It appears that process (54) alone cannot explain the fact 4>n, = 1.4 at 1470 A., since the process (54) followed by process (59) gives 4>n2 1. Of course, the N atom may be formed in the (2Z)) state, in which case it may react with N20 to produce N2 + NO. Such a mechanism is, however, still incapable of explaining the high N quantum yield. The process (53) followed by (57) and (58) would explain 4>n, = 1.4. [Pg.189]

AlO is another product from photolysis of HAIOH which is not observed in the Infrared studies (1 ) since the AlO 917 cm l band (13) is obscured by the presence of other product bands. In multiple-collision studies, Gole and Oblath observe emission from the B(2z" ) state of AlO postulating that the state is col-llsionally activated by HAIOH since single-collision studies display no AlO emission. [Pg.354]

A simple bookkeeping argument that counts the states on the left and right of this equation affirms this possibility. For a given j, there are 2jx + 1 possible values of %, for j2 there are 2j2 + 1 possible m2 s. Hence, there are (2j + l)(2ja + 1) terms on the right. An exhaustive representation of tf/jm is possible if its multiplicity is also equal to (2jx + 1)(2j2 + 1). Let us, therefore, count the available number of m-values compatible with fixed jx and j2. Table 7-1 facilitates this procedure. We first note that in the expansion (7-35), mx + m2 must equal m. To see this, apply fz — /Xz + 2z to both sides of it this multiplies the left side by m, each term on the right by mx + m2). On rearranging, the equation reads... [Pg.403]

Whether or not 0+(2D) ions undergo fast reaction with N2 at low energies is of great interest since they satisfy both rules. Unfortunately, the experimental evidence is not decisive. The resonance potential (8) in this case would be repulsive but weak since the ionization potential of O to this state, 16.94 e.v., is considerably greater than that of N2, 15.56 e.v. The resonance potential may be strong enough to inhibit reaction of 0+(2Z)) ions at low energies. [Pg.31]

Finally, brief mention should be made of various mixed sandwich species for which magnetic data are available. Thus the (Cp)7l/(Bz) species are known for M = Mn, and Cr, the former being diamagnetic with a 12+ ground state, and the latter giving a moment of 1.70 B.M., virtually the spin-only value, thus indicating a 2Z+(a S4) ground level (113). [Pg.108]

HS-HS] pairs. The third component (relative intensity z=44.0%) with parameters <5Hs(bpym, S) and AEq(HS) (bpym, S) can be unambiguously assigned to the HS state in [HS-LS] pairs, because the measured effective magnetic field at the iron nuclei of 81 kOe clearly originates from a spin quintet ground state of iron(II) (S=2). As a result, the complete distinction of dinuclear units becomes possible. It follows from the area fractions of the subspectra intensities that at 4.2 K the sample (bpym, Se) contains 2z=88.0% [HS-LS],7=4.0% [HS-HS] and (x-z)=8.0% [LS-LS] pairs. [Pg.196]

Fig. 2-10.—The interaction of Bpin angular momentum and orbital angular momentum to form the total angular momentum for the states 2Z>j and 2D . |

In the homonuclear AX case, e.g. a proton-proton two-spin system, the 180" pulse affects both nuclei. The doublet vectors are reflected at the x z plane. Inverting the precession states of all other coupling nuclei, the 180" pulse also inverts rotation of both AX components (Fig. 2.38(c)). At time 2z, the vectors will be aligned antiparallel along + x (Fig. 2.38(c)) the resultant is zero and no signal will be detected. [Pg.74]

CASSCF calculations for the lowest 2Z+ and 2n state of CCCN in this geometry. [Pg.246]

Assuming that these lactonium salts exist in the EZ form 21, the two transition states yielding 21 and 21 can be illustrated by 2j> and ]] respectively. In 2Z> the bond to be broken is antiperiplanar to the non polar ( -Cj bond whereas in 21> the bond to be broken is antiperiplanar to the C- -0- polar bond. In other words, in 26 the electron pair orbital of the C5 - O5... [Pg.37]

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