Химическая кинетика (задачи, примеры, задания). Пурмаль А.П - 157 стр.

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157
XeO
4
T
d
r
Xe-O
=1,736
ν
1
=776
ν
2
=267(2)
ν
3
=879(3)
ν
4
=306(3)
………
BH
3
CO C
3v
∠HBC = 104
o
37′
∠BCO = 180
o
∠HBH = 113
o
52′
r
B-H
=1,194
r
C-O
=1,131
r
B-C
=1,540
ν
1
=2385
ν
2
=2166
ν
3
=1083
ν
4
=707
ν
5
=2456(2)
ν
6
=1100(2)
ν
7
=819(2)
ν
8
=314(2)
Δ
f
H
0
0
=
−177
CH
3
OH C
s
ν
1
=3682; ν
7
=1075
ν
2
=2999; ν
8
=1034
ν
3
=2844; ν
9
=2970
ν
4
=1478; ν
10
=1465
ν
5
=455; ν
11
=1145
ν
6
=1336; ν
12
=270
−202
CH
3
CO
……… ……… −18 ± 4
CH
3
CN C
3v
∠CCH = 109,7
o
r
C-H
=1,107
r
C-N
=1,159
r
C-C
=1,468
ν
1
=2954
ν
2
=2278
ν
3
=1111
ν
4
=920
ν
5
=3009(2)
ν
6
=2257(2)
ν
7
=1047(2)
ν
8
=1042(2)
66 ± 7
C
2
H
4
D
2h
∠HCH = 117
o
22′
∠CCH = 121
o
19′
r
C-H
=1,086
r
C=C
=1,337
ν
1
=3026; ν
7
=949
ν
2
=1623; ν
8
=3136
ν
3
=1342; ν
9
=1374
ν
4
=1023; ν
10
=1130
ν
5
=3102; ν
11
=769
ν
6
=1222; ν
12
=756
52,5 
XeO4           Td                 ν1=776       ………
          rXe-O=1,736           ν2=267(2)
                                ν3=879(3)
                                ν4=306(3)
BH3CO         C3v                ν1=2385       ΔfH00 =
        ∠HBC = 104o37′           ν2=2166        −177
         ∠BCO = 180o             ν3=1083
        ∠HBH = 113o52′            ν4=707
          rB-H=1,194           ν5=2456(2)
          rC-O=1,131           ν6=1100(2)
          rB-C=1,540            ν7=819(2)
                                ν8=314(2)
CH3OH         Cs           ν1=3682; ν7=1075     −202
                           ν2=2999; ν8=1034
                           ν3=2844; ν9=2970
                           ν4=1478; ν10=1465
                            ν5=455; ν11=1145
                            ν6=1336; ν12=270
CH3CO       ………                   ………          −18 ± 4
CH3CN         C3v                ν1=2954        66 ± 7
        ∠CCH = 109,7o            ν2=2278
          rC-H=1,107             ν3=1111
          rC-N=1,159              ν4=920
          rC-C=1,468           ν5=3009(2)
                               ν6=2257(2)
                               ν7=1047(2)
                               ν8=1042(2)
C2H4          D2h           ν1=3026; ν7=949     52,5
        ∠HCH = 117o22′     ν2=1623; ν8=3136
        ∠CCH = 121o19′     ν3=1342; ν9=1374
          rC-H=1,086       ν4=1023; ν10=1130
          rC=C=1,337        ν5=3102; ν11=769
                            ν6=1222; ν12=756




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