Giao an Hoa 10 CB Theo Chuan

Th by ngy 20 thng 08 nm 2011

Gio n ho hc lp 10 c bn- Nm hc 2011-2012

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Tit th 1: N TP U NM (tit 1)

Kin thc c c lin quan

Kin thc mi trong bi cn hnh thnh

Nhng khi nim c bn: Nguyn t ho hc, phn ng ho hc, cht tinh khit, ho tr, n cht, hp cht, nguyn t ...

Cng c kin thc v cc khi nim c bn, k nng lp CTHH, xc nh ho tr, phn bit cc loi hp cht v c, cn bng phng trnh ho hc

I. MC TIU:

1.Kin thc: Gip HS nh li cc kin thc hc lp 8 v 9

*Cc khi nim: n cht, hp cht, nguyn t, nguyn t ha hc, ha tr, phn ng ho hc, ...

*S phn loi cc hp cht v c.

2.K nng: Rn cho HS k nng gii cc dng bi:

*Tm ha tr, lp cng thc hp cht

*Phn bit cc loi hp cht v c

*Cn bng phng trnh ho hc

3.Thi : To nn mng c bn ca mn ho hc

II. TRNG TM:

*Tm ha tr, lp cng thc hp cht

*Phn bit cc loi hp cht v c

*Cn bng phng trnh ho hc

III.PHNG PHP GING DY: Din ging pht vn- kt nhm.

IV. CHUN B GIO C:

*Gio vin: ch (powerpoint cng tt)

*Hc sinh: n li kin thc c

V. TIN TRNH BI DY:

1.n nh lp: Lm quen lp, kim tra s s, ng phc...

2.Kim tra bi c: (0 pht)

3.Bi mi:

a. t vn : Chng ta lm quen vi mn ho hc chng trnh lp 8, 9. By gi chng ta s n li mt s kin thc c bn cn phi nm tip tc nghin cu v mn ho hc

b. Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: I. Mt s khi nim c bn

Mc tiu: Cng c kin thc v cc khi nim c bn

Tr chi ch

Hc sinh ln lt tr li cc t hng ngang tm ra mt t cha kho c ghp t cc ch ci c c cc hng ngang

* Hng ngang 1: C 13 ch ci: Cht khng ln bt c mt cht no khc ( vd: Nc ct) gi l g?

Ch trong t cha kha: H, C

* Hng ngang 2: C 7 ch ci: y l loi cht c to nn t 2 hay nhiu nguyn t ho hc

Ch trong t cha kha: H

* Hng ngang 3: C 6 ch ci: y l ht i din cho cht, gm mt s nguyn t lin kt vi nhau v th hin y tnh cht ca cht

Ch trong t cha kha: P, H

* Hng ngang 4: C 8 ch ci: : y l khi nim :L ht v cng nh v trung ha v in

Ch trong t cha kha: N,

* Hng ngang 5: C 14 ch ci: L tp hp cc nguyn t cng loi c cng s p trong ht nhn

Ch trong t cha kha: A

* Hng ngang 6: C 6 ch ci: L con s biu th kh nng lin kt ca nguyn t hoc nhm nguyn t

Ch trong t cha kha: O

* Hng ngang 7: C 14 ch ci: Hin tng cht bin i m vn gi nguyn l cht ban u

Ch trong t cha kha: N,G

* Hng ngang 8 : C 14 ch ci: Dng biu din cht gm 1,2 hay 3 KHHH v ch s mi chn k hiu.

Ch trong t cha kha: O,A

Gi t cha kha: Qu trnh lm bin i t cht ny thnh cht khc

ch

C

H

T

T

I

N

H

K

H

I

T

H

P

C

H

T

P

H

N

T

N

G

U

Y

N

T

N

G

U

Y

N

T

H

O

A

T

R

I

H

I

N

T

N

G

V

T

L

Y

C

N

G

T

H

C

H

O

A

H

O

C

cha kha: phn ng ha hc

(Phn ng ho hc l qu trnh bin i cht ny thnh cht khc)

Hot ng 2: Ho tr

Mc tiu: Cng c kin thc v ho tr, rn luyn k nng xc nh ho tr

v lp cng thc ho hc

GV: Nhc li N ho tr

- Ho tr ca H, O l bao nhiu?

GV: Ly Vd vi cng thc ho hc

x

ab

y

AB

th quy tc ho tr c vit nh th no?

HS: Tnh ha tr ca cc nt trong cc cthc: H2S; NO2

II. Ho tr

-Ha tr l con s biu th kh nng lin kt ca nt nt ny vi nt ca nt khc.

-Ha tr ca mt nt c xc nh theo ha tr ca nt Hidro (c chn lm n v) v ha tr ca nt Oxi (l hai n v).

-Qui tc ha tr: gi a,b l ha tr ca nguyn t A,B. Trong cng thc AxBy ta c: AaxBby

a*x = b*y

Vd: Ala2O23 ta c 2*a = 3*2 a = 3

Hot ng 3: Phn bit cc loi hp cht v c

Mc tiu: Cng c kin thc v cc loi hp cht v c, rn k nng phn bit cc loi hp cht

-Hs lm vic c nhn: Mt s hc sinh ln bng, hc sinh khc nhn xt, b sung

- Gv: Nhc li khi nim oxit, axit, baz

III. Phn bit cc loi hp cht v c

Ghp ni thng tin ct A vi ct B sao cho ph hp

Tn hp cht

Ghp

Loi cht

1. axit

a. SO2; CO2; P2O5

2. mui

b. Cu(OH)2; Ca(OH)2

3. baz

c. H2SO4; HCl

4. oxit axit

d. NaCl ; BaSO4

5. oxit baz

Hot ng 4: Cn bng phn ng ho hc

Mc tiu: Rn k nng cn bng phng trnh ho hc

Hon thnh PTHH sau, cho bit cc PT trn thuc loi phn ng no?

CaO + HCl CaCl2 + H2O

Fe2O3 + H2 Fe + H2O

Na2O + H2O NaOH

Al(OH)3 t Al2O3 + H2O

Hs lm vic theo nhm, c i din nhm ln bng

Nhm khc nhn xt, gv nhn xt, gii thch

IV. Cn bng phn ng ho hc

Hon thnh PTHH, xc nh loi phn ng:

CaO + 2HCl CaCl2 + H2O ( P/ th)

Fe2O3 + 3H2 2Fe + 3H2O( P/ oxi ha)

Na2O + H2O 2NaOH( P/ ha hp)

2Al(OH)3 t Al2O3 + 3H2O( P/ phn hy)

4. Cng c:

- Lp CTHH ca Al ho tr III v nhm OH ho tr I

- Cn bng phn ng ho hc sau: Fe(OH)3

o

t

Fe2O3 + H2O

5. Dn d: V nh xem li cc khi nim, cng thc lin quan n dung dch

Rt kinh nghim:

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Tit th 2: N TP U NM (tip)



- Khi nim v mol, cng thc tnh

- Nng dung dch

Rn luyn k nng tnh mol, nng mol, nng phn trm

I. MC TIU:

1.Kin thc: Gip HS nh li cc kin thc hc lp 8 v 9: Cc cng thc tnh, cc i lng ha hc: mol, t khi, nng dung dch.

2.K nng: Rn cho HS k nng gii cc dng bi:

*Tnh lng cht, khi lng, ...

*Nng dung dch.

3.Thi : To nn mng c bn ca mn ho hc

II. TRNG TM:

*Tnh lng cht, khi lng, ...

*Nng dung dch


IV. CHUN B GIO C:

*Gio vin: La chn bi tp, gio n

*Hc sinh: n bi c

V. TIN TRNH BI DY:

1.n nh lp: Kim tra s s, ng phc...

2.Kim tra bi c: Trong qu trnh n tp, kt hp ly im kim tra ming

3.Bi mi:

a. t vn : t nn tng vng chc cho mn ho hc cn nm c nhng khi nim, nhng cng thc tnh n gin nht, c bn nht, nn chng ta cn n li tht k phn ny.

b. Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Khi nim v mol

Mc tiu: Cng c khi nim v mol v cng thc tnh

Gv pht vn hs v mol, cng thc tnh, cho v d

Gv thng tin cho hs cng thc tnh s mol iu kin thng

Hs lm vic c nhn: Tnh s mol ca 28 gam Fe; 2,7 gam nhm; 11,2 lt kh oxi (ktc)

Hs ln bng trnh by

Gv nhn xt, nhc li cho hs nh v t khi cht kh:Cng thc:

B

A

B

A

M

M

d

=

;

29

M

d

A

kk

A

=

V. Khi nim v mol :

1/ nh ngha :

Mol l lng cht cha 6,023.1023 ht vi m (nguyn t, phn t, ion).

Vd : 1 mol nguyn t Na(23g) cha 6,023.1023 ht nguyn t Na.

2/ Mt s cng thc tnh mol :

* Vi cc cht :

m

n

M

=

* Vi cht kh :

- Cht kh iu kin tiu chun (OoC, 1atm)

V

n

22,4

=

- Cht kh toC, p (atm)

Hot ng 2: nh lut bo ton khi lng

Mc tiu: Cng c, rn k nng tnh khi lng theo nh lut bo ton khi lng

Gv cho phn ng tng qut, yu cu hs vit biu thc cho LBTKL

Hs lm vic theo nhm, i din hs ln bng, nhm khc b sung

Gv nhn xt, gii thch

VI. nh lut bo ton khi lng

Khi c p: A + B C + D

p dng LBTKL ta c:

mA + m B = mC + mD

msp = mtham gia

Vd: cho 6,50 gam Zn p vi lng va dung dch cha7,1 gam axit HCl thu c 0,2 gam kh H2. Tnh khi lng mui to thnh sau p?

Gii

Zn + 2HCl ZnCl2 + H2

6,5g 7,1g xg 0,2g

p dng LBTKL ta c:

6,5 + 7,1 = x + 0,2 x = 13,4g

Hot ng 3: Nng dung dch

Mc tiu: Cng c v rn k nng tnh nng mol, nng phn trm

- Gv pht vn hs v cng thc tnh nng %, nng mol/lit, hng dn hs tm ra cng thc lin h gia 2 loi nng (thng tin v ct tnh mdd)

- Hs lm vic theo nhm

- Gv gii thch, kt lun

- Gv kt lun

VII. Nng dung dch :

1/ Nng phn trm (C%).

=

ct

dd

m

C%100%

m

2/ Nng mol (CM hay [ ])

=

M

ct

dd

n

Chay[]

V

Vdd : th tch dung dch(lit)

3/ Cng thc lin h :

mdd = V.D (= mdmi +mct)

=

M

10.C%.D

C

M

lu : V (ml) ; D (g/ml)

4. Cng c:

Bi tp1)Tnh s mol cc cht sau:

a) 3,9g K; 11,2g Fe; 55g CO2; 58g Fe3O4

b) 6,72 lt CO2 (ktc); 10,08 lt SO2 (ktc); 3,36 lt H2 (ktc)

c) 24 lt O2 (27,30C v 1 atm); 12 lt O2 (27,30C v 2 atm); 15lt H2 (250C v 2atm).

Bi tp2)Tnh nng mol ca cc dung dch sau:

a) 500 ml dung dch A cha 19,88g Na2SO4.

b) 200ml dung dch B cha 16g CuSO4.

c) 200 ml dung dch C cha 25g CuSO4.2H2O.

Bi tp3) Tnh nng phn trm ca cc dung dch sau:

a) 500g dung dch A cha 19,88g Na2SO4.

b) 200g dung dch B cha 16g CuSO4.

c) 200 g dung dch C cha 25g CuSO4.2H2O.

5. Dn d:

- Lm bi tp: Ha tan 8,4 g Fe bng dung dch HCl 10,95%(va )

a. Tnh th tch kh thu c (KTC)

b. Tnh khi lng axit cn dng

c. Tnh nng % ca dd sau phn ng

- c trc bi 1: Thnh phn nguyn t

Rt kinh nghim:

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Th nm ngy 01 thng 09 nm 2011

CHNG I: NGUYN T

Tit th 3: Bi 1: THNH PHN NGUYN T



- Thnh phn cu to nguyn t

- Du in tch electron, proton

- S tm ra electron, ht nhn, proton, ntron

- C th c im cc loi ht trong nguyn t: in tch, khi lng...

I. MC TIU:

1.Kin thc: Bit c :

( Nguyn t gm ht nhn mang in tch dng v v nguyn t mang in tch m ; Kch thc, khi lng ca nguyn t.

( Ht nhn gm cc ht proton v ntron.

( K hiu, khi lng v in tch ca electron, proton v ntron.

2.K nng:

( So snh khi lng ca electron vi proton v ntron.

( So snh kch thc ca ht nhn vi electron v vi nguyn t.

3.Thi : Kch thch s hng th vi b mn, pht huy kh nng t duy ca hc sinh

II TRNG TM; Nguyn t gm 3 loi ht: p, n, e (k hiu, khi lng v in tch)

III.PHNG PHP GING DY: Din ging pht vn

IV. CHUN B GIO C:

*Gio vin: M hnh th nghim m phng ca Tom-xn pht hin ra tia m cc v ca R--pho khm ph ra ht nhn nguyn t

*Hc sinh: Chun b bi mi trc khi n lp.

V. TIN TRNH BI DY:


2.Kim tra bi c: (5 pht) Kim tra vic lm bi tp nh

3.Bi mi:

a. t vn : Nguyn t c to nn t nhng loi ht no? Chng ta hc lp 8. Hm nay chng ta s tm hiu r hn v in tch, khi lng, kch thc ca chng

b. Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1; Thnh phn cu to ca nguyn t

Mc tiu: Bit s tm ra electron, ht nhn nguyn t, proton, ntron, c im ca tng loi ht (Hiu thnh phn cu to ca nguyn t, so snh khi lng electron vi p, n

-Gv:Electron do ai tm ra v c tm ra nm no?

-Hs tr li

-Gv: Trinh chiu m hnh s th nghim tm ra tia m cc, yu cu hs nhn xt c tnh ca tia m cc

- Gv yu cu hs cho bit khi lng, in tch ca electron (Gv kt lun

- Ht nhn c tm ra nm no, do ai?

- Gv trnh chiu m hnh th nghim bn ph l vng tm ra ht nhn nt.

- Hs nhn xt v cu to ca nguyn t

- Gv kt lun

- Proton c tm ra vo nm no, bng th nghim g?

- Gv thng tin v khi lng, in tch ( Gi tr in tch p bng vi electron nhng tri du; qe = 1- th qp = 1+

- Gv thng tin, yu cu hs so snh khi lng ca electron vi p v n

- Hs kt lun

- Cc em c th kt lun g v ht nhn nguyn t ?

- Gv kt lun

I. THNH PHN CU TO CA NGUYN T:

1. Electron (e):

S tm ra electron: Nm 1897, J.J. Thomson (Tm-xn, ngi Anh ) tm ra tia m cc gm nhng ht nh gi l electron(e).

Khi lng v in tch ca e:

+ me = 9,1094.10-31kg.

+ qe = -1,602.10-19 C(coulomb) = -1 (vt m, k hiu l e0).

2. S tm ra ht nhn nguyn t:

Nm 1911, E.Rutherford( R-d-pho, ngi Anh) dng tia

a

bn ph mt l vng mng chng minh rng:

-Nguyn t c cu to rng, phn mang in tch dng l ht nhn, rt nh b.

-Xung quanh ht nhn c cc e chuyn ng rt nhanh to nn lp v nguyn t.

-Khi lng nguyn t hu nh tp trung vo ht nhn ( v khi lng e rt nh b).

3. Cu to ht nhn nguyn t:

a) S tm ra proton:

Nm 1918, Rutherford tm thy ht proton(k hiu p) trong ht nhn nguyn t:

mp = 1,6726. 10-27kg.

qp = +1,602. 10-19Coulomb(=1+ hay e0,tc 1 n v .tch dng)

b) S tm ra ntron:

Nm 1932,J.Chadwick(Chat-uch) tm ra ht ntron (k hiu n) trong ht nhn nguyn t:

mn

;

mp .

qn = 0 .

c) Cu to ca ht nhn nguyn t:

Trong ht nhn nguyn t c cc proton v ntron.

pe

=

Hot ng 2: Kch thc v khi lng ca nguyn t

Mc tiu: Bit s chnh lch kch thc gia ht nhn v nguyn t v so snh, Bit n v o kch thc nguyn t, n v o khi lng nguyn t

- Gv thng tin

-Nguyn t H c bn knh khong 0,053nm(ng knh khong 0,1nm, dng knh ht nhn nguyn t nh hn nhiu, khong 10-5nm(Em hy xem ng knh nguyn t v ht nhn chnh lch nhau nh th no?

- Hs tnh ton, tr li

- Gv minh ho v d phng i nt

- Gv thng tin, yu cu hs nghin cu bng 1/8

II/ KCH THC V KHI LNG CA NGUYN T:

1. Kch thc nguyn t:

Ngi ta biu th kch thc nguyn t bng:

+ 1nm(nanomet)= 10- 9 m

+ 1A0 (angstrom)= 10-10 m

Nguyn t c kch thc rt ln so vi kch thc ht nhn (

1

5

10

10.000

10

nm

nm

-

-

=

ln).

de,p

10-8nm.

2. Khi lng nguyn t:

- Do khi lng tht ca 1 nguyn t qu b, ngi ta dng n v khi lng nguyn t u(vC).

1 u = 1/12 khi lng 1 nguyn t ng v cacbon 12 = 1,6605.10-27kg.(xem bng 1/trang 8 sch GK 10).

- m nguyn t = mP + mN (B qua me)

4. Cng c:

Cho hc sinh c li bng 1/8 sch gio khoa.

1, 2/trang 9 SGK v 6/trang 4 sch BT

5. Dn d:

3,4,5/trang 9/SGK v 1.1,1.2, 1.5/3 v 4 sch BT.

Lm cu hi trc nghim.

Chun b bi 2

Rt kinh nghim:

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Th hai ngy 5 thng 9 nm 2011

Tit th 4: HT NHN NGUYN T- NGUYN T HO HC-NG V (tit 1)



- Cu to ca ht nhn nguyn t

- Nguyn t ho hc

- S khi, in tch ht nhn, s hiu nguyn t

- ng v

- nh ngha NTHH mi

- K hiu nguyn t

- in tch ht nhn l c trng ca nguyn t

I. MC TIU:

1.Kin thc:Hiu c :

( Nguyn t ho hc bao gm nhng nguyn t c cng s n v in tch ht nhn.

( S hiu nguyn t (Z) bng s n v in tch ht nhn v bng s electron c trong nguyn t.

( K hiu nguyn t :

A

Z

X.X

l k hiu ho hc ca nguyn t, s khi (A) l tng s ht proton v s ht ntron.

2.K nng: Xc nh s electron, s proton, s ntron khi bit k hiu nguyn t ngc li.

3.Thi : Pht huy kh nng t duy ca hc sinh

II. TRNG TM:

( c trng ca nguyn t l in tch ht nhn (s p) ( nu c cng in tch ht nhn (s p) th cc nguyn t u thuc cng mt nguyn t ha hc

( Cch tnh s p, e, n


IV. CHUN B GIO C:

*Gio vin: Gio n

*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.

V. TIN TRNH BI DY:



Tng s ht trong nguyn t mt nguyn t X l 34. Trong s n hn s p la 1. Tm s ht mi loi trong nguyn t?

3.Bi mi:

a) t vn : Ta bit ht nhn nguyn t to nn t cc ht proton v ntron v c kch thc rt nh b. Hm nay chng ta s tm hiu v nhng vn lin quan xung quanh s n v in tch ht nhn

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Ht nhn nguyn t

Mc tiu: Hiu v ht nhn nguyn t; Bit cch tnh v rn luyn tnh nguyn t khi trung bnh, tnh cc loi ht da vo s khi v s hiu

- Gv: in tch ht nhn nguyn t c xc nh da vo u?

- Hs tr li

- Gv: S khi A c xc nh nh th no?

- Hs tr li

- Gv ly vd cho hs tnh s khi

I/ HT NHN NGUYN T:

1.in tch ht nhn:

-Ht nhn c Z proton

in tch ht nhn l +Z.

-S n v in tch ht nhn Z = s proton = s electron .

nguyn t trung ha v in .

2.S khi (A): = S proton(Z) + S ntron(N)

A = Z + N

S n v in tch ht nhn Z v s khi A c trng c bn cho ht nhn v nguyn t.

BT: Tng s ht c bn trong nguyn t ca mt nguyn t l 60, trong tng s ht mang in nhiu hn s ht khng mang in l 20. Tm s khi A?

Hot ng 2: Nguyn t ho hc

Mc tiu: Bit nh ngha mi v nguyn t ho hc, hiu k hiu nguyn t

- Gv: Trong phn n tp u nm, chng ta c nhc n nguyn t ho hc, em no c th nhc li nh ngha?

- Hs tr li(Gv kt lun

- Gv thng tin

- Gv ly mt s v d hs xc nh s khi, s hiu nguyn t :

23633956

11291926

;;;

NaCuKFe

- Hs vn dng tnh s n ca cc nguyn t trn

II/ NGUYN T HA HC:

1. nh ngha:

Nguyn t ha hc gm nhng nguyn t c cng in tch ht nhn .

2. S hiu nguyn t (Z):

S n v in tch ht nhn nguyn t ca 1 nguyn t c gi l s hiu ca nguyn t , k hiu l Z.

3. K hiu nguyn t:

Nguyn t X c s khi A v s hiu Z c k hiu nh sau:

A

Z

X

4. Cng c:

Nu cc nh ngha v: nguyn t ha hc?

Tr li cc cu hi: 1, 2/trang 13 v 4/14 sch gio khoa v 1.15/trang 6 sch BT.

5. Dn d:

Chun b phn ng v, khi lng nguyn t

Lm cu hi trc nghim.

Rt kinh nghim:

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Tit th 5: HT NHN NGUYN T- NGUYN T HO HC-NG V (tit 2)



- Cu to ca ht nhn nguyn t

- Nguyn t ho hc

- S khi, in tch ht nhn, s hiu nguyn t

- ng v

- Tnh ton v ng v

- Nguyn t khi, khi lng nguyn t trung bnh

I. MC TIU:

1.Kin thc:Hiu c : Khi nim ng v, nguyn t khi v nguyn t khi trung bnh ca mt nguyn t.

2.K nng: Tnh nguyn t khi trung bnh ca nguyn t c nhiu ng v.

3.Thi : Pht huy kh nng t duy logic ca hc sinh

II. TRNG TM:

Khi s n trong ht nhn nguyn t ca cng mt nguyn t khc nhau s tn ti cc ng v.

Cch tnh nguyn t khi trung bnh


IV. CHUN B GIO C:

*Gio vin: Gio n


V. TIN TRNH BI DY:



- Xc nh s e, s p, s n, s khi, in tch ht nhn ca:

23633956

11291926

;;;

NaCuKFe

- Tng s ht c bn trong nguyn t ca mt nguyn t l 36, trong tng s ht mang in nhiu hn s ht khng mang in l 12. Tm s khi A?

3.Bi mi:

a) t vn : Ta bit cch tnh s khi ca nguyn t = Z+ N; Z ca mt nguyn t lun khng i, khi N thay i th th no? Nguyn t khi ca n s c tnh ra sao?

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1:ng v

Mc tiu: Cng c v ng v

- Gv ly vd cc ng v ca H(Nhng nguyn t nh th no c gi l ng v ca mt nguyn t ?

- Hs tr li

- Gv kt lun

III/ NG V:

ng v l nhng nguyn t c cng s proton, nhng khc v s ntron nn s khi khc nhau.

Vd : Nguyn t hiro c 3 ng v :

Proti

1

1

H

teri

2

1

H

Triti

3

1

H

Hot ng 2:Nguyn t khi-Nguyn t khi trung bnh

Mc tiu: Bit cch tnh nguyn t khi trung bnh

- n v khi lng nguyn t c tnh nh th no? K hiu?

- Hs tr li

- Gv thng tin

- Gv thng tin v a ra biu thc tnh

IV/ NGUYN T KHI V NGUYN T KHI TRUNG BNH CA CC NGUYN T HA HC:

1. Nguyn t khi A(khi lng tng i ca nguyn t): Cho bit khi lng ca nguyn t nng gp bao nhiu ln n v khi lng nguyn t.

Do khi lng ca e qu nh nn nguyn t khi coi nh bng s khi.

2. Nguyn t khi trung bnh

A

:

Do 1 nguyn t thng c nhiu ng v, nn dng nguyn t khi trung bnh:

1122

...

100

nn

AxAxAx

A

+++

=

Hot ng 3:Vn dng

Mc tiu: Rn k nng tnh nguyn t khi trung bnh

- Gv cho hs ghi , yu cu hs trnh by tng gii bi ton

- Mt hs ln bng

- Gv cho hs ghi

- Hs tho lun tm cch gii

- i din mt nhm ln bng

- Nhm khc nhn xt, b sung

- Gv nh gi

BT1: Clo c 2 ng v:

35

17

Cl

(chim 75,77%)

v

37

17

Cl

(chim 24,23%)

-Hy tm

A

Cl =?

Gii:

A

Cl =

100

37

*

23

,

24

35

*

77

,

75

+

= 35,5

BT2: Cho

A

Cu =63,54 . Tm %

65

29

Cu

?

63

29

Cu

?

-Gi%

65

29

Cu

l x th %

63

29

Cu

l 100-x

100

)

100

(

63

65

x

x

-

+

=63,54

=>x = 27% = %

65

29

Cu

%

63

29

Cu

= 100-27 = 73%

4. Cng c:

- Biu thc tnh nguyn t khi trung bnh

- Cu to nguyn t ?

- Mi lin h gia cc loi ht trong nguyn t ?

5. Dn d:

- Lm bi tp 3,6,7,8/14 SGK

- c phn t liu Trang 14- 15

- Tr li cu hi trc nghim cho trc

*Chun b Bi 3: LUYN TP: THNH PHN NGUYN T

(1) Thnh phn cu to nguyn t, S khi , nguyn t khi , nguyn t ho hc, S hiu nguyn t,k hiu nguyn t, ng v, nguyn t khi TB

(2)Xc nh s e, p, n v nguyn t khi khi bit k hiu nguyn t

(3)Xc nh nguyn t khi TB ca nguyn t ho hc

Rt kinh nghim:

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Tit th 6: LUYN TP: THNH PHN NGUYN T



- Thnh phn nguyn t v c im cc loi ht

- Nguyn t ho hc, k hiu nguyn t, ng v, nguyn t khi, nguyn t khi trung bnh

H thng ho v nguyn t

I. MC TIU:

1.Kin thc: Cng c kin thc v:

- Thnh phn cu to nguyn t, ht nhn nguyn t, kch thc, khi lng, in tch ca ht nhn

- nh ngha nguyn t ho hc, k hiu nguyn t, ng v, nguyn t khi, nguyn t khi trung bnh

2.K nng: Rn luyn k nng xc nh s electron, s proton, s ntron v nguyn t khi khi bit k hiu nguyn t

3.Thi : T gic trong hc tp, hot ng nhm

II.PHNG PHP GING DY: Pht vn- kt nhm.

III. CHUN B GIO C:

*Gio vin: Gio n, bi tp cho hs lm trc

*Hc sinh: n bi c, lm bi tp trc khi n lp.

IV. TIN TRNH BI DY:



- Lm bi tp 8/14 SGK

- Kim tra v mt s hs

3.Bi mi:

a) t vn : Chng ta nghin cu v thnh phn nguyn t. By gi s cng c li nhng kin thc hc v vn dng vo lm bi tp

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Cu hi trc nghim

Mc tiu: Cng c kin thc v cu to nguyn t, nguyn t ho hc, ng v, ...

Cu 1: Pht biu no sau y khng ng?

1. Trong mt nguyn t lun lun c s proton bng s electron v bng s n v in tch ht nhn

2. Tng s proton v s electron trong ht nhn c gi l s khi

3. S khi A l khi lng tuyt i ca nguyn t

4. S proton bng s n v in tch ht nhn

5. ng v l cc nguyn t c cng s proton nhng khc nhau v s ntron

A. 2, 3 B. 3, 4, 5 C. 1, 3 D. 2, 5

Cu2: C cc ng v sau:

1233537

1111717

;;;;

HHHClCl

. Hi c th to ra bao nhiu phn t hiroclorua c thnh phn ng v khc nhau?

A. 8 B. 12 C. 6 D. 9

Cu 3: Nhng iu khng nh sau y c phi bao gi cng ng ?

a) S hiu nguyn t bng s n v in tch ht nhn nguyn t

b) S proton trong nguyn t bng s ntron

c) S proton trong ht nhn bng s electron lp v nguyn t

d) Ch c ht nhn nguyn t oxi mi c 8 proton

e) Ch c ht nhn nguyn t oxi mi c 8 ntron

f) Ch c ht nhn nguyn t oxi mi c t l gia s proton v ntron l 1:1

Cu hi trc nghim:1, 2, 3/trang 9 SGK; 1.15/trang 6 SBT;1, 2, 3/trang 13 SGK.

Hs: Tho lun tr li

Cu 1: A

Cu 2: C

Cu 3:

a) ng

b) Sai

c) ng

d) ng

e) Sai

f) Sai

Hot ng 1: Cu hi t lun

Mc tiu: Rn luyn k nng xc nh s ht, in tch ... trong nguyn t khi bit k hiu nguyn t, tnh phn trm ng v, s nguyn t ca mt ng v...

Hs lm vic theo nhm, i din ln bng, nhm khc nhn xt( Gv nh gi, gii thch

Cu 1: C cc loi nguyn t sau:

3537

1717

;

ClCl

121314

666

;;

CCC

a/ Xc nh s ntron, s proton, s e v s khi ca mi loi nguyn t trn?

b/ nh ngha ng v?

Cu 2: Cho cc nguyn t:

10648411109634039106

5293654729191947

;;;;;;;;

ABCDGHELJ

.

a/ nh ngha: A v D; B v H; E v L; G v J? Gii thch?

b/ Mt nguyn t X c s hiu Z, s khi A c k hiu nh th no?

Cu 3: BT 6, 7/trang 14 SGK.

Cu 1:

a)

KHNT

S p

S n

S e

S khi

35

17

Cl

17

18

17

35

37

17

Cl

17

20

17

37

12

6

C

6

6

6

12

13

6

C

6

7

6

13

14

6

C

6

8

6

14

b) Hs t gii

Cu 2:

a) Cc cp nguyn t l ng v. V chng c cng s proton nhng khc nhau v s khi

b)

A

Z

X

Cu 3: 4hs ln bng

4. Cng c: Cng c xen trong cc bi tp

5. Dn d: Nm vng cc kin thc hc, chun b bi Cu to v nguyn t

Rt kinh nghim:

..............................................................................................................................................................

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Th su ngy 16 thng 9 nm 2011

Tit th 7 : CU TO V NGUYN T (tit 1)




- c im electron

- S chuyn ng ca electron trong nguyn t theo quan nim c v mi

- Lp v phn lp electron

I. MC TIU:

1.Kin thc: Bit c:

- Cc electron chuyn ng rt nhanh xung quanh ht nhn nguyn t khng theo nhng qu o xc nh, to nn v nguyn t.

- Trong nguyn t, cc electron c mc nng lng gn bng nhau c xp vo mt lp (K, L, M, N, O, P, Q).

- Mt lp electron bao gm mt hay nhiu phn lp. Cc electron trong mi phn lp c mc nng lng bng nhau.

2.K nng:

Xc nh c th t cc lp electron trong nguyn t, s phn lp (s, p, d) trong mt lp.

3.Thi : Kch thch s yu thch mn hc

II. TRNG TM:

- S chuyn ng ca cc electron trong nguyn t

- Lp v phn lp electron


IV. CHUN B GIO C:

*Gio vin: M hnh mu hnh tinh nguyn t


V. TIN TRNH BI DY:



Tng s ht trong nguyn t ca mt nguyn t l 36. Trong , tng s ht mang in nhiu hn s ht khng mang in l 12. Tm A?

3.Bi mi:

a) t vn : V nguyn t c to nn bi ht no?

- Hs tr li

( Cc electron lp v nguyn t chuyn ng nh th no? By gi chng ta s tm hiu xem.

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: S chuyn ng ca cc electron trong nguyn t

Mc tiu: Phn bit c s chuyn ng ca electron trong nguyn t theo quan nim c v mi

- Gv thng tin v trnh chiu m hnh nguyn t ca Bo hs quan st

- Theo quan nim hin i th cc electron chuyn ng nh th no?

- Hs tr li

- Gv trnh chiu m hnh nguyn t hin i cho hs quan st

I.S CHUYN NG CA CC ELECTRON TRONG NGUYN T:

1.Quan nim c (theo E.Rutherford, N.Bohr, A.Sommerfeld): Electron chuyn ng quanh ht nhn nguyn t theo nhng qu o hnh bu dc hay hnh trn (Mu nguyn t hnh tinh).

2.Quan nim hin i: Cc electron chuyn ng rt nhanh quanh ht nhn nguyn t trn nhng qu o khng xc nh to thnh nhng m my e gi l obitan.

Hot ng2: Lp electron v phn lp electron

Mc tiu: Bit trong nguyn t c bao nhiu lp e, mi lp e c bao nhiu phn lp v th t mc nng lng ca cc lp electron

Cc electron chuyn ng khng theo qu o nht nh nhng khng phi hn lon m vn tun theo quy lut nht nh

- Gv thng tin v lp v phn lp

II.LP ELECTRON V PHN LP ELECTRON:

1. Lp electron:

- Gm nhng e c mc nng lng gn bng nhau.

- Cc electron phn b vo v nguyn t t mc nng lng thp n mc nng lng cao( t trong ra ngoi ) trn 7 mc nng lng ng vi 7 lp electron:

Mc nng lng n

1

2

3

4

5

6

7

Tn lp

K

L

M

N

O

P

Q

2.Phn lp electron:

- Mi lp chia thnh cc phn lp

Cc e trn cng mt phn lp c mc nng lng bng nhau.

C 4 loi phn lp: s, p, d, f.

Lp th n c n phn lp ( vi n

4).

4. Cng c:

K tn cc lp, phn lp e trong nguyn t, s phn lp trong mt lp? ( Cu hi trc nghim

5. Dn d:

Sch GK: Cu 1 ( 4/trang 22.

Sch BT: Cu 1.25 ( 1.31/trang 8 v 9

Chun b phn III

Rt kinh nghim:

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Th su ngy 16 thng 9 nm 2011

Tit th 8 : CU TO V NGUYN T (tit 2)



- Lp electron, phn lp electron

- S electron ti a trn mi phn lp, mi lp electron

- S phn b electron trn cc phn lp, cc lp

I. MC TIU:

1.Kin thc: Bit c: S electron ti a trong mt lp, mt phn lp.

2.K nng: Xc nh c s lp electron trong nguyn t, s phn lp (s, p, d) trong mt lp.

3.Thi : Kch thch s yu thch mn hc

II. TRNG TM: S electron ti a trn mt phn lp, mt lp


IV. CHUN B GIO C:

*Gio vin: Gio n in t


V. TIN TRNH BI DY:



- S chuyn ng ca electron trong nguyn t theo quan nim mi v c khc nhau nh th no?

- Cho bit cc k hiu phn lp, lp, s phn lp trong mt lp?

3.Bi mi:

a) t vn : Cc electron ti a trn mi phn lp v mi lp nh th no?

b)Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: S electron ti a trong mt lp, phn lp

Mc tiu: Bit v nm vng v s electron ti a trn mt lp, phn lp electron

- Gv thng tin v s electron ti a trong mt phn lp

- Gv cho hs bit s electron ti a trong lp th n (n

4) l 2n2

- Gv yu cu hs cho bit s phn b e trn cc phn lp v s e ti a trn mt lp

- Gv trnh chiu khung trng, hs ln lt pht biu s phn b e( Trnh chiu m hnh nguyn t mt s nguyn t

III.S ELECTRON TI A TRONG MT PHN LP, LP:

1.S electron ti a trong mi phn lp:

Phn lp

s

p

d

f

S electron ti a trn 1 phn lp

2

6

10

14

(Phn lp c s electron ti a gi l phn lp electron bo ha.

2.S electron ti a trong lp th n l 2n2 e (n

4)

Hot ng : Vn dng

Mc tiu: Rn luyn k nng xc nh s lp electron, xc nh s ht, s phn b e trong nguyn t

Hs tho lun lm bi tp

i din 2 nhm ln bng trnh by

Nhm khc nhn xt

Gio vin nh gi, din gii

Bi 1: Xc nh s lp e ca cc nguyn t

14

7

N,

24

12

Mg.

Bi 2: Nguyn t agon c k hiu l

40

18

Ar.

a) Hy xc nh s p, s n v s e trong nguyn t.

b) Hy x/nh s phn b e trn cc lp e.

4. Cng c: C th cho hc sinh phn b e trong lp v ca nguyn t : 20Ca, 16S.

5. Dn d:

Sch GK: Cu 5/trang 22.

Sch BT: Cu 1.32 ( 1.35/trang 8 v 9

c bi c thm, chun b bi Cu hnh electron nguyn t

Rt kinh nghim:

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Tit th 9: CU HNH ELECTRON NGUYN T



- K hiu nguyn t

- Lp, phn lp, s electron ti a

- Th t tng mc nng lng

- Cu hnh electron v cch vit

- c im lp electron ngoi cng

I. MC TIU:

1.Kin thc: Bit c:

- Th t cc mc nng lng ca cc electron trong nguyn t.

- S phn b electron trn cc phn lp, lp v cu hnh electron nguyn t ca 20 nguyn t u tin.

- c im ca lp electron ngoi cng: Lp ngoi cng c nhiu nht l 8 electron (ns2np6), lp ngoi cng ca nguyn t kh him c 8 electron (ring heli c 2 electron). Hu ht cc nguyn t kim loi c 1, 2, 3 electron lp ngoi cng. Hu ht cc nguyn t phi kim c 5, 6, 7 electron lp ngoi cng.

2.K nng:

- Vit c cu hnh electron nguyn t ca mt s nguyn t ho hc.

- Da vo cu hnh electron lp ngoi cng ca nguyn t suy ra tnh cht ho hc c bn (l kim loi, phi kim hay kh him) ca nguyn t tng ng.

II. TRNG TM:

- Th t cc mc nng lng ca cc electron trong nguyn t.

- S phn b electron trn cc phn lp, lp v cu hnh electron nguyn t.

- c im cu hnh ca lp electron ngoi cng.


IV. CHUN B GIO C:

*Gio vin: S phn b mc nng lng ca cc lp v phn lp ( hoc bng qui tc Kleckowski); cu hnh e ca 20 nguyn t u.


V. TIN TRNH BI DY:


2.Kim tra bi c: (5 pht) Xc nh s lp e, s e mi lp trong cc nguyn t:

815111718

;;;;Ar

OPNaCl

3.Bi mi:

a) t vn : Da vo s electron ti a ca tng lp, tng phn lp ta c th vit cu hnh e ca nguyn t. Cu hnh e c biu din nh th no, hm nay chng ta s cng tm hiu.

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Th t cc mc nng lng trong nguyn t

Mc tiu: Bit th t mc nng lng trong v nguyn t

- Gv: Trong 7 lp e ca nguyn t, lp no c mc nng lng thp nht?

- Hs tr li

- Gv thng tin v v th t mc nng lng cc phn lp

- Gv lu hs v s chn mc nng lng dn n nng lng phn lp 4s nh hn 3d

- Cho hs xem s phn b mc nng lng ca cc lp v phn lp

I. TH T CC MC NNG LNG TRONG NGUYN T:

Cc electron sp vo cc lp v phn lp t mc nng lng thp n mc nng lng cao theo th t sau: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s,

Khi in tch ht nhn tng, c s chn mc nng lng nn mc nng lng ca 4s thp hn 3d.

Hot ng 2: Cu hnh electron ca nguyn t

Mc tiu: Bit v hiu cch vit cu hnh electron nguyn t, bit xc nh h ca nguyn t da vo cu hnh electron

- Gv: S biu din electron phn b trn cc phn lp, lp theo th t t trong ra ngoi gi l cu hnh e nguyn t (GV yu cu hs cho bit quy c v cc bc vit cu hnh electron

- Gv vit cu hnh e ca H, He, O

- Hs vit cu hnh e ca Ar, Ca, Br

- Gv nhn xt v vit cu hnh gn theo nguyn t kh him c cu hnh gn ging

- Gv thng tin v nguyn t s, p, d, f

- Hs xc nh nguyn t s, p, d, f cho cc vd trn

- Hng dn hs xem cu hnh e ca 20 nguyn t u trong SGK

II. CU HNH ELECTRON CA NGUYNT:

1. Cu hnh e ca nguyn t:

Cu hnh electron: Biu din s phn b e trn cc lp v phn lp

V d: Cu hnh e ca cc nguyn t:

1H: 1s1

2He: 1s2

8O: 1s2 2s2 2p4 hay

[

]

He

2s2 2p4

18

Ar

: 1s2 2s2 2p6 3s2 3p6

20Ca: 1s2 2s2 2p6 3s2 3p6 4s2 hay

[

]

Ar

4s2

35Br: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 hay

[

]

Ar

3d10 4s2 4p5

Phn lp cui cng l h ca nguyn t :

+ H, He, Ca: l nguyn t s v e cui cng in vo phn lp s .

+ O, Ar, Br: l nguyn t p v e cui cng in vo phn lp p.

+ Ngoi ra cn c nguyn t d, nguyn t f.

2/ Cu hnh electron ca 20 nguyn t u ( xem sch GK)

Hot ng 3: c im electron lp ngoi cng

Mc tiu: Bit xc nh tnh cht ho hc c bn nguyn t ho hc da vo c im lp electron ngoi cng

- Gv: Da vo v d trn cho bit lp e ngoi cng c ti a bao nhiu e?

- Hs tr li

- Gv thng tin v c im lp e ngoi cng, yu cu hs vn dng cho cc v d trn

III/ C IM LP ELECTRON NGOI CNG:

i vi nguyn t ca tt c cc nguyn t, lp e ngoi cng c nhiu nht l 8 e

Cc nguyn t u c khuynh hng t trng thi bo ha bn vi 8 e lp ngoi cng( tr He, 2e ngoi cng).

Lp e ngoi cng quyt nh tnh cht ha hc ca mt nguyn t:

+ Nu tng s e ngoi cng < 4 (1,2,3e) => Nguyn t CHO e

l kim loi.

+ Nu tng s e ngoi cng > 4 (5,6,7e)

Nguyn t NHN e

l phi kim.

+ Nu tng s e ngoi cng = 4

Nguyn t c th l kim loi hoc phi kim.

+ Nu tng s e ngoi cng = 8 ( tr He , 2e ngoi cng)

Nguyn t bn v mt ha hc

l kh him.

( Vy: khi bit cu hnh e ca nguyn t c th d on c cc loi nguyn t.

4. Cng c:

Vit li th t s tng mc nng lng phn b e vo cc lp v nguyn t?

Vit cu hnh e v xc nh cc nguyn t sau thuc kim loi, phi kim, kh him?Ti sao?

20Ca ;

29

Cu

; 36Kr

5. Dn d:

Cu hi trc nghim: 1,2,3/ trang 27, 28 sch GK v 1.46/trang 10 sch BT.

Lm vo tp: Bi 4 ( 6 / trang 28 sch GK v 1.41/trang 10 sch BT.

Rt kinh nghim:

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Tit th 10:

LUYN TP: CU TO V ELECTRON CA NGUYN T(tit1)



- S chuyn ng ca electron trong nguyn t

- Lp, phn lp electron v s electron ti a

- Cu hnh electron nguyn t

- Rn luyn k nng vit cu hnh electron

- Xc nh tnh cht c bn ca nguyn t

I. MC TIU:


- S chuyn ng ca electron trong nguyn t

- Lp, phn lp v s electron ti a trn mt lp, phn lp

- Cu hnh electron v c im electron lp ngoi cng

2.K nng:

- Rn luyn k nng vit cu electron nguyn t

- Xc nh tnh cht c bn ca nguyn t

3.Thi : Pht huy kh nng t duy logic

II.PHNG PHP GING DY: Din ging pht vn- kt nhm.

III. CHUN B GIO C:

*Gio vin: Gio n, bi tp


IV. TIN TRNH BI DY:


2.Kim tra bi c: Kt hp trong qu trnh luyn tp

3.Bi mi:

a) t vn : Chng ta nghin cu v lp v nguyn t v cu hnh electron, by gi chng ta s tin hnh vn dng kin thc hc vo thc t bi tp

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Kin thc cn nm vng

Mc tiu: Cng c li nhng kin thc c bn v v nguyn t

-Gv pht vn hs v phn kin thc hc:

+ Th t mc nng lng?

+ C bao nhiu loi phn lp, s electron ti a trn mi phn lp?

+ Vi n

4 th s electron ti a trn mt lp c tnh nh th no?

+ Da vo u ta bit c h ca nguyn t?

+ c im lp electron ngoi cng?

+ Gv thng tin v s to thnh ion

A/ KIN THC CN NM VNG:

1/ Th t cc mc nng lng: 1s2s2p3s3p4s3d4p5s4d5p6s4f5d6p7s

2/ S e ti a trong:

Lp th n (=1,2,3,4) l 2n2e.

Phn lp: s2 , p6 , d10 , f14 .

3/ Electron c mc nng lng cao nht phn b vo phn lp no th chnh l h ca nguyn t.

4/ Lp e ngoi cng quyt nh tnh cht ha hc ca nguyn t, s bo ha bn vi 8e( Tr He, 2e ngoi cng).

Hot ng 2: Bi tp v cu hnh e

Mc tiu: Rn luyn k nng vit cu hnh electron v xc nh tnh cht c bn ca nguyn t

4 nhm tho lun lm 4 bi tp (5)

( i din mi nhm ln bng trnh by, nhm khc nhn xt

( Gv nhn xt, ging gii

BT4/30SGK:

Cu hnh e:

226262

122334

sspsps

a) C 4 lp electron

b) Lp ngoi cng c 2 e

c) Nguyn t l kim loi

BT6/30SGK:

a) 15e

b) 15

c) lp th 3

d) C 3 lp e, Lp th nht c 2e, lp th 2 c 8e, lp th 3 c 5e

e) l phi kim v c 5e lp ngoi cng

BT8/30SGK:

a)

21

12

ss

b)

223

122

ssp

c)

226

122

ssp

d)

22623

12233

sspsp

e)

22625

12233

sspsp

g)

22626

12233

sspsp

Hot ng 3: Bi tp v ng v

Mc tiu: Vn dng kin thc v ng v gii bi tp tm nguyn t khi trung bnh, nguyn t khi ca mt ng v cha bit

BT1: Brm c 2 ng v, trong ng v 79Br chim 54,5%. Xc nh ng v cn li, bit nguyn t khi trung bnh ca Br l 79,91.

BT2: Clo c 2 ng v l

3537

1717

;

ClCl

. T l s nguyn t ca 2 ng v ny l 3:1. Tnh nguyn t lng trung bnh ca clo?

- Phn nhm chn, l tho lun 2 bi tp

- Gio vin ch nh i din bt k ca 2 nhm ln bng

- Hc sinh khc theo di, nhn xt

- Gio vin nhn xt, nh gi

BT1: Phn trm ng v th hai l

100 - 54,5 = 45,5%

Gi M l nguyn t khi ca ng v th 2, ta c:

_

79.54,5.45,5

79,91

100

M

A

+

==

( M= 81(u)

BT2: Nguyn t khi trung bnh ca Clo:

_

35.337.1

35,5

31

A

+

==

+

(u)

4. Cng c: Cu hnh electron ca nguyn t M sau khi i 1e l

22626

12233

sspsp

. Hy vit cu hnh electron ca nguyn t, cho bit in tch ht nhn, s proton, ntron ca nguyn t M v tnh cht ho hc c bn ca nguyn t M?

5. Dn d: Lm bi tp

- SGK: 1,2,3,5,7,9/30

- SBT: 1.51(1.57/11,12

- Gv hng dn bi tp v nh: Clo c 2 ng v l

3537

1717

;

ClCl

. Hy tnh s nguyn t

35

17

Cl

c trong 5,85 g NaCl, bit rng nguyn t khi trung bnh ca clo l 35,5.

Rt kinh nghim:

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Tit th 11: LUYN TP: CU TO V NGUYN T(tit 2)




- c im ca cc loi ht trong nguyn t

- Rn luyn k nng vit cu hnh electron

- Rn luyn k nng tnh ton ho hc v thnh phn nguyn t

I. MC TIU:

1.Kin thc: Cng c kin thc v thnh phn cu to nguyn t v vit cu hnh electron

2.K nng:

- Rn luyn k nng vit cu hnh electron nguyn t

- Rn luyn k nng tnh ton ho hc v cc loi ht, s khi,....

3.Thi : Pht huy kh nng t duy ca hc sinh


III. CHUN B GIO C:

*Gio vin: Gio n, chn bi tp

*Hc sinh: n bi c, lm bi tp trc khi n lp

IV. TIN TRNH BI DY:



3.Bi mi:

a) t vn : Thnh phn cu to nguyn t gm cc loi ht no? K hiu, c im?

- Hc sinh tr li

( l nhng iu chng ta cn nm vng p dng gii quyt cc bi ton sau

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Bi tp v tng s ht c 2 d kin

Mc tiu: Hs bit cch tnh ton cc loi ht, s khi, ...trong nguyn t da vo c im ca cc loi ht bng cch gii h 2 phng trnh

Bi tp1: Nguyn t X c tng s ht bng 60. Trong s ht notron bng s ht proton. X :

a

r

40

18

A

b

K

39

19

c

Ca

40

20

d

Sc

37

21

HD:-Trong nguyn t c cc loi ht no?

- Hs tr li

- Tng s ht l 2Z + N

- Hs gii, trnh by( Gv nhn xt

Bi tp2 Mt nguyn t X c tng s cc ht bng 115. S ht mang in nhiu hn s ht khng mang in l 25. Tim Z, A, vit cu hnh e?

HD: S ht mang in gm c e v p, ht khng mang in l n ( lp phng trnh th 2 ri gii tng t bi 1

Bt1:

Tng s ht = S p + s e + s ntron = 60

2Z + N = 60 (1)

M: S n = S p ( N = Z, thay vo (1) ta c:

3Z = 60 ( Z = 60/3 = 20

Vy X l Ca (p n c)

Bt2: Tng s ht = S p + s e + s ntron = 115

2Z + N = 115 (1)

M s ht mang in nhiu hn s ht khng mang in 25 nn: 2Z N = 25 (2)

T (1) v (2) ta c hpt:

2Z + N = 115 (1)

2Z N = 25 (2)

( 4Z = 140 ( Z = 140/4 = 35

( N = 115 2.35 = 45

Vy A = Z + N = 35 + 45 = 80

Cu hnh e:

226261025

12233344

sspspdsp

Hot ng 1: Bi tp v tng s ht c 1 d kin

Mc tiu: Hs bit cch tnh ton cc loi ht, s khi, ...trong nguyn t da vo c im ca cc loi ht bng cch kt hp phng trnh v bt phng trnh

Bi 1: Tng s ht proton, ntron v electron trong nguyn t ca mt nguyn t X l 13 . S khi ca nguyn t X l bao nhiu?

HD: Kt hp iu kin nguyn t bn:

1

1,5

N

Z

kt hp vi phng trnh tng s ht gii

Bi 2:Tng s ht proton, ntron v electron ca mt nguyn t mt nguyn t X l 21. S hiu nguyn t ca nguyn t X l bao nhiu?

HD: Tng t bi 1

BT1: Tng s ht: 2Z + N = 13( N = 13- 2Z (1)

Li c: 1

1,5

N

Z

(2)

Kt hp (1) v (2) ta tm c: 3,7

4,3

Z

Z l mt s nguyn dng nn ta chn Z = 4

( N = 13 2.4 = 5

Vy s khi A = 4 + 5 = 9

BT2: Tng s ht: 2Z + N = 21( N = 21- 2Z (1)

Li c: 1

1,5

N

Z

(2)

Kt hp (1) v (2) ta tm c: 6

7

Z

Z l mt s nguyn dng nn ta chn Z = 6

hoc Z = 7

4. Cng c: Lm bi tp s 4/28 SGK

5. Dn d: n li kin thc chng I chun b kim tra 1 tit

Rt kinh nghim:

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Tit th 12: KIM TRA 1 TIT- LN 1

I. MC TIU KIM TRA:

- Kim tra kh nng tip thu kin thc ca hc sinh v thnh phn nguyn t; ht nhn nguyn t-nguyn t ho hc-ng v; cu to v nguyn t; cu hnh e nguyn t

- Kim tra k nng gii bi ton xc nh loi ht trong nguyn t; in tch ht nhn; tnh nguyn t khi trung bnh; s khi; vit cu hnh e nguyn t

II. NI DUNG KIN THC V K NNG CN KIM TRA:

1. Kin thc:

1.1/. Thnh phn nguyn t: c im cc loi ht cu to nn nguyn t

1.2/. Ht nhn nguyn t- nguyn t ho hc - ng v:

1.2.1. c trng ca ht nhn nguyn t

1.2.2. ng v- nguyn t khi- nguyn t khi trung bnh

1.3/. Cu to v nguyn t:

1.3.1. Cu to v nguyn t

1.3.2. S e ti a trn mt lp, phn lp

1.4/. Cu hnh e nguyn t:

2. K nng:

2.1. Xc nh s ht p, e, n, s khi, in tch ht nhn, s n v in tch ht nhn,...

2.2. Xc nh nguyn t khi, nguyn t khi trung bnh, % cc ng v

2.3. Vit cu hnh e nguyn t

2.4. Xc nh loi nguyn t

III. HNH THC KIM TRA: Va trc nghim, va t lun

IV. LP MA TRN KIM TRA:

Ni dung

Nhn bit

Thng hiu

Vn dng

Cp thp

Cp cao

TN

TL

TN

TL

TN

TL

TN

TL

Thnh phn nguyn t

Cu to nguyn t

Cu to nguyn t

Tng s ht

S cu

S im

1cu5

(0,5)

1cu7

(0,5)

1cu2

(2)

Ht nhn nguyn t -NTHH - ng v

Nhn ra ng v

Ht nhn nguyn t

Tnh % ng v

Tnh % ng v

2.2

S cu

S im

1cu4

(0,5)

1cu3 (0,5)

1cu (0,5)

1/2cu3

(1)

1/2cu3

(1)

Cu to v nguyn t

S e ti a trn phn lp, lp

S cu

S im

2cu1,8

(1)

Cu hnh e nguyn t

Xc nh s e lp ngoi cng

Vit cu hnh e, xc nh KL, PK

S cu

S im

1cu2

(0,5)

1cu1

(2)

V. KIM TRA V HNG DN CHM:

1. kim tra: (km theo)

2. Hng dn chm:

* 1:

- Phn trc nghim: 0,5/1cu

Cu

1

2

3

4

5

6

7

8

p n

A

B

C

A

D

C

C

C

- Phn t lun:

Cu 1: Mi nguyn t 0,5

2262

12

:1223

Mgssps

( Kim loi v lp e ngoi cng c 2e

22623

15

:12233

Psspsp

( Phi kim v lp e ngoi cng c 5e

22625

17

:12233

Clsspsp


2262682

28

:1223334

Nisspspds

(Kim loi v lp e ngoi cng c 2e

Cu 2: Ta c: 2Z + N = 54 (1) (0,5)

li c: 2Z N = 14 (2) (0,5)

T (1) v (2) ta c hpt:

254

214

ZN

ZN

+=

-=

(

17

20

Z

N

=

=

(0,5)

S khi A = Z + N = 17 + 20 = 37 (0,25)

Cu hnh e:

22625

12233

sspsp

(0,25)

Cu 3:

a) Tnh thnh phn phn trm:

Gi x l %

63

Cu

( %

65

Cu

l 100-x

Ta c:

63.65.(100)

63,54

100

73

Cu

xx

A

x

+-

==

=

Vy %

63

Cu

l 73%; %

65

Cu

l 27% (1)

b) S mol Cu2O =

14,4

0,1()

144

mol

=

(0,25)

C 1 mol Cu2O c 2 mol Cu

0,1 mol Cu2O c 2.0,1 = 0,2 mol Cu (0,25)

Tng s nguyn t Cu = 0,2. 6,02.1023=1,204.1023 (nguyn t) (0,25)

M

63

Cu

chim 73% nn s nguyn t

63

Cu

=

23

20

1,204.10.73

878,92.10

100

=

(nguyn t) (0,25)

* 2:

- Phn trc nghim: 0,5/1cu

Cu

1

2

3

4

5

6

7

8

p n

B

C

C

D

B

C

D

B

- Phn t lun:

Cu 1: Mi nguyn t 0,5

2261

11

:1223

Nassps


22621

13

:12233

Alsspsp


22624

16

:12233

Ssspsp


2262662

26

:1223334

Fesspspds

(Kim loi v lp e ngoi cng c 2e

Cu 2: Ta c: 2Z + N = 58 (1) (0,5)

li c: 2Z N = 18 (2) (0,5)

T (1) v (2) ta c hpt:

258

218

ZN

ZN

+=

-=

(

19

20

Z

N

=

=

(0,5)

S khi A = Z + N = 19 + 20 = 39 (0,25)

Cu hnh e:

226261

122334

sspsps

(0,25)

Cu 3:

a) Tnh thnh phn phn trm:

Gi x l %

35

Cl

( %

37

Cl

l 100-x

Ta c:

35.37.(100)

35,5

100

75

Cl

xx

A

x

+-

==

=

Vy %

35

Cl

l 75%; %

37

Cl

l 25% (1)

b) S mol FeCl2 =

12,7

0,1()

127

mol

=

(0,25)

C 1 mol FeCl2 c 2 mol Cl

0,1 mol FeCl2 c 2.0,1 = 0,2 mol Cl (0,25)

Tng s nguyn t Cl = 0,2. 6,02.1023=1,204.1023 (nguyn t) (0,25)

M

35

Cl

chim 75% nn s nguyn t

35

Cl

=

23

20

1,204.10.75

903.10

100

=

(nguyn t) (0,25)

VI. KT QU KIM TRA V RT KINH NGHIM:

1. Kt qu kim tra:

Lp

0( theo chiu tng ca Z => tnh PK tng dn Si < P < S

_ N, P, As thuc cng nhm A => theo chiu tng ca Z => tnh PK tng dn As < P < N

4. Cng c:

Cu 1: Nguyn t ca mt nguyn t c cu hnh e lp ngoi cng l

22

33

sp

. Hy xc nh v tr v tnh cht ho hc c bn ca nguyn t ?

Cu 2: Mt nguyn nm chu k 3, nhm VIA ca BTH. Hy xc nh cu to nguyn t ca nguyn t ?

5. Dn d:

- Hc bi

- Lm bi tp SGK, SBT

- n li ton b chng II

Rt kinh nghim:

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Tit th 19: LUYN TP

BNG TUN HON, S BIN I TUN HON CU HNH ELECTRON NGUYN T V TNH CHT CA CC NGUYN T HO HC (tit 1)



- Cu to BTH

- S bin i tun hon cu hnh e nguyn t cc nguyn t ho hc

- Quy lut bin i bn knh nguyn t, m in, tnh kim loi, tnh phi kim ca cc nguyn t trong chu k, nhm A

- Quy lut bin i ho tr, tnh axit- baz, ho tr cao nht vi oxi v hiro ca mt s nguyn t trong chu k, nhm

- nh lut tun hon

Cng c kin thc v bng tun hon

I. MC TIU:


- Bng tun hon

- S bin i tun hon cu hnh electron nguyn t

- S bin i tun hon tnh cht (Tnh kim loi, phi kim, m in, bn knh nguyn t) ca nguyn t v tnh axit, baz ca hp cht

- nh lut tun hon

2.K nng: H thng ho kin thc

3.Thi : Tch cc, ch ng

II.PHNG PHP GING DY: Pht vn

III. CHUN B GIO C:

*Gio vin: Gio n, cu hi trc nghim


IV. TIN TRNH BI DY:


2.Kim tra bi c: Kim tra trong qu trnh luyn tp

3.Bi mi:

a) t vn : Tng hp kin thc chng II

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Kin thc cn nm vng v bng tun hon

Mc tiu:Cng c, h thng ho kin thc v bng tun hon

Gio vin pht vn vi hc sinh tr li mt s cu hi sau:

- Cc nguyn t ho hc c xp vo BTH theo nhng nguyn tc no?

- Hng v ct tng ng vi thnh phn no trong BTH?

- nguyn t cho ta bit nhng thng tin no?

- C tt c bao nhiu chu k?

- Chu k no l chu k nh, chu k ln?

- Nhng nguyn t nm trong mt chu k c c im g?

- Nhng nguyn t nh th no c xp vo cng mt nhm?

- Phn loi nhm?

- Nguyn t s thuc nhm no?

- Nguyn t p thuc nhm no?

- Xc nh s th t nhm da vo u?

- Nhm B gm nhng nguyn t thuc h g?

- Nhng nguyn t f nm u trong BTH?

- Cch xc nh s TT cc nguyn t nhm B?

A.KIN THC CN NM VNG:

1,Cu to bng tun hon:

a.Nguyn tc sp xp cc nguyn t trong BTH: 3 nguyn tc:

- Cc nguyn t c sp xp theo chiu tng dn ca in tch ht nhn nguyn t.

- Cc nguyn t c cng s lp e trong nguyn t c xp thnh 1 hng (chu k)

- Cc ngyn t c s e ho tr trong nguyn t nh nhau c xp thnh 1 ct (Nhm).

b. nguyn t: Mi nguyn t c xp vo 1 gi l nguyn t

c.Chu k:

-Mi hng l 1 chu k

-C 3 chu k nh : 1,2,3

-C 4 chu k ln: 4,5,6,7

( Nguyn t cc nguyn t thuc 1 chu k c s lp e nh nhau

d.Nhm:

*Nhm A: Gm chu k nh v chu k ln ,t IA ( VIIIA.-Nguyn t s thuc nhm IA,IIA.

-Nguyn t p thuc nhm IIIA ( VIIIA.

*Nhm B: (IIIB (VIIIB;IB,IIB)

-Nguyn t d,f thuc chu k ln

Hot ng 2: Kin thc cn nm vng v s bin i tun hon

Mc tiu: Cng c kin thc v s bin i tun hon cu hnh e, tnh KL, tnh PK, bn knh nguyn t, gi tr m in; Nm ni dung nh lut tun hon

Gio vin pht vn vi hc sinh tr li mt s cu hi sau:

- S e lp ngoi cng ca nguyn t cc nguyn t bin i nh th no trong mt chu k?

- Trong mt chu k, tnh KL v PK, bn knh nguyn t, gi tr m in bin i nh th no?

( H thng thnh bng

- Gv: Pht vn hs v cng thc oxit cao nht, hp cht kh vi hiro

(S bin i tnh axit, baz?

2.S bin i tun hon:

a.Cu hnh electron nguyn t:

S e ngoi cng ca nguyn t cc nguyn t mi chu k tng t 1(8 thuc cc nhm t IA( VIIIA.Cu hnh e ca nguyn t cc nguyn t bin i tun hon

b.S bin i tun hon tnh KL, PK,Rnguyn t,gi tr A ca cc nguyn t c tm tt trong bng sau:

Rnguyn t

KL

PK

A

Chu k

Gim

Gim

Tng

Tng

Nhm

Tng

Tng

Gim

Gim

Gv yu cu hs nu nh lut tun hon

3.nh lut tun hon:

- Tnh cht ca cc nguyn t v n cht cng nh thnh phn v tnh cht ca cc hp cht to nn t cc nguyn t bin i tun hon theo chiu tng dn ca THN nguyn t.

Hot ng 3: Vn dng

Mc tiu: Cng c kin thc v BTH

Gio vin c cu hi, hc sinh tr li, gii thch (Gio vin nhn xt, kt lun

Hs tho lun 3( Hai hs ln bng, hs khc nhn xt, b sung

Gv nh gi

Cu 1: Tm cu sai trong nhng cu di y:

A. Trong chu k, cc nguyn t c xp theo chiu in tch ht nhn tng dn

B. Trong chu k, cc nguyn t c xp theo chiu s hiu nguyn t tng dn

C. Nguyn t cu cc nguyn t trong cng mt chu k c s e bng nhau

D. Chu k thng bt u l kim loi kim, kt thc l mt kh him (tr chu k 1 v chu k 7 cha hon thnh)

Cu 2: S hiu nguyn t Z ca cc nguyn t X, A, M, Q ln lt l 6, 7, 20, 19. Nhn xt no sau y ng?

A. X thuc nhm VA B. A, M thuc nhm IIA

C. M thuc nhm IIB D. Q thuc nhm IA

Cu 3: S hiu nguyn t Z ca cc nguyn t X, A, M, Q ln lt l 6, 7, 20, 19. Nhn xt no sau y ng?

A. C 4 nguyn t trn thuc mt chu k B. M, Q thuc chu k 4

C. A, M thuc chu k 3 D. Q thuc nhm IA

Cu 4: Trong bng tun hon, nguyn t X c STT 16, nguyn t X thuc:

A. Chu k 3, nhm IVA B. Chu k 4, nhm VIA

C. Chu k 3, nhm VIA D. Chu k 4, nhm IIIA

Cu 5: Theo quy lut bin i tnh cht n cht ca cc nguyn t trong bng tun hon th:

A. Phi kim mnh nht l it B. Kim loi mnh nht l Liti

C. Phi kim mnh nht l flo D. Kim loi yu nht l cesi

Cu 6: Cc nguyn t c sp xp theo chiu tnh phi kim tng dn (t tri sang phi) nh sau:

A. F, Cl, S, Mg C. Cl, F, Mg, S

B. Mg, S, Cl, F D. S, Mg, Cl, F

Cu 7: Cc nguyn t halogen c sp xp theo chiu bn knh nguyn t gim dn (t tri sang phi) nh sau:

A. I, Br, Cl, F C. F, Cl, Br, I

B. I, Br, F, Cl D. Br, I, Cl, F

Cu 8: Hai nguyn t A, B nm 2 chu k lin tip trong mt nhm A. Tng s ht proton trong 2 nguyn t A, B l 24. Tm A, B? p n: O(Z=8) v S(Z=16)

Cu 9: Hai nguyn t A, B nm k tip nhau trong cng mt chu k; tng s n v in tch ht nhn trong hai ht nhn ca 2 nguyn t l 25. Xc nh A,B?

Cu 10: Vit cu hnh e ca ion: O2-; Mg2+; Zn2+; Fe2+

4. Cng c:

- Nguyn tc sp xp cc nguyn t trong BTH

- Cu to BTH

- S bin i tun hon cu hnh e, bn knh nguyn t, tnh cht

- nh lut tun hon

5. Dn d:

- Hc bi, nm k kin thc v BTH

- Lm bi tp: 5,6,7,8,9/54SGK

Rt kinh nghim:

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Tit th 20: LUYN TP

BNG TUN HON, S BIN I TUN HON CU HNH ELECTRON NGUYN T V TNH CHT CA CC NGUYN T HO HC (tit 2)



- Hp cht oxit cao nht, hp cht kh vi hiro ca nguyn t

- Bng tun hon

- Cng c kin thc v hp cht cc nguyn t

- Rn luyn k nng gii ton ho

I. MC TIU:

1.Kin thc: Cng c kin thc v hp cht oxit cao nht v hp cht kh vi hiro ca cc nguyn t ho hc

2.K nng: Rn luyn k nng:

- Xc nh ho tr ca nguyn t da vo cng thc oxit cao nht v hp cht kh vi hiro

- Gii bi ton xc nh nguyn t

3.Thi : Tch cc trong hot ng nhm


III. CHUN B GIO C:

*Gio vin: Gio n


IV. TIN TRNH BI DY:


2.Kim tra bi c: Kt hp trong bi luyn tp

3.Bi mi:

a) t vn : Hy vit cng thc hp cht kh vi hiro, cng thc oxit cao nht ca cc nguyn t tng ng c cng thc oxit cao nht v hp cht kh vi hiro nh sau: RH4, R2O5, RO2, RH? ( vo bi

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Bi ton tng s ht kt hp v tr nguyn t trong BTH

Mc tiu: Rn luyn k nng gii ton ho kt hp tng s ht trong nguyn t v k nng xc nh v tr nguyn t trong bng tun hon

BT5/54SGK: Tng s ht trong mt nguyn t ca mt nguyn t thuc nhm VIIA l 28.

a) Tnh nguyn t khi

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

DH: Gii ging nh mt bi tng s ht bnh thng, so kt qu vi v tr bi cho chn kt qu ng

Hs ln bng, hs khc nhn xt

Gv nh gi

BT5/54:

Tng s ht= 2Z + N = 28 (N= 28 2Z (1)

Kt hp iu kin:

11,51,5

N

ZNZ

Z

(2)

T (1) v (2) ta c:

2821,5

ZZZ

-

(

89,3

Z

Nu Z=8:

224

122

ssp

thuc nhm VIA (loi)

Nu Z=9:

225

122

ssp

thuc nhm VIIA (chn)

(N = 28- 2.9= 10

a) Nguyn t khi = A= 19

b) Cu hnh e:

225

122

ssp

Hot ng 1: Bi ton xc nh nguyn t da vo v tr trong BTH

Mc tiu: Rn luyn k nng gii bi ton xc nh nguyn t cha bit da vo pthh v v tr nguyn t trong bng tun hon

BT9/54SGK: Khi cho 0,6 gam mt kim loi nhm IIA tc dng vi nc to ra 0,336 lt kh hiro ktc. Xc nh kim loi ?

HD: Kim loi Nhm IIA c ho tr II, Gi kim loi l M v vit phng trnh ging nh mt nguyn t bnh thng bit tm ra khi lng nguyn t v xc nh nguyn t

HS ln bng, hs khc nhn xt

BT9/54:

S mol kh hiro to thnh:

0,336

0,015

22,4

nmol

==

Kim loi thuc nhm IIA nn c ho tr II

M + 2H2O ( M(OH)2 + H2

M(g) 2(g)

0,6(g) 2.0,015(g)

(

20,6.2

40

0,60,030,03

M

M

===

Vy kim loi l Canxi

Hot ng 1: Bi ton xc nh nguyn t da vo cng thc oxit cao nht

v hp cht kh vi hiro

Mc tiu: Rn luyn k nng chuyn i gia cng thc oxit cao nht v hp cht kh vi hiro, gii bi ton da vo thnh phn phn trm nguyn t trong phn t

BT7/54SGK: Oxit cao nht ca mt nguyn t l RO3, trong hp cht ca n vi hiro c 5,88%H v khi lng. Xc nh nguyn t khi ca nguyn t ?

HD: Da vo cng thc oxit cao nht xc nh v tr ca nguyn t (Xc nh hp cht kh vi hiro v gii

BT8/54SGK: Hp cht kh vi hiro ca mt nguyn t l RH4. Oxit cao nht ca n cha 53,3% oxi v khi lng. Tm nguyn t khi ca nguyn t ?

HD: Da vo hp cht kh vi hiro xc nh v tr nguyn t suy ra cng thc oxit cao nht v gii

Hs ln bng, hs khc lm vo v, nhn xt(gv nh gi

BT7/54:

Oxit cao nht ca R l RO3 nn R thuc nhm VIA

Do hp cht vi hiro ca R l RH2

Ta c:

2

2

5,8825,882.1002.5,88

32

10021005,88

H

R

RHR

M

M

MM

-

====

+

Vy R l lu hunh

BT8/54:

Hp cht kh vi hiro ca R l RH4 nn R thuc nhm IVA. Do , cng thc oxit cao nht l RO2

Ta c:

2

2.

53,33253,332.10032.53,3

28

1003210053,3

O

R

ROR

M

M

MM

-

====

+

Vy nguyn t khi ca R l 28

4. Cng c:

- Mun xc nh nguyn t cn xc nh i lng no?

- Chuyn i qua li gia cng thc oxit cao nht v hp cht kh vi hiro

5. Dn d: n tp ton b chng II, chun b kim tra 1 tit

Rt kinh nghim:

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Th ngy thng nm 2011

Tit th 21: KIM TRA 1 TIT- LN 2

I. MC TIU KIM TRA:

- Kim tra kh nng tip thu kin thc ca hc sinh v bn tun hon, s bin i tun hon cu hnh e, tnh cht cc nguyn t v hp cht

- Kim tra k nng vn dng kin thc vo gii bi tp

II. NI DUNG KIN THC V K NNG CN KIM TRA:

1. Kin thc:

1.1/. ngha bng tun hon:

- Quan h gia v tr nguyn t v cu hnh e nguyn t

- Quan h gia v tr v tnh cht nguyn t v hp cht

1.2/. S bin i tun hon:

- Cu hnh e lp ngoi cng

- Tnh cht nguyn t

- Bn knh nguyn t

- Tnh cht ca hp cht

2. K nng:

2.1 Xc nh v tr nguyn t trong BTH

2.2. Xc nh cu to nguyn t

2.3. Xc nh tn nguyn t trong cng mt chu k, 2 chu k lin tip

2.4. Xc nh loi nguyn t da vo phn ng ho hc

2.5. Vit cu hnh e ca ion

III.HNH THC KIM TRA: T lun (10 cu)

IV.LP MA TRN KIM TRA:

Mc

Ni dung

Bit

Hiu

Vn dng

thp

Vn dng

cao

1) Xc nh v tr nguyn t trong BTH

Cu 1

2) Xc nh cu to nguyn t

Cu 2

3) Sp xp nguyn t theo chiu tng gim tnh cht

Cu 3

4) Cho cu hnh e nguyn t, vit CT oxit, hp cht kh vi hiro, tnh cht

Cu 4

5) Sp xp cc nguyn t theo chiu s e lp ngoi cng tng dn

Cu 5

6) Xc nh nguyn t nm 2 trong 1 chu k

Cu 6

7) Xc nh tn kim loi + dd axit

Cu 7

8) Cho CT oxit cao nht... (Tm R

Cu 8

9) Xc nh tn 2 nguyn t 2 chu k lin tip

Cu 9

10) Vit cu hnh e nguyn t khi bit cu hnh e ca ion

Cu 10

V. KIM TRA V HNG DN CHM:

3. kim tra:

Cu 1: Nguyn t ca mt nguyn t c s proton trong ht nhn l 16. Hy xc nh v tr ca nguyn t trong bng tun hon?

Cu 2: Nguyn t ca mt nguyn t A c cu hnh electron lp ngoi cng l

23

33

sp

. Hy cho bit:

a) S proton, s electron trong nguyn t A ?

b) S lp electron trong nguyn t?

c) S electron lp ngoi cng trong nguyn t?

Cu 3: Sp xp cc nguyn t sau theo chiu tnh kim loi tng dn (t tri qua phi):

816617

;;;

ABCD

.

Cu 4: Nguyn t ca mt nguyn t R c cu hnh electron l

22625

12233

sspsp

. Hy cho bit:

a) Cng thc oxit cao nht ca R, tnh cht?

b) Cng thc hp cht hiroxit ca R, tnh cht?

Cu 5: Sp xp cc nguyn t sau theo chiu s electron lp ngoi cng gim dn:

131967

;;;

XYZT

Cu 6: Cho 2 nguyn t A, B nm 2 lin tip trong cng mt chu k ca bng tun hon. Hy xc nh s th t ca 2 nguyn t, bit rng tng s proton trong 2 nguyn t A, B l 23?

Cu 7: Ho tan hon ton 4,6 gam mt kim loi kim vo nc thu c dung dch A. trung ho ht dung dch A cn 100 ml dung dch HCl 2M. Xc nh tn nguyn t kim loi?

Cu 8: Cng thc oxit cao nht ca mt nguyn t l RO3. Trong hp cht kh vi hiro, R chim 94,12% v khi lng. Hy xc nh tn nguyn t R?

Cu 9: Hai nguyn t A v B thuc cng mt phn nhm chnh v hai chu k lin tip trong bng tun hon. Tng s ht proton trong hai ht nhn nguyn t A v B l 32. Xc nh tn A, B v vit cu hnh electron nguyn t?

Cu 10: Hy vit cu hnh electron nguyn t ca:

Ion M2+ c cu hnh electron phn lp ngoi cng l 3d5

Ion X- c cu hnh electron phn lp ngoi cng l 2p6

4. Hng dn chm: Mi cu 1 im

Cu 1: th 16, chu k 3, nhm VIA

Cu 2:

a) S proton, s electron trong nguyn t A =15

b) S lp electron trong nguyn t A = 3

c) S electron lp ngoi cng trong nguyn t = 5

Cu 3: AY

Cu 6: Gi s ZB>ZA

Ta c:

23

1

BA

BA

ZZ

ZZ

+=

-=

(

11

12

A

B

Z

Z

=

=

Cu 7: Ta c:

0,1.20,2()

HCl

nmol

==

Gi A l kim loi cn tm

PT: 2A + 2H2O ( 2AOH + H2 (1) v AOH + HCl ( ACl + H2O (2)

Theo pt (2):

0,2

AOHHCl

nnmol

==

; Theo pt (1):

0,2

AAOH

nnmol

==

(

4,6

23

0,2

A

M

==

(g/mol) Vy kim loi l Natri (Na)

Cu 8:

Cng thc oxit cao nht l RO3 nn R thuc nhm VIA

( Hp cht kh vi hiro: RH2

Ta c: %R =94,12( %H= 100-94,12 = 5,88

(

%94,12

32

%25,882

RR

R

H

MM

R

M

HM

===

Vy R l lu hunh (S)

Cu 9: Gi s ZB>ZA ; Hai nguyn t cng nm trong mt nhm A v 2 chu k lin tip 3,4 hn km nhau 8 proton. Do ta c:

2262

226262

3212:1223

8

20:122334

ABA

BA

B

ZZZssps

ZZ

Zsspsps

+==

-=

=

Cu 10:

M:

2262652

1223334

sspspds

X:

225

122

ssp

VI.KT QU KIM TRA V RT KINH NGHIM:

3. Kt qu kim tra:

Lp

0( Qu trnh

2

Cu

+

nhn thm 2 electron gi l qu trnh kh

2

Cu

+

(s kh

2

Cu

+

).

Phn ng (2): Cht oxh l CuO, cht kh l Hiro.

Tm li:

+ Cht kh ( cht b oxh) l cht nhng electron.

+ Cht oxh ( Cht b kh) l cht thu electron.

+ Qu trnh oxh ( s oxh ) l qu trnh nhng electron.

+ Qu trnh kh (s kh ) l qu trnh thu electron.

2.Xt phn ng khng c oxi tham gia

2x1e

VD3: 2

0

Na

+

0

2

Cl

( 2

11

NaCl

+-

(3)

Phn ng ny c s thay i s oxi ha, s cho nhn electron:

0

Na

(

1

Na

+

+ 1e

0

Cl

+ 1e (

1

Cl

-

VD4:

0

2

H

+

0

2

Cl

( 2

11

HCl

+-

(4)

Trong phn ng (4) c s thay i s oxi ha ca cc cht, do cp electron gp chung lch v Clo.

VD 5:

35

43

NHNO

-+

(

1

2

NO

+

+ 2H

2

O

Phn ng (5) nguyn t N-3 nhng e, N+5 nhn e

( c s thay i s oxh ca mt nguyn t.

Hot ng 2: Phn ng oxi ho- kh

Mc tiu: Hiu th no l phn ng oxi ho- kh

- Nhn xt g v s oxi ho ca nguyn t trc v sau p trong cc pthh cc vd trn?

- Hs: u c s thay i s oxi ho ca nguyn t

(Nhng phn ng nh vy gi l phn ng oxi ho- kh

3.Phn ng oxi ho- kh

N: Phn ng oxh kh l phn ng ha hc, trong c s chuyn electron gia cc cht phn ng, hay p oxh kh l phn ng ha hc trong c s thay i s oxh ca mt s nguyn t.

Hot ng 3: ngha ca phn ng oxi ho- kh trong thc tin

Mc tiu: Bit c tm quan trng ca phn ng oxi ho kh trong thc tin

- Gv: Phn ng oxi ho kh c tm quan trng trong i sng v sn xut ( C th trong i sng, sn xut?

- Hs tr li

II. ngha ca phn ng oxi ho- kh trong thc tin (SGK)

4. Cng c: Lm BT 1,2/82 SGK

5. Dn d:

- Bi tp v nh : 3, 4, 5, 6 (SGK)

- Son phn: Lp pthh ca phn ng oxi ho- kh

Rt kinh nghim:

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Tit th 30: PHN NG OXI HO - KH (tit 2)


Kin thc mi cn hnh thnh

- Cht kh, cht oxi ho

- S kh, s oxi ho

- Phn ng oxi ho- Kh

- Cc bc lp PTHH ca phn ng oxi ho- kh

I. MC TIU:

1.Kin thc:Hiu c cc bc lp phng trnh phn ng oxi ho - kh,

2.K nng: Lp c phng trnh ho hc ca phn ng oxi ho - kh da vo s oxi ho (cn bng theo phng php thng bng electron).


II. TRNG TM: Cch lp phng trnh ha hc ca phn ng oxi ha - kh

III.PHNG PHP GING DY: Din ging pht vn - kt nhm

IV. CHUN B GIO C:

*Gio vin: Gio n, my chiu, giy A4


V. TIN TRNH BI DY:



Xc nh cht kh- cht oxi ho, s kh- s oxi ho trong cc phn ng sau?

1) 4NH3 + 5O2

,

o

txt

4NO + 6H2O

2) 2NH3 + 3Cl2 ( N2 + 6HCl

3.Bi mi:

a) t vn : Da vo bi c(Vi cc phn ng oxi ho kh nh th ny th chng ta c th nhm cn bng nhng i vi mt s phn ng oxi ho kh, vd nh phn ng:

8Al + 30HNO3 ( 8Al(NO3)3 + 3N2O + 15H2O th vic nhm cn bng l mt vic rt kh khn. V vy ngi ta nghin cu v tm ra mt cch cn bng p dng chung cho cc phn ng oxi ho kh m hm nay c s gii thiu vi cc em, l cch lp PTHH ca p oxi ho kh ( Cn bng theo phng php thng bng electron)

b) Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Lp phng trnh ho hc ca phn ng oxi ho kh (cn bng theo phng php thng bng electron)

Mc tiu: Hiu c cc bc lp phng trnh ho hc ca phn ng oxi ho kh

Gio vin trnh chiu tng bc lp PTHH ng thi yu cu hc sinh thc hin cc bc tng ng cn bng phn ng

NH3 + Cl2 ( N2 + HCl

II. Lp PTHH ca phn ng oxi ho kh:

Bc 1: Xc nh s oxh ca cc nguyn t tm cht oxi ho v cht kh:

Bc 2: Vit qu trnh oxh v qu trnh kh, cn bng mi qu trnh

Bc 3: Tm h s thch hp cho cht oxh v cht kh sao cho tng s electron cho bng tng s electron nhn

Bc 4: t h s ca cc cht oxh v kh vo s phn ng, t tnh ra h s cc cht khc. Kim tra cn bng s nguyn t ca cc nguyn t v cn bng in tch hai v hon thnh PTHH

V d: Lp PTHH ca phn ng oxi ho kh sau:

NH3 + Cl2 ( N2 + HCl

Bc 1:

310011

322

NHClNHCl

-++-

++

S oxh ca N tng t -3 ln 0: Cht kh

S oxh ca Cl gim t 0 xung -1: Cht oxh

Bc 2:

Qu trnh oxh:

30

2

26

NNe

-

+

Qu trnh kh:

01

2

22

CleCl

-

+

Bc 3:

Qu trnh oxh:

30

2

26

NNe

-

+

x 1

Qu trnh kh:

01

2

22

CleCl

-

+

x 3

3001

22

236

NClNCl

--

++

Bc 4: 2NH3 + 3Cl2 ( N2 + 6HCl

Hot ng 2: Vn dng

Mc tiu: Rn luyn k nng cn bng phn ng oxi ho kh theo phng php thng bng electron

Hc sinh tho lun nhm lp PTHH ca cc phn ng oxi ho kh:

1) Mg + AlCl3 (MgCl2 + Al

2) KClO3 ( KCl + KClO4

3) KClO3 ( KCl + O2

4) FeS2 + O2 ( Fe2O3 + SO2

5) MnO2 + HCl ( MnCl2 + Cl2 + H2O

Gv trnh chiu kt qu ca tng nhm, i din nhm trnh by, cc nhm khc nhn xt ( Gv ging gii, ch cho hc sinh cc loi p oxi ho kh

1)Phn ng n gin

2)Phn ng t oxi ho, t kh

3)Phn ng oxi ho kh ni phn t

4, 5) Phn ng oxi ho kh phc tp

Lp PTHH ca cc phn ng oxi ho kh sau:

1)

0320

32

MgAlClMgClAl

++

++

Mg l cht kh;

3

Al

+

(trong AlCl3) l cht oxi ho

02

2

MgMge

+

+

x 3

30

3

AleAl

+

+

x 2

0320

3232

MgAlMgAl

++

++

Phng trnh s l:

3Mg + 2AlCl3 (3MgCl2 + 2Al

2)

517

34

KClOKClKClO

+-+

+

5

Cl

+

(trong KClO3) va l cht kh va l cht oxh

51

6

CleCl

+-

+

x 1

57

2

ClCle

++

+

x 3

517

413

ClClCl

+-+

+

Phng trnh s l: 4KClO3 ( KCl + 3KClO4

3)

_1

50

2

3

KClOKClO

+

+

5

Cl

+

(trong KClO3) l cht oxi ha;

2

O

-

(trong KClO3) l cht kh

_1

5

6

CleCl

+

+

x 2

20

2

24

OOe

-

+

x 3

_1

520

2

2623

ClOClO

+-

++

Phng trnh s l: 2KClO3 ( 2KCl + 3O2

4)

2103242

22232

FeSOFeOSO

+-+-+-

++

21

,

FeS

+-

(trong FeS2) l cht kh;

0

2

O

l cht oxi ho

23

1

FeFee

++

+

14

2210

SSe

-+

+

2134

2

211

FeSFeSe

+-++

++

x 4

02

2

42

OeO

-

+

x 11

210342

22

4114822

FeSOFeSO

+-++-

+++

Phng trnh s l:

4FeS2 + 11O2 ( 2Fe2O3 + 8SO2

5)

4120

2

222

MnOHClMnClClHO

+-+

+++

4

Mn

+

(trong MnO2) l cht oxi ho;

1

Cl

-

(trong HCl) l cht kh

42

2

MneMn

++

+

x 1

10

2

22

ClCle

-

+

x 1

4120

2

2

MnClMnCl

+-+

++

Phng trnh s l:

MnO2 + 4HCl ( MnCl2 + Cl2 + 2H2O

4. Cng c: Cc bc lp PTHH ca phn ng oxi ho kh

5. Dn d:

- Bi tp v nh : 7, 8/83 (SGK)

- Son bi: Phn loi phn ng trong ho hc v c

Rt kinh nghim:

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Tit th 31: PHN LOI PHN NG TRONG HO HC V C


Kin thc mi cn hnh thnh

- Phn ng ho hp, phn ng phn hu, phn ng th, phn ng trao i

- Phn ng oxi ho- Kh

- Cc phn ng ho hc c chia thnh 2 loi: phn ng oxi ho - kh v khng phi l phn ng oxi ho - kh.

I. MC TIU:

1.Kin thc:Hiu c:

Cc phn ng ho hc c chia thnh 2 loi: phn ng oxi ho - kh v khng phi l phn ng oxi ho - kh.

2.K nng: Nhn bit c mt phn ng thuc loi phn ng oxi ho - kh da vo s thay i s oxi ho ca cc nguyn t.


II. TRNG TM: Phn loi phn ng thnh 2 loi.

III.PHNG PHP GING DY: Din ging pht vn - kt nhm

IV. CHUN B GIO C:

*Gio vin: Gio n, my chiu, giy A4


V. TIN TRNH BI DY:



Lp PTHH ca cc phn ng oxi ho kh sau:

1) KMnO4 (K2MnO4 + MnO2 + O2

2) NH3 + CuO ( Cu + N2 + H2O

3.Bi mi:

a. t vn : Phn ng 1 trong bi c, ngoi l mt phn ng oxi ho kh th n l loi phn ng no chng ta hc? Chng ta hc nhng loi phn ng ho hc no? Hs tr li ( By gi chng ta s tm hiu xem tng loi phn ng .

b. Trin khai bi

HOT NG THY V TR

NI DUNG KIN THC

Hot ng 1: Phn ng c s thay i s oxi ho v phn ng khng c s thay i s oxi ho

Mc tiu: Hiu c cc phn ng ho hc c chia thnh 2 loi: phn ng c s thay i s oxi ho v khng thay i s oxi ho

Chng ta bit v phn ng ho hp, phn hu, th, trao i. By gi chng ta s xt tng loi phn ng

- Gv cho 2 p, yu cu 2 hs ln bng xc nh s oxh cc nt (C nhn xt g v s oxh cc nt trc v sau p 2 phng trnh




I. PHN NG C S THAY I S OXH V PHN NG KHNG C S THAY I S OXH

1. Phn ng ha hp:

VD 1:

0012

22

2

22

HOHO

+-

+

S oxh ca hiro tng t 0 ( +1

S oxh ca oxi gim t 0 ( -2

VD2:

2242242

2

3

CaOCOCaCO

+-+-++-

+

S oxh ca cc nguyn t khng thay i.

( Nhn xt: Trong phn ng ha hp, s oxi ha ca cc nguyn t c th thay i hoc khng thay i.

2. Phn ng phn hy:

VD1:

5210

3

2

2KClO2KCl3O

+--

+

S oxh ca Oxi tng t -2 ln 0;

S oxi ha ca clo gim t +5 xung -1.

VD2:

2212212

2

2

Cu(OH)CuOHO

+-++-+-

+

S oxi ha ca cc nguyn t khng thay i.

( Nhn xt: Trong phn ng phn hy, s oxh ca c th thay i hoc khong thay i.

3. Phn ng th:

VD1:

0

o12

332

Cu2AgNOCu(NO)2Ag

++

++

S oxh ca ng tng t 0 ln +2;

S oxh ca H gim t +1 xung 0.

VD2:

0120

2

2

Zn2HClZnClH

++

++

S oxh ca tt ca Zn km tng ln t 0 ln +2;

S oxh ca hiro gim t +1 xung 0.

( Nhn xt: Trong ha hc v c, phn ng th bao gi cng c s thay i s oxh ca cc nguyn t.

4. Phn ng trao i:

VD1:

1521111152

33

AgNONaClAgClNaNO

++-+-+-++-

++

S oxi ha ca tt c ca tt c cc nguyn t khng thay i.

VD2:

1212122111

2

2

2NaOHCuClCu(OH)2NaCl

+-++-+-++-

++

S oxh ca tt c cc

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