1 (439)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 431
When the applied magnetic �eld lies along the bisector of the ONO angle,
g = hν
µBB0
= (6.6261 × 10−34 J s) × (9.302 × 109 Hz)
(9.2740 × 10−24 J T−1) × (331.94 × 10−3 T)
= 2.002
P12D.4 �e splitting due to the two N atoms results in 2NI + 1 = 2× 2× 1+ 1 = 5 lines
with relative intensity ratio 1:2:3:2:1, a quintet . Adjacent lines are separated by
a(N) = 0.148 mT. Each of these lines is further split into a 1:4:6:4:1, a quintet
by coupling to four H atoms results (2NI + 1 = 2 × 4 × 1
2 + 1 = 5). Adjacent
lines in the quintet are separated by a(H) = 0.112 mT. Because the two hyper-
�ne constant are similar the multiplets overlap extensively, giving a complex
multiplet.
Figure 12.14 shows how the complete multiplet is formed from �ve overlapping
1:4:6:4:1 quintets, a–e, centred at the positions of the lines of the 1:2:3:2:1 quintet,
and with intensities in proportion. For example, quintet a is centred at the �eld
of the le�-most line of the 1:2:3:2:1 a quintet, and has relative intensity 1. Quintet
b is centred at the �eld of the next le�-most line of the 1:2:3:2:1 a quintet, and
has relative intensity 2.�e dotted lines show the relationship of the multiplets
a–e to the lines in the 1:2:3:2:1 a quintet.
a
b
c
d
e
complete multiplet
1:2:3:2:1 quintet
Figure 12.14
P12D.6 �e McConnell equation is given in [12D.5–521], a = Qρ, where a is the hy-
per�ne coupling constant observed for a H atom in a given molecule, Q is
a constant, and ρ is the spin density on the adjacent carbon atom. For the
benzene radical anion, Q = 2.25 mT; the same value is used for each of the
anions considered here. For example, at position 3 in 1
ρ = a
Q
= 0.011 mT
2.25 mT
= 4.89 × 10−3
�e spin densities for each carbon, computed in the same way, are shown in
Fig. 12.15.