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278 8 ATOMIC STRUCTURE AND SPECTRA
Solutions to exercises
E8B.1(b) Hydrogenic orbitals are written in the form [8A.12–307], Rn , l(r)Yl ,m l (θ , ϕ),
where the appropriate radial function Rn , l is selected from Table 8A.1 on page
306 and the appropriate angular function Yl ,m l is selected from Table 7F.1 on
page 286. Using Z = 2 for the 1s and Z = 1 for the 3s gives
ψ1s(r) = R1,0Y0,0 = 2(2/a0)3/2e−2r/a0 × (4π)−1/2
ψ3s(r) = R3,0Y0,0
= (243)−1/2(2/3a0)3/2[6 − 6(2r/3a0) + (2r/3a0)2]e−r/3a0 × (4π)−1/2
�e overall wavefunction is simply the product of the orbital wavefunctions
Ψ(r1 , r2) = ψ1s(r1)ψ3s(r2)
E8B.2(b) For a subshell with angular momentum quantum number l there are 2l + 1
values of m l , each of which corresponds to a separate orbital. Each orbital
can accommodate two electrons, therefore the total number of electrons is 2 ×
(2l + 1). �e subshell with l = 5 can therefore accommodate 2(10 + 1) = 22
electrons.
E8B.3(b) All con�gurations have the [Kr] core. �e table shows the ‘accepted’ con�gu-
rations for the ground states.
Y Zr Nb Mo Tc
5s24d1 5s24d2 5s14d4 5s14d5 5s24d5
Ru Rh Pd Ag Cd
5s14d7 5s14d8 4d10 5s14d10 5s24d10
E8B.4(b) 1s22s22p6 = [Ar]
E8B.5(b) Across the period the energy of the orbitals generally decreases as a result of
the increasing nuclear charge.�e second ionization energy corresponds to an
electron being removed from the ionM+. For Li+ this would involve removing
an electron from the 1s orbital, which is much lower in energy (and therefore
harder to ionize) than the 2s. For Be+ it is a 2s electron which is ionized,
and as this element has the lowest nuclear charge of the remaining elements in
Period 2 it is expected to have the highest orbital energy and hence the lowest
ionization energy.
Solutions to problems
P8B.2 �e electronic con�guration of the Y atom is [Kr] 4d15s2 as opposed to 4d25s1.
�is is due to the 5s orbital being much larger and di�use than the 4d orbitals,