asdcasd ksadj asd as

5/21/2018 asdcasd ksadj asd as

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Bi ging mn Csl thuyt Ho hc

PHN I: CU TO CHT

CHNG I - CU TO NGUYN T

I. Mu

1. Cc ht cbn to thnh nguyn t:* Nguyn tc:

- Kch thc khong 1 0A ( 10-10m).- Khi lng: 10-23kg.

* Nguyn tgm:

- Ht nhn ( in tch +Z) gm:+ Proton (p), mp =1,672. 10

-27kg, tch in dng + 1,602. 10-19C.

+ Notron(n), mn= 1,675. 10-27kg, khng mang in .

Ht nhn ca cc nguyn tu bn (trcc nguyn tphng x).

- Electron(e) ,me = 9,1. 10-31kg , tch in m - 1,602. 10-19C.

Trong bng hthng tun hon (HTTH), sTT nguyn t= in tch ht nhn = se.

VD: Ca c sTT= 20 => Z=se=20.

2. Thuyt lng t

nh sng l mt sng in tlan truyn trong chn khng vi vn tc c = 3.108m/s, c

c trng bng bc sng hay tn sdao ng:

c= .

Thuyt sng ca nh sng gii thch c nhng hin tng lin quan vi struyn sngnhgiao thoa v nhiu xnhng khng gii thch c nhng dkin thc nghim vshp thv spht ra nh sng khi i qua mi trng vt cht.

Nm 1900, M.Planck a ra gi thuyt: Nng lng ca nh sng khng c tnh chtlin tc m bao gm tng lng ring bit nhnht gi l lng t. Mt lng tca nhsng (gi l phtn) c nng lng l:

E=h

Trong : E l nng lng ca photon : tn sbc xh = 6,626 .10-34J.s - hng sPlanck.

Nm 1905, Anhstanh da vo thuyt lng t gii thch tha ng hin tngquang in. Bn cht ca hin tng quang in l cc kim loi kim trong chn khngkhi b, khi bchiu sng spht ra cc electron; nng lng ca cc electron khng

ph thuc vo cng ca nh sng chiu vo m ph thuc vo tn s nh sng.Anhstanh cho rng khi c chiu ti bmt kim loi, mi photon vi nng lng h struyn nng lng cho kim loi. Mt phn nng lng E0 c dng lm bt electron

ra khi nguyn tkim loi v phn cn li strthnh ng nng 221 mv ca electron:


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20 2

1mvEh +=

Nhng bc x c tn s b hn tn s gii hnh

E00 = s khng gy ra hin tng

quang in.

Sdng cng thc trn ta c thtnh c vn tc ca electron bt ra trong hin tngquang in.3. Cc m hnh nguyn t:

*M hnh nguyn tRutherford:Mi nguyn tc mt ht nhn mang in tch dng

v cc e quay xung quanh.

*M hnh nguyn tBohr:

- Trong nguyn tmi electron quay xung quanh nhn ch theo nhng quo trnng tm c bn knh xc nh.

- Mi quo ng vi mt mc nng lng xc nh ca electron. Quo gn nhnnht ng vi mc nng lng thp nht, quo cng xa nhn ng vi mc nng

lng cng cao. Nng lng ca electron trong nguyn tH2c xc nh nhsau:

22

4

2

0

nn

1.

h

me.

8

1E =

Trong h = 6,626 .10-34J.s - hng sPlanck

m - khi lng ca e

o - hng sin mi trong chn khng o = 8,854.10-12 C2/Jmn - l cc snguyn dng nhn cc gi tr1,2,3...,,

- Khi e chuyn tquo ny sang quo khc th xy ra shp thhoc gii phngnng lng. Khi e chuyn tquo c mc nng lng thp sang mc nng lng

cao hn th n hp thnng lng. Khi electron chuyn tmt mc nng lng cao

sang mc nng lng thp hn th xy ra spht xnng lng. Nng lng ca bc

xhp thhoc gii phng l:

E = En- En = h= ch.

*Kt quv hn chca thuyt Bohr

Kt qu:- Gii thch c quang phvch ca nguyn thyro

- Tnh c bn knh ca nguyn thydro trng thi cbn a= 0,529 A0

Hn ch- Khng gii thch c cc vch quang phca cc nguyn tphc tp


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- Khng gii thch c stch cc vch quang phdi tc dng ca in trng, t

trng

- Githuyt c tnh c on.

* Cc m hnh trn u khng gii thch c 1 svn thc nghim t ra. Nguyn

nhn l do:- Khng cp n tnh cht sng ca electron

- Do coi quo chuyn ng ca electron trong nguyn tl quo trn c bn

knh xc nh.

II. Quan im hin i vcu to nguyn t:

1. Lng tnh sng ht ca cc ht vi mNm 1924 nh vt l hc ngi Php Louis De Broglie a ra githuyt: mi ht

vt cht chuyn ng u c thcoi l qu trnh sng c c trng bng bc sng

v tun theo hthc :

mv

h=

Trong : m - Khi lng ca ht, kg

v - Vn tc chuyn ng ca ht , m/s

h - Hng sPlanck, h= 6,63.10-34J.s

- i vi ht vm: m kh ln (h =const) kh nh -> tnh cht sng c thbqua.

- i vi ht vi m : m nh(h =const) kh kh ln -> khng thbqua tnh chtsng.

V d1: Mt ht c khi lng m = 0,3 kg, vn tc chuyn ng V= 30m/s th ca

ht l?

Gii:

p dng hthc Louis De Broglie

m

3434

1073603030

10636

=== .,.,.,

mv

h

ca ht v cng nhnn bqua tnh cht sng ca ht.2. Nguyn l bt nh Heisenberg

* Pht biu nguyn l

Khng thxc nh ng thi chnh xc ctov vn tc ca ht, do khng th

vc chnh xc quo chuyn ng ca ht.

x. vx m

h

y l hthc bt nh Heisenberg

Trong x- bt nh (sai s) vtotheo phng x


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vx- bt nh (sai s) vvn tc theo phng x

Nu x cng nh th vx cng ln, ngha bt nh vtocng nh th btnh vvn tc cng ln.

Ty rt ra mt kt lun quan trng l khng thdng chc cin m tmt

cch chnh xc quo chuyn ng ca ht vi m nhthuyt ca Bohr m phi sdng mt mn khoa hc mi l: chc lng t.

III. Khi nim vchc lng t

1.Hm sng:

Trng thi chuyn ng ca e trong nguyn tc m tbng mt hm ca to

x,y,z v thi gian t, c gi l hm sng (x,y,z,t).

Trong trng hp t khng i th khng phthuc vo thi gian, c gi l trng

thi dng ca electron. Khi chphthuc vo 3 bin x,y,z.* Tnh cht ca hm sng:

- C thl m, dng hay l 1 hm phc.- 2mt xc sut tm thy electron ti 1 im trong phn khng gian xung quanhht nhn.

- 2dv m txc sut tm thy electron thi im t trong yu tthtch dv baoquanh im c tox,y,z

V electron c mt trong khng gian v hn nn xc sut tm thy n bng 1:

1dv2

=+

L iu kin chun ha hm sng.

2. Phng trnh Schrodinger:

tm ra hm sng ta phi gii phng trnh sng, cn gi l phng trnh

Schrodinger. l phng trnh vi phn ca hm sng i vi ht vi m (eleclectron)chuyn ng trong trng thV:

EVm8

h2

2

=

+

Trong d:2

2

2

2

2

2

zyx

+

+

= - Ton tLaplace

V- L thnng ca ht

E - Nng lng ton phn ca ht

C thvit di dng tng qut hn: H=E, trong H l ton tHamilton ca h

nghin cu.Gii phng trnh sngtm c E, t bit c chuyn ng ca e.


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3. Obitan nguyn tv my electron.

- Mi gi trnghim gi l 1 obitan nguyn t, k hiu l AO.My e c quy c l min khng gian gn ht nhn nguyn t, trong xc sut c

mt electron khong 90%. Mi m my electron c xc nh bng mt bmt gii

hn gm nhng im c cng mt xc sut. m my s l hnh cu. m my pc dng hnh quti, m my d c dng hnh hoa bn cnh.

IV. H1 e ( nguyn tH v ion tng t).

1. Phng trnh sng:

- Hgm 1 e v 1 ht nhn in tch +Ze:Thnng ca h:

V=r

Ze0

2

4

Trong r: khong cch gia ht nhn v e.0 : hng sin mi ca chn khng.

thnng V chthuc vo r => trng to ra l trng xuyn tm ( trng c ixng tm) gi l trng Culng.

Phng trnh Schrodinger c dng:4 0

EZe-m8

h

2

2

2

=

r

- gii phng trnh sng trn a vhta cu: (x,y,z)(r,,)

x

yz

sp

p

z

y

x

p

x

y

z

dx2-y2

z

y

x

zy

xdxz

z

y

xdxy

z

y

xdyz


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- Li gii phng trnh sng Schrodingersthu c l nng lng ton phn ca e(E), hm sng m ttrng thi chuyn ng ca e () v khi gii sxut hin 3 slng tn, l ,m.

2. Nng lng:

* Kt qugii phng trnh sng thu c nng lng ton phn ca e:

2

2

nn

z13,6.E = (eV)

n: c gi trnguyn dng, gi l slng tchnh.

* Nhn xt:

- Ee phthuc vo n+ n cng ln -> Ee cng ln v ngc li.

+ n gin onEe gin on -> nng lng ca e trong nguyn tc phn

thnh tng mc, mi mc ng vi 1 gi trca n.+ Khi n=1 E1min -> mc E1gi l trng thi cbn. Vy trng thi cbn l

trng thi c mc nng lng thp nht.

3. Hm sng:

(x,y,z) = (r,,)

- Khi gii phng trnh sng, dn n vic t hm sng (r,,) thnh tch ca haihm:

(r,,)= Rn,l (r).Ym,l(,)Trong : R(r) - L hm xuyn tm phthuc vo hai tham sn, lY (,) - L hm gc phthuc vo hai tham sl l, m.

l - l slng tph: l = 0,1,2,...,n-1 -> ng vi 1 gi trca n c n gi tr

ca l.

m - l slng tt: m = 0, 1,2,...,l -> ng vi 1 gi trca l c2l + 1 gi trca m.

- Nhvy hm sng thu c phthuc vo 3 slng tl n,l,m : n,l,mhay nicch khc: Mt hm sng (1AO) c c trng bng 3 slng tn,l,m.

* Nhn xt:

- Cc e c cng 1 mc nng lng c th c n trng thi khc nhau, mi trng thic c trng bi slng tl.

- ng vi mi 1 trng thi c thc 2l+1 cch nh hng khc nhau trong khng gian.VD: n=1 ( mc nng lng K) -> l =0, m=0 => n,l,m= 100

100=1AO => mc nng lng K c 1 AO.n=2 (mc L)l =0,1; m=0, 1.


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n =2, l =0 => 200=1AO.

n =2, l =1 => m=0 => 210=1AO.

m=1 => 211=1AO.

m=-1 => 21-1=1AO.

Mc L c 4 AOVy: Mt mc nng lng n c n2hm sng => c n2AO.

Mt gi trca l c (2l+1) hm sng -> c (2l+1) AO

4. Gii thiu mt smy e

Hnh dng ca cc my e gn ging hnh dng ca cc AO tng ng nhng chkhc:

khi biu din hm sng th c du (+) hay (-) cn my e th khng c du.

Gi trca l: 0 1 2 3

K hiu: s p d f

Vy vi n1 => c ns= AO ns => my ns.

n2 => c np= AO np => my np.m=0 (z) ->

znp = AO npz => my npz

m=1 (x) ->xnp

= AO npx=> my npx

m=-1 (y) ->ynp

= AO npy => my npy.

My np gm 3 m my ng vi 3 gi trca .a. My ns- ns c tnh cht i xng cu, khng phthuc vo , .- My s: Mt my e phn bng hng v l 1 khi cu.

z

x

y

My s

z

x

y

nsAO

b. My p

- Mi hm ns l 2 mt cu i xng nhau qua gc ta c phn (+) v phn (-) theochiu ca trc ta .

- Mi my p: C dng hnh qut, cc i ca my e phn bdc theo trc ta .5. Chuyn ng ring ca e trong nguyn t:

Chuyn ng ton bca e trong nguyn tgm 2 chuyn ng:


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- Chuyn ng xung quanh nhn ( chuyn ng obitan) c c trng bng 3 slng tn,l,m.

- Chuyn ng ring(chuyn ng tquay) c c trng bng slng ttspinms; ms chnhn 2 gi trl +1/2 hoc 1/2.

* Vy chuyn ng ca ton be trong nguyn tc c trng bi 4 slng tn,l,m v ms trong :

- n c trng cho kch thc my e.- l c trng cho hnh dng my e.- m c trng cho hng my e.

V. Hnhiu e

Hnhiu e -> e kho st chu tc dng ca:

- Lc ht ht nhn.- Lc y ca cc e cn li. trng thto ra khng xuyn tm, nng lng ca e trong trng ny khng

nhng phthuc vo n m cn phthuc vo l. kho st hny -> phi a

hvh1e -> dng phng php gn ng.

1. Phng php gn ng 1e. Khi nim in tch ht nhn hiu dng

* Phng php gn ng 1e:

- Coi e kho st chuyn ng trong 1 trng Z do ht nhn v tt ccc e cn li gyra. Z c gi l in tch ht nhn hiu dng.

- Z = Z- A, A l hng schn ca cc e cn li.- Coi cc e cn li chn bt nh hng ht nhn 1 i lng A- Coi trng to ra do Z l trng xuyn tm.* Kt quca bi ton 1 e c thp dng cho bi ton nhiu e ( bng cch sdng

phng php gn ng trn): Cc biu thc tnh E, u ging nhau, chkhc chnoc Z th c thay thbng Z.

2. p dng kt qubi ton 1e cho hnhiu e.

a. Nng lng:

- H1 e :2

2

nn

Z13,6.E = => E=f(n).

-Hnhiu e:2

2

ln,n

Z'13,6.E = => E=f(n,Z) =f(Z,n,l).

Nhn xt:

- Vy trong h1 e => E chphthuc vo slng tchnh n, cn trong hnhiu e thE phthuc vo n v Z (hoc Z, n v l).

- Trong hnhiu e, mt mc nng lng btch thnh n phn mc, mi phn mc ctrng bi 1 gi trca l.


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l c trng cho lc y ca cc e cn li, l cng ln En,lcng ln.

- Trong hnhiu e, nng lng c hin tng suy bin.b.Hm sng

Hnh dng AO v my e hon ton khng i (nhtrong h1e) nhng mt phn be

theo khong cch ti nhn l khc nhau do Z Z.3. ngha ca 4 slng t:

*Khi nim lp, phn lp e:

- Lp e: Trong nguyn tnhiu electron, nhng electron c cng gi trslng t

chnh to thnh mt lp. Cc lp c k hiu nhsau:

n 1 2 3 4 5 6 7

Lp K L M N O P Q

n cng ln th lp electron cng xa nhn v electron c nng lng cng cao.

- Phn lp e: Trong cng mt lp cc electron c chia thnh n phn lp, mi phnlp trong cng mt lp c c trng bng mt gi trca l. k hiu cc phn

lp dng cc k hiu sau y:

l 0 1 2 3

K hiu s p d f

chphn lp electron thuc lp no vit thm h s c gi trbng s lng tchnh n ca lp trc k hiu phn lp.

V d:

Lp K ng vi n = 1 chgm c mt phn lp c c trng bi l = 0 v n=1, 1s

Lp L ng vi n=2 gm c hai phn lp c c trng

==

2p1l

2s0l

Lp M ng vi n=3 gm c 3 phn lp c c trng

==

=

3d2l

3p1l

3s0l

ngha ca 4 slng t:a. Slng tchnh n.- Xc nh lp e trong nguyn tVD: n =1 -> ng vi lp K

n=2-> ng vi lp L

-

Xc nh kch thc ca my e: n cng ln -> kch thc my e cng ln v mt my e cng long.


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- i vi nguyn tH hay ion 1 e, n xc nh mc nng lng ca e trong nguyn thoc ion:

2

2

nn

Z13,6.E =

- i vi nguyn tnhiu e -> Ee =f(n,l) n chxc nh mc nng lng trung bnhca cc e trong cng 1 lp: 2

2

ln,nZ'13,6.E =

b. Slng tphl- xc nh hnh dng ca m my e

My s hnh cu, my p - quti, my d dng phc tp.

c. Slng tt:- m xc nh s nh hng ca AO hay cc my electron trong khng gian xung

quanh ht nhn.

V d: ng vi l=0 (my s) => m=0; my s chc 1 snh hng xung quanhht nhn (my s c hnh cu).

l=1 (my p) => ma= -1, 0 ,+1 my p c 3 snh hng khc nhau xung quanh

ht nhn.

d. Slng ttspin ms: c trng cho schuyn ng ring ca e.

VI. Sphn be trong nguyn tnhiu e.

1.Nguyn l ngoi trPauli

Trong mt nguyn tkhng thtn ti hai electron c cng gi trca 4 slng t.

VD: Lp K; n=1 => l=0 => m=0=> ms =+2

1v ms =-

2

1

lp K c nhiu nht 2 e: e thnht c ga trn =1, l=0, m =0 v ms =+2

1; e th2

c gi trn =1, l=0, m =0 v ms=-2

1

Hqu:Da vo nguyn l pauli c thtnh c selectron ti a trong mt lng

t, mt phn lp hay mt lp.

+ Selectron ti a trong mt lng t l 2e (v trong mt lng tcc e c 3 slng tging nhau, slng tthtms phi khc nhau, nhn gi trl +1/2 v -1/2)

+ Selectron ti a trong mt phn lp l 2(2l+1).

Phn lp s p d f

S lng t 1 3 5 7

Se ti a 2 6 10 14

VD: Tnh se ti a phn lp np ( n c gi trbt k).

VD n =2, cn p ng vi l=1. T:

n=2 -> l=1 => m=-1 => ms=+1/2 v ms=-1/2 => ng vi AO 2py c nhiu nht 2e.

n=2 -> l=1 => m=0 => ms=+1/2 v ms=-1/2 => ng vi AO 2pz c nhiu nht 2e.


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n=2 -> l=1=> m=+1 => ms=+1/2 v ms=-1/2 => ng vi AO 2px c nhiu nht 2e.

Vy phn lp p c nhiu nht 6e.

- Se nhiu nht cc phn lp: Mt lp e ng vi 1 gi trca n c ti a 2n2e.VD: Tnh se ti a lp L ( n=2).

n=2 -> l=0 => m=0 => ms=+1/2 v ms=-1/2 c ti a 2e.l=1 => m =-1 => ms=+1/2 v ms=-1/2 c ti a 2e.

m=0 => ms=+1/2 v ms=-1/2 c ti a 2e.

m=+1 => ms=+1/2 v ms=-1/2 c ti a 2e.

Vy lp L (n=2) c nhiu nht l 8e=2n2.

2.Nguyn l vng bn:

Trong nguyn tcc electron chim trc ht cc AO c mc nng lng thp nht.

Nng lng ca cc AO trong nguyn tc xp theo thttng dn nhsau:

1s < 2s < 2p < 3s < 3p < 4s 3d < 4p < 5s 4d < 5p < 6s < 4f 5d < 6p < 7sTha mn quy tc Klechkowsky:

3. Quy tc Hund

- lng t: Mi AO c c trng bng 3 s lng tn,l, m; mi AO c biudin bng 1 vung c gi l 1 lng t, k hiu l

- Quy tc Hund: Trong mt phn lp cha selectron ti a cc electron c khuynhhng phn bu vo cc lng tsao cho s electron c thn vi spin song

song l cc i.

* Quy lut phn b cc e trong nguyn t: phi tun theo nguyn l Pauli, nguyn l

vng bn v quy tc Hund.

4. Cch vit cu hnh e ca nguyn ttrng thi cbn.


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a. Cu hnh dng ch:

* vit cu hnh e dng chcn bit:

- Se trong nguyn t(bng Z).- Thtin cc electron theo nguyn l vng bn .- Bit selectron ti a trong mt phn lp: Phn lp s c ti a 2e, phn lp p - 6e,

phn lp d - 10e, phn lp f- 14e.

* Cch vit:

- Vit di dng k hiu cc phn lp.- in e theo thtnng lng tng dn v cc e mi phn lp vit di dng sm

( tng tt ccc smcc phn lp = se = Z).VD: Vit cu hnh e nguyn tca Mn (Z=25) dng ch.

Mn (Z=25) -> se = 25 : 1s22s22p63s23p64s23d5

Hay: 1s22s22p63s23p63d54s2

* Ch : - Khi vit cu hnh e ca nguyn tth se =Z nhng khi vit cu hnh e ca ion

th phi ch se s Z (in tch ht nhn ca ion v nguyn tnhnhau nhng se

th phikhc nhau): Se < Z ( i vi ion dng) v se > Z ( i vi ion m).

VD: Mn3+(Z=25) -> se =22: 1s22s22p63s23p63d4.

Vit cu hnh e ca nguyn thay ion: khi in e vo nguyn t lun in theo th t

nng lng theo nguyn l vng bn nhng khi mt e ( trthnh ion) th mt e lp

ngoi cng trc ( mt t lp ngoi ri ti lp trong) : in (n-1)d sau ns, khi mt nstrc (n-1)d.

6. Cu hnh e nguyn tdng lng t:

* Cch vit:

- Vit cu hnh e dng ch.- Da vo cu hnh e dng chvit cu hnh e dng lng t( mi lng tcha

ti a 2e).

- Mi e c k hiu bng 1 mi tn quay ln (vi ms=+1/2) , quay xung quay ln (vi ms=-1/2).

- Nu 1 c 2e -> 2e phi c spin ngc chiu nhau => 2e ghp i, spin =0. Nu 1 c 1e -> gi l e c thn.

- Vi nhng phn lp cha bo ha e-> vic fn be phi tun theo quy tc Hund.VD: Vit cu hnh e di dng ca N (z=7)

N(Z=7) 1s22s22p3

* Ch : Trong mt snguyn t, vit cu hnh e theo nguyn l vng bn trng thi

cbn c cu hnh ns2(n-1)d4hoc ns2(n-1)d9th c schuyn 1e ns sang (n-1)d thnh


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ns1(n-1)d5hoc ns1(n-1)d10. Nguyn nhn l do hiu nng lng (E(n-1)d- Ens) nhv cc

phn lp d10v d5l cc phn lp bo ha v na bo ha l cc phn lp bn => khi

(n-1)d c se gn bng 10 (hoc gn bng 5) th 1 e ns schuyn sang (n-1)d to

thnh cc phn lp bn.


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Chng II

NH LUT TUN HON - BNG TUN HON CC NGUYN T

HO HCI. Mu

Nm 1869, Mendeleep khm ph ra nh lut tun han : Tnh cht ca ccnguyn tv hp cht ca chng bin thin tun hon theo chiu tng ca khi lng

nguyn t. Trn c s Mendeleep xp 63 nguyn t thnh bng tun hon

(HTTH) theo chiu tng ca khi lng nguyn t.

Vi cch sp xp cho php suy on sxut hin cc nguyn tmi nhng cn mt

shn ch:

- Khng gii thch c nguyn nhn ca tnh tun hon.- Khng gii thch c skhc nhau vsnguyn tgia cc hng.- C trng hp ngoi l, khi lng nguyn t ca nguyn tng trc ln hn

khi lng nguyn tca nguyn tng sau.

Ar(Z=18): 39,948 > K (Z= 19) : 39,698

Co (Z=27): 58,933> Ni(Z=28): 58,70

Te (Z= 52): 127,60> I (Z=53): 126,9015

II. nh lut tun hon. HTTH theo thuyt cu to hin i.

Ngy nay, di nh sng ca thuyt cu to nguyn t, nh lut tun hon v HTTH l

hqutnhin ca cc quy lut tun hon trong cu to ve ca cc nguyn t.1. nh lut tun hon:

Tnh cht ca cc nguyn tv hp cht ca chng bin thin tun hon theo chiu tng

ca in tch ht nhn.

Tnh tun hon l do sbin i tun hon trong cu to ca cc nguyn ttheo chiu

tng ca in tch ht nhn. => Tnh cht nguyn tv hp cht ca chng do in tch

ht nhn quyt nh.

2. Bng tun hon cc nguyn tha hc

thhin c tnh tun hon trong cu to nguyn tv tnh cht cu cc nguyn t-> xp cc nguyn tthnh HTTH theo nguyn tc sau:

* Nguyn tc xp:

- Xp theo chiu tng dn ca in tch ht nhn.- m bo tnh tun hon vcu hnh e nguyn tca cc nguyn t.

+ Cc nguyn tc cng slp e xp theo 1 hng (chu k).

+ Cc nguyn tc se lp ngoi cng ging nhau hoc 2 phn lp e ngoicng

ging nhau c xp vo 1 ct.(nhm)

a. Chu k: L tp hp cc nguyn tc cng slpve v t theo 1 hng ngang.- Slp e = sthtchu k.


15/40


- Gm 7 hng ngang ng vi 7 chu k, khng phn bit chu kln nh, chn l:Chu k1: Chc 2 nguyn t(xy dng phn lp 1s)

Chu k2, 3: Mi chu kc 8 nguyn t(xy dng cc phn lp 2s, 2p, 3s, 3p)

Chu k4, 5: Mi chu kc 18 nguyn t

Chu k6: C 32 nguyn tChu k7: Cn ang ddang.

* Nhn xt: - Trong 1 chu k i tu ti cui chu k, se lp ngoi cng tng dn t1-

>8.

- Sbin thin tun hon trong cu to nguyn tca cc nguyn tkhng lp li 1 cch

n gin m c smrng tchu k 4 -> snguyn ttrong cc chu ktng.

b. Nhm: L tp hp cc nguyn tc cu hnh e ha trtng tnhau xp thnh mt

ct.

Gm 8 nhm: nh stI->VIII, mi nhm c chia thnh 2phn nhm:* Nhm A(phn nhm chnh): Nguyn tca cc nguyn tnhm A c c im:

- Nguyn tang c in e vo phn lp ns hoc np ( n l lp ngoi cng).- VD: Z=3; 1s22s2

Z= 9 1s22s22p5

- Se lp ngoi cng ca nguyn t= sthtnhm cha n.- nhn bit 1 nguyn tthuc nhm A no -> da vo cu hnh e nguyn t:

Nhm IA: gm cc nguyn tc cu trc e lp ngoi cng l ns1

Nhm IIA: ns2

Nhm IIIA: ns2np1

Nhm IVA: ns2np2

Nhm VA: ns2np3

Nhm VIA: ns2np4

Nhm VIIA: ns2np5

Nhm VIIIA: ns2np6

Nhm B( phn nhm ph): Nguyn tca cc nguyn tnhm B c c im:- Nguyn tang c in e vo phn lp (n-1)d hoc (n-2)fVD Z=21 : 1s22s22p63s23p64s23d1=> thuc nhm B

- Se lp ngoi cng ca hu ht cc nguyn tnguyn tnhm B l 2 (ns2), ca mts t l 1 (ns1) v ca 1 trng hp Pd (Z=46) khng cha e no lp ngoi cng

(5s0). Vy se lp ngoi cng ca cc nguyn tnguyn tnhm B < 3=> hu ht

cc nguyn tnhm B l kim loi.

- nhn bit 1 nguyn tthuc nhm B no -> da vo cu hnh e nguyn t:Nhm IIIB: c hai phn lp ngoi cng l (n-1)d1ns2

Nhm IVB: c hai phn lp ngoi cng l (n-1)d2ns2


16/40


Nhm VB: c hai phn lp ngoi cng l (n-1)d3ns2

Nhm VIB: c hai phn lp ngoi cng l (n-1)d4ns2(n-1)d5ns1Nhm VIIB: c hai phn lp ngoi cng l (n-1)d5ns2

Nhm VIIIB: c hai phn lp ngoi cng l (n-1)d6,7,8ns2

Nhm IB: c hai phn lp ngoi cng l (n-1)d10

ns1

Nhm IIB: c hai phn lp ngoi cng l (n-1)d10ns2

* Nhn xt:

+ Nu vit cu hnh e nguyn tda vo dy nng lng theo nguyn l vng bn => th

tt ccc nguyn tnhm B u c 2 e lp ngoi cng ns2. Tuy nhin thc nghim xc

nh rng 1 snguyn tnguyn tnhm B c 1 e ns -> (n-1)d, tr1 trng hp Pd

2e 5s2u chuyn vo 4d. Cc trng hp ny xy ra khi phn lp (n-1)d gn na bo

ho (d5) hoc bo ha (d10) l cc phn lp bn v nng lng e phn lp (n-1) v ns xp

xnhau.+ Cu hnh e nguyn tca 1 snguyn tm sin e cui cng xy ra (n-2)f cng

hi khc so vi cch in e theo nguyn l vng bn.

VD Z=64: 1s22s22p63s23p64s23d104p65s24d105p66s24f8( theo nguyn l vng bn)

Trong thc t: 1s22s22p63s23p64s23d104p65s24d105p66s24f75d1.

1e 4f chuyn sang 5d: t ti cu hnh bo ha na se f7bn.

* Nguyn ts, p, d, f:

-

Nguyn tm sin e cui cng vo nguyn txy ra ns gi l cc nguyn ts.nh ngha tng ti vi cc nguyn tp, d,f.

+ Cc nguyn tnhm IA, IIA l cc nguyn ts.

+ Cc nguyn tnhm IIIA->VIIIA l cc nguyn tp.

+ Cc nguyn tnhm B l cc nguyn td (ring nhm IIIB c cnguyn tf).

+ Cc nguyn t f m sin e cui cng vo nguyn t xy ra 4f -> gi l cc

nguyn tlantanoit hoc cc nguyn thlantan, cn cc nguyn tf m sin e cui

cng 5f -> cc nguyn tActinoit (hactini).

3. Mt sdng bng HTTHa. Dng bng ngn

- Chu knhc 1 hng, chu kln c 2 hng.- C 8 nhm, mi nhm chia thnh 2 phn nhm : phn nhm chnh (A) v phn nhm

ph (B).

b. Dng bng di:- Cc nguyn ttrong mi chu kc xp thnh 1 hng.- Ton bng c 16 nhm, nh stIA, IIA...VIIIA v IB,IIB.. VIIIB. Hlantan v h

actini gm cc nguyn tf c xp vo nhm IIIB.


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III. Sbin i tun hon 1 stnh cht ca cc nguyn t

1.Hiu ng chn:

Trong nguyn tH c 1 e -> e ny bton bin tch ht nhn ht

Trong nguyn tnhiu e, ngoi lc ht ca ht nhn i vi cc e, cn lc y gia cc e

c in tch cng du. Lc y ny lm gim lc ht ca ht nhn i vi cc e. Trongtrng hp ny ngi ta ni cc e chn ln nhau. Nhvy trong nguyn tnhiu e, mi e

u bchn bi cc e cn li v chnh n li chn cc e khc trong nguyn tnhng s

chn ny khng hon ton.

A =

=

1

1

n

i

iA A: Hschn tng cng ca cc e cn li i vi e xt.

n: se c trong nguyn t.

Ai: Hschn ca e thi i vi e kho st.

V chn khng hon ton => Ai tc dng chn ca e thi i vi e kho st tng v Z i vi e kho st

gim v ngc li.

* Nhn xt vquy lut chn:

- Cc e cng xa nhn ( cc e c n ln v l ln) th bchn cng nhiu v tc dng chnca n i vi e cng t ( e xa nhn nht l e c gi trn v l ln nht).

- Cc e trong cng 1 lp (cng n) chn ln nhau km v theo chiu ns-np-nd-nf tcdng chn gim dn, khnng bchn tng dn (A tng).

- Cc e trong cng 1 phn lp ( cng n,l) chn ln nhau cng km, c bit cc e trongcng 1 phn lp y 1 na, se c spin song song nhau chn ln nhau km nht ( v

lc ny mi e chim 1 AO trong phn lp xa nht).

- Cc phn lp bo ha e thuc lp bn trong mt e dy c chn mnh e bn ngoi.2. Quy lut bin thin nng lng ca cc AO ha trEAO

Theo cng thc:

EAOha tr=( )

2

2

2

2

613613n

Z

n

AZ ',, =

khi Z tng th EAO gim. n tng -> EAO tng.

a. Trong 1 chu k

- Trong 1 chu k, ttri qua phi EAOha trgim v n=const, Z tng -> Z tng: Z tng

do A tng chm hn Z, A tng chm v ttri qua phi trong 1 chu kcc e c in

vo cng mt lp nn tcdng chn ln nhau km (s nh).

- Trong mt chu k ttri qua phi, hiu nng lng gia cc AO np v ns tng dn:nsnp EEE = tng .

b. Trong 1 nhm: ttrn xung trong 1 nhm th:


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- EAOha trtng dn do:+ Z tng nhiu t8-> 18, cu trc bo ha e chn mnh vi e ngoi => Z tng chm

hn, n tng nhanh hn.

- Hiu nsnp EEE = gim do n tng nhanh.

3. Nng lng ion ha ca nguyn tI (eV, kJmol-1)a. Nng lng ion ha thnht: I1

Nng lng ion ho thnht I1 l nng lng ti thiu cn thit tch 1e ra khi

nguyn ttrng thi cbn v thkh thnh ion tch in +1 cng trng thi kh v

cbn.

X(K,CB)- 1e X+(K,CB) I1> 0, I1- l nng lng ion ho thnht

Nng lng ion ho thhai ng vi qu trnh bt electront hhai

X+(K,CB)- 1e X2+

(K,CB) I2> 0

Ta lun c I2> I1

b. Electron no btch khi nguyn tkhi bion ha:

Khi nguyn tbion ha th e lin kt yu nht vi ht nhn sbbt ra trc tin, l

e lp ngoi cng (ng vi e c n ln nht) c nng lngln nht.

VD Ti (Z=22) 1s22s22p63s23p64s23d2e lp ngoicng l 4s2-> cc e ny btch trc tin khi bion ha, sau mi

n cc e 3d2.

c. Cc yu tnh hng n nng lng ion ho:Khi nguyn tmt e tc l xy ra hin tng ion ho th e schuyn tcc AO ra xa

v cng. Khi nng lng ion ho ng vi qu trnh mt e c tnh bng cng thc:

I = E- Ee = 22

n

Z'.,613 (eV)

Trong : El nng lng ca electron xa v cng = 0

Ee l nng lng ca electron btch

Tbiu thc ta thy rng nng lng ion ho I phthuc vo n v Z. I cng ln khi n

nhv Z ln. Z phthuc vo Z v A do I sphthuc vo n, l v Z.

d. Quy lut bin i I1trong 1 chu k:Trong mt chu kkhi i ttri sang phi ni chung I1tng dn v t gi trcc i

kh tr. Tnguyn tkh trca chu k trc n nguyn tu tin ca chu k tip

theo I1li gim xung t ngt ri li tng dn nhchu ktrc. Qu trnh lp i lp litchu kny sang chu kkhc v c gi l stun hon ca I1.


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Gii thch: Trong 1 chu k, ttri qua phi th Ee gim -> I1tng do n=const , Z tng

dn (Z tng mnh hn A).

I1min u chu kns1

I1max cui chu kns2np6

V vi ns1

lp bn trong bo ha-> chn tt-> I1min.ns2np6bo ha -> kh tch e ->I1max.

V n = const Z tng dn, tuy nhin I1 tng khng u v thy xut hin nhng cc

tiu nh(bn trong mt chu k) nhng nguyn tc phn lp ngoi cng c phn

by (ns2) hoc y mt na electron (np3).

Cc cc tiu nhxut hin bn trong mi chu k, nn y l stun hon ni chu k

ca nng lng ion ho thnht.

Trong mt phn nhm chnh khi i ttrn xung gi trI1gim dn, cn trong mt

phn nhm phsbin thin ny chm v khng u.

e. Quy lut bin i I1trong nhm:

Trong mt phn nhm chnh khi i ttrn xung gi trI1gim dn, cn trong mt

phn nhm phsbin thin ny chm v khng u.

4. i lc i vi e ca nguyn tA (eV,kJ)

L khnng kt hp electron ca nguyn tto thnh ion m, n ng vi qu trnh:X(k,cb)+ 1e -> X

-1(k,cb)

Vy: Nng lng kt hp electron l nng lng thot ra hay thu vo khi kt hp

thm 1e vo nguyn ttrung ho trng thi kh, cbn.

- Khc vi nng lng ion ho th nng lng kt hp electron c thm, dng hocbng 0. i lc vi electron cng ln th nng lng kt hp electron cng nh.

- Trong mt chu ki lc vi electron tng dn v t cc i cc nguyn tnhmVIIA

5. m in ca nguyn t()

I1

Z

He

HLi

Be

B

C

N

O

F

Ne

Na

Mg

Al

Si

P

S

Cl

A r

K


20/40


nh ngha: l i lng c trng cho khnng ht cp e lin kt ca nguyn ttrong

phn t.

- Cng thc tnh theo phng php Miliken: =

2

XX AI +

Ix,AX: nng lng ion ha v i lc e ca nguyn tX.

Tng (Ix+AX) cng ln => khnng ht e ca nguyn tcng ln.

6. Tnh kim loi v phi kim:- Tnh kim loi: L tnh dnhng electronMt nguyn tc se lp ngoi cng < 4 l kim loi trB (Z = 5) v H (Z=1), l

cc nguyn tnhm IA, IIA, IIIA. Mt snguyn tkim loi c se lp ngoi cng l 4nhGe, Sn, Pb; c se ho trl 5 nhBi, Sb.

- Tnh phi kim: L tnh dnhn e

Cc nguyn tc se lp ngoi cng > 3 l nguyn tphi kim l cc nguyn tnhm IVA, VA, VIA, VIIA trSn, Pb, Ge (IVA), Sb, Bi(VA) lcc kim loi.- Quy lut bin i:Trong mt chu kkhi i tu n cui chu ktnh kim loi gim dn, cn tnh phikim tng dn.Trong mt phn nhm chnh t trn xung tnh kim loi tng dn cn tnh phi kim

gim dn. Trong phn nhm phi ttrn xung tnh kim loi gim dn.7. Sxi ha ca nguyn t

Khi tng tc ho hc lun xy ra sdi chuyn electron ho trtnguyn tny sang

nguyn tkhc. Chnh sdi chuyn ny xc nh soxi ho ca nguyn ttham giatng tc.

Nguyn tmt e ho trstch in dng v do n sc soxi ho (+). Soxiho (+) = se ho trmt. Nguyn tthu e ho trstch in m, do c soxi ho(-). Nhvy, soxi ho = se ho trnhng hoc nhn.Soxi ho cao nht ca mt nguyn t= sthtnhm ca n trF, O, cc nguyn

tnhm IB, VIIIB.Soxi ho m thng chc cc nguyn tphi kim tIVA n VIIA. Soxi ho

m thp nht ca cc phi kim = sthtnhm - 8V d: soxi ho m thp nht ca cc halogen nhm VIIA l = 7-8 = -1

soxi ho m thp nht ca cc nguyn tnhm VIA l = 6-8 = -2soxi ho cao nht ca cc nguyn t nhm VA, VB l +5soxi ho cao nht ca cc nguyn tnhm VIIA (trF), VIIB l +7

IV. Mi lin hgia cu to nguyn tvi vtr v tnh cht ca cc nguyn ttrong bngtun hon

1. Bit cu to velectron suy ra vtr v tnh cht

V d1: Nguyn tc Z = 22

Bit Z vit cu hnh electron: 1s22s22p63s23p63d24s2

- Lp ngoi cng c n= 4 nn nguyn tchu k4

- Cc electron cui cng ang c in phn lp d nn l nguyn td v thuc

nhm B


21/40


- Tng se phn lp ns + se phn lp (n-1)d = 4 nn nguyn tthuc nhm IVB

- V c se lp ngoi cng = 2 < 4 nn l kim loi

V d2: Nguyn tc Z = 35

Cu hnh electron ca nguyn tl: 1s22s22p63s23p64s23d104p5

- Nguyn tthuc chu k4 v c 4 lp electron- y l nguyn tp v cc e cui cng ang in phn lp 4p

- Se lp ngoi cng = 7 > 3 l phi kim, thuc nhm VIIA

2. Bit vtr trong bng tun hon suy ra cu to velectron

V d1: Nguyn tX, chu kIII, nhm VIIA

- V nguyn tchu kIII nn c 3 lp electron

- V nhm VIIA nn cu trc lp e ngoi cng l ns2np5

- Cu hnh electron: 1s22s22p63s23p5

V d2: Nguyn tA chu kIV, nhm VIIB- V chu kIV nn c 4 lp velectron n = 4- Nhm VIIB, cu trc lp velectron ngoi cng l: (n-1)d5ns2: 3d54s2

- Cu trc velectron: 1s22s22p63s23p63d54s2


22/40


CHNG IIILIN KT HO HC V CU TO PHN T

A - Lin kt ho hcI. Cc c trng cbn ca lin kt ho hc

1. Nng lng lin kt (E):(y quy c E ng vi qu trnh ph vlin kt)Nng lng lin kt l nng lng ng vi qu trnh ph v lin kt, do nng

lng lin kt cng ln th lin kt cng bn.i vi phn t2 nguyn t, nng lng lin kt ng vi qu trnh:A-B(K,CB)A(K,CB)+ B(K,CB), EA-B> 0.V d: HCl(K,CB)H(K,CB)+ Cl(K,CB), EHcl= 431 kJ.mol-1

i vi phn tnhiu nguyn tkiu ABn, nhCH4-> dng k hiu nng lng linkt trung bnh.

CH4(k,cb)-> C(k,cb)+ 4H(k,cb) ; HCEH = 4

=> 41641 == HE HC (kJ.mol

-1)

2. di lin kt:L khong cch gia hai tm ca hai nguyn t tham gia lin kt. di lin kt

thng c tnh bng nanomet (nm) hoc anstron (A0).

VD: lH-H= 0,740

A => l tng -> bn lin kt gim v ngc li.

3. Gc lin kt:L gc to bi mt nguyn tlin kt trc tip vi hai nguyn tkhc trong phn t.

V d: Oxy to hai lin kt vi hai nguyn tH trong phn tH2O. Gc lin kt HOHtrong phn tnc l 104,50

II. Phn loi lin ktDa vo chnh lch v gia 2 nguyn ttham gia lin kt ( ).

1.Lin kt ion- iu kin to thnh: Khi hai nguyn ttham gia lin kt c schnh lch vm

in l 2. Khi cp e ho trschuyn hn tnguyn tc m in nhhn sang nguyn tc m in ln hn, do shnh thnh ra cc ion ngc du.Sau cc ion ngc du s lin kt vi nhau bng lc ht tnh in ca cc ionngc du.

Vy bn cht ca lin kt ion l lc tnh in gia cc ion tri du. xy ra bi 1 kim loi in hnh v 1 phi kim in hnh.

VD: Na 1e = Na+

Cl +1e =Cl-

hnh thnh lin kt ion trong NaCl. c im lin kt:

O

H H


23/40


- Lin kt ion khng c tnh nh hng: Mi ion c th to ra mt in trngxung

quanh n, do lin kt ion c hnh thnh theo mi hng

- Lin kt ion khng c tnh bo ho, v vy mi ion c th ht c nhiu ion xungquanh n- Lin kt ion l lin kt bn, nng lng ca lin kt ion ckJ trln.

Do cc c im trn iu kin thng cc hp cht lin kt ion l cc cht rn.Gm v scc ion m v dng lin kt vi nhau theo nhng trt tnht nh

V d: Cc mui, cc oxit v hydroxit kim loi2.Lin kt cng ha tr:- Lin kt cng ha trphn cc: Nu 2 cp e lin kt lch vpha nguyn t

ca nguyn tc ln hn.VD : HCl, SO2..- Lin kt cng ha trkhng phn cc: Khi 0= -> cp e lin kt khng blch.

VD H2.,...

III. Lin kt cng ho tr. Phng php cp electron lin kt1. Sto thnh phn thydro thai nguyn tH:

Nm 1927 hai nh Bc hc Heitler v London p dng chc lng tgii biton tnh nng lng lin kt trong phn tH2. Kt qucho thy:

- Lin kt gia hai nguyn tH c hnh thnh khi 2 electron ca hai nguyn tcspin ngc du

- Khi hnh thnh lin kt, mt my electron khu vc khng gian gia hai htnhn tng ln. Do in tch m ca my electron sc tc dng ht hai ht nhn vlin kt chng li vi nhau.

- Nu hai electron c spin cng du th mt my electron khu vc gia haiht nhn gim xung, mt electron khu vc ngoi hai ht nhn tng ln v cc myny c tc dng y nhau lm hai ht nhn tch xa nhau.=> khng hnh thnh lin kt.

Nhvy, phn tH2c hnh thnh l do sghp i ca 2e c spin ngc chiunhau. Sau , ngi ta khi qut ho cc kt qutrn v mrng thnh phng php

cp e lin kt p dng cho mi phn t.2. Ni dung cbn ca phng php cp e lin kt

+ +


24/40


Mi lin kt cng ho trc hnh thnh l do sghp i ca 2e c thn cspin ngc du ca 2 nguyn t tham gia lin kt. Khi xy ra sxen phgia hai my electron lin kt

Khi hai my electron xen phnhau cng mnh th lin kt cng bn (xenphcng mnh khi cc my electron tham gia xen phc nng lng cng xp

xnhau) Lin kt cng ho trl lin kt c hng. Hng ca lin kt ny l hng cxen phcc my electron ca 2 nguyn ttham gia lin kt l ln nht

V d: Sxen phcc my s v s, s v p, p v p

s-s s-p

p-p

3.Ho trca nguyn ttheo phng php cp electron lin kt

Ttin 1 ca phng php cp electron lin kt ta thy rng: Ho trc thc camt nguyn tc tnh bng se c thn. Da vo iu ny ta c thgii thch cho tr1 ca nguyn tH, ho tr3 ca nguyn tN, ho tr2,4 ca C.

V d1: H c 1e: 1s1H c 1e c thn nn c ho tr1

N: 1s22s22p3

N c 3e c thn nn c ho tr3C: 1s22s22p2

C c 2e c thn nn c ho tr2

trng thi cbn C c 2e c thn nn c ho tr2 (v dphn tCF2), tuy nhin Ccn c ho tr4 (trong phn tCH4, CCl4). iu ny c gii thch nhsau: Khi ccung cp nng lng th 1e ghp i phn lp 2s schuyn sang trng ca phnlp 2p, lm cho se c thn ca nguyn tC tng ln, trng thi ny ca nguyn tCc gi l trng thi kch thch. Trng thi kch thch l trng thi ca nguyn t cc khi nhn thm nng lng, khi xy ra hin tng tch cc cp e chuyn 1e

sang obitan cn trng thuc cng mt lp.Nng lng tiu tn chuyn nguyn tttrng thi cbn sang trng thi kch thch c b bng nng lng gii phng khi hnhthnh mi lin kt ho hc.


25/40


C*:trng thi kch thch se c thn ca C l 4 v do c ho tr4

V d2: Xc nh cc ho trc thc ca P bng phng php cp e lin kt

P (Z=15): 1s22s22p63s23p3

... Trng thi cbn c 3e c thn c hotr3

Khi c cung cp thm nng lng th 1e c ghp i ocbitan 3s schuyn sang1 obitan trng 3d, lm se c thn ca nguyn tP l 5, trng thi ny ca nguyn tP l trng thi kch thch.

Trng thi kch thch c 5e c thn c ho tr5

Xc nh cc trng thi cng ho trc thc ca S

S (Z=16): 1s22s22p63s23p4

Trng thi cbn c 2e c thn c hotr2

3s 3p 3d

* Trng thi kch thch * c 4e c thn c hotr 4

** Trng thi kch thch ** c 6e c thn c ho tr6

Ch :

Sdch chuyn electron i hi tiu tn nng lng, nn sdch chuyn chxy ra trong cng mt lp, tphn lp ny sang phn lp khc (ns np, nd,np nd). Nng lng tiu tn c n b bng nng lng to lin kt. Sdch chuyn electron tlp ny sang lp khc khng thxy ra c, v tiutn nng lng qu ln khng thn b bng nng lng gii phng khi hnhthnh lin kt

Cc nguyn tphn nhm chnh, c bit vi nguyn tchu kII, lp ngoicng (n=2) c 4 obitan nn cc nguyn tny chc ho trln nht l 4. Ccnguyn tthuc chu kIII c 3 phn lp ngoi cng l s, p, d nn cc nguyntny c thc ho tr> 4 (P c ho tr3, 5; S c ho tr2, 4, 6; Cl c ho tr1, 3, 5, 7)


26/40


Phng php tm ho tr ca nguyn tbng phng php cp e lin kt chng i vi cc nguyn tnhm A (trcc kh him).

T cc v d trn ta thy rng s lin kt cng ho tr ca 1 nguyn t l c hn,ngha l c tnh bo ho (iu ny khc vi trng hp lin kt ion).

4. Tnh nh hng ca lin kt cng ho trCc nguyn t trong phn t lin kt vi nhau theo hng c s xen ph cc my

electron l ln nht, chnh l tnh nh hng ca lin kt cng ho tr. T tnh nhhng ny ta c thdon c cu hnh hnh hc ca phn t.

V d: Phn tH2S

H: 1s1S: 1s22s22p63s23p4

Hai e c thn 3p ca S sto lin kt vi 2e c thn ca hai nguyn tH, sxen phl cc i th cc AO phi nh hng theo trc lin kt nhnhau. Do hnhdng cc my xen phsquyt nh cu hnh hnh hc ca phn t, do gc HSH =90o, nhng v yu tkhc nh lc y tnh in (sphn cc ca S, H) xut hin nnthc tgc HSH l 92o2, gii thch tng tgc HSeH = 91o, HTeH = 90o.

5. Lin kt cho - nhnLin kt cho - nhn cng l lin kt cng ho tr. Lin kt cho - nhn c hnh thnh

do mt cp e khng phn chia ca mt nguyn tvi 1 AO ho trtrng ca 1 nguyn t

khc. Cp e dng chung to lin kt chdo mt nguyn t(ion) cung cp.V d:

N: 1s22s22p3

3e c thn ca N sghp i vi 3e c thn ca 3 nguyn tH to thnh 3 lin ktN-H. Nhvy, trong phn tNH3cn 1 cp e cha phn chia

N :

3pz

3py

1s

1s

+

+

H

H

H


27/40


H+: 1so c AO ho trtrng 1s

Khi NH3tin gn ti H+n sbri vo trng tc dng ca H+, H+sht cp e cha

phn chia ca N to thnh mt lin kt N-H thtc to ra bi mt cp e cha chiaca N v 1AO ho tr trng ca H+. Lin kt ny l lin kt cho - nhn, c k hiu

bng mt mi tn () di tnguyn tcho cp e n nguyn tnhn cp e.

N

H

H

H

H N

H

H

H

H

Shnh thnh phn tBF4-, H3O

+u c gii thch tng t, 1 lin kt B-F v 1lin kt H-O c hnh thnh do scho- nhn.

Bng thc nghim ngi ta xc nhn c rng 4 lin kt N-H trong phn tNH4+hoc phn tBF4

-ging ht nhau nn lin kt cho- nhn l lin kt cng ho tr.6. Cng ho trcc i

Lin kt cng ho trc to thnh do:- Cc e c thn cha ghp i- Mt cp e cha phn chia- lng ttrngSlin kt cng ho trm nguyn tc khnng to thnh (cn gi l cng ho tr

cc i) bng s lng tca nguyn tc khnng tham gia lin kt (= sAO hotr) (bao gm lng tc 1e c thn, 1 cp e khng phn chia hay lng ttrng).

V d1:Cc nguyn tchu kII c 4 lng tho trc khnng tham gia lin kt l 2s,

2px, 2py, 2pz nn cng ho trcc i ca chng l 4.

V d2: B (Z=5): 1s22s22p1

B: B*:

1s 2s 2pB trng thi kch thch c 3e c thn, sto c ba lin kt B-F, hnh thnh phn

tBF3

B

Ngoi ra, nguyn tB cn 1 obitan trng, sto c lin kt B-F thtl lin ktcho - nhn, hnh thnh phn t BF4-: C cng ho trcc i l 4

F

F

F


28/40


C (Z = 6): 1s22s22p2:

C*: C 4e c thn, nn to c 4lin kt C-H

N (Z= 7): 1s22s22p3 C 3e c thn to c 3 lin

kt N-H, hnh thnh c phn tNH3, tuy nhin trong nguyn tN cn 1 cp e chaphn chia, c thto thnh phn tNH4+: N c cng ho trcc i l 4

IV. Thuyt lai ha

1.iu kin ra i ca thuyt lai haThuyt lai ho ra i nhm gii quyt c hai kh khn ca phng php cp electronlin kt. Cho php gii thch c cu trc hnh hc v bn ca lin ktVD Xt shnh thnh CH4C(Z=6) 1s22s22p2

trng thi kch thch C*

theo phng php cp e lin kt th 4e c thn ca C*sto thnh 4 lin kt C-H, trong c:- 3 lin kt p-s: 3AO 2p ca C xen phvi 3 AO 1s ca 3 nguyn tHto 3 gc

lin kt HCH =900.- 1 lin kt C-H th4 to thnh do sxen phcc AO ha trca 2s ca C v 1s ca H

(s-s) khng c hng xc nh trong khng gian (v mc xen phcc AO s vinhau l nhnhau theo mi hng). Nu coi lin kt ny phi cch u 3 lin kt kiath gc lin kt HCH thtphi bng 125014.. Kt quny cn dn n bn ca1 lin kt C-H ( do xen phs-s) ny khc vi bn ca 3 lin kt C-H cn li (do

xen php-s). (phng php cp e khng gii thch c skhc nhau ny)- Tuy nhin thc nghim chng trng 4 gc lin kt HCH u bng 109028 (bnggc tdin u) v bn ca 4 lin kt C-H u bng nhau.

- Gii gii quyt 2 kh khn ny ca phng php cp e lin kt-> phi dng thuytlai ha: githit rng khi to lin kt th 1 AO 2s v 3AO 2p ca C lai ha (trn ln)vi nhau to thnh 4AO lai ha sp3ging ht nhau hng ti 4 nh ca hnh tdinu, chng xen phvi 4AO 1s ca 4 nguyn tH. Vy 4 lin kt C-H figing nhau v cn phi bng gc ca hnh tdin u m nguyn tC nm tmca hnh ny.

Cu hnh phn tCH4theo thuyt lai ha2. Ni dung thuyt lai ha:

Lai ha AO l sthp cc AO ha trca 1 nguyn tto thnh 1 stng ngcc AO mi c cng nng lng nh hng xc nh trong khng gian v c dng

to lin kt bn hn. Cc AO tham gia thp c thc 1e, 2e hoc l 1 lng ttrng.3. Cc kiu lai haa.Lai ha sp


29/40


1AOs + 1AOp 2AO lai ho sp. 2AO lai ho ny nh hng thng hng vi nhauv to vi nhau mt gc bng 180o

V d: Dng lai ho ny gp trong nguyn tBe ca phn tBeF2, BeH2, BeCl2, nn ccphn tny c dng thng.

b. Lai ho kiu sp2:1AOs + 2AOp 3AO lai ho sp2. 3AO lai ho ny nm trong cng mt mt phng

v to vi nhau gc bng 120o

V d: Kiu lai ho ny gp trong nguyn tB ca phn tBF3, BCl3c. Lai ho kiu sp3:1AOs + 3AOp 4 AO lai ho sp3. 4 AO lai ho ny nh hng ttm ti 4 nh

ca tdin u, gc to thnh gia cc AO lai ho l 109o28.

V d: Gp trong nguyn tO ca phn tH2O, nguyn tN ca phn tNH3v ionNH4

+

4. iu kin lai ho bn:Lai ho ca nguyn tl bn khi tho mn cc iu kin sau y:

p

p+

+ +

-

- +

-

120o+

+

+

- +

-


30/40


? Cc AO nguyn t tham gia lai ho phi c nng lng xp x nhau. Nh vy,trong mt chu ki tu n cui chu kth hiu cc mc nng lng Enp- Ens lndn ln. Do i tu n cui chu khiu qulai ho km dn.

V d: chu kII hiu qulai ho ca cc AO(2s) v AO(2p) i vi cc nguyn tu chu k nhBe, B, C rt tt. i vi nguyn tBe c lai ho sp v gc gia cc AO

lai l 180

o

, i vi B c lai ho sp

2

v gc gia cc AO lai l 120

o

, i vi C c lai hosp3v gc gia cc AO lai l 109o28?Nng lng ca cc AO tham gia lai ho phi thp. Do cc AO lp thhai (2s,

2p) tham gia lai ho c hiu quhn, cn cc AO lp thba (3s, 3p) hiu qulai hokm hn, lp tht(4s, 4p) lai ho khng ng k.

V d: trong dy H2O - H2S- H2Se- H2Te hiu qulai ho gim dn nn gc lin ktgim dn theo dy 104o5 - 92o2- 91o- 90o

? xen phca cc AO lai ho vi cc AO nguyn tkhc tham gia lin kt philn.5.Don kiu lai ho v cu trc hnh hc

chn kiu lai ho cho nguyn t trung tm da vo n l tng s lin kt canguyn ttrung tm vi scp e ha trkhng phn chia.

Nu tng bng 2 th nguyn ttrung tm c lai ho dng spNu tng bng 3 th nguyn ttrung tm c lai ho dng sp2Nu tng bng 4 th nguyn ttrung tm c lai ho sp3-Khi bit c kiu lai ha ca nguyn ttrung tm trong phn t-> cha xc nh

c cu hnh hnh hc ca phn t. V cu hnh hnh hc ca phn tphthuc vo:+ Dng lai ha ca ca nguyn ttrung tm.+ Slin kt ca nguyn ttrung tm vi cc nguyn txung quanh.

+Scp e ha trca nguyn ttrung tm cha lin kt. Mun bit cu hnh hnh hc ca 1 phn t=> phi bit 3 yu ttrn.

Cth:n=2: -> lai ha sp: cu trc thng -> gc 1800n=3: -> lai ha sp2: nu c 3 lin kt + 0 cp e ha trcha lin kttam gic

nu c 2 lin kt + 1 cp e ha trcha lin kt -> cu trcgc.n=4: -> lai ha sp3: nu c 4 lin kt + 0 cp e ha trcha lin kttdin

nu c 3 lin kt + 1 cp e ha trcha lin kt -> thp tamgic.

nu c 2 lin kt + 2 cp e ha trcha lin kt -> cu trcgc.

V d:BeH2; Nguyn ttrung tm Be (Z=4)n= slin kt + scp e ha trcha lin kt

=2 +0 =2 => Be c lai ha sp => phn tc dng ng thng.Tng tta c:

CH4- C c lai ho sp3- cu trc hnh hc l tdin u

NH3- N c lai ho sp3- cu trc hnh hc l thp tam gic

H2O - O c dng lai ho sp3- cu trc dng gc

6. nh gi u khuyt im ca phng php lin kt ha tr:- Gii thch n gin, dhiu, cho php gii thch cu trc hnh hc ca

nhiu phn t


31/40


- Khng gii thch c s tn ti ca mt sion nh: H2+, O2

+, NO+,khng gii thch c tnh thun t, nghch tca phn tO2

- Khng c tnh nh (khng chng minh c)V. Phng php MO-LCAO ( phng php MO - thp tuyn tnh cc AO- Phng phpobitan phn t)

Nguyn tc: Mrng hm sng cho phn t.1.Nhng githit cbn ca phng php MO-LCAO- Coi mi electron chuyn ng trong 1 trng gy ra bi cc ht nhn v tt ccc

electron cn li, c gi l trng thp (khc vi trng ca nguyn t, trngny khng i xng cu).

- Trng thi ca electron c m tbi mt hm sng hay cn gi l obitan phntMO.

- Cc electron ca lp trong ca mi nguyn tkhng thuc vton bphn tm chthuc v nguyn t m thi, nn khng to thnh MO, c ngha l ch c ccelectron ho trmi tham gia to thnh cc MO.

Bi ton a vgii phng trnh Shrodinger c dng: H= ETrong H l ton tHamilton, E l gi trring ca nng lng, - l hm sng m

ttrng thi ca 1e trong phn t.Gii phng trnh ny stm c nng lng ca e trong phn t(E) v hm sng tng ng. Sau xy dng gin nng lng v vit cu hnh e ca phn t:Vic phn be vo cc MO ging ht nhquy lut phn bcc e vo AO, tc l tuntheo nguyn l vng bn, nguyn l loi trPauli v quy tc Hund.

2. Ni dung cbn ca sto thnh MO bng phng php LCAO( Phng phpthp tuyn tnh cc AO).

Theo phng php ny th MO c xc nh bng phng php thp tuyn tnh

cc AO nguyn tda trn githuyt sau y:

Xt phn tgm hai nguyn t1 v 2:

Khi electron chuyn ng gn nguyn t1 th n chu tc dng chyu canguyn t1, do obitan phn tc dng tng tnhAO ca nguyn t1 l1nhng chu snhiu lon gy ra bi nguyn t2 nn phi c mt hsbsung vo biu thc xc nh MO

Khi electron chuyn ng gn nguyn t2 th n chu tc dng chyu canguyn t

2, do

obitan phn t

c d

ng t

ng t

nh

AO c

a nguyn t

2 l

2ng thi chu snhiu lon gy ra bi nguyn t1 Ty thy rng hm tt nht m ttrng thi ca electron trong phn t

sl thp tuyn tnh ca cc AO: = C1. 1+ C2. 2 a = C1. 1+ C2. 2vo phng trnh Schrodinger, gii ra ta stm c

v nng lng E tng ng

V d: Xt trng hp n gin nht l ion H2+. Hgm hai ht nhn ging ht nhau

AO ho trca nguyn t1 l 1s c c trng bi hm sng 1, AO ho trcanguyn t2 l 1s c c trng bi hm sng 2. Khi = C1. 1+ C2. 2l hm

m t trng thi ca electron trong phn t H2+. Thay biu thc vo phng trnhSchrodinger, gii phng trnh sxc nh c biu thc ca MO


32/40


- MO thnht l += ( )212

1+ vi nng lng tng ng l E+, nng lng

ny thp hn nng lng ca cc AO

- MO thhai l ( )212

1= vi nng lng tng ng l E-, nng lng ny

cao hn nng lng ca cc AO.+l MO lin kt s; -l MO phn lin kt MO*, s

*

S t hp ca cc AO thnh MO thng c biu din di dng gin nnglng

H2+chc 1e duy nht c in vo s. iu c ngha l khi to thnh ion H2+

nng lng gim xung mt gi trl a, ngha l hH2+bn hn hH v H+ringbit.

Cu hnh electron ca H2+: s

1

Tnh bc lin kt:2

*NN

=

Trong : - Bc ca lin kt hay bi ca lin ktN - l selectron trn cc MO lin ktN* - l selectron trn cc MO phn lin kt

21201 /2 *NN === 0 ion H2+c tn ti.

Gin nng lng cc MO i vi phn tH2l: s2=> 12

02 =

=

i vi phn tHe2: s2s*

2 => 02

22 =

= =>phn tHe2khng tn ti.

Nhvy, bng phng php MO th c thgii thch c stn ti ca ion H2+v

nhiu phn tkhc.Tphng php MO-LCAO th cnAO thp vi nhau bng phng php thp

tuyn tnh th thu c n MO, i vi sto thnh phn tA2sc: n/2MO lin kt c

nng lng thp hn nng lng ca cc AO em thp v n/2 MO phn lin kt (MO*)c nng lng cao hn nng lng ca cc AO em thp.3. Cch kho st cu trc phn ttheo phng php MO

s

s*

AOH AOH

1s a 1s

E+

E-

b

MO


33/40


*iu kin cc AO c khnng thp vi nhau l:

- Cc AO tham gia thp vi nhau phi c nng lng xp xnhau- Cc AO tham gia thp phi xen phnhau r rt cho nn chcn xt sthp

ca cc AO ho trlp ngoi cng

- Cc AO tham gia thp ca hai nguyn tphi c tnh i xng ging nhaui vi trc lin kt

* Phn loi cc MO:

Da vo tnh i xng ca cc MO hay tnh i xng ca phn xen phgia cc AOho tr, phn loi cc MO nhsau:

- MO: c to ra khi phn xen phca cc AO c tnh i xng trcns(1) + ns(2) sv s*

npz(1) + npz(2) zv z*npz(1) + ns(2) v *ns(1) + npz(2) v *

- MO: c to ra nu phn xen phni trn c mt ct cha trc ni hai ht nhn

npx(1) + npx(2) xv x*npy(1) + npy(2) yv y*

* Vgin nng lng ca cc MO:

- Cc AO thp c nng lng cng thp th mc nng lng ca cc MO thuc cng thp

- Khi hai AO tham gia thp xen phnhau cng mnh th stch cc mc nnglng cng ln, ngha l schnh lch gia MO lin kt v MO phn lin kt cng ln

* Tnh selectron ho trca phn tv xp cc electron vo cc MO theo quytc sau y:

- Nguyn l Pauly: Mi MO xp ti a 2e- Nguyn l vng bn: Cc electron c xp ln lt vo cc MO c nng

lng tthp n cao- Quy tc Hund: Nu c nMO c mc nng lng bng nhau th cc electron c

khuynh hng chim u vo cc MO sao cho selectron c thn l ln nht

* Xc nh cc c trng cbn ca phn t

- Tnh bc lin kt:

2

*NN

=

Trong : - Bc ca lin kt hay bi ca lin kt


34/40


N - l selectron trn cc MO lin ktN* - l selectron trn cc MO phn lin kt

Tbc lin kt suy ra bn ca lin kt v di lin kt. cng ln th lin ktcng bn v di lin kt cng ngn v ngc li. = 0 th khng to lin kt.

- Xc nh ttnh ca phn t:

Khi trong phn tc electron c thn th phn t bttrng ngoi ht v cht c gi l cht thun t. Khi s electron c thn cng ln th tnh thun tcngmnh.

Nu trong phn tkhng c electron c thn th cht b t trng ngoi y vcht c gi l cht nghch t.4. Kho st mt sphn tA2theo phng php MO (A thuc chu kII)

a. Tm cc MO

Nguyn tA c 4AO ho trl 2s, 2px, 2py, 2pz, cc phn lp trong skhng thamgia thp

Cc AO c tnh i xng ging nhau sthp li vi nhau v to nn cc MO tngng nhsau:

AO(2s) + AO(2s) MOsv MOs*2px + 2px xv x*2py + 2py yv y*

2pz + 2pz zv z*

b. Vgin nng lng cc MO

Ngoi ra, c thvit di dng dy thtcc mc nng lng tthp n cao nhsau:


35/40


(KK) s s*zx=yx*=y*z* (dy 1). Bng thc nghim chng trng dy 1 chng i vi cc nguyn tcui chu k: O2F2Ne2

Cn i vi cc nguyn tu chu ktLi2n N2tun theo dy 2:

(KK) s s* x=y zx*=y*z* (dy 2). iu ny c gii thch do sy nhaugia cc mc nng lng s* v z, v i vi cc nguyn tu chu kmc nnglng cc AO2sv AO2pxp xnhau.

c. Sp xp cc electron ho trca phn tvo cc MO theo nhng nguyn l thchhp

Phn t Selectron hotr

Cu hnh electron

Li2(Z=3) 2 (KK) s2 1

Be2(Z=4) 4 (KK) s2s*2 0B2(Z=5) 6 (KK) s2s*2x1=y1 1C2(Z=6) 8 (KK) s2s*2x2=y2 2

N2(Z=7) 10 (KK) s2s

*2x2=y

2z2 3

O2(Z=8) 12 (KK) s2s

*2z2x

2=y2x

*1=y*1 2

F2(Z=9) 14 (KK) s2s*2z2x2=y2x*2=y*2 1

5. Cc phn tkhng i xng AB (B >A)

* Nguyn tc:

Phn tc cu to bi hai nguyn tkhc nhau, v d: CO, NO, CN-

Cng chn cc AO tham gia thp to thnh cc MO phn ttheo nguyn tc trn,nhng ch n mc nng lng ca cc AO. Cc AO tham gia t hp phi c mcnng lng xp xnhau.

Nu gia hai nguyn tA v B m m in ca B ln hn m in ca A thmc nng lng AO ca B thp hn mc nng lng AO tng ng ca A nn MO lin

kt thu c snm gn mc nng lng AO ca B, cn MO phn lin kt thu c snm gn mc nng lng AO ca A. Ni chung phn tAB chu k 2 c dy nng lngcc MO tng tcc phn tA2u chu k 2.

V d1: Vit cu hnh electron ca phn tCO

2s(C)+ 2s(O) sv s*2px(C)+ 2px(O) xv x*2py(C)+ 2py(O) yv y*2pz(C)+ 2pz(O) z v z*

Tng electron ho trl 10: (KK) s2s*2x2=y2z2 32

28 ==

Tng tkho st cc phn tCN, CN-, NO v NO+


36/40


V d2: Phn tHFTham gia thp to MO gm AO(1s) ca H v 2pz ca F to thnh MOv MO*

do tnh i xng ging nhau. Ngoi ra, nguyn tF cn c 2 obitan ho tr2px v 2py

khng c obitan tng ng ca nguyn tH cng thp. Cc AO ny vn nh chti nguyn tF v c mc nng lng nhtrong nguyn t, trong phn tchng c

gi l cc MO khng lin kt.

Cu hnh e: HF s2nx2=ny2 1

2

02=

=

Ghi ch:

- Kt quthu c ph hp vi phng php cp electron lin kt- Tuy nhin c u im hn so vi phng php cp electron lin kt l c th

gii thch c stn ti ca lin kt c bi l thp phn v dnhF2+, H2

+, O2+, O2

-

- Gii thch c tnh thun tv nghch tca nhiu phn t

*

2px 2py 2pz


37/40


B- Cu to phn tI. phn cc ca phn t:

1. Phn tc cc v khng cc

Phn tkhng phn cc l phn tc cu to hon ton i xng nn trng tm cain tch (+) v trng tm ca in tch (-) ca phn ttrng ln nhau

V d: Phn tgm hai nguyn tging nhau nhH2, O2, N2, hoc phn tc cuto i xng nhCH4, BF3, BF4

-

Phn tc cc l phn tc cu to khng i xng, do trng tm ca in tch(+) v trng tm in tch (-) khng trng nhau

V d: Phn tHCl, HF, H2O, NH3

2.M men lng cc ca phn t()Mi phn tc cc l mt lng cc in gm hai in tch ngc du (+q) v (-q)

t cch nhau mt khong l l

nh gi phn cc ca phn tngi ta a ra mt i lng l momen lng

cc = q.l (C.m) hay (D : debye), 1D = 3,33.10-30

Cm

Trong : q l gi trtuyt i ca in tch , C (Coulomb)l l di lng cc, m

Momen lng cc l mt i lng c hng. Ngi ta quy c chiu hng ttrng tm ca in tch (+) n trng tm ca in tch (-).

Momen lng cc c trng cho phn cc ca phn t: cng ln th phn tcng phn cc: cc phn tcng ho trc trong khong t0 n 4D, cc phn tionc trong khong t4 - 11D.

Mmen lng cc phthuc vo nhiu yu tnh: schnh lch m in gia ccnguyn ttham gia lin kt, tnh i xng ca phn t, cc cp electron tdoV d:+ Phn tHCl c = 1,04D, phn tHI c = 0,44D+ Phn tCO2c cu trc thng:

C OO

1 1 c tng=0, cc lin kt C=O phn cc mnh nhng phn tCO2khngphn cc do =0.

+ -

q q


38/40


V d: Hai phn tNH3v NF3u c cu to l thp tam gic, ng lphn ccca hai phn tny phi bng nhau nhng thc tNH3= 1,46D v NF3= 0,2D, iu nyc gii thch nhsau:

4321NH 3 +++= 4321NF3 ++=

Trong phn tNH3momen lng cc ca cp electron tdo cng chiu vi momenlng cc ca cc mi lin kt N-H do tng momen lng cc ca phn tl ln.Cn trong phn tNF3hng ca cp electron tdo ngc chiu vi ca cc milin kt N-F do tng ca phn tNF3nhhn.3. Sphn cc ho phn t

Di tc dng ca in trng ngoi cc phn tbbin dng v thay i cu trc, do momen lng cc phn tbthay i. l hin tng phn cc ho phn t.

* Hin tng phn cc ho phn t

Di tc dng ca in trng ngoi gy ra bi hai tin, cc phn tc cc csp xp li theo phng ca in trng, l sphn cc nh hng

Mt khc, mi momen lng cc cng bko di ra lm tng trsca momen lngcc phn t, l sphn cc bin dng.

i vi cc phn tkhng cc: Khi t trong in trng gia hai bn tin th ccmy electron bht vbn (+) ca tin, cn ht nhn bht vpha bn (-), kt qutrong phn txut hin mt momen lng cc cm ng, y l hin tng phn cc hocm ng.

Cc mi lin kt yu:

Ngoi cc mi lin kt ho hc nhlin kt cng ho tr, lin kt ion c nng lng

cvi trm kJ/mol trln, cn gp nhiu loi lin kt yu hn c nng lng cvi chckJ/mol l lin kt hydro v lc Van der Waals. Cc lin kt yu ny c vai tr quantrng trong qu trnh chuyn trng thi nhbay hi, nng chy, chuyn dng th hnh

N

HH

H

1

2

3

4

N

FF

F

1

2

3

4

+- - +

-+

+-

++++++

-----

- +

- +

- +

- +


39/40


1. Lin kt hydro:

L lin kt ph, nguyn tH sau khi lin kt vi nguyn tX c m in ln li ckhnng lin kt phvi mt nguyn tkhc cng c m in ln

* Cchto lin kt hydro: nguyn tH khi lin kt vi nguyn tX c m inln nhF, O, N th cp electron ho trsblch vpha nguyn tX, nguyn tH chcn li ht nhn tch in dng, do n c khnng lin kt vi nguyn tkhc cngc m in ln v lin kt ny c gi l lin kt hydro.

V d: ++ FHFH ...

* Nng lng ca lin kt H c8- 40 kJ.mol-1. Nng lng ca lin kt hydro cngln khi m in ca nguyn tlin kt vi n cng ln v kch thc cng nh

* nh hng ca lin kt hydro:

Nng lng ca lin kt hydro nhso vi cc lin kt khc nn hu nhn chnhhng n tch cht l hc ca cc cht nh nhit si, nhit nng chy hay khnng ho tan gia cc cht.

V d:

- Do lin kt hydro gy ra hin tng lin hp phn t: (HF)n: n= 2-4; (H2O)n:

n= 2-3. Do hin tng lin hp phn tlm cho cc cht trnn kh bay hi, do lmtng nhit si, nhit bay hi. HF c nhit si, nhit bay hi cao hn nhiu sovi HCl, HBr, HI. H2O c nhit si v nhit bay hi cao hn nhiu so vi H2S,H2Se, H2Te.

- Do lin kt H lm gim khnng in ly ca nhiu cht: HF l axit yu, chtin ly yu trong HCl, HBr, HI l axit mnh

- Gy ra sbt thng vtkhi ca nc: thng thng khi nhit tng th tkhi ca cc cht gim xung, nhng i vi nc t < 4oCth tkhi ca nc tng

theo nhit v t gi trcc i 4oC

v sau tkhi li gim dn theo nhit .

2. Lc gia cc phn t:

Thc nghim cho thy, gia cc phn tca mt cht (kccc phn tkhng phncc) lun tn ti lc tng tc, gi l lc Van der Waals. Lc Vander Waals givai trquan trng trong qu trnh chuyn trng thi tp hp.

Bn cht ca lc Van der Waals gm c ba loi lc sau:

* Lc nh hng:


40/40


Tn ti trong cc phn t phn cc. Cc phn tphn cc ht ln nhau bng cc intch ngc du ca lng cc phn t, do cc phn tny nh hng vi nhau theomt hng xc nh. cng ln th lc nh hng cng ln.

* Lc cm ng:

Xut hin gia cc phn tc cc v khng cc. Khi phn tkhng cc tin gn nphn t c cc th di tc dng ca in trng gy ra bi phn t phn cc th ccphn tkhng cc bcm ng in v to thnh lng cc cm ng

* Lc khuch tn:

Do schuyn ng khng ngng ca electron v chuyn ng dao ng ca ht nhngy nn sbt i xng tm thi vsphn btrng tm in tch (+) v in tch (-) tonn momen lng cc tm thi trong phn t. Lng cc tm thi lun xut hin, trit

tiu, i du.Sxut hin lng cc ny v smt i xy ra mt cch nhp nhng tothnh mt lc ht thng xuyn gi l lc khuch tn.

* c im ca lc Van der Waals

Khng c tnh chn v bo hoNng lng nhhn 40 kJ/mol-1Lc Van der waals cng ln khi phn tc momen lng cc ln, c kch thc vkhi lng ln.

Ti liu tham kho:

1.Nguyn nh Chi, CSL Thuyt Ha Hc, NXB GD, 2004.2.Nguyn Hnh, , CSL Thuyt Ha Hc, Tp 2, NXB GD 1997.3. L Mu Quyn, CSL Thuyt Ha Hc - Phn Bi Tp, NXB KHKT,

2000.

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