Mot So Bai Tap Tinh The

8/3/2019 Mot So Bai Tap Tinh The

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MT S VN V TINH TH.

PGS TS L Kim Long, i hc Quc gia H Ni

t vn :

Trong chng trnh ho hc ph thng, phn trng thi rn ca cht v c th v tinh

th l mt phn kh l th v tru tng. Sch gio khoa nu c mt s tng cbn gio vin cn tham kho thm ti liu mi gip hc sinh hnh dung c v p dng

cc kin thc vo gii cc bi ton lin quan. Chuyn ny nhm cung cp cc kin thc

c th v cu trc tinh th v vn dng cho cc bi tp lin quan

L thuyt:

* Cu trc tinh th: Mng li tinh th (cu trc tinh th) l mng li khng gian ba

chiu trong cc nt mng l cc n v cu trc (nguyn t , ion, phn t ...).

- Tinh th kim loi

- Tinh th ion

- Tinh th nguyn t ( Hay tinh th cng ho tr)

- Tinh th phn t.

* Khi nim v c s:

L mng tinh th nh nht m bng cch tnh tin n theo hng ca ba trc tinh th ta

c th thu c ton b tinh th.

Mi c s c c trng bi cc thng s:

1. Hng s mng: a, b, c, , ,

2. S n v cu trc : n

3. S phi tr

4. c kht.

I. Mng tinh th kim loi:

1. Mt s kiu mng tinh th kim loi.

1.1. Mng lp phng n gin:

- nh l cc nguyn t kim loi hay ion dng kim

loi.

- S phi tr = 6.

- S n v cu trc: 1

1


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3/16

2. S phi tr, hc t din, hc bt din, c kht ca mng tinh th, khi lng

ring ca kim loi.

2.1. c kht ca mng tinh th

2.2. Hc t din v hc bt din:

a. Mng lp phng tm mt:

- Hc t din l 8

- Hc bt din l: 1 + 12.1/4 = 4

b. Mng lc phng:

- Hc t din l 4

- Hc bt din l: 1 + 12.1/4 = 2

2.3. c kht ca mng tinh th

a) Mng tinh th lp phng tm khi

3

A

B

C A

A

B

Lp ph ng tm khi Lp ph ng tm mt Lc ph ng cht kht

Hc t din Hc b t din


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2

3a

a

a

=4r

S qu cu trong mt c s : 1 + 8. 1/8 = 2

Tng th tch qu cu 34

2. .3

r 34 32. .( )3 4

a

Th tch ca mt c s a3 a3

b) Mng tinh th lp phng tm din

a

a

a 2 = 4.r

S qu cu trong mt c s : 6. 1/2 + 8. 1/8 = 4

Tng th tch qu cu 34

4. .3

r 34 2

4. .( )3 4

a


c) Mng tinh th lc phng cht kht

S qu cu trong mt c s: 4. 1/6 + 4. 1/12 + 1 = 2

Tng th tch qu cu 34

2. .3

r 34

2. .( )3 2

a

Th tch ca mt c s3 2 . 6

. .2 2

aa a 3 2a

4

== 68%

= = = 74%

= = = 74%

=


5/16

a

a

a

a

aa 32

a 63

2a 63

a =2.r cs

b=

a

Nhn xt: Bng tng qut cc c im ca cc mng tinh th kim loi

Cu trc Hng s

mng

S

ht

(n)

S

phi

tr

S

hc

T

S hc O c

kht (%)

Kim loi

Lp

phng

tm khi

(lptk:bcc)

= = =

90o

a=b=c

2 8 - - 68 Kim loi

kim, Ba,

Fe , V, Cr,

Lp

phng

tm din(lptd: fcc)

= = =

90o

a=b=c

4 12 8 4 74 Au, Ag, Cu,

Ni, Pb, Pd,

Pt,

Lc

phng

c kht

(hpc)

= = 90o

=120o

abc

2 12 4 2 74 Be, Mg, Zn,

Tl, Ti,

2.4. Khi lng ring ca kim loi

a) Cng thc tnh khi lng ring ca kim loi

D = 33. .

4 . A

M P

r N (*) hoc D = (n.M) / (NA.V1 )

M : Khi lng kim loi (g) ; NA: S Avogaro, n: s nguyn t trong 1 c s.

P : c kht (mng lp phng tm khi P = 68%; mng lp phng tm din,

lc phng cht kht P = 74%)r : Bn knh nguyn t (cm), V1 : th tch ca 1 mng.

5


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b) p dng:

Bi 1: Tnh khi lng ring ca tinh th Ni, bit Ni kt tinh theo mng tinh th lp

phng tm mt v bn knh ca Ni l 1,24 0A .

Gii:

a

a

a 2 = 4.r

a =04 4.1,24 3,507( )

2 2r A= = ;P= 0,74

Khi lng ring ca Ni:

8 3 23

3.58,7.0,74

4.3,14.(1, 24.10 ) .6,02.10=9,04 (g/cm3)

Bi 2: ( HSG QG 2007) Thc nghim cho bit pha rn, vng ( Au) c khi lng

ring l 19,4 g/cm3 v c mng li lp phng tm din. di cnh ca mng n v

l 4,070.10-10 m. Khi lng mol nguyn t ca vng l: 196,97 g/cm3.

1. Tnh phn trm th tch khng gian trng trong mng li tinh th ca vng.

2. Xc nh tr s ca s Avogadro.

Gii:

a

a

a 2 = 4.r

- S nguyn t trong 1 c s:

8.1/8 + 6.1/2 = 4.

- Bn knh nguyn t Au:

4.r = a 2 r= a 2 /4= 1,435.10-8

cm

Th tch b chim bi cc nguyn t:

Vnguynt= 4/3. .r3 = 4.4/3.3,14.(1,435.10-8 )3 = 5.10-23 cm3.

Th tch 1 n v:

V1 = a3 = (4,070.10-8 )3 = 6,742.10-23 cm3.

Phn trm th tch khng gian trng:

(V1 - Vnguynt).100 / Vnguyn t = 26%.

Tr s ca s Avogadro: NA = (n.M)/ ( D.V) = 6,02.1023.

Bi 3: ng kt tinh theo kiu lp phng tm din.

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a. Tnh cnh ca hnh lp phng ca mng tinh th v khong cch ngn nht gia hai

tm ca hai nguyn t ng trong mng, bit nguyn t ng c bn knh bng 1,28A0.

b. Tnh khi lng ring ca ng theo g/ cm3. Cho Cu = 64.

Gii: Bn knh nguyn t Cu l: r = 1,28.10-8 cm.

T cng thc: 4.r = a 2 a= 4.r / 2 = (4.1,28.10-8 )/1,41 = 3,63.10-8 cm.Khong cch ngn nht gia 2 tm ca hai nguyn t ng trong mng.

2.r = 2,56.10-8 cm.

Khi lng ring: D = (n.M) / (NA.V1 ) = 8,896 g/cm3.

Bi 4: ( HSG QG 2009) Mu trong c th ngi c mu v cha hemoglobin

( cht vn chuyn oxi cha st). Mu ca mt s ng vt nhuyn th khng c mu m

c mu khc v cha kim loi khc ( X). T bo n v ( mng c s) lp phng tmdin ca tinh th X c cnh bng 6,62.10 -8 cm. Khi lng ring ca nguyn t ny l

8920 kg/m3.

a. Tnh th tch ca cc nguyn t trong mt t bo v phn trm th tch ca t bo b

chim bi cc nguyn t.

b. Xc nh nguyn t X.

Gii:S nguyn t trong mt t bo: 8.1/8 + 6.1/2 = 4.

Tnh bn knh nguyn t: r = 1,276.10-8 cm.

Th tch b chim bi cc nguyn t V nguyn t = 4.4/3. .r3 = 3,48.10-23 cm3.

Th tch 1 mng c s V 1 = a3 = 4,7.10-23 cm3.

Phn trm th tch t bo b chim bi cc nguyn t: 74%.

Khi lng mol phn t: M = 63,1 g/mol. Vy X l ng.

Bi 5: Xc nh khi lng ring ca Na, Mg, K.

Gii: Xc nh khi lng ring ca cc kim loi trn theo cng thc:

D = 33. .

4 . A

M P

r N Sau in vo bng v so snh khi lng ring ca cc kim loi , gii

thch kt qu tnh c.

Kim loi Na Mg Al

Nguyn t khi (v.C) 22,99 24,31 26,98Bn knh nguyn t ( 0A ) 1,89 1,6 1,43

Mng tinh th Lptk Lpck Lptm7


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c kht 0,68 0,74 0,74Khi lng ring l thuyt (g/cm3) 0,919 1,742 2,708

Khi lng ring thc nghim

(g/cm3)0,97 1,74 2,7

Nhn xt: Khi lng ring tng theo th t: DNa < DMg < DAl. L do s bin i cutrc mng tinh th kim loi, c kht tng dn v khi lng mol nguyn t tng dn.

II. Mng tinh th ion:

* Tinh th hp cht ion c to thnh bi nhng cation v anion hnh cu c bn knh xc

nh

*Lc lin kt gia cc ion l lc ht tnh in khng nh hng

* Cc anion thng c bn knh ln hn cation nn trong tinh th ngi ta coi anion nhnhng qu cu xp kht nhau theo kiu lptm, lpck, hoc lp phng n gin. Cc cation c

kch thc nh hn nm cc hc t din hoc bt din.

Bi 1: Tinh th NaCl c cu trc lp phng tm mt ca cc ion Na +, cn cc ion Cl-

chim cc l trng tm mt trong mng c s ca cc ion Na+, ngha l c 1 ion Cl-

chim tm ca hnh lp phng. Bit cnh a ca mng c s l 5,58 0A . Khi lng mol

ca Na v Cl ln lt l 22,99 g/mol; 35,45 g/mol. Cho bn knh ca Cl- l 1,81 0A . Tnh :

a) Bn knh ca ion Na+. b) Khi lng ring ca NaCl (tinh th).

Gii:

Cc ion Cl -xp theo kiu lp phng tm mt, cc cation Na+ nh hn chim ht s hc

bt din. Tinh th NaCl gm hai mng lp phng tm mt lng vo nhau. S phi tr ca

Na+ v Cl- u bng 6.8

Na

Cl


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S ion Cl- trong mt c s: 8.1/8 + 6.1/2 = 4

S ion Na+ trong mt c s: 12.1/4 + 1.1 = 4

S phn t NaCl trong mt c s l 4

a. C: 2.(rNa+ + rCl-) = a = 5,58.10

-8

cm rNa+ = 0,98.10

-8

cm;b. Khi lng ring ca NaCl l:

D = (n.M) / (NA.V1 ) D = [ 4.(22,29 + 35,45)]/[6,02.1023.(5,58.10-8)3 ]

D = 2,21 g/cm3;

Bi 2: Phn t CuCl kt tinh di dng lp phng tm din. Hy biu din mng c s

ca CuCl.

a) Tnh s ion Cu+ v Cl - ri suy ra s phn t CuCl cha trong mng tinh th c s.

b) Xc nh bn knh ion Cu+.

Cho: D(CuCl) = 4,136 g/cm3 ; rCl-= 1,840

A ; Cu = 63,5 ; Cl = 35,5

Gii:

Cc ion Cl -xp theo kiu lp phng tm mt, cc cation Cu+ nh hn chim ht s hc

bt din. Tinh th CuCl gm hai mng lp phng tm mt lng vo nhau. S phi tr ca

Cu+ v Cl- u bng 6

S ion Cl- trong mt c s: 8.1/8 + 6.1/2 = 4

S ion Cu+ trong mt c s: 12.1/4 + 1.1 = 4; S phn t CuCl trong mt c s l 4.

Khi lng ring caCuCl l:

D = (n.M) / (NA.a3) a = 5,42.10-8 cm ( a l cnh ca hnh lp phng)

C: 2.(rCu+ + rCl-) = a = 5,42.10-8 cm rCu+ = 0,87.10-8 cm;III. Ti nh th nguyn t:

* Trong tinh th nguyn t, cc n v cu trc chim cc im nt mng l cc nguyn t,

lin kt vi nhau bng lin kt cng ho tr nn cn gi l tinh th cng ho tr.

* Do lin kt cng ho tr c tnh nh hng nn cu trc tinh th v s phi tr c quyt

nh bi c im lin kt cng ho tr,khng ph thuc vo iu kin sp xp khng gian

ca nguyn t.

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* V lin kt cng ho tr l lin kt mnh nn cc tinh th nguyn t c cng c bit

ln, nhit nng chy v nhit si cao, khng tan trong cc dung mi. Chng l cht

cch in hay bn dn.

Bi 1:

a) Hy v s m t cu trc ca mt t bo s ng ca kim cng.b) Bit hng s mng a = 3,5 0A . Hy tnh khong cch gia mt nguyn t C v mt

nguyn t C lng ging gn nht. Mi nguyn t C nh vy c bao quanh bi my

nguyn t khong cch ?

c) Hy tnh s nguyn t C trong mt t bo s ng v khi lng ring ca kim cng.

Gii:

a. * Cc nguyn t C chim v tr cc nh, cc tm mt v mt na s hc t din. S

phi tr ca C bng 4 ( Cacbon trng thi lai ho sp2).

* Mi t bo gm 8.1/8 + 6.1/2 + 4 = 8 nguyn t

* Khong cch gia mt nguyn t Cacbon v mt nguyn t cacbon lng ging gn nhtl: 2r = d/4; vi d l ng cho ca hnh lp phng d = 3.a .

2.r = 4/3.a = 1,51.10-8 cm;

b. Mi nguyn t cacbon c bao quanh bi 4 nguyn t cacbon bn cnh.

c. Khi lng ring ca kim cng:

D =VNA

Mn

.

.

= 3823 )10.5.3.(10.02,6011,12.8

= 3,72 g/cm3

Bi 2: (HSG QG 2008) Silic c cu trc tinh th ging kim cng.

10

a = 3,55 A

Lin kt C-C di 1,54 A


11/16

1. Tnh bn knh nguyn t silic. Cho khi lng ring ca silic tinh th bng 2,33g.cm-3;

khi lng mol nguyn t ca Si bng 28,1g.mol-1.

2. So snh bn knh nguyn t ca silic vi cacbon (rC = 0,077 nm) v gii thch.

Gii:

a. T cng thc tnh khi lng ring

D =VNA

Mn

.

.

V1 = ( 8.28,1)/(2,33.6,02.1023) = 16,027 cm3.

a= 5,43.10-8 cm; d = 3.a = 5,43.10-8 .1,71 = 9.39.10-8 cm;

Bn knh ca nguyn t silic l: r = d/8 = 1,17 .10-8cm;

b. C rSi (0,117 nm) > rC( 0,077 nm). iu ny ph hp vi quy lut bin i bn knh

nguyn t trong mt phn nhm chnh.

IV. Mt s kiu mng tinh th kim loi.

IV.1. Mng lp phng n gin:

- nh khi lp phng l cc nguyn t kim loi hay ion

dng kim loi; S phi tr = 6.

IV.2. Mng lp phng tm khi:

- nh v tm khi hp lp phng l nguyn t hay

ion dng kim loi; S phi tr = 8.

IV.3. Mng lp phng tm din

- nh v tm cc mt ca khi hp lp phng l cc

nguyn t hoc ion dng kim loi; S phi tr = 12.

IV.4. Mng su phng c kht (mng lc phng):

- Khi lng tr lc gic gm 3 mng c s. Mi mng

c s l mt khi hp hnh thoi. Cc nh v tm khi hp hnh thoi

l nguyn t hay ion kim loi;

- S phi tr = 12.

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V. nh hng ca lin kt kim loi n tnh cht vt l ca kim loi

Do cu trc c bit ca mng tinh th kim loi m cc kim loi rn c nhng tnh cht

vt l chung: tnh dn in, tnh dn nhit, tnh do, nh kim. Cc tnh cht vt l chung

u do electron t do trong kim loi gy ra.

Ngoi ra c im ca lin kt kim loi: Mt nguyn t (hay c kht), mt electron t do, in tch ca cation kim loi cng nh hng n cc tnh cht

vt l khc ca kim loi nh: cng, nhit nng chy, nhit si, t khi.

VI. c kht ca mng tinh th, khi lng ring ca kim loi.

VI. 1. c kht ca mng tinh th

a) Mng tinh th lp phng tm khi

2

3a

a

a

=4r

S qu cu trong mt c s : 1 + 8. 1/8 = 2

Tng th tch qu cu 34

2. .3

r 34 32. .( )3 4

a


b) Mng tinh th lp phng tm din

a

a

a 2 = 4.r

S qu cu trong mt c s : 6. 1/2 + 8. 1/8 = 4

Tng th tch qu cu

344. .3

r

34 24. .( )3 4

a


12

== = 68%

= = = 74%


13/16

c) Mng tinh th lc phng cht kht

S qu cu trong mt c s: 4. 1/6 + 4. 1/12 + 1 = 2

Tng th tch qu cu 34

2. .3

r 34

2. .( )3 2

a

Th tch ca mt c s3 2 . 6

. .2 2

aa a

3

2a

a

a

a

a

aa 32

a 63

2a 63

a =2.r cs

b=

a

VI.2. Khi lng ring ca kim loi

a) Cng thc tnh khi lng ring ca kim loi

D = 33. .

4 . A

M P

r N (*)

M : Khi lng kim loi (g) ; NA: S Avogaro

P : c kht (mng lp phng tm khi P = 68%; mng lp phng tm din,

lc phng cht kht P = 74%)

r : Bn knh nguyn t (cm)

b) p dng:

V d 1: Tnh khi lng ring ca tinh th Ni, bit Ni kt tinh theo mng tinh th

lp phng tm mt v bn knh ca Ni l 1,24 0A .

a

a

a 2 = 4.r

a =

04 4.1,24

3,507( )2 2

r

A= = ;P= 0,74

Khi lng ring ca Ni:

8 3 23

3.58,7.0,74

4.3,14.(1,24.10 ) .6,02.10=9,04 (g/cm3)

13

= = = 74%


14/16

V d 2: Xc nh khi lng ring ca Na, Mg, K

Kim loi Na Mg AlNguyn t khi (v.C) 22,99 24,31 26,98Bn knh nguyn t ( 0A ) 1,89 1,6 1,43

Mng tinh th Lptk Lpck Lptm

c kht 0,68 0,74 0,74Khi lng ring l thuyt (g/cm3) 0,919 1,742 2,708Khi lng ring thc nghim

(g/cm3)0,97 1,74 2,7

Bi 1. ng (Cu) kt tinh c dng tinh th lp phng tm din.

a) Tnh cnh lp phng a( 0A ) ca mng tinh th v khong cch ngn nht gia hai tm

ca hai nguyn t ng trong mng, bit rng nguyn t ng c bn knh bng 1,28 0A .

b) Tnh khi lng ring d ca Cu theo g/cm3. (Cho Cu = 64).

HD: a)04

2 4 2 2. 2 2.1,28 3,622

ra r a r A= = = = =

b) S nguyn t Cu trong mng tinh th:1 1

8. 6. 48 2

+ =

3

3 23 8 3

4. 4.64

8,96 /6,02.10 .(3, 62.10 )Cu

Cu

Mm g

d g cmV a cm= = = =

Bi 2. Phn t CuCl kt tinh di dng lp phng tm din. Hy biu din mng c s

ca CuCl.

a) Tnh s ion Cu+ v Cl - ri suy ra s phn t CuCl cha trong mng tinh th

c s.

b) Xc nh bn knh ion Cu+.

Cho: d(CuCl) = 4,136 g/cm3 ; rCl = 1,84 0A ; Cu = 63,5 ; Cl = 35,5

Bi 3. T nhit phng n 1185K st tn ti dng Fe vi cu trc lp phng tm

khi, t 1185K n 1667K dng Fe vi cu trc lp phng tm din. 293K st c

khi lng ring d = 7,874g/cm3.

a) Hy tnh bn knh ca nguyn t Fe.

b) Tnh khi lng ring ca st 1250K (b qua nh hng khng ng k do s dn n

nhit).

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15/16

Thp l hp kim ca st v cacbon, trong mt s khong trng gia cc nguyn

t st b chim bi nguyn t cacbon. Trong l luyn thp (l thi) st d nng chy khi

cha 4,3% cacbon v khi lng. Nu c lm lnh nhanh th cc nguyn t cacbon vn

c phn tn trong mng li lp phng ni tm, hp kim c gi l martensite cng

v dn. Kch thc ca t bo s ng ca Fe khng i.

c) Hy tnh s nguyn t trung bnh ca C trong mi t bo s ng ca Fe vi hm

lng ca C l 4,3%.

d) Hy tnh khi lng ring ca martensite. (cho Fe = 55,847; C = 12,011;

s N = 6,022. 1023 )

HD: a) S nguyn t Fe trong mt mng c s lp phng tm khi l: 20

83

23 3 23

2.55,847 2.55,8472,87.10 2,87

6,022.10 . 6,022.10 .7,874Fem

d a cm AV a

= = = = =

033 4 1,24

4

aa r r A= = =

b) nhit 1250 st tn ti dng Fe vi cu trc mng lp phng tm din.

Ta c:0

2 2. 2 2.1,24 3,51a r A= = =;

323 8 3

4.55,8478,58 /

6, 022.10 .(3, 51.10 )Fe

gd g cm

cm

= =

c) S nguyn t trung bnh ca C trong mi t bo s ng ca Fe l:

% . 4,3.2.55,8470,418

12, 011 % .12, 011 95, 7.12, 011C Fem C m

Fe= = =

d) Khi lng ring ca martensite:3

23 8 3

(2.55,847 0,418.12,011)8,20 /

6, 022.10 .(2,87.10 )

gg cm

cm+

=

Bi 4. a) Hy v s m t cu trc ca mt t bo s ng ca kim cng.

b) Bit hng s mng a = 3,5 0A . Hy tnh khong cch gia mt nguyn t C v mt

nguyn t C lng ging gn nht. Mi nguyn t C nh vy c bao quanh bi my

nguyn t khong cch ?

c) Hy tnh s nguyn t C trong mt t bo s ng v khi lng ring ca kim cng.

Bi 5. Tinh th NaCl c cu trc lp phng tm mt ca cc ion Na+, cn cc ion Cl-

chim cc l trng tm mt trong mng c s ca cc ion Na+, ngha l c 1 ion Cl-

chim tm ca hnh lp phng. Bit cnh a ca mng c s l 5,58 0A . Khi lng mol

ca Na v Cl ln lt l 22,99 g/mol; 35,45 g/mol.15


16/16

Tnh :

a) Bn knh ca ion Na+. b) Khi lng ring ca NaCl (tinh th).

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Mot So Bai Tap Tinh The

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