sigma(v) sigma' (v) 1 2 3 4 5 6 7 8 9 10 Chapter 15: E.9: a) NO2 : E ...

Chapter 15:
E.9:
a) NO 2 : E, C 2 , s v , s v ’ > C 2v
b) N 2 O: E, C 4 , C 2 , s v , > C 4v
c) CHCl 3 : E,C 3 , 3s v > C 3v
d) CH 2 =CH 2 : E, C 2 , 2C’ 2 , s h > D 2h
e) cis- CHBr=CHBr: E, C 2 , s v , s v ’ > C 2v
f) trans-CHCl=CHCl: E, C 2 , s h , i > C 2h
E.5:
Must determine how x and y individually transform then determine how xy transforms.
E C 2 C 4 s v s d
x x -x y x -y
y y -y -x -y -x
xy xy xy -xy -xy xy
x 1 1 -1 -1 1
then from the C 4v character table it is found that this set is of B 2 .
sigma' (v)
P.12:
7
6
8
5
1
9
10
4
2
3
sigma(v)
*all numbers in table are subscripts to pi, except in the “X” column.
1 2 3 4 5 6 7 8 9 10 X
E 1 2 3 4 5 6 7 8 9 10 10
C2 5 6 7 8 1 2 3 4 10 9 0
sigma(v) 4 3 2 1 8 7 6 5 10 9 0

sigma’(v) 8 7 6 5 4 3 2 1 9 10 2
From the character set (10,0,0, 2) we can get to 3A 1 + 2A 2 + 2B 1 + 3B 2
* once again all number are subscripts to pi.
A 1 = 1+4+5+8
A 1 = 2+3+6+7
A 1 = 9+10
A 2 = 1+5-4-8
A 2 = 2+6-3-7
B 1 = 1-5+4-8
B 1 = 2-6+3-7
B 2 = 1-5-4+8
B 2 = 2-6-3+7
B 2 = 9-10