Химическая кинетика (задачи, примеры, задания). Пурмаль А.П - 158 стр.

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158
N
2
O
4
D
2h
ONO = 135,4
o
r
N-N
=1 782
r
N-O
=1,190
ν
1
=1380; ν
7
=425
ν
2
=808; ν
8
=672
ν
3
=266; ν
9
=1758
ν
4
=79; ν
10
=270
ν
5
=1712; ν
11
=1264
ν
6
=482; ν
12
=751
9,2 ± 1,7
CH
3
NO
2
C
s
ONO = 125,3
o
NCH = 107,2
o
r
C-N
=1,489
r
N-O
=1,224
r
C-H
=1 088
ν
1
=3048; ν
9
=599
ν
2
=2965; ν
10
=3048
ν
3
=1488; ν
11
=1582
ν
4
=1449; ν
12
=1413
ν
5
=1384; ν
13
=1097
ν
6
=1153; ν
14
=476
ν
7
=921; ν
15
=(200)
ν
8
=647
74,7
SF
6
O
h
r
S-F
=1,564
ν
1
=770, ν
6
=947(3)
ν
2
=640 (2)
ν
3
=522 (3)
ν
4
=345 (3)
ν
5
=615 (3)
1221 ± 1
H
2
SO
4
C
2v
O-S-O = (100
o
)
O=S=O = (125
o
)
SOH = (105
o
)
r
S=O
=(1,42)
r
O-H
=(0,97)
r
S-O(H)
=(1,53)
ν
1
=3500; ν
9
=3610
ν
2
=1223; ν
10
=1159
ν
3
=1138; ν
11
=883
ν
4
=834; ν
12
=400
ν
5
=550; ν
13
=1450
ν
6
=380; ν
14
=450
ν
7
=450; ν
15
=568
ν
8
=390;
741 ± 8
N
2
O
5
C
2v
O=N=O =134 ± 9
o
NON =95 ± 3
o
r
N=O
=1,21
r
N-O
=1,46
ν
1
=1728; ν
9
=577
ν
2
=1338' ν
11
=1728
ν
3
=743; ν
12
=1247
ν
4
=614; ν
13
=860
ν
5
=353; ν
14
=743
ν
6
=(85); ν
15
=353
ν
7
=(614);
ν
8
=?; ν
10
=?
11,3 ± 1,5
 N2O4           D2h             ν1=1380; ν7=425    9,2 ± 1,7
          ∠ONO = 135,4o          ν2=808; ν8=672
            rN-N=1 782          ν3=266; ν9=1758
            rN-O=1,190           ν4=79; ν10=270
                              ν5=1712; ν11=1264
                                ν6=482; ν12=751
CH3NO2          Cs              ν1=3048; ν9=599      −74,7
          ∠ONO = 125,3o       ν2=2965; ν10=3048
          ∠NCH = 107,2o       ν3=1488; ν11=1582
            rC-N=1,489        ν4=1449; ν12=1413
            rN-O=1,224        ν5=1384; ν13=1097
            rC-H=1 088         ν6=1153; ν14=476
                               ν7=921; ν15=(200)
                                     ν8=647
 SF6            Oh            ν1=770, ν6=947(3)    −1221 ± 1
            rS-F=1,564        ν2=640 (2)
                              ν3=522 (3)
                              ν4=345 (3)
                              ν5=615 (3)
H2SO4            C2v           ν1=3500; ν9=3610    −741 ± 8
         ∠O-S-O = (100o)      ν2=1223; ν10=1159
         ∠O=S=O = (125o)       ν3=1138; ν11=883
          ∠SOH = (105o)         ν4=834; ν12=400
            rS=O=(1,42)        ν5=550; ν13=1450
            rO-H=(0,97)         ν6=380; ν14=450
           rS-O(H)=(1,53)       ν7=450; ν15=568
                                     ν8=390;
 N2O5           C2v             ν1=1728; ν9=577    11,3 ± 1,5
         ∠O=N=O =134 ± 9o     ν2=1338' ν11=1728
          ∠NON =95 ± 3o        ν3=743; ν12=1247
            rN=O=1,21           ν4=614; ν13=860
            rN-O=1,46           ν5=353; ν14=743
                                ν6=(85); ν15=353
                                    ν7=(614);
                                   ν8=?; ν10=?

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