t1 20 hoa 10 nc hai

GIO N HA 10 NNG CAO Hong T. Thanh Hi

Ngy son :10/08/2013 Bi: N TP U NM Tit: 01

I. MC TIU

1. Kin thc:

- Gip HS h thng cc kin thc ho hc c hc THCS c lin quan n chng trnh lp 10 .

-Phn bit c cc khi nim c bn v tru tng :nguyn t , nguyn t ho hc, phn t, n cht , hp cht , nguyn cht v hn hp .

2. K nng - Rn luyn k nng lp cng thc , tnh theo cng thc v PTHH, t khi ca cht kh .

- Rn luyn k nng chuyn i gia : M, n, m,V. . .

3. Thi

Gip HS nm li kin thc c mt cch h thng .

II. CHUN B - GV: h thng bi tp v cu hi gi - HS: n tp kin thc 8,9.III. HOT NG DY HC 1. n nh t chc : im danh.

2. Kim tra bi c: Kim tra trong ni dung n tp.

3. Ging bi mi

Tin trnh tit dy

Thi gianHot ng ca gio vinHot ng ca hc sinhNi dung

15Hot ng 1: n tp cc khi nimc bn.

- GV: Yu cu HS nhc li cc khi nim: nguyn t , nguyn t ho hc, phn t , n cht , hp cht , nguyn cht v hn hp?

- GV: a ra s phn bit cc khi nim .

GV: Yu cu HS a ra cc mi quan h gia cc i lng khi lng , khi lng mol , s phn t cht (A) v cht kh ktc.

- GV: a ra s mi quan h gia khi lng , khi lng mol , s phn t cht (A) v cht kh ktc.

- GV: T mi quan h gia n v V trong s ta c :

VA=VBnA = nB cng k t0,

- GV: Yu cu HS nhc li nh ngha v t khi ca cht kh .

- GV: Tnh MKK=?

- HS: Pht biu , a ra v d

- HS: Ghi cc cng thc :

-n = n/M

-n =V/22,4

-n = s nt, pt/N(N=6.1023)

- HS: Ghi cng thc.

- HS thc hin. I. n tp cc khi nim c bn :

1. Cc khi nim v cht :

n cht(1 ng t)

Ngt Ngt

Phn t Hp cht(2,3ngt)

Nguyn cht (1 cht)

Cht

Hn hp (2,3 cht)

2. Mi quan h gia khi lng , khi lng mol , s phn t cht (A) v cht kh ktc

m

n A

V

3.T khi ca kh A so vi kh B:

dA/B== =

(mA , mB , kl o cng T, p)

==29 g/mol

205Hot ng 2: Mt s bi tp p dng .

- GV: luyn tp mt s dng bi tp c bn c hc lp 8,9.

Bi 1:

a.Hy in vo trng ca bng sau cc s liu thch hp

S p

S n

S e

Nt 1

19

20

?

Nt 2

?

18

17

Nt 3

19

21

?

Nt 4

17

20

?

b. Trong 4 nguyn t trn , nhng cp nguyn t no thuc cng mt nguyn t ho hc ? v sao?

c. T 4 ng t trn c kh nng to ra c nhng n cht v hp cht ho hc no ?

Bi 2.

Xc nh khi lng mol ca cht hu c X , bit rng khi ho hi 3g X thu c th tch hi ng bng th tch ca 1,6g oxi trong cng iu kin .

Bi 3:

Xc nh dA/H2 , bit ktc 5,6lt kh A c khi lng 7,5 g

Bi 4:

Mt hn hp kh A gm SO2 v O2 c dA/CH4 = 3. Trn V lt O2 vi 20 lt hn hp A thu c hn hp B c dB/CH4= 2,5.Tnh V.

Hot ng 3: Cng c

GV:Nhc HS ni dung s luyn tp tit hai v yu cu HS n tp cc ni dung sau :

1. Cch tnh theo CT v tnh theo PTP trong bi ton ho hc

2. Cc CT v dd : tan , nng C% , nng CM

HS:in vo bng

- HS: Ngt 2,4 thuc cng 1 ngt ho hc v c cng s lp p l 17(Cl). Ngt 1,3 thuc cng 1 ngt ho hc v c cng s p l 19(K).

HS:n cht K , Cl2 , hp cht: KCl.

- HS: VX=VOXInx=nOXI

MX = 60

HS: nA= 0,25 MA= 30 dA/H2 =15

- HS:A =48

B ==16.2,5

V=20 lit

II. Bi tp p dng .Bi 1

a.

S p

S n

S e

Nt 1

19

20

19

Nt 2

17

18

17

Nt 3

19

21

19

Nt 4

17

20

17

b. Ngt 2,4: Cl. Ngt 1,3 :K

c. n cht K. Cl2

Hp cht :KCl.

Bi 2. MX= 60

Bi 3. dA/H2 = 15

Bi 4. V = 20 lt

4.Dn d HS chun b cho tit hc tip theo (5)- BTVNBi 1:

Mt hn hp kh A gm 0,8 mol O2 , 0,2 mol CO2 v 2 molCH4.

a. Tnh khi lng mol TB ca hn hp A .

b. Cho bit kh A nng hn hay nh hn khng kh ? bao nhiu ln ?

c. Tnh % th tch v % khi lng mi kh trong A .

Bi 2.

Phi ly bao nhiu gam tinh th CuSO4 .5H2O v bao nhiu gam nc iu ch c 200 gam dd CuSO4 40%.

Bi 3:

C bao nhiu gam tinh th NaCl tch ra khi lm lnh 600gam dd NaCl bo ho t 90oC xung OoC. Bit rng : SNaCl(OoC) = 35gam , SNaCl(90oC) = 50 gam.

Bi 4.

Cho m gam CaS td hon ton m1 gamdd axit HBr 8,5 % thu c m2 g dd trong mui c nng 9,6% v 672 ml kh ktc.

a. Tnh m, m1, m2 .

b. Cho bit dd HBr dng hay d , nu d hy tnh nng % ca HBr sau p.

Bi 5.

Ngm 1 l Al ( lm sch lp oxt ) trong 250ml dd AgNO3 0,24 M sau mt thi gian ly ra ra nh lm kh thy khi lng Al tng thm 2,97 g.

a. Tnh khi lng Al p

b. Tnh nng mol cc cht trong ddsau p.- Chun b bi mi: n tp cc khi nim c bn v dd v s dng thnh tho cc CT tnh tan , nng C %, nng mol CM khi lng ring ca dd

IV.RT KINH NGHIM, B SUNGNgy son: 10-8-2013 Bi : N TP U NM

Tit : 02

I. MC TIU:

1. Kin thc:

-Gip HS h thng cc kin thc ho hc c bn hc c THCS c lin quan n chng trnh lp 10 .

-n tp cc khi nim c bn v dd v s dng thnh tho cc CT tnh tan , nng C %, nng mol CM khi lng ring ca dd .

2. K nng:

- Rn luyn k nng lp CT , tnh theo CT v PTHH .

- Rn luyn k nng chuyn i gia :C% , CM , D

3. Thi :

Gip HS nm li kin thc c mt cch h thng .

II. CHUN B:

- GV: h thng bi tp v cu hi gi - HS: n tp kin thc 8,9.

III. CC HOT NG DY V HC:

1.n nh t chc: im danh.

2.Kim tra bi c: kt hp vi lm bi tp

3. Ging bi mi

Tin trnh tit dy


10 Hot ng 1: n tp cc khi nim v cng thc v dung dch :

- GV: Yu cu nhm HS h thng li cc khi nim v cng thc thng dng khi gii cc bi tp v dd.

- GV: a CT v cc ni dung trn:- HS: Tho lun nhm (3 ph)

- HS: Ghi cc CT vo v hc .

I. n tp cc khi nim v cc cng thc v dung dch:

1. Dung dch :

mdd = mct +mdm2. tan :

S= mct /mdm.100

3. Phn loi dd: da vo gi tr tan :

-Nu mct = S: dd bo ho.

- Nu mct< S: dd cha bo ho

- Nu mct >S: dd qu bo ho.

4. Cc loi cng thc tnh nng dd:

C% =.100%

CM=

5.Mi quan h gia C% v CM, D:

CM =

30Hot ng 2: Hng dn gii 1 s dng bi tp :

GV: Ghi cc bi tp ln bng:

Bi 1:

C bao nhiu gam tinh th NaCl tch ra khi lm lnh 600g dd NaCl bo ho t 90oC xung Ooc. Bit rng : SNaCl(OoC) = 35g

SNaCl(90oC) = 50g.

- GV: Gi 1 HS nhc li tan ca NaCl thay i nh th no khi gim nhit ca dd ?

- GV: Lm th no tnh c khi lng cht tan v khi lng ca dung mi nc trong 600g dd NaCl bo ho 90oC ?

- GV: Nu gi m l khi lng NaCl tch ra khi lm lnh dd t 90oC xung OOC th ti OoC mct v mdm l bao nhiu ?

- GV: Hng dn HS gii tm ra nghim m

- GV: Nhn xt ng thi nhc li cc bc lm chnh .

Bi 2:

Cho mg CaS td hon ton m1g dd axit HBr 8,58% thu c m2 g dd , trong mui c nng 9,6% v 672 ml kh ktc.

a.Tnh m, m1, m2b. Tnh nng % ca HBr sau p-GV:Gi HS vit phng trnh phn ng v tnh s mol H2S?

- GV: Tnh s mol cc cht phn ng , t tnh m, m1, m2 ?

- GV: Tnh C% HBr d?

Bi 3:

Cho 500ml ddAgNO3 1M(D = 1,2g/ml) vo300ml dd HCl 2M (D = 1,5 g/ml). Tnh C%, CM cc cht trong dd thu c . Bit p xy ra ht .

- GV: Tnh s mol AgNO3 , HCl ban u?

- GV: Vit PTP?

- GV: Xc nh thnh phn ca cht tan trong dd sau p?

- GV: Tnh CM.C%?

- GV: Nhn xt ng thi nhc li cc bc lm chnh .

- HS: suy ngh 3ph.

- HS: tan gim.

- HS: C th gii nh sau :

50g NaCl + 100g H2O

150g dd.

Vy suy ra :

200g NaCl + 400g H2O

600gdd .

mct = 200-m (OoC)

Mdm = 400 g(OoC)

V kt hp S(OOC) = 35g =

=.100 = 35

m = 60g

C2: 35 100

140?400

- HS: Chun b 3ph.

- HS: Vit p :

CaS +2HBrCaBr2 + H2S

0,03 0,06 0,03 0,03

- HS thc hin.

HS: Chun b 3ph.

II. Hng dn gii 1 s dng bi tp :

Bi 1:

50g NaCl + 100g H2O

150g dd.

Vy suy ra :

200g NaCl + 400g H2O

600gdd .

mct = 200-m (OoC)

Mdm = 400 g(OoC)

V kt hp S(OOC) = 35g =

=.100 = 35

m = 60g

C2: 35 100

140?400

Bi2: CaS +2HBrCaBr2 + H2S

0,03 0,06 0,03 0,03

nH2S = 0,03 mol

m = mCaS = 0,03.72 = 2,16g

m(CaBr2) = 200. 0,03= 6g

m2 = 6.100/9,6 = 62,5g

m1 = 62,5+ 4.0,03 - 2,16 =1,36 g

- Chng t HBr d .

m(HBr d) = 0,4g

C% = 0,64%

Bi 3:

nAgNO3 = 0,5 mol

nHCl = 0,6 mol

AgNO3+HClAgCl+HNO30,5 < 0,6 0,5 0,5

-Dd sau p gm HNO3, HCl d .

CM(HNO3) = 0,5/0,8

= 0,625M

CM(HCld ) = 0,1/0,8 = 0,125M

Khlng dd sau pbng:

m(ddAgNO3) + m(ddHCl) m(AgCl) = 600 + 450 71,75 = 978,25g

C%(HNO3) = 3,22%

C%(HCld) = 0,37%

4.Dn d HS chun b cho tit hc tip theo (5)

- BTVNBi 1: Ho tan 15,5g Na2O vo nc c 0,5 lt ddA.a. Tnh nng mol dd A.

b. Tnh th tch dd H2SO4 20% (D = 1,14 g/ml) cn dng trung ho dd A .

c. Tnh nng mol ca cht tan trong dd sau khi trung ho .

Bi 2: Kh hon ton 10,23 g hh CuO , pbO bng CO nhit cao . Kh CO2 sinh ra dn qua bnh cha Ca(OH)2 thu c 5g kt ta , cn li l dd X . un nng dd X thu thm 6g kt ta na .Tnh % theo khi lng mi o xt trong hh . Bi 3:Ho tan a gam kl M va trong 200g ddHCl 7,3% thu c dd A trong nng mui to thnh l 11,96 % (theo khi lng ) . Tnh A v xc nh kim loi M .(Mn : 55)- Chun b bi mi: Thnh phn nguyn tIV.RT KINH NGHIM, B SUNGNgy son :11/08/2013 Chng I : NGUYN T

Tit: 03 Bi: THNH PHN NGUYN T

I. MC TIU:

1. Kin thc:

HS bit:

- Nguyn t gm ht nhn mang in tch dng v v electron nguyn t mang in tch m, kch thc khi lng nguyn t .

- Ht nhn gm cc ht prton v ntron

- K hiu , khi lng v in tch ca prton , electron v ntron .

2. K nng:

- So snh khi lng ca electron vi proton v n tron .

- So snh kch thc ca ht nhn vi electron v vi ng t .

- Hiu v s dng cc n v o lng v khi lng , in tch v kch thc ca ngt nh : u , vt , n m , AO.

3. Thi :

Gip HS tp pht hin v gii quyt vn qua cc th nghim kho st v cu trc nguyn t .

II. CHUN B: - GV: Gio n , chun b tranh v phng to m phng th nghim : s tm ra electron , m hnh th nghim khm ph ra ht nhn nguyn t .

- HS: Xem trc bi mi .

III. HOT NG DY HC 1.On nh t chc: (1 ph)

2.Kim tra bi c: kt hp vi lm bi tp .

3. Ging bi mi

* Vo bi: t vn : T trc CN n th k XIX ngi ta cho rng cc cht u c to nn t nhng ht cc k nh b khng th phn chia c na gi l nguyn t . Ngy nay, ngi ta bit rng nguyn t c cu to phc tp. Nhng cu to nguyn t phc tp nh th no? Bi hc hm nay s gii p cu hi .

Tin trnh tit dy:


15Hot ng 1: Thnh phn cu to nguyn t

- Gv da trn s nhc li thnh phn cu to nguyn t. Vy ai l ngi pht hin ra cc loi ht ? Chng ta ln lt nghin cu cc loi ht trn.

- GV: Hng dn HS tm hiu th nghim minh ho hnh (1.3 SGK) theo phng php dy hc t v gii quyt vn .

- GV: s dng tranh v phng to hnh 1.1, 1.2 SGK m t th nghim ca Tm-Xn (1879) phng in vi 1 ngun in (khong 15 KV) gia hai in cc bng kim loi gn vo 2 u ng thu tinh kn trong cn rt t khng kh (gn nh chn khng)thy thnh ng thu tinh pht sng mu lc nhtchng t iu g ?

- GV: Ngi ta gi chm tia l nhng tia m cc (pht ra t m cc)

- GV: Trn ng i ca tia m cc nu ta t mt chong chng nh thy chong chng quay chng t iu g ?

- GV: Ht vt cht trong tia m cc c mang in hay khng ? mang in dng hay m ? lm th no chng minh c iu ny ?- GV: Minh ho qua th nghim m phng hoc m t tia m cc lch v pha bn cc dng .Hin tng tia m cc b lch v bn cc dng chng t iu g ? Vy tia m cc l chm ht mang in dng hay m ?

- GV kt lun: Ngi ta gi nhng ht to thnh tia m cc l electron (k hiu l e).

- GV thng bo : electron c mt mi cht , n l mt trong nhng thnh phn cu to nn nguyn t ca mi nguyn t ho hc .

- GV: yu cu HS c v ghi khi lng v in tch e vo v .- GV: e c in tch m v c gi tr qe= - 1,602. 10-19 C, l in tch nh nht nn c dng lm in tch n v (tv): qe = 1-- HS quan st tranh v, nhn xt.

- HS: Phi c chm tia khng nhn thy c pht ra t cc m p vo thnh ng.

- HS: Chm tia ny chuyn ng rt nhanh.

- HS: C th t ng phng tia m cc gia 2 bn in cc mang in tch tri du tia m cc mang in m.

- HS: Tia m cc l chm ht electron.

- HS: c SGK v ghi khi lng v in tch e vo v .I- Thnh phn cu to nguyn t

1. Electron:a. S tm ra electron:

- Th nghim ( SGK)

- Nhn xt:+ Chm tia khng nhn thy pht ra t cc m gi l tia m cc .

+ Tia m cc l chm ht mang in tch m v mi ht u c khi lng c gi l electron, k hiu l e.

b) Khi lng v in tch ca electron.

me = 9,1094. 10-31kg = 0,00055 u

qe = -1,602. 10-19 C

* V in tch 1,602.10-19C l in tch nh nht nn c dng lm vt, c qui c l 1 -

10Hot ng 2: S tm ra ht nhn nguyn t.

- GV : M t th nghim ca Rdpho 1911 ( hnh 1.3 SGK) : s dng cht phng x Rai phng ra 1 chm ht nhn anpha mang in tch dng , c khi lng gp khong 7500 ln khi lng ca e , qua khe h nh v pha tm bia bng vng mng , xung quanh l mn hunh quang hnh vng cung , ph ZnS quan st cc ht anpha bn v cc pha (mng s lo sng khi c ht anpha bn vo).

- GV thng bo kt qu th nghim :

+ Hu ht cc ht anpha xuyn qua tm vng mng .

+ Mt s t ht anpha (khong 1/1000) b bt tr li .Kt qu ny chng t iu g ?

- GV b sung v rt ra kt lun.

- HS: Nghin cu cc thit b ca th nghim v mc ch ca chng.

- HS: Ghi kt qu th nghim .

-HS: Chng t ngt c cu to rng .

HS: Ghi kt lun SGK.

2. S tm ra ht nhn nguyn t :

* Th nghim: (SGK).

* Nhn xt:- Hu ht cc ht anpha xuyn qua tm vng mng .

- Mt s t ht anpha (khong 1/1000) b bt tr li hoc b lch hng ban u.

* Kt lun :

+ Nguyn t c cu to rng .

+Ht nhn nguyn t mang in tch dng nm tm ca nguyn t v c kch thc nh b so vi kch thc ca nguyn t . Xung quanh ht nhn c cc ht e chuyn ng to nn lp v nguyn t.

10Hot ng 3: Cu to ca ht nhn nguyn t:

- GV t vn : ht nhn nguyn t cn phn chia na khng?

- GV: M t th nghim ca Rdpho nm 1918 : khi bn ph ht nhn ngt nit bng ht anpha, ng thy xut hin ht nhn nguyn t oxi v 1 loi ht c khi lng 1,6727 .10-27kg mang 1 vt dng l proton.

- GV kt lun: Ht proton l 1 thnh phn cu to ca ht nhn nguyn t .

- GV: Khi lng v in tch ca p l bao nhiu?

- GV: Nm 1932, Chat-uych (cng tc vin ca Rdpho) dng ht anpha bn ph ht nhn ngt Be thy xut hin 1 ht mi khng mang in tch l ht ntron.

- GV: Khi lng v in tch ca ntron l bao nhiu ?

- GV: Hy kt lun v cu to nhn nguyn t?

Hot ng 4: Kch thc v khi lng nguyn t :

- GV: yu cu HS c SGK v tr li cc cu hi sau:

1. n v kch thc nguyn t? Cch chuyn i?

2. Cho bit ng knh ca nguyn t, ht nhn, e v p.

3. So snh ? Yu cu HS rt ra nhn xt.

- HS: Ghi kt lun v nhn xt .

- HS: Ghi khi lng , in tch ca ht p.

- HS: Nghe v ghi thng tin .

- HS: Ghi khi lng, in tch ca ht ntron.

-HS: Ghi kt lun v nhn xt

HS: n v o kch thc ngt v cc ht l nm, A0 .

HS: Ghi gi tr ca KNT, KHN, K cc ht.

3. Cu to ca ht nhn nguyn t :

a. S tm ra proton:

- Th nghim (SGK)

- Nhn xt:

+ Ht proton l 1 thnh phn cu to ca ht nhn nguyn t .

+ qp =+ 1,602. 10-19C (1+)

+ mp = 1,6726. 10-27kg = 1 u

b. S tm ra ntron:

+ Ht ntron l 1 thnh phn cu to ca ht nhn nguyn

+ qn = 0

+mn = 1,6748. 10-27kg = 1 u

Kt lun :

Thnh phn nguyn t gm :

- Ht nhn nm tm nguyn t gm cc ht proton v ntron .

- V nguyn t gm cc e chuyn ng rt nhanh xung quanh ht nhn .

- Khi lng ca nguyn t hu ht tp trung ht nhn , khi lng ca cc e l khng ng k so vi khi lng ca nguyn t.

II. Kch thc v khi lng nguyn t:

1. Kch thc:1nm = 10-9m = 10 A01 A0 = 10-10m = 10-8 cm

- dNT khong 10-10m (rt nh).

- KHNNT khong 10-5 nm (rt nh).

- K ca e , p vo khong10-8nm

= = 104 (ln)

6Hot ng 5: Khi lng nguyn t :

- GV gii thiu: biu th khi lng ca nguyn t v cc tiu phn ca n, ngi ta dng n v khi lng nguyn t, k hiu l u (actomicmas unit):

1 u = 1/12 khi lng ca 1 nguyn t ng v cacbon -12 (c gi tr l 19 9265. 10-27 kg).

1u = 1,6605. 10-27kg

- GV: Cn phn bit khi lng nguyn t tuyt i v tng i :

+ Khi lng tuyt i l khi lng thc ca 1 nguyn t , bng tng s lng tt c cc ht trong nguyn t.

+ Khi lng tng i ca 1 nguyn t l khi lng tnh theo n v khi lng nguyn t (u)

- GV ch : khi lng nguyn t dng trong bng HTTH chnh l khi lng tng i gi l nguyn t khi.HS: Khi lng tuyt i l khi lng thc ca 1 ngt :

M = me + mp + mnHS: Khi lng tng i ca 1 ngt l khi lng tnh theo n v khi lng ngt (u) 2. Khi lng:

- dng n v u (vC)

* 1u = 1/12 mC

1u =

= 1,6605.10-27kg

*me=9,1094.10-31kg 0,00055 u

mp = 1,6726.10-27kg 1 u

mn = 1,6748.10-27kg 1 u

- KLNT 10-26kg

- Khi lng ca 1 ngt H l:

1,6738.10-27kg 1u (nh nht)

- Khi lng ca 1 ngt C l:

19,9265.10-27kg 12 u

2 * Hot ng 6 cng c +Lu : C th tnh khi lng nguyn t C : mC=2.1023g

Lp v: c ht e, kl, t ?

+Nguyn t

Ht nhn :c ht p, n ,kl, t ?

+Hy tm tt qu trnh pht minh nguyn t:

1897: Tomxon(A) tm ra e.

1909: Milikan(M) x t e.

1911: R pho tm ra ht nhn nguyn t.

1918: Rpho (A) tm ra p.

1932: Chatuych (A) tm ra n.

4.Dn d HS chun b cho tit hc tip theo (1) BTVN: 1-5/ SGK + SBT

IV.RT KINH NGHIM, B SUNGNgy son: 15/08/2013

Tit: 04 Bi 2: HT NHN NGUYN T - NGUYN T HO HC

I. MC TIU:

1. Kin thc: HS bit c :

- Cc khi nim THN , s khi , NTK v cch tnh .

- T hiu c nh ngha ca cc ngt ho hc v k hiu ngt ca ngt.

2. K nng:

- Xc nh c s ht mi loi trong 1 ngt khi bit KHNT v s khi ca ngt v ngc li.

- Bit cch vit khnt.

3. Thi :

Tp pht hin v gii quyt cc vn qua cc th nghim kho st v cu trc nguyn t .

II. CHUN B:

- GV: tranh v phng to m phng cu to ht nhn ca 1 nguyn t , cc phiu hc tp . - HS: Nm vng cc c im cc ht cu to nn nguyn t , xem trc bi mi.

III. HOT NG DY HC 1.n nh t chc: (1 ph)

2.Kim tra bi c:. (3 ph)

* Yu cu HS trnh by tm tc thnh phn cu to ca nguyn t v cho bit in tch , khi lng ca cc ht c bn ?

3. Ging bi mi: Thi gianHot ng ca gio vinHot ng ca hc sinhNi dung

12Hot ng 1: in tch ht nhn nguyn t

- Ht nhn nguyn t c mang in khng? Nu c th in tch ht nhn do loi ht no quyt nh? C tri s l bao nhiu theo n v in tch nguyn t?- GV: S n v in tch ca ht nhn phi bng s ht no trong ht nhn ?

- GV: Nu ht nhn c Z proton th in tch ca ht nhn phi bng Z+ v s n v in tch ht nhn bng Z .

- GV : Nguyn t trung ho v in , vy c nhn xt g v s p v s e trong nguyn t ?

- Yu cu HS nu kt lun v s n v in tch ht nhn Z vi s p, s e.

- GV cho v d: phiu hc tp s 1.

1. Nguyn t C c 6 proton, nguyn t Al c 13 electron, hy cho bit s n v in tch ht nhn, in tch ht nhnv s electron trong mi nguyn t ?

2. Nguyn t Nit c 7 electron cho bit in tch ht nhn, s proton ca ngt Nit?

- Ht nhn nguyn t c mang in tch dng do ht proton quyt nh v mt proton mang mt n v in tch dng nn tr s ca in tch ht nhn bng s n v in tch ht nhn, bng s proton.

- HS: S p = s e

- HS lm phiu hc tp s 1 vo v.

1. Nguyn t C c 6 proton

+ S electron : 6

+ S n v thn: 6

+in tch ht nhn: 6+2. Nguyn t Al c 13 proton

+ S proton: 7

+ S n v thn: 7

+in tch ht nhn :7+I. HT NHN NGUYN T1. in tch ht nhn- Nguyn t c Z proton s v THN l Z , THN l Z+.

+ Nguyn t trung ho v in:

s n v THN = s p = s e

V d: 1. Nguyn t C c 6 proton

+ S electron : 6

+ S n v thn: 6

+in tch ht nhn: 6+2. Nguyn t Al c 13 proton

+ S proton: 7

+ S n v thn: 7

+in tch ht nhn :7+

7Hot ng 2: S khi

- Yu cu hc sinh c SGK v cho bit s khi ca ht nhn l g?

- GV cho v d: phiu hc tp s 2.

1. Ht nhn ngt oxi c 8 proton v 9 ntron. S khi ca ngt ny l bao nhiu?

2. Ngt Clo c in tch ht nhn l 17+. S khi nguyn t bng 35. Tm s ntron?

3. S khi ca nguyn t Kali l 39. bit ht nhn nguyn t cha 20 ntron. Hy cho bit s proton?

4. lp v ca nguyn t S c 16e, bit s khi l 33. tnh s p,n ca nguyn t?

- Nhn xt v nguyn t khi tnh theo vC v s khi ca ht nhn? Gii thch.

- GV thng bo: s khi A v s in tch ht nhn Z l nhng c trng rt quan trng ca ngt. Da vo A,Z ta bit c cu to ngt.

- GV thng bo : A, Z l nhng i lng rt quan trng ca nguyn t , da vo nhng i lng ny ta bit c cu to nguyn t .

- HS: Ghi nh ngha v cng thc

-Hc sinh lm phiu hc tp s 2 vo v

1. A = Z + N = 8 + 9 = 17

2. N = A Z = 35 17 = 18

3. Z = A Z = 33 - 16 = 19

s proton: 19

4. S p = s e = 16

N = A Z = 33 -16 =17

- HS: Khi lng ca proton v ntron gn bng 1 u , m e c khi lng nh hn rt nhiu (c th b qua). Vy c th ni s khi bng NTK.

2. S kh (A): S khi ca ht nhn bng tng s proton (Z) v tng s notron(N).

A = Z + N hay A = P + N

V d:1. A = Z + N = 8 + 9 = 17

2. N = A Z = 35 17 = 18

3. Z = A Z = 33 - 16 = 19

s proton: 19

4. S p = s e = 16

N = A Z = 33 -16 =17

8Hot ng 3: Nguyn t ho hc

- GV yu cu hc sinh c sch gio khoa, cho bit nguyn t ho hc l g?

- GV: Tt c cc nguyn t c cng s n v THN l 11 u thuc cng ngt Na chng u c 11p v 11e

- Nh vy, s proton v s electron ca tt c cc nguyn t ca cng mt nguyn t ho hoc c quan h nh th no ?

- Hy phn bit khi nim nguyn t v nguyn t?

- GV: cho n nay ngi ta bit khong 92 ngt ho hc c trong t nhin v khong 18 ngt nhn to c tng hp trong cc phng th nghim ht nhn.- HS: Ghi nh ngha.

+ s proton v s electron

bng nhau .

+ Ni nguyn t l ni n mt loi ht vi m gm cc ht nhn v lp v, cn ni nguyn t l ni n tp hp cc nguyn t c cng in tch ht nhn nh nhau.II. NGUYN T HO HC :

1. nh ngha: Ngt ho hc l nhng ngt c cng in tch ht nhn.

* Nhn xt:- Nguyn t ca cng 1 ngt c cng s p v s e.

- Nhng ngt c cng THN u c tnh cht ho hc ging nhau .

5Hot ng 4: S hiu nguyn t

- GV: Cho HS c SGK v a ra nh ngha .

- GV: s hiu ngt cho bit g?

Tm s hiu ngt ca nguyn t Na?

- GV ly thm VD :

S hiu nguyn t ca st l 26.

Nguyn t Fe ng th 26 trong bng tun hon c 26 proton trong ht nhn , c 26 electron trong v ca nguyn t ,c s n v in tch ht nhn l 26

- HS: Ghi nh ngha.

- HS: S hiu nguyn t k hiu l Z, cho bit : s p trong ht nhn v s e trong nguyn t .

2. S hiu nguyn t :

a. Khi nim : S hiu nguyn t l s n v THN nguyn t ca 1 nguyn t .

b. K hiu : Z

- s hiu nguyn t c trng cho cho 1 nguyn t ho hc v cho bit s p trong ht nhn v s e trong nguyn t .

V d:

S hiu ngt Au l 79:

THN: 79+ , s p = s e = 79.

5Hot ng 5: K hiu nguyn t

- c SGK v gii thch k hiu nguyn t

- GV: Nguyn t Na c 11p, 11e, 12n .Hy vit k hiu nguyn t Na?

- GV: V s in tch ht nhn Z v s khi A c coi l nhng s c trng c bn ca nguyn t , ngi ta thng t k hiu cc ch s c trng bn tri k hiu nguyn t X vi s khi A pha trn , s n v in tch ht nhn Z pha di

- GV: K hiu ngt oxi :. Hy cho bit s ht mi loi trong nguyn t oxi?- Tr li cu hi.

- HS:

- HS : Nguyn t O c 8p, 8e, v 8n. 3. K hiu nguyn t :

- X:k hiu nguyn tho hc .

- Z: s hiu nguyn t.

- A: s khi.

V d:

Trong ht nhn nguyn t Br :

- s hiu nguyn t l 35 ,

- THN: 35+- s p: 35

- s n: 80 35 =45

- KLNT: 80 vC.

4Hot ng 6 : cng c Kin thc cn nm c :

- S lin quan gia THN vi s p, s e.

- Cch tnh s khi ht nhn.

- Khi nim nguyn t ho hc .

- Mi lin h s p , s e, s v THN trong ngt.

Z=p=e

A=Z+n

4.Dn d HS chun b cho tit hc tip theo (1) - BTVN: 3,5/SGK:1.18 n 1.24/SBT

- Chun b bi mi;

IV.RT KINH NGHIM, B SUNGNgy son: 15-08-2013

Tit: 05

Bi 3: NG V NGUYN T KHI V NGUYN T KHI TRUNG BNH

I. MC TIU:

1. Kin thc:

HS Bit c:

- Khi nim ng v, nguyn t khi v nguyn t khi trung bnh ca 1 nguyn t. 2. K nng:

Gii c bi tp: tnh nguyn t khi trung bnh ca nguyn t c nhiu ng v, tnh t l % khi lng mi ng v v 1 s bi tp khc lin quan.

3. Thi :

Hiu c s tn ti cc ng v trong t nhin.

II. CHUN B:

-GV: chun b tranh v phng to m phng cc ng v ca Hiro, chun b cc phiu hc tp.

-HS: Nm vng c im cc ht cu to nn nguyn t, xem trc bi mi.

III. HOT NG DY HC 1.n nh t chc: (1 ph)

2.Kim tra bi c: (4 ph)

Hy cho bit s n v THN, s p, s n, s e, s khi, in tch ht nhn ca cc nguyn t c k hiu sau:

3517 Cl ; 19578 Pt

3. Ging bi mi: Thi gianHot ng ca gio vinHot ng ca hc sinhNi dung

15 Hot ng 1 : vo bi- S dng phiu hc tp s 1.

a. Hy tnh s p, s n ca cc nguyn t sau ?

16 8O ; 17 8O ; 18 8O

b. Hy cho bit c im chung ca cc ng t trn ?

c. nh ngha ng v ?

- GV: Da vo cu b dn dt HS n nh ngha ng v ?

- GV: Tng t Hiro trong t nhin cng l hn hp 3 ng v, l:

11 H; 21 H ; 31 H ;

(99,984%) (0,016%) (rt t)

( Xc nh s ht mi loi ?

GV: C khong 340 ng v tn ti trong t nhin, hn 2400 ng v nhn to.

- S dng phiu hc tp s 2:Cho 2 ng v hiro:11H; 21 H v ng v Cl: 3517Cl, 3717Cl. Hi c bao nhiu loi phn t HCl ?

GV: Nu 1 s ng dng ca ng v phng x trong i sng, y hc.

Hs:Trli:

HS: nh ngha ng v.

HS: Tr li:

HS: C 4 loi:

11H 3517Cl 11H 3717Cl

21 H 3517Cl 21 H 3717Cl

HS: c t liu trong SGK.

I. NG V:

1. nh ngha:

Vd:

Vy: Cc ng v ca cng 1 nguyn t ha hc l nhng nguyn t c cng s proton nhng khc nhau v s ntron nn s khi (A) ca chng khc nhau.

2. Cc ng v ca hiro:

11 H: gi l Proti.

20 Hot ng 2

I: Nguyn t khi v nguyn t khi trung bnh:- S dng phiu hc tp s 3:

a. Phn bit NTK vi:

- n v khi lng nguyn t.

- S khi ht nhn.

b. Nguyn t khi TB l g ? vit CT tnh nguyn t khi Tb v gii thch.

c. Tnh NTKTB ca nguyn t Ni, bit rng trong t nhin cc ng v ca Ni tn ti theo t l:5828Ni(67,76%);

6028Ni(26,16%);

6128Ni (2,42%);

6228Ni (3,66%);

GV: Hng dn HS gii bi tp 5/SGK.

(Lu : Nu c 2 ng v:)

=A.x+ B.(1-x)

HS: Tr li:

- NTK l khi lng ca ngt tnh ra u, n cho bit khi lng ca ngt nng gp bao nhiu ln n v khi lng ngt.

- C th coi NTK s khi ca ht nhn. (?)

HS: Tr li:

- NTK ca cc ngt c nhiu ng v l NTKTB ca hn hp cc ng v c tnh n t l phn trm s ngt ca mi ng v.

HS:

HS:

-Gi x l s nguyn t ca 63Cu

-Gi (1-x) l s nguyn t ca 65Cu .

Ta c: 63x + 65(1-x) = 63,54

( x=0,73 (73%)

I: Nguyn t khi v nguyn t khi trung bnh:

1. Nguyn t khi: (l khi lng tng i ca 1 nguyn t)

- NTK ca 1 nguyn t cho bit khi lng ca nguyn t nng gp bao nhiu ln n v khi lng nguyn t.

- C th coi NTK s khi ca ht nhn (v khi lng ca 1 nguyn t bng tng khi lng ca p, n e trong nguyn t . P,n u c khi lng 1u, e c khi lng nh hn rt nhiu (0,00055u)

2. NTK trung bnh:

- NTK ca cc ngt c nhiu ng v l NTKTB ca hn hp cc ng v c tnh n t l phn trm s ngt ca mi ng v.

-CT:

Trong : A, B l NTK ca ng v A, B. a, b l t l phn trm s ngt ca ng v A, B.

- Hay CT:

VD: Ngt Clo c 2 ng v bn: 3517Cl chim 75,77% v 3717Cl chim 24,23%. Tnh Cl.

/:

Cl =

5

Hot ng 3: cng c

+ ng v? NTK ? cch tnh NTKTB, hoc tnh % s ngt, s khi tng ng v ?

HS tr li

4.Dn d HS chun b cho tit hc tip theo (1) BTVN: 1, 3,4/sgk. Chun b bi mi: S chuyn ng ca eletron trong nguyn t, obitan ngt.

IV.RT KINH NGHIM, B SUNGNgy son: 15/08/2013Tit: 06 Bi: S CHUYN NG CA ELECTRON TRONG NGUYN T - OBITAN NGUYN T I. MC TIU:

1. Kin thc:

Bit c: M hnh ngt ca Bo, Rdpho. M hnh hin i v s chuyn ng ca eletron trong nguyn t. Obitan ngt, hnh dng cc obitan ngt s, px, py, pz. 2. K nng: Trnh by c hnh dng ca cc obitan ngt s, p v s nh hng ca chng trong khng gian.

3. Thi : Gio dc hc sinh hiu r qui lut ca khoa hc. Hnh thnh cho hc sinh khi nim nng lng tn ti, nng lng tch e gip hc sinh lin h trong thc t n cc lc ko, p dng trong lao ng.

II. CHUN B:

-Chun b ca Thy: Tham kho ti liu, chun b m hnh mu hnh tinh nguyn t ca Rzfo v Bohr ( nguyn t H v O), hnh nh cc obitan s,p, d.

-Chun b ca Tr: Hc bi c, xem trc bi mi.

III. HOT NG DY HC 1.n nh t chc: 1/.

2.Kim tra bi c: 9/ 1. HS1 cha bi tp 3-SGK.

2. HS2 cha bi tp 4-SGK.

3. HS3 + c lp: Tnh ngt khi TB ca Ar v K, bit rng trong t nhin:

- Ar c 3 ng v: 3618Ar (0,3%), 3818Ar (0,06%), 4018Ar (99,6%)

- K c 3 ng v: 3919K ( 93,08%) , 4019K (0,012%) , 4119K (6,9%)

T kt qu trn hy gii thch v sao Ar c s hiu ngt l 18, cn K l 19 nhng Ar li c NTKTB ln hn.

3. Ging bi mi: Vo bi: Nh bit v e ca ngt gm cc e chuyn ng xung quanh ht nhn. Vy s chuyn ng e trong nguyn t nh th no ? Trng thi chuyn ng ca e c ging s chuyn ng ca cc vt th ln hay khng ?


7 phtHot ng 1 : M hnh hnh tinh nguyn t:

- GV: Treo s mu hnh tinh nguyn t ca Rdpho v Bo v thng bo: m hnh ny cho rng trong nguyn t, e chuyn ng trn nhng qu o trn hay bu dc xc nh xung quanh ht nhn, nh cc hnh tinh quay xung quanh mt tri. Thnh cng ca thuyt Bo l gii thch c quang ph nguyn t H. Tuy nhin m hnh ny khng phn nh ng trng thi chuyn ng ca e trong nguyn t.

- GV: T l thuyt vt l hin i, l thuyt c hc lng t ta bit trng thi chuyn ng ca e (l nhng ht vi m) c nhng khc bit hn v bn cht so vi s chuyn ng ca cc vt th v m m ta thng quan st thy hng ngy. M hnh nguyn t ca Bo v c bn da trn nhng nh lut ca c hc c in t ra khng y gii thch tnh cht ca nguyn t.- HS: c SGK, quan st s mu hnh tinh nguyn t, sau ( kt lun.

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

1. M hnh hnh tinh nguyn t:

Trong nguyn t, e chuyn ng trn nhng qu o trn hay bu dc xc nh xung quanh ht nhn, nh cc hnh tinh quay quanh Mt Tri.

15 phtHot ng 2 : M hnh hin i v s chuyn ng ca e trong nguyn t, obitan nguyn t:

- GV: Dng tranh m my nguyn t H, gip HS tng tng ra hnh nh xc sut tm thy e.

-Trong nguyn t, cc e chuyn ng rt nhanh xung quanh ht nhn khng theo 1 qu o xc nh. Ngi ta ch ni n kh nng quan st thy e ti 1 im no trong khng gian ca ngt. Tc l ni n xc sut c mt e ti 1 thi im quan st c.

- i vi nguyn t H mt xc sut c mt e ln nht vng gn ht nhn, cng xa ht nhn mt xc sut c mt e cng nh dn. Ngi ta xc nh khong khng gian e chuyn ng xung quanh ht nhn ngt H l 1 khi cu, bn knh khong 0,053nm, trong xc sut c mt e khong 90%.

- i vi nhng nguyn t nhiu e, s chuyn ng ca cc e to thnh nhng khong khng gian c hnh dng khc nhau, my e khc nhau.

- Lu : ni m my e nhng khng phi do nhiu e to thnh m l nhng v tr ca 1 e.

- GV: Electron c th c mt khp ni trong khng gian nguyn t, nhng kh nng khng ng u. Tp hp tt c nhng im m ti xc sut tm thy e ln nht l hnh nh obitan nguyn t.

-Ch : e c th tn ti ngoi khu vc khng gian qui c trn vi xs c mt khong 10%. Nh vy v nguyn tc AO khng c gii hn.

- HS: Quan st hnh v m my nguyn t H, c SGK sau c th tng tng ra hnh nh xc sut tm thy e.

- HS: Ghi kt lun.

- HS: Hiu khi nim obitan nguyn t.

- HS: Ghi nh ngha obitan nguyn t.

2. M hnh hin i v s chuyn ng ca e trong nguyn t, obitan nguyn t:

a. S chuyn ng ca e trong nguyn t:

Trong nguyn t, cc e chuyn ng rt nhanh xung quanh ht nhn khng theo 1 qu o xc nh no. b. Obitan nguyn t: (AO)

- Obitan nguyn t l khu vc khng gian xung quanh ht nhn m ti xc sut c mt (xc xut tm thy) electron khong 90%.

10 phtHot ng 3 : Hnh dng obitan nguyn t:- GV: Treo tranh v hnh nh cc obitan s, p, d ( hy nhn xt hnh nh obitan nguyn t H ?

- GV: Phn tch: khi chuyn ng trong nguyn t, cc e c th chim nhng mc nng lng khc nhau c trng cho trng thi chuyn ng ca n. Nhng e chuyn ng gn ht nhn hn, chim nhng mc nng lng thp hn, tc l trng thi bn hn nhng e chuyn ng xa ht nhn c mc nng lng cao hn.

-GV: Da trn s khc nhau v trng thi ca e trong nguyn t, ngi ta phn loi thnh cc obitan s, p, d, f.

- GV: Da vo tranh v, phn tch hnh nh cc obitan.- HS: Quan st hnh nh cc obitan s, p, d ( hy nhn xt hnh nh obitan ngt H.

II. Hnh dng obitan nguyn t:

- Da trn s khc nhau v trng thi ca e trong nguyn t, ngi ta phn loi thnh cc obitan: s, p, d, f.

* Obitan s : c dng hnh cu, tm l ht nhn nguy t.

* Obitan p: gm 3 obitan: px, py, v pz c dng hnh s 8 ni. Mi obitan c s nh hng khc nhau trong khng gian.

* Obitan d, f c hnh dng phc tp hn

- Hnh nh cc AO s, p (AOs, AOp)

5Hot ng 4: Cng c

Cng c bng bi tp 1,2,3/SGK trang 20.

HS lm bi tp

4.Dn d HS chun b cho tit hc tip theo (1) - BTVN: 4, 5,6 /sgk. 1.35 ( 1. 38 , 1.39, 1.40 /SBT.- Chun b bi luyn tpIV.RT KINH NGHIM, B SUNGNgy son: 1-9-2013

Tit: 07 LUYN TPTHNH PHN CU TO NGUYN T KHI LNG

NGUYN T OBITAN 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 cc ht, nh ngha nguyn t ha hc, k hi nguyn t.

2. K nng: Rn luyn k nng xc nh s ht mi loi, khi lng nguyn t.

3. Thi : T duy h thng phi hp cc ni dung kin thc v k nng gii bi tp.

II. CHUN B:

-GV: H thng bi tp v cu hi gi .

-HS: n tp cc kin thc v thnh phn nguyn t thng qua hot ng gii bi tp.III. HOT NG DY HC 1.n nh t chc: 1/.

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

3. Ging bi mi

Tin trnh tit dy


10Hot ng 1: Kin thc cn nm vng:-GV: Kim tra 4 HS:

1. Cho bit thnh phn cu to nguyn t v khi lng, in tch ca cc ht cu to nn nguyn t ?

2. Mi quan h cc ht trong nguyn t vi s n v THN Z, VD ?

3. Trnh by KHNT, nh ngha nguyn t ha hc, VD ?

4. Cch tnh s khi ca ht nhn ntn ? nu ni s khi bng NTK th c ng khng ? ti sao?HS: Nguyn t bao gm ht nhn mang in tch dng (p,n) v lp v bao gm 1 hay nhiu e mang in tch m:

mp = mn 1u; me = 0,00055u

qp = 1+ ; qn = o; qe = 1-

HS: Z = s p = s e

HS: Tr li theo SGK.

HS: A = Z +N

Mt cch gn ng v tr s s khi bng NTK. V khi lng ngt c coi nh bng tng khi lng cc ht p v n trong ht nhn ngt. Mi ht p, n c khi lng gn bng 1u.A. Kin thc cn nm vng:

1. Cho bit thnh phn cu to nguyn t v khi lng, in tch ca cc ht cu to nn nguyn t ?

2. Mi quan h cc ht trong nguyn t vi s n v THN Z, VD ?

3. Trnh by KHNT, nh ngha nguyn t ha hc, VD ?

4. Cch tnh s khi ca ht nhn ntn ? nu ni s khi bng NTK th c ng khng ? ti sao?

30Hot ng 2 : Bi tp p dng:- GV: Lu :

- Tng ht = Z + E +N

= 2Z + N

- Cc ng v bn (Z khi lng ca ngt coi bng khi lng ca ht nhn (b qua khlng ca e).

HS: Tr li:

MO=15,842. MH MC=11,906. MH

(

HS:Chun b 2 pht v tr li:

Tng ht = Z + E +N = 34

= 2Z + N =34

V N Z = 1

( Z=P=E=11; N=12;

THN = 11+

HS:Chun b 2 pht v tr li:

Gi:

Ta c: Z + E +N = 13

-> 2Z + N =13

V: Cc ng v bn (Z 4/SGK v 1.40 -> 1.44; 1.47 /SBT.

IV.RT KINH NGHIM, B SUNGNgy son : 2/09/2013

Tit: 10 - 11

NNG LNG CA ELECTRON TRONG NGUYN T

CU HNH ELECTRON CA NGUYN T

I. MC TIU:

1. Kin thc: Bit c:

+ Mc nng lng obitan nguyn t v trt t sp xp.

+ Cc nguyn l v qui tc phn b e trong nguyn t: Nguyn l vng bn, nguyn l Pauli, qui tc Hun.

+ Cu hnh e v cch vit cu hnh e trong nguyn t.

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

+ c im ca lp e ngoi cng.

2. K nng:

+ Vit c cu hnh e di dng lng t ca 1 s nguyn t ha hc.

+ Da vo cu hnh e lp ngoi cng ca nguyn t suy ra tnh cht c bn ca nguyn t l kim loi, phi kim hay kh him.

3. Thi : T con ng l thuyt HS c th d on c cu to nguyn t v t suy ra tnh cht ca nguyn t .

II. CHUN B:

-Chun b ca Thy: Gio n, chun b tranh v: trt t cc mc nng lng obitan nguyn t, bng cu hnh e v s phn b e trn cc obitan ca 20 nguyn t u tin.

-Chun b ca Tr: Hc bi c, xem trc bi mi.

III. HOT NG DY HC 1.n nh t chc: 1/.

2.Kim tra bi c: 9/ Cu 1: Khi nim v k hiu lp, phn lp electron ? (Tr li theo ni dung bi hc). Cu 2: Bi tp 1, 6 /SGK

3. . Ging bi mi: Tin trnh tit dy


10Hot ng 1 : Nng lng ca electron trong nguyn t- GV: Cc e trong cng lp e, cng phn lp e c mc nng lng nh th no ?

- Mi e u c 1 nng lng xc nh, cc e trn cng lp c nng lng gn bng nhau, cn cc e trn cng phn lp c nng lng bng nhau.

- Mi phn lp e tng ng vi 1 gi tr nng lng xc nh ca e. Ni cch khc cc e trn cng 1 phn lp thuc cng mc nng lng. Ngi ta gi mc nng lng ny l mc nng lng obitan nguyn t, gi tc l mc nng lng AO.

- GV: Phn lp 2p c bao nhiu obitan v mc nng lng ca chng ntn ?- HS: Tr li theo SGK.

- HS: Phn lp 2p c 3 obitan: 2px; 2py; 2pz, tuy c s nh hng trong khng gian khc nhau nhng c cng mc nng lng obitan.I. Nng lng ca electron trong nguyn t1.Mc nng lng obitan ng t- Trong nguyn t cc e trn mi obitan c 1 mc nng lng xc nh. Ngi ta gi mc nng lng ny l mc nng lng obitan nguyn t ( gi tt l mc nng lng AO).

- Cc e trn cc AO khc nhau ca cng 1 phn lp c nng lng nh nhau.

5Hot ng 2: Trt t cc mc nng lng obitan nguyn t- GV: Cho HS nghin cu hnh 1.12 v rt ra trt t cc mc nng lng obitan nguyn t.

- GV: Gii thiu cho HS cch s dng qui tc ng cho theo quy tc Kleccopski

- HS: Rt ra trt t cc mc nng lng obitan nguyn t:

Thc nghim v l thuyt cho thy khi s hiu nguyn t Z tng th cc mc nng lng AO tng dn theo th t sau:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p6s 4f 5d 6p 7s 5f 6d

2. Trt t cc mc nng lng obitan nguyn t:

- Cc mc nng lng AO tng dn theo th t sau:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p6s 4f 5d 6p 7s 5f 6d

- Nhn xt: Khi THN tng c s chn mc nng lng, mc 4s tr nn thp hn 3d.

8Hot ng 3: Cc nguyn l v qui tc phn b electron trong nguyn t:

- GV: Thng bo v tiu s v thnh tch khoa hc ca Pau-li.

- Cho HS nghin cu SGK, tr li cu hi lng t l g ?

- GV: a ra cch k hiu lng t.

- Cc vung ny ging nhau c v vi cao khc nhau ch s khc nhau v mc nng lng ca cc phn lp- GV: Cho HS nghin cu SGK cho bit:

- Ni dung nguyn l Pau-li ?

- Cch k hiu e trong 1 lng t.

- GV: Cho HS tnh s electron ti a trong 1 phn lp.

- GV: Gii thiu cho HS cch phn b e trn cc obitan s, p, d, f.

- Biu din trng thi e dng k hiu:

+ vd: 2p4 , s 2 ng bn tri ch lp n=2, s 4 pha trn bn phi ch s e trn phn lp 2p.

- Cc phn lp: s2, p6, d10, f14 c s e ti a gi l phn lp bo ha. Cn phn lp cha s e ti a gi l phn lp cha bo ha., vd:- HS: Tr li theo SGK.

- HS: Trnh by ni dung nguyn l Pau-li.

- HS: Da vo d kin: s obitan trn 1 phn lp v s e ti a trong 1 obitan tnh tng s e ti a trong 1 phn lp (lp n c n2 obitan, c ti a 2n2 electron).II. Cc nguyn l v qui tc phn b electron trong nguyn t:

1. Nguyn l Pau-li:

a. lng t:

biu din obitan nguyn t 1 cch n gin, ngi ta dng vung nh, c gi l lng t (1 lng t ng vi 1 AO)

VD: Cc lng t ng vi n=1, n=2:

1s 2s 2p

b. Nguyn l Pau-li:

Trn 1 AO ch c th c ti a l 2 electron v 2 electron ny chuyn ng t quay khc chiu nhau xung quanh trc ring ca mi electron.

- Mi e biu th bng 1 mi tn.

- Khi AO c 2 e th 2e gi l e ghp i.

- Khi AO ch c 1e gi l e c thn.

2 electron ghp i

1e c thn

c. S electron ti a trong 1 lp v trong 1 phn lp:

- S e ti a trong 1 lp e: lp n c ti a 2n2 electron (v c n2 obitan).

- S e ti a trong 1 phn lp e:

Phn lp

AO

S e ti a

s

p

d

f

1

3

5

7

2

6

10

14

- Cc phn lp: s2, p6, d10, f14 c s e ti a gi l phn lp bo ha.

- Phn lp cha s e ti a gi l phn lp cha bo ha., vd:

5Hot ng 4 : Nguyn l vng bnGV: Cho HS nghin cu SGK ri cho bit:

- Ni dung nguyn l vng bn?

- Biu din s phn b electron ca nguyn t H, He, Li, Be theo lng t v cch dng k hiu.

HS: Trnh by ni dung nguyn l vng bn theo SGK.2. Nguyn l vng bn:

trng thi c bn, trong ngt cc e chim ln lt nhng obitan c mc nng lng t thp n cao.

Vd: S phn b e ca cc ngt H, He, Li, Be nh sau:

H(Z=1): 1s1

He(Z=2): 1s2

Li(Z=3): 1s22s1

Be(Z=4): 1s22s2

7Hot ng 5 : Quy tc Hun- GV: Cho HS nghin cu SGK ri cho bit: ni dung qui tc Hun ?

- GV: Gii thiu cch biu din s phn b electron ca nguyn t C, N

- GV: trnh cng knh, ngi ta ch biu din s cao thp ca cc lng t khi cn th hin mc nng lng khc nhau ca tng phn lp electron.

- Cc e c thn trong 1 ngt c k hiu bng cc mi tn cng chiu, thng c vit hng ln trn.

* Cng c: (3 pht)

- Trng thi nng lng ca e trong nguyn t ?

- Cc nguyn l, qu tc phn b e trong nguyn t?- HS: Trnh by ni dung quy tc Hun theo SGK.3. Quy tc Hun:

Trong cng 1 phn lp, cc e s phn b trn cc obitan sao cho s e c thn l ti a v cc e ny phi c chiu t quay ging nhau.

- VD: S phn b e trn cc obitan ca ngt C, N:

.

30Hot ng 6: tit 11cu hnh electron nguyn t- GV: Cho HS c SGK v cho bit: cu hnh electron l g ?

- GV: Gii thiu qui c cch vit cu hnh electron:

- STT ca lp c vit bng s.

- Phn lp c k hiu bng ch ci thng: s, p, d, f

- S electron vit trn k hiu ca cc phn lp nh s m.

- GV: Tip tc gii thiu cc bc vit cu hnh electron:

+ Xc nh s electron ca nguyn t.

+ Cc electron phn b theo th t tng dn cc mc nng lng AO, theo nguyn l Pau-li, nguyn l vng bn, qui tc Hun.

- GV: Hng dn HS vn dng vit cu hnh electron ca cc nguyn t: Na, Mg, Ar, Fe...

- GV: Cho nhm HS1 vit cu hnh electron nguyn t ca cc nguyn t c Z =1->10, nhm HS2 vit cu hnh electron nguyn t ca cc nguyn t c Z =11->20 . Sau xc nh s e lp ngoi cng, nhn xt v s e lp ngoi cng khi s hiu nguyn t tng dn.- HS: Tr li theo ni dung trong SGK.

- HS: Vn dng vit cu hnh electron ca cc nguyn t: Na, Mg, Ar, Fe

- HS: Nhm 1, 2 vit cu hnh electron ( nhn xt.III. CU HNH ELECTRON NGUYN T:

1. Cu hnh electron nguyn t:

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

- Quy c cch vit cu hnh e nguyn t:

+ STT lp e c vit bng cc ch s: 1, 2, 3

+ Phn lp c k hiu bng ch ci thng: s, p, d, f.

+ S e c ghi bng ch s pha trn, bn phi k hiu ca phn lp: s2, p4

- Cch vit cu hnh e nguyn t:

+ Xc nh s e ca nguyn t.

+ Cc e c phn b theo th t tng dn cc mc nng lng AO, theo cc nguyn l v quy tc phn b e trong nguyn t.

+ Vit cu hnh e theo th t cc phn lp trong 1 lp v theo th t ca cc lp electron.

VD: C hnh e ca Na, Fe:

Na(Z=11): 1s22s22p63s1Fe(Z=26): 1s22s22p63s23p64s23d6Phi sp xp ( cu hnh electron: 1s22s22p63s23p63d64s2

Hay [Ar]3d64s2(Ar: l kh him gn nht ng trc Fe).

2. Cu hnh electron nguyn t ca 1 s nguyn t: ( SGK)

10Hot ng 7 : c im ca e lp ngoi cng:- GV: t cc cu hi sau:

+ Da vo cu hnh e nguyn t ca ngt Cl, Na, cho bit e no gn ht nhn nht, xa ht nhn nht, electron no lin kt vi ht nhn mnh nht, yu nht ?

- Da vo bng cu hnh e ca 20 nguyn t u, cho nhn xt v s lng e lp ngoi cng ?

- Trong bng trn, nguyn t no l kim loi, phi kim , kh him ?

- Trong nguyn t, e no quyt nh tnh cht ha hc ca 1 nguyn t, v sao ?- HS: Tr li cc cu hi.3. c im ca e lp ngoi cng:

Cc e lp ngoi cng quyt nh tnh cht ha hc ca mt nguyn t.

Trong nguyn t:

- Lp ngoi cng c ti a 8e.

- Nguyn t c 8e lp ngoi cng (tr He c 2 e) u rt bn vng, l cc nguyn t kh him.

- Nguyn t c 1, 2, 3 e lp ngoi cng l cc nguyn t kim loi (tr H, He v B).

- Nguyn t c 5, 6, 7e lp ngoi cng l cc nguyn t phi kim.

- Nguyn t c 4e lp ngoi cng c th la nguyn t kim loi hay phi kim.

4. CNG C:(4/) thng qua cc bi tp sau:

1. Vit cu hnh e nguyn t ca cc nguyn t sau bng 2 cch: Ne, Ca, Al, Mn, Br.

a. Nguyn t no l kim loi, phi kim, kh him.

b. Cho bit s lp e, s e c thn ca nguyn t cc nguyn t trn.

2. BTVN: 1-> 7/SGK v 1.48 -> 1.54 / SBT.

*Lu : c 2 TH phi t chuyn cu hnh cho bn:

- (n-1)d4ns2 ( (n-1)d5ns1

- (n-1)d9ns2 ( (n-1)d10ns1 Vd: 29Cu: 1s22s22p63s23p63d104s1 (3d94s2).

IV.RT KINH NGHIM, B SUNGNgy son: 5/9/2013 Tit: 12, 13

Bi: LUYN TP CHNG I

I. MC TIU:

1. Kin thc: Cng c kin thc v:

+Thnh phn cu to nguyn t, c im cc ht cu to nn nguyn t.

+ Nguyn t ha hc, nhng c trng ca nguyn t ha hc.

+ S chuyn ng ca electron trong nguyn t.

+ S ph b electron trn cc phn lp theo th t lp.

+ c im ca lp electron ngoi cng.

2. K nng:

+ Vn dng kin thc thnh phn cu to nguyn t, c im ca cc ht cu to nn nguyn t lm bi tp v cu to nguyn t.

+ Vn dng cc nguyn tc, nguyn l vit cu hnh electron nguyn t ca cc ng t.

+ Da vo c im lp electron ngoi cng phn loi cc nguyn t ha hc.

3. Thi : T duy 1 cch h thng phi hp cc ni dung kin thc v k nng gii bi tp.

II. CHUN B:

-Chun b ca Thy: Gio n, cc phiu hc tp.

-Chun b ca Tr: n li bi c.

III. HOT NG DY HC 1. On nh t chc: 1/.

2. Kim tra bi c

3. Ging bi mi: Tin trnh tit dy


10Hot ng 1: Kin thc cn nm vng

- GV: chia HS trong lp thnh cc nhm (khong 5-> 6 HS 1 nhm), trao i v bi tp lun phin cho nhau, t chm im (lm bi 1-> 5 Tr.34-SGK, mi bi 1 im).

- GV: Theo di, hng dn HS lm vic, sau GV ly vi quyn v HS chm nhn xt kt qu lm vic ca HS v gii p nhng vn kh m HS cn lng tng.

- GV: Da vo bng tm tt kin thc SGK, cng c cho HS theo tng nhm kin thc:- Trao i v cho nhau t chm.

- Nhm trng bo co tnh hnh hc tp ca cc bn trong nhm.A. Kin thc cn nm vng: (SGK)

7Hot ng 2 :Nhm kin thc v thnh phn cu to nguyn t:- GV: Hng dn HS n li nhng kin thc trng tm sau:

+ Nguyn t c thnh phn cu to nh th no v c im ca cc ht cu to nn nguyn t ?

+ THN, nguyn t ha hc v ng v ?

+ Kch thc v khi lng nguyn t ?- Tr li theo ni dung hc.2. Cng c l thuyt:

a. Nhm kin thc v thnh phn cu to nguyn t:

10 Hot ng 3 : Nhm kin thc v thnh phn cu to nguyn t:GV: Hng dn HS n li cc khi nim sau:

+ Nguyn t ha hc, ng v ?

+ Nguyn t khi, NTKTB, CT tnh ?

+ Chuyn ng ca electron trong nguyn t. Obitan ngt ?

+ Lp v phn lp electron. Cch k hiu lp v phn lp electron ?

+ S lng obitan trong 1 phn lp v trong 1 lp ?

+ Cc nguyn l v qui tc phn b electron ca nguyn t vo cc mc nng lng ?

+ Cch vit cu hnh electron nguyn t ca cac nguyn t ?- Tr li theo ni dung hc.b. Cng c nhm kin thc v nguyn t ha hc, v nguyn t:

17Hot ng 4 : Rn luyn k nng vn dng l thuyt gii bi tp.

- GV: Cho HS ln bng gii bi tp tiu biu hoc bi tp m nhiu HS cha gii c.

GV: Pht phiu hc tp s 1: in vo cc trng ca cc s sau:a.

c tnh ht

V nt

Ht nhn

E

P

N

in tch (q)

Khi lng (m)

b.

Lp

n=1

(K)

n=2

(L)

n=3

(M)

n=4

(N)

S phn lp

K hiu phlp

S e ti a phn lp

S e ti a lp

- GV: Ly vi phiu kim tra, nhn xt.

- GV: Pht phiu hc tp s 2: Ghp thng tin ct bn tri vi cc thng tin ct bn phi sao cho ng nht.

1. Nguyn t

A. Khng mang in

2. AO

B. Dng hnh khi cu

3.S khi

C. Trung ha in

4. NTK trung bnh

D. A=Z +N

5. Obitan s

E.

6. Obitan p

G. Hnh nh xc sut electron ln nht

H. Dng hnh s 8 ni


* Cng c: Nhc HS n tp cc kin thc hc, lm tip cc bi tp cn li SGK, SBT, tit sau n tp tip.- Hon thnh phiu hc tp 1.

- Hon thnh phiu hc tp 2.

B. Bi tp:

- Phiu hc tp s 1

- Phiu hc tp s 2.

40Hot ng 5 (40 pht): tit 13

- GV: Pht phiu hc tp s 3:

1. Mc nng lng ca cc obitan 2px, 2py v 2pz c khc nhau khng?v sao ?

2. Cc electron thuc lp K hay L lin kt vi ht nhn chc ch hn ?v sao ?

(GV: Minh ha qua hnh v sau)

K L M


- GV: Pht phiu hc tp s 4:

1.Trong nguyn t, nhng e ca lp no quyt nh tnh cht ha hc ca 1 nguyn t ? vd ?

2. Lp v ca 1 nguyn t c 20e. Hi:

a. Nguyn t c bao nhiu lp e ?

b. Lp ngoi cng c bao nhiu e ?

c. Nguyn t l kloi hay phi kim ?

- GV: Ln lt ghi cc bi tp sau ln bng:


- GV: Cho HS ghi bi tp 1 ( 4.

1. 3 nguyn t X, Y, Z sx theo th t THN tng dn. Nguyn t ca cc nguyn t ny u c 1e lp th 4. Vit cu hnh e v xc nh tn ca X, Y, Z.

2. Nguyn t R mt i 3e to ra cation R3+ c cu hnh e nguyn t phn lp ngoi cng l 2p6. Vit cu hnh e nguyn t v s phn b e theo obitan ca nguyn t R.

3. Vit k hiu ch phn lp e vi e cui cng :

a. Lp th 2, phn lp s v e l c thn.

b. Lp th 2, phn lp p, lng t th 2 (t tri sang) v e l c thn.

c. Lp th 3, phn lp p, lng t th 2 (t tri sang) v e l ghp i.

4. Trong t nhin nguyn t Clo c 2 ng v: 35Cl v 37Cl c t l s nguyn t tng ng l 3:1. Nguyn t Cu c 2 ng v, trong 6329Cu chim 73% s nguyn t. ng v clo to c hp cht CuCl2, trong % khi lng Cu chim 47,228%. Xc nh ng v th 2 ca Cu.- Hon thnh phiu hc tp 3.

- Tr li: ging nhau v l cc obitan thuc cng 1 phn lp.

- Tr li: Cc electron thuc lp K lin kt vi ht nhn chc ch hn v gn ht nhn hn v mc nng lng thp hn.

- Hon thnh phiu hc tp 4.

- Trong nguyn t, nhng e lp ngoi cng quyt nh tnh cht ha hc ca 1 nguyn t

- Tr li:

+ 4 lp electron.

+ Lp ngoi cng c 2 electron

+ L kim loi.

-X: 1s22s22p63s23p64s1 (K)

-Y: 1s22s22p63s23p63d54s1 (Cr)

-Z: 1s22s22p63s23p63d104s1 (Cu)

-R: 1s22s22p63s23p1

- Phiu hc tp s 3.

- Phiu hc tp s 4.

- Ghi bi tp 1 ( 4.

4. CNG C:(4/)

- BTVN:

1. Nguyn t A khng phi l kh him, nguyn t ca n c phn lp ngoi cng l 3p. Nguyn t ca nguyn t p c phn lp ngai cng l 4s.

- Nguyn t no l kim loi, phi kim ?

- Xc nh cu hnh electron c th c ca A v B v cho bit tn ca A. Bit tng s electron 2 phn lp ngoi cng ca chng l 7.

2. Hp cht MX2 c cc c im nh sau:

+ Tng s ht trong phn t l 140, trong s ht khng mang in km hn s ht mang in l 44.

+ NTK ca M nh hn NTK ca X l 11.

+ Tng s ht trong ion X- nhiu hn trong ion M2+ l 19.

Xc nh CTPT MX2. (p n: MgCl2 )

- Nhc HS v nh n tp tit sau kim tra vit.

IV.RT KINH NGHIM, B SUNGNgy son : 10/9/2013

Tit 14 KIM TRA 1 TIT ( Bi vit s 1)

I. MC TIU:

1/ Kin thc:

-Cc ni dung ca chng nh: cu to nguyn t, thnh phn nguyn t.

-Cu hnh electron nguyn t.

-nh gi kt qu hc tp ca HS qua vic lm bi kim tra.

2/ K nng:

-Rn luyn k nng lm bi c lp, t ch.

-Lm bi tp, nh li l thuyt hc trong chng I.

-Rn luyn k nng trnh by bi lm khi kim tra v thi c.

3/ Thi :

-Rn luyn s kin tr, chu kh hc tp.

-C thc hc tp ng n.

-C thc vn ln, t rn luyn bn thn lm ch kin thc.

II. CHUN B:

1/ Chun b ca gio vin:

- kim tra, p n, biu im.

2/ Chun b ca hc sinh:

-Chun b bi c ca chng.

-Giy lm bi, giy nhp, bt mc, my tnh... lm bi.

III. HOT NG KIM TRA:

GV pht kim tra v gim st vic lm bi ca HS.

Kt Qu

LpS sGiiKhTBYuKm

10

10

IV.RT KINH NGHIM, B SUNGNgy son: 15/9/2013

Tit: 15, 16

Chng II: BNG TUN HON CC NGUYN T HA HC

-NH LUT TUN HON

Bi: BNG TUN HON CC NGUYN T HA HC

I. MC TIU:

1. Kin thc: Nguyn tc xy dng BTH, cu to BTH.

2. K nng: Nm c mi quan h chc ch gia cu hnh electron nguyn t vi v tr ca nguyn t trong BTH.

3. Thi : H iu cu to BTH v ngha ca BTH.

II. CHUN B:

- Chun b ca Thy: Gio n, chun b hnh v nguyn t c phng to, BTH dng di.

- Chun b ca Tr: n li cch vit cu hnh electron nguyn t ca cc nguyn t.


2.Kim tra bi c:

3. Ging bi mi: Tin trnh tit dy:

* Tit 16 sau hot ng 3. Thi gianHot ng ca gio vinHot ng ca hc sinhNi dung

14Hot ng 1: nguyn tc sp xp cc nguyn t trong bng tun hon:

- GV: Gi 3 HS vit cu hnh e ca cc nguyn t hng 1:Z=1, Z=2 v hng 2:Z= 3,Z=4, ct dc s 1.

- GV: Da vo BTH yu cu HS nhn xt:

- THN ca cac nguyn t trong cng 1 hng ngang, trong cng 1 ct dc.

- S lp electron ca cc nguyn t trong cng 1 hng ngang, trong cng 1 ct dc.

- S electron ha tr ca cc nguyn t trong cng 1 hng ngang, trong cng 1 ct dc.

(Nn chn cc nguyn t thuc chu k nh, nhm A)

- GV: Ghi tm tt kin ca HS ln bng.

- GV: Da vo cc nhn xt yu cu HS rt ra nguyn tc xy dng BTH.- HS: Vit cu hnh e:

- 1s1 ; 1s2- 1s22p1 ; 1s22p2- HS: Da theo SGK, HS tr li cu hi.

- HS: a ra 3 nguyn tc xy dng BTH.

I. NGUYN TC SP XP CC NGUYN T TRONG BNG TUN HON:

- Cc nguyn t c xp theo chiu tng dn ca THN nguyn t.

- Cc nguyn t c cng s lp electron trong nguyn t c xt thnh 1 hng.

- Cc nguyn t c cng s electron ha tr trong nguyn t c xp thng 1 ct.

Electron ha tr l nhng e c kh nng tham gia hnh thnh lin kt ha hc. Chng thng nm lp ngoi cng hoc c phn lp st lp ngoi cng nu phn lp cha bo ha.

10Hot ng 2: cu to bng tun hon

- GV: Treo hnh v nguyn t.

- GV: Da vo s nguyn t c phng to trn bng yu cu HS nhn xt v thnh phn ca nguyn t .

- GV: nhn mnh nhng thnh phn khng th thiu trong 1 nguyn t, nh KHHH, s hiu ngyn t, NTKTB. Ngoi ra, c th c 1 s thng tin khc nh cu hnh electron ca nguyn t, m in,cc s oxi ho ca 1 nguyn t

- GV: nguyn t l n v nh nht cu to nn BTH. Mi nguyn t chim khong 1 , BTH c khong 110 nguyn t.- HS: Tr li:

- Mi nguyn t c xp vo 1 ca BTH ( gi l nguyn t.

- STT ca = Z

II. CU TO BNG TUN HON:

1. nguyn t:

- Mi nguyn t c xp vo 1 ca BTH ( gi l nguyn t.

- STT ca ng bng s hiu nguyn t ca nguyn t .

VD: nguyn t ca H, Al:

20 Hot ng 3: Chu k

- GV: Yu cu HS da vo BTH trong SGK cho bit c bao nhiu dy nguyn t c xp thnh hng ngang ?

- GV: Yu cu HS nhn xt s lng cc nguyn t trong mi chu k, vit cu hnh electron ca 1 s nguyn t tiu biu trong chu k, nhn xt s lp electron ca cc nguyn t trong cng chu k.

- GV: Ch s bt thng khi xy dng lp v electron ca nguyn t cc nguyn t thuc chu k 4 v chu k 5.

- GV: Nhng kin GV vit ln bng 1 cch h thng HS d dng so snh v rt ra kt lun.

- GV: T bng tng kt cho HS rt ra nhn xt:

+ S lng cc nguyn t trong mi chu k.

+ S lp electron ca nguyn t cc nguyn t trong mi chu k.

- GV b sung: Cc chu k 1, 2, 3 l chu k nh. T chu k 4 tr i l chu k ln. Ring chu k 7 cha hon chnh.

GV: Ghi kt lun trn bng.- HS: Tr li: c 7 hng ngang, mi hng ngang l 1 chu k c nh s th t bng ch s Arp t 1 n 7.

- HS: Tr li:

Cc chu k 1, 2, 3, mi chu k gm 8 nguyn t, tr chu k 1. Chu k 4, 5 c 18 nguyn t. Chu k 6 c 32 nguyn t. Chu k 7 cha hon thnh.

2. Chu k:

- Chu k l dy cc nguyn t m nguyn t ca chng c cng s lp electron, c xp theo chiu THN tng dn.

- BTH gm c 7 chu k. Cc chu k 1, 2, 3 l chu k nh, mi chu k gm 8 nguyn t, tr chu k 1 ch c 2 nguyn t. Cc chu k 4, 5, 6, 7 l chu k ln. Chu k 4, 5 c 18 nguyn t. Chu k 6 c 32 nguyn t. Chu k 7 cha hon thnh.

- STT ca chu k trng vi s lp electron ca nguyn t cc nguyn t trong chu k .

20Hot ng 4: Nhm

- GV: Da vo BTH dn dt HS tr li cc cu hi:

+Nhm nguyn t l g ?

+Cc nhm nguyn t c chia lm my loi ?

+C bao nhiu nhm A ? c im cu to nguyn t ca cc nguyn t thuc nhm A ?

+ C bao nhiu nhm B ? c im cu to nguyn t ca cc nguyn t thuc nhm B ?

- Th no l cc nguyn t s, p, d, f ?

- Cho bit v tr cc nguyn t s, p, d, f trong BTH ?

- GV: Trnh by thm v cc nguyn t xp cui bng.

- Hng dn HS cch phn loi theo khi.

- Lu : STT nhm bng s e ha tr:

+ Nhm A: STT nhm bng s e lp ngoi cng.

+ Nhm B: STT nhm bng s e ha tr (bng tng s e phn lp s ngoi cng v phn lp d st n, sgv: nhng nguyn t d c phn lp d bo ha th STT nhm bng s e lp ngoi cng)

- GV: Cho HS lm bi tp vn dng: Vit cu hnh e nguyn t cc nguyn t Br (Z=35), Mn(Z=25), Ba (Z=56) v xc nh v tr ca nguyn t trong BTH.

- HS: Tr li cc cu hi theo ni dung ca SGK.

-HS: Tr li:

- 1s22s22p63s23p63d104s24p5 Chu k 4, nhm VIIA

- 1s22s22p63s23p63d54s2 Chu k 4, nhm VIIB

1s22s22p63s23p63d104s24p6

4d105s25p66s2 Chu k 6, nhm IIA

3. Nhm nguyn t:

- Nhm nguyn t l tp hp cc nguyn t m nguyn t c cu hnh electron tng t nhau, do c tnh cht ha hc gn ging nhau v c xp thnh 1 ct.

- STT nhm = s electron ha tr (tr 1 s trng hp ngoi l)

- Phn loi theo nhm:

+ Nhm A: C 8 nhm t: IA ( VIIIA.

+ Nhm B: C 8 nhm t: IB ( VIIIB.

Mi nhm 1 ct, ring nhm VIIIB gm 3 ct.

- Phn loi theo khi:

+ Khi cc nguyn t s: gm nguyn t nhm IA, IIA.

Nguyn t s l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp s.

VD: Na, Mg: vit cu hnh e.

+ Khi cc nguyn t p: gm nguyn t t nhm IIIA n VIIIA (tr He).

Nguyn t p l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp p.

VD: O, S, Ne: vit cu hnh e.

+ Khi cc nguyn t d: gm nguyn t thuc cc nhm B.

Nguyn t d l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp d.

VD: Cu, Fe: vit cu hnh e.

+ Khi cc nguyn t f: gm nguyn t thuc h Lantan v Actini.

Nguyn t f l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp f.

VD: Ce, Ac: vit cu hnh e.

Vy:

- Cc nhm A bao gm cc nguyn t s v nguyn t p.

- Cc nhm B bao gm cc nguyn t d v nguyn t f.

23Hot ng 5: Cng c

+ Cng c l thuyt:

- BTH c xy dng da trn nguyn tc no ?

- Mi quan h gia cu hnh e v v tr cc nguyn t trong BTH ?

+ Bi tp:

Bi 1: Nguyn t X c phn lp e ngoi cng l 3p1. Hy ch ra iu sai khi ni v nguyn t X:

a. Ht nhn nguyn t X c 16 proton.

b. Lp ngoi cng ca nguyn t X c 6e.

c. Trong BTH X nm nhm IVA.

d. X l 1 nguyn t phi kim.

Bi 2: Hy ch ra cu tr li ng: Cation R+ c cu hnh e phn lp ngoi cng l 2p6. V tr ca R trong BTH l:

a. Chu k 2, nhm IVA. b. Chu k 2, nhm VIIIA.

c. Chu k 3, nhm IA. c . Chu k 2, nhm VIB. e. Tt c u sai.


Bi dy: S BIN I TUN HON CU HNH

Tit: 17.

ELECTRON NGUYN T CC NGUYN T HA HC

I. MC TIU:

1. Kin thc:

*. Hiu c:

- S bin i tun hon cu hnh electron nguyn t ca cc nguyn t ha hc.

- Mi lin quan gia cu hnh electron nguyn t ca cc nguyn t vi v tr ca chng trong BTH.

*. Bit c: c im cu hnh electron ha tr ca nguyn t cc nguyn t nhm B.

2. K nng:

- Da vo cu hnh electron ca nguyn t nhm A, suy ra cu to nguyn t, c im cu hnh electron lp ngoi cng.

- Da vo cu hnh electron xc nh nguyn t s, p, d.

3. Thi : Hiu c ngha ca BTH.

II. CHUN B:

- Chun b ca Thy: Gio n, BTH dng di.

- Chun b ca Tr: n bi cu to BTH cc nguyn t ha hc.

III. HOT NG DY HC 1. On nh t chc: (1 ph)

2.Kim tra bi c: ( 5ph) Vit cu hnh e nguyn t cc nguyn t Br (Z=19), Mn(Z=25), Ba (Z=28) v xc nh v tr ca nguyn t trong BTH.

3. Bi mi: NI DUNG


10GV: Chun b 8 phiu hc tp, mi phiu ghi sn Z ca khong 3 hoc 4 nguyn t nhm A pht cho 8 nhm HS v yu cu vit cu hnh electron nguyn t. Sau cho HS ln bng in vo bng 1:

HS:

- Hon thnh cc phiu hc tp .

- C i din nhm ln bng in vo bng sau:I. Tm hiu cu hnh electron nguyn t ca cc nguyn t nhm A.

12Hot ng 2: T chc cho HS rt ra nhn xt

GV: T cu hnh electron nguyn t va xy dng hy nhn xt v c im cu hnh electron lp ngoi cng ca nguyn t cc nguyn t theo chu k, theo nhm.

Nguyn t s nhm no, nguyn t p nhm no ?HS: Rt ra nhn xt.

HS:

- Nguyn t s: gm nguyn t nhm IA, IIA.

- Nguyn t p: gm nguyn t nhm IIIA n VIIIA.

*Nhn xt:

- Nguyn t ca cc nguyn t trong cng 1 nhm c s electron lp ngoi cng bng nhau v bng STT ca nhm. chnh l nguyn nhn lm cho cc nguyn t trong cng 1 nhm c tnh cht ha hc tng t nhau.

- Sau mi chu k, cu hnh electron lp ngoi cng ca nguyn t cc nguyn t nhm A c lp li. chnh l nguyn nhn ca s bin i tun hon tnh cht cc nguyn t.

*Kt lun: S bin i tun hon v cu hnh electron lp ngoi cng ca nguyn t cc nguyn t khi THN tng dn chnh l nguyn nhn ca s bin i tun hon v tnh cht ca cc nguyn t.

10GV:

- Da vo BTH hy nhn xt v tr ca cc nguyn t nhm B trong BTH.

- Yu cu HS vit cu hnh electron nguyn t ca 1 s nguyn t: Z=22, Z=25, Z=30 ( nhn xt v c im xy dng lp v electron nguyn t ca cc nguyn t nhm B. Lu 1 s trng hp c bit nh nguyn t Cr, Cu c s chuyn 1 electron phn lp ns vo phn lp (n-1)d nhanh chng t cu hnh bn bo ha hoc na bo ha.

GV: Thng bo s electron ha tr ca cc nguyn t nhm B, vd.

+ STT nhm B bng s e ha tr (bng tng s e phn lp s ngoi cng v phn lp d st n, sgv: nhng nguyn t d c phn lp d bo ha th STT nhm bng s e lp ngoi cng)

HS: Cc nguyn t B u thuc chu k ln.

- 1s22s22p63s23p63d24s2- 1s22s22p63s23p63d54s2- 1s22s22p63s23p63d104s2HS: Cu hnh electron ca nguyn t cc nguyn t ny c dng: (n-1)dans2II. Tm hiu cu hnh electron nguyn t ca cc nguyn t nhm B:

- Cc nguyn t B u thuc chu k ln. Chng l cc nguyn t d v nguyn t f ( gi l cc kim loi chuyn tip.

- Cu hnh electron ca nguyn t cc nguyn t ny c dng:

(n-1)dans2 (a=1-> 10, tr 1 s trng hp ngoi l)

Vd: Cu hnh electron ca Cr, Cu: 1s22s22p63s23p63d104s1

Cr: 1s22s22p63s23p63d54s2

- Cc nguyn t d. f c s electron ha tr nm lp ngoi cng hoc c phn lp st lp ngoi cng cha bo ha, khi phn lp st ngoi cng bo ha th s electron ha tr c tnh theo s electron lp ngoi cng.

Vd: Ag: [Kr]4d105s1, s electron ha tr bng 1.

7V.CNG C

1. Cng c l thuyt:

- Nhn xt cu hnh electron nguyn t ca cc nguyn t nhm A ?

- Cu hnh electron nguyn t ca cc nguyn t nhm B ?

2. Bi tp v nh: 1( 5 (SGK), 2.9 ( 2.13 (SBT)

+ Bi tp:Bi 1: in vo ch trng nhng t, cm t cn thit:

Chu k bao gm cc nguyn t c sp xp theo chiu THN tng dn. Nguyn t ca cc nguyn t trong cng chu k c cng s lp electron. S th t ca chu k trng vi s lp electron nguyn t ca nguyn ttrong chu k . Trong mi chu k, s electron lp ngoi cng tng dn. M u mi chu k bao gi cng l nguyn t c 1 electron lp ngoi cng v kt thc ca mi chu k bao gi cng l nguyn t c 8 electron lp ngoi cng (tr chu k 1). Nh vy, theo chiu tng dn ca THN, cu hnh electron nguyn t ca cc nguyn t bin i tun hon.

Bi 2: Mnh no sau y khng ng ?

a. Nguyn t ca cc nguyn t trong cng 1 nhm A bao gi cng c s electron l`p ngoi cng bng nhau.

b. S th t nhm bng s e lp ngoi cng ca nguyn t nguyn t trong nhm .

c. Cc nguyn t trong cng 1 nhm c tnh cht ha hc tng t nhau.

d. Trong 1 nhm, nguyn t ca 2 nguyn t thuc hai chu k lin tip hn km nhau 1 lp electron.

e. Tnh cht ha hc ca cc nguyn t trong cng 1 nhm A bin i tun hon.

p n: b, e khng ng.


Tit: 18 Bi : S BIN I MT S I LNG VT L

CA CC NGUYN T HA HC

I. MC TIU:

1. Kin thc: Hiu c: khi nim v qui lut bin i tun hon ca bn knh nguyn t, nng lng ion ha, m in ca 1 s nguyn t, trong 1 chu k, trong 1 nhm A.

2. K nng: Da vo qui lut chung suy on c s bin thin tnh cht c bn trong chu k, nhm A. C th s bin thin v: m in, bn knh nguyn t, nng lng ion ha th nht.

3. Thi : Gip HS tin tng vo khoa hc tm ra qui lut bin i mt s i lng vt l ca cc nguyn t ha hc.

II. CHUN B:

- Chun b ca Thy: Gio n, chun b hnh v 2.1, 2.2, 2.3 v bng 2.2, 2.3.

- Chun b ca Tr: n bi c, xem trc bi mi.


2.Kim tra bi c: (5 pht) Nguyn nhn no lm cho tnh cht ca cc nguyn t bin i 1 cch tun hon ? cho v d ?

3. Tin trnh tit dy: Thi gianHot ng ca gio vinHot ng ca hc sinhNi dung

10GV: Cho HS xem bng 2.2 v nu qui lut bin i bn knh nguyn t ca cc nguyn t theo chu k v theo nhm ?

GV: Da vo c im cu to ca cc nguyn t trong 1 chu k v trong 1 nhm, GV hng dn HS gii thch qui lut bin i bn knh nguyn t theo chu k, theo nhm.

GV: Nu kt lun bin i bn knh nguyn t ?HS: Xem bng v a ra qui lut bin i bn knh nguyn t ca cc nguyn t theo chu k v theo nhm.

HS: Gii thch qui lut bin i bn knh nguyn t theo chu k, theo nhm.I. Bn knh nguyn t:

+ Trong 1 chu k theo chiu tng dn ca THN, bn knh nguyn t ca cc nguyn t gim dn.

Gii thch: Trong 1 chu k, nguyn t cc nguyn t c cng s lp electron, nhng khi THN tng, lc ht gia ht nhn vi cc electron lp ngoi cng cng tng theo, do bn knh nguyn

t ni chung cng gim dn.

+ Trong 1 nhm, theo chiu tng dn ca THN, bn knh nguyn t ca cc nguyn t tng dn.

Gii thch: Trong 1nhm,theo chiu t trn xung di, s lp electron tng, do bn knh nguyn t tng nhanh

*Kt lun: Bn knh nguyn t ca cc nguyn t nhm A bin i tun hon theo chiu tng ca THN.

5GV: Cho HS tm hiu SGK bit nng lng ion ha l g ?

GV: B sung: nng lng ion ha ni n trn l nng lng ion ha th nht (I1 ). Ngoi ra cn c I2, I3, I4(I1

t1 20 hoa 10 nc hai

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