12.34 The placement of the outer electron for boron in a 2p orbital compared 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.
12.37 Nitrogen 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 combinations 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
r
r
r
r
r
2
0 1
2
3
1
2
4
2
2
2
2
2
πε
A
A
A
B
B
2 2
1
1
1
4
3
1 2
2 3
1 3
r
r
r
r
r
B
A B
(
−
−
−
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
ψ r r r r
1 2 3 )
−
∇ + ∇ + ∇
Z
2
1
2
2
2
3
2
1 2 3
2m
r r r
(
)(
)
ψ
−
∇ + ∇
−
+
Z
2
1
2
2
2
1 2
2
0
2
4
1
1
m
r r
e
r
r
[
] ( )
ψ
πε
,
A1
A2 2
B 1
B 2
,
( )
+
+
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
1
1
12
1 2
r
r
r
r r
ψ
−
+
+
+
+
+
e
r
r
r
r
r
2
0
3
1
2
4
1
1
1
1
1
1
πε A1
A2
A
B
B
r r
r
r
r
r
B
A B
3
1 2
2 3
1 3
1
1
1
1
−
−
−
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
−
+
+
+
−
−
e
r
r
r
r
r
2
0
1 2
4
1
1
1
1
1
1
πε A1
A2
B1
B2
r r AB
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
p
m
p
m
e
e
1
2
2
2
2
2
+
ANSWERS TO PROBLEMS
471
9781405124362_5_end.qxd 4/29/08 9:17 Page 471