Химическая кинетика (задачи, примеры, задания). Пурмаль А.П - 156 стр.

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156
CF
4
T
d
r
C-F
=1,322
ν
1
=908
ν
2
=435(2)
ν
3
=1272(3)
ν
4
=632(3)
−933
CCl
4
T
d
r
C-Cl
=1,766
ν
1
=459
ν
2
=218(2)
ν
3
=776(3)
ν
4
=312(3)
−103
SiH
4
T
d
r
Si-H
=1,480
ν
1
=2185
ν
2
=972(2)
ν
3
=2189(3)
ν
4
=913(3)
35 ± 1
N
2
O
3
O O
′
N−N′
O′′
C
s
∠ONN′ = 105,1
o
∠O′N′N = 112,7
o
∠O′′N′N = 117,5
o
r
N-N
′
=1,864, r
N-O
=
1,142, r
N
′
-O
′
=1,202
r
O
′′
-N
′
=1,217
ν
1
=1832 ν
8
=337
ν
2
=1652 ν
9
=63
ν
3
=1305
ν
4
=773
ν
5
=414
ν
6
=241
ν
7
=160
83,3 ± 1
SO
3
F C
3v
∠OSF = 108,6
o
r
S-O
=1,46
r
S-F
=1,64
ν
1
=1056
ν
2
=839
ν
3
=534
ν
4
=1178(2)
ν
5
=604(2)
ν
6
=369(2)
………
SO
4
C
2v
1434; 1267; 925;
777; 611; 498; 490
H
2
O
3
(C
2
)
ν
3
≈855
ν
4
=500
ν
8
=755
………
FClO
3
C
3v
∠OClO = 116,5
o
∠OClF = 100,8
o
r
Cl-O
=1,404
r
Cl-F
=1,619
ν
1
=1061
ν
2
=707
ν
3
=549, ν
6
=405(2)
ν
4
=1315(2)
ν
5
=589(2)
−21,4 ± 2 
    CF4                     Td                    ν1=908          −933
                       rC-F=1,322               ν2=435(2)
                                               ν3=1272(3)
                                                ν4=632(3)
    CCl4                   Td                     ν1=459          −103
                      rC-Cl=1,766               ν2=218(2)
                                                ν3=776(3)
                                                ν4=312(3)
    SiH4                   Td                    ν1=2185         35 ± 1
                      rSi-H=1,480               ν2=972(2)
                                               ν3=2189(3)
                                                ν4=913(3)
    N2O3                     Cs            ν1=1832     ν8=337   83,3 ± 1
               ′
O          O        ∠ONN′ = 105,1o         ν2=1652     ν9=63
    N−N′           ∠O′N′N = 112,7o         ν3=1305
           O′′     ∠O′′N′N = 117,5o        ν4=773
                   rN-N′=1,864, rN-O =     ν5=414
                   1,142, rN′-O′=1,202     ν6=241
                      rO′′-N′=1,217        ν7=160
    SO3F                    C3v                  ν1=1056         ………
                    ∠OSF = 108,6o                 ν2=839
                        rS-O=1,46                 ν3=534
                        rS-F=1,64              ν4=1178(2)
                                                ν5=604(2)
                                                ν6=369(2)
    SO4                   C2v               1434; 1267; 925;
                                           777; 611; 498; 490
    H2O3                  (C2)                  ν3 ≈855          ………
                                                ν4=500
                                                ν8=755
    FClO3                 C3v              ν1=1061              −21,4 ± 2
                   ∠OClO = 116,5o          ν2=707
                   ∠OClF = 100,8o          ν3=549, ν6=405(2)
                     rCl-O=1,404           ν4=1315(2)
                     rCl-F=1,619           ν5=589(2)

                                         156

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