cac dang bai tap thuong gap trong chuong nguyen tu

Ch 3

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CC DNG BI TP THNG GP CHNG NGUYN TDng 1: Xc inh tn nguyn t v vit cu hnh electron nguyn t nguyn t khi bit tng s ht proton, electron, ntron l a ht ( vi a l s ht cho trc)

Phng php gii:

Gi E, N, P ln lt l s ht electron, ntron, proton c trong mt nguyn t ca nguyn t .

i vi cc nguyyn t thuc nguyn t c Z 82 c gi l nguyn t bn nn chng ta c cng thc:

1 1,5 (1)Mt khc tng s ht electron, ntron, proton trong 1 nguyn t l a ht nn:

E + P + N = 2P + N = a

> N = a 2P (2)

Thay (2) vo (1) ta c:

1 1,5

> P (3)

V P N* nn ta ly cc gi tr nguyn dng. Nu P ch c 1 gi tr th ta suy ra ngay nguyn t .

Nu P c nhiu gi tr khc nhau th ta phai lp bng bin lun nh sau:

P

A

Vi A = N + P v N = a 2P

A l s khi ca nguyn t. T hng th A bng hoc gn bng vi nguyn t khi ca nguyn t nguyn t v g tr i s nn ta ly cp P v A tha mn ri suy ra tn nguyn t .

Ch :

Khi lm bi tp trc nghim th chng ta ch cn nh cng thc (3) d lm nhanh bi tp dng ny.

VD1: Xc inh tn nguyn t v vit cu hnh electron nguyn t khi bit tng s ht proton, electron, ntron l 21 ht VD2: Xc inh tn nguyn t v vit cu hnh electron nguyn t khi bit tng s ht proton, electron, ntron l 13 ht.VD3: Tng s ht proton , ntron , electron ca nguyn t mt nguyn t kim loi l 34.

a) Xc nh tn nguyn t da vo bng tun hon cc nguyn t ho hc).

b) Vit cu hnh electron ca nguyn t ca nguyn t .

c) Tnh tng obitan v s electron trong nguyn t ca nguyn t trng thi c bn.

Dng 2: Xc inh tn nguyn t v vit cu hnh electron nguyn t nguyn t khi bit in tch ca ht nhn hay in tch ca lp v electron ca nguyn t nguyn t l Q.Phng php gii:

S proton hay s electron ca nguyn t nguyn t :

P = E = T gi tr ca P suy ra tn nguyn t v vit cu hinh electron.VD1: Xc inh tn nguyn t v vit cu hnh electron nguyn t nguyn t khi bit in tch ca ht nhn l + 17.622culng.

Dng 3: Xc inh tn nguyn t v vit cu hnh electron ca 1hoc xc nh s khi ca n khi bit:

Tng s ht c bn ( p, n, e ) ca nguyn t nguyn t R l a ht v s ht mang in nhiu hn s ht khng mang in l b ht ( vi a, b l s ht cho trc)Phng php gii:

Gi s ht electron, ntron, proton trong mt nguyn t nguyn t R ln lt l P, N, E.( P, N, E > 0).

Tng s ht c bn c trong 1 nguyn t R:

P + N + E = a (1)

Hiu s ht mang in so vi s ht khng mang in:

2P - N = b (2)

Gii h (1) v (2) ta c:

P = (I)

N = 2P - b = a - 2P (II)

T gi tr ca P = > tn nguyn t v A = P + N v vit cu hnh e ca n d dng.

Ch :

Khi lm bi tp trc nghim th chng ta ch cn quan tm dn cng thc (I), (II) lm nhanh bi tp dng ny.

VD1: Tng s ht c bn ( p, n, e ) ca nguyn t nguyn t R l 28 ht v s ht mang in nhiu hn s ht khng mang in l 8 ht. Xc inh tn nuyn t Rv s khi A ca n.

VD2: Tng s ht c bn ( p, n, e ) ca nguyn t nguyn t R l 82 ht v s ht mang in nhiu hn s ht khng mang in l 22 ht. Xc inh tn nuyn t Rv s khi A ca n.VD2:

Mt nguyn t X c tng 3 loi ht c bn l 40 trong s ht mang in nhiu hn s ht khng mang in l 12. X l:

A. Ca

B. Ni

C. Al

D. Si Tng s ht c bn ( p, n, e ) ca nguyn t nguyn t R l a ht v s ht mang in gp c ln s ht khng mang in ( vi a, c l s ht cho trc).

Phng php gii:

Gi s ht electron, ntron, proton trong mt nguyn t nguyn t R ln lt l P, N, E.( P, N, E > 0).

Tng s ht c bn c trong 1 nguyn t R:

P + N + E = a (1)

Mt khc ta c: P + E = 2P = c . N (2)

Gii h (1) v (2) ta c:

P = (III) N = (IV)

T gi tr ca P = > tn nguyn t v A = P + N v vit cu hnh e ca n d dng.

Ch :

Khi lm bi tp trc nghim th chng ta ch cn nh cng thc (IV) hoc (III) lm nhanh bi tp dng ny.VD1: Tng s ht c bn ( p, n, e ) ca nguyn t nguyn t R l 34 ht v s ht mang in gp ln s ht khng mang in. Xc nh tn nguyn t Rv s khi A ca n.Dng 4: Xc nh c bao nhiu phn t loi A2B c to thnh t n ng v ca A v m ng v ca B.

Phng php gii:

Gi s ng v ca A ln lt ng vi cc ng v l:

.

Phn t A2B dc vit li dng: ABALy 1 ng v ca B kt hp vi n ng v ca A.

Chn ng u ta c cc lai phn t:

. c n loii phn t.Chn ng u ta c cc lai phn t:

. c (n - 1) loi phn t.

.

.

Chn ng u ta c cc lai phn t: c 1 loi phn t.

Vy khi 1 dng v ca B kt hp vi n ng v ca A th c tt c:1 + 2 + .. + n = loi phn t.

V B c m ng v nn c tt c :S = (V) loi phn t A2B c to thnh.VD1: Hidro c 3 ng v l: ; ; ; Oxi c 3 ng v l: ; ; .Hy xc nh c bao nhiu phn t loi H2O c to thnh t cc lai ng v trn.VD2 : Trong t nhin Oxi c 3 ng v: ; ; ; Cacbon c 2 ngv l: ; Hi c th c bao nhiu loi phn t kh cacbonic hp thnh t cc ng v trn. Xc nh c bao nhiu phn t loi A3B c to thnh t n ng v ca A v m ng v ca B.

Chng minh tng t ta c cng thc:

C tt c :

S = m (

EMBED Equation.3 + + +.. + 1) (VI)loi phn t A3B c to thnh.

Ch :

Khi lm bi tp trc nghim th chng ta ch cn nh cng thc (V) v (VI) lm nhanh bi tp dng ny.

Dng 5:

Vit cu hnh electron ca 1 nguyn t nguyn t R khi bit s electron hoc s proton hoc in tch ht nhn.

Phng php gii:

Khi nim:

Cu hnh electron nguyn t biu din s phn b e trn cc phn lp thuc cc lp khc nhau.

Quy c cch vit cu hnh e nguyn t:

s th t lp e c vit bng cc ch s (1, 2, 3..)

phn lp c k hiu bng cc ch ci thng (s, p, d, f).

s e dc ghi bng ch s pha trn, bn phi k hiuca phn lp ( s2, p2..).

Cch vit cu hnh electron nguyn t:

Xc inh s electron ca nguyn t. Cc electron c phn b theo th t tng dn cc mc nng lng AO, theo cc nguyn l v quy tc phn b electron trong nguyn t. Theo s cc mc nng lng sau:

7s 7p 7d7f7g7h 6s

6p6d6f6g6h

5s

5p5d5f 5g

4s

4p4d4f 3s3p3d 2s 2p 1s

Vit cu hinh electron theo th t cc phn lp trong mt lp v theo th tca cc lp electron.

VD1:

Vit cu hnh electron ca nguyn t nguyn t Na (Z =11), Fe (Z =26), Ca (20), Zn (Z = 30)Ch :

Khi vit cu hnh electron ca nguyn t cc nguyn t th phi lu 2 trng hp gi bo ha sau:

TH1 : Trng hp bn bo ha:

Khi vit m electron cui cung in vo AOd nh sau:(n-1)d4 ns2 th 1 electron phn lp ns nhy sang phn lp (n-1)d t cu hnh bn vng hn nn phai vit lai cu hinh ng ca cc nguyn t nguyn t ny dng (n-1)d5 ns1 th mi ng vi thc t.

VD 1: Vit cu hinh electron ca nguyn t Cr (Z = 24)

TH2 :

Trng hp vi bo ha:

Khi vit m electron cui cung in vo AOd nh sau:

(n-1)d9 ns2 th 1 electron phn lp ns nhy sang phn lp (n-1)d t cu hnh bn vng hn nn phai vit lai cu hinh ng ca cc nguyn t nguyn t ny dng (n-1)d10 ns1 th mi ng vi thc t.

VD 2: Vit cu hinh electron ca nguyn t Cu (Z = 29).

Vit cu hnh electron ca ion R+n khi bit s proton hoc in tch ht nhn.

Phng php gii: Vit cu hnh electron ca nguyn t R. Bt dn t 1, 2, . n electron trong cu hnh electron ca R theo th t t ngoi vo ( t phi sang tri theo th t sp xp trong cu hnh nguyn t R theo quy tc ht lp ngoi ri mi vo n lp trong.Ch :

i vi nguyn t nguyn t R c cu hnh 2 phn lp ngoi l (n-1)da ns2 th khi vit cu hinh cho ion R ta cng bt ln lt 1, 2, . n electron t phn lp ns2 trc n ht ri mi bt electron phn lp (n-1)da.VD1:

Vit cu hnh electron ca ion K+, Fe2+, Fe3+, Cr3+, Cr2+, Pb2+

T cu hnh electron lp ngoi cng hy bin lun tm tn nguyn t:VD1:Hy bin lun tm tn nguyn t R khi bit cu hnh electron lp ngoi cng nh sau: (n-1)da ns1.Phng php gii:V phn lp ns ch c 1 electron nn c 3 kh nng xy ra:

nu a = 0 th nguyn t R l nguyn t S > cu hnh ca R ri > Z v tn nguyn t R. nu a = 5 th nguyn t R l nguyn t d > cu hnh ca R ri > Z v tn nguyn t R.

nu a = 10 th nguyn t R l nguyn t d > cu hnh ca R ri > Z v tn nguyn t R.T cu hnh electron lp ngoi cng ca nguyn t nguyn t R hy vit cu hnh electron y ca n v xc nh tn nguyn t R:VD1: Cu hnh electron lp ngoi cng ca 1 nguyn t nguyn t R l 4s1 hy xc nh tn tt c cc nguyn t R c th c.Dng 6:S ht trong c bn trong A nhiu hn trong B:TH1 : Trong phn t AaB c tng s ht (p, n, e) l d1 ht, trong s ht mang in nhiu hn s ht khng mang in l d2 ht. S khi ca nguyn t A ln hn s khi ca nguyn t B l d3. Tng s ht (p, n, e) trong nguyn t A nhiu hn trong nguyn t B l d4 ht. Vit cu hnh electron ca cc nguyn t A, B. Vit cng thc phn t hp cht .

Phng php gii:

Phn t AaB trung ha v in c to yhnh do s kt hp a nguyn t A v 1 nguyn t B.

Gi s proton, s ntron trong ht nhn ca nguyn t A, ln lt l: x, z (x, z > 0).Gi s proton, s ntron trong ht nhn ca nguyn t B ln lt l: y, t (y, t > 0).

V cc nguyn t lun trung ha v in nn ta c: s ht electron trong nguyn t A, B ln lt l x, y.

S proton v s ntron khng b thay ddooir khi xy ra phn ng ha hc kt hp a nguyn t A vi 1 nguyn B.s dng cc iu kin u bi ra ta c h cc phng trnhbc 1 sau:

+ Trong phn t AaB c tng s ht (p, n, e) l d1 ht:

2a.x + 2y + a.z + t = d1 (1)+ Trong phn t AaB s ht mang in nhiu hn s ht khng mang in l d2 ht:

2a.x + 2y - a.z - t = d2 (2)

+ S khi ca nguyn t A ln hn s khi ca nguyn t B l d3:

x - y + z - t = d3 (3)+ Tng s ht (p, n, e) trong nguyn t A nhiu hn trong nguyn t B l d4 ht:

2x - 2y + z - t = d4 (4)

Gii h (1), (2), (3), (4) thu c:x =

y = x - (d4 - d3 )VD1:

Trong phn t A2B c tng s ht (p, n, e) l 140 ht, trong s ht mang in nhiu hn s ht khng mang in l 44 ht. S khi ca nguyn t A ln hn s khi ca nguyn t B l 23. Tng s ht (p, n, e) trong nguyn t A nhiu hn trong nguyn t B l 34 ht. Vit cu hnh electron ca cc nguyn t A, B. Vit cng thc phn t hp cht .

TH2 : S ht trong c bn trong A nhiu hn trong B:Trong phn t AaBb c tng s ht (p, n, e) l d1 ht, trong s ht mang in nhiu hn s ht khng mang in l d2 ht. S khi ca nguyn t A ln hn s khi ca nguyn t B l d3. Tng s ht (p, n, e) trong nguyn t A nhiu hn trong nguyn t B l d4 ht. Vit cu hnh electron ca cc nguyn t A, B. Vit cng thc phn t hp cht .

Phng php gii:


Gi s proton, s ntron trong ht nhn ca nguyn t A, ln lt l: x, z (x, z > 0).

Gi s proton, s ntron trong ht nhn ca nguyn t B ln lt l: y, t (y, t > 0).




2a.x + 2y + a.z + bt = d1 (1)

+ Trong phn t AaB s ht mang in nhiu hn s ht khng mang in l d2 ht:

2a.x + 2y - a.z - bt = d2 (2)

+ S khi ca nguyn t A ln hn s khi ca nguyn t B l d3:

x - y + z - t = d3 (3)

+ Tng s ht (p, n, e) trong nguyn t A nhiu hn trong nguyn t B l d4 ht:

2x - 2y + z - t = d4 (4)

Gii h (1), (2), (3), (4) thu c:X =

y = x - (d4 - d3 )

TH1 : S ht trong c ban trong B nhiu hn trong A:Phng php gii:







2a.x + 2y + a.z + t = d1 (1)


2a.x + 2y - a.z - t = d2 (2)

+ S khi ca nguyn t B ln hn s khi ca nguyn t A l d3:

-x + y - z + t = d3 (3)

+ Tng s ht (p, n, e) trong nguyn t B nhiu hn trong nguyn t A l d4 ht:

-2x + 2y - z + t = d4 (4)

Gii h (1), (2), (3), (4) thu c:x =

y = x + (d4 - d3 )

TH2 : S ht trong c ban trong B nhiu hn trong A:

Trong phn t AaBb c tng s ht (p, n, e) l d1 ht, trong s ht mang in nhiu hn s ht khng mang in l d2 ht. S khi ca nguyn t B ln hn s khi ca nguyn t A l d3. Tng s ht (p, n, e) trong nguyn t B nhiu hn trong nguyn t A l d4 ht. Vit cu hnh electron ca cc nguyn t A, B. Vit cng thc phn t hp cht .Phng php gii:







2a.x + 2y + a.z + t = d1 (1)


2a.x + 2y - a.z - t = d2 (2)

+ S khi ca nguyn t B ln hn s khi ca nguyn t A l d3:

-x + y - z + t = d3 (3)

+ Tng s ht (p, n, e) trong nguyn t B nhiu hn trong nguyn t A l d4 ht:

-2x + 2y - z + t = d4 (4)

Gii h (1), (2), (3), (4) thu c:x =

y = x + (d4 - d3 )Dng 7: Xc inh tn nguyn t v vit cu hnh electron ca n hoc xc nh s khi ca nguyn t khi bit:TH1 : Tng s electron trn phn lp p l m ht.(m l s cho trc)Phng php gii:

V s e t a trn phn lp p khi c lp y l 6 e nn ta vit c di dng:

n.6 + a = m

dng phong php th cc gi tr ca n sao cho n.6 m ri ly 1 gi tr ca n n.6 l gi tr gn vi m nht ri i xc nh gi tr ca a(sao cho a < 6 ) . khi a l s e cui cng c in vo phn lp p th ( n + 1) > cu hnh e ca nguyn t R > tn RTH1 : Tng s electron trn phn lp s l m htPhng php gii:V s e t a trn phn lp s khi c lp y l 2 e nn ta vit c di dng:

n.2 + 1 = m (nu m l s l)> cn 1 e cui cng c in vo phn lp s th (n + 1).

2.n = m> 2 e cui cng c in vo phn lp s th n.

> cu hnh e ca nguyn t R > tn R.Dng 7: Tnh bn knh nguyn t gn ng ca R toC, bit rng t nhit khi lng ring ca R l d g/cm3. gi thit rng trong tinh th cc nguyn t Rl nhng hnh cu chim b% th tch tinh th, phn cn li l khe rng gia cc qu cu. Cho bit nguyn t khi ca R l MR.Phng php gii:Th tch 1 mol nguyn t nguyn t R l:

V = V trong V cha 6,02.1023 nguyn t R => th tch 1 nguyn t R : Mt khc: bn knhnguyn t tnh theo cng thc:

T suy ra bnh knh nguyn t gn ng ca R.VD1:Tnh bn knh gn ng ca nguyn t canxi, bit th tch ca 1 mol canxi tinh th bng 25,87 cm3. Cho bit: Trong tinh th, cc nguyn t canxi ch chim 74% th tch, cn li l khe trng.

Chn cu tr li ng:

A. 2,17 . 10-8 dm

B. 1,20 . 10-8 cm

C. 1,97 . 10-8 cm

D. 2,17 . 10-8 cm

BI TP TNG HPBI TP CHNG NGUYN T PHN T

Cu 1.Vit cu hnh e ca Fe, Fe2+, Fe3+ Cu 2.Oxi c 3 ng v l , , . Cacbon c 2 ng v l 12C v 13C. Xc nh cc loi phn t CO2 c th to thnh. Tnh M CO2.

Cu 3.Hidro c nguyn t khi l 1,008. Hi c bao nhiu nguyn t ca ng v 2H trong 1ml nc ( cho rng trong nc ch c ng v 1H v 2H, cho = 18, khi lng ring ca nc l 1g/ml

Cu 4. Trong t nhin ng v 37Cl chim 24,23,% s nguyn t clo.Tnh thnh phn phn trm v khi lng 37Cl c trong HClO4 ( vi hidro l ng v 1H, oxi l ng v 16O). Cho khi lng nguyn t trung bnh ca Clo l 35,5

Cu 5. Nguyn t X c 2 ng v A v B.T l s nguyn t ca 2 ng v A v B l 27: 23. ng v A c 35p v 44n. ng v B nhiu hn ng v A 2 nowtron. Xc nh nguyn t khi trung bnh ca X.

Cu 6. Tng s ht p, n, e ca nguyn t nguyn t X l 22. Xc nh nguyn t X.

Cu 7.Tng s ht p, n, e trong mt nguyn t nguyn t X l 155 trong s ht mang in nhiu hn s ht khng mang in l 33 ht. Xc nh nguyn t X.

Cu 8. Tng s ht p, n, e trong mt nguyn t nguyn t X l 115 trong s ht mang in nhiu hn s ht khng mang in l 25 ht. Xc nh nguyn t X.

Cu 9.Nguyn t X c tng s ht l 60, trong s ht n bng s ht p. Xc nh nguyn t X.

Cu 10. Oxit B c cng thc l X2O. Tng s ht c bn p, n, e trong B l 92, trong s ht mang in nhiu hn s ht khng mang in l 28. Xc nh B.

Cu 11.Trong t nhin Cu c 2 ng v l 63Cu v 65Cu . Khi lng nguyn t trung bnh ca Cu l 63,54. Thnh phn phn trm v khi lng ca 63Cu trong CuCl2 l bao nhiu ( bit M Cl = 35,5)

Cu 12. M l kim loi to ra 2 mui MClx, MCly v 2 oxit MO0,5x, M2Oy. T l v khi lng ca clo trong 2 mui l 1: 1,172, ca oxi trong 2 oxit l 1: 1,35. Xc nh nguyn t khi ca M.

Cu 13. Hp cht Y c cng thc M4X3. Bit

-Tng s ht trong phn t Y l 214 ht

-Ion M3+ c s e bng s e ca ion X4--Tng s ht p, n, e ca nguyn t nguyn t M nhiu hn tng s ht ca nguyn t nguyn t X trong Y l 106. Xc nh hp cht Y

Cu 14. Trong t nhin Cu c 2 ng v l 63Cu v 65Cu. Nguyn t trung bnh ca Cu l 63,546. S nguyn t 63Cu c trong 32g Cu l bao nhiu bit NA = 6,022.1023 Cu 15. Nguyn t X c 3 ng v l X1 chim 92,23%, X2 chim 4,67% v X3 chim 3,1%. Tng s khi ca 3 ng v bng 87. S nowtron trong X2 nhiu hn trong X1 1 ht. Nguyn t khi trung bnh ca X l 28,0855. Tm X1, X2, X3 Cu 16. Mg c 3 ng v : 24Mg ( 78,99%), 25Mg (10%), 26Mg( 11,01%).

a. Tnh nguyn t khi trung bnh.

b. Gi s trong hn hp ni trn c 50 nguyn t 25Mg, th s nguyn t tng ng ca 2 ng v cn li l bao nhiu.

Cu 17. Trong t nhin Clo c 2 ng v l 35Cl v 37Cl c nguyn t khi trung bnh l 35,5. Tnh s nguyn t ca ng v 37Cl, trong 3,65g HCl.

Cu 18. Trong t nhin Brom c 2 ng v l 79Br v 81Br c nguyn t khi trung bnh l 79,92. Thnh phn phn trm v khi lng ca 81Br trong NaBr l bao nhiu. Cho MNa = 23

Cu 19. Tnh bn knh nguyn t gn ng ca nguyn t Au 200C bit nhit khi lng ring ca vng l 19,32g/em3 vi gi thit trong tinh th cc nguyn t Au l nhng hnh cu chim 75% th tch tinh th, phn cn li l khe rng. Cho khi lng nguyn t ca Au l 196,97.

Cu 20. Tnh bn knh nguyn t gn ng ca nguyn t Fe 200C bit nhit khi lng ring ca Fe l 7,87g/em3 vi gi thit trong tinh th cc nguyn t Fe l nhng hnh cu chim 74% th tch tinh th, phn cn li l khe rng. Cho khi lng nguyn t ca Fe l 55,85.

Cu 21. Tnh bn knh nguyn t gn ng ca nguyn t Ca bit V ca mt nguyn t gam Ca bng 25,87 em3. Bit trong tinh th cc nguyn t Ca chim 74% th tch, cn li l khe rng.

Cu 22. Nguyn t khi ca B l 10,81. B gm 2 ng v 10B v 11B. C bao nhiu phn trm ng v 11B trong axit boric H3BO3.

Cu 23.Hp cht A c cng thc l Mxa trong M chim 46,67% v khi lng. M l kim loi, X l phi kim chu k 3. Trong ht nhn M c n- p = 4, ca X c n = p, , trong n, n, p, p, l s nowtron v s proton. Tng s proton trong Mxa l 58. Xc nh tn, s khi ca M, s TT ca nguyn t X trong bng tun hon. Vit cu hnh e ca X.

Cu 24. : Hp cht Z c to bi hai nguyn t M, R c cng thc MaRb trong R chim 6.667% khi lng . Trong ht nhn nguyn t M c n=p + 4, cn trong ht nhn ca R c n , = p , , trong n, p, n,, p , l s ntron v proton tng ng ca M v R. Bit rng tng s ht proton trong phn t Z bng 84 v a+b =4. Tm cng thc phn t ca Z.

Cu 25. Mt hp cht B c to bi mt kim loi ha tr 2 v mt phi kim ha tr 1. Tng s ht trong phn t B l 290. Tng s ht khng mang in l 110, hiu s ht khng mang in gia phi kim v kim loi trong B l 70. T l s ht mang in ca kim loi so vi phi kim trong B l 2: 7. Tm A, Z ca kim loi v phi kim trn.

Cu 26. Mt hp cht ion cu to t ion M2+ v ion X-. Tng s gt p, n, e trong phn t MX2 l 186 ht, trong s ht mang in nhiu hn s ht khng mang in l 54 ht. S khi ca ion M2+ nhiu hn nhiu hn trong X- l 21 ht. Tng s ht p, n, e trong M2+ nhiu hn trong X- l 27 ht. Xc nh v tr ca M, X trong bng tun hon,

Cu 27. Mt hp cht ion cu to t ion M+ v ion X2-.Trong phn t M2X c tng s ht l 140 ht, trong ht mang in nhiu hn ht khng mang in l 44 ht. S khi ca ion M+ nhiu hn ion X2- l 23. Tng s ht trong M+ nhiu hn trong X2- l 31

a. Vit cu hnh e ca X2- v M+b. Xc nh v tr ca M v X trong bng HTTH.

Cu 28. Tng s p, n, e trong nguyn t ca 2 nguyn t M v X ln lt l 82 v 52. M v X to thnh hp cht Mxa, trong phn t ca hp cht tng s proton ca cc nguyn t bng 77. Hy vit cu hnh e ca M v X t xc nh v tr ca chng trong bng HTTH. CTPT ca Mxa.

Cu 29. X, Y, R, A, B theo th t l 5 nguyn t lin tip trong HTTH c tng s in tch ht nhn l 90 ( X c in tch ht nhn nh nht )

a. Xc nh s in tch ht nhn ca X, Y, R, A, B, gi tn cc nguyn t .

b. Vit cu hnh e ca X2-, Y-, R, A+, B2+. So snh bn knh ca chng.

Cu 30.Trong t nhin nguyn t Clo c 2 ng v 35Cl v 37Cl c phn trm s lng tng ng l 75% v 25%. Nguyn t Cu c 2 ng v trong 63Cu chim 73% s lng. Bit Cu v Cl to c hp cht CuCl2 trong Cu chim 47,228% khi lng. Xc nh ng v th 2 ca Cu.

Cu 31.Phn t X c cng thc abc. Tng s ht mang in v khng mang in trong phn t X l 82. Trong dod s ht mang in nhiu hn s ht khng mang in l 22, hiu s khi gia b v c gp 10 ln s khi ca a, tng s khi gia b v c gp 27 ln s khi ca a. Tm CTPT ng ca X .

Cu 32. Mt hp cht c to thnh t cc ion M+ v . Trong phn t M2X2 c tng s ht p, n, e bng 164, trong s ht mang in nhiu hn s ht khng mang in l 52. S khi ca M nhiu hn s khi ca X l 23. Tng s ht p, n, e trong ion M+ nhiu hn trong ion l 7 ht

a. Xc nh cc nguyn t M, X v cng thc phn t M2X2b. Cho hp cht M2X2 tc dng vi nc. Vit phng trnh phn ng xy ra v trnh by phng php ha hc nhn bit sn phm.Cu 33: X v Y l 2 phi kim. Trong nguyn t X v Y c tng s ht mang in nhiu hn s ht khng mang in ln lt l 14 v 16.

Hp cht Z c cng thc l XYnc c im l:

+ X chim 15,0486 % v khi lng

+ Tng s proton l 100

+ Tng s ntron l 106Cu 34: Cho H c 3 ng v 1H1, 1H2, 1H3 vi t l % tng ng l:99,1%; 0,6%; 0,3%

O c 3 ng v 8O16, 8O17, 8O18 vi t l % tng ng l: 97,3%; 2%; 0,7%.

C bao nhiu phn t H2O c to thnh t cc ng v trn?

Nu cho mt mol phn t H2O hp th vo bnh P2O5 thy khi lng bnh tng m gam, tinh m

Cu 35:Cho tng s ht p, n, e trong phn t MX2 l 178 ht, trong hy nhn ca M s nowtron nhiu hn s proton 4 ht, cn trong ht nhn ca X s nowtron bng s proton. S proton trong ht nhn ca M nhiu hn s proton trong ht nhn ca X l 10 ht. Xc nh cng thc ca MX2.

Cu 36: Hp cht M2X c tng s cc ht trong phn t l 116, trong s ht mang in nhiu hn s ht khng mang in l 36. Khi lng nguyn t X nhiu hn M l 9. Tng s ht p, n, e trong X2- nhiu hn trong M+ l 17 ht. Xc nh s khi ca M v X.

Cu 37: Tng s ht p, n, e trong phn t MX3 l 196, trong s ht mang in nhiu hn s ht khng mang in l 60. Khi lng nguyn t ca X ln hn ca M l 8. Tng s ht p, n, e trong X- nhiu hn trong M3+ l 16. Xc nh M v X.

BNG H THNG TUN HON CC NGUYN T HA HC

Cu 1: Cc nguyn t no c cu hnh e lp ngoi cng l 4s1. Tm v tr cc nguyn t trong bng HTTH.

Cu 2: Hy sp xp c gii thch cc ht vi m cho di y theo chiu gim dn bn knh ht: Rb+ (z=37), Y3+( z=36), Br_ (z=35), Se2- (z=34), Sr2+(z=38)

-Cho cc ht vi m: Na, Na+, Mg, Mg2+, Al, Al3+, F-, O2-. Sp xp cc nguyn t theo chiu gim dn bn knh ht.

Cu 3:Mt nguyn t R c hp cht kh vi H l RH3. Oxit cao nht ca R cha 43,66% khi lng ca R

a. Xc nh nguyn t R.

b. Cho 28,4g oxit trn ha tan vo 80 ml dd NaOH 25% ( d=1,28). Tnh C% ca dd mui sau phn ng.

Cu 4: A, B l hai nguyn t cng mt phn nhm v thuc 2 chu k lin tip trong bng HTTH. Tng s proton trong hai ht nhn nguyn t A, B bng 32. Hy vit cu hnh e ca A, B v cc ion m A, B c th to thnh.

Cu 5: Hai nguyn t A v B hai phn nhm chnh lin tip trong HTTH. B thuc nhm VA. trng thi n cht , A v B khng phn ng vi nhau. Tng s p trong ht nhn nguyn t A v B l 23.

a.Vit cu hnh e ca A, B

b.T cc n cht A, B v cc ha cht cn thit, hy vit cc PTPU iu ch 2 axit trong A v B c s oxi ha dng cao nht.

Cu 6: Hai nguyn t A v B hai phn nhm chnh lin tip trong HTTH. A thuc nhm VIA. trng thi n cht , A v B phn ng vi nhau. Tng s p trong ht nhn nguyn t A v B l 25.Vit cu hnh e ca A v B. Xc nh v tr ca A v B trong bng tun hon.

Cu 7: : Hai nguyn t A v B cng phn nhm chnh trong HTTH v thuc 2 chu k lin tip trong HTTH. B v D l hai nguyn t k cn nhau trong cng mt chu k.

a. Nguyn t A c 6e lp ngoi cng. Hp cht X ca A vi H cha 11,1% H. Xc nh phn t lng ca X suy ra A, B.

b. Hp cht Y c cng thc AD2 trong 2 nguyn t A v D u t c cu bn ca kh him. Xc nh tn ca D.

c. Hp cht Z gm 3 nguyn t B, A, D c t l khi lng mA: mB: mD = 1: 1: 2,2. Hn hp gm 2 lit hi ca Y v mt lit hi ca Z c d/H2 = 51,5. Xc nh cng thc phn t ca Z.

Cu 8:Hp cht X c to bi 2 nguyn t A, B v c khi lng phn t l 76. A v B c s oxi ha cao nht trong cc oxit l +nO v +mO v c s oxi ha m trong cc hp cht vi H l nH v mH tha mn cc iu kin = v = 3. Hy thit lp cng thc phn t ca X. Bit rng A c s oxi ha cao nht trong X.

Cu 9: Hp cht M c to nn t cation X+ v anion Y3-, mi ion u do 5 nguyn t ca hai nguyn t phi kim to nn. Bit tng s proton trong X+ l 11 v trong Y3- l 47. Hai nguyn t trong Y3- thuc 2 chu k lin tip trong bng tun hon v c s th t cch nhau 7 n v.

a.Hy xc nh CTPT ca M.

b. M t bn cht cc lin kt trong phn t M.

Bai 10 : Hp cht khi cua nguyn t R vi hidro co cng thc la RH3 . Trong oxit cao nht cua R , oxi chim 56.34 % v khi lng.

Xac inh tn nguyn t R va cng thc ca hp cht trn.

A. NH3

B. PH3

C. BiH3

d. gi tr khc

Vit phng trinh iu ch axit cua R ng vi hoa tri cao nht t R.

Cu 11 : Mt hp cht khi cua mt nguyn t phi kim R vi hiro co cng thc la RH3 . Trong oxit cao nht , no chim 25.926 % v khi lng .

a) Xac inh nguyn t R va cng thc oxit cao nht ca R:

A. N2O5

B. P2O5

C. Bi2O5

d. gi tr khc

b) Vit phng trinh iu ch axit M trong o R co hoa tri cao nht t R.

Cu 12 : Mt hp cht khi cua mt nguyn t phi kim R vi hiro co cng thc la RH . Trong oxit cao nht cua no, oxi chim 61.202 % v khi lng .

b) Xac inh nguyn t R va cng thc cua hp cht trn.

A. HCl

b. HBr

C. HF

D. HI

c) Vit phng trinh iu ch axit RH t mui natri ca R.

Cu 13: Nguyn t X l phi kim thuc chu k 2 ca bng h thng tun hon, X to c hp cht kh vi hiro v c cng thc oxit cao nht l XO2. Nguyn t X to vi kim loi M hp cht c cng thc M3X, trong M chim 93,33% v khi lng. M l

A. Mg B. Zn

C. Fe

D. Cu

Cu 14: Nguyn t R c ho tr trong oxit bc cao nht bng ho tr trong hp cht kh vi hiro. Phn t khi ca oxit ny bng 1,875 ln phn t khi hp cht kh vi hiro. R l nguyn t no sau y

A. Cacbon

B. Silic

C. Lu hunh

D. Nit

Cu 15: Hai nguyn t X, Y c hiu in tch ht nhn gia X v Y l 16. Phn t Z gm 5 nguyn t ca hai nguyn t X v Y c 72 proton. Cng thc phn t ca Z l:

A. Cr2O3

B. Cr3O2

C. Al2O3

D. Fe2O3Bai 16 : Mt cation X3+ c tng s ht l 79 trong s ht mng in nhiu hn s ht khng mang in l 19. K hiu ha hc ca cation trn l:

a. Al3+

b. Fe3+

c. Cr3+

d. gi tr khc

Bai 17: Tng s hat proton, ntron, electron cua hai nguyn t A va B la 76 trong o s hat mang in nhiu hn s hat khng mang in la 24. S khi cua A nho hn cua B la 3. Tng s hat cua A nho hn cua B la 4. Tn cua A, B l:

a. Mg v Al

b. Na v Si

c. Ca v Mg

d. Na v Fe

Bai 18 ( thi H Khi B 2003).Tng s hat proton, ntron, electron trong 2 nguyn t kim loai A va B la 142, trong o tng s hat mang in nhiu hn tng s hat khng mang in la 42. S hat mang in cua B nhiu hn cua A la 12. Tn ca 2 kim loai A va B l:

a. Mg v Al

b. Na v Fe

c. Ca v Mg

d. Ca v Fe

Bai 19 Khi lng phn t ca 3 mui RCO3, RCO3, RCO3 lp thnh 1 cp s cng vi cng sai bng 16. Tng s ht proton, ntron ca ba ht nhn nguyn t ba nguyn t trn l 120.

*Ba nguyn t trn l:

A. Mg, Ca, FeB. Be, Mg, CaC. Be, Cu, Sr

D. Mg, Ca, Cu

E. Tt c u khng xc nh c

Cu 20:

Y l phi kim thuc chu k 3 ca bng HTTH, Y to c hp cht kh vi hirov cng thc oxit cao nht l YO3

Y: to hp cht (A) c cng thc MY2 trong M chim 46,67% v khi lng M l:

A.Mg

B.Zn

C.Fe

D.Cu

E.

Cu 21: Hroxit cao nht ca mt nguyn t R c dng HRO4. R cho hp cht kh vi hiro cha 2,74% hiro theo khi lng. R l nguyn t no sau y

A. Phot pho

B. Clo

C. Brm

D. Ioi

BI TP CHNG NGUYN T - HTTH Bi 1: Nguyn t ca nguyn t X c cu to bi 115 ht. Trong , ht mang in nhiu hn ht khng mang in l 25 ht. Vit cu hnh ca nguyn t ?

HD: 2Z + N = 115 v 2Z N = 25. Mc nng lng : 1s2s2p3s3p4s3d4p5s

Bi 2: Tng s ht trong nguyn t mt nguyn t l 13.

Xc nh tn nguyn t.

Vit cu hnh electron ca nguyn t .

Bi 3: Mt nguyn t R c tng s ht l 34, trong s ht mang in nhiu gp 1,8333 ln s ht khng mang in. Tm s ht p, n, e v s khi ca R?

Bi 4: Mt nguyn t X c tng s ht l 62 v c s khi nh hn 43. Tm s p,n v khi lng mol nguyn t?

Bi 5: Mt nguyn t R c tng s ht l 52. Xc nh nguyn t khi ca nguyn t ?

Bi 6: R c 2 loi ng v l R1 v R2. Tng s ht trong R1 l 54 ht v trong R2 l 52 ht. Bit R1 chim 25% v R2 chim 75%. Tnh khi lng nguyn t trung bnh ca R.

Bi 7: Cho bit khi lng nguyn t trung bnh ca Clo l 35,5. Clo c 2 ng v l Cl v Cl. Hm lng % ca Cl l bao nhiu?

Bi 8: Agon tch t khng kh l mt hn hp ca 3 ng v Ar (99,6%); Ar (0,063%); Ar(0,337%).

Tnh th tch ca 20 gam Agon ktc?

Bi 9: Tm nguyn t khi trung bnh ca Br, bit trong t nhin c 2 ng v Br(54,5%) v Br (45,5%)?

Bi 10: Nguyn t R c tng s ht l 115 v c s khi l 80. Tm in tch ht nhn ca R?

Bi 11: Hy xc nh in tch ht nhn, s p, s e ca nguyn t cc nguyn t sau?

Li;

F;

Mg;

Ca;

Na

Bi 12: Trong anion X3- tng s ht l 111, s e bng 48% s khi. Tm s p, n, e v s khi ca X3-?

Bi 13: Tng s ht trong nguyn t R l 76, s ht mang in nhiu hn s ht khng mang in l 20. Tm s p, n, e v s in tch ht nhn ca R?

Bi 14: Tnh bn knh gn ng ca nguyn t Ca, bit th tch mt nguyn t gam canxi bng 25,87cm3. (Trong tinh th, cc nguyn t Ca ch chim 74% th tch, cn li l cc khe trng)

Bi 15:

a) Mt cation R3+ c tng s ht l 37. T s ht e i vi n l 5/7. Tm s p, e, n trong R3+?

b) Nguyn t X c s khi nh hn 36 v tng s cc ht l 52. Tm cc s p, n, e v cho bit X l g?

Bi 16: Hn hp hai ng v c nguyn t khi trung bnh l 40,08. Hai ng v ny c s n hn km nhau l 2. ng v c s khi nh hn chim 96% v ng v c s khi ln chim 4%. Tm s khi mi ng v?

Bi 17: C bao nhiu nguyn t Hidro trong 0,46g C2H5OH?

Bi 18: Nguyn t X c 2 ng v vi t l s nguyn t l 27/23. Ht nhn nguyn t X c 35p. ng v th nht c 44 ntron. ng v th hai c nhiu hn ng v th nht 2 ntron. Tnh nguyn t khi trung bnh ca X?

Bi 19: Bit rng nguyn t st gm 26p, 39n, 26e.

a. Tnh khi lng e c trong 1 kg st?

b. Tnh khi lng st cha 1 kg electron?

Bi 20: Mt nguyn t X gm hai ng v l X1 v X2. ng v X1 c tng s ht l 18. ng v X2 c tng s ht l 20. Bit rng % cc ng v trong X bng nhau v cc ht trong X1 cng bng nhau. Xc nh khi lng nguyn t trung bnh ca X?

Bi 21: Nguyn t nhm c bn knh 1,43 A0 v c khi lng nguyn t l 27 .v.C.

a. Tnh khi lng ring ca nguyn t Al?

b. Trong thc t, th tch tht chim bi cc nguyn t ch 74% ca tinh th, cn li l cc khe trng. nh khi lng ring ng ca Al. Bit th tch hnh cu: V =

EMBED Equation.3 R3Bi 22:

a. ng trong thin nhin gm hai loi ng v l Cu v Cu vi t s Cu: Cu = 105:245. Tnh khi lng nguyn t trung bnh ca Cu?

b. Mg c hai ng v l X v Y. ng v X c khi lng nguyn t l 24. ng v Y nhiu hn X 1 notron. Tnh khi lng nguyn t trung bnh ca Mg, bit s nguyn t trong hai ng v t l X:Y = 3:2.

Bi 23: Nguyn t X c tng s ht l 49, trong s ht mang ing bng 53,125% s ht mang in. Tm s khi v in tch ht nhn?

Bi 24: Nguyn t X c bn knh 1,28A0 v khi lng ring l 7,89 g/cm3. Bit rng cc nguyn t ch chim 74% th tch, cn li l cc khe trng. Tnh khi lng mol nguyn t ca X?

Bi 25: Bn knh nguyn t v khi lng mol nguyn t Fe ln lt l 1,28A0 v 56g/mol. Tnh khi lng ring ca Fe. Bit rng trong tinh th, cc tinh th Fe chim 74%, th tch cn li l phn rng?

Bi 26: Nguyn t Au c bn knh v khi lng mol nguyn t ln lt l 1,44A0 v 197g/mol. Bit khi lng ring ca Au l 19,36g/cm3. Hi cc nguyn t Au chim bao nhiu % th tch trong tinh th?

Bi 27: Mt hn hp gm hai ng v c s khi trung bnh 31,1 v t l % ca cc ng v ny l 90% v 10%. Tng s ht trong hai ng v l 93 v s ht khng mang in bng 0,35 ln s ht mang in. Tm s Z v s notron ca mi ng v?


Bi 29: Tng s ht trong nguyn t R l 21. Tm s p, n, e v s in tch ht nhn ca R?


Bi 31: Tng s ht trong nguyn t R l 36, s ht mang in l 24. Tm s p, n, e v s in tch ht nhn ca R?

Bi 32: Tng s ht trong nguyn t R l 34.

Cho bit s hiu nguyn t v s khi ca nguyn t?

Vit cu hnh e ca nguyn t ?

Cho bit nguyn t l kim loi hay phi kim?

Bi 34: Vit cu hnh e nguyn t cc nguyn t c Z = 8; Z = 16; Z = 36; Z = 28.

a. Cho bit s e, s lp e, s e lp ngoi cng?

b. Cho bit cc nguyn t l kim loi hay phi kim?

Bi 35: Hy vit cu hnh e y v cho bit s hiu nguyn t ca cc nguyn t c cu hnh e ngoi cng nh sau:

a. 2s1

b. 2s22p3

c. 2s22p6

d. 3s2e. 3s23p1

f. 3s23p4

g. 3s23p5

h. 3d34s2Bi 36: Vit cu hnh e ca Fe, Fe2+; fe3+; S; S2- bit Fe th 26 v s ca S l 16 trong bng tun hon?

Bi 37: Cation R+ c cu hnh e phn lp ngoi cng l 2p6a. Vit cu hnh e nguyn t ca nguyn t R?

b. Vit s phn b e vo cc obitan nguyn t?

Bi 38: Cho 5,9 gam mui NaX tc dng ht vi dung dch AgNO3 thu c 14,4g kt ta.

a. Xc nh nguyn t khi ca X v vit cu hnh e?

b. Nguyn t X c hai ng v bn, xc nh s khi ca mi ng v, bit rng:

% ca cc ng v bng nhau.

ng v th nht c n notron v ng v th 2 c n+2 notron.

Bi 39: Nguyn t ca mt nguyn t X c s e mc nng lng cao nht l 4p5. T s gia s ht mang in v khng mang in l 0,6429. Tm s in tch ht nhn v s khi ca X?

Bi 40: cu hnh e ngoi cng ca mt nguyn t l 5p5. T l s notron v in tch ht nhn bng 1,3962. S notron trong nguyn t X gp 3,7 ln s notron ca nguyn t Y. Khi cho 1,0725g Y tc dng vi lng d X thu c 4,565 gam sn phm c cng thc XY.

a. Vit cu hnh e y ca X?

b. Xc nh s hiu nguyn t, s khi, tn gi ca X, Y?

c. X, Y cht no l kim loi, l phi kim?

Bi 41: cho dung dch cha 8,19g mui NaX tc dng vi lng d dung dch AgNO3 thu c 20,09g kt ta.

a. Xc nh nguyn t khi, gi tn X, vit cu hnh e?

b. X c hai ng v t nhin, trong ng v th nht c s nguyn t nhiu hn ng v th hai l 50%. Ht nhn ng v th nht t hn ht nhn ng v th hai l 2 notron. Xc nh s khi ca mi ng v?

Bi 42: X l kim loi ha tr II. Cho 6,082 gam X tc dng ht vi dung dch HCl d thu c 5,6 lt kh H2 ktc.

a. Tm nguyn t khi ca X v cho bit tn ca X?

b. Vit cu hnh e ca X?

Bi 43: Vit k hiu ca cc nguyn t A, B, E, F bit:

a. Nguyn t A c tng s ht c bn l 24, s ht khng mang in chim 33,33%?

b. Nguyn t B c tng s ht c bn l 34, s n nhiu hn s p mt ht?

c. Nguyn t E c tng s ht c bn l 18, s ht mang in nhiu hn s ht khng mang in l 6 ht?

d. Nguyn t F c s khi l 207, s ht mang in m l 82?

Bi 44: Cho hp cht MX2. Trong phn t MX2, tng s ht c bn l 140 v s ht mang in nhiu hn s ht khng mang in l 44. S khi ca X ln hn s khi ca M l 11. Tng s ht c bn trong X nhiu hn trong M l 16. xc nh k hiu nguyn t M, X v cng thc phn t MX2?

Bi 45: Nguyn t vng (Au) c bn knh v khi lng mol ln lt l 1,44A0 v 1,97g/mol. Hi cc tinh th vng chim bao nhiu % th tch trong tinh th, bit khi lng ring thc ca Vng l 19,36g/cm3?

Bi 46: Mt hp cht M2X. Tng s ht trong M2X l 140 ht, trong s ht mang in nhiu hn s ht khng mang in l 44. S khi ca ion M+ ln hn s khi ca ion X2- l 13. Tng s ht c bn trong ion M+ nhiu hn trong ion X2- l 31.

a. Vit cu hnh e ca cc ion M+; X2- v nguyn t M?

b. Xc nh cng thc phn t MX2?

Bi 47: Cho hp cht MX3. Trong phn t MX3, tng s ht c bn l 196 v s ht mang in nhiu hn s ht khng mang in l 60. S ht mang in trong nguyn t M t hn s ht mang in trong nguyn t X l 16 ht.

a. Xc nh hp cht MX3?

b. Vit cu hnh e ca M v X?

Bi 48: Khi cho 10,12g natri kim loi tc dng hon ton vi mt phi kim B th thu c 45,32 gam mui natri.

a. Tm khi lng mol ca B v tn gi ca B?

b. Bit B c hai ng v l B v B trong B chim 50% v s nguyn t khi v s khi A1 ln hn s khi A2 l 2 n v.

Tm cc s khi A1, A2.

Vit phn ng ca B vi Na, Zn, Cu, P?

Bi 49: Nguyn t R c tng s ht trong nguyn t l 52, s ht khng mang in gp 1,059 ln s ht mang in dng. Xc nh v tr ca R trong bng HTTH?

Bi 50: Tng s ht c bn trong nguyn t ca mt nguyn t l 34.

a. Xc nh khi lng nguyn t ca nguyn t ?

b. Vit cu hnh e ca nguyn t ?

c. Nguyn t l kim loi hay phi kim?

d. Cho bit v tr ca nguyn t trong bng HTTH?Bi 51: X l kim loi c ha tr II. Ha tan hon ton 6,082g X vo dung dch HCl d thu c 5,6 lt H2 ktc.

a. Tm khi lng nguyn t v tn nguyn t X.

b. X c 3 ng v, bit tng s khi 3 ng v l 75. S khi ca ng v th nh bng trung bnh cng s khi hai ng v kia. ng v th nht c s p bng s notron. ng v th 3 chim 11,4% s nguyn t v c s notron nhiu hn ng v th hai l 1 n v.

Tm s khi v s notron ca mi ng v?

Tm % v s nguyn t 2 ng v cn li?

c. Mi khi c 50 nguyn t ca ng v th nh th c bao nhiu nguyn t cc ng v cn li?

Bi 52: C 3 ng v ca nguyn t X, m tng s ht trong 3 nguyn t ng v l 75. Trong ng v 1, s p bng s n, ng v 2 c s n km thua ng v 3 l 1.

a. Xc nh s khi ca mi ng v?

b. Trong X, s nguyn t ca cc ng v th nht, 2, 3 ln lt theo t l 115:3:2. Tm khi lng mol trung bnh ca X?

Bi 53: Nguyn t X c s ht khng mang in bng 53,125 s ht mang in v tng ht l 49. Nguyn t Y c s ht mang in ln hn s ht khng mang in l 8 v s ht khng mang in bng 52,63% s khi. Tm s p,n, nguyn t khi v xc nh X, Y?

Bi 54: Nguyn t nguyn t X c s e mc nng lng cao nht l 4p5, t s gia s ht khng mang in v s ht mang in l 0,6429.

a. Tm s in tch ht nhn, s ko6i1 ca X?

b. Nguyn t nguyn t R c s notron bng 57,143% s p ca X. Khi cho R tc dng vi X th thu c hp cht RX2 c khi lng gp 5 ln khi lng R phn ng. Vit cu hnh e nguyn t ca R v phn ng gia R vi X?

Bi 55: Cho 7,2 gam hn hp 2 mui cacbonat ca hai kim loi kim th thuc 2 chu k lin tip tc dng vi dung dch HCl d thy thot ra kh B. Cho kh B hp th ht vo dung dch Ba(OH)2 d thu c 15,76g kt ta. Xc nh 2 mui cacbonat v tnh thnh phn % ca chng?

Bi 56: Cho 3,1 gam hn hp 2 kim loi kim hai chu k lin tip tc dng ht vi nc, ta thu c 1,12 lt kh dktc. Xc nh 2 kim loi v % theo khi lng ca chng trong hh?

Bi 57: Ha tan 2,84gam hh hai mui cacbonat ca hai kim loi kim th thuc 2 chu k lin tip tc dng vi dung dch HCl d thu c dung dch A v kh B. C cn dd A thu c 3,17g mui khan.

a. Tnh th tch kh B ktc?

b. Xc nh tn hai kim loi?

Bi 58: Khi cho 3,33 gam mt kim loi kim tc dng vi nc th c 0,48g hidro thot ra. Hy cho bit tn kim loi kin ?

Bi 59: Khi cho 0,6 gam mt kim loi thuc phn nhm chnh nhm II tc dng vi nc th c 0,336 lt kh hidro thot ra ktc. Gi tn kim loi ?

Bi 60: cho 2 nguyn t kim loi hai chu k lin tip v u thuc phn nhm chnh nhm IIA ca bng HTTH. Bit rng 4,4gam hh hai kim loi ny tc dng vi dung dch HCl d th thu c 3,36 lt kh H2 ktc. Xc nh tn hai kim loi ?

Bi 61: Hai nguyn t A, B ng k tip nhau trong cng mt chu k ca bng HTTH c tng s in tch ht nhn l 25. Xc nh v tr ca A, B trong bng HTTH?

Bi 62: Ha tan hon ton 17 gam hh hai kim loi kim A, B thuc hai chu k lin tip nhau vo nc c 6,72 lt kh ktc. Xc nh tn hai kim loi kim v thnh phn % v khi lng ca mi kim loi trong hh?

Bi 63: Ha tan hon ton 20 gam hh hai kim loi kim th A, B thuc hai chu k lin tip vo dd HCl d thu c 15,68 lt k ktc. Xc nh tn hai kim loi kim th v thnh phn % v khi lng ca mi kim loi trong hh?

Bi 64: Ha tan hon ton 14,2g hai mui cacbonat ca hai kim loi A, B lin tip nhau trong nhm IIA bng lng va dd H2SO4. Sau p thu c 3,36 lt kh ktc. Xc nh CTPT ca hai mui v % v k.l ca mi mui trong hh?

Bi 65: nguyn t nguyn t X c s khi nh hn 36 v tng s ht c bn l 52. Tm s p, n v suy ra X?

Bi 66: Hn hp hai ng v c nguyn t khi trung bnh l 40,08. hai ng v ny c s n hn km nhau l 2. ng v c s khi nh hn chim 96%, cn li l % cc nguyn t c s khi ln hn. Xc nh s khi ca mi ng v?

Bi 67: Mt nguyn t R c tng s ht l 95, trong s ht khng mang in bng 0,5833 s ht mang in. Tm s ht p, n, e v s khi ca R?

Bi 68: Mt nguyn t R c tng s cc ht l 115. S ht mang in nhiu hn s ht khng mng in l 25 ht. Tnh s p, s khi v xc nh v tr ca trong bng HTTH ri gi tn R?

Bi 69: Mt s nguyn t c cu hnh e nh sau:

a. 1s22s22p1

b. 1s22s22p6

c. 1s22s22p63s23p5d. 1s22s22p63s23p63d104s2

e. 1s22s22p63s23p63d104s24p1xc nh v tr ca chng trong bng HTTH. Nguyn t no l kim loi? phi kim? Kh him?Bi 70: Vit cu hnh e cc nguyn t c s th t: 19, 35, 52, 24, 83 v cho bit v tr ca chng trong bng HTTH? Tnh kim loi, phi kim ca mi nguyn t?

Bi 71: Ha tan 28,4 gam 2 mui cacbonat ca 2 kim loi ha tr II bng dung dch HCl d thu c 6,72 lt kh ktc v dd A.

a. Tnh khi lng mui c trong dd A?

b. Xc nh hai kim loi, bit chng hai chu k lin tip nhau trong phn nhm chnh nhm II?

c. Tnh % theo k.l mi mui trong hh u?

Bi 72: em m gam hh hai kim loi kim tc dng vi HCl d thu c 2,24 lt kh thot ra ktc. C cn sn phm thu c 11,7 gam mui khan.

a. Tnh m?

b. Xc nh tn hai kim loi kim v khi lng tng kim loi, bit chng cch nhau 1 chu k trong bng HTTH?

Bi 73: Tng s ht p, n, e ca nguyn t mt nguyn t thuc phn nhm chnh nhm VII l 28.

a. Tnh khi lng nguyn t?

b. Vit cu hnh e?

Bi 74: Cho A v B l 2 nguyn t thuc cng phn nhm v 2 chu k lin tip nhau trong HTTH. Tng s p trong 2 ht nhn nguyn t A v B l 32. Xc nh tn A, B v vit cu hnh e ca chng?

Bi 75: A v B l 2 nguyn t lin tip nhau trong cng mt chu k. Tng s p trong hai ht nhn l 49. Vit cu hnh e v xc nh v tr ca A, B trong bng HTTH?

Bi 76: X v Y l 2 nguyn t thuc cng 1 nhm v hai chu k lin tip nhau trong bng HTTH. Tng s ht p trong 2 ht nhn ca 2 nguyn t X v Y l 30. Vit cu hnh e ca X, Y?

Bi 77: Hp cht c cng thc MX2 trong chim 46,67% v khi lng. Trong ht nhn M c s n nhiu hn s p l 4 ht. Trong ht nhn X c s n bng s p. Tng s p trong MX2 l 58.

a. Tm s khi ca M v X?

b. Xc nh CTPT ca MX2?

Bi 78: Mt hp cht B c to bi mt kim loi ha tr II v mt phi kim ha tr I. Tng s ht trong phn t B l 290. Tng s ht khng mang in l 110. Hiu s ht khng mang in gia phi kim v kim loi l 70. T l s ht mang din ca kim loi so vi phi kim trong B l 2/7. Tm A, Z ca kim loi v phi kim trn?

Bi 79: Mt nguyn t gm 2 ng v c s nguyn t t l vi nhau l 27:23. ht nhn ng v th nht cha 35p v 44n. ht nhn ng v 2 cha nhiu hn 2n. Xc nh khi lng nguyn t trung bnh ca nguyn t trn?

Bi 80: cho 0,345 gam mt kim loi c ha tr khng i tc dng vi nc thu c 168ml kh H2 ktc. Tm tn kim loi v v tr ca n trong bng HTTH?

Bi 81: Ha tan mt oxit kim loi ha tr II bng mt lng va dd H2SO4 10% thu c dd mui c nng 15,17%. Tm cng thc ca oxit kim loi ?

Bi 82: Oxit cao nht ca mt nguyn t ng vi cng thc R2O5. Hp cht ca n vi hidro l 1 cht c thnh phn khi lng l 82,35%R v 17,65% H. Tm nguyn t ?

Bi 83: Oxit cao nht ca mt nguyn t ng vi cng thc RO3. Trong hp cht ca n vi hidro c 5,88% H v khi lng. Tm nguyn t ?

Bi 84: Hp cht kh vi H ca mt nguyn t ng vi cng thc RH4. Oxit cao nht ca n cha 53,3%O. Gi tn nguyn t ?

Bi 85: Hp cht kh vi H ca mt nguyn t ng vi cng thc RH3. Oxit cao nht ca n cha 25,93%R. Gi tn nguyn t ?

Bi 86: Cho cc nguyn t A, B, C, D c s hiu nguyn t ln lt l 3, 11, 12, 13. Xc nh v tr ca chng trong bng HTTH v sp xp chng theo chiu tnh kim loi tng dn?

Bi 87: Cho cc nguyn t A, B, C, D c s hiu nguyn t ln lt l 6, 9, 14, 17. Xc nh v tr ca chng trong bng HTTH v sp xp chng theo chiu tnh phi kim loi tng dn?

Bi 88: Cho bit R c Z = 35.

Xc nh v tr ca R trong bng HTTH, CT oxit cao nht, hidroxit cao nht, hp cht vi H v tnh cht ca cc hp cht ny?

So snh tnh cht ca hp cht ca R vi cc hp cht ca 2 nguyn t trn v di R trong cng nhm.

Bi 89: Cho X (Z = 15); Y (Z = 20); M (Z = 25)

Xc nh cu hnh e, v tr cc nguyn t => Tnh kim loi, phi kim, cng thc oxit cao nht v hidroxit tng ng; cng thc hp cht vi Hidro?

Bi 90: Oxit ca 1 nguyn t nhm IIB cha 19,75% v khi lng. Hy xc nh tn nguyn t v vit cu hnh e nguyn t ca nguyn t ?

Bi 91: Oxit ca mt nguyn t ng vi cng thc l R2O5. Hp cht ca nguyn t vi H c 8,82% H v khi lng. Xc nh R?

Bi 92: M thuc nhm IIIA. Trong oxit bc cao nht ca M, oxi chim 47,05% khi lng. x thuc nhm VIA, trong oxit bc cao nht, X chim 40% v khi lng. Xc nh tn ca nguyn t M v X. Vit CTPT ca cc oxit trn?

Bi 93: Trong oxit bc cao nht ca R (thuc nhm A) Oxi chim 56,338% v khi lng. Xc nh CTPT ca oxit trn?

Bi 94: Hp cht ion c to bi cc ion M2+ v X2-. Bit rng trong phn t MX tng s ht l 84. S n v s p trong ht nhn nguyn t M v X bng nhau. S khi ca X2- ln hn s khi ca M2+ l 8.

a. Vit cu hnh e ca M2+; X2-; ?

b. Xc nh v tr ca M v X trong bng HTTH?

Bi 95: Mt nguyn t R c tng s cc ht l 48.

a. Cho bit tn v xc nh v tr ca Rtrong bng HTTH?

b. Vit CTHH ca oxit v hidro ng vi ha tri cao nht ca R, cho bit tnh cht ca cc cht ny?

Bi 96: Hp cht c dng AB3, tng s ht p trong phn t l 40, trong thnh phn ht nhn A cng nh B u c s ht p bng s ht n. A thuc chu k 3 ca bng HTTH. Xc nh tn gi ca A, B?

Bi 97: Mt hh X gm 2 mui cacbonat kim loi kim A, B thuc hai chu k lin tip trong bng HTTH c tng khi lng l 41,9 gam. Xc nh A, B v s mol ca cacbonat trong hh X bit rng khi cho X tc dng vi H2SO4 d v cho kh CO2 to ra p ht vi nc vi trong d ta thu c 3,5g kt ta.

Bi 98: Ha tan 7,83gam mt hh X gm 2 hai kim loi kim A, B thuc hai chu k k tip nhau trong bng HTTH c 1 lt dd C v 2,8 lt kh H2 thot ra ktc. Xc nh A, B v s mol A, B trong C?

Bi 99: Nguyn t ca mt nguyn t X c tng s ht bng 34. Trong o, hat mang ien nhieu hn hat khong mang ien la 10.

a. Xc nh v tr ca X trong bng HTTH?

b. Vit p iu ch trc tip X?

Bi 100: Cho 2 gam hh hai kim loi hai chu k lin tip v thuc phn nhm chnh nhm II tc dng ht vi dd H2SO4 10% ri c cn thu c 8,72 gam hh 2 mui khan.

a. Xc nh 2 kim loi?

b. Tnh khi lng dd H2SO4 dng?

Bi 101: A v B l nguyn t hai chu k lin tip v thuc cng 1 phn nhm chnh, B di A. Cho 8gam B tan hon ton trong 242,4g Nc thu c 4,48 lt kh H2 ktc v dd M.

a. Xc nh A, B v vit cu hnh e ca hai nguyn t?

b. Tnh C% ca dd M?

Bi 102: Y l hidroxit ca nguyn t M thuc nhm IA hoc IIA hoc IIIA. Cho 80g dd 50% ca Y p ht vi dd HCl ri c cn thu c 5,85 gam mui khan. Xc nh Y?

Bi 103: Cho hp cht XY2 tha mn:

Tng s ht p ca hp cht bng 32.

Hiu s ca X v Y bng 8 ht.

X v Y u c s p = s n trong nguyn t.

Xc nh nguyn t X, Y v suy ra hp cht XY2?

Bi 104: Mt nguyn t kim loi M chim 52,94% v khi lng trong oxit cao nht ca n.

a. Xc nh M?

b. Cho 20,4g oxit ca M tan hon ton trong 246,6 gam dung dch 17,86% ca hp cht vi hidro v phi kim X thuc nhm VIIA, to thnh dung dch A. Gi tn X? Tnh C% ca dd A?

BI TP TRC NGIM CHNG NGUYN TCu 1: K hiu ha hc biu th y c trng cho nguyn t ca nguyn t ha hc v n cho bit: A. s A v s Z B. s A C. nguyn t khi ca nguyn t D. s hiu nguyn tCu 2: Mt ng v ca nguyn t photphoP c s proton l: A. 32 B. 15 C. 47 D. 17Cu 3: Nguyn t F c s khi l: A. 10 B. 9 C. 28 D. 19Cu 4: Nguyn t khi trung bnh ca R l 79,91; R c 2 ng v. Bit 81R( 54,5%). S khi ca ng v th nht c gi tr l A. 79 B. 81 C. 82 D. 80Cu 5: Nguyn t X c 2 ng v X1 v X2. ng v X1 c tng s ht l 18. ng v X2 c tng s ht l 20. Bit rng % cc ng v bng nhau v cc loi ht trong X1 cng bng nhau. Nguyn t khi trung bnh ca X l: A. 15 B. 14 C. 12 D. p n khc, c th l:....Cu 6: Nguyn t F khc vi nguyn t P . l nguyn t P : A. h n nguyn t F 13p B. hn nguyn t F 6e C. hn nguyn t F 6n D. h n nguyn t F 13eCu 7: Nguyn t Cu c nguyn t khi trung bnh l 63,54 c 2 ng v X v Y, bit tng s khi l 128. S nguyn t ng v X = 0,37 s nguyn t ng v Y. Vy s ntron ca ng v Y hn s ntron ca ng v X l: A. 2 B. 4 C. 6 D. 1Cu 8: Cho 10 gam mt mui cacbonat ca kim loi ha tr II vo dd HCl d thu c 2,24 lt CO2(ktc).Vy kim loi ha tr II l: A. Be B. Ca C. Ba D. MgCu 9: Ht nhn nguyn t Cu c s ntron l: A. 94 B. 36 C. 65 D. 29Cu 10: Nguyn t ca nguyn t no sau y c s ht e ln nht ? A. F B. Sc C. K D. CaCu 11: ng c 2 ng v 63Cu v 65Cu. T l % ca ng v 63Cu l bao nhiu. Bit rng nguyn t khi trung bnh ca Cu l 63,5. A. 90% B. 50% C. 75% D. 25%Cu 12: Nhng nguyn t Ca, K, Sc c cng: A. s hiu nguyn t B. s e C. s ntron D. s khiCu 13: Nguyn t khi trung bnh ca R l 79,91; R c 2 ng v. Bit 79R( 54,5%). Nguyn t khi ca ng v th 2 c gi tr l bao nhiu? A. 81 B. 85 C. 82 D. 80Cu 14: Cho 5,85 gam mui NaX tc dng vi dd AgNO3 d ta thu c 14,35 gam kt ta trng. Nguyn t X c hai ng v 35X(x1%) v 37X(x2%). Vy gi tr ca x1% v x2% ln lt l: A. 25% & 75% B. 75% & 25% C. 65% & 35% D. 35% & 65%Cu 15: Cc ht cu to nn nguyn t ca hu ht cc nguyn t l A. proton,ntron B. ntron,electron C. electron, proton D. electron,ntron,protonCu 16: ng c hai ng v, chng khc nhau v: A. S electron B. S P C. Cu hnh electron. D. S khiCu 17: Nguyn t ha hc l nhng nguyn t c cng: A. s ntron v proton B. s ntron C. s proton D. s khi.Cu 18: Nguyn t khc vi nguyn t l nguyn t Li c: A. nhiu hn 1p B. t hn 2p C. t hn 2n D. nhiu hn 1nCu 19: ng c 2 ng v 63Cu v 65Cu. T l % ca ng v 65Cu l bao nhiu. Bit rng nguyn t khi trung bnh ca Cu l 63,5(u) A. 25% B. 50% C. 75% D. 90%Cu 20: Trong phn t M2X c tng s ht p,n,e l 140, trong s ht mang in nhiu hn s ht khng mang in l 44 ht. S khi ca M ln hn s khi ca X l 23. Tng s ht p,n,e trong nguyn t M nhiu hn trong nguyn t X l 34 ht. CTPT ca M2X l A. K2O B. Rb2O C. Na2O D. Li2OCu 21: Trong phn t MX2 .Trong M chim 46,67% v khi lng. Ht nhn M c s ntron nhiu hn s proton l 4 ht. Trong nhn X s ntron bng s proton. Tng s proton trong phn t MX2 l 58. CTPT ca MX2 l A. FeS2 B. NO2 C. SO2 D. CO2Cu 22: Nguyn t c s lp electron ti a l A. 8 B. 5 C. 7 D. 4Cu 23: Nguyn t ca nguyn t no sau y c s electron c thn l ln nht? A. Cl(Z=17) B. Ca(Z=20) C. Al(Z=13) D. C(Z=6)Cu 24: Nguyn t ca nguyn t R c tng s ht p,n,e bng 18. v tng s ht khng mang in bng trung bnh cng ca tng s ht mang in.Vy s electron c thn ca nguyn t R l A. 1 B. 2 C. 3 D. 4Cu 25: Nguyn t no sau y cha nhiu ntron nht? A. 24Mg(Z=12) B.23Na(Z=11) C.61Cu(Z=29) D.59Fe(Z=26)Cu 26: Nguyn t S(Z=16) nhn thm 2e th cu hnh e tng ng ca n l: A. 1s2 2s2 2p6 3s1 B. 1s2 2s2 2p6 C. 1s2 2s2 2p6 3s3 D. 1s2 2s2 2p6 3s2 3p6Cu 27: Nguyn t Na(Z=11) b mt i 1e th cu hnh e tng ng ca n l: A. 1s2 2s2 2p6 B. 1s2 2s2 2p6 3s1 C. 1s2 2s2 2p6 3s3 D. 1s2 2s2 2p6 3s2 3p6 4s1Cu 28: Nguyn t ca nguyn t R c tng s ht p,n,e bng 34, hiu s ht ntron v electron bng 1. Vy s e c thn ca R l: A. 3 B. 4 C. 2 D. 1Cu 29: Nguyn t K(Z=19) c s lp electron l A. 3 B. 2 C. 1 D. 4Cu 30: Lp th 4(n=4) c s electron ti a l A. 32 B. 16 C. 8 D. 50Cu 31: Lp th 3(n=3) c s phn lp l A. 7 B. 4 C. 3 D. 5Cu 32: Nguyn t ca nguyn t R c 4 lp e, lp ngoi cng c 1e. Vy s hiu nguyn t ca nguyn t R l A. 15 B. 16 C. 14 D. 19Cu 33: Cu hnh e sau: 1s2 2s2 2p6 3s2 3p6 4s1 l ca nguyn t no sau y: A. F(Z=9) B. Na(Z=11) C. K(Z=19) D. Cl(Z=17)Cu 34: Nguyn t ca nguyn t no sau y l phi kim. A. D(Z=11) B. A(Z=6) C. B(Z=19) D. C(Z=2)Cu 35: Nguyn t ca nguyn t R c 3 lp e, lp ngoi cng c 3e. Vy s hiu nguyn t ca nguyn t R l: A. 3 B. 15 C. 14 D. 13Cu 36: Pht biu no sau y l ng. A. Nhng e c mc nng lng bng nhau c xp vo mt phn lp. B. Tt c u ng. C. Nhng e c mc nng lng gn bng nhau c xp vo mt lp. D. Lp th n c n phn lp( nCu 37: Nguyn t P(Z=15) c s e lp goi cng l A. 8 B. 4 C. 5 D. 7Cu 38: Nguyn t ca nguyn t no sau y c s e c thn khc vi 3 nguyn t cn li. A. D(Z=7) B. A(Z=17) C. C(Z=35) D. B(Z=9)Cu 39: Nguyn t ca nguyn t R c phn lp ngoi cng l 3d1. Vy s hiu nguyn t ca nguyn t R l: A. 21 B. 15 C. 25 D. 24Cu 40: Lp ngoi cng c s e ti a l A. 7 B. 8 C. 5 D. 4Cu 41: Nguyn t C(Z=6) trng thi c bn c bao nhiu e lp ngoi cng? A. 5 B. 8 C. 4 D. 7Cu 42: S e ti a trong phn lp d l: A. 2 B. 10 C. 6 D. 14Cu 43: Nguyn t ca nguyn t A v B u c phn lp ngoi cng l 2p. Tng s e hai phn lp ngoi cng hai nguyn t ny l 3. Vy s hiu nguyn t ca A v B ln lt l: A. 1 & 2 B. 5 & 6 C. 7 & 8 D. 7 & 9Cu 44: Nguyn t ca nguyn t no sau y khi nhn thm 1e th t cu hnh e ca Ne(Z=10). A. Cl(Z=17) B. F(Z=9) C. N(Z=7) D. Na(Z=11)Cu 45: Cu hnh e no sau y l ng: A. 1s2 2s2 2p6 3s2 3p6 B. 1s2 2s2 2p6 3s2 3p3 4s2 C. 1s2 2s2 2p6 3s2 3p7 D. 1s2 2s2 2p6 3s2 3p5 4s1Cu 46: Cu hnh e sau: ....... 4s2 l ca nguyn t no sau y: A. Na(Z=11) B. Cl(Z=17) C. K(Z=19) D. Ca(Z=20)Cu 47: Lp th 3(n=3) c s obitan l A. 9 B. 10 C. 7 D. 18Cu 48: Nguyn t ca nguyn t no sau y l kh him. A. C(Z=11) B. D(Z=2) C. B(Z=5) D. A(Z=4)Cu 49: Nguyn t ca nguyn t no sau y l kh him. A. A(Z=4) B. B(Z=5) C. D(Z=18) D. C(Z=20)Cu 50: Cu hnh electron no sau y l ca kim loi: A. 1s2 2s2 2p6 3s2 3p1 B. 1s2 2s2 2p6 3s2 3p5 C. 1s2 2s2 2p6 3s2 3p4 D. 1s2 2s2 2p6 3s2 3p3

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