12.34 The placement of the outer electron for boron in a 2p orbital com-
pared to the 2s orbital for beryllium results in a lower ionization
energy despite the increase in the nuclear charge.
12.35 Spin is a relativistic term that has no classical analog. Since
Schrödinger’s equation is based upon classical physics, spin does
not appear.
12.36 Two components represent spin (up or down) and the other two
components represent positive or negative energy, or equivalently
matter and antimatter.
12.37Nitrogen has seven electrons. The first four fill the 1s and 2s orbitals.
The remaining three electrons fill the 2p orbitals, with the electrons
occupying three different 2p orbitals to minimize the interactions
between the electrons.CHAPTER 13
13.1
13.2 (a)(b)13.3
=Eψ(r 1 ,r 2 )13.4
=Eψ(r 1 r 2 r 3 )13.5 The wavefunction must be written considering all possible com-
binations of the electrons in each state.13.6 Two atoms will have an attractive interaction due to the London
dispersion interaction in which random fluctuations give rise to
transient dipole moments.−+++++
e
rr rrr2(^4012312)
22222
πε AAABB221114
rrrrr 3122313 BAB−−−− (
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
ψrrr r 123 )− ∇ +∇ +∇
Z^2
1
2
2
2
3
2
2 m()()ψrr r^123−∇+∇ − +
Z^2
1
2
2
2
122(^240)
11
mrr
e
rr[]()ψ
πε,
A1 A 2 2B1B2++− (),
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
111
12
rrrψrr^12− +++++
e
rrrrr2(^40312)
111111
πε A1 A2 A B B rrrrrrBA312 2313B111 1
−−−−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
−+++−−
e
rrrrr2(^4012)
11111 1
πε A1 A2 B1 B2 rrAB⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
p
mp
eem1
2
2
2
22+
ANSWERS TO PROBLEMS 471
9781405124362_5_end.qxd 4/29/08 9:17 Page 471