Bt chuong 2

2 - -

2.1 4

ă mol hydro

S: 24 6 x 10 4 (J/mol) 2.2

4

S: l = 0, m=0 l = 1, m = -1, 0, +1 l = 2, m = -2, -1, 0, +1, +2 l = 3, m = -3, -2, -1, 0, +1, +2, +3 16

2.3 Cr, Fe, Cu bi t bi t r â ầ 14 1 24 2 29 S: 1s 2s 2p 3s 3p 3d 4s 8O: 14Si: 16S: 24Cr: 26Fe: 29Cu: 2.4 â

a) 1s22s22px12py1; b) 1s22s22px22py1; c) 1s22s22px12py12pz13s2; d) 1s22s22px22py12pz13s2 S:

:

b) 1s 2 2s 2 2p x 1 2p y 1 2p z 1 ; c) 1s 2 2s 2 2p x 2 2p y 2 2p z 1 ; d) 1s 2 2s 2 2p x 2 2p y 2 2p z 2 2.5 3 1 3 2

3 2 -1 4 1 4 2 1 s = +1/2

S: a) 6; b) 10; c) 2; d) 2; e) 1

2.6 s

a) 1,0 ,0, -1/2; b) 3,3,0,1/2; c) 3,2,3,+1/2; d) 1,1,2,+1/2

S:

1

ă  3 0 0 1/2

 3 2 2 -2, -1, 0 +1),+1/2

 1 0 0 1 2

2.7 ă ầ bi t

r â ầ 2 1 19 9 13

S:

F < O < S < Al < Ca < K

2.8 ă ầ 2+, K+, Al3+; b) Se2-,

Br-, Te2- Bi â a Ca, K, Al lầ 2 19 13 a Se, Br, Te lầ 34

35, 52

S: a) Al 3+

< Ca 2+ < K + b) Br - < Se 2- < Te 2-

2.9 ă ă ầ ă

Mg, Al, Si, P bi t â a Na, Mg, Al, Si, P lầ 11, 13, 13, 14, 15

S: Na < Al < Mg < Si < P

2.10 ă ầ bi t

â a lầ 5, 11, 9, 8

S: Na < B < O < F

2.11 ầ 2 ă

â 1= 1086; IE2 = 2352; IE3 = 4620; IE4 = 6222; IE5 = 37831;

IE6 = 47277

S: 1s 2

2s 2 2p 2

2.12 ầ 3 ă

â 1= 578; IE2 = 1817; IE3 = 2745; IE4 = 11576; IE5 = 14841; IE6 = 18379 S: 1s 2 2s 2 2p 6 3s 2 3p 1 2.13 ù 4 S: 8 2.14 - â 2, O2+, O2-, O2

c) so

S: O 2 :  1s 21s *22s 22s *22p 22py 2 = 2pz 22py *1 = 2pz *1  2

O 2 + :  1s 21s *22s 22s *22p 22py 2 = 2pz 22py *1  2 5

O 2 - :  1s 21s *22s 22s *22p 22py 2 = 2pz 22py *2 = 2pz *1  1 5

O 2 2- :1s 21s *22s 22s *22p 22py 2 = 2pz 22py *2 = 2pz *2  1

O 2 > O 2 + > O 2 - > O 2 2- O 2 < O 2 + < O 2 - < O 2 2- 2.15 2+ 2 2+

2

S: 2 + 2 5 2 2 2 + > O 2

2 + 2 5 2 3 2 + < N 2 2.16 - 22+ 2 S: B 2 2+ :  1s 21s *22s 22s *2  4-4)/2 = 0 F 2 2- :  1s 21s *22s 22s *22p 22py 2 = 2pz 22py *2 = 2pz *22p *2  10-10)/2 = 0 0