hỌc tỐt hÓa 10 - nguyỄn bÁ nhiỀu

7/26/2019 HC TT HA 10 - NGUYN B NHIU

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g gp PDF bi GV. Nguyn Thanh T


2/444

'THa m l c

Li ni u 3

CU TO NGUYN T 5

H'THNG TUN HON CC NGUYNT HA HC 38

LIN KT HA HC - NH LUT TUN HON

MENLEP 62

PHN NG OXI HA KH 141

VN TC PHN NG - CNBNG HA HC 214

HALOGEN 244

OXI -LU HUNH 343

4




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3/444

CU TO NGUYN T . ... .*>

THNH PHN CU TO, KCH THC,KHI LNG NGUYN T

A' TM TT GIO KHOA

Nguyn t l ht v cng .nh, i in cho nguyn t v khngb phn chia Dh trong phn ng ha hc.

/- Thnh phn cu to

* Nguyn t ca mi nguyn t iu c 2 phn : ht nhn v vnguyn t.

- Ht nhn . \

Gm 2 loi ht chinh l proton v ntron. '

* Proton (p) : mang in dutng, c in tch q = + 1 ,6 .10 ! 9

coulomb, c quy c l 1 + ' - * Ntron (n) : khng mang in, hoi nng hn proton.

Vy ht nhn mang in tch duong.

2 - V nguyn t.

To rin txri cc ht nh mang in m, gi l elctron : e (haydin t), c in tch q =* 1 ,6 . 1 0 - *9 coulomb, irc quy c l 1-

II. Kch thuc - Khi lng ca nguyn i :1- Kick thuvc I

. * Nguyn t* dc xem nh mt khi cu, ng knh 1

( : c l angstrom, 1 = lO-lo 0 i)




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4/444

* Ht nhn nguyn t' cng xem nh mt khi cu, ung knh

nh hn 1 0 . 0 0 0 ln, khoug 1 0 - 4 .

* Electron chuyn dng qanh ht nhn, t thnh v nguyn t.

Gia v nguyn t v ht nhn c. khong khng. V nguyn t ccu to rng.

-K h l yng i

Tn Khu

Khi lng m in tch q nvi Coulomb (C)n v kg n v : .v.c

Proton p 1,6726.10"27 1 .v.c + 1,602,10*19c

Electron e 9,1095.10~31 = 0,549.10 3.v.c -1,602.10_19c

Notron n, N 1,6750.10-27 * 1 .v.c 0

* in tch ca proton v electron c tr s tuyt i bng nhau.Nhng khi lung ca proton ln gp 1836 ln khi lng caelectron.

* Khi luxng ca nguyn t bng tng s khi ltig ca proton,

notron v electron. Nhng v khi lung elcctron qu nh so vi khilung proton, nn ta xem nh kh lung nguyn t gn bng tngs khi Iirqmg proton v notron.

* Ch : MOL

V k thut tnh ton trong cc bi p phc v cho chung u,nn ngh ua phn MOL trong choiig II ra hc truc.

Cc nh ngha

I- Mol : Mol l lng cht cha 6,023.1023 ht vi m.(ht vi m = ht v cng nh b).

Tr s N = 6.023.1023 gi l s' Avogadro.

6




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5/444

L khi lng thc s ca mt nguyn t (hay phn t), rtnh b. " .

Thi d : Cacbon c mc = 1,992.10~23 gam

3- Khi ltrng twong .

L khi lirng gp N ln khi lng tuyt i.

Thi d : Mc = N . mc = 6,023.1


6/444

"Trong cng mt iu kin v nhit v p sut, nhng th tbng nhau ca mi cht kh u cha cng mt s phn t."

* Ta suy ra :

Th tch moi phn t l th tch chim bi mt moi phn t kTrong iu kin tiu chun (0c, 1 atm), th tch mol phn

ca mi cht kh u bng 22,4 lit.

0* PHUONG PHP GII TON

* Vh 1 : Lin quan gia khi lng tuyt i (KLT) khi lung tuong i (KLt) hc khi luxyng mol (M). Theo ngha ta c :

KLt = 6,023.1023 KLT hoc M = 6,023 . 102 3 KLT

Khc nhau

KLt tnh bng gam hay .v.c.

M tinh bng g/mol.

Th d 1 :

Bi tkhi lng tng di ca nguyn t Nit bng 14 .v.

Tnh khi lng tuyt di ca hgyn t Nit ?

GII

Ta c : 14 = 6,023.102 3 . KLT =6- KLT =*2,324 . 10- 2 3 ga

Th d 2:

Bit khi lng tuyt i ca phn t khi Cl Cl2 b

11,789.10-23 gam. Tnh khi lng mi phn t, Clo ?

GII

Ta c : M - 6,023.1023 KLT

= 6,023.1023 X11,789.10~23 71 g/mol.




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7/444

* Van 2.Lin quan gia s Avogadro N mol M, khi liKngm v s nguyn t, s phn t, s ln.

Theo nh ngha :

* 1moi nguyn t c 6,023.1023 nguyn t.

* 1 mi phn t, c 6,023.1023 phn tv * 1 mol Ipn c 6,023.1023 lon,

Mt khc, khi lir-ng mol M l khi lutyng cua 1 moi, nn

M = ~ hay m = Mnn

m l khi lirng c n moi.

Th d 1 :

Tnh s nguyn t lu hunh s cha trng 12,8 gam uu hunh, bit nguyn t lng s bng 32 .v.c.

GII

Ta c : 32 gam S cha 6,023.-1023 nguyn t s.

12,8 gam s cha X ?

2,8X6,023 23 AS nguyn t s : X - " 2,4092 10" nguyn t*

* .

th d 2 :

diu kin chun 5,6ml kh mtancha bo nhiu phn tmtan ?

> GII.

Mt th tch mol phn t mtan hay 22400ml ktc cha

6,023.1023 phn t. Vy 5,6ml khi mtn cha :

5 , 6 X 6 , 0 2 3 X 1 023 : - ss 1 ,5 . l


8/444

GII..

Ta c : m = M =*n = ^ '= = 5 mol.

S phn t HjO tuong ng :

u X 6 ,0 2 3 X 1 023 = 9 ,0 3 4 5 X 1 023

S' nguyn t H = 2 s phn t H,0 = 18,069 Xl23.

c- BI TP P DNG V NNG CAO

Bi 1

Cho .v.c - 1,66.10- 2 4 gam1. Tnh khi lng mol nguyn t ca Na, Mg, p, s bit r

khi lng tut i ca chng ln lt nh sau

mNa = 38,1634.10~24g ; mMg = 40,358. l< r2 4 g;

n ip = 5 1 , 4 1 7 . 1 0 24g ; m s = 5 3 , 2 2 6 . 1 0 ^ 4g

2) Tnh khi lng tuyt di ca K, Al, N, Cl bit rng nguyt lng ca chng ln lt bng :

Mk = 39,102 .v.c; MA[ = 26,982 .v.c

= 14,007 .v.c; MC| = 35,5 .v.c

GII

1 - Ta bit rng :

khi lng tuyt iKL moi nguyn t = ----- *~~ ~

1,66 . 10 24hay KL, mi nguycn t 6,0 23.1023 khi lung tuyt di.

M 6,023 . 102 3 m

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9/444

* $ ) m = . 3,14 . (1 ,44.10' 8 ) 3 .26,16 = 32,704 . l< r23g

Khi lung mol nguyn t :

M . Nm * 6,023. 0 2 3 .32,704.10- 2 3 * 196,976 g/mol

M * 197 g/mol.Khi lng nguyn t : M = 197 .v.c = nip + ttijj = ip + 1"18

Vy : rp 79 .v.c ng vi 79 ht proton.

Bi 4 :

Nguyn t lng (Ma bc bng 107,87 .v.c v khi lng ringca bc bang 10,5 g/'cm3. Cc nguyn t bc chim 74% th tchca tinh th, cn li l phn rng. Th bn kinh nguyn t ca bc theo angstrom ( , . '

%>

Khi lutng tut ca bc :

/ rttheo cng thc : m - Vd * *jjrR3 d, R l bn kinh, on v ,

Ta c : 17,91. l< r2 3 - . 3,14. (R . 1(T8 ) 3 . o 3 > ^

r 3 _ 17,91. O~33,-3 .100

74 .4.3,14 .10,5 . 10- 2 4

R ,766.




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10/444

HT NHN NGUYN T NGUYN T HA HC NG V'

r TM TT GIO KHOA,

I Ht nhn nguyn t.

I- in tch ht nhn z .

Theo quy uc, n tch ca mi prton bng 1 + v mi electrbng 1 -, nn trong ht nhn c z proton th c z+ in tch. in tch ht nhn.

Mt khc, nguyn Hr trung ha v in, nn :

s din tch ht nhn Z = s proton p s elecon e.

2yS'ch A ;

Bng tng s s ht proton v s ntron. Ngha l :

A = z + N hay A = p + n (N, n : s notron).

II Nguyn t ha hc .

1- 3nh ngha v tnh ch :

L nhng oi nguyn liu c bn cu to nn cc cht. HocL mt loi nguyn t c cng in tch ht nhn.

Hin c vo khong 109 ngilyn t ha hc v O8 0 nguyt ca chng.

Tnh cht ca mt nguyn t ha hc l tinh cht ca tt cc nguyn t ca Ijguyen t .

2- S' hiu nguyn t.

L s din tch ht nhn z ca nguyn t ca mt nguycn t.

14




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11/444

(S hiu nguyn t cn gi l s th t hay bc s nguyn t)

3- K hiu nguyn hr.

Nguyn t ca nguyn t X c s hiu nguyn t v s khi _\ 1 ! _\__ '___ A, c k hiu nh sau :

A

III Bng v.. , ;

1- nh ngha.

L-nh ng Ilgun t. ca mt ngyn t, c cng s proton p

nhung khe s ntrn n, do khc nhau s khi A.

2- Tnh ch.

- Cc nguyn t ng v c cng tinh cht ha hc nhung khcnhau tnh cht l hc.

3- Khi lng nguyn t trung bnh ca nguyn t'.

L khi lQHg nguyn t trung bnh ca hn hp cc ng v(cn gi l nguyn t lng trung bnh), ki hiu l M

khi lng hn hp cc ng v

tng s nguyn t ng v

B* PHLPNG PHP GII TON.

* Vn 1. \

Phn bit in tch ht nhn, s proton, selectron v s hiunguyn t.

* Ging nhau : c cng tr s tuyt i. (Z = p = e)

* Khc nhau :- in tch ht nhn z + bng in tch proton p +- in tch electron : bng z -




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12/444

- s hiu nguyn t' z khng c n v.

Thi d :

Nguyn t Na c s hiu nguyn t bng 1 1 . Ngha l Na c - in tch ht nhn * in tch proton 1 1 +

- in tch electron = 11 - V* Vh 2 .

Phn bit s khi A v khi lng nguyn t M

* Ging nhau : c cttg tr s tuyt i.

* Khc nhau :

- S khi A : khng c dn v, bng tng s s ht proton vht ncrtron : A = p + n

- Khi lutyng ngycn t M : bng tng s khi lngproton vkhi lung ntron, tnh bng .v.c : M *lip + mn

Th d :

Nguyn t AI c s hiu nguyn t 13 v c 14 ntron vy :

, S khi A - z + N p + n - 13 + 14 = 27 ht* Khi lung nguyn t.

M Bp + mn = 13 + 14 = 27 .v.c.* Vn 3.

Nguyn t lng trung bnh ca nguyn t.

* Mt nguyn t X c nhiu ng v; mi ng vchimms nguycn t. t XJ,X,, x3... l s nguycn t .'v M3 l

nguyn t lng ca ng v 1 ,2 ,3 .. .

Nguyn t lutng trung hnh ca nguyn t X l :

khi hnng hn hp cc ng VItng s nguycn t ng v (

hy :

16




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13/444

* N X| ,X2 X3 . . . l % s nguyn t ng v 1, 2, 3... thi

_ M 1X1 + M3X2+ M;jX3+ ...

M '


14/444

khi lng hn hp ng v 4997,5 - -M .t A' M2 "" 4- K J^ o

tiig s nguyn t ng v 125

* p dng cng thc

MX + M2x2 + M3x3M * ---------------------

Xj + x2 + x3

36.0,34 + 38,0,06 + A . 99,639,98 --------- "

100

. * A 4 0

Thi d 3:

Ngun t lng ca Ne bng 20,18 v Ne c 2 ng v c skhi bng 20, 22. Tnh phn trm s nguyn t ca mi ng v ?

GII

t Xl % s nguyn t ca ng v 20'(^Ne)

p dng cng thc : M = Mjx + M2 (1 x)

Ta c : 20,18 = 20x + 22 (1 - x)

X - 0 ,9 1 h a y 9 1 %

* Vn 4 iT s' N/Z hay p ?

* Cc nguyn t ha hc c s z > 82 (Z = 82 l ca Pb)uc tnh phng x, ht nhn nguyn t khng bn v t phn h

* Trong s cc ht nhn nguycn t ca nguyn t c z < 83 th207

ch " Pb c t s s notron v -S proton cao nht l :o2

. n N 207 - 82___

p = z = ~ 8 2 *

v t s thp nht l: = 1 ( ca c, o... tr I H)p z 1

18




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15/444

i S f l

* Vy vi nhng nguyn t c z < 82 thi ti s N/Z trongkhong: ,1 ^ - < l,>24-" | ' :

* Mt khc, gi l tng 's phn t ong nguyn t.

Ta c : = p + e + n = 2p + n'?:'* 2 => 2 = 5 - 2

Vy : 1 - - 2 < 1,524p

s S< > < f 3,524

Th d:Xc nh s proton, s electron v s' ntron c mt nguyn

t, bit rng tt> s phn t trong nguyn tic l 58 v s khi ca nguyn t nh hn 40.

GII

V tng s phn t bng 58 nn nguyn t khng phi lH (Z - 1) v Pb (Z = 82). Do :

s sX < p < 3,524 3

58 58=> < p p as z * 16

nx = 17

2- Ta c :

* Ax = p + nx - 16 + 17 = 33. ,

* Ay =: p + Dy 16 + 17' + 3 * 36.

* A2 = "" (Ay 4" Ay) ^ 2 A2 * 36 ^ Aj

* "t Ay 41Az f" At = 135

36 4Athy 33+36+----- L + Ax = 135

At

s 32

Az = 34

* Nguyn t lipng ttng bnh ca nguyn t A :

M = 0A(? h 12 = 32,088315

* t z v t l phn trm s nguyn t ca 2 ng v z, T. Tac :

- _ 33 .0,74 + 36.0,02 + 34z + 32t M = ----------- - 7 --------------- 1 = 32,088.100=> 34z + 32t = 3183,66 (1)

v : 0,74 + 0,02 + z + t = 100

* z + t = 99,24 (2)

23




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20/444

(3) s> - + A2 + 3 = 202 (3')

(1 ) => A, + A2 + A 3 = 618 (1/)

Suy ta : Aj = 208

Thy A4 v (3) vo (4) :

207- 202 + 4p = 333' p = z - 82

T h a y ' ; A4 v p vo (2) : A3 = 204

Thay A j; vo (1') : 206

2) Nguyn t lung trung bnh ca nguyn t Y :

- 208.50,9 + 206.23,3 + 204.0,9 + 207.24,9

M = --------------

-------------------------------

100M 207,249

Bi 3 :

Mt nguyn t X c tng s' ht bng 49 v s' ht khng min bng 53,125% s ht mang in.

1 - Tnh in tch ht nhn ca nguyn t X.

2 - Nguyn t' A gm 4 ng v X Y z T

- ng v X ni trn, chim 0,74% s nguyn t.

- ng v Y nhiu hn X 3 ntron v chim 0,02%.

- Tng s nguyn t lng ca 4 ng v X Y z T bngI35.v

- S khi ca ng v z bng trung bnh cng s khi ca v Y v T .

- Khi lung ca 315 nguyn t ca nguyn t' A b

10.107,72&VC.Hy tnh phn trm s' nguyn t ca ng v z, T ?

22




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21/444

1- Tng s h t : p + e + n = 49

. 5 3 , 1 2 5 . ^ ,V p = e v n = JQQ (p + e)

53 125nn : p + e + n - 2p + JQQ (p + p) = 3,0625p = 49

p = z = 16

nx = 17

2- Ta c :

* A p * ^ 17 * 33.

* Ay = p+ ny = 16 + 17 + 3 = 36.

* Az - 2 (Ay + Ax) => 2AZ = 36 + AT

* Aj{ + Ay + Ajr + Aj = 135

36 + Athay 33+36+- 1 + Ar = 135

=> Af 32

Az - 34

* Nguyn t lng trung bnh ca nguyn t A :

M ;= 1 0 1 0 I *-.72-> 32,088

* t z v t l phn ttm s nguyn t ca 2 ng v z, T.c :

= 33 .0,74 + 36 02 + 34z +_32t s 3 2 88/

100=> 34z + 32t - 3183,66 (1)

v : 0,74 + 0,02 +z +t - 100

=*> z + t = 99,24 (2)

GII. .




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22/444

T.(l) v (2),ta c :

z = 3,99 v t 95,25

; Bi 4 : - '

Mt nguyn t X ca nguyn t' R c tng s ht bng 54 c s khi nh hn 38.

1- Xc nh s proton v s ntron ca nguyn t X ca R.

2- Nguyn t' R c 2 ng vX, Y m s ntron hn km n2 ht. Tng s khi ca 2 ng v gp 3 ln s' in tch ht nhca Crm (c z = 24). . v.-v

Khi cho 1,43 gam Zn tc dng vi lng R d, thu c 2,9gam mui khan ZttRy Hy tinh s khi v phn trm s nguyn

ca mi loi ng v. (Zn * 65)

GII.

Ta c : - p + c + n - 2p + n = 54.n 54 ,*> - = - 2p p

v s 54


23/444

Ax = p + nx

Ay = p + nY

* Mt khc :Zn + R,

65

l>43g

Ax - & - nx

37 - Ay - 2

AY = 35

ZnR2

65 + 2R

2,992

65 + 2R =65x2,992

1,43

R *. 35,5 = M l nguycn t lirng trung binh

t X l %s nguycn t ca ng v X.

t 1 - X l %s nguycn t ca dng v Y.

Ta c : M = 3 7 x + 35(1 - x) = 35,5

=> X = 0,25 hay 25%

Bi 5. Mt nguyn t R c 3 ng v X, Y, z. .

Tng s phn t ca 3 ing v bng 129.- S n.tron ca ng v X v Y hon kcm nhau 1 ht,

- ng v Z c s proton bng s notron.

1- Xc nh in tch ht nhn nguycn t v s khi ca 3 ngv X, Y, z ?

2- Bit 752.875V1020 nguyn t ca R'c khi lng m gam.T l s nguyn t cc dng v .nhu-sau z :

Y = 2769 : 141 v V : X = 611*: 390.Xc nh nguyn t lutng trung bnh ca R v tinh m ?

25




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24/444

1- Ta c : ;

* (p + e + nx) + (p + e + ny) + (p + e + 1 12) * 129

vi p = e v p > nz nn :

7p + nx + nY = 129

5nx ~ Iiy = 1

* T (1) v (2) : 7p + 2nx = 130

n* 65=* _ = - 3,5

p p

GII

* Da vo iu kin : 1 < < 1,524p

(1)

(2)

ta c :

hay :

Bin lun.:

1 $ - 3,5 < 1,524p

12,9 < p < 14,4

p 13 14

nx 19,5 16

Ax 32,5 30

Kt lun loi ng

Suy ra : Az = p + nz

=>'

p = 14 v nx 16

Ax - 30

Ay = p + tiy = 14 + 15 = 29.

2. t X, y; z l s nguyn t ca 3 dng v X, Y, .Ta c :

* z : = 2769 : 141923

s> z = y47 y

26

t sS iiir




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25/444

* : X 611 : 390 => X = ~ y

Nguyn t irng trung binh ca R :

Axx + AYy + Azz 30x + 29y + 28zMr = ------------ ---------- -= ----------------------

x + y + z x + y + z

3 0 . ^ y + 29y + 2 8 . ^ y

R ' , I 0 7 y + y + y

47 7 47

C mt moi'nguycn t hay 28,107 gam R c 6,023.1023 nguynt. Vy khi lirng ttia 752.875 . l 2 0 nguyn t l 3513,375gam.

27




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26/444

v NGUYN T

A' TM TT GIO KHOA

/. S chuyn ng ca Electron trong nguyn t'.

Electron l phn t mang in tch m, chuyn ng trong nguynt khng theo mt qu o xc nh. V chuyn dng rt nhanh nnelectron to thnh quanh ht nhn mt vng khng gian mang inm gM my electron hay orbitan nguyn t. y kh nng hindiu ca elcctron l ti a, n 90% (Ngha l trong 100 giy thi cn 90 giy l electron c mt).

t. Lp electron (hay tng, hay mc nng lng).* Trong nguyn t, cc electron lin kt vi ht nhn khng u

* Electron gn ht nhn nht th b ht nhn ht mnh nht,nghla l chng mc nng lutng thp nht.

* Ngc li, elcctron xa ht nhn nht, th b ht yu nht, cmc nng lng cao nht.

Ty theo mc nng lung cao, thp m electron uc phn bthnh tng lp.

* C tt c 7 lp u-c nh du theo th t t trong ht nhn

ra ngoi l : 1,2.,. 7 hay K, L... Q.Q --------- n = 7

p ------------- 6

o -------- ----- -5 K L M N 0 p Q- ' A V 4

N ------- ------ 4 > .'\V

> : : :tK -------------- n = 1 n = 12. 3 '4 5 6




B

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O

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27/444

s e ti a trong mi lp utyc xc nh bi cng thiiv 2 n2vi w *s 4 ( l l s th t ca lp).

Vy : lp K (n - 1 ) c ti a 2c.

lp L (n 2) c ti a 8 e

lp M (n= 3) c ti a 18e

lp N (n = 4) c ti a 32e

Cc lp 5, 6 , 7 cng ti a 32e v s nguyn t tm thy hnnay chira .

II. Phn lp electron (hay ph tng, hay phn mc nnglng). I

Mi lp electron to thnh bi'mt hoc nhiu phn lp.

Lp th n c n phn lp. Ta ch xct n n < 4 , nn c 4' phnlp, gi tn theo th t s, p, d, f. Cr' ^ * '

Electron phn lp no th gi tn theo phn lp .

S electron ti a trong phn lp nh sau :

* Phn lp s c ti a 2e, k hiu s2

* Phn lp p c ti a 6 e, ki hiu p6

* Phn lp c ti a 10c, ki hiu d1 0

Phn lp f c ti a 4, k hiu f1 4 .

Tm li :

VLp S ti a /

lp

Phn lp S e ti a / phn lp

1 hay K 2 s s2

2 hay L 8 s p S2 P6

3 hay M 18 s p s2 p6 10

4 hay N 32 s p d f s2 p6 I Ofl4




B

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O

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28/444

ly Orbitan

*Orbitan l vng khng gian quanh ht nhn, trong kh nnghin din ca electron l ti a.

* S v hnh dng ca orbitan ty thuc vo phn lp. Ta ch

xt phn lp s v p.* Mi orbitan cha ti a 2 electron, do :

- Phn lp s c 1 orbitan s. Orbitan s c hnh cu, tm htnhn nguyn t.

- Phn lp P c 3 orbitan p (v phn lp p c ti a 6 e). Mi

orbitan s orbitan p

* Orbitan no ch c 1 electron, gi l electron c thn.

V Cu trc electron trong nguyn t

1- Nguyn li vng bn :

Trong nguyn t, cc electron irc phn b theo mc nng lngt thp n cao (tc l t gn ht nhn nht n xa ht nhn nht).

2- Cu hnh electron

Dng biu din s phn b electron theo lp v phn lp.

a) Biu din bng ch v s.

* Lp tnyc ghi bng s.* Phn lp duc ghi bng ch ci s, p, d, f km theo s lp.

* S e ghi bn phi pha trn ch ci.

30




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O

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29/444

Th d :

iiNa: Is2 2s2 2p6 3s 1 vit gn l N a : 2/8/1

b) Biu, din bng orbitan. '

Mi orbitan biu din bng mt vung hay mt vpng trn; mi

eletron bng mt mi tn v hi elecon bng hatmi tn ngirc

: : : / :

chiu.

Thi d

i7c l : liU E l I n m l l s a T

Is 2 2s2 2p6 3s2 3P5

vit gn 1 7 CI : 2/8/7

VI c im ca lp electron ngoi cng.

* Ti a bng 8 .

* Nguyn t no c 8e ngoi cng th rt bn, khng tham giacc phn ng ha hc. l nguyn t kh him.

* Ngun t kim loi c 1, 2, 3 electron ngoi cng.

* Nguyn tphi kim c 5, 6, 7 electron ngoi cng.

* Nguyn t c 4e ngoi cng c th l phi kim ( chu k nh)

hoc kim loi ( chu k lti)* Cc eecon ngoi cng quyt nh tinh cht ha hc ca nguyn

t.

B. PHITNG PHP GII TON.

Vn 1 .

l n h bn ca lp v phn lp

* Lp bn khi cha ti a s clecon (gi l bo h)* Lp ngoi cng bn khi cha ng 8 e (hoc 2e ch trng hp

nguyn t c 1 lp l Hci H). ;

* Phn lp bn khi bo ha hoc bn bo ha.

31




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O

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-

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30/444

Th d : s2 hoe s* , p6 hoc p3 ; d 1 0 hoc d5 ,

v f1 4 hoc f7 (bn)

Vn 2.

Dng quy tc Klechkowski vit mc nng lumtg v

hnh electron :Ngoi 2 lp K v L c mc nng lng khc nhau r rt,

lp cn li mc nng lng k tip c th chng ln lcn nhau,I s phn lp nhiu hn.

V vy ta dng quy tc Klcchktyski sau y, sp xp cc electtheo th t mc nng lng tng dn, vit cu hnh electron mt nguy nt. t . .. .

Lp 7 . 7s 7p I

6 { 1 I V X N5 5s 5p 5d 5f

V V X( 4 4s 4p 4(1 4f1 4

N X . ' "3 , I 3s>3p 3d1 0 (u mi ten ny

I2 , ) 2 s 2p6 l ui mi tn k

1

J .

is2

1

Ht nhn

Th d : vit cu hnh elcctron ca :

* 1 0Ne: ls 2 2 s2 2p6 hay 2 / 8 hay[NcJ,

* 1 7 C1 : Is2 2s2 2p6 3s2 3P5 hay [Nc] 3s2 3p5 hay 2

* gAr : ls2 2s2 2p6 3s2 3p6 hay 2/8/8 hay [Ar]

* l9K : Is2 2s2 2p6 3s2 3p6 4s1 hay 2/8/8/1hay [Ar] 4s!.

Ch : Vi trung hp ngoi l c mc nng lung, thi d v Cu sau y :

32




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31/444

24'

29

Cr : thay v : ls2 2s2 2p6 3s2 3p6 4s2 3d4 (Khng bn)

th ls2 2s2 2p 3s2 3p6 4s 3d5 (kh bn)

9Cu : thay v : ls2 2s2 2p6 3s2 3p6 4s2 3d? (khng bn)

thi: Is2 2s2 2p6 3s2 3p6 4s13d10 (kh bn)

Ch 2 :Ngoi quy tc trn, ngui ta cn dng thm quy tc Hund :Trong mi phn lp, cc electron c phn b sao cho s electrondc thn l nhiu nht.

Th d : trung hp p3 th.

(khng thch hp) v t \ 1 1 I t (thch hp)

C * BI TP P DNG V NNG CAO.

Bi 1 :

Trong nhng nguyn t c z t n 40, nguyn t no c :

lp electron ngoi cng bng 8 v l. s. ti a ?

GII

Vit c hnh elcctron ca 40 nguycn t ny, ta thy ch cNeon (Ne : z = 10) (kh him) l tha iu kin bi : Is2 2s22p6 hay 2 /8 .

Cc nguyn t khi hii cn li chi t mt trong 2 yu cu bi. l :

He : z - 2 : C Mp clcctron : ti a 2e Ar : z 18 : C 3 lp electron : 2/8/8 : lp ngoi cngbng-

8

nhng khng phi ti a (l 18) Kr : (Z = 36) : c 4 lp electron : 2/8/18/8 : lpi ngo cng

bng 8 nhung khng phi ti (l 321




B

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N

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O

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-

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32/444

Ch bit mc nng lung ngoi cng ca cc nguh t

sau : ) 3s2 b) 4p6 c) 3d5 d) 5ps

1- Nguyn t no l kim hi ? L phi kim ? L kh him

2- Vit cu knh electron y ca tng nguyn t. Sus hiu nguyn t ca nguyn t tng ng. Da vo quy tc vit uc cu hnh electron ny ?

' GII

1 . a) v c) L m loi d) L phi kim b) L khi

2) a ) : Is2 2s2 2p6 .3s2 (Z =12)

b) :Is2 2s2 2p6 3s2 3p6 3' 4s2 4p6 : (Z = 36)c) :Is2 2s2 2p6 3s2 3p6- 3^ 5 4s2 : (Z = 25)

d) : Is2 2s2 2p6 3s2 3p6 j P 4s2 4p6 4d 5s2 5p5 : (Z * 53

Da vo nguynli vng bn m in khai c th l quyKlcchkowski : Gc elcctron ln lut chim cc mc nng lunthp n cao. Khi no mc nng lng thp cha selecti a thi electron mi vo cc mc nng lutjiig cao hon.

Bi 3.

A l nguyn t' m nguyn t c phn lp ngoi cng lB l nguyn t m nguyn t cung c phn lp 3p, hai phnny cch nhau mt electron. B c 2 electron lp ngoi cng

1- Xc nh s' hiu nguyn t ca A v B. Nguyn tkim loi ? L phi kim ? L khi him ?

2 - X v Y l 2 ng y ca nguyn t' A c tng s' k

72. Hiu s' s' ntron ca X, bng 1/10 in tch ht nhnB. T l s nguyn t ca X v Y bng 32,75/98,25. Tnh s ca X, Y v nguyn t lung trung binh ca A ?

34




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33/444

1 - Theo bi, mc nng lng :

* Ca l : Is2 2s2 2p6 3s2 3px (x : s electron ca 3p)

* Ca B i : ls^ 2s2 2p6 3s2 3p6 4s2 (v B phi c phn lp 3p

y 6 e, mi c phn lp 4s ngoi cng cha 2e). Phn lp 3pca A v B cch nhau mt electron nn 6 - X = 1 => X = 5

* : ls2 2s2 2p6 3s2 3p5 c s hiu nguycn t z = p = 17* B : ls 2 2s2 2p6 3s2 3p6 4s2 : c s hiu nguyn t z = p = 20= A l phi kim; B l kim loi

2- Theo bi :* S khi X = p + nx

* S khi Y = p + nY

* Tng s X + Y = 2p + nx + nY = 72

11X + nY = 38 (1)

* .. _! - _ 2 0 x - y - 1 0 Z b - i 0

Iix nY = 2 (2)

T (1) v (2) ta c : nx = 20 v Oy = 18

Suy ra s khi : X = 37 v Y = 35

T l s nguyn t ca 2 ng v X, Y :

X '+Y




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34/444

ta c 37x + 35y 37x + 35. 3xM = ----- --- .as -

X + y X + 3x

M = 35,5

Bi 4 :

Nguyn t ca mt nguyn t A c cu hnh electron ngocng 5p5. Ti s s' ntron v in tch ht nhn bng 1,396S ntron ca nguyn t ny bng -3,7 ln s ntron ca nguyt thuc nguyn t B. Khi cho 4,29gam B tc dng vi lirng

, thu c 18,26 gam sn phm c cng thc AB.

J - Xc nh s hiu nguyn t v khi lng mol ngyn 9 r% n r%

GII .

jl 1 ) * Mc nng lng y ca nguycn t ca A :

ls 2 2s2 2p6 3s2 36 4s2 3(l 0 4p6 5s2 4(1, 5P5 hay 2/8/18/18/7

2- Cht no kim loi ? L phi kim ?

p

S khi : p + - 53. + 74 - 127

S hiu nguyn t : z = 53 = p* - = 1,3962 => n * 13962 X 53 74

n

Khi lung mi nguycn t M = 127 g/nol

* S notron ca nguycn t ca nguyn t B

S phn ng : t ------- - 9 - aB

M' M' + 127

4^9 18,26 13,97 0,11

M' =39

4,29 gam 18,26 gantiM' M' + 127 1271




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35/444

Khi lng mol nguyn t M' * 39 g/mol

S khi tuong ng : p' + n = 39

S hiu nguycn t Z' = p' = 19

Cu hnh electron ca nguyn t ca B :

ls2 2s2 2p6 3s2 3p6 4s : (2/8/8/1)

2) A l phi kim, B l kim loi.

3'i




B

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-

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36/444

H THNG TUN HON

, CC NGUYN T H HC- -

- TM TT GIO KHOA/. Nguyn tc sp xp :

H thong tun hon (HTTH) hay Bng tun hon hin nay gmkhong 109 nguyn t rnyc sp xp nhau theo chiu tng dn cin tch ht. nhn.

IIM t brig H thng tun hon :

C 2 loi bng tun hon : bng ngn v bng di. Ta ch xdng bng ngn.

Cc nguyn t uc sp xp trong cc theo chu ki v nhm.

1- Chu k v hng.

Chu ki gm tt c nhng nguyn t m nguyn tir ca chng ccng s lp electron, s th* t ca chu k bng s lp clcctton, do c 7 chu k chia lm 10 hng. C 2 loi chu k :


37/444

* Chu ki 7 : cha y , hiu c 23 nguyn t c z * 87 n = 109 ;

Trng 3 ch k ln 4, 5, 6 :

- Hng trn c 10 nguyn t kim loi. *

* Hng dui c 8 nguyn t va kim loi va phi kimV 2 - Nhm v phn nhm .

Nhm gm tt c nhng nguyn t c ha tr cao nht i viOxi bng nhau v bang s th t M+




B

I

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N

GT

O

N

-

L

-

H

A

C

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2

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1

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0

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38/444

- dng on cht: kim loi kint l lm lo in hnh.

- dng hp cht: kim loi c ha tr 1.

y) Phn nhm chnh nhm VII (hay nhm Halogen)

*Gm 5 nguyn t F, Cl, Br,/I v At.* Cc ngyn t Halogen c 7 electron ngoi cng, thiu mt eso vi nguyn t kh him gn nht. Do trong phn ng ha hc,chng d dtg nhn thm-mt ena t c cu bn ca kh him :

' x + c ----- * X -

* Vy :

- dng on cht: phn t Halogen x 2 l nhngphi kim in

hnh . . -

- dng hp cht vi kim loi, Halogen c- h r 1.

b) Phn nhm ph.

Gm cc nguyn t chu ki ln v u l nguyn t kim loi.Mi nhm c mt phn nhm ph, ricng nhm vu i c 3 phn nhm

ph, ni lcti hng trn v hng diri ca 3 chu k ln.

3 - Cc nguyn t' xp ngoi bng tun hon.

*Trong chu ki VI c 15 nguyn t kim loi z = 5 7 n z * 71,c tinh cht rt ging nha, h tr 3, to thnh h Lantan. Do ,chng uc xp vo nhm III v nm ricng mt hng cui bng.

* Trong chu ki VII cng c 15 nguyn t kim loi, C...2 89n z = 103, c tnh cht rt ging nhau, ha tr 3, taotfyahh h

Actini. Do , chng uqrc xp vo nhm III vnffi ricrig tnt irang cui bng.

III. Nhn x t v s* bin d) cu trc electron ca

nguyn t' ca cc ngyn t trong bng tun hon.* Vi cu trc electron ca lp ngoi cng quyt nh tinh chtha hc ca cc nguyn t v khi in tch ht nhn tng, s clectronlp ngoi cng trong nguycn t ca cc nguycn t bin i tun hon.

40




B

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GT

O

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-

L

-

H

A

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P

2

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1

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0

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39/444

* Do , khi in tch ht nhn tng : s election lp ngoicng bin i tun hon lm cho tinh cht ha hc ca cc nguynt bin i tun hon theo.

B - p h n g p h p g i i t o n v g i o k h o a

Van 1 : Chu k

* By chu ki khi u bng 7 nguyn t nhm I (gm H v nguyn t kim loi kim) v kt thc bng 6 nguyn t khi him nhorn VIII (chu VII chua kt thc).

* S th t n ca chu ki ng vi s lp electron.

* Chu k no cng khi a bng phn lp ns1 v kt thc bng

phn lp np6 . Do , s electron ngoi cng bin di t 1 n 8 (trchu k I c ' 2 nguyn t).

* Khi kt hp vi Oxi to thnh xit, cc nguyn t c ha trtng dn t 1 n 7 (tr nguycn t khi him nhm VIII). Cngt h c t ng qu t R 2O x , X l t h t nhm ,

* Khi kt hp vi Hir, cc nguyn t phi kim to thnh cchp cht kh, c ha tr gim dn t nhm IV n VII, cng thctng qut RHg _ x, X l th t nhm.

* Nguc li, cc nguyn t kim loi kt hp vi hir to thnhcc hirua, thng l nhng tinh thkhng mu (mt s dc bay hoithnh cht kh). Ha tr cc nguyn t kim loi trong hirua tng dnt nhm I n nhm III. Cng thc tng qut RHX

Vh 2: Nhm

* Nguyn t c cc nguyn t trong phn nhm chnh c selectron lp. ngoi cng bng s th t ca nhm

* dng hp cht : A

- Nguyn t ca cc nguyn t kim loi trong phn nhm chinhm I, II, III khi kt hp vi phi kim, hir, c ha tr 1, 2, 3.




B

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N

GT

O

N

-

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40/444

Nguyn t ca cc nguyn t phi kim trng phn nhm chnhm V, VI, VII khi kt hp vi kim loi, hir, c ha tr 3, 2, 1 .

- Nguyn t ca cc nguyn t nhm IV gm kim loi, p; ' u c ha tr 4,

ti 3 : Cng thc' Oxit bc cao nh v hp cht vi Hirca nguyn t 'R

Nhm X I II IH IV V VI y n

Oxit bc

cao nht

Cng thc RO RO ^ 3 r o 2 R25 r o 3 r 27

Ha tr R 1 3 4 5 6 7

m p

cht vi

Hir

Cng thc RH r h 2 r h 3 r h 4

'.

-X

001

\j\

RH8-6

r*1

00

Ha tr R 1 2 3 4 3 2 , 1

Th d 1:

Mt nguyn t R phn nhm chnh nhm vu chim 38,79%v khi tng trong oxit bc cao nht. Xc nh nguyn t lngcua R ?

GII .Nguyn t R nhm VII nn cng thc oxit bc cao nht l

R2 O7 . Theo bi :

mR 2R 38,79 . , , ,= _ = ' ..=>R as 3 5 5 .v.c l nguyn t lmyng ClomOxit 2R + 7.16 100

Th d 2 * :

Mt nguyn t' R phn nhm chnh nhm V, chim 91,176%

trong hp cht kh vi H. Xc nh nguyn t lng ca R. GII .

Nguyn t R nhm V ncn hp cht kh vi H c cng thcRH3. Theo bi :

42




B

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N

GT

O

N

-

L

-

H

A

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P

2

3

1

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41/444

mR R 91 176 = - = 7 ' => R 31 .vx l nguyn t irgug p.li R H3 100

Th dif 3 :

Oxit bc cao nht ca nguyn t R c dng RjO . Trong hp

chat kh vi H, H chim 2,5% khi lng xc nh nguyn t lungca R ?

: GII.

Vi oxit bc cao nht l R20 nn R nhm I, hp cht kh vi

H c crig thc RH Theo bi. .

1|I 1 2 5 ; = ~ ~ = R 39 l nguyn t lung KmRH R + 1 100

Vn 4.

Xc nh ngun t lng kim loi kim, kim thV

Loi ton thng gp na xc nh nguyn t lng kim loikim hay kim th, bng phn ng ca kim loi vi H20 hay axit,

gii phng kh H2 . Bit irc khi lng kim loi v lng kh Hr

sinh ra, ta dng xc nh utyc nguyn t lmjoig kim loi.

Th d 1 :Khi cho 1,17 gam kim loi kim vo nc d, th c 336 m

'khi Hir ktc. Xc nh nguyn t lng kim loi kim.

? GII

t M l k hiu nguyn t v nguyn t lutng, ta c :

M + H20 = MOH + ^H 2 t

M U lit

1,17 gam 336ml = 0,336 lit




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42/444

M - = 390,336

l nguyn t lng Kali.

Th d 2 :

Cho 15,07 gam kim loi M tc dng vi axit clohiric d, thuuc 0,22 gam khi Hirv dng dch c mi trong M c hatr 2. Xc nh nguyn t ung kim loi M.

GII

Kim loi M ha tr 2 tc dng' vi xit HC1 theo phtrang trnhphn ng :

M + 2HC1 - MC12 + H2 t

, M 2 gam

15,07 gam 0,22 gam

_ . . 15,07*2Ta c :M = - f f f - =137

*22 l nguyn t lung ca Bari (Ba).

Vn 5. S phng x y s bin i cc nguyn t' hahc.

Trong sch giao khoa lp 10, on n l phn c thm nhungthirc t ttong thi TSH c nu ra.

1- nh ngha s phng x.

* L s t chuyn ha dng v khng bn ca nguycn t hahc ny thnh ng v ca nguyn t khc, kcm theo s phng racc ht s cp... (nh proton, notron, positon)

* Cc ng v phng x l dng V c khi ung gim dn theo

thi gian.2- Chu ki phng x (hay chu k bn r, chu k thot bin).

L thi gian cn thit cho khi lng cht phng x gimphn na.




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43/444

3- Cc trirng hp phng x.

a) Phng x t nhin.

L hin tung phng x trcn cc ng v thin nhin (c sz > 82) khng do tc dng bn ngoi.

* ; >b) Phng x nhn to.

L hin tng phng x xy ra do s bn ph ht nhn nguycnt bng cc ht a, gii phng ht positon.

4 Ht a l ht nhn ca nguyn t Hcli. He)

Ht positon l ht s cp, c khi lung bng khi lutyngelectron, nhung mang in dng. K, hiu e+ hay p.

__ 27 . 4 30 _ 1 * V n A + 2 He 15p + 0 n

4. Phn irng ht nhn.

L s bin i ht nhn do t phn r ht nhn hoc do tuungtc gia cc ht nhn vi nhau ha gia ht nhn vi cc ht c bn.

Sau phn ng ht nhn, nguyn t ca nguyn t ny bin thnhngun t ca nguyn t kia. .

Tt c cc nguyn t


44/444

' GII.

2 g m -* 1 gam -* 0,5 gam - 0,25 gam

T 2 gam ch gtn cn 0,25 gam, phi mt 3 ln 19,4 nm. Nghal : 19,4 X 3 * 58,2 nm .

c . BI TP P DNG V NNG CAO.

Bi I ,

Ba nguyn t A (Z = 20), B (Z = 35) v c z = 47). Vit mcnng lng v suy ra v tri ca mi nguyn t trong HTTH 7 Don tnh cht ca chung ?

GII .* Cu hnh electron :

- A (Z - 20) : Is2 2s2 2p6 3s2 3p6 4s2 : 2/8/8/2

- B (Z = 35) : Is2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p5 : 2/8/18/7

- G (Z - 47) : Is2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 5s2 4d9

(hay 5s1 4d10)

* V tr ttong HTTH v tnh cht :- A chu k 4, phn nhm chinh nhm II : Kim loi mnh.

- B chu k 4, phn nhm chnh nhm V : phi kim mnh

- c chu k 5, phn nhm ph nhm I : Kim loi

Bi 2.

Nguyn t ca mt nguyn t X c tng s phn t bng. 108.

1- Cho bit nguyn t' X chu k my trong HTTH ?

2- Xc nh v tr ca X, bit rng X nhm V.

3- Tnh phn trm khi lng ca X trong Oxit bc cao nht.

GII :

. 46 " ' .




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45/444

1- T CO : s p + e + n * 2p + n = 108.

Da vo iu kin : 1 4 < 1,524 hay < p 4 p 3)524 V

, N>8 108 . 'Suy t : < p 4 2 hay 30,65


46/444

* Nu khng dng bng tun hon th khng xc nh utyc vtr ca R trong HTTH, v khng bit c s th t .

Mt nguyn t' R c cng thc hp cht vi H l RH, Trongot bc cao nht, R chim 74,2% v khi lng. '

Xc nh nguyn t lung ca R. Qua bng tun hon, cho bitR l kim loi hay phi kim ?

Qua bng tun hon, y chinh l nguyn t lirng c nguynt Na l kim li in hnh.

Bi 5.

Oxit bc cao nht ca mt nguyn t R c phn t lng bng

80 dv.c.

1- Xc nh nguyn t lirng ca R ?

GII

Cng thc RH cho thy R c th l : 1

* Ph kim nhm VII : oxit bc cao nht R2 0 7

* Kim loi nhm I : oxit bc cao nht l R2

* Nu R l phi kim :

1 V l,v khng c phi kim nhm 7 c nguyn t lutyng nh th

* Nu R l kim loi :

_ . mR 2R 74,2 . ' ' wTa c : s= = "7 R = 23 : ng.

nij^ 2R + 16 100




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47/444

2- T ng s phn t trong nguyn t ca nguyn t R bng banhiu ? (da theo bng tun hon)

GII

- Ta bin lun 2 trung hp :

* Trng hp R nhm l t, UI, V, vu '

Cng thc oxit bc cao nht l R2 Ox vi X 4,3, 5, 7

* Phn t ltrqrng Oxit : 2R + 16x = 80yrP-

Nhm X 1

R = 40 - 8 x

3 5 . 7

^NTL R 32 16 0 < 0

Kt lun / / / /

Trung hp R nhm chn II, IV, VICng thc Oxit bc cao nht ROy vi X = 2y

Phn t lung oxit : R + 16y 80

' y 1 2 3

Nhm X 2 4 6

NTLcaR 64 48 32

Kt lun / / ng

y l nguyn t s ,nhm VI, c nguyn t lirng 32 .v.c

2) S th t ong HTTH cua s : z V 16 p

Tong s phn t ttong nguyn t :

s =p.,-t n .+ e * A + p - .32 + 16

Vy s 48Bi 6.

Mt nguyn t R chim 95,238% v khi lng trong hp chtvi Hir.

49




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48/444

1- Xc nh nguyn t lung ca Jt Vit cu hnh electron suy ra v tr ca R trong HTTH, bit trong nguyn t ca R

proton bng 50 ntron.

2-Nguyn t ca nguyn t'R' c s khi bng 4 ln s eleccua R v hiu s santron vi s proton bng lXc nh ngu

t Urng, Vtr trorig HTTHca R'. Bit phn tram khi uca Oxi trong xt ca R' bng 33,3%. lp png thc ca Oxit

GII ;

1) Bin lun 2 trng hp :

) Nu R l nguyn t phi kim : Cng thc hp cht khHir l RHg X (x l th t nhm t IV i VII). Theo b

mR R 95,238

X

R + 8 -

4

= => R loU - Z.UX X 100

5 6 7

R 80 60 40 20

Kt lun / / . / /

Khng cnguyn t ph . kim no tha.

b) N R l nguyn t kim oi ; Cng thc hp ch t v

l RHX(x l th t nhm t I n III). Ta c :

?JL = _ R = 20x 'raRH R + * 100

X 1 2 3

R 20 40 60

Kt ln / /

Nguyn t phi tm c ngycn t lung 40 .V.C (hay s khi l

Vi p = n nn A = p + n = 2p * 40 => p = 20 = z

-Cu hnh electron cua R : ls 2 2s2 2p6 3* 3p6 4s2 hay 2/8/8/2




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49/444

Vy R phn nhm chnh nhm II, chu k *IV : Ca.

2) S khi ca nguyn t R' : A' = 4e (ca R) = 4 . 20 80

Nguyn t lung tung ng M = 80 .v.c

Ta c : A' p' + n' - 80 v n' - p ' 10

Suy ra : n ' 45 v p ' 35. Cu hinh electron ca R' :

ls2 2s2 2p6 3s2 3p6 3d1 0 4s2 4p5 hay 2/8/18/7

R' chu ki IV, phn nhm chinh nhm VII l Br

Cng thc oxt ca Brom c dng Br->Ox .

Theo bi :

m ox i 1 6 x 3 3 , 3 m ' = .......... " = X 5mBro 160+16x 1 0 0

Cng thc oxit l Br2 0 5

Bi 7.. ' ' c

Cho 4,68 gam mt kim loi kim tc dng vi 27,44ml H2 0,

thu c 1,344 lt kh Hr ktc v dung dch X.

1- Xc nh nguyn t lng kim loi kim.2- Tnh nng phn trm cht tan trong dung dch X.

; GII

* 1,344 lt H2 ktc ttrng ung

1 , 3 4 4 * 0*12 gam H2.

1) t M l k hiu v l nguyn t lutyng turn loi kim.Phuong trinh phn ng vi H,0 :




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50/444

M + H20 = MOH + - H 2 t

M 11,2 lit

4,68g 1,344 lit

Ta C : , M = 39 l nguyn t iutng K a li'1,344

2) K + H20 = KOH + H2t

Dung dch X c :

- Cht tan l KOH, khi lrnyng: ^ = 6,72 gam

- Khi Iwyng * mddmK + m ^o nij^ = 4,68 + 27,44 - 0,12

mdd = 32 gam

Nng phn trm KOH trong dung dch X :100. mKOH _ 100.6,72 21

mdd 32Bi 8, /

Cho 15,07 gam mt kim lo M (c ha tr n) tc dng v tanhon ton trong nuc, gii phng 0,22 gam kh H2 v dung dch

Y c khi lung ring 1,03 g/ml v th tch 60,68ml.1- Xc nh nguyn t lng kim oi. M - V tr trong. HTTH.

2- Tnh nng phn trm dung dch Y v th tch nc thingban u ?

GII

1) Phuong* trnh phn ng :

M + nH2 0 M(OH)n + - H 2 t

M n gam

15,07 gam 0,22 gam




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51/444

Ta c : M -15,07. n

0 , 2 2* 68,5(1

Bin lun ; n 1 2 3

M 68,5 137 205,5

Kt lun / /

Nguyn t M Ba c nguyn t lung 137 .v.c, chu ki VI,phn nhm chnh ahm II

2) Khi lng dung dch Y :

m dd * V 6 0 ,6 8 X 1 ,0 3 * 6 2 4 g a m

Ba.+ 2H20 = Ba(OH) 2 + H2 t

Dung dch Y c eht tan l B(OH) 2 khi lng l :

022171 x ' = = 18,81 gam

Nng phn trm Ba(OH) 2 trong dung dch Y :

c _ 100.mB.(OH)z = ioo^ lj i8J _ 30 096

6 2 , 5

p dng nh lut bo ton khi lung. Ta c :

Ba + ^o " mdd + H2 t

1 5 ,0 7 + m j ^ o - 6 2 ,5 + 0 ,2 2

mHO - 47,65 gam tung ng 47,65ml nuc

Bi 9.

Cho m gam lm loi phn nhm chnh nhm It tc dnghonton vi 100ml dung dch HCI, thu c 0,8 gam kh Hir v 125

gam dung -dch X trong mui c nng 30,4 %.

53




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52/444

v 1 - Tnh khi lung mui. Suy ra nguyn t lng v khi lngm gam c kim loi dng. Cho Cl = 35,5; H = 1.

2 - Tnh khi ung ring dung dch axit HCl ban u.

3 - Cho bit trong dung dch X c HCl cn d hay khng ?

Nu c th hng % xit d l bao nhiu ? Bit nng % ungdch axit ban u l 28,27%.

' GII

1 ) t M l k hiu v nguyn t lng kim o ha tr 2

Phutyng trinh phn ng : M"+ 2HC1 = MClj + H2 T

Ntg %Tk1 0 0 .mMCi 1 0 0 . mMCi

c = =* 30,4 ^m d dx 125

* Khi lung mui Clorua : * 38 gam100

Theo phn ng,t c :

M + 71 2 _ 38x2 _ =* M + 71 * - 7 7 - = 95

38 0,8 0,8 .

M = 24 l nguyn t lng Mg ,' \ ' 24 2* v : = .. * m 9,6 gam

m 0 , 8

2 ) p dng nh lut bo ton khi lung :

m Mg + m ddHCI = mddX + m H,) ' i

9 , 6 + n i d d n Q = 1 2 5 + 0 , 8

mddHCl =116,2 gam




B

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O

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53/444

0 83) S mol axit HC1 tham gia phan ng : ^ ~ x 2 * 0 , 8 moi

S mol axit HC1 ban :

116,2 28,27 _ , ,c.>. x ~. r r ~ 8 8 0,9 mol > 0 ,8 .moi36,5 100

Vy axit cn d : 0,9 - 0,8 * 0,1 mol ong dung dch X.

Nng % axit HC d :

- 100x36,5x0,1 , o , ?

Bi 10.

A l nguyn t phn nhm chinh VII, X l hp cht kh ca

vi Hir. Cho 8,76 gam dung dch 12,5%(ca X tan trong rtc)tc ng va vi 0,72 gm Natri hrua NaH, thu c dungdch v 0,06 gam kh Hir. Cho Na = 23, H = 1, 0 = 16

1- Xc nh nguyn t lng ca A. Suy ra phn trm ca trong oxit bc cao nht.

2- Tnh nng phn trm dung dch '.

, GII

1) V A phn nhm chnh nhm VII ncn cig thc hp chtX ca A vi H l HA.

, . . . 8,76x12,5S gam X ttong .dung;dch : = 1,095 gam

Phung trnh phn ng :

NaH + HA = NaA + H2 t

24 A + 1

0,72 1,095

Ta c. 24 JC1,095

0,72

55




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54/444

A J , l nguyn t lng o : C1

Cng thc oxit bc cao nht ca Cl. C12 0 7

n, 3 J_!? ol 2 X 35,5x 100


55/444

Phuwng php 1.

1 ) t cng thc 2 mui Ccbonat kim loi kim l M2 C3 v

M2 'C 0 3 (M v M' l k hiu v nguin t lu-ng 2 kim loi, M < M )

t a, b l s mol tng ng ca mi mui. 1 .

Phuung ttnh phn ngM2C03+ H2S04 M2S04+ 2 + 2a a a a

M'23 + 24 M2SO4+. CO2 + 2 b b b b

1 344S mo kh C0 2 sinh ra : a + b = 006 (1)

V' . 22,4

Cch a :So gam hai mui Cacbonat :

(2M + 60) a + (2 M + 60) b 7,16

=> 2(Ma + M'. b) + 60(a + b) = 7,16 ^

thay ( 1 ) vo :

>;^2(Ma + M b) + 60. 0,06 - 7,16

Ma + M\ b - 1,78 (2 )S gam hai mui khan sunfat

(2M + 96)a + (2 M' + 96)b = 2(Ma + M' b) + 96(a + b)

= (2.1,78) + (96.0,06) = 9,32 gam

Cch b :

p dng nh lut bo ton khi lung :

l^Cacbonat + ^RjSO^ m sunfat + m C02 + m HjO7,16, *i>8 (aH,b) = m + 44(a + b) + 18(a + b)

7,16 + 36(a + b) - 7,16 + (36. 0,06)

m 9,32 gam

57




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56/444

Cch c

C 1 moi hn hp mui Cacbonat to ra 1 mol hn hp \ sunfat, khi lng tng ln : (2M + 96) - (2 M + 60) * 36 gam

C 0,06 ol hn hp th khi lng tng :36X 0,Q6 =

' Vy khi lng hn hp mui khan sunfat l 7,16 + 2,16 = 9,32 gam

2) Cch .

T ( 1 ) => a = 0,06 - b

Thay vo (2) : M' b + M{0,06 - b) 1,78

T (1) =*> 0 < b < 0,06

Bin lun :

* h > 0 : (3) => 1,78 - 0,006M > 0 (v M' - M > 0)

M < 29,67

(M' - M)b 1.78 - 0,06*'

1.78 - 0,061m ' -T mb

* b < 0,006 : (3) *

1,78 - 0,06M < 0,06 (M' - M)

1,78 < 0,06

M > 29,6

M < 29,67 < M'Suy ra

V 2 kim loi kim 2 ch k lin tip, nn ta chn

M 23 l nguycn t lqtig !

M' ^ 39 l nguyn t luqng

Thay tr s M, M' vo (3) :




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57/444

. 1,78-0,06.23 ____ .b ~~~ ~ ~ ~ * 0,025 mol

3 9 - 2 3

a 0,06 - 0,025 0,035 mol

S gam Na2 C03 : 106a =106 . 0,035 * 3,71 gam

So gam K2 C03 : 138b - 138 . 0,025 - 3,45 gam

Phn ttm khi lirng NajCOj : - = 51,82

Phn trtn khi lng K2 C0 3 : 100 - 51,82 * 48,18

Cch b.

Vi M < M' ncn : M(a + b) < Ma + M' b < M' (a + b)

L lun nh cch a, ta c cc p s trn.

Phteong php 2.

1) t cng thc 2 mui Cacbonat kim loi kim l M,C 0 3.M

l k hi v nguyn t lirng trung bnh ca 2 kim loi kim M' vM : M' < M < M" '

a v b l s mol 2 mui Ccbonat tirng ng.

Phung trnh phn ng :

hay :

m 2 c o 3 + H2 S0 4

(a + b) mol

S mol kh C02

sinh ra

m 2 s o 4 + C0 2 t+ H20

a + b a


58/444

; /

Cch a :

S gam bai mui Cacbonat : (2M + 60)(a + b) 7,16

liay ( 1 ) Vo : (2M + 60) 0,06 * 7,16

M - 59,67 - 30 - 29,67 (4)

S gam mui sunfat:(2M + 96) (a + b) * (2 X29,67 + 96) 0,06 * 9,32 gam

Cch b :L un nh phung php 1.

Cch c :L lun nh phutmg php 1 .

2) T (4),ta c : M' < M = 29,67 < M

Suy ta M' * 23 l nguyn t lng Na < 29,67

M 39 ngyn t lrnjtig Kali > 29,67S gam hai mui Na2 C03 v K2 C0 3 :

106a + 138b = 7,16 (5)

Gii ij phuong trinh ( 1 ) v (5) suy ra : a = 0,035 mol Naj

i . __ 100x106.0,035 _Phn trm Na,CO, : ------- ---------= 51,82z i 7,16

Bi 12.

*phn ng ht nhn phng tia a c dng :

i x rte + t z t Y 2 2Vit phuong trnh phn ng ht nhn ca u , Ra,

yZ 0 0

GII

p dng phn ng ht nhn tng qut trcn, ta c :

* Q? u * Hc +-3'Th (Xcmtrong bng HTTmt yu




B

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59/444

* 20206 Ra *H e + 2*2 Rn (Xem trong bng HTTH )88 2 8 0

* 2*Po - He + 2o06 Pb (Xem oag bng HTTH)84 2 82

Bi 13.

Hon thrih cc phung trnh phn ng ht nhn :

11 29442pU + Ne * 4 n + ?

2/ h + ? 2 2 He + * n

3 / n + ? ^ C + h

GII

Ta c :' ' 242 , 22 XT ' .1 , 260

9 4 ' " w " 6 - * V 104Ko

2 / h * ] u 2 ^ . 1 c ^ n

31 ~ 6 C t H

B/ W.

Chu M bn r ca Urani l 4.5 t nm. Hi sau bao nhiu tnm th 2,4 gam Urani gim xung cn 0,15 gam.

\ GII

C gim khi lng mt mu Urani xung phn nu phi mt 4.S tnm. T 2,4 gain cn 0,15 gam, phi mt 4 X 4,5 t nm = 3 li nm.

61




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60/444

LIEN KET HOA HOC

NH LUT TUN HON MENDELEE

LIN KT HA HC v

A. TM TT GIO KHOA

Khng k trng hp cc kim loi v cht bn dn, vt cht xem nh to thnh bi cc loi lin kt sau :

* Lin kt to thnh phn t gm c :

- Lin kt cng ha tr.- Lin kt phi tr.

- Lin kt lon.

* Lin kt gia cc phn t (hay lin kt lin phn t') gm

- Lin kt Van Der Waals.

- Lin kt Hir.

I. Lin kt cng ha tr1. Nguyn tc

gii thch s to thnh phn t n cht hay hp cht cc nguja t phi kim.vi nhau, hoc phi kimi- km loi rhai ng Lewis v TLangmuir cho rng :

* "Cc nguyn t tm cch gp chung cc electron vi nhat c cu bn ca khi him" (C 8 electron ngoi cng ho

electron trung hp H ).* C s gp chung l nguyn t thiu bao nhiu electron s

chung by nhiu.




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O

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61/444

*i electron gp chung thuc v c 2 nguyn t v c biudin bang 1 gch ni gia 2 nguyn t.

2- S to thnh v biu din lin kt cng Ha tr.

a. Lin kt n:

*Cn gi l lin kt sigma ( - ) , bn, do 2 orbitan cng trcxen ph nhau.

a. h d : H2 , Cl2 Br-

Cng thc cu to (CTCT) H - c u hay H-Cl

b. Lin kt i

Gm c 2 iin kt n hp li .

- Lin kt sigma ( ) bn.

- Lin kt pi (i) km bcn do 2 orbitan c trc sng song nth tch xen ph l nhau nh hon, so vi lien kt sigma

a, Phn t n chtnh Oxi : o

C1 c l ' Cl2*1 b *cng thc electron : lC * C1 *

A


62/444

*r\ A* 'Cng thc electron : 0 J o ^

Cng thc cu to (CTCT) : hay 0 * 0

. Ch :

Cch biVdin Ox theo qui tc bt tnh trn, khng ph hp

vi thc nghim (do sch gio khoa khng ncu, nhng sch giokhoa thc nghim lp 10 nm 1990 th nu trang 46).

, Phn t hp chtnh c2C,H4...

Cng thc electron : *0 C',jOf

Cng thc cu to : ) = c = ) hay ' o - c ,= oc- Lin kt bar

Gm c mt lin kt sigma


63/444

3. Lin ki cng ha tr c cc v khng c ce.

a) Lin kt cng ha tr khng c cc- .. ' * / _

L lin kt xy ra trong cc phn t n cht, nhH2 0 2 , Cl2 , N2 Br2... Trong phn t dn cht, i elctron ni lin

nguyn t khng b lch v pha no. Ta gi l licn kt cng hc

tr : ng cc.b) Lin kt cng ha tr c cc.

L lin kt xy ra trong cc phn t hp cht nh HCHBr2 H2 0... Trong phn t hp cht, i electton ni lin 2 ngu

t k h c n h a u s b h t VC p h a n g u y n t n o t n h p h i k im m t

hon. Lc ht cng ln th phn cc cng tnnh.

Ta biu din lien kt cng ha tr c cc nh sau : Th d troi

phn t hir clorua :H : Cl hay H C1 # hay H+_ C l ^ ( c l delta)

(Ki hiu u : i e gp chung lch v pha Cl)

H. Lin kt phi tri,

I. Nguyn tc.

Lin kt phi tr l trung hp c bit ca lin kt cng hNguyn t B nhn clcctron t nguyn t A, bin thnh on(t c cu bn ca khi him gn nht).

- Hai ion dng v ion m ht nhau bng lc tnh into th> phn t hp cht.

y . 66




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65/444

2. Thi to thnh v biu din tin kt on.

a) Phn t NaCl:

Nguyn t Na

* hay

Na* + 1- *+*f Phng trnh phn ng :

2 e2 ta + l.

b) Phn t KO

-> 2NaCl

Nguyn t K Nguycn t o

hay 2K* + *': . *

Phutmg trnh phn ng :

4.1e1 1

4 K + 2 *=. 2K20

lon K+ lon o 2" lon K1

2K+ + [ O ]1-> K20




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66/444

a) Nguyn t cho l A t utc c cu bn ca kh him bngcc lin kt cng ha tr, m vn cn d cc i electron.

Nguyn t nhn B thiu ng 2 electron ngoi cng.

* Do , lin kt phi tri ln lun di km theo lin kt

ha tr v hnh thnh sau licn kt ny.b) Nu ngoi nguyn t A, cn c nguycn t A' cng c kh

nng to lin kt phi tri nh A, thi u tin to ni s thuc vguyn t no c tinh phi kim yu hn.

Thi d : S to thnh phn t khi s o ,

T* ** t >*; 0 > S : O* - : : S : O

* * *^ t i .

Biu din: /,r \p , J f0sO = hay o = S'*

Tng t, c th gii thieh tip s tp thnh phn t SO3

Ch :

Tuy khc nhau v cch to thnh, nhung lin kt phitr hoton c tnh cht ging licn kt cng ha tr i.

III. Lin kt lon.

. Nguyn tc.

gii thch s to thnh phn t hp cht gia nhng nguynt kim loi v ph kim, ng Kossl cho rng "Khi 2 nguycn t A, Bva chm nhau :

- Nguyn t A nhung clectron cho nguyti t B, bin thnh onduung (t cu hnh electron ca kh him gn nht).

- Nguyn t B nhn clcctron t nguycn t , bin thnh ion m

(t c cu bn ca kh him gn nht).- Hai ion dtrng v ion m ht nhau bng lc tnh into thtih

phn t hp cht.




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O

N

-

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67/444

2. Th to thnh v biu din lin k t on.

a) Phn n NaCl

* hay

Na* + *X- * *f, Phng trnh phn ng :

2 e25T 2NaCl

b) Phn t K2O

Nguyn t K Nguyn ,t o

hay 2K + 'O

Phng ttinh phn ng :

4.1e

4 K + 0 2 2K20

' 2K+ +

lon K+ ln o2- lon K+

M 1 *-

mK ,0

6 '




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N

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O

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-

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68/444

IV. S ir chuyn tip gia lin kt cng ha U vki on. .. *

Lin kt cng ha tt cung nh lin kt lon u lien quan s thay i cu hnh electron ca lp v ngoi cng.

* in kt cng ha tr : 2 nguyn t gp chung electron.

* lin kt ion : c s chuyn electron t nguycn t cho snguyn t nhn.

C 2 truiig hp u c mt i electron linf, chng ch knhau v v tri m thi. ^

Th d V tr ca i electron lin

Lin kt

cng ha

tr

Khng

cc

Cl Cl Cch u 2 nguyn t

c cc H Cl B lch v pha nguyn t n

c tnh phi kim mnh hn

Lin kt on Na+ C~ Lch hn v pha nguyn

phi kim

V. m in"L i lirng c trung minh ha kh nng ht i electron

kt ca mt nguyn t' do ng Pauling lp ra.

m in ca mt riguycn t :

* Cng n th tnh phi kim cng mnh, tnh kim loi cng

* Cng nh thi tnh phi kim cng yu, tnh kim loi cng m

m in ln nht bng 4, l ca Fl (phi kim in hn

v m in nh nht bng 0,7, l ca Cs, Fr (kim loi hnh) .

! ' . Vy : .li: I Trong cng chu k : m in tng dn t tri sang ph

68




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69/444

* Trong cng phn nhm chinh : m din tng dn t duiln trn.

VI. Ha tr ca nguyn t.

* Electron ha i nhng electron lp v ngoi cng,c khnng tham gia s to thnh lin kt ha hc.

1. in ha tr.

in ha tr ca mt nguyn trong hp cht c licn kt ionbng s din tch ca ion .

Thi d : N - e -* N+ v C1 + e -* Cl~

Trong hp cht NaCl, in ha tr ca Na bng +1 v ca Clc

bng - 1 .2. Cng ha tr.

Cng ha tr ca mt nguyn t l s electron m nguycn t Cnguyn t gp chhg to nn licn kt vi nguyn t ichc. Gitt l ha tr.

Th d:

* Trong phn t H2S : H - s - H : Cng ha tr c H bng 1v ca s bng 2 .

Ch :

Trong hp cht va c lin .kt'cng ha tr va c licn kt phitr, ha tr ng vi lin kt phi tr bng 2 .

Th d : o . ' . y* Trong phn t S02: 0 = s

- Nguycn t Oxi gp chung 2e cng nh nguycn t o nhn 2eu c ha tr bng 2 .

- Nguyn t s va gp chung 2c va cho mn 2c, c ha trbng 4.

6




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O

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-

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70/444

VI. Lin kt Van er Waals.

1. nh ngha:

L lin kt gia phn t vi phn t, xy ra do lc ht giaduong ca phn t ny vi cc m ca phn t kia.

2 . l n h c h t .

* Lin kt Van der Waals yu hon nhiu so vi lin kt cha tr v tin kt lon.

* Lin kt Van der Waals cng mnh khi :

- Cht rn cng kh nng ch.

- Cht lng cng kh si v d ng c.

- Cht kh cng d ha ng.

3. Cc yu t' nh hmg trn lin kt Van er Waals.

a) phn cc ca phn t.

Phn t cng b phn cc th licn kt cng mnh.

b) Khong cch gia cc phn t.

Khong cch cng nh th lin kt cng mnh.

c) Khi lng phn t (hay phn t lung)

Khi luxng phn t cng ln th lin lct cng mnh.

VU. Lin kt Hidr.

1. nh ngha :

* L trung hp c bit ca lin kt Van dcr Waals, trong

- 1 cc ca phn t ny l Hir.

- 1 cc ca phn t kia l nguyn t c m in ln,

knh nguyn t nh. l F, o, N, Cl...* Lin kt Hir biu din bng cc du chm gia 2 phn

Th d : p . F . F.' H H H

70




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N

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O

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-

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71/444

2. Tnh cht.

Ivtotig t nh lien kct Van dcr Waals, nhng mnh hoti :

- Cht rn kh nng chy.

- Cht lng c si cao bt thung, ha tan v hn trong ncd ng c.

- Cht khi d ha lng.




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O

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72/444

B. PHNG PHP GII TON

Y h 1 .

D on lin kt cng ha tr hay lin kt ion trong cng thphn t mt cht.

Cch 1:

*Nu thy trong cng thc phri t 1 cht c mt nguyn t kloi v ph kim, th cht c lin kt, ion.

Th d : A12 0 3 MgCl, K2 S...**

* Nu thy trong cng thc phn t 1 cht ch c mt nguynphi kim, th cht c lin kt cng ha tr.

Th d ; HC1, C02

,CC14...* Nu thy l phn t n cht, thi c licn kt cng ha tr.

Th d : H2, 0 2 , N2, co,...

Cch ny khng c chnh xc lm, v lin kt ioi| ch xyd dng gia :

- Kim loi mnh nh kim, kim th...

- Phi kim mnh nh halogen, o, S...Cch 2 :

* Lp hiu s m in gia 2 nguyn t A v B.

*Nu hiu s m in > 1,7 : gia 2 nguyn t c lin klon.

Thi d : Br K = 2,8 - 0,8 = 2 (trong KBr)

* Nu hiu s m in


73/444

Thi d : N - H 3 - 2,1 - 6,9 < 1,7 (trong phn t NH3)

Vn 2 :

So snh phn cc ca 2 lin kt cng ha tr

Ta bit rng kh hiu s m din s - H - 2,5 - 2,1.

* Nu hiu s m in bng khng th lin kt, cng ha trkhng cc.

Th d :

* Lin kt p - H trong PHj khng cc v p - H = 2,1 - 2,1' - 0

* Lin kt Br - Br ttong Br, khng cc v Br Br = 2,8 - 2,8 * 0

Vh 3 :

Vit cng thc electron v cng thc cu to ca mtcht clin kt cng ha tr. ;

* Truc ht, d on ha tr ca cc nguyn t ph kim trn cs bit h tt ca H bng 1 v Oxi bng 2.

* Sp xp cc nguyn t gn nhau, biu din selcctronngoicng bng cc du (.) hoc (x) cho d phn bit.

* Cng thc electron ca hp cht cng ha tr l cng thc trong cc nguyn t lin kt nhau c 8 c hoc 2 c ngoi cng, sau khi gp chung cc e cn thiu.




B

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O

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-

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74/444

* Cng thc cu to l cng thc trong d cc nguyn t Hnkt nhau bng cc gch ni. Mi gch ni l 1 i clcctron gpchung.

Thi d :

Hp cht NHj : N c ha t 3v H c ha tr 1 .

H. *H - H N ; H vH - N - H I

H H H

Cng thc clcctron Cng thc cu to

* Hp cht C2 H4 : c c ha tr 4

c H * H r C C : h v H c = c . . # * , , L

H H H H H H

Vh 4 :

Vit cng thc electron v cng thc cu to ca hp chi clin kt phi tri lin kt cho nhn)

* Lin kt phi tri ch xy ra trong cc hp cht c lin kcng ha tr. V

* N g u y c n t " c ho " p h i l n g u y c n t g p s c l c c tr o n c n

thiu, t c cu bn ca khi hicm, m vn cn dir 2, 4 hocelcctron cha s dng. Do , nguyi'n t "cho" thung l nguyn tca nguyn t nhm V, VI, VI] Iihir N, p, s. Cl; Br, I...

* Nguycn t "nhn" l nguyn t (hoc lon H+) cn thiuelectron, thng d l - nguyn t. Oxi.

* Khi hai nguycn t khc loi nhau, cng c kh nng to lienkt phi tri vi nguyn t th ba, u tin s thuc v nguyn t noc m in nh hn.




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O

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-

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75/444

1 ) Phn t S0 2 v S0 3

)o

SO : X* S: * * Oi; S : O 0 = s * >*

CT electron CT Cu to

' ; ' 0 ; J b

S03 : ;& . S;^ ;O; s : 0 = s ^

'9: \ : \ >CT clcctron CT Cu to

2 ) Phn t HN03, - I

'* ^H* * * N* - H ; ; N * r- H O N

> , ; \

3) Phn t HC104

,M * 0 H' * CI: *0 * -^ H C : :

Trcng cc th d trcn :

c ha tr 4 (SO,) hoc 6 ( SO3 )

N c ha tr 5 C1 c ha tri 7




B

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O

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-

L

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76/444

* Kh mt nguyn t (hay mt nhm nguyn t) thay i electron ngoi cng bng cch bt hoc nhn thm t c cca kh him, n bin thnh lon.

* C 2 loi lon :

1) lon rt nguyn t :

Do mt ngyn t to ragm c :

j[ lon dng (gi l 'Cation)

Cc nguyn t kim loi u c kh nng nhirng s e ngoi cbin thnh lon dutmg

M - xe -* Mx+ (x l s elcctron ngoi -cng ca M)

b) lon m (gi l anion).

Cc nguyn t phi kim u c kh nng nhn thm ! s e scho lp ngoi cng bo ha, bin thnh lon m.

R + xc -* Rx~ (x l s elcctron nhn them)

% lon a nguyn t :

Do nhiu nguyn t hp li, to thnh ging lon n nguyn

a) lon dttng .Rt it gp loi ny, cht tiu biu nht l NH4 (lon amni)

b) lon m.

Chim s ligng nhiu hon. Cc lon thung gp l :

. NO3 HCO3 GO^- HCOJ* H S--H S O J- so f'S|-> PO ^...

Vn 6 :

Vit cng thc-electron v cng thc cu to ca hp cht* Lin kt lon/xy ra trng cc hp cht c mt ngycn t

|iim loi v phi kim.




B

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N

GT

O

N

-

L

-

H

A

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1

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77/444

* Da vo v tri trong h thng tun hon ca kim loi v phikim* bit xem :

Kim loi nhung bao nhiu electron, bin thnh icm duong.

Phi kim nhn bao nhicu electron, bin thnh ion m.

* Cn nh hp cht c lin kt lon khi hiu s m in gia2 nguyn t * 1,7.

Thi d Phn t A1F3 :

Al* + 3 *Fi - [Al]3+ +.3 f r F; r ** **

Cng thc electron :

[Al]3+ [ : F: ] / ' '

Cng thc cu to : F - A I - FF

Vn 7;

So snh lin kt Van der Waals. Nhit si v ha lng :

* Ta bit lin kt Van dcr Waals ph thuc vo 3 yu t :

* phn cc ca cc phn t.

* Khong cch gia cc phn t.* Khi lutng phn t (hay phn t lirng).

i)p d, mun so snh lin kt Vandcr Waals gia cc cht khcnhau, ta ln lt a v cc ycu t li trcn.

Lin kt Van dcr Waals cng mnh kh :

* phn cc ctig ln.

* Khng cch gia cc phn t cng gn.

* Khi luxyng phn t cng ln.

T suy ra nhit si (hay nhit ha lng) cng tng ln.

Th d:




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O

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78/444


79/444


80/444

* Trong b cht kh : o , CH4 v NH3 th sH ha tn v

trong nuc v NH3 c licn kt Hir vi nuc. '

: ; v H - N -H . - 0 - H - N - H O - H

L ' 1 " / I I

' H H H , . . H* Trong ba cht lng : C2 H5 - o - C2 H5, C6 H6 v C2 H5CO

thi C2 H5COOH ha tan v hn trong nuc v C2 H5COOH c lin

Hir vi nuc.

w ./ " 1 V

C,HS- C- o - H-O - H-.-O=c - 0 - H -0 - H O=c - o: I ' I ; ( - , . I

H C2H5 H C2Hsp ; I TP P DNG V NNG CA>1. Ln kt cng ha tr v lin kt phi tr.

Bi 1.

Hi nguyn t X (Z - 1) v Y ( z = 16) cth to thnh cht no ? Vit cng thc electron v cng thc cu to ca cht n.

GII

X (Z * 1) l H v Y (Z =16) l s (2/8/6). Gngthcto bs l H2S : ; ;

H s ; H * * H- s - HBi 2 .

Mt nguyn t R c cu hnh electron lp ngoi cng l

v nguyn t'R' c cu hnh electron Is1.

1. Xc nh v tr ca R v K trong HTTH ?

2. Vit cng thc electron v cng thc cu toca phnn cht v hp cht to thnh bi 2 nguyn t trn.

GII1 - Cu hnh electron y ca :

80




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81/444

* R l : is2 2s2 2p6 3s2 3p6 3d1 0 4s2 4p5 : 2/8/18/7 * z ,= 35

R l nguyn t Br phn nhm chinh nhm VII chu ki IV

* R' l ls : z = l H.

2- Cng thc electron v cng thc cu to.

* n cht

H2 : H : H V H - H

Br2 : B r r B r ; , Br - Br

* Hp cht HBr :

H : B r ; H--.Br

Bi 3.Mt nguyn t R c cu hnh electron ngoi cng l 3s2 3p3

1-r Xc nh s in tch ht nhn v v tr c R trong HTTH,

2 - Vit cng thc electron v cng thc cu to ca hp ch&kh vi Hir v oxit bc cao nht ca R.

GII

1) Cu hnh electron y ca R :

ls 2 2s2 2p6 3s2 3p3 => 2/8/5 => z =15

Nguyn t' R..l P phn nhm chinh nhm V, chu k III

2- Cng thc electron v cng thc cu to ca :

* Hp cht kh vi H / PH3

H : p ; H H P-H4* I

H H ; * Oxit bc cao nht : P9 O5

8

:0 :: p r ; p : ;* * **0: O:

0= P I = Ob




B

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O

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L

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Y

N

H

N



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Bi 4'Hi nguyn t' X v Y c :

- Tng s in tch tch ht nhn bng 15.

- Hiu s irt tch ht nhn bng .1- Xc nh-v tr ca X v Y trong HTTH ?

hỌc tỐt hÓa 10 - nguyỄn bÁ nhiỀu

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