He Thong Ly Thuyet Bai Tap Hoa 10 LTDH

8/6/2019 He Thong Ly Thuyet Bai Tap Hoa 10 LTDH

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng cao

Chng 1 + 2: NGUYN T - BNG TUN HON HA HC

A/KIN THC CN NH

I/ THNH PHN NGUYN T

1/ Cu to nguyn t - c tnh cc ht: Nguyn t c cu to gm:- Ht nhn gia nguyn t, gm cc ht proton (p) (mang in tch dng) v cc ht ntron (n) (khngmang in).- V nguyn t gm cc ht electron (e) (mang in tch m) chuyn ng xung quanh ht nhn.

Ht Khi lng (m) in tch (q)Tht Tng i Tht Tng iProton 1,6726.10-27 kg 1u +1,602.10-19C 1+

Ntron 1,6748.10-27 kg 1u 0 0Electron 9,1094.10-31 kg 1

1836u

-1,602.10-19C 1-

* Kt lun.

+ Khi lng nguyn t bng khi lng ht nhn nguyn t (v khi lng ca e rt b so vi khi

lng cc ht ntron v proton, c th27

31

1,6726.101836

9,1094.10p

e

m

m

= ).

+ Nguyn t trung ha v in, nn s p = s e.2/ Kch thc v khi lng nguyn ta/ Kch thc nguyn t: rt nh, c tnh bng n v nanomet (nm). 1 nm = 10-9m = 10

ng knh So snhNguyn t 10 -1 nm

lanD

D

hatnhan

nguyentu 4

5

1

1010

10==

ht nhn 10-5 nmlan

D

D

electron

nguyentu 7

8

1

1010

10==

Electron (hay proton) 10-8 nmlan

D

D

electron

hatnhan 3

8

5

1010

10==

V vy electron chuyn ng xung quanh ht nhn trong khng gian rng ca nguyn t.

b/ Khi lng nguyn t: rt nh, c tnh bng u (hoc vC).Vi 1u = 12

1.

12 Cm =

112

. 19,9265.10-27 kg 1u = 1,6605.10-27 kg.

II/ HT NHN NGUYN T. NGUYN T HA HC

1/ in tch ht nhn (Z+). in tch ht nhn chnh l tng in tch ca proton.Z = s proton = s electron = E (Nguyn t trung ha v in)

2/ S khi ht nhn (A). S khi ca ht nhn bng tng s proton (Z) vi tng s ntron (N).A = Z + N

3/ S hiunguyn t (Z). S hiu nguyn t l s n v in tch ht nhn nguyn t ca mt nguyn t.4/ K hiu nguyn t ca nguyn t X.XAZ

Trong : A: s khi Z: s hiu nguyn t X: k hiu ha hc ca nguyn t

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III/ NG V. NGUYN T KHI. NGUYN T KHI TRUNG BNH

1/ ng v: ng v l nhng nguyn t ca cng mt nguyn t c cng s proton, khc s ntron.V d. Nguyn t H c 3 ng v H11 , H

2

1 , H3

1

Ch . Cc ng v bn c Z 82.2/ Nguyn t khi. Nguyn t khi trung bnh

a/ Nguyn t khi (M). Nguyn t khi l khi lng tng i ca nguyn t, bng s khi ht nhnM = A

b/ Nguyn t khi trung bnh (M ). Nguyn t khi trung bnh ca nguyn t c nhiu ng v c tnhbng h thc

cba

cCbBaAM

++

++=

Vi a, b, c: l s nguyn t (hoc % s nguyn t) ca mi ng v.A, B, C: l nguyn t khi (hay s khi) ca mi ng v.

M rng:

( )1

1

.k

i i

i

k

i

i

n M

M

n

=

=

=

Vi:ni : % hay s mol hay th tch ca cht th i ( khi ni l th tch th ch s dng cho cht kh)

Mi : Khi lng mol ca cht th iNu trong hn hp ch c hai cht , ta c th gi x l s mol (% hay th tch) ca cht th nht trong1 mol hn hp, khi suy ra s mol ca cht th hai l (1 x) mol.

1 2. (1 ).M x M x M = +

Lu : Mmin < M< Mmax

1 2

2M M

M+

= 1 2

1 2

1 250%

n n

V V

x x

=

= = =

( th tch ca kh khng p dng cho th tch dung dch)

M1 = M2 1 2M M M= = , , ,n V x

M n cht M hp cht S ng cho:

V1 (hay n1) M1 M2 M

M 2

1

2 1

M MV

V M M

=

V2 (hay n2) M2 M M1* Ch :- Phn bit nguyn t v nguyn t:+ Nguyn t l loi ht vi m gm ht nhn v cc ht electron quanh ht nhn.+ Nguyn t l tp hp cc nguyn t c cng in tch ht nhn.

- Tnh cht ha hc nguyn t l tnh cht ha hc cc nguyn t ca nguyn t .- Mi quan h gia cc ht c bn cu to nn nguyn t:+ S ht c bn = 2.Z + N (mang in: 2.Z, khng mang in: N).+ S ht mang in = s electron + s proton = 2.Z .

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng cao+ S ht ht nhn = s proton + s ntron = Z + N.+ iu kin bn ca ht nhn nguyn t l

N1 1,33

Z vi Z 20

1 1,5N

Z vi Z 82

Tng ht = Z + E + N = 2.Z + N m : Z N 1,5.Z

Nn: 2.Z + Z 2.Z + N 2.Z + 1,5.Z 3.Z Tng ht 3,5.Z 3,5 3

hat hat Z

- T k hiu nguyn t XAZ => s p v s n trong ht nhn cng nh s electron v nguyn t v ngcli.- Tt c cc nguyn t c cng s in tch ht nhn Z u thuc cng mt nguyn t ha hc.- Cng thc tnh th tch ca mt nguyn t:

33

4RV = (R l bn knh nguyn t)

III/ S chuyn ng ca e trong nguyn t. Obitan nguyn t.

1/ S chuyn ng ca electron trong nguyn tTrong nguyn t, cc electron chuyn ng rt nhanh xung quanh ht nhn khng theo qu o xc

nh no v to thnh m my electron.2/ Obitan nguyn t (AO)a/ nh ngha: Obitan nguyn t l khu vc m my electron xung quanh ht nhn m xc sut c mtelectron khong 90%.b/ Hnh dng obitan nguyn t: Da trn s khc nhau v trng thi electron trong nguyn t ta c:- Obitan s: dng hnh cu.- Obitan p: gm 3 obitan px, py, pz c hnh dng s 8 ni, nh hng theo 3 trc Ox, Oy, Oz ca h ta .

IV/ Lp v phn lp e:1/ Lp electron: Lp electron gm cc electron c mc nng lng gn bng nhau. Cc lp electron xptheo th t mc nng lng t thp n cao (t gn nhn ra ngoi):

Lp th n 1 2 3 4 5 6 7Tn lp K L M N O P QC s obitan l n2 1 4 9 16C s electron ti a l 2n2 2 8 18 32

2/ Phn lp electron- Mi lp electron chia thnh cc phn lp s, p, d, f gm cc electron c mc nng lng bng nhau:

Phn lp s p d f C s obitan 1 3 5 7C s electron ti a 2 6 10 14

- Trong 1 lp electron th s phn lp = s th t lp:

Lp th 1 2 3 4C phn lp 1s 2s2p 3s3p3d 4s4p4d4f

- Phn lp electron cha electron ti a gi l phn lp electron bo ha.

V/ Nng lng Cu hnh e trong nguyn t :1/ Cc nguyn l v quy tc phn b electron trong nguyn t

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng caoa/ Nguyn l Pauli: Trn 1 obitan c ti a 2e v 2e ny chuyn ng t quay khc chiu nhau:

1 obitan c 2e: 2e ghp i

1 obitan c 1e: 1e c thn

b/ Nguyn l vng bn: trng thi c bn, trong nguyn t cc electron chim ln lt cc obitan c mcnng lng t thp n cao.

c/ Quy tc Hund: Trong 1 phn lp, cc electron s phn b trn cc obitan sao cho s electron c thn lti a v c chiu t quay ging nhau.

V d: 7 N

1s2 2s2 2p3

d/ Trt t cc mc nng lng nguyn t: Trong nguyn t, cc electron trn cc obitan khc nhau, nhngcng 1 phn lp c mc nng lng nh nhau. Cc mc nng lng nguyn t tng dn theo trnh t:1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

* e nh ta dung quy tac Klechkowsky

1s 2s 3s 4s 5s 6s 7s

2p 3p 4p 5p 6p 7p

3d 4d 5d 6d 7d

4f 5f 6f 7f

2/ Cu hnh electron nguyn t: Cu hnh electron nguyn t biu din s phn b electron trn ccphn lp ca cc lp electron khc nhau.a/ Cch vit cu hnh electron nguyn t

- Xc nh s electron trong nguyn t.- Phn b cc electron theo trt t mc nng lng AO tng dn.- Vit cu hnh electron theo th t cc phn lp electron trong mt lp.

V d: 26Fe. Vit theo trt t mc nng lng AO tng dn: 1s2 2s2 2p6 3s2 3p6 4s2 3d6

- Sau vit li theo th t cc phn lp electron trong 1 lp: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 4s2

- Vit gn: [Ar] 3d6 4s2

* Ch : Khi vit cu hnh electron d nh trt t cc mc nng lng, ta vit theo th t lp vi 2 phnlp s, p nh sau:

1s 2s2p 3s3p 4s ... 4p 5s ... 5p 6s ... 6p 7s ... 7p- Sau thm 3d vo gia lp 4s ... 4p- Thm 4d vo gia lp 5s ... 5p- Thm 4f 5d vo gia lp 6s ... 6p- Thm 5f 6d vo gia lp 7s ... 7p- Ta s c 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

b/ c im ca lp electron ngoi cng

- Cc electron lp ngoi cng quyt nh tnh cht ha hc ca mt nguyn t.- S electron lp ngoi cng ti a l 8e+ Cc nguyn t kim loi c: 1e, 2e, 3e lp ngoi cng.+ Cc nguyn t phi kim c: 5e, 6e, 7e lp ngoi cng.

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng cao+ Cc nguyn t kh him c: 8e (He c 2e) lp ngoi cng.+ Cc nguyn t c 4e lp ngoi cng c th l kim loi (Ge, Sn, Pb) c th l phi kim (C, Si).

VI. Mt s vn b sung:

Xc nh v tr ca nguyn t X trong bng tun hon ha hc:1. Phn nhm chnh ( nhm A ) : khi cc e cui cng c in vo phn lp s hay p, c th: nsa npb

(vi iu kin a,b s nguyn v a 1 , 0 b 6)

S th t nhm = a + b trong :

3 .4 . / .5 .8 .

a b K L

a b K L P K

a b P K

a b Kh h

+ + =

+ + =

2. Phn nhm ph ( nhm B ) : khi cc e cui cng c in vo phn lp d hay f, c th: (n-1)da nsb(vi iu kin a,b s nguyn v b = 2 , 1 a 10) Nu a + b < 8 S th t nhm = a + b Nu a + b = 8 hay 9 hay 10 S th t nhm = 8

Nu a + b > 10 S th t nhm = (a + b) 10 Cc nguyn t nhm B u thuc kim loi chuyn tip.

Ngoi tr: b = 2 , a = 4 b = 1 , a = 5 (bn bo ha gp) b = 2 , a = 9 b = 1 , a = 10 (bo ha gp)

VII. S bin i tun hon tnh cht ca cc nguyn t ha hc:

1. Bn knh nguyn t:

Trong cng mt chu k: i t tri sang phi theo chiu tng dn ca in tch ht nhn, bnknh nguyn t ca cc nguyn t gim dn. Nguyn nhn l do s lp e ngoi cng nhnhau, khi i t ny sang sau lin k vi n th e lp ny tng ln v in tch ht nhncng tng lm lc ht gia ht nhn v e tng bn knh nguyn t gim.

Trong cng mt phn nhm chnh (nhm A): i t trn xung di theo chiu tng dn cain tch ht nhn, bn knh nguyn t ca cc nguyn t tng dn.Nguyn nhn l do s lpe tng dn khi i t trn xung v e lp ngoi cng ging nhau lm lc ht gia ht nhn vie lp ngoi cng gim dn, mc d in tch ht nhn c tng.

Ch : . .n e m en mM M M ++ cation anion

Khi mt ngt mt e to thnh ion dng (cation) th kch thc gim i rt nhiu bn knh ca cation bao gi cng nh hn bn knh ca nguyn t tng ng.Cng mt nguyn t, nu in tch ion cng ln th bn knh cng nh.Vd: 2 3Fe Fe Fer r r+ +> >

Khi mt ngt nhn thm e to thnh ion m (anion) th kch thc ion tng ln v enhn thm vo lm tng tng tc y e e. Bn knh ca anion bao gi cng ln hn bn knh ca nguyn t tng ng.Vd: ClCl Cl r r r +> >

2. Nng lng ion ha: I

Ni mt cch tm tt: nng lng ion ha l nng lng ti thiu cn tiu th tch mt e ra khinguyn t th kh v bin thnh ion dng.

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C th, nng lng ion ha ca mt nguyn t hay mtphn t l nng lng cn thit tch mtin t t nguyn t hay phn t trng thi c bn. Mt cch tng qut hn, nng lng ion hath n l nng lng cn thit tch in t th n sau khi tch (n-1) in t u tin. Trng thic bn chnh l trng thi m ti , nguyn t khng chu nh hng ca bt k mt t trngngoi no c. Tc l mt nguyn t kim loi trng thi c bn s c dng kh, v cu hnh electronca n cng l cu hnh c bn: tun theo nguyn l Pauli, Nguyn l vng bn v qui tc Hund.

Theo t in Gio khoa Vt l ca cc tc gi V Thanh Khit,..., Nh Xut bn Gio dc- nm 2007

th nng lng ion ho c nh ngha nh sau: nng lng ion ha ca mt nguyn t, phn thoc ion l nng lng cn thit tch lectron lin kt yu nht ra khi mt ht trng thi c bnsao cho ion dng c to thnh cng trng thi c bn. l nng lng ion ho th nht. Ccgiai on ion ho tip theo s ng vi cc nng lng ion ho th hai, th ba,..

Nguyn t cng d nhng e (tnh kim loi cng mnh) th gi tr I cng nh. Phn bit nng lng ion ha th nht I1 , th hai I2 ,

M M+ + 1e , I1 > 0M+ M2+ + 1e , I2 > I1

Quy tc Koopmans: Nng lng ion ha th nht I1 ca nguyn t l bng i ca nng lng caobitan m e b tch chim.

Trong mt chu k, theo chiu tng dn ca in tch ht nhn, lc lin kt gia ht nhn v e lpngoi cng tng, lm cho nng lng ion ha ni chung cng tng theo.

Trong cng mt phn nhm chnh, theo chiu tng dn ca in tch ht nhn, bn knh nguyn ttng,lc lin kt gia e lp ngoi cng v ht nhn gim, do nng lng ion ha ni chung gim.

i nt v ion : Ion l mt nguyn t hay nhm nguyn t b mt hay thu nhn thm c mt hay nhiu in

t. Mt ion mang in tch m,khi n thu c mt hay nhiu in t, c gi l anion, vmt ion mang in tch dng khi n mt mt hay nhiu in t, c gi l cation. Qu trnhto ra cc ion gi l ion ha.

Cc nguyn t hay nhm nguyn t b ion ha c biu din di dng cc s vit nh lntrn, bn phi k hiu ca nguyn t hay nhm nguyn t, th hin s lng in t m nthu c hay mt i (nu ln hn 1) v du + hay ty theo n mt hay thu c (cc) int. Trong trng hp mt hay thu c ch mt in t th khng cn ghi gi tr s. V d H +

hay O2-. Cc kim loi c xu hng to ra cc cation (mt i in t) trong khi cc phi kim loi c xu

hng to ra anion, v d natri to ra cation Na+ trong khi clo to ra cc anion Cl-. Cc ion ln u tin c l thuyt ha bi Michael Faraday khong nm 1830, miu t

cc thnh phn ca phn t m chuyn ng v pha ant hay catt. Tuy nhin, c ch m ccchuyn ng ny c th din ra khng c miu t cho n tn nm 1884 khi Svante

August Arrhenius trong lun n tin s ca mnh trong trng i hc tng hp Uppsala miu t chng. L thuyt ca ng ban u khng c chp nhn (ng nhn c hc vtin s vi im thp nht c vt qua) nhng lun n tin s ca ng ot gii Nobelv ha hc nm 1903.

T ion c t tn bi Michael Faraday, t ting Hy Lp , ng tnh t thi hin ti ca , "chuyn ng", v th l "ngi i li". Danh php ny da trn xu hng ca cc anion chuyn ng v pha ant, v ca cc cation chuyn ng v pha catt. V th, anion( ) v cation ( ) c ngha l "(mt th) i ln" v "(mt th) i xung", mt cch tng ng, v ant, , v catt, , c ngha l "i ln" v "i xung", tng ng t , "ng".

3. i lc e: E

i lc e cng l nng lng gii phng khi mt nguyn t th kh kt hp mt e vo binthnh ion m.

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7/23

H thng kin thc ha hc lp 10 Chng trnh C bn v Nng caoM + 1e M- , E < 0

Trong cng mt chu k, ni chung i lc e cng m theo chiu tng dn ca in tch htnhn. Ngoi tr cc kh him li c i lc e dng.

Trong cng mt phn nhm chnh, theo chiu tng dn ca in tch ht nhn, phn ln i lce km m dn (gi tr tuyt i ca E gim dn). i lc e ca cc nguyn t nhm II.A,II.B cgi tr dng.

4. m in: (c l khi)

m in ca mt nguyn t c trng cho kh nng ht e ca nguyn t khi to thnhlin kt ha hc.

m in ca mt nguyn t cng ln th tnh phi kim ca nguyn t cng mnh. Ngcli, m in cng nh, tnh kim loi ca nguyn t cng mnh.

Trong cng mt chu k, i t tri sang phi theo chiu tng dn ca in tch ht nhn, min ca cc nguyn t ni chung tng dn.

Trong cng mt phn nhm chnh, i t trn xung di theo chiu tng dn ca in tch ht

nhn, m in ca cc nguyn t ni chung gim dn. Nguyn t flo (F) c m in ln nht l 3,98 .

5. Tnh kim loi, tnh phi kim:

Tnh kim loi l tnh cht ca mt nguyn t m nguyn t ca n d nhng e tr thnhion dng.

M n.e Mn+ Nguyn t ca nguyn t cng d nhng e, tnh kim loi cng mnh.

Tnh phi kim l tnh cht ca mt nguyn t m nguyn t ca n nhn thm e tr thnh

ion m. M + n.e Mn- Nguyn t ca nguyn t cng d nhn e, tnh phi kim cng mnh.

Trong cng mt chu k, i t tri sang phi theo chiu tng dn ca in tch ht nhn, tnhkim loi gim dn, tnh phi kim tng dn (ca cc nguyn t).

Trong cng mt phn nhm chnh, i t trn xung di theo chiu tng dn ca in tch htnhn, tnh kim loi tng dn, tnh phi kim gim dn (ca cc nguyn t).

Trong Bng THHH, kim loi chim phn di bn tri v phi kim chim phn trn bn phi,gii hn ny khng r rt l ng cho k t gc trn bn phi.

6. S bin i v ha tr ca cc nguyn t:

Ha tr cao nht ca mt nguyn t vi oxi, vi hidro ca cc phi kim bin i tun hon theochiu tng dn ca in tch ht nhn.

Trong mt chu k, i t tri sang phi theo chiu tng dn ca in tch ht nhn, ha tr caonht ca cc nguyn t vi Oxi tng ln lt t 1 7, cn ha tr vi hidro ca cc phi kimgim t 4 1.

Bng bin i tun hon ha tr ca cc nguyn t nhm A.

Nhm I.A II.A III.A IV.A V.A VI.A VII.AHp cht vi Oxi Na2O

K2O

MgO

CaO

Al2O3

Ga2O3SiO2

GeO2P2O5

As2O5SO3

SeO3Cl2O7

Br2O7Ha tr cao nht viOxi

I

II

III

IV

V

VI

VII

Tng qut ha tr cao R2O RO R 2O3 RO2 R2O5 RO3 R2O7

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nht vi OxiHp cht kh vi hidro SiH4

GeH4PH3

AsH3H2S

H2SeHClHBr

Ha tr vi hidro IV III II ITqut ha tr vi hidro RH4 RH3 RH2 RH

Ch : Ho tr cao nht vi oxi + ho tr vi hiro = 8 ( ch p dng cho nguyn t

nhm A )

Tnh cht nguyn t nhm IA v IIA:

a) Nhm IA ( nhm kim loi kim )

- Tc dng vi nc k thng cho kim tng ng v gii phng hiro.2M + 2H2O 2MOH + H2

- Tc dng mnh vi oxi cho ra oxit baz kim, cc oxit ny tc dng mnhvi nc cho kim

4M + O2 2M2O ( ch to ra peoxit v supeoxit )M2O + H2O 2MOH

- Tc dng vi phi kim cho mui.

b) Nhm IIA ( kim loi kim th )

- k thng tc dng vi nc ( tr Mg tc dng chm vi nc lnh, Be khngp )

R + 2H2O R(OH)2 + H2- Tc dng mnh vi oxi cho oxit, oxit tc dng mnh vi nc cho dung dch

kim

2R + O2 2RORO + H2O R(OH)2

7. S bin i tnh axit bazo ca oxit v hidroxit:

Oxit v hidroxit ca kim loi th hin tnh bazo. Oxit v hidroxit ca phi kim th hin tnh axit. Tnh axit bazo ca chng mnh yu ph thuc vo mnh yu ca kim loi v phi kim

tng ng. Hidroxit kim loi M(OH)n c tnh bazo v: M l nguyn t kim loi, c xu hng nhng e,

tng ng vi tc dng y e mnh s phn cc lin kt M O tng v s phn cclin kt O H gim lin kt M O phn cc mnh hn, km bn, d t cho ion OH-

Th hin tnh baz. Hidroxit phi kim R(OH)n c tnh axit v: R l nguyn t phi kim, c xu hng nhn e, tng

ng vi tc dng ht e mnh s phn cc lin kt R O gim v s phn cc lin ktO H tng lin kt O H phn cc mnh hn, km bn, d t cho ion H+ Th hin tnh axit mnh.

Trong cng mt chu k , i t tri sang phi theo chiu tng dn ca in tch ht nhn, tnhbazo ca oxit v hidroxit tng ng gim dn, ng thi tnh axit ca chng tng dn.

8. Quy tc bo ha sm v na bo ha sm:

Cu hnh bn ca phn lp d ng vi trng thi bo ha (10e) hay na bo ha (5e). V vy,khi v ngoi ca nguyn t, phn lp d c 9 hoc 4 e th c s nhy e t phn lp s ca lp

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng caolin bn ngoi phn lp d t trng thi bo ha hay na bo ha bn vng. Hin tngny gi l bo ha sm v na bo ha sm.

Hin tng ny thng xy ra i vi mt s nguyn t thuc nhm I.B v VI.B trong bngtun hon.

Th d: Cu ( Z = 29 ): 1s2 2s2 2p6 3s2 3p6 3d9 4s2

thc t l: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Bo ha sm

Th d: Cr ( Z = 24 ): 1s2

2s2

2p6

3s2

3p6

3d4

4s2

thc t l: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Na bo ha sm

9. Bng tun hon cc nguyn t nhm A:

IA IIA IIIA IVA VA VIA VIIA VIIIA

H1Hi

He 2H

Li3Li

Be 4B

B 5B

C 6Ch

N 7Nu

O 8ng

F 9Fi

Ne 10Nu

Na11

No

Mg 12Mng

Al 13n

Si 14Sn

P 15Phi

S 16S

C17 Cho

Ar 18Ai

K19Khng

Ca 20Cn

Ga 31G

Ge 32G

As 33n

Se 34St

Br 35Bom

Kr 36Khng

Rb37

Rt

Sr 38S (r)

In 49t

(nhiu)

Sn 50Sang

Sb 51Sn

Te 52T

I 53

Xe 54X

Cs55C(sa)

Ba 56Ba

T81 Tin(l)

Pb 82Ph(bun)

Bi 83B

Po 84Ph

At 85t

Rn 86Ra

Fr87F(ri)

Ra 88Ra

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0939.889.444

Chu k 4 v 5:T II.AIII.Ath ZIII = ZII+11

Chu k 6:T II.AIII.Ath ZIII = ZII+25

nh s th ttng dn t trisang phi


10/23


***Cc v d:

Vd 1:Tng s ht proton, notron, v e ca mt nguyn t X trong mt nguyn t bng 21.a) Xc nh s proton, notron v s khi ca nguyn t.

b) Vit cu hnh e ca nguyn t v biu din s sp xp chng trn cc obitan. Suy ra v tr ca nguynt trong bng tun hon.

Gii:a.

Cch 1: Ta c: 2.Z + N = 21 N = 21 2.Z Z < 82 1 1,5N

Z hay

1

1,5

N

ZN

Z

21 2.1

21 2.1,5

Z

ZZ

Z

21 2.21 2. 1,5.

Z Z

Z Z

21 3.21 3,5

Z

Z

76

Z

Z

6 Z 7

K bng:Z 5 6 7 8

N = 21 2.Z 11 9 7 5A = Z + N 16 15 14 13

Chn77

Z

N

=

= E = 7 v A = 14 (vC)

Cch 2:

Ta c: Tng ht = 21 21 213,5 3

Z 6 Z 7

K bng:Z 5 6 7 8

N = 21 2.Z 11 9 7 5A = Z + N 16 15 14 13

Chn77

Z

N

=

= E = 7 v A = 14 (vC)

b.X l Nito : N

N ( Z = 7 ): 1s2 2s2 2p3

N th 7 trong bng tun hon v thuc chu k 2.N c 3 e cui cng in vo phn lp p nn N l nguyn t p v c 5e thuc lp 2

N thuc nhm V.A

Vd 2:Hai nguyn t A v B c phn lp ngoi cng l 3p v 4s tng ng. Bit tng s e ca 2 phn lp l5 v hiu s l 3. Hy vit cu hnh e ca hai nguyn t v nh gi tr Z ca A v B.Gii:

Gi x l s e ca phn lp 3py l s e ca phn lp 4s

Theo , ta c: 53x yx y+ = = 41xy = =

Do , cu hnh e ca nguyn t A v B l:A: 1s2 2s2 2p6 3s2 3p4 ZA = 16

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B: 1s2 2s2 2p6 3s2 3p6 4s1 ZB = 19

Vd 3: Da vo bng khi lng ca proton, notron,v e. Hy tnh khi lng ca nguyn t: Cl gm 17p,18n v 17e.Gii: Khi lng 1 nguyn t Clo l:

mCl = (17.1,6726.10-24

) + (18.1,6748.10-24

) + (17.9,1094.10-28

) = 58,596.10-24

gVd 4: Bit khi lng mt nguyn t st 93,6736.10-24g c 26 proton, t khi l 7,9. Bit cc nguyn t sttrong tinh th ch chim 74% th tch.

a) Tnh t s khi lng ca cc e trong nguyn t so vi khi lng ca ton nguyn t. T c thcoi khi lng nguyn t thc t bng khi lng ht nhn c khng?

b) Xc nh bn knh nguyn t gn ng ca st.Gii:

a) T s khi lng e so vi khi lng ngun t st l:28

4

24

26.9,1094.102,5284.10

93,6736.10e

ngtu

m

m

= = 1=

c th xem thc t khi lng nguyn t bng khi lng ht nhn

b) Khi lng mol ca st: MFe = 93,6736.10-24

.6,02.1023

= 56,391 gTh tch 1 mol Fe:

56,3917,138

7,9M

VD

= = = cm3

Th tch ca mt mol st trong mng tinh th l:74%

7,318. 5,41532100%

= cm3

Vy th tch ca mt nguyn t st trong mng tinh th: 23235,41532

0,9.106,02.10

= cm3

Gi s nguyn t Fe l mt khi cu:23

3 3 24 83334 3. 3. 3.0,9.10

. 2,150.10 1, 29.10

3 4. 4. 4.

V VV R R R

= = = = = = cm3

Vd 5: Khi lng nguyn t ca clo l 35,5. Clo c hai ng v l 3517Cl v37

17Cl. Tnh phn trm v s

nguyn t ca mi ng v.Gii:

Gi x l % ca s nguyn t ng v th nhty l % ca s nguyn t ng v th hai y = 100 x

35. (100 ).3735,5

100

x xM

+ = = x = 75%

3517Cl chim 75% v37

17Cl chim 25%

Vd 6: Hidro iu ch t nc nguyn cht c khi lng nguyn t l 1,008. Hi c bao nhiu nguyn t11H v 21H trong 1g nc.Gii:

Gi x l thnh phn phn trm v s nguyn t ca ng v 11H v (100-x) l thnh phn phn trm

s nguyn t ca ng v 21H

1. (100 ).21,008

100H

x xM

+ = = x = 99,2%

1

1

2

1

:99,2%

:0,8%

H

H

Ta c : 1g nc th c1

18mol H2O

M 1mol H2O c 6,02.1023 ht phn t H2O

S phn t H2O c trong 1g nc l:1

18.6,02.1023 ht

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S nguyn t 11H c trong 1g nc l:1

18.6,02.1023.2.

99,2100

= 663,53.1020 ht

S nguyn t 21H c trong 1g nc l:1

18.6,02.1023.2.

0,8100

= 5,35.1020 ht

Vd 7: Cho bit tng s e trong anion 23AB l 42. Trong cc ht nhn A cng nh B s proton bng s

notron.a. Tnh s khi ca A v B.b. Vit cu hnh e v s phn b e trong cc obitan ca cc nguyn t A, B.

Gii:

Ta c: ZA + 3.ZB = 42 2 = 40 ZB Mg2+ > F- > O2-. B. Mg2+ > Na+ > F- > O2-.C. F- > Na+ > Mg2+ > O2-. D. O2-> F- > Na+ > Mg2+.Cu 20: Trong hp cht ion XY (X l kim loi, Y l phi kim), s electron ca cation bng s electron

ca anion v tng s electron trong XY l 20. Bit trong mi hp cht, Y ch c mt mc oxi ha duy

nht. Cng thc XY lA. LiF. B. NaF. C. AlN. D. MgO.Cu 21: Hai nguyn t X v Y thuc cng mt nhm A chu k 2 v 3 c s n v in tch ht nhn hn

km nhau lA. 8. B. 18. C. 2. D. 10.

Cu 22: Hai nguyn t A, B 2 nhm A lin tip trong h thng tun han. B thuc nhm V. trng thin cht, A v B khng phn ng vi nhau. Tng s proton trong ht nhn nguyn t A v B l 23. Tnca A v B lA. cacbon, photpho. B. oxi, photpho. C. nit, lu hunh. D. nit, oxi.

Cu 23: Hai nguyn t A, B c phn lp electron ngai cng ln lt l 2p, 3s. Tng s electron ca haiphn lp ny l 5 v hiu s electron ca chng l 1. S th t A, B trong bng HTTH ln lt l :

A. 5, 10 B. 7, 12C. 6, 11 D. 5, 12Cu 24:Nguyn t ca nguyn t X c tng s electron trong cc phn lp p l 7. Nguyn t ca nguyn t

Y c tng s ht mang in nhiu hn tng s ht mang in ca X l 8. Cu hnh electron lp ngoicng ca Y lA. 3s2 3p4. B. 3s2 3p5. C. 3s2 3p3. D. 2s2 2p4.

Cu 25: Mt nguyn t thuc nhm VA c ha tr cao nht vi oxi v ha tr trong hp cht vi hidro lnlt lA. III v V. B. V v V. C. III v III. D. V v III.

Cu 26: Nguyn t X l phi kim c ho tr cao nht vi oxi l a; ho tr trong hp cht kh vi hidro l b.Quan h gia a v b l

A. a = b. B. a + b = 8. C. a b. D. a - b = 8.Cu 27: Nguyn t chu k 4, nhm VIB c cu hnh electron ha tr l

A.4s24p4. B.6s26p2. C.3d54s1. D.3d44s2.Cu 28: Tng s ht trong ion M3+ l 37. V tr ca M trong bng tun han l

A. chu k 3, nhm IIIA. B. chu k 3, nhm IIA.C. chu k 3, nhm VIA. D. chu k 4, nhm IA.

Cu 29:Nguyn t X c tng s ht c bn (p, n, e) l 82. S ht mang in nhiu hn s ht khng mangin l 22. V tr ca X trong bng tun hon lA. chu k 4, nhm VIB. B. chu k 4, nhm VIIIB.C. chu k 4, nhm IIA. D. chu k 3, nhm IIB.

Cu 30: Anion X- v cation Y2+ u c cu hnh electron lp ngoi cng l 3s23p6. V tr ca cc nguyn ttrong bng tun hon cc nguyn t ha hc l

A. X c s th t 17, chu k 4, nhm VIIA; Y c s th t 20, chu k 4, nhm IIA.B. X c s th t 18, chu k 3, nhm VIA; Y c s th t 20, chu k 4, nhm IIA.C. X c s th t 18, chu k 3, nhm VIIA; Y c s th t 20, chu k 3, nhm IIA.D. X c s th t 17, chu k 3, nhm VIIA; Y c s th t 20, chu k 4, nhm IIA.

Cu 31: Tnh phi kim ca cc nguyn t trong nhm VIA theo th t: 8O, 16S, 34Se, 52Te, bin i theo chiuA. tng. B. gim. C. khng thay i. D. va tng va gim.

Cu 32: Cc nguyn t thuc nhm IIA trong bng tun hon c tnh cht no sau y?A. D dng cho 2e t cu hnh bn vng. B. D dng nhn 2e t cu hnh bn vng.C. D dng nhn 6e t cu hnh bn vng. D. L cc phi kim hot ng mnh.

Cu 33: Ion Y

c cu hnh e: 1s

2

2s

2

2p

6

3s

2

3p

6

. V tr ca Y trong bng tun hon lA. chu k 3, nhm VIIA. B. chu k 3, nhm VIIIA.C. chu k 4, nhm IA. D. chu k 4, nhm VIA.

Cu 34:Nguyn t cc nguyn t trong mt nhm A ca bng tun han th c cngA. s ntron. B. s lp electron. C. s proton. D. s e lp ngoi cng.

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng caoCu 35: Trong nguyn t ca nguyn t R c 18 electron. S th t chu k v nhm ca R ln lt l

A. 4 v VIIIB. B. 3 v VIIIA. C. 3 v VIIIB. D. 4 v IIA.Cu 36: Da vo quy lut bin i tnh cht ca bng tun hon th kim loi mnh nht (tr nguyn t

phng x) v phi kim mnh nht lA. franxi v iot. B. liti v flo. C. liti v iot. D. xesi v flo.

Cu 37: Trong mt chu k ca bng tun hon, s bin i tnh axitbaz ca cc oxit cao nht v cchidroxit tng ng theo chiu tng ca in tch ht nhn l

A. tnh axit v bazo u tng. B. tnh axit tng dn, tnh bazo gim dn.C. tnh axit v bazo u gim. D. tnh axit gim dn, tnh bazo tng dn.Cu 38: Cho cc nguyn t M (Z=11), X (Z=17), Y(Z=9) v R (Z=19). m in ca cc nguyn t tng

dn theo th tA. M


21/23


Bi 1Ho tan hon ton 1,2 gam mt kim loi thuc nhm IIA trong bng h thng tunhon, bng dd axit HCl. Sau p thu c 1,12 lt kh H2 (ktc). Kim loi l :

A. Be B. Mg C. Ca D. BaBi 2Ho tan hon ton 2,4 gam mt kim loi R thuc nhm IIA vo 100 ml dd axit HCl 1,5

M. Sau p thy vn cn mt phn R cha tan ht.Cng 2,4 gam R trn nu cho tc dng vi 125 ml dd axit HCl 2 M. Sau p thy vn cnd axit. R l

A. Be B. Mg C. Ca D. Ba

Bi 3Cho 0,425 gam hn hp 2 kim loi kim thuc hai chu k lin tip p vi mt lng nc cd. Sau p thu c dd A v 168 cm3 kh H2 (ktc). trung ho ht dd A cn phi dngva ht V ml dd H2SO4 1M. Hai kim kim v gi tr V l :

A. Li, Na v V = 60 ml B. Na, K v V = 30 ml C. Mt kt qu khc.Bi 4Hp cht kh vi hiro ca mt nguyn t c dng RH3. Oxit cao nht ca nguyn tny cha 25,92 % R.a)Nguyn t R l : A. N B. P C. As D. Sbb) So snh tnh phi kim ca R vi O, F, P.Bi 5Cho 13,7 gam hn hp gm Ba v Ca tc dng ht vi nc th thot ra V lt kh H2(ktc)Xc nh V.Bi 6Oxit cao nht ca nguyn t R c khi lng phn t l 108. Hy bin lun xc nhnguyn t R.

A. Si B. N C. P D. Mt kt qu khc.Bi 7Oxit cao nht ca nguyn t R c khi lng phn t l 80. Hy bin lun xc nhnguyn t R.

A. Si B. N C. P D. Mt kt qu khc.Bi 8T l gia khi lng phn t hp cht kh vi hiro ca nguyn t R so vi oxit cao

nht ca n l 17: 40. Hy bin lun xc nh nguyn t R.

Bi 9T l gia khi lng phn t hp cht kh vi hiro ca nguyn t R so vi oxit caonht ca n l 1: 2,75. Hy bin lun xc nh nguyn t R.Bi 10Cho 6,2 gam hn hp Na v mt kim loi kim X tc dng ht vi 104 gam nc thu c

110 gam dd. Xc nh kim loi X bit MX < 40.Bi 11Cho 2,74 gam mt kim loi thuc nhm IIA vo cc cha nc. Kt thc p thy khi lngdung dch thu c tng 2,7 gam. Kim loi l

A. Ca B. Sr C. Ba D. RaBi 12Cho 3,6 gam hai kim loi thuc nhm IIA v thuc hai chu k k tip vo cc cha

dd axit HCl. Kt thc p thy khi lng dung dch thu c tng 3,2 gam. Hai kim loi l

A. Be v Mg B. Mg v Ca C. Ca v Ba D. Ra v Ba

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H thng kin thc ha hc lp 10 Chng trnh C bn v Nng caoBi 13Cho 20 gam kim loi Ca tc dng ht vi nc th sinh ra V lt kh H2 o 27 0 C v 1

atm. Tnh VA. V = 8,96 lt B. V = 12,3 lt C. V = 17,44 lt D. Mt kt qu khc.

Bi 14Cho 0,56 gam kim loi kim tc dng ht vi nc th sinh ra 0,873 lt kh H2 o 0 0

C v 780 mm Hg. Kim loi kim l

A. Li B. Na C. K D. CsBi 15Ho tan ht mu hp kim Ba Na vo nc c dd A v c 6,72 lt kh H2 bay ra(ktc). Cn dng bao nhiu ml dd HCl 1M trung ha hon ton 1/10 dd A.

A. 60 ml B. 40 ml C. 600 ml D. 750 mlBi 16Ho tan 46 gam mt hn hp gm Ba v 2 kim loi kim A v B thuc hai chu k lintip vo nc th c dd D v 11,2 lt kh o ktc.Nu thm 0,18 mol Na2SO4 vo dd D th sau p th vn cha kt ta ht ion Ba.Nu thm 0,21 mol Na2SO4 vo dd D th sau p cn d Na2SO4. Xc nh 2 kim loikim A, B.

Bi 17Mt nguyn t X c cu hnh e phn lp ngoi cng l 4s1.

a) Vit cu hnh e y v suy ra s hiu nguyn t v tn nguyn t.b) xc nh ng X, ngi ta ly 2,8 gam oxit ca X cho tc dng va vi 50

ml dd axit HCl 1,4 M. Gi tn ng X.Bi 18Ho tan ht 4 gam mt kim loi vo 96,2 gam nc, c dd baz c nng 7,4 % vV lt kh H2 (ktc). Tm kim loi v th tch V.

A. Na v V = 11,2 lt B. Ca v 2,24 lt C. K v V = 3,3 6 lt D. mtkt qu khc

Bi 19Khi cho 3,1 gam hn hp 2 kim loi kim A, B tc dng vi 47 gam nc thy c x lt khthot ra (ktc). Dung dch thu c c tng nng phn trm ca cht tan l 9,6%.a) Tnh gi tr x : A. 1,12 B. 2,24 C. 3,36 D. mt kt qukhc.b) Nu A, B l 2 kim loi kim thuc hai chu k lin tip, hy xc nh 2 kim loi trn.Bi 20Gi tn nguyn t trong chu k 3 v vit cu hnh e ca nguyn t bit rng nnglng ion ho (I) c cc gi tr sau ( tnh theo kJ/mol )

I1 I2 I3 I4 I5 I6

1012 1903 2910 4956 6278 22230Bi 21Gi tn nguyn t trong chu k 3 v vit cu hnh e ca nguyn t bit rng nnglng ion ho (I) c cc gi tr sau ( tnh theo kJ/mol )

I1 I2 I3 I4 I5 I6577 1816 2744 11576 14829 18375

Bi 22A, B l hai nguyn t cng nm trong mt nhm A v thuc hai chu k lin tip trongbng HTTH. Tng cc ht mang in ca A v B l 160. Vit cu hnh e ca A v B.Bi 23Hp cht Z c to bi 2 nguyn t M v R c cng thc M aRb, trong R chim 6,667% v kki lng. Trong ht nhn nguyn t M c n = p + 4, cn trong ht nhn ca R cn = p. Bit rng tng s ht proton trong phn t Z bng 84 v a + b = 4. TmCTPT ca Z.

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CC CNG THC TNH TON

1.m

nM

= m = n.M 2. Vkh = n . 22,4 =>

22,4kh

Vn = 3. .M M

M

n nC n C V V

V C= => = => =

4..100% %.

%100%

ct dd ct

dd

m C mC m

m= = 5. % .100%AA

hh

mm

m=

6. Tnh theo sn phm: li thuyetthuc te thuc tethuc te li thuyet

li thuyet

H.mm m*100% m = m = *100%

m 100% HH = => =>

Tnh theo cht tham gia:li thuyet li thuyetthuc te

li thuyet thuc te

thuc te

m mH.m*100% m = m = *100%

m 100% HH = => =>

7. S mol A o t0C, P atm...A

pVn

R T=

P: p sut kh A o t0C (tnh bng atm) V: th tch kh A o t0C (tnh bng lt) n: s mol cht kh AT: nhit tuyt i (tnh bng 0K) T = t + 273 (T tnh bng 0K) R = 22,4/273 = 0,082

8. T khi hi ca cht kh A. BA

M

M

d BA = 29A

M

d kkA = Mkk = 29

9. Khi lng ring:( )

( )

dd g

dd ml

mD

V= mdd = Vdd . D

10. Mi lin quan gia nng phn trm v nng mol/l

M

DCCM

%10= D: khi lng ring; M: khi lngmol cht tan

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