Good question. However the statement itself might not require a reference as it has none in IUPAC gold book for example.(?)
doi:10.1351/goldbook.S05695.
Dear Valeria,
it is a good question .
Please read this link :
https://www3.nd.edu/~ndrlrcdc/Compilations/QY/qy.pdf
Here the introduction of this article:
Ground state molecular oxygen, O2, is well known to be highly reactive and essential for animal respiration.
Its properties have been actively studied for more than 200 years. However the properties of its lowest electronically
excited state have only been actively investigated during the last three decades. This is illustrated in the book
entitled Singlet Molecular Oxygen1 where Paul Schaap collects 28 Benchmark papers covering the period
1924-1973 which tell the story of the ‘discovery’ of singlet oxygen and many of the key steps taken towards present
day understanding. There is strong evidence for the involvement of singlet oxygen, a powerful oxidant, in photosensitized
oxidations, in photodynamic inactivation of viruses and cells, in phototherapy for cancer, in photocarcinogenisis,
in the photodegradation of dyes and polymers and in the dye sensitization of the photodegradation of
polymers. Much current interest in the chemical reactions of singlet oxygen stems from its potential as a
photochemo-therapeutic agent.
The ground electronic state of molecular oxygen, which has zero angular momentum about the internuclear
axis and contains two unpaired p electrons, has the group theoretical symbol 3
Σg
−
. The two electronically excited
singlet states which arise from this same electron configuration but with spin pairing of the two electrons are the..........................................
and scheme 1
Rate Constants Quantum Yields and
Fractional Probabilities
1 S0 + hν → S1 rate = Ia
2 S1 → S0 + hνF k F φF = kSD
____
kF
3 S1 → S0 kic
E
A
F
A
G
kSD = (τS
0 )
−1
4 S1 → T1 kisc φT = kSD
_
k
___isc
5 S1 + O2 → T1 + 1O2* kS∆
O2 f∆
S = kS∆
O2 / kSQ
O2
6 S1 + O2 → T1 + 3O2 kisc
O2 fT
O2 = ( kS∆
O2 + kisc
O2 ) / kSQ
O2
7 S1 + O2 → S0 + 3O2 kSd
O2
E
A
A
F
A
A
G
kSQ
O2
PS
O2 = kSQ
O2 [O2]/( kSD + kSQ
O2 [O2] )
8 S1 + O2 → products kSr
O2 1 − PS
O2 =1/(1+ kSQ
O2 τS
0 [O2] )
φT
O2 = φT ( 1 − PS
O2 ) + fT
O2 PS
O2
9 T1 → S0 + hνP kTp
10 T1 → S0 kTd
E
F
G
kTD = (τT
0 )
−1
11 T1 + O2 → S0 + 1O2* kT∆
O2 f∆
T = kT∆
O2 / kTQ
O2
12 T1 + O2 → S0 + 3O2 kTd
O2
E
A
F
A
G
kTQ
O2
13 T1 + O2 → products kTr
O2 PT
O2 = kTQ
O2 [O2]/( kTD + kTQ
O2 [O2] )
14 1O2* → 3O2 + hνP k∆P
15 1O2* → 3O2 k∆d
E
F
G
k∆ = τ∆
−1 fP
∆ = k∆P / (k∆P + k∆d)
16 1O2*+M → products kr
M
17 1O2*+M → 1M + 3O2 kq
M
E
F
G
kM fr
M = kr
M / (kr
M + kq
M)
f∆
S = fraction of S1 quenched by O2 which gives 1O2*
fT
O2 = fraction of S1 quenched by O2 which gives T1
f∆
T = fraction of T1 quenched by O2 which gives 1O2*
PS
O2 = proportion of S1 quenched by O2
PT
O2 = proportion of T1 quenched by O2..................................................
please search and obtain via library or web the references cited in this pdf. There are interesting to read.
Good luck
RG
Dear Valeria,
a fine question throughout !
The guys at the University of Leeds could bring together some infos as for this topic, including the history of the problem - all that was discussed already before the World War II ... Please confer the link below:
http://www1.chem.leeds.ac.uk/delights/texts/expt_27.html
If you'll get troubles with reading publications in German, please contact me directly ...
Respectfully yours,
Evgeni
