Химическая термодинамика (задачи, примеры, задания). Захаров И.В - 94 стр.

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94
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)
………
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
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
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 
    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)
    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)
    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; ν4=707
                   ∠HBH = 113o52′             ν5=2456(2)
                     rB-H=1,194               ν6=1100(2)
                     rC-O=1,131                ν7=819(2)
                     rB-C=1,540                ν8=314(2)
    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)

                                         94

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