca A2 bng cu hnh electron ca phn t.2. Ion A3B

2- v A3C2- ln lt c 42 v 32 electron

2.1. Tm 2 nguyn t B v C ( s hiu nguyn t, tn, k hiu )2.2. Dung dch mui ca A3B

2- 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 171. 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

Cl

Cl

.

.

(c)

TeCl

Cl

Cl

Cl

(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)

7/31/2019 2011 -Cau Tao Chat

25/38



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 1,13 205

Da 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.

HNG DN GIICu 1.1. a. CH4 > NH3 > H2O Gii 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 i e-lin kt v pha n nhiu hn lm tng ln gc lin kt.c. So snh nhit nng chy ca cc cht:

MgO > NaCl > KClGii 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 < CH3COOHGii 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 C2H5O H O

CH3 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 t trongphn t(dng hnh hc ca phn t)

Momen lng cc (o phn cc) l mt i lng c ln v c chiu.

7/31/2019 2011 -Cau Tao Chat

26/38



Trong phn t, nu cc lin kt phn cc c sp xp i xng nhau, momen lng cc c cng 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 > 0

Cn 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

Khng cc v momen lng cclin kt b trit tiu C cc v lng cc lin ktkhng trit tiu C cc v lng cc lin ktkhng trit tiu

Cu 3

a) 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(2Z

65)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 thng

Trng thi lai ho : sp

Cl ClBe

7/31/2019 2011 -Cau Tao Chat

27/38



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 BeCl2 kia to lin kt cho-nhn. Vy BeCl2 ckhuynh hng polime ho:

Cu 4

1.

Xe

OO

O

Xe lai ha sp 3 phn t

dng thp tam gic

B BHHH

H H H

B lai ha sp 3, phn t B 2H6 gm

2 t din lch c 1 cnh chung,

lin kt BHB l lin kt 3 tm v

ch c 2 electron, 1 electron ca

H v 1 electron ca B.

Al

ClCl

Al

ClCl

Cl

Cl

Al lai ha sp3

, phn t Al 2Cl6 gm 2 t din lchc 1 cnh chung, c 2 lin kt cho nhn -c

to thnh do cp e khng lin kt ca Cl v

obitan trng ca Al.Trong Al 2Cl6 nguyn t Al

t -c cu trc bt t vng bn.

2. O 2 : ( )lk

s

2 ( *s

) 2 )( lkz

2 )( lk

x

2 = )( lky

2 )( *

x

1= )( *y

OO

O 22 : ( )lk

s

2 (*

s ) 2 )( lk

z

2 )( lkx

2 = )( lk

y

2 )( *x

2= )( *

y

2OO

2

O

2 c electron c thn nn thun t. O2

2 khng c electron c thn nn ngch t.3.

N

HH

H

Cc vect momen l-ng ccca cc lin kt v cpelectron khng lin kt cng

chiu nn momen l-ng ccca phn t ln nht.

N

FF

F

Cc vect momen l-ng ccca cc lin kt v cpelectron khng lin kt ng-c

chiu nn momen l-ng ccca phn tb hn NH3.

BF

F

F

Phn t dng tam gic uCc vect momen l-ng ccca cc lin kt trit tiu ln

nhau(tng bng khng) phnt khng phn cc.

........

Cl

Be

Cl

Cl

Be

Cl

Cl

Be

Cl

Cl

Be

Cl

Cl

Be

Cl

C phn t Al2Cl6 v nguyn t Al t cutrc bt t vng bn.Khng c phn t B2F6 v: phn t BF3 bndo c lin kt pi khng nh ch c tothnh gia obitan trng ca B vi cpelectron khng lin kt ca F v kch thc

ca nguyn t B b so vi nguyn t F nntng tc y gia 6 nguyn t F ln lmcho phn t B2F6 tr nn km bn.

7/31/2019 2011 -Cau Tao Chat

28/38



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

-3n(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) = 1:1Cng 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

900H2OH2O

H2O

H2O

Bt din u

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

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 cch gia 2 cp elin kt trong phn t H2O l nh nht nn lc y tnh in mnh nht gc lin 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 nngcho 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 : 1s

22s22p63s23p3 nn e cui cng c b bn s lng t l:n =3, l=1, m = +1, s = +1/2

Thay PX = 15; NX = 16 vo (1), (2) ta c nPY = 85; nNY = 90nn: 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

2s2

2p6

3s2

3p5

,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 .

7/31/2019 2011 -Cau Tao Chat

29/38



b. ClA: 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 = POCl3PCl3 + 3H2O = H3PO3 + 3 HCl

2. Nhn xt:a. Nhn chung nng lng ion ho tng dnGii thch: T tri sang phi trong mt chu k, in tch ht nhn ca cc nguyn t tng dn v s e ngoi

cng cng tng thm c in vo lp n ang xy dng d. Kt qu cc e b ht v ht nhn mnh hn lmbn knh nguyn t gim, dn n lc ht ca nhn vi e ngoi cng tng lm e cng kh b tch ra khinguyn t lm nng lng ion ho tng

b.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 ny

N c cu hnh e: 1s2

2s2

2p3

phn lp p bn bo ho, y cng l mt cu hnh bn nn cng cn cung cpnng lng cao hn ph v cu hnh nyCu 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 Oxi3. 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, Y nx, ny l s ntron X, Y

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

AxAx + nAy

=> Ax = 31

= 15,0486100

(4)

7/31/2019 2011 -Cau Tao Chat

30/38



Trong nguyn t X : 2Px Nx = 14Px = 15

Nx = 16Thay Px, Nx vo (1) , (2)

n (Ny Py) = 5 ( 5)2Py ny = 16 (6) => Ny = 2Py - 16n(Py 16) = 5

Py = 5 16nn

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)- C v Si u c bn electron ha tr tuy nhin khc vi CO2 (O = C = O) SiO2 khng phi l mt

phn t n gin vi lin kt Si =O. nng lng ca 2 lin kt i Si=O km xa nng lng ca bn lin ktn 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 bntrong 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 cc

=> X l photpho

hoc Z + H2

7/31/2019 2011 -Cau Tao Chat

31/38



Cu 9

1. 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 : 1s22s22p63s23p3 ; As : 1s22s22p63s23p63d104s24p3

P v As u c 5 electron ha tr v c 3 electron c thn trong XH3X

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 hn so

vi As nn lc y mnh hn.2.3.

N FF

F

Si

ClH

ClClsp3

S

FF

O

O Osp3 sp3 sp2

B

F

FF

Si

F

F

F

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 C2.

Phn t, ion Trng thi lai ho cu nguyn t trung tm Cu trc hnh hcCS2 sp ng thngSO2 sp

2 GcSO 23

sp3 Chp y tam gic u

SO 24 sp

3 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 1obitan p thun chng cha bo ho xen ph pi vi 1 obitan p cha bo ho cn li cu oxi. Oxi dng 1obitan p bo ho xen ph v obitan p trng ca cacbon to lin kt pi kiu pp.

7/31/2019 2011 -Cau Tao Chat

32/38



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 t oxicho sang obitan trng cu nguyn t cacbon lm gim phn cc ca lin kt nn lm gim moment lngcc.+ S hnh thnh lin kt trong phn t Fe(CO)5Fe (Z = 26) [Ar] 3d6 4s2 4p0

Fe* [Ar] 3d8 4s0 4p0

trng thi kch thch, nguyn t Fe dng 1 obitan 3d trng t hp vi 1obitan 4s v 3 obitan 4p to thnh 5obitan lai ho dsp3 trng hng ra 5 nh ca hnh lng chp y tam gic u tm l nguyn t Fe.CO dng cp electron t do cha lin kt trn nguyn t cacbon to lin kt phi tr vi cc obitan lai hotrng ca st to ra phn t phc trung ho Fe(CO)5Fe* [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 S2O3

2- v trn nguyn t S trong SO32- cn c mt cp

electron t do cha lin kt c kh nng cho vo obiatn 3p trng ca nguyn t S to lin kt cho nhn.

2-

S O

S

OO

2-

S O

OO

S

Cu 11 AlCl3 + Cl AlCl

4

(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 hc

Cl Cl

Al Al

Cl Cl Cl ClClTam gic phng T din

7/31/2019 2011 -Cau Tao Chat

33/38



(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.(a). S sai bit gc ho tr trong phn t SCl2 v OCl2 l do s khc bit v m in ca oxi

v lu hunh. ca nguyn t trung tm cng nh th cc cp electron lin kt b y nhiu vpha cc nguyn t lin kt, nn chng chim vng khng gian nh xung quanh nguyn t 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 t linkt. 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 n.t X l: 1s2 2s2 2p6 3s2 3p6 3d10 4s24p6 4d10 5s25p5. Vy ZX = 53 = s proton X

3962,1X

X

Z

N NX = 74 AX = ZX + NX = 53 + 74 = 127

Ta c : 7,3Y

X

N

N NY = 20

Y + X XYMY MXY

4,29 18,26

3926,18

127

29,426,18

29,4

Y

YY

XY

Y MMM

M

M(g/mol)

Vy : ZY + NY = 39 ZY = 19. Cu 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 XCl3 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 = 54,3

2

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

3 = 44,36 ()3

7/31/2019 2011 -Cau Tao Chat

34/38



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: np1Mt 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 -

2

1= 3,5

Vy 4 s lng t ca :

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: 1s2

2s2

2p6

3s2

( Mg ).B: 1s22s22p63s23p1 ( Al ).Cu 14.1) C2H5OH v CH3COOH to c lin kt hidr gia cc phn t nn c nhit si cao hn cccht kia.

t0 si ca CH3COOH > t0 si ca C2H5OH v c

OHHCCOOHCH 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 H nyvi 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 sp2

d c cu hnh vung

2

NCNC

Ni

NCNC

ICl3 : I trng thi lai ha sp3d, c cu lng thp tam gic

XeF4 : 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 + Cl2Cu 16 :

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)

7/31/2019 2011 -Cau Tao Chat

35/38



2.2.1

2.2.

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 OGia 2 nguyn t oxy c 1 lin kt cng ha tr bnh thng v 2 lin kt c bit 3e (3lectronny do 1 nguyn t a ra 1, nguyn t kia a ra 2 gp chung), trong ch c 1 electronc 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 :

2s2

2s*2

z2

x2

y2

x*1

y*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 RbH3

2-

* 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 A3B

2- l SO32-

A3C2- l CO3

2-

SO32- + 2H+ SO2 + H2O

CO32- + 2H+ CO2 + H2O

D l SO2 ; E l CO2- Dng hnh hc phn t :SO2 : nguyn t S trng thi lai ha sp

2 nn phn t c cu to gc

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

SO2 + Br2 + 2H2O = H2SO4 + 2HBr- Kh SO2 c th kt hp vi O2 to SO3do lu hunh trong SO2 cn cp electron t do. CO2khng c kh nng ny do cc-bon trong phn t khng cn electron c thn.

7/31/2019 2011 -Cau Tao Chat

36/38



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:8O

2 > 9F> 11Na

+ > 12Mg2+ > 13Al

3+ .- 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 > O2-> 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 lin kt lnh 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 ca mtt din u. Mi nguyn t hidro lin kt chnh vi mt nguyn t oxi v lin kt hidro vi mt nguyn toxi khc. Cu trc ny tng i xp nn c t khi nh. Khi tan thnh nc lng cu trc ny b ph vnn th tch gim v do t khi tng. Kt qu l nc nh hn nc.

O

H

H

H

H

O

O

H

O

H

H

O

HH

H

Cau truc t dien cua tinh the nc a

Cu 181. 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 t trungtm s y nhau nhiu hn, lm tng gc lin kt. V vy gc lin kt trong phn t H2O ln hn gclin 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, s ele ctronlp ngoi cng tng dn, lm cho lc ht gia ht nhn vi lp ngoi cng mnh dn dn 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 = 3

Cu hnh MO ca phn t CO: (2s)2 (2s

*)2 (x)2 = (y)

2 (z)2 bc lin kt = 3

Lin 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 ttng t F n I. Tnh kh tng t HF n HI v m in gim t F n I lm cho kh nng nhng electron tngt F1 n Cl1.

7/31/2019 2011 -Cau Tao Chat

37/38



II-BI TP T GII

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 axit Lewis.

c/ Vit cc phn ng ca SnCl4 vi Br

-

( tl mol 1:1 v 1:2 ) v vi etilen iamin .Hy s dng thuyt sy 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 phn t .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 HNH vFNF. Gii thch?

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

-1290

C v ha rn ti -2090

C.c/ So snh tnh baz v tnh kh ca NH3 v NF3 v So snh tnh bn nhit ca NH3 v NF3 .Cu 5. Cho bit hnh dng ca NH3

v H3O+. Trn c s ca thuyt lai ha hy gii thch hnh dng ( cu

trc) ca NH3 v ca H3O+. So snh gc to bi gia HNH v HOH.

Cu 6. Cho Xe (Z=54) ; P ( Z = 15) ; F ( Z = 9) v O ( Z = 8 ).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. Momen lngcc ( = 0 hay 0 ) ca mi phn t sau: SF4, HClO2 , HClO , ICl

-4 , IF7 , BrF5, HNO3 v C2H6.

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: S 0 ,298 puK = -197,9 J/K v H0

,298 puK = -91,8 kJ

Da vo cu trc v c im lin kt trong cc phn t N2 , H2 v NH3 hy gii thch ti sao phnng (1) li ta nhit.Cu 9. Cc hp cht halogen - halogen.

Tt c cc hp cht gia cc halogen cc cng thc chung l XXntrong X c m in lnhn v n l cc sphi 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 t trong ccphn t .

4/ Gii thch ti sao Iot c th to vi flo hp cht IF7 song li khng to c vi clo hp cht ICl7.Trn c s ca l thuyt v sc y ca cc cp electron ha tr ( VSEPR). Hy xc nh cu trc khng gianca phn t IF7 .

5/ Ti nhit khong 3500C, IF7 phn hy thnh IF5 v F2. Trn c s ca thuyt VB hy xc nhtrng thi lai ha ca Iot trong phn t IF5 v cu trc phn t ca IF5 .

7/31/2019 2011 -Cau Tao Chat

38/38


Cu 10. Hy gii thch lin kt ha hc trong cc phn t sau theo thuyt lai ha, dng m hnh VSEPRa/ 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 cho bittrng 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 cu trckhng gian ca phn t X. Hy cho bit trng thi lai ha ca photpho trong phn t v m t cc lin kt

to thnh trong phn t .3/ Cho PF3 tc dng vi Cl2thu 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 PCl2F3 v cho bit trng thi lai ha ca P trong

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 tch 1-.1/ Photphopentaclorual cht rn mu trng, km bn nhit. ti trng thi rn c thnh phn l P2Cl10. Hyxc nh cc ion trong phn t P2Cl10. Cho bit trng thi lai ha ca photpho trong cc hp cht .

2/ Hp cht ion A c cng thc n gin l PBr2F3.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 vcho 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 50

0C - 600C sau lm lnh n -400C, 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 phn t, lu

hunh trng thi lai ha g?5/ Tin hnh in phn SF4 trong HF lng ngi ta thu c kh H2 v SF6. Vit cc phng trnh

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 .

Documents

2011 -Cau Tao Chat