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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 279
and so the average distance between electrons is larger and so the average re-
pulsion between electrons is much smaller. [Kr] 4d15s2 is therefore the lower
energy, more stable con�guration.
�eAg atomhas the con�guration [Kr] 4d105s1, as opposed to [Kr] 4d95s2.�e
number of both spin up and spin down electrons is the same in both of these
con�gurations, but the number of parallel spin pairs within the 4d subshell
is larger for the con�guration [Kr] 4d105s1, which means this con�guration
is lower in energy, as a parallel spin pair is lowest in energy when between
two electrons within the same subshell. Also this has fewer electrons in higher
energy subshells.
P8B.4 �e electronic con�guration of Fe is [Ar] 3d64s2, of Fe2+ it is [Ar] 3d6, and of
Fe3+ it is [Ar] 3d5. Hence, the outermost electron in both ions is in a 3d orbital,
with the only di�erence between these con�gurations that Fe2+ has one of the
�ve 3d orbitals doubly occupied.
�ere is more repulsion between the 6 3d electrons in Fe2+ than the 5 3d elec-
trons in Fe3+, and there is no compensating increase in the atomic number to
draw the electrons towards the nucleus, and so the Fe2+ ion is expected to be
the larger of the two ions.
8C Atomic spectra
Answers to discussion questions
D8C.2 �e selection rules are given in [8C.8–335]. In part these can be rationalised
by noting that a photon has one unit of (spin) angular momentum and that
in the spectroscopic transition this angular momentum must be conserved.
�e selection rule for l , ∆l = ±1, can be understood as a single electron in
the atom changing angular momentum by one unit in order to accommodate
the angular momentum from the photon. �is selection is derived in How
is that done? 8C.1 on page 327 by considering the relevant transition dipole
moment.�e selection rule for the total spin, ∆S = 0, stems from the fact that
the electromagnetic radiation does not a�ect the spin directly.
�e selection rules for multi-electron atoms are harder to rationalise not least
because the change in the overall angular momentum (L and J) is a�ected both
by changes in the angular momenta of individual electrons and by the way in
which these couple together.
D8C.4 �is is discussed in Section 8C.2(b) on page 329.
Solutions to exercises
E8C.1(b) �e spectral lines of a hydrogen atom are given by [8A.1–304], ν̃ = R̃H(n−21 −
n−22 ), where R̃H is the Rydberg constant and ν̃ is the wavenumber of the tran-
sition.