giao an hoa 10 cb theo chuan
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Th by ngy 20 thng 08 nm 2011
Gio n ho hc lp 10 c bn- Nm hc 2011-2012
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Th by ngy 20 thng 08 nm 2011
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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Th by ngy 20 thng 08 nm 2011
Tit th 2: N TP U NM (tip)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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
III.PHNG PHP GING DY: Din ging pht vn- kt nhm.
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
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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:
1.n nh lp: Kim tra s s, ng phc...
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)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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
III.PHNG PHP GING DY: Din ging pht vn- kt nhm.
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:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (8 pht)
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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Th hai ngy 5 thng 9 nm 2011
Tit th 5: HT NHN NGUYN T- NGUYN T HO HC-NG V (tit 2)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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
III.PHNG PHP GING DY: Din ging pht vn- kt nhm.
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:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (8 pht)
- 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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Th hai ngy 12 thng 9 nm 2011
Tit th 6: LUYN TP: THNH PHN NGUYN T
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (8 pht)
- 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)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- Thnh phn cu to nguyn t
- 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
III.PHNG PHP GING DY: Din ging pht vn
IV. CHUN B GIO C:
*Gio vin: M hnh mu hnh tinh nguyn t
*Hc sinh: Chun b bi mi trc khi n lp.
V. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (7 pht)
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)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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
III.PHNG PHP GING DY: Din ging pht vn
IV. CHUN B GIO C:
*Gio vin: Gio n in t
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
V. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (8 pht)
- 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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Th by ngy 24 thng 9 nm 2011
Tit th 9: CU HNH ELECTRON NGUYN T
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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.
III.PHNG PHP GING DY: Din ging pht vn- kt nhm.
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.
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
V. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
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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Th by ngy 24 thng 9 nm 2011
Tit th 10:
LUYN TP: CU TO V ELECTRON CA NGUYN T(tit1)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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:
1.Kin thc: Cng c kin thc v:
- 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
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
IV. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
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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Th by ngy 1 thng 10 nm 2011
Tit th 11: LUYN TP: CU TO V NGUYN T(tit 2)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- Thnh phn cu to nguyn t
- 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
II.PHNG PHP GING DY: Din ging pht vn- kt nhm.
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:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (0 pht)
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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Th by ngy 1 thng 10 nm 2011
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
( Phi kim v lp e ngoi cng c 7e
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
( Kim loi v lp e ngoi cng c 1e
22621
13
:12233
Alsspsp
( Kim loi v lp e ngoi cng c 3e
22624
16
:12233
Ssspsp
( Phi kim v lp e ngoi cng c 6e
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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Th by ngy 29 thng 10 nm 2011
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)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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:
1.Kin thc: Cng c kin thc v:
- 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
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
IV. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
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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Th hai ngy 31 thng 10 nm 2011
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)
Kin thc c c lin quan
Kin thc mi trong bi cn hnh thnh
- 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
II.PHNG PHP GING DY: Din ging pht vn- kt nhm.
III. CHUN B GIO C:
*Gio vin: Gio n
*Hc sinh: Chun b bi mi trc khi n lp.
IV. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
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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Th by ngy 03 thng 12 nm 2011
Tit th 30: PHN NG OXI HO - KH (tit 2)
Kin thc c c lin quan
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).
3.Thi : Tch cc, ch ng
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
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
V. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (7 pht)
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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Th by ngy 17 thng 12 nm 2011
Tit th 31: PHN LOI PHN NG TRONG HO HC V C
Kin thc c c lin quan
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.
3.Thi : Tch cc, ch ng
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
*Hc sinh: Hc bi c, chun b bi mi trc khi n lp.
V. TIN TRNH BI DY:
1.n nh lp: Kim tra s s, ng phc...
2.Kim tra bi c: (7 pht)
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
- 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
- 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
- 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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