Chapter 2 Atomic Structure
CHAPTER 2: ATOMIC STRUCTURE
2.1
a.
=
h
6.626  10 34 J s
=
 2.426  10 11 m
mv (9.110  10 –31kg)  (0.1)  (2.998  108 m s –1 )
b.
=
h
6.626  1034 J s
=
 6  10 34 m
3
mv 0.400 kg  (10 km/hr  10 m/km  1hr/3600s)
c.
=
h
6.626  10 34 J s
35
m
=
-1  9.1 10
mv 8.0 lb  0.4536 kg/lb  2.0 m s
d.
h
6.626  10 34 J s
=
=
mv 13.7 g  kg/10 3 g  30.0 mi hr -1  1 hr/3600s  1609.3 m/mi
 3.61 10 33 m
2.2
 1
hc 1
1 

E  RH  2 – 2 ; RH  1.097 10 7 m–1  1.097 10 5 cm–1 ;  
E __
nh 
2

nh  4
1 1 
 12 
E  RH  –   RH    20,570cm –1  4.085  10 –19 J
 4 16 
 64 
  4.862 10 –5 cm = 486.2 nm
nh  5
 1
 21 
1 
E  RH  –
 23,040cm –1  4.577  10 –19 J ;
 RH 


 100 
 4 25 
  4.34110 –5 cm = 434.1 nm
nh  6
 1
 8
1 
E  RH  –
 RH    24,380cm –1  4.841 10 –19 J ;

 36 
 4 36 
  4.102 10 –5 cm = 410.2 nm
2.3
 1
 45 
1 
  RH    25,190cm–1  5.002 10 –19 J
E  RH 
–
4
196 
49 


1
  __  3.970 10 –5 cm = 397.0 nm

2.4
hc
(6.626  1034 J s)(2.998  108 m/s)
 5.178  1019 J
9
(383.65 nm)(m/10 nm)
383.65 nm
E
379.90 nm
(6.626  10 34 J s)(2.998  108 m/s)
E

 5.229  10 19 J
9

(379.90 nm)(m/10 nm)


hc
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1
2
Chapter 2 Atomic Structure

1
 1
1 E  2
1
1 E
1 
and nh   
 
E  RH  2  2  ;

2
4 RH
4 RH 
nh  nh
2

1
1 5.178 10 19 J  2
9
For 383.65 nm: nh   
18 
4 2.1787 10 J 

1
1 5.229 10 19 J  2
 10
For 379.90 nm: nh   
18 
4 2.1787 10 J 
2.5
The least energy would be for electrons falling from the n = 4 to the n = 3 level:
 1 1 
 7 
E  RH  2  2   2.1787 10 18 J    1.059 10 19 J
3 4 
144 
The energy of the electromagnetic radiation emitted in this transition is too low for
humans to see, in the infrared region of the spectrum.
2.6
E  102823.8530211 cm 1  97492.221701 cm 1  5331.6313201 cm 1
E  hc  (6.626  1034 J s)(2.998  108 m s)(100cm m)(5331.6313201 cm 1 )
E  1.059  1019 J
This is the same difference found via the Balmer equation in Problem 2.5 for
a transition from n  4 to n  3. The Balmer equation does work well for hydrogen.
2.7
a.
We begin by symbolically determining the ratio of these Rydberg constants:
RHe + 
RHe +
RH

2 2  He + (Z He+ )2 e4
(4 0 )2 h2
 He+ (Z He+ )2
 H (Z H )
2

4  He+
H
RH 
2 2  H (Z H )2 e4
(4 0 )2 h2
(since Z  2 for He  )
The reduced masses are required (in terms of atomic mass units):
1
1
1
1
1




4
H me m proton 5.4857990946  10 m 1.007276466812 mu
u
1
H
 1823.88 mu1;  H  5.482813  104 mu
1
He 
1
He +

1
1
1
1



4
me mHe2 5.4857990946  10 m 4.001506179125 mu
u
 1823.13 mu1;  He +  5.485047  10 4 mu
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Chapter 2 Atomic Structure
3
The ratio of Rydberg constants can now be calculated:
RHe+ 4  He+ 4(5.485047  104 mu )


 4.0016
RH
H
5.482813  104 mu
b.
RHe +  4.0016RH  (4.0016)(2.1787  1018 J)  8.7184  1018 J
c.
The energy difference is first converted to Joules:
E  hc  (6.626  1034 J s)(2.998  108 m s)(100cm m)(329179.76197 cm 1 )
E  6.5391 1018 J
The Rydberg equation is applied, affording nearly the identical Rydberg constant for He+:
1
1
1 1
E  RHe + ( 2  2 )  6.5391 1018 J  RHe + (  )
1 4
n
n
l
h
RHe +  8.7188  1018 J
2.8
a.
–
h2
2
E ;
8 2 m x 2
  A sin rx  B cos sx

= Ar cos rx – Bs sin sx
x
2 
= –Ar 2 sin rx – Bs 2 cos sx
2
x
–
h2
 – Ar sin rx – Bs cos sx   E  A sin rx + B cos sx 
8 m
2
2
2
If this is true, then the coefficients of the sine and cosine terms must be independently
equal:
h 2 Bs 2
h 2 Ar 2

EA
;
= EB
8 2 m
8 2 m
r 2  s2 
8 2 mE
2
; r  s  2mE
2
h
h
b.
  A sin rx ; when x  0,   A sin 0  0
when x  a,   A sin ra  0
n
 ra  n ; r  
a
c.
E
r 2 h 2 n 2 2 h 2
n 2h 2


8 2 m
a 2 8 2 m 8ma 2
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Chapter 2 Atomic Structure
d.
a
a
x 
a
 dx  A sin 
 dx  A
 a 
n
0
0



2
2 n
2
2
a
nx  nx 
sin 2 
 d 

 a   a 

0
a
a 2 1 nx  1 2nx 
A  
sin
–


  1
n
2  a  4  a  0

aA 2 1 na 1
1

–
sin2n

–
0

sin0
1

2 a

4
4
n 
A
2.9
2
a
a.
3pz
4dxz
b.
3pz
4dxz
c.
3pz
4dxz
For contour map, see Figure 2.8.
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Chapter 2 Atomic Structure
2.10
a.
4s
5dx2-y2
b.
4s
5dx2-y2
c.
4s
5dx2-y2
z
x
Dashed circles represent radial
nodes. Electron density is inside
the smallest node, between the
nodes, and outside the largest
node.
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5
6
Chapter 2 Atomic Structure
2.11
a.
5s
4dz2
b.
5s
4dz2
c.
5s
4dz2
z
x
Dashed circles represent radial nodes. Electron
density is inside the smallest node, between the
nodes, and outside the largest node.
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Chapter 2 Atomic Structure
2.12
2.13
7
4 f z(x 2 y 2 )orbital
a.
no radial nodes (The number of radial nodes = n – l –1. Since n = 4 and l = 3 for f
orbitals, n – l –1 = 4 – 3 – 1 = 0 radial nodes.)
b.
3 angular nodes (The number of angular nodes = l = 3.)
c.
The angular nodes are solutions for z(x2–y2) = 0. These solutions are z = 0 (xy plane), and
the planes where x = y and x = –y, both perpendicular to the xy plane.
d.
There are 8 lobes, 4 above and 4 below the xy plane. Down the z axis, this orbital
looks like a dx2-y2 orbital, but the node at the xy plane splits each lobe in two. For
an image of this orbital, please see http://winter.group.shef.ac.uk/orbitron/ or another
atomic orbital site on the Web.
A 5 f xyz orbital has the same general shape as the 4 f z(x 2 – y 2 ), with the addition of a radial node
and rotation so the lobes are between the xy, xz, and yz planes.
2.14
a.
1 radial node (= n – l –1, as in problem 2.12)
b.
3 angular nodes (= l)
c.
The angular nodes are solutions for xyz = 0. These solutions are the three planes where
x = 0 (yz plane), y = 0 (xz plane), and z = 0 (xy plane).
d.
The diagram is similar to that in problem 2.12, with a radial node added and rotated
by 45° around the z axis. There are 8 lobes, one in each octant of the coordinate system,
both inside and outside the radial node, for a total of 16 lobes. For an image of this
orbital, please see http://winter.group.shef.ac.uk/orbitron/ or another atomic orbital site
on the Web.
a.
no radial nodes (= n – l –1, as in problem 2.12)
b.
3 angular nodes (= l)
c.
The angular nodes are solutions for z(5z2 – 3r2) = 0
Solutions are z = 0 (the xy plane) and 5z2 – 3r2 = 0, or 5z2 = 3r2
Because r2 = x2 + y2 + z2 we can write
5z2 = 3(x2 + y2 + z2)
2z2 = 3x2 + 3y2
z2 = 3/2 (x2 + y2)
This is the equation for a (double) cone.
d.
For an image of this orbital, please see http://winter.group.shef.ac.uk/orbitron/ or another
atomic orbital site on the Web.
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8
Chapter 2 Atomic Structure
2.15
a.
2.16
2.17
5d
4f
7g
n
5
4
7
l
2
3
4
ml
–2,–1,0,1,2
–3,–2,–1,0,1,2,3
–4,–3,–2,–1,0,1,2,3,4
b.
n
l
ml
3
2
–2
3
2
–1
3
2
0
3
2
1
3
2
2
10 possible combinations
c.
For f orbitals (l = 3) possible values of ml are –3, –2, –1, 0, 1, 2, and 3.
d.
2. (No more than 2 electrons can occupy any orbital!)
a.
For a 5d electron, l = 2 and n = 5.
b.
At most there can be ten 4d electrons, half of which can have ms =  .
c.
f electrons have quantum number l = 3 and can have ml = –3, –2, –1, 0, 1, 2, or 3.
d.
For l = 4 (g electrons), ml can be –4, –3, –2, –1, 0, 1, 2, 3, or 4.
a.
The l quantum number limits the number of electrons. For l = 3, ml can have
seven values (–3 to +3) each defining an atomic orbital, and for each value of ml the
3d
ms
±2
±2
±2
±2
±2
1
2
value of ms can be 
1
1
or  ; there can be two electrons in each orbital. At most,
2
2
therefore, there can be 14 electrons with n = 5 and l = 3.
2.18

b.
A 5d electron has l = 2, limiting the possible values of ml to –2, –1, 0, 1, and 2.
c.
p orbitals have l = 1 and occur for n  2.
d.
g orbitals have l = 4. There are 9 possible values of ml and therefore 9 orbitals.
a.
   

b.
One exchange
E = e
W:
The first configuration is favored. 
This configuration is stabilized by
e (which is negative), and the second
is destabilized by c (which is positive).
  and    The first configuration is favored. It is
3 exchanges
(1–2, 1–3, 2–3)
E = 3e
2.19

One pair in same orbital
E = c
1 exchange, 1 pair
E = e + c
     X:
2 exchanges (1–2, 3–4)
E = 2e
stabilized by 2e more than the second
configuration, and the second
configuration is also destabilized by c.
   
3 exchanges (1–2, 1–3, 2–3)
E = 3e, so this state has lower energy
than W (because e is negative).
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Chapter 2 Atomic Structure
2.20
Y:
    
Z:
    
6 exchanges (1–2, 1–3, 1–4, 2–3, 2–4, 3–4)
E = 6e, so this state has lower energy than Y.
(because e is negative).


2.21
Three states are possible:     (three exchanges, E  3e );
4 exchanges (1–2, 1–3,
2–3, 4–5); E = 4e
9
    (one exchange, E  e );     (one
exchange and one pair, E  e   c ). Energy ranking:
c
Energy
2e
2.22
2.23
2.24
2.25
a.
Figure 2.12 and the associated text explain this. Electron-electron repulsion is minimized
by placing each electron in a separate orbital when the levels are close enough to allow it.
At Cr, the second 4s electron has an energy higher than the lower five 3d electrons, and
therefore the configuration is 4s1 3d5.
b.
Ti is 4s2 3d2, since both 4s electrons have energies below that of the 3d electrons at that
Z. For ions, the 3d levels move down in energy, and are below the 4s levels for all
transition metal M2+ ions, so Cr2+ is [Ar] 3d4.
a.
V
1s2 2s2 2p6 3s2 3p6 4s2 3d3
b.
Br
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
c.
Ru3+
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d5
d.
Hg2+
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 4f14 5d10
e.
Sb
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3
a.
Rb–
[Kr] 5s2
b.
Pt2–
[Xe] 6s2 4f14 5d10
Electrons fill into atomic orbitals to maximize the nuclear attraction they experience. The more
attraction an electron experiences, the more it is stabilized, and the more its energy is reduced (the
more negative its energy). The order of orbital filling from n = 1 to n = 2 to n = 3 is rationalized
by increasing the most probable distances these orbitals offer their electrons. Note that the most
probable distances (the distance r where the radial probability is maximized) are
roughly the same for all orbitals of the same n. As the most probable distance increases, the
nuclear charge an electron experiences is attenuated to a greater extent by shielding by other
electrons with shorter most probable distances. The 1s orbital fills first since its electrons have the
most unhindered exposure to the charge of the nucleus; these electrons are subsequently
Copyright © 2014 Pearson Education, Inc.
10
Chapter 2 Atomic Structure
stabilized to the greatest extent. Electrons fill the n = 2 orbitals before the n = 3 orbitals due to the
shorter most probable distance (and greater nuclear charge) offered by the former orbitals.
Within a shell (orbitals with the same n), orbital energies are governed by the varying penetration
abilities of orbitals with different l values. Electrons experience an increased nuclear charge (and
are stabilized more) when residing in orbitals that penetrate towards the nucleus, affording
electrons a reasonable probability of being found much closer to the nucleus than the most
probable distance. Examination of the radial probability functions for n = 3 (Figure 2.7) reveals
that a 3s electron has a higher probability of being found closest to the nucleus (inner maximum
in graph at say 0.5r) relative to a 3p or 3d electron. A 3d electron has the lowest probability of
being found at this distance from the nucleus; this orbital penetrates poorly and has no inner
maximum. The order of filling within this shell reflects these relative penetration abilities; the 3s
electrons are lower in energy than the 3p electrons, while the 3d electrons are the highest in
energy of the electrons in the third shell.
2.26
2.27
2.28
a.
When a fluorine atom gains an electron, it achieves the same electron configuration of
the noble gas Ne, with all subshells filled.
b.
Zn2+ has the configuration [Ar] 3d10, with a filled 3d subshell.
c.
This configuration [Kr] 5s1 4d 5 has each electron outside the Kr core in a separate
orbital and has the maximum number of electrons with parallel spins (6). In Mo, this
configuration has a lower energy than [Kr] 5s2 4d 4. The greater stability of the
s1d 5 configuration is also found in the case of Cr (directly above Mo), but W (below
Mo) has an s2d 4 configuration. In Cr and Mo, the ns and (n-1)d subshells are sufficiently
close in energy so that electron pairing is avoided.
a.
Ag+ has the configuration [Kr] 4d10, with a filled 4d subshell.
b.
In Cm, the s2d1f 7 configuration enables the last eight electrons each to occupy a
separate orbital with parallel spin. This minimizes c in comparison with the
alternative configuration, in which one f orbital would be occupied by an electron pair.
c.
The ion Sn2+ has the configuration [Kr] 5s2 4d10, with both the 5s and 4d subshells
filled. This is an example of the “inert pair” effect, in which heavier elements of
groups 13-16 often form compounds in which the oxidation states of these elements
are 2 less than the final digit in the group number (other examples include Tl+ and
Bi3+). The reasons for this phenomenon are much more complex than the ions’ electron
configurations; see N. N. Greenwood and A. Earnshaw, Chemistry of the Elements,
2nd ed., Butterworth-Heinemann, London, 1997, pp. 226-227 for useful comments.
a.
Ti2+ has the configuration [Ar] 3d2; Ni2+ has the configuration [Ar] 3d8. In each case it
should be noted that all the electrons outside the noble gas cores in these transition
metal ions are d electrons, which occupy orbitals that are lower in energy than the 4s
orbitals (see Figure 2.12(b) and the associated discussion).
b.
The preferred configuration of Mn2+ is [Ar] 3d5. As in the examples in part a, the 3d
orbitals of Mn2+ are lower in energy than the 4s. In addition, the configuration
minimizes electron-electron repulsions (because each d electron is in a separate
orbital) and maximizes the stabilizing effect of electrons with parallel spins
(maximum e).
Copyright © 2014 Pearson Education, Inc.
Chapter 2 Atomic Structure
2.29
(1s2) (2s2 2p6) (3s2 3pn)
1
0.85
0.35
15 – (2 × 1 + 8 × 0.85 + 4 × 0.35) =
16 – (2 × 1 + 8 × 0.85 + 5 × 0.35) =
17 – (2 × 1 + 8 × 0.85 + 6 × 0.35) =
18 – (2 × 1 + 8 × 0.85 + 7 × 0.35) =
a.
Z
P
S
Cl
Ar
Z*
r
4.8
5.45
6.1
6.75
106 pm
102 pm
99 pm
98 pm
11
The size of the atoms decreases slightly as Z increases, even though the number of
electrons in the atom increases, because Z* increases and draws the electrons closer.
Ar has the strongest attraction between the nucleus and the 3p electron, and the smallest
radius.
b.
Z
O2–
–
F
Na+
Mg2+
(1s2)
(2s2 2p6)
0.85
0.35
8 – (2 × 0.85 + 7 × 0.35) =
9 – (2 × 0.85 + 7 × 0.35) =
11 – (2 × 0.85 + 7 × 0.35) =
12 – (2 × 0.85 + 7 × 0.35) =
Z*
r
3.85
4.85
5.85
6.85
126 pm
119 pm
116 pm
86 pm
These values increase directly with Z, and parallel the decrease in ionic size.
Increasing nuclear charge results in decreasing size for these isoelectronic ions, although
–
the change between F and Na+ is smaller than might be expected.
c.
Cu
(1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s1)
4s S = 2 + 8 + (8×0.85) + (10×0.85) =
3d S = 2 + 8 + (8×1.00) + (9×0.35) =
25.3
21.15
Z* = 29 – 25.3 = 3.7
Z* = 29 – 21.15 = 7.85
The 3d electron has a much larger effective nuclear charge and is held more tightly;
the 4s electron is therefore the first removed on ionization.
d.
(1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s2 4p6) (4d10) (4fn)
Ce S = 2 + 8
+ 8
+ 10 + 8 + 10
= 46
[Xe] 6s2 4f1 5d1
4f1
Z* = 58 – 46 = 12
S= 2 + 8
+
[Xe] 6s2 4f3
8
+ 10 +
8
+ 10 + (2×0.35) = 46.7
4f3
Z* = 59 – 46.7 = 12.3
Nd S = 2 + 8
+
[Xe] 6s2 4f4
8
+ 10 +
8
+ 10 + (3×0.35) = 47.05
4f4
Z* = 60 – 47.05 = 12.95
Pr
The outermost electrons experience an increasing Z*, and are therefore drawn in to
slightly closer distances with increasing Z and Z*.
Copyright © 2014 Pearson Education, Inc.
12
Chapter 2 Atomic Structure
2.30
Sc:
Ti:
3d electron:
Electron
(1s2)(2s2, 2p6)(3s2, 3p6)(3d1)(4s2)
configuration
(1s2)(2s2, 2p6)(3s2, 3p6)(3d2)(4s2)
Contribution
of other d
electrons
None [only one electron
in (3d1)]
One other 3d electron:
Contribution to S = 1 × 0.35 = 0.35
Contribution
of electrons
to left of (3dn)
18 × 1.00 = 18.00
18 × 1.00 = 18.00
Total S
Z*
18.00
21 – 18.00 = 3.00
18.35
22 – 18.35 = 3.65
4s electron:
Electron
(1s2)(2s2, 2p6)(3s2, 3p6)(3d1)(4s2)
configuration
(1s2)(2s2, 2p6)(3s2, 3p6)(3d2)(4s2)
Contribution
of other (4s2)
electron
Contribution to S = 0.35
Contribution to S = 0.35
Contribution
of (3s2, 3p6)(3dn)
electrons
9 × 0.85 = 7.65
10 × 0.85 = 8.50
Contribution
of other electrons
10 × 1.00 = 10.00
10 × 1.00 = 10.00
Total S
Z*
0.35 + 7.65 + 10.00 = 18.00
21 – 18.00 = 3.00
0.35 + 8.50 + 10.00 = 18.85
22 – 18.85 = 3.15
In scandium, Slater’s rules give the same value for the effective nuclear charge Z* for the 3d
and 4s electrons, consistent with the very similar energies of these orbitals. In Ti the effective
nuclear charge is slightly less for 4s than for 3d, by 0.50. This difference in the energies of 4s
and 3d increases across the row of transition metals; by nickel, as calculated in the example in
Section 2.2.4, the 4s orbital has an effective nuclear charge 3.50 less than the 3d orbitals.
This is consistent with the experimental observation that transition metal cations have electron
configurations in which there are no valence s electrons; the 3d orbitals are lower in energy
than the 4s orbital, so d orbitals are the ones that are occupied. A similar phenomenon is
observed for the second and third row transition metals.
Copyright © 2014 Pearson Education, Inc.
Chapter 2 Atomic Structure
2.31
13
N
Z* = 7 – (2 × 0.85 + 4 × 0.35) = 3.9
IE = 1.402 MJ/mol r = 75 pm
P
Z* = 15 – (2 × 1 + 8 × 0.85 + 4 × 0.35) = 4.80
IE = 1.012 MJ/mol r = 106 pm
As
Z* = 33 – (2 × 1 + 8 × 1 + 18 × 0.85 + 4 × 0.35) = 6.3 IE = 0.947 MJ/mol r = 120 pm
The effect of shielding alone is not sufficient to explain the changes in ionization energy.
The other major factor is distance between the electron and the nucleus. Wulfsberg
(Principles of Descriptive Inorganic Chemistry, Brooks/Cole, 1998) suggests that Z*/r2
correlates better, but As is still out of order:
N
P
As
2.32
radius(pm)
75
106
120
Z*/r2
6.93 × 10–4
4.27 × 10–4
4.375 × 10–4
Zr through Pd have the electron configurations shown here:
40
41
42
43
44
45
46
Zr
Nb
Mo
Tc
Ru
Rh
Pd
5s2 4d2
5s1 4d4
5s1 4d5
5s2 4d5
5s1 4d7
5s1 4d8
4d10
The lower d line crosses the upper s line between elements 40 and 41, the upper d line crosses the
upper s line between 43 and 44, and the upper d line crosses the lower s line between 45 and 46.
This graph fits the experimental configurations well.
2.33
Li
IE
5.39
Z*
1.30
r(pm, covalent) 123
Z*/r2
8.59
r(pm, ionic)
90
16.0
Z*/r2
>
Na
5.14
2.20
154
9.28
116
16.3
>
K
4.34
2.20
203
5.34
152
9.52
>
Rb
4.18
2.20
216
4.72 (all × 10–5 pm–2)
166
7.98 (all × 10–5 pm–2)
The Z*/r2 function explains the order, except for lithium, which seems to require more energy for
removal of the electron than predicted. Apparently the very small size and small number of
electrons on lithium result in the electron being held more tightly than in the other alkali metals.
The Z*/r2 function also predicts larger differences between the IE values than are observed.
Copyright © 2014 Pearson Education, Inc.
14
Chapter 2 Atomic Structure
2.34
C+ (1s2) (2s2 2p1)
B (1s ) (2s 2p )
2
2
1
Z* = 6 – (2 × 0.85 + 2 × 0.35) = 3.6
Z* = 5 – (2 × 0.85 + 2 × 0.35) = 2.6
The energies change by a factor of approximately three, but Z* changes only by 38%. Based on
the data in Table 2.8 and the relative ionic and covalent sizes in problem 2.33, C+ has a radius of
about 58 pm, B a radius of about 83 pm. Z*/r2 values are then 10.7 × 10–4 and 3.8 × 10–4, a ratio
of 2.8. The ionization energies have a ratio of 2.99. This function of size and effective charge
explains this pair quite well.
2.35
The maxima at 4 electrons correspond to the isoelectronic species Li , Be , and B+ ( 1s 2 2s2 )
changing to Li , Be + , and B2+ , respectively ( 1s 2 2s1 ). The energy requirement to remove
electrons from species with completely filled subshells is generally relatively high. The minima at
5 electrons correspond to the isoelectronic species Be  , B , and C+ ( 1s 2 2s2 2 p1) changing to
Be , B+ , and C2+ , respectively ( 1s 2 2s2 ). The removal of an electron from a singly occupied
subshell requires relatively less energy. These data support the stability of the 1s 2 2s2 electronic
configuration.
2.36
a.
He and H cannot form +3 cations; the first element under consideration is therefore Li.
The third ionization energy of Be 2+ is expected to be a local maximum on the basis of
removing an electron from a 1s 2 configuration. The third ionization energy of B2+
( 1s 2 2s1 ) will be a local minimum on the basis of removing an electron from a singly
occupied subshell. The third ionization energy of C2+ ( 1s 2 2s2 ) will be a local maximum
on the basis of the removal of an electron from a filled subshell. The corresponding
energies for N 2+ ( 1s 2 2s2 2 p1) and Ne 2+ ( 1s 2 2s 2 2 p 4 ) will be local minima on the
graph on the basis of removal of an electron from a singly filled subshell, and the
formation of a 2p subshell with three unpaired electrons (and no coulombic repulsion),
respectively. Finally, there will be a maximum at Mg 2+ , which has a filled n = 2 subshell
and a configuration matching that of neutral Ne.
b.
2.37
The graph will share the common features of local maxima at 2 and 4 electrons, and local
minima at 3, 5, and 8 electrons, as detailed above. The magnitudes of these third
ionization energies are all greater than the corresponding second ionization energies for
the same electronic configurations.
Because the second ionization energies involve removing an electron from an ion with a
single positive charge, the electron configurations of the cations must be considered:
He+
Li+
Be+
B+
C+
1s1
1s2
1s2 2s1
1s2 2s2
1s2 2s2 2p1
N+
O+
F+
Ne+
1s2 2s2 2p2
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p5
Copyright © 2014 Pearson Education, Inc.
Chapter 2 Atomic Structure
15
The peaks and valleys now match the peaks and valleys for electron configurations matching
those of the atoms in the ionization energy graph in Figure 2.13. For example, because there is a
minimum in the ionization energy for Li (electron configuration 1s2 2s1), there should also be a
minimum in the graph of second ionization energy for Be+, which also has the configuration 1s2
2s1. The maximum in the ionization energy for Be (configuration 1s2 2s2) would be matched by a
maximum in the second ionization graph for B+, which has the same configuration as Be.
Overall, the maxima and minima in the two graphs are:
Maximum or Minimum
Configuration
1st IE
2nd IE (kJ/mol)
Maximum
Minimum
Maximum
Minimum
1s2
1s2 2s1
1s2 2s2
1s2 2s2 2p1
1s2 2s2 2p2
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p5
He
Li
Be
B
C
N
O
F
Li+
Be+
B+
C+
N+
O+
F+
Ne+
Maximum
Minimum
2.38
2.39
2.40
(7298)
(1757)
(2427)
(2353)
(2856)
(3388)
(3374)
(3952)
a.
Fe (7.87 eV) > Ru (7.37 eV) They have the same Z* (6.25), but Ru is larger, so Z*/r2
is smaller than for Fe.
b.
P (10.486 eV) > S (10.36 eV) Z* is smaller for P (4.8) than for S (5.45), but S has one
electron paired in the 3p level, which increases its energy and makes it easier to remove.
c.
Br (11.814 eV) > K (4.341 eV) Z* for K is 2.2; for Br it is 7.6, a very large difference. K
is also nearly twice as large as Br. Both factors contribute to the difference in IE.
d.
N (14.534) > C (11.260) Increasing Z* (3.9 for N, 3.25 for C) and decreasing size (75
pm for N, 77 pm for C) both work in the same direction.
e.
Cd (8.993 > In (5.786) Indium starts a new 5p subshell, so the last electron is easily
removed in spite of a larger Z* (2.30 for In, 1.65 for Cd).
f.
The smaller F has a higher ionization energy (17.422) than Cl (12.967). An electron is
more easily removed from the 3p subshell of Cl than from the smaller 2p subshell of F.
a.
S (2.077 eV) has a smaller EA than Cl (3.617 eV) because Cl has a larger Z* (6.1 vs.
5.45) and a slightly smaller radius. Both increase the attractive power for an electron.
b.
I (3.059) has a smaller EA than Br (3.365) because it is larger, with the same Z* (7.6).
c.
B (8.298) has a smaller IE than Be (9.322) because it is starting a new p subshell.
d.
S (10.360) has a smaller IE than P (10.486) because S is losing one of a pair of p
electrons.
a.
The maximum at Mg comes at a completed subshell (3s2) and the minimum at Al is at
the first electron of a new subshell (3p), increasing the energy and making removal of an
electron easier. The maximum at P is at a half-filled subshell (3p3) and the minimum
Copyright © 2014 Pearson Education, Inc.
16
Chapter 2 Atomic Structure
at S is at the fourth 3p electron, which must pair with one of the others; this also raises
the energy of the electron and makes its removal easier.
b.
The reasons for the minima and maxima in the electron affinity graph are the same as in
the ionization energy graph. The maxima and minima are shifted by one in the two
graphs because the reactants in the process defining electron affinity have a negative
charge, one more electron than a neutral atom. For example, minima occur for ionization
energy at Al ([Ne]3s23p1  and for electron affinity at Mg (([Ne]3s23p1 Al and
–
Mg have identical electron configurations.
2.41
The Bohr equation predicts that the energy levels of 1-electron systems should be
proportional to Z 2. Z 2 = 4 for He+ and 9 for Li2+, so ratios of the ionization energies to that of H
are 4:1 and 9:1.
2.42
In both the transition metals and the lanthanides (problem 2.29.d), the gradual change in Z* pulls
the outer electrons closer. The increase in Z* is 0.65 for each unit increase in atomic number,
so the increase in attraction is relatively small, and the change in radius must also be small.
2.43
a.
Se2– > Br > Rb+ > Sr2+ These ions are isoelectronic, so the sizes are directly dependent
on the nuclear charges. For example, Sr2+ has the greatest nuclear charge and is the
smallest ion.
b.
Y3+ > Zr4+ > Nb5+ These ions are also isoelectronic, so the increasing nuclear charge
results in decreasing size.
c.
Co4+ < Co3+ < Co2+ < Co The smaller number of electrons with constant nuclear charge
results in a smaller size.
a.
F– has the largest radius. All three choices in this isoelectronic set have 10 electrons, and
F– has the fewest protons (9) to attract these electrons.
b.
Te2– has the largest volume. In general, atomic and ionic sizes increase going
down a column of main group elements as the number of shells increases, with
electrons closer to the nucleus shielding outer electrons from the full effect of the
nuclear charge.
c.
Mg, with a filled 3s subshell, has the highest ionization energy. The situation is
similar to that in the second period, where Be has a higher ionization energy than
Li and B.
d.
Fe3+ has the greatest surplus of positive charge (26 protons, 23 neutrons), making
it the smallest and the most difficult from which to remove an electron.
e.
Of the anions O–, F–, and Ne–, F– has the configuration of the noble gas Ne, from
which it is more difficult to remove an electron than from the other ions; F,
therefore, has the highest electron affinity.
2.44
–
Copyright © 2014 Pearson Education, Inc.
Chapter 2 Atomic Structure
2.45
2.46
2.47
17
a.
V is the smallest; the effective nuclear charge on the outer electrons increases across
the row of transition metals.
b.
These species are isoelectronic; each has 18 electrons. S2–, which has the fewest protons,
is therefore the largest.
c.
The ionization energy is smallest for the largest alkali metals (weaker attraction between
the nucleus and outermost electrons), so Cs has the lowest value.
d.
The electron affinity is the greatest for the smallest halogens (stronger attraction between
nucleus and outermost electrons), so Cl has the highest value.
e.
The Cu2+ ion has the greatest surplus of protons and is therefore the smallest and the most
difficult from which to remove an electron.
a.
7 orbitals
b.
See images.
c.
The number of radial nodes increases as n increases: 4f orbitals have no radial nodes,
5f have one radial node, and so forth.
a.
9 orbitals
b.
See images.
c.
As in problem 2.46.c, with no radial nodes for 5g, one for 6g, etc.
Copyright © 2014 Pearson Education, Inc.
18
Chapter 3 Simple Bonding Theory
CHAPTER 3: SIMPLE BONDING THEORY
3.1
a.
Structures a and b are more likely than c, because the negative formal charge is on the
electronegative S. In c, the electronegative N has a positive charge.
S
CH3
C
1–
S
CH3
C
CH3
S
1–
N
S
b
1+
C
CH3
N
S
1–
CH3
CH3
c
–
b.
a.
S
N
a
3.2
1–
The same structures fit (OSCN(CH3)2 . The structure with a 1– formal charge on O is
most likely, since O is the most electronegative atom in the ion.
2–
1+
Se
The formal charges are large, but match electronegativity.
C
N
C
N Negative formal charge of 1– on Se, a low electronegativity atom.
1–
Se
1–
Se
b.
the
C
N Negative formal charge on N, the most electronegative atom.
BBest resonance structure of the three.
b is better than a, because the formal charge is on
more electronegative O.
O
H
a and b are better than c, because one of
the formal charges is on the more
electronegative O.
1–
S
H
C
a
c.
O 1–
S
b
1–
1–
O
S
–
SNO : a has a 1– formal charge on S. Not very likely, doesn’t
match electronegativity (negative formal charge is not on most
electronegative atoms). b has 1– formal charge on O, and
is a better structure.
–
1–
1–
S
1+
1+
2–
N
S
O
1–
S
S
1–
O
N
a
1–
1–
S
c
b
N
N
O
S
a
Overall, the S=N–O structure is better based on formal charges, since it has only a negative
charge on O, the most electronegative atom in the ion.
Copyright © 2014 Pearson Education, Inc.
C
S
b
–
NSO : a has 2– formal charge on N, 1+ on S. Large formal
charges, not very likely. b has 1– formal charges on N and O,
1+ on S, and is a better structure.
O
C
S
a
3.3
S
1–
O
C
C
b
1–
O
Chapter 3 Simple Bonding Theory
I
3.4
1+
O
A
N
O
B
N
1+
1+
O
C
N
II
1–
C
N
C
N
1–
1–
1+
C
N
O
III
1+
O
N
C
1–
1+
N
1–
2–
1+
N
O
19
N
C
N
O
1–
1+
N
O
C
N
2+
1–
1–
O
C
N
2–
1+
N
N
C
1–
2+
N
C
N
Structure IB is best by the formal charge criterion, with no formal charges, and is expected
to be the most stable. None of the structures II or III are as good; they have unlikely charges (by
electronegativity arguments) or large charges.
3.5
N
1+
N
1–
1–
O
N
2–
1+
N
O
N
1+
N
1+
O
The first resonance structure, which places the negative formal charge on the most
electronegative atom, provides a slightly better representation than the second structure, which
has its negative formal charge on the slightly less electronegative nitrogen. Experimental
measurements show that the nitrogen–nitrogen distance (112.6 pm) in N2O is slightly closer to
the triple bond distance (109.8 pm) in N2 than to the double bond distances found in other
nitrogen compounds, and thermochemical data are also consistent with the first structure
providing the best representation. The third resonance structure, with greater overall magnitudes
of formal charges, is the poorest representation.
O
3.6
H
O
O
1+
H
N
O
1–
1+
N
O 1–
O
3.7
Molecule,
Including
Usual Formal
Charges
1–
C
1+
O
1–
N
O
H
F
Atom
Group
Number
Unshared
Electrons
C
4
2
O
6
2
N
5
4
O
6
4
H
1
0
F
7
6
  A 
2 

  A   B 
 2.544 
2 
  0.83
2.544  3.61 
3.61


2 
  1.17
2.544  3.61 
 3.066 
2 
  0.92
3.066  3.61 
3.61


2 
  1.08
3.066  3.61 
 2.300

2 
  0.71
2.300  4.193 
4.193


2 
  1.29
2.300  4.193 
–
Number of
Bonds
Calculated
Formal
Charge
3
–0.49
3
0.49
2
–0.84
2
–0.16
1
0.29
1
–0.29
Surprisingly, CO is more polar than FH, and NO is intermediate, with C and N the negative
–
atoms in CO and NO .
Copyright © 2014 Pearson Education, Inc.
20
Chapter 3 Simple Bonding Theory
3.8
a.
Cl
SeCl4 requires 10 electrons around Se. The lone pair of electrons in an
equatorial position of a trigonal bipyramid distorts the shape by
bending the axial chlorines back.
Cl
Se
Cl
Cl
b.
–
I3 requires 10 electrons around the central I and is linear.
S
c.
PSCl3 is nearly tetrahedral. The multiple
bonding in the P–S bond compresses
Cl
the Cl—P—Cl angles to 101.8°,
Cl
significantly less than the tetrahedral angle.
P
Cl
–
d.
IF4 has 12 electrons around I and has a square planar shape.
e.
PH2 has a bent structure, with two lone pairs.
–
2–
F
F
f.
TeF42– has 12 electrons around Te, with a square planar shape.
g.
N3 is linear, with two double bonds in its best resonance structure.
N
SeOCl4 has a distorted trigonal bipyramidal shape with the extra
repulsion of the double bond placing oxygen in an equatorial
Cl
position.
Cl
h.
+
H
3.9
PH4 is tetrahedral.
Cl
Se
Cl
+
H
H
P
H
–
a.
ICl2 has 10 electrons around I and is linear.
b.
H3PO3 has a distorted
tetrahedral shape.
c.
F
–
N
i.
F
Te
–
H
–
BH4 is tetrahedral.
H
H
B
H
Copyright © 2014 Pearson Education, Inc.
N
O
Chapter 3 Simple Bonding Theory
21
O
d.
e.
.
f.
POCl3 is a distorted tetrahedron. The Cl—P—Cl angle is
compressed to 103.3° as a result of the P—O
double bond.
P
Cl
Cl
Cl
O
–
IO4 is tetrahedral, with significant double bonding;
all bonds are equivalent.
I
O
O
O
IO(OH)5 has the oxygens arranged octahedrally, with hydrogens
on five of the six oxygens.
HO
SOCl2 is trigonal pyramidal, with one lone pair
and some double bond character in the S–O bond.
h.
ClOF4 is a square pyramid. The double bonded O
and the lone pair occupy opposite positions.
i.
The F—Xe—F angle is nearly linear (174.7°), with the
two oxygens and a lone pair in a trigonal planar configuration.
Formal charges favor double bond character in the Xe–O bonds.
The O—Xe—O angle is narrowed to 105.7° by lp-bp repulsion.
OH
O
H
S
Cl
Cl
OH
I
HO
O
g.
O
–
F
O
Xe
O
F
3.10
a.
SOF6 is nearly octahedral around the S.
O
F
F
F
IS
F
F
O
F
b.
of
POF3 has a distorted tetrahedral shape, with F—P—F angles
101.3°.
c.
ClO2 is an odd electron molecule, with a bent shape, partial double bond
character, and an angle of 117.5°.
d.
F
F
P
F
Cl
OI
I
O
N
NO2 is another odd electron molecule, bent, with partial double bond
I
I
O
character and an angle of 134.25°. This is larger than the angle of ClO2 O
because there is only one odd electron on N, rather than the one pair and single electron
of ClO2.
Copyright © 2014 Pearson Education, Inc.
22
Chapter 3 Simple Bonding Theory
S
e.
S2O42– has SO2 units with an angle of about 30°
between
their planes, in an eclipsed conformation.
O
2–
S
O
O
O
H
f.
N2H4 has a trigonal pyramidal shape at each N, and a gauche
conformation. There is one lone pair on each N.
N
N
H
H
g.
h.
i.
ClOF2+ is a distorted trigonal pyramid with one lone pair
and double bond character in the Cl—O bond.
CS2, like CO2, is linear with double bonds.
S
C
F
S
+ H
Cl
O
F
–
O
–
The structure of XeOF5 is based on a pentagonal bipyramid,
with a lone pair and the oxygen atom in axial positions. See K.
O. Christe et al., Inorg. Chem., 1995, 34, 1868 for evidence in
support of this structure.
F
F
F
Xe
F
F
3.11
All the halate ions are trigonal pyramids; as the central atom increases
in size, the bonding pairs are farther from the center, and the lone pair forces a smaller angle. The
decreasing electronegativity Cl > Br > I of the central atom also allows the electrons to be pulled
farther out, reducing the bp-bp repulsion.
3.12
a.
AsH3 should have the smallest angle, since it has the largest central atom. This
minimizes the bond pair—bond pair repulsions and allows a smaller angle. Arsenic is
also the least electronegative central atom, allowing the electrons to be drawn out
farther and lowering the repulsions further. Actual angles: AsH3 = 91.8°, PH3 = 93.8°,
NH3 = 106.6°.
b.
Cl is larger than F, and F is more electronegative and should pull the electrons
farther from the S, so the F—S—F angle should be smaller in OSF2. This is consistent
with the experimental data: the F—S—F angle in OSF2 is 92.3° and the Cl—S—Cl angle
in OSCl2 is 96.2°.
–
c.
NO2 has rather variable angles (115° and
132°)
in different salts. The sodium salt
(115.4°) has a slightly smaller angle than
O3 (116.8°). The N–O electronegativity
difference should pull electrons away from
N, reducing the bp-bp repulsion and the angle.
3.13
–
O
O
O
O
O
–
O
N
N
O
O
–
d.
BrO3 (104°) has a slightly smaller angle than ClO3 (107°), since it has a larger
central atom. In addition, the greater electronegativity of Cl holds the electrons closer
and increases bp-bp repulsion.
a.
N3 is linear, with two double bonds. O3 is bent (see solution to 3.12.c), with
one double bond and a lone pair on the central O caused by the extra pair of electrons.
–
Copyright © 2014 Pearson Education, Inc.
O
O
–
Chapter 3 Simple Bonding Theory
b.
23
Adding an electron to O3 decreases the angle, as the odd electron spends part of its
time on the central O, making two positions for electron repulsion. The decrease in angle
is small, however, with angles of 113.0 to 114.6 pm reported for alkali metal ozonides
(see W. Klein, K. Armbruster, M. Jansen, Chem. Commun., 1998, 707) in
comparison with 116.8° for ozone.
O
3.14
Cl
O
Cl
O
CH3 H3Si
H3C
110.9°
111.8°
SiH3
144.1°
As the groups attached to oxygen become less electronegative, the oxygen atom is better able to
attract shared electrons to itself, increasing the bp-bp repulsions and increasing the
bond angle. In the case of O(SiH3)2, the very large increase in bond angle over O(CH3)2 suggests
that the size of the SiH3 group also has a significant effect on the bond angle.
3.15
C3O2 has the linear structure O=C=C=C=O, with zero formal charges.
+
1+
N5+ with the same electronic structure has formal charges of 1–, 1+,
1+, 1+, 1–, unlikely because three positive charges are adjacent to
each other. Changing to N=N=N–NN results in formal charges of
1–, 1+, 0, 1+, 0, a more reasonable result with an approximately
trigonal angle in the middle. With triple bonds on each end, the
formal charges are 0, 1+, 1–, 1+, 0 and a tetrahedral angle. Some
contribution from this would reduce the bond angle.
1–
N
N
N
1+
N
1–
1+
N
N
N
+
1+
N
OCNCO+ can have the structure OC–N–CO, with formal
charges of 1+, 0, 1–, 0, 1+ and two lone pairs on the central N.
N
This would result in an even smaller angle in the middle, but has
C
C
positive formal charges on O, the most electronegative atom.
O
=
=
O C N–CO has a formal charge of 1+ on the final O.
Resonance would reduce that formal charge, making this structure
1–
and a trigonal angle more likely. The Seppelt reference also
N
mentions two lone pairs on N and cites “the markedly
C
C
higher electronegativity of the nitrogen atom with respect to the
1+
O
central atom in C3O2, which leads to a higher localization of
electron density in the sense of a nonbonding electron pair.”
Therefore, the bond angles should be OCCCO > OCNCO+ > N5+. Literature values are 180°,
130.7°, and 108.3 to 112.3° (calculated), respectively.
3.16
a.
H
H
C
H
H
N
C
H
H
N
H
H
In ethylene, carbon has p orbitals not involved in sigma bonding. These orbitals interact
to form a pi bond between the carbons, resulting in planar geometry. (Sigma and pi
Copyright © 2014 Pearson Education, Inc.
N
N
+
O
1+
+
O
1+
24
Chapter 3 Simple Bonding Theory
bonding are discussed further in Chapter 5.) In hydrazine each nitrogen has a steric
number of 4, and there is sigma bonding only; the steric number of 4 requires a threedimensional structure.
b.
–
In ICl2 the iodine has a steric number of 5, with three lone pairs in equatorial positions;
–
the consequence is a linear structure, with Cl atoms occupying axial positions. In NH2
the two lone pairs require a bent arrangement.
3.17
3.18
3.19
c.
Resonance structures of cyanate and fulminate are shown in Figures 3.4 and 3.5. The
fulminate ion has no resonance structures that have as low formal charges as structures A
and B shown for cyanate. The guideline that resonance structures having low formal
charges tend to correspond to relatively stable structures is followed here. Hg(CNO)2,
which has higher formal charges in its resonance structures, is the explosive compound.
a.
PCl5 has 10 electrons around P, using 3d orbitals in addition to the usual 3s and 3p.
N is too small to allow this structure. In addition, N would require use of the 3s, 3p, or
3d orbitals, but they are too high in energy to be used effectively.
b.
Similar arguments apply, with O too small and lacking in accessible orbitals beyond the
2s and 2p.
a.
The lone pairs in both molecules are equatorial, the position that
minimizes 90° interactions between lone pairs and bonding pairs.
b.
In BrOF3 the less electronegative central atom allows electrons in
the bonds to be pulled toward the F and O atoms to a greater
extent, reducing repulsions near the central atom and enabling a
smaller bond angle. In BrOF3 the Feq–Br–O angle is approximately
4.5° smaller than the comparable angle in ClOF3.
F
F
X
O
F
a.
The CH 3 —N— CH 3 angle is expected to be larger than the CH 3 —P— CH 3 angle;
bp-bp repulsion will be more intense at the N due to the higher electronegativity of
N relative to P. The angles are 108.2° ( CH 3 —N— CH 3 ) and 103.4° ( CH 3 —P— CH 3 ).
b.
N(CH 3 )3 is expected to exert a greater steric influence on Al(CH 3 )3 relative to
P(CH 3 )3 on the basis of a shorter Al—N bond distance (204.5 pm) than Al—P bond
distance (253 pm). Therefore, (CH 3 )3NAl(CH 3 )3 has a more acute CH 3 — Al—CH 3
angle (114.4°) than (CH 3 )3PAl(CH 3 )3 (117.1°).
c.
On the basis of the steric argument applied in part b, (CH 3 )3NAl(CH 3 )3 should have a
longer Al—C distance. However, while this distance is slightly longer in
Copyright © 2014 Pearson Education, Inc.
Chapter 3 Simple Bonding Theory
25
(CH 3 )3NAl(CH 3 )3 relative to (CH 3 )3PAl(CH 3 )3 (1.978 pm vs. 1.973 pm), these lengths
are not statistically different when their standard deviations are considered.
Data for (CH 3 )3NAl(CH 3 )3 from T. Gelbrich, J. Sieler, U. Dümichen, Z. Kristallogr.,
2000, 215, 127. Data for (CH 3 )3PAl(CH 3 )3 from A. Almenningen, L. Fernholt, A.
Haaland, J. Weidlein, J. Organomet. Chem., 1978, 145, 109.
2–
3.20
IF32– has three lone pairs and three bonds. Overall, this ion is predicted to
be T-shaped, with bond angles slightly less than 90°.
3.21
a.
There are three possibilities:
b.
The third structure, with the lone pair and double bonds in a facial arrangement, is
least likely because it would have the greatest degree of electron-electron repulsions
involving these regions of high electron concentrations.
F
I
F
F
The second structure, which has fewer 90° lone pair–double bond repulsions than the
first structure, is expected to be the most likely. Experimental data are most consistent
with this structure.
3.22
–
c.
One possibility: XeO2F3
a.
Three unique arrangements of the nonbonding pairs and the oxygen atom are possible in
XeOF3– when the octahedral electron-group geometry is considered: (1) trans
nonbonding pairs; (2) one pair trans and one cis the oxygen atom; and (3) both pairs cis
the oxygen atom.
F
F
F
Xe
1
O
F
F
Xe
2
F
O
F
F
F
Xe
O
3
A square-planar structure (1) is expected to minimize lp-lp repulsions relative to
structures 2 and 3 with unfavorable 90° lp-lp interactions. A low temperature Raman
spectroscopic study coupled with quantum-chemical calculations of shock-sensitive salts
of XeOF3– confirms this prediction (D. S. Brock, H. P. A. Mercier, G. J. Schrobilgen, J.
Am. Chem. Soc., 2010, 132, 10935).
Copyright © 2014 Pearson Education, Inc.
26
Chapter 3 Simple Bonding Theory
b.
3.23
3.24
3.25
This anion is notable as the first example of a VSEPR arrangement ( AX 3YE 2 ) that
features a doubly bond atom (oxygen) positioned approximately 90° to relative the
domains of two nonbonding pairs.
I(CF3)Cl2 is roughly T-shaped, with the two Cl atoms opposite each other and
the CF3 group and two lone pairs in the trigonal plane. The experimental
Cl—I—C angles are 88.7° and 82.9°, smaller than the 90° expected if there
were no extra repulsion from the lone pairs. Repulsion between the lone pairs
and the larger CF3 group put them in the trigonal plane, where there is more
room.
Cl
I
CF3
Cl
a.
CF3 has a greater attraction for electrons than CH3, so the P in PF2(CF3)3 is more
positive than the P in PF2(CH3)3. This draws the F atoms in slightly, so the P—F
bonds are shorter in PF2(CF3)3 (160.1 pm vs. 168.5 pm).
b.
Al—O—Al could have an angle near 109°, like water, or could have double bonds
in both directions and a nearly linear structure. In fact, the angle is
about 140°. The single-bonded picture is more probable; the high
O
electronegativity of O compared to Al draws the bonding pairs
I
Al
AlI
closer, opening up the bond angle. A Lewis structure with zero
formal charges on all atoms can be drawn for this molecule with four electrons on each
Al.
Al
c.
CAl4 is tetrahedral. Again, a Lewis structure with zero formal
charges can be drawn with four electrons on each Al.
a.
Al
Al
C
Al
The Te—X(axial) distances are expected to be longer than the Te—X(equatorial)
distances on the basis of the increased lp-bp and bp-bp repulsion that the electron groups
in the axial positions experience relative to those in the equatorial positions. The
observed bond distances exhibit these features for both TeF4 (Te—F(axial), 189.9 pm;
Te—X(equatorial), 184.6 pm) and TeCl4 (Te—Cl(axial), 242.8 pm; Te—Cl(equatorial),
228.9 pm).
b.
These angles should both be smaller in TeF4 , on the basis of reduced bp-bp repulsion at
the Te atom in TeF4 due to the higher electronegativity of F relative to Cl. The angles
were determined as 164.3° (F(axial)—Te—F(axial)), 176.7° (Cl(axial)—Te—Cl(axial)),
99.5° (F(equatorial)—Te—F(equatorial)), and 102.5° (Cl(equatorial)—Te—
Cl(equatorial). The equatorial nonbonding pair in these complexes has a greater influence
in TeF4 than in TeCl4 .
3.26
Octahedral
SeCl62–
TeCl62–
ClF6–
Distorted
SeF62–
IF6–
The distorted structures have the smallest outer atoms in
comparison with the size of the central atom. In these cases,
there apparently is room for a lone pair to occupy a position
that can lead to distortion. In the octahedral cases there may
be too much crowding to allow a lone pair to distort the shape.
Copyright © 2014 Pearson Education, Inc.
Chapter 3 Simple Bonding Theory
F
F
3.27
O
27
Xe
N
CCH3
Xe
F
O
F
In F2OXeNCCH3, the nitrogen–xenon bond is weak; see the reference for details on bond
distances and angles.
3.28
a.
O is more electronegative than N and can draw the electrons more strongly away from
the S. The more positive S in OSCl2 consequently attracts bonding pairs in S–Cl bonds
closer to sulfur, increasing bp-bp repulsions and increasing the Cl—S—Cl angle (96.2° in
OSCl2, 93.3° in NSCl2–).
b.
Because the sulfur in OSCl2 attracts the S–Cl bonding pairs more strongly, these bonds
–
are shorter: 207.6 pm in OSCl2, 242.3 pm in NSCl2 .
Cl
3.29
The larger, less electronegative Br atoms are equatorial.
Br
Br
P
Cl
Cl
3.30
3.31
a.
In PCl3(CF3)2, the highly electronegative CF3 groups occupy axial positions.
b.
The axial positions in SbCl5 experience greater repulsions by bonding pairs, leading to
longer Sb–Cl (axial) bonds (223.8 pm) than Sb–Cl (equatorial) bonds (227.7 pm).
The pertinent group electronegativity ranking is CF3 > CCl3 > CH 3 . Therefore, ClSO2CF3 is
expected to possess the lowest concentration of electron density near the S of the S—C bond, and
ClSO2CH 3 the highest concentration of electron density. The bp-bp repulsion that influences the
Cl—S—C angles should decrease as ClSO 2CH 3 > ClSO 2CCl3 > ClSO2CF3. Therefore,
ClSO2CF3 should exhibit the smallest Cl—S—C angle, and ClSO2CH 3 the largest Cl—S—C
angle. The angles measured in the gas phase for these molecules are 101° ( ClSO2CH 3 ; M.
Hargittai, I. Hargittai, J. Chem. Phys., 1973, 59, 2513) , 96° ( ClSO 2CCl3 ; N. V. Alekseev, Z.
Struki. Khimii, 1967, 8, 532), and 95.4° ( ClSO2CF3; R. Haist, F. Trautner, J. Mohtasham, R.
Winter, G. L. Gard, H. Oberhammer, J. Mol. Struc., 2000, 550, 59).
3.32
The FSO 2 X molecule with the smallest O—S—O angle is expected to be that with the greatest
concentration of electron density at the S atom from the S—X bond. This molecule should exert
the greatest amount of bp-bp repulsion between the S—F and S—X bonds, maximally hindering
expansion of the O—S—O angle within this series. The pertinent group electronegativity ranking
is F > OCH 3  CH 3; the S— CH 3 bond should possess the greatest electron density at the S
Copyright © 2014 Pearson Education, Inc.
28
Chapter 3 Simple Bonding Theory
atom. While FSO2CH 3 exhibits a smaller O—S—O angle (123.1°, I. Hargittai, M. Hargittai, J.
Mol. Struc., 1973, 15, 399) than found in FSO2 (OCH 3 ) (124.4°, I. Hargittai, R. Seip, K. P.
Rajappan Nair, C. O. Britt, J. E. Boggs, B. N. Cyvin, J. Mol. Struc., 1977, 39, 1.), the O—S—O
angle of FSO2 F (123.6°, D. R. Lide, D. E. Mann, R. M. Fristrom, J. Chem. Phys., 1957, 26, 734)
is smaller than expected on the exclusive basis of group electronegativity arguments. It is
noteworthy that O—S—O angles ranging from 122.6 to 130°, with rather large standard
deviations (see K. Hagen, V. R. Cross, K. Hedberg, J. Mol. Struc., 1978, 44, 187), have also been
reported for FSO2 F .
3.33
a.
Because Te is less electronegative than Se, the highly electronegative C5F4 N groups
draw electron density away from the Te atom more effectively than from the Se atom,
rendering more effective lp-bp repulsion in compressing the C—group 16 atom—C angle
in Te(C5F4 N)2 .
b.
As the electronegativity of the group bound to these atoms increases, lp-bp repulsion is
expected to have increasing impact in compressing the C—group 16 atom—C angle. The
pertinent group electronegativity ranking is C5F4 N > C6 F5 on the basis of theoretical
calculations ( B. Hoge, C. Thösen, T. Hermann, P. Panne, I. Patenburg, J. Flourine
Chem., 2004, 125, 831).
3.34
PF4+ has the bond angle expected for a tetrahedron, 109.5°. In PF3O the
multiple bond to oxygen results in distortion away from the oxygen, leading to
a smaller F–P–F angle. By the LCP approach the F…F distances should be
approximately the same in these two structures. They are similar: 238 pm in
PF4+ and 236 pm in PF3O.
3.35
As more (less electronegative) CH3 groups are added, there is greater concentration of
electrons near P, and greater electron-electron repulsion leads to longer axial P–F bonds.
Reported P–F distances:
161
3.36
PF4(CH3)
pm
PF3(CH3)2
164 pm
PF2(CH3)3
168 pm
Bond angles and distances:
F2C=CF2
F2CO
CF4
–
F3CO
Steric Number
3
3
4
4
C—F (pm)
133.6
131.9
131.9
139.2
FCF angle (°)
109.2
107.6
109.5
101.3
F—F (pm)
218
216
216
215
The differences between these molecules are subtle. The LCP model views the F
ligands as hard objects, tightly packed around the central C in these examples. In this
approach, the F…F distance remains nearly constant while the central atom moves to minimize
repulsions.
3.37
The calculation is similar to the example shown in Section 3.2.4.
Copyright © 2014 Pearson Education, Inc.
Chapter 3 Simple Bonding Theory
Cl
O
55.9°
C
289 pm
x
C
C
Cl
Cl
Cl
Cl
111.8°
Cl
Cl
144.5 pm
289 pm
sin 55.9° = 0.828 =
x = 174.5 pm
3.38
29
144.5
x
CF3+ is expected to be trigonal planar with a 120° F—C—F angle.
F
60o
C
C
x
F
120o
F
215 pm
F
107.5 pm
sin 60  0.866 
x  124.1 pm
107.5
x
The C—F bond length predicted via quantum chemical calculations is 124.4 pm, with a F…F
distance of 216 pm (R. J. Gillespie and P. L. A. Popelier, Chemical Bonding and Molecular
Geometry, Oxford, New York, 2001, p. 119).
3.39
By the LCP approach, from the structures of HOH and FOF, the hydrogen radius would be
76 pm (half of the H…H distance) and the fluorine radius (half of the F…F distance) would be
110 pm. Because the LCP model describes nonbonded outer atoms as being separated by the
sums of their radii, as if they were touching spheres, the H…F distance in HOF would therefore
be the sum of the ligand radii, 76 + 110 = 186 pm, in comparison with the actual H…F distance
of 183 pm. If the covalent O–H and O–F bonds in HOF are similar to the matching distances in
HOH and FOF, the H–O–F angle must be smaller than the other angles because of the H…F
distance.
O
An alternative explanation considers the polarity in HOF. Because of the high
+
electronegativity of fluorine, the F atom in HOF acquires a partial negative
–
H
charge, which is attracted to the relatively positive H atom. By this approach,
F
electrostatic attraction between H and F reduces the bond angle in HOF, giving it the smallest
angle of the three compounds.
3.40
The electronegativity differences are given in parentheses:
a.
C–N
N is negative (0.522)
d.
O–Cl
O is negative (0.741)
b.
N–O
O is negative (0.544)
e.
P–Br
Br is negative (0.432)
c.
C–I
C is negative (0.185)
f.
S–Cl
Cl is negative (0.280)
The overall order of polarity is O–Cl > N–O > C–N > P–Br > S–Cl > C–I.
Copyright © 2014 Pearson Education, Inc.
30
Chapter 3 Simple Bonding Theory
3.41
a.
b.
O
VOCl3 has a distorted tetrahedral shape,
with Cl—V—Cl angles of 111°, and
Cl—V—O angles of 108°.
V
Cl
Cl
Cl
PCl3 has a trigonal pyramidal shape with Cl—P—Cl angles of 100.4°.
F
c.
SOF4 has a distorted trigonal bipyramidal shape.
The axial fluorine atoms are nearly linear with
the S atom; the equatorial F—S—F angle is 100°.
F
S
F
P
Cl
Cl
Cl
O
F
O
d.
SO3 is trigonal (triangular), with equal bond angles of 120°.
S
O
e.
O
ICl3 would be expected to have two axial lone pairs, causing
distortion to reduce the Cl (axial)–I–Cl (equatorial) angles to
< 90°. However, reaction of I2 with Cl2 yields dimeric I2Cl6,
which readily dissociates into ICl and Cl2.
Cl
I
F
f.
SF6 is a regular octahedron.
F
F
S
Cl
F
F
F
F
g.
h.
F
IF7 is a rare example of pentagonal bipyramidal geometry.
F
F
O
F
F
Xe
O
CF2Cl2, like methane, is tetrahedral.
O
P4O6 is described in the problem. Each P
has one lone pair, each O has two.
I
F
The structure of XeO2F4 is based on an octahedron, with oxygens in
trans positions because of multiple bonding.
C
Cl
Cl
j.
F
F
F
i.
Cl
P
O
O
P
O
F
P
O
Copyright © 2014 Pearson Education, Inc.
P
O
F
F
Chapter 3 Simple Bonding Theory
3.42
a.
PH3 has a smaller bond angle than NH3, about 93°. The larger
central atom reduces the repulsion between the bonding pairs.
b.
H2Se has a structure like water, with a bond angle near 90°.
The larger central atom increases the distance between the
S–H bonding pairs and reduces their repulsion,
resulting in a smaller angle than in water.
P
H
H
H
H
Se
H
F
c.
SeF4 has a lone pair at one of the equatorial positions of a trigonal
bipyramid, and bond angles of about 110° (equatorial) and 169° (axial).
Seesa
w shape.
F
F
d.
PF5 has a trigonal bipyramidal structure.
e.
IF5 is square pyramidal, with slight distortion
away from the lone pair.
f.
F
XeO3 has a trigonal pyramidal shape, similar to
NH3, but with Xe–O double bonds.
Cl
g.
h.
i.
BF2Cl is trigonal planar, with
FBCl larger than FBF.
P
31
F
Se
F
F
F
F
I
F
F
F
F
F
O
O
Xe
O
B
F
F
SnCl2 has a bond angle of 95° in the vapor phase, smaller than the
trigonal angle. As a solid, it forms polymeric chains with bridging
chlorines and bond angles near 80°.
Sn
Cl
Cl
KrF2 is linear: F—Kr—F . VSEPR predicts three lone pairs on krypton in equatorial
positions, with the fluorine atoms in axial positions.
O
j.
IO2F52– has a steric number of 7 on iodine, with oxygen atoms
occupying axial positions.
F
F
F
F
O
3.43
Polar: VOCl3, PCl3, SOF4, ICl3, CF2Cl2
3.44
Polar: PH3, H2Se, SeF4, IF5, XeO3, BF2Cl, SnCl2
Copyright © 2014 Pearson Education, Inc.
F
I
2–
32
Chapter 3 Simple Bonding Theory
3.45
a.
The H–O bond of methanol is more polar than the H–S bond of methyl mercaptan.
As a result, hydrogen bonding holds the molecules together and requires more
energy for vaporization. The larger molecular weight of methyl mercaptan has a
similar effect, but the hydrogen bonding in methanol has a stronger influence.
b.
CO and N2 have nearly identical molecular weights, but the polarity of CO leads to
dipole-dipole attractions that help hold CO molecules together in the solid and liquid
states.
c.
The ortho isomer of hydroxybenzoic acid can form
intramolecularhydrogen bonds, while the meta and para
isomers tend to form dimers and larger aggregates in their
hydrogen bonding. As a result of their better ability to
form hydrogen bonds between molecules (intermolecular
hydrogen bonds), the meta and para isomers have higher
melting points (ortho, 159°; meta, 201.3°; para, 214-5°).
C
Intramolecular
hydrogen bond
O
H
O
d.
The London (dispersion) forces between atoms increase with the number of electrons,
so the noble gases with larger Z have larger interatomic forces and higher boiling points.
e.
Acetic acid can form hydrogen-bonded dimers in the gas
phase, so the total number of particles in the gas is half
the number expected by using the ideal gas law.
f.
3.46
H
O
O
H3 C
H
O
C
C
O
H
Acetone has a negative carbonyl oxygen; chloroform has a
CH3
positive hydrogen, due to the electronegative character of the
H3C
C
chlorines. As a result, there is a stronger attraction between
the different kinds of molecules than between molecules of the
O
same kind, and a resulting lower vapor pressure. (This is an unusual
case of hydrogen bonding, with no H–N, H–O, or H–F bond involved.)
CH3
O
Cl
H
C
Cl
g.
CO has about 76 kJ contribution to its bond energy because of the electronegativity
difference between C and O; attraction between the slightly positive and negative ends
strengthens the bonding. Although this is not a complete explanation, it covers most of
difference between CO and N2. In spite of its high bond energy, N2 is thought by some
to have some repulsion in its sigma bonding because of the short bond distance.
a.
The trend in these angles is counter-intuitive on the basis of electronegativity arguments.
Electronegativity decreases as P > As > Sb, and the C—group 15 atom—C angle is
expected to decrease as P > As > Sb on the basis of less bp-bp repulsion at the group 15
atom as P > As > Sb. Both As(CF3 )3 and Sb(CF3 )3 are expected to exhibit more acute
C—group 15 atom—C angles relative to P(CF3 )3 .
b.
On the basis of the argument above, the C—Sb—C angle of Sb(CF3 )3 should be
reinvestigated. This angle is predicted to be smaller than the newly determined
C—As—C angle of As(CF3 )3 .
Copyright © 2014 Pearson Education, Inc.
Cl
Chapter 4 Symmetry and Group Theory
33
CHAPTER 4: SYMMETRY AND GROUP THEORY
4.1
4.2
a.
Ethane in the staggered conformation has 2 C3 axes (the C–C line), 3 perpendicular C2
axes bisecting the C–C line, in the plane of the two C’s and the H’s on opposite sides of
the two C’s. No h, 3d, i, S6. D3d.
b.
Ethane in eclipsed conformation has two C3 axes (the C–C line), three perpendicular C2
axes bisecting the C–C line, in the plane of the two C’s and the H’s on the same side of
the two C’s. Mirror planes include h and 3d. D3h.
c.
Chloroethane in the staggered conformation has only one mirror plane, through both C’s,
the Cl, and the opposite H on the other C. Cs.
d.
1,2-dichloroethane in the trans conformation has a C2 axis perpendicular to the
C–C bond and perpendicular to the plane of both Cl’s and both C’s, a h plane through
both Cl’s and both C’s, and an inversion center. C2h.
a.
Ethylene is a planar molecule, with C2 axes through the C’s and perpendicular to the C–C
bond both in the plane of the molecule and perpendicular to it. It also has a h plane and
two d planes (arbitrarily assigned). D2h.
b.
Chloroethylene is also a planar molecule, with the only symmetry element the mirror
plane of the molecule. Cs.
c.
1,1-dichloroethylene has a C2 axis coincident with the C–C bond, and two mirror planes,
one the plane of the molecule and one perpendicular to the plane of the molecule through
both C’s. C2v.
cis-1,2-dichloroethylene has a C2 axis perpendicular to the C–C bond, and in the plane of
the molecule, two mirror planes (one the plane of the molecule and one perpendicular to
the plane of the molecule and perpendicular to the C–C bond). C2v.
trans-1,2-dichloroethylene has a C2 axis perpendicular to the C–C bond and
perpendicular to the plane of the molecule, a mirror plane in the plane of the molecule,
and an inversion center. C2h.
4.3
a.
Acetylene has a C axis through all four atoms, an infinite number of perpendicular
C2 axes, a h plane, and an infinite number of d planes through all four atoms. Dh.
b.
Fluoroacetylene has only the C axis through all four atoms and an infinite number of
mirror planes, also through all four atoms. Cv.
c.
Methylacetylene has a C3 axis through the carbons and three v planes, each
including one hydrogen and all three C’s. C3v.
d.
3-Chloropropene (assuming a rigid molecule, no rotation around the C–C bond) has
no rotation axes and only one mirror plane through Cl and all three C atoms. Cs.
e.
Phenylacetylene (again assuming no internal rotation) has a C2 axis down the long axis
of the molecule and two mirror planes, one the plane of the benzene ring and the other
perpendicular to it. C2v
Copyright © 2014 Pearson Education, Inc.
34
Chapter 4 Symmetry and Group Theory
4.4
a.
Napthalene has three perpendicular C2 axes, and a horizontal mirror plane (regardless
of which C2 is taken as the principal axis), making it a D2h molecule.
b.
1,8-dichloronaphthalene has only one C2 axis, the C–C bond joining the two rings, and
two mirror planes, making it a C2v molecule.
c.
1,5-dichloronaphthalene has one C2 axis perpendicular to the plane of the molecule, a
horizontal mirror plane, and an inversion center; overall, C2h.
d.
1,2-dichloronaphthalene has only the mirror plane of the molecule, and is a Cs molecule.
a.
1,1-dichloroferrocene has a C2 axis parallel to the rings, perpendicular to the Cl–Fe–Cl
h mirror plane. It also has an inversion center. C2h.
b.
Dibenzenechromium has collinear C6, C3, and C2 axes perpendicular to the rings, six
perpendicular C2 axes and a h plane, making it a D6h molecule. It also has three v
and three d planes, S3 and S6 axes, and an inversion center.
c.
Benzenebiphenylchromium has a mirror plane through the Cr and the biphenyl bridge
bond and no other symmetry elements, so it is a Cs molecule.
d.
H3O+ has the same symmetry as NH3: a C3 axis, and three v planes for a C3v molecule.
e.
O2F2 has a C2 axis perpendicular to the O–O bond and perpendicular to a line connecting
the fluorines. With no other symmetry elements, it is a C2 molecule.
f.
Formaldehyde has a C2 axis collinear with the C=O bond, a mirror plane including all the
atoms, and another perpendicular to the first and including the C and O atoms. C2v.
g.
S8 has C4 and C2 axes perpendicular to the average plane of the ring, four C2 axes through
opposite bonds, and four mirror planes perpendicular to the ring, each including two S
atoms. D4d.
h.
Borazine has a C3 axis perpendicular to the plane of the ring, three perpendicular C2 axes,
and a horizontal mirror plane. D3h.
i.
Tris(oxalato)chromate(III) has a C3 axis and three perpendicular C2 axes, each splitting
a C–C bond and passing through the Cr. D3.
j.
A tennis ball has three perpendicular C2 axes (one through the narrow portions of each
segment, the others through the seams) and two mirror planes including the first rotation
axis. D2d.
a.
Cyclohexane in the chair conformation has a C3 axis perpendicular to the average plane
of the ring, three perpendicular C2 axes between the carbons, and three v planes, each
including the C3 axis and one of the C2 axes. D3d.
b.
Tetrachloroallene has three perpendicular C2 axes, one collinear with the double bonds
and the other two at 45° to the Cl—C—Cl planes. It also has two v planes, one defined
by each of the Cl–C–Cl groups. Overall, D2d. (Note that the ends of tetrachlorallene are
staggered.)
4.5
4.6
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
35
c.
The sulfate ion is tetrahedral. Td.
d.
Most snowflakes have hexagonal symmetry (Figure 4.2), and have collinear C6, C3, and
C2 axes, six perpendicular C2 axes, and a horizontal mirror plane. Overall, D6h. (For high
quality images of snowflakes, including some that have different shapes, see K. G.
Libbrecht, Snowflakes, Voyageur Press, Minneapolis, MN, 2008.)
e.
Diborane has three perpendicular C2 axes and three perpendicular mirror planes. D2h.
f.
1,3,5-tribromobenzene has a C3 axis perpendicular to the plane of the ring, three
perpendicular C2 axes, and a horizontal mirror plane. D3h.
1,2,3-tribromobenzene has a C2 axis through the middle Br and two perpendicular
mirror planes that include this axis. C2v
1,2,4-tribromobenzene has only the plane of the ring as a mirror plane. Cs.
4.7
g.
A tetrahedron inscribed in a cube has Td symmetry (see Figure 4.6).
h.
The left and right ends of B3H8 are staggered with respect to each other. There is a
C2 axis through the borons. In addition, there are two planes of symmetry, each
containing four H atoms, and two C2 axes between these planes and perpendicular to
the original C2. The point group is D2d.
i.
A mountain swallowtail butterfly has only a mirror that cuts through the head, thorax, and
abdomen. Cs
j.
The Golden Gate Bridge has a C2 axis and two perpendicular mirror planes that include
this axis. C2v
a.
A sheet of typing paper has three perpendicular C2 axes and three perpendicular mirror
planes. D2h.
b.
An Erlenmeyer flask has an infinite-fold rotation axis and an infinite number of v
planes, Cv.
c.
A screw has no symmetry operations other than the identity, for a C1 classification.
d.
The number 96 (with the correct type font) has a C2 axis perpendicular to the plane
of the paper, making it C2h.
e.
Your choice—the list is too long to attempt to answer it here.
f.
A pair of eyeglasses has only a vertical mirror plane. Cs.
g.
A five-pointed star has a C5 axis, five perpendicular C2 axes, one horizontal and five
vertical mirror planes. D5h.
h.
A fork has only a mirror plane. Cs.
i.
Wilkins Micawber has no symmetry operation other than the identity. C1.
Copyright © 2014 Pearson Education, Inc.
36
4.8
4.9
4.10
Chapter 4 Symmetry and Group Theory
j.
A metal washer has a C axis, an infinite number of perpendicular C2 axes, an infinite
number of v mirror planes, and a horizontal mirror plane. Dh.
a.
D2h
f.
C3
b.
D4h (note the four knobs)
g.
C2h
c.
Cs
h.
C8v
d.
C2v
i.
Dh
e.
C6v
j.
C3v
a.
D3h
f.
Cs (note holes)
b.
D4h
g.
C1
c.
Cs
h.
C3v
d.
C3
i.
Dh
e.
C2v
j.
C1
Hands (of identical twins): C2
Baseball: D2d
Eiffel Tower: C4v
6 × 6: C2v
Dominoes:
Atomium: C3v
3 × 3: C2
5 × 4: Cs
Bicycle wheel: The wheel shown has 32 spokes. The point group assignment depends on how
the pairs of spokes (attached to both the front and back of the hub) connect with the rim. If the
pairs alternate with respect to their side of attachment, the point group is D8d. Other arrangements
are possible, and different ways in which the spokes cross can affect the point group assignment;
observing an actual bicycle wheel is recommended. (If the crooked valve is included, there is no
symmetry, and the point group is a much less interesting C1.)
4.11
a.
Problem 3.41*: a. VOCl3: C3v
b. PCl3: C3v
c. SOF4: C2v
d. SO3: D3h
e. ICl3: C2v
f. SF6: Oh
g. IF7: D5h
h. XeO2F4: D4h
i. CF2Cl2: C2v
j. P4O6: Td
* Incorrectly cited as problem 4.30 in first printing of text.
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
b.
Problem 3.42*: a. PH3: C3v
b. H2Se: C2v
c. SeF4: C2v
d. PF5: D3h
e. IF5: C4v
f. XeO3: C3v
g. BF2Cl: C2v
h. SnCl2: C2v
i. KrF2: Dh
a. CO2: Dh
b. SO3: D3h
c. CH4: Td
d. PCl5: D3h
e. SF6: Oh
f. IF7: D5h
a. CO2: Dh
b. COF2: C2v
c. NO2 : C2v
d. SO3: D3h
e. SNF3: C3v
f. SO2Cl2: C2v
g. XeO3: C3v
h. SO42–: Td
i. SOF4: C2v
j. ClO2F3: C2v
k. XeO3F2: D3h
l. IOF5: C4v
37
2–
j. IO2F5 : D5h
4.12
a.
Figure 3.8:
g. TaF83–: D4d
b.
4.13
4.14
Figure 3.15:
–
a.
px has Cv symmetry. (Ignoring the difference in sign between the two lobes, the point
group would be Dh.)
b.
dxy has D2h symmetry. (Ignoring the signs, the point group would be D4h.)
c.
dx2–y2 has D2h symmetry. (Ignoring the signs, the point group would be D4h.)
d.
dz2 has Dh symmetry.
e.
fxyz has Td symmetry.
a.
The superimposed octahedron and cube
show the matching symmetry elements.
C3
The descriptions below are for the
elements of a cube; each element also
applies to the octahedron.
E
Every object has an identity
operation.
8C3
Diagonals through opposite
corners of the cube are C3 axes.
6C2
Lines bisecting opposite edges are C2 axes.
6C4
Lines through the centers of opposite faces are C4 axes. Although there are only
three such lines, there are six axes, counting the C43 operations.
3C2
(=C42) The lines through the centers of opposite faces are C4 axes as well as
C2 axes.
* Incorrectly cited as problem 3.41 in first printing of text.
Copyright © 2014 Pearson Education, Inc.
C 2, C 4
38
4.15
Chapter 4 Symmetry and Group Theory
i
The center of the cube is the inversion center.
6S4
The C4 axes are also S4 axes.
8S6
The C3 axes are also S6 axes.
3h
These mirror planes are parallel to the faces of the cube.
6d
These mirror planes are through two opposite edges.
b.
Oh
c.
O
a.
There are three possible orientations of the two blue faces.
If the blue faces are opposite each other, a C3 axis connects the centers of the
blue faces. This axis has 3 perpendicular C2 axes, and contains three vertical
mirror places (D3d).
If the blue faces share one vertex of the octahedron, a C2 axis includes this
vertex, and this axis includes two vertical mirror planes (C2v).
The third possibility is for the blue faces to share an edge of the octahedron. In
this case, a C2 axis bisects this shared edge, and includes two vertical mirror
planes (C2v).
b.
There are three unique orientations of the three blue faces.
If one blue face is arranged to form edges with each of the two remaining blue
faces, the only symmetry operations are identity and a single mirror plane (Cs).
If the three blue faces are arranged such that a single blue face shares an edge
with one blue face, but only a vertex with the other blue face, the only symmetry
operation is a mirror plane that passes through the center of the blue faces, and
the point group is Cs.
If the three blue faces each share an edge with the same yellow face, a C3 axis
emerges from the center of this yellow face, and this axis includes three vertical
mirror planes (C3v).
c.
4.16
If there are four different colors, and each pair of opposite faces has the identical color,
the only symmetry operations are identity and inversion (Ci).
Four point groups are represented by the symbols of the chemical elements. Most
symbols have a single mirror in the plane of the symbol (Cs), for example, Cs! Two
symbols have D2h symmetry (H, I), and two more ( , S) have C2h. Seven exhibit C2v
symmetry (B, C, K, V, Y, W, U). In some cases, the choice of font may affect the point
group. For example, the symbol for nitrogen may have C2h in a sans serif font ( ) but
otherwise Cs (N). The symbol of oxygen has D∞h symmetry if shown as a circle but D2h
if oval.
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
4.17
4.18
a.
on-deck circle Dh
e.
home plate
C2v
b.
batter’s box
D2h
f.
baseball
D2d (see Figure 4.1)
c.
cap
Cs
g.
pitcher
C1
d.
bat
C v
a.
D2h
d.
D4h
g.
D2h
b.
C2 v
e.
C5h
h.
D4h
c.
C2 v
f.
C2v
i.
C2
y
N
4.19
N
S
F
SNF3
F2
F
F
F3
x
F1
(top view)
Symmetry Operations:
F3
F1
F2
N
N
F1
N
F3
F2
F3
after E
F1
F2
after C3
after v (xz)
Matrix Representations (reducible):
1 0 0
E: 0 1 0



0 0 1

 2
cos
3

2
C3: sin
3


 0

2
3
2
cos
3
 sin
0
  1
0 –
2
 
3
0 
  2
 
1  0
 

3
2
1
–
2
–
0

0

0


1

1 0 0
v(xz): 0 1 0



0 0 1
Characters of Matrix Representations:
3
0
(continued on next page)
Copyright © 2014 Pearson Education, Inc.
1
39
40
Chapter 4 Symmetry and Group Theory
Block Diagonalized Matrices:
Irreducible Representations:
E
2
1
2 C3
–1
1
3v
0
1
Coordinates Used
(x, y)
z
Character Table:
C3v
A1
A2
E
E
1
1
2
2 C3 3v
1
1
1
–1
–1
0
Matching Functions
z
Rz
(x, y), (Rx, Ry)
x2 + y2, z2
(x2 – y2, xy)(xz, yz)
H
Cl
C
4.20
a.
b.
c.
H
C2h molecules have E, C2, i, and h operations.
E:
C2:
i:
 1 0 0 


 0 1 0 
 0 0 1 
 1 0 0 


 0 1 0 
 0 0 1 
 1 0 0 


 0 1 0 
 0 0 1 
C
Cl
h:
 1 0 0 


 0 1 0 
 0 0 1 
These matrices can be block diagonalized into three 1 × 1 matrices, with the
representations shown in the table.
Bu
Au
(E)
1
1
(C2)
–1
1
(i)
–1
–1
(h)
1
–1
from the x and y coefficients
from the z coefficients
The total is  = 2Bu +Au.
d.
Multiplying Bu and Au:
1 × 1 + (–1) × 1 + (–1) × (–1) + 1 × (–1) = 0, proving they are orthogonal.
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
H
4.21
a.
D2h molecules have E, C2(z), C2(y), C2(x), i, (xy), (xz), and
(yz) operations.
H
C
H
b.
 1 0 0 


E:  0 1 0 
 0 0 1 
 1 0 0 


C2(y):
 0 1 0 
 0 0 1 
 1 0 0 


i:  0 1 0 
 0 0 1 
 1 0 0 


(xy):  0 1 0 
 0 0 1 
 1 0 0 


(xz):
 0 1 0 
 0 0 1 
c.
d.
e.

1
2
3
E
3
1
1
1
C2(z)
–1
–1
–1
1
C2(y)
–1
–1
1
–1
H
 1 0 0 


 0 1 0 
 0 0 1 
C2(z):
 1 0 0 


C2(x): 0 1 0


 0 0 1 
C
C2(x)
–1
1
–1
–1
 1 0 0 


(yz):
 0 1 0 
 0 0 1 
i
–3
–1
–1
–1
(xy)
1
1
1
–1
(xz)
1
1
–1
1
(yz)
1
–1
1
1
matching B3u
matching B2u
matching B1u
1 × 2 = 1 × 1 + (–1) × (–1) + (–1) × 1 + 1 × (–1) + (–1) × (–1) + 1 × 1
+ 1 × (–1) + (–1) × 1 = 0
1 × 3 = 1 × 1 + (–1) × 1 + (–1) × (–1) + 1 × (–1) + (–1) × (–1) + 1 × (–1)
+ 1 × 1 + (–1) × 1 = 0
2 × 3 = 1 × 1 + (–1) × 1 + 1 × (–1) + (–1) × (–1) + (–1) × (–1) + 1 × (–1)
+ (–1) × 1 + 1 × 1 = 0
4.22
a.
h = 8 (the total number of symmetry operations)
b.
A1 × E = 1 × 2 + 2 × 1 × 0 +1 × (–2) + 2 × 1 × 0 + 2 × 1 × 0 = 0
A2 × E = 1 × 2 + 2 × 1 × 0 +1 × (–2) + 2 × (–1) × 0 + 2 × (–1) × 0 = 0
B1 × E = 1 × 2 + 2 × (–1) × 0 +1 × (–2) + 2 × 1 × 0 + 2 × (–1) × 0 = 0
B2 × E = 1 × 2 + 2 × (–1) × 0 +1 × (–2) + 2 × (–1) × 0 + 2 × 1 × 0 = 0
c.
E: 4 + 2 × 0 + 4 + 2 × 0 + 2 × 0 = 8
A1 : 1 + 2 × 1 + 1 + 2 × 1 + 2 × 1 = 8
A2 : 1 + 2 × 1 + 1 + 2 × 1 + 2 × 1 = 8
B1 : 1 + 2 × 1 + 1 + 2 × 1 + 2 × 1 = 8
B2 : 1 + 2 × 1 + 1 + 2 × 1 + 2 × 1 = 8
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41
42
Chapter 4 Symmetry and Group Theory
d.
1 = 2A1 + B1 + B2 + E:
A1: 1/8[1 × 6 + 2 × 1 × 0 + 1 × 2 + 2 × 1 × 2 + 2 × 1 × 2] = 2
A2: 1/8[1 × 6 + 2 × 1 × 0 + 1 × 2 + 2 × (–1) × 2 + 2 × (–1) × 2] = 0
B1: 1/8[1 × 6 + 2 × (–1) × 0 + 1 × 2 + 2 × 1 × 2 + 2 × (–1) × 2] = 1
B2: 1/8[1 × 6 + 2 × (–1) × 0 + 1 × 2 + 2 × (–1) × 2 + 2 × 1 × 2] = 1
E: 1/8[2 × 6 + 2 × 0 × 0 + (–2) × 2 + 2 × 0 × 2 + 2 × 0 × 2] = 1
2 = 3 A1 + 2A2 + B1:
A1: 1/8[1 × 6 + 2 × 1 × 4 + 1 × 6 + 2 × 1 × 2 + 2 × 1 × 0] = 3
A2: 1/8[1 × 6 + 2 × 1 × 4 + 1 × 6 + 2 × (–1) × 2 + 2 × (–1) × 0] = 2
B1: 1/8[1 × 6 + 2 × (–1) × 4 + 1 × 6 + 2 × 1 × 2 + 2 × (–1) × 0] = 1
B2: 1/8[1 × 6 + 2 × (–1) × 4 + 1 × 6 + 2 × (–1) × 2 + 2 × 1 × 0] = 0
E: 1/8[2 × 6 + 2 × 0 × 4 + (–2) × 6 + 2 × 0 × 2 + 2 × 0 × 0] = 0
4.23
C3v
1 = 3A1 + A2 + E:
A1: 1/6[1 × 6 + 2 × 1 × 3 + 3 × 1 × 2] = 3
A2: 1/6[1 × 6 + 2 × 1 × 3 + 3 × (–1) × 2] = 1
E: 1/6[2 × 6 + 2 × (–1) × 3 + 3 × 0 × 2] = 1
2 = A2 + E:
A1: 1/6[1 × 5 + 2 × 1 × (–1) + 3 × 1 × (–1)] = 0
A2: 1/6[1 × 5 + 2 × 1 × (–1) + 3 × (–1) × (–1)] = 1
E: 1/6[2 × 5 + 2 × (–1) × (–1) + 3 × 0 × (–1)] = 2
Oh
3 = A1g + Eg + T1u:
A1g: 1/48[6 + 0 + 0 + 12 + 6 + 0 + 0 + 0 + 12 + 12] = 1
A2g: 1/48[6 + 0 + 0 – 12 + 6 + 0 + 0 + 0 + 12 – 12] = 0
Eg: 1/48[12 + 0 + 0 + 0 + 12 + 0 + 0 + 0 + 24 + 0] = 1
T1g: 1/48[18 + 0 + 0 + 12 – 6 + 0 + 0 + 0 – 12 – 12] = 0
T2g: 1/48[18 + 0 + 0 – 12 – 6 + 0 + 0 + 0 – 12 + 12] = 0
A1u: 1/48[6 + 0 + 0 + 12 + 6 + 0 + 0 + 0 – 12 – 12] = 0
A2u: 1/48[6 + 0 + 0 – 12 + 6 + 0 + 0 + 0 – 12 + 12] = 0
Eu: 1/48[12 + 0 + 0 + 0 + 12 + 0 + 0 + 0 – 24 + 0] = 0
T1u: 1/48[18 + 0 + 0 + 12 – 6 + 0 + 0 + 0 + 12 + 12] = 1
T2u: 1/48[18 + 0 + 0 – 12 – 6 + 0 + 0 + 0 + 12 – 12] = 0
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
4.24
The dxy characters match the characters
of the B2g representation:
dxy
y
x
dxy
43
The dx2-y2 characters match the characters
of the B1g representation:
dx2–y2


E
1
E
1
C4
–1
C4
–1
C2
1
C2
1
C2
–1
C2
1
C2
1
C2
–1
i
1
i
1
S4
–1
S4
–1
h
1
h
1
v
–1
v
1
d
1
d
–1
y
x
dx2–y2
3–
4.25
Chiral: 4.5: O2F2, [Cr(C2O4)3] 4.6: none 4.7: screw, Wilkins Micawber 4.8: recycle
symbol 4.9: set of three wind turbine blades, Flying Mercury sculpture, coiled
spring
4.26
a.
C4v

A1
A2
B1
B2
E
Point group: C4v
E
18
1
1
1
1
2
2C4
2
1
1
–1
–1
0
O
C2
–2
1
1
1
1
–2
2v
4
1
–1
1
–1
0
2d
2
1
–1
–1
1
0
b.
 = 4 A1 + A2 + 2B1 + B2 + 5E
c.
Translation: A1 + E (match x, y, and z)
Rotation: A2 + E (match Rx, Ry, and Rz)
Vibration: all that remain: 3 A1 + 2B1 + B2 + 3E
d.
F
F
F
Xe
F
z
Rz
(x, y), (Rx, Ry)
The character for each symmetry operation for the Xe–O stretch is +1.
This corresponds to the A1 irreducible representation, which matches
the function z and is therefore IR-active.
Copyright © 2014 Pearson Education, Inc.
O
F
F
Xe
F
F
44
Chapter 4 Symmetry and Group Theory
4.27
For SF6, the axes of the sulfur should point at three of the fluorines. The fluorine
axes can be chosen in any way, as long as one from each atom is directed toward the sulfur atom.
There are seven atoms with three axes each, for a total of 21.
Oh

T1u
T1g
A1g
A2u
Eg
E
21
3
3
1
1
2
8C3
0
0
0
1
1
–1
6C2
–1
–1
–1
1
–1
0
6C4
3
1
1
1
–1
0
3C2
–3
–1
–1
1
1
2
i
–3
–3
3
1
1
2
6S4
–1
–1
1
1
–1
0
8S6
0
0
0
1
1
–1
3h
5
1
–1
1
1
2
6d
3
1
–1
1
–1
0
T2u
T2g
3
3
0
0
1
1
–1
–1
–1
–1
–3
3
1
–1
0
0
1
–1
–1
1
(x,y,z)
(Rx, Ry, Rz)
(2z2 – x2 – y2,
x2 – y2)
(xy, xz, yz)
Reduction of  gives  = 3T1u + T1g + A1g + Eg + T2g + T2u. T1u accounts for translation and also
infrared active vibrational modes. T1g is rotation. The remainder are infrared-inactive vibrations.
4.28
a.
cis-Fe(CO)4Cl2 has C2v symmetry.
O
C
The vectors for CO stretching have the representation :
C2v

A1
A2
B1
B2
E
4
1
1
1
1
C2
0
1
1
–1
–1
v(xz)
2
1
–1
1
–1
v(yz)
2
1
–1
–1
1
Cl
Cl
Fe
C
O
z
x
y
n(A1) = 1/4[4 × 1 + 0 × 1 + 2 × 1 + 2 × 1] = 2
n(A2) = 1/4[4 × 1 + 0 × 1 + 2 × (–1) + 2 × (–1)] = 0
n(B1) = 1/4[4 × 1 + 0 × (–1) + 2 × 1 + 2 × (–1)] = 1
n(B2) = 1/4[4 × 1 + 0 × (–1) + 2 × (–1) + 2 × 1] = 1
 = 2 A1 + B1 + B2, all four IR active.
Cl
b.
OC
OC
trans-Fe(CO)4Cl2 has D4h symmetry.
D4h

A2u
Eu
E
4
1
2
2C4
0
1
0
C2
0
1
–2
2C2
2
–1
0
2C2
0
–1
0
Fe
CO
CO
Cl
i
0
–1
–2
2S4
0
–1
0
h
4
–1
2
Copyright © 2014 Pearson Education, Inc.
2v
2
1
0
2d
0
1
0
z
(x,y)
CO
CO
Chapter 4 Symmetry and Group Theory
45
Omitting the operations that have zeroes in :
n(A2u) = 1/16[4 × 1 + 2 × 2 × (–1) + 4 × (–1) + 2 × 2 × 1] = 0
n(Eu) = 1/16[4 × 2 + 2 × 2 × 0 + 4 × 2 + 2 × 2 × 0] = 1 (IR active)
Note: In checking for IR-active bands, it is only necessary to check the
irreducible representations having the same symmetry as x, y, or z, or a
combination of them.
c.
Fe(CO)5 has D3h symmetry.
The vectors for C–O stretching have the following representation :
D3h

E
A2
E
5
2
1
2C3
2
–1
1
3C2
1
0
–1
h
3
2
–1
2S3
0
–1
–1
3v
3
0
1
O
C
OC
OC
Fe
CO
C
O
(x, y)
z
n(E) = 1/12 [ (5 × 2) + (2 × 2 × –1) + (3 × 2)] = 1
n(A2) = 1/12 [(5 × 1) + (2 × 2 × 1) + (3 × 1 × –1) + (3 × –1) + (3 × 3 × 1)] = 1
There are two bands, one matching Eand one matching A2. These are the only
irreducible representations that match the coordinates x, y, and z.
4.29
In 4.28a, the symmetries of the CO stretching vibrations of cis-Fe(CO)4Cl2 (C2v symmetry) are
determined as 2 A1 + B1 + B2. Each of these representations matches Raman-active functions: A1
(x2, y2, z2) ; A2 (xy), B1 (xz); and B2 (yz), so all are Raman-active.
In 4.28b, the symmetries of the CO stretching vibrations of trans-Fe(CO)4Cl2 (D4h symmetry) are
A1g + B1g + Eu. Only A1g (x2 + y2, z2) and B1g (x2 – y2) match Raman active functions; this complex
exhibits two Raman-active CO stretching vibrations.
In 4.28c, the symmetries of the CO stretching vibrations of Fe(CO)5 (D3h symmetry) are 2 A1 +
E + A2. Only A1 (x2 + y2, z2) and E(x2 – y2, xy) match Raman-active functions; this complex
exhibits four Raman-active CO stretching vibrations.
4.30
a.
The point group is C2h.
b.
Using the Si–I bond vectors as a basis generates the representation:
C2h

Ag
Bg
Au
Bu
E
4
1
1
1
1
C2
0
1
–1
1
–1
i
0
1
1
–1
–1
h
0
1
–1
–1
1
z
x, y
 = Ag + B g + A u + B u
The Au and Bu vibrations are infrared active.
c.
The Ag and Bg vibrations are Raman active.
Copyright © 2014 Pearson Education, Inc.
x2 + y2, z2
xz, yz
46
Chapter 4 Symmetry and Group Theory
4.31
trans isomer (D4h):
cis isomer (C2v):
The simplest approach is to consider if the number of infrared-active I–O stretches is different for
these structures. (Alternatively, one could also determine the number of IR-active I–F stretches, a
slightly more complicated task.)
trans:
D4h

A1g
A2u
E
2
1
1
C2
2
1
1
2C4
2
1
1
C2 C2
0
0
1
1
–1
–1
i
0
1
–1
2S4
0
1
–1
h
0
1
–1
2v
2
1
1
2d
2
1
1
z
There is only a single IR-active I–O stretch (the antisymmetric stretch), A2u.
cis:
C2v

A1
B1
C2
0
1
–1
E
2
1
1
 (xz)
2
1
1
 (yz)
0
1
–1
z
x
There are two IR-active I–O stretches, the A1 and B1 (symmetric and antisymmetric). Infrared
spectra should therefore be able to distinguish between these isomers. (Reversing the x and y
axes would give A1 + B2. Because B2 matches y, it would also represent an IR-active vibration.)
Because these isomers would give different numbers of IR-active absorptions, infrared spectra
should be able to distinguish between them. The reference provides detailed IR data.
4.32
a.
As
One way to deduce the number of Raman-active vibrations of AsP3 is to first
P
determine the symmetries of all the degrees of freedom. This complex exhibits P
P
C3 symmetry, with the C3 axis emerging from the As atom. The (E) is 12; the
x, y, and z axes of the four atoms do not shift when the identity operation is carried out.
Only the As atom contributes to the character of the C3 transformation matrix; the P
atoms shift during rotation about the C3 axis. The general transformation matrix for
rotation about the z axis (Section 4.3.3) affords 0 as (C3) for The As atom and
one P atom do not shift when a v reflection is carried out, and (v) = 2 (see the v(xz)
transformation matrix in Section 4.33 for the nitrogen atom of NH3 as a model of how the
two unshifted atoms of AsP3 will contribute to the character of the v transformation
matrix).
C3v

A1
A2
E
E
12
1
1
2
C3
0
1
1
–1
v
2
1
–1
0
z
Rx
(x, y), (Rx, Ry)
x2 + y2, z2
(x2 – y2, xy), (xz, yz)
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
47
Reduction of the reducible representation affords 3 A1 + A2 + 4 E.
On the basis of the character table, the translational modes of AsP3 have the
symmetries A1 + E, and the rotational modes have the symmetries A2 + E. The six
vibrational modes of AsP3 subsequently have the symmetries 2 A1 + 2 E. These
vibrational modes are Raman-active, and four absorptions are expected (and observed)
since the sets of E modes are degenerate. B. M. Cossairt, M.C. Diawara, C. C. Cummins,
Science 2009, 323, 602 assigns the bands as: 313 (a1), 345 (e), 428 (a1), 557 (e) cm-1.
Alternatively, the set of six bonds may be selected as the basis for a representation
focused specifically on stretches of these bonds. This approach generates the following
representation:
C3v

E
6
C3
0
v
2
This representation reduces to 2 A1 + 2 E, the same result as obtained by first considering
all degrees of freedom, then subtracting the translational and rotational modes.
b.
As
As2P2 exhibits C2v symmetry, and (like AsP3) will have six vibrational modes
P
P
(3N – 6). Inspection of the C2v character table indicates that all vibrational
As
modes will be Raman active. Since each irreducible representation has a
dimension of 1, the number of Raman absorptions expected is 6 (that is, there will be no
degenerate vibrational modes).
This prediction can be confirmed via deduction of the symmetries of the vibrational modes.
As in part a, we will first determine the symmetries of all the degrees of freedom. The (E)
of the transformation matrix is 12. In As2P2, the C2 axis does not pass through any atoms,
and all four atoms shift upon rotation; (C2) = 0. Two atoms do not shift upon reflection
through each of the v planes. The contribution to the character of the transformation matrix
for each of these unshifted atoms is 1, and (v(xz)) = (v(yz)) = 2.
C2v

A1
A2
B1
B2
E
12
1
1
1
1
C2
0
1
1
–1
–1
v(xz)
2
1
–1
1
–1
v(yz)
2
1
–1
–1
1
z
Rz
x, Ry
y, Rx
x2, y2, z2
xy
xz
yz
Reduction of the reducible representation affords 4 A1 + 2 A2 + 3 B1 + 3 B2.
On the basis of the above character table, the translational modes of As2P2 have the
symmetries A1 + B1 + B2, and the rotational modes have the symmetries A2 + B1 + B2. The
six anticipated Raman-active vibrational modes of As2P2 subsequently have the
symmetries 3 A1 + A2 + B1 + B2.
(continued on next page)
Copyright © 2014 Pearson Education, Inc.
48
Chapter 4 Symmetry and Group Theory
As in part a, an alternative approach is to select the set of six bonds as the basis for a
representation focused specifically on stretching vibrations. This approach generates the
following representation:
C2v

C2
2
E
6
v(xz)
2
v(yz)
2
This representation reduces to 3 A1 + A2 + B1 + B2, the same result as obtained by first
considering all degrees of freedom, then subtracting the translational and rotational
modes.
c.
Td

A1
A2
E
T1
T2
P
The issue here is whether or not P4 (Td) exhibits 4 Raman-active vibrations as
P
does AsP3. We will first determine the symmetries of all the degrees of
freedom. The (E) of the transformation matrix is 12. Only one P4 atom is
fixed upon rotation about each C3 axis; (C3) = 0. All four atoms are shifted upon
application of the C2 and S4 axes; (C2) = (S4) = 0. Two atoms do not shift upon
reflection through each of the d planes. The contribution to the character of the
transformation matrix for each of these unshifted atoms is 1, and (d) = 2.
E
12
1
1
2
3
3
8C3
0
1
1
–1
0
0
3C2
0
1
1
2
–1
–1
6 S4
0
1
–1
0
1
–1
6d
2
1
–1
0
–1
1
P
P
x2 + y2 + z2
(Rx, Ry, Rz)
(x, y, z)
(2z2 – x2 – y2, x2 – y2)
(xy, xz, yz)
Reduction of the reducible representation affords A1 + E + T1 + 2 T2. Since the
symmetries of the translational modes and rotational modes are T2 and T1, respectively,
the symmetries of the vibrational modes are A1 + E + T2, all of these modes are Ramanactive so three Raman absorptions are expected for P4, and P4 could potentially be
distinguished from AsP3 solely on the basis of the number of Raman absorptions.
The alternative approach, using the set of six P–P bonds as the basis for a representation
focused specifically on bond stretches, generates the following representation:
Td

E
6
8C3
0
3C2
2
6 S4
0
6d
2
This representation reduces to A1 + E + T2, the same result as obtained by first
considering all degrees of freedom, then subtracting the translational and rotational
modes.
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
4.33
49
The possible isomers are as follows, with the triphenylphosphine ligand in either the axial (A) or
equatorial (B) sites.
Ph3P
OC
CO
CO
Fe
CO
Fe
OC
CO
PPh3
CO
(A)
CO
C3v
(B)
C2v
Note that the triphenylphosphine ligand is approximated as a simple L ligand for the sake of the
point group determination. Rotation about the Fe–P bond in solution is expected to render the
arrangement of the phenyl rings unimportant in approximating the symmetry of these isomers in
solution. The impact of the phenyl rings would likely be manifest in the IR ν(CO) spectra of these
isomers in the solid-state.
For A, we consider each CO bond as a vector to deduce the expected number of carbonyl
stretching modes. The irreducible representation is as follows:
C3v

A1
A2
E
E
4
1
1
2
C3
1
1
1
–1
v
2
1
–1
0
z
Rx
(x, y), (Rx, Ry)
x2 + y2, z2
(x2 – y2, xy), (xz, yz)
Reduction of the reducible representation affords 2 A1 + E. These stretching modes are IR-active
and three ν(CO) absorptions are expected for A.
For B, a similar analysis affords the following irreducible representation:
C2v

A1
A2
B1
B2
E
4
1
1
1
1
C2
0
1
1
–1
–1
v(xz)
2
1
–1
1
–1
v(yz)
2
1
–1
–1
1
z
Rz
x, Ry
y, Rx
x2, y2, z2
xy
xz
yz
Reduction of the reducible representation affords 2 A1 + B1 + B2. These stretching modes are IRactive, and four ν(CO) absorptions are expected for A.
The reported ν(CO) IR spectrum is consistent with formation of isomer A, with the
triphenylphosphine ligand in the axial site.
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50
Chapter 4 Symmetry and Group Theory
4.34
As in 4.33, we consider the triphenylphosphine ligand as a simple L group for point group
determination. The point groups for isomers A, B, and C are as follows:
PPh3
CO
OC
PPh3
Fe
PPh3
CO
Fe
OC
CO
PPh3
(B)
C2v
PPh3
CO
PPh3
CO
(A)
Fe
OC
CO
D3h
(C) Cs
For A, the set of irreducible representations for the three CO stretching vibrational modes is 2 A1
+ B1. These modes are all IR-active in the C2v character table, and three ν(CO) IR absorptions are
expected for isomer A.
For B, the set of irreducible representations for the three CO stretching vibrational modes is A1 +
E. Only the E mode is IR-active in the D3h point group, and one ν(CO) IR absorption is
expected for isomer B.
For C, the set of irreducible representations for the three CO stretching vibrational modes is
2 A These modes are all IR-active in the Cs point group, and three ν(CO) IR absorptions
are expected for isomer C.
The single ν(CO) IR absorption reported for Fe(CO)3(PPh3)2 supports the presence of the D3h
isomer B.
The trans isomer B is reported in R. L. Keiter, E. A. Keiter, K. H. Hecker, C. A. Boecker,
Organometallics, 1988, 7, 2466, and the authors observe splitting of the absorption associated
with the Emode in CHCl3. The forbidden A1 stretching mode was observed as a weak
absorption.
4.35
The IR spectrum exhibits two ν(CO) absorptions. The two proposed metal carbonyl fragments
will be considered independently for analysis.
CO
OC
Ti
CO
OC
OC
Ti
OC
CO
CO
C4v
D4h
The reducible representation for the four vectors associated with the CO bonds in the C4v
fragment is as follows:
C4v

E
4
2C4
0
C2
0
2v
2
2d
0
Reduction of this representation affords A1 + B1 + E. The A1 and E modes are IR-active, and a
titanium complex with a square pyramidal titanium tetracarbonyl fragment is supported by the IR
spectral data.
Copyright © 2014 Pearson Education, Inc.
Chapter 4 Symmetry and Group Theory
51
For the D4h fragment, the reducible representation for the set of vectors associated with the CO
bonds is:
D4h

E
4
C2
0
2C4
0
2C2
2
i
0
2C2
0
2v
2
h
2S4
0
4
2d
0
Reduction of the reducible representation affords A1g + B1g + Eu. Only the E modes are IR-active,
and a complex with a square planar titanium tetracarbonyl fragment is expected to exhibit a single
IR ν(CO) stretching absorption. This D4h possibility can therefore be ruled out on the basis of the
spectrum.
4.36
(OPh)3
P
A reasonable product is the C4v molecule Mo(CO)5(P(OPh)3), with CO
replacing a triphenylphosphite ligand.
OC
The reducible representation for the five vectors associated with the CO bonds
in this molecule is:
C4v

E
5
C2
1
2C4
1
2v
3
OC
Mo
CO
CO
C
O
2d
1
Reduction of this representation affords 2 A1 + B1 + E. The A1 and E modes are IR-active, and
three IR ν(CO) stretching absorptions are expected. The reported IR spectrum features three
strong ν(CO) absorptions, and one “very weak” absorption attributed to the forbidden B1 mode in
D. J. Darensbourg, T. L. Brown, Inorg. Chem., 1968, 7, 959.
4.37
I has C2 symmetry, with a C2 axis running right to left, perpendicular to the Cl, N, Cl, N and
Cl, P, Cl, P faces.
II also has C2 symmetry, with the same C2 axis as I. (Lower left corner occupied by Cl, not C.)
III has only an inversion center and Ci symmetry.
4.38
An example for each of the five possible point groups:
F
Td :
C
F
F
4.39
H
C3v:
F
C
F
F
H
C2v:
F
C
F
F
F
Cs :
H
Br
Br
F
C1 :
C
H
Br
Cl
C
H
a.
The S–C–C portion is linear, so the molecule has a C3 axis along the line of these three
atoms, three v planes through these atoms and an F atom on each end, but no
other symmetry elements. C3v
b.
The molecule has only an inversion center, so it is Ci. The inversion center is equivalent
to an S2 axis perpendicular to the average plane of the ring.
c.
M2Cl6Br4 is Ci.
d.
This complex has a C3 axis, splitting the three N atoms and the three P atoms (almost as
drawn), but no other symmetry elements. C3
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52
Chapter 4 Symmetry and Group Theory
F
e.
The most likely isomer has the less electronegative Cl atoms in
equatorial positions. Point group: C2v
Cl
Cl
P
F
4.40
The structures on the top row are D2d (left) and Cs (right). Those on the
bottom row are C2h (left) and C4v (right).
4.41
a.
C3v
b.
D5h
c.
Square structure: D2d (bottom ligand on lower left Re should be CO instead of L);
Corner structure: Cs
d.
D3d
4.42
F
Web of Science and SciFinder Scholar should be helpful, but simply searching for these
symmetries using a general Internet search should provide examples of these point groups. Some
examples:
a.
S6 :
Mo2(SC6H2Me3)6 (M. H. Chisholm, J. F. Corning, and J. C. Huffman,
J. Am. Chem. Soc., 1983, 105, 5924)
Mo2(NMe2)6 (M. H. Chisholm, R. A. Cotton, B. A. Grenz, W. W. Reichert,
L. W. Shive, and B. R. Stults, J. Am. Chem. Soc., 1976, 98, 4469)
[NaFe6(OMe)12(dbm)6]+ (dbm = dibenzoylmethane, C6H5COCCOC6H5) (F.
L. Abbati, A. Cornia, A. C. Fabretti, A. Caneschi, and D. Garreschi, Inorg.
Chem., 1998, 37, 1430)
b.
T
Pt(CF3)4, C44
c.
Ih
C20, C80
d.
Th
[Co(NO2)6]3–, Mo(NMe2)6
In addition to examples that can be found using Web of Science, SciFinder, and other Internet
search tools, numerous examples of these and other point groups can be found in I. Hargittai and
M. Hargittai, Symmetry Through the Eyes of a Chemist, as listed in the General References
section.
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
53
CHAPTER 5: MOLECULAR ORBITALS
5.1
There are three possible bonding interactions:
pz
5.2
dz2
py
dyz
px
dxz
a.
Li2 has a bond order of 1.0 (two electrons in a  bonding orbital; see
Figures 5.7 and 5.1). Li2+ has a bond order of only 0.5 (one electron in a
 bonding orbital). Therefore, Li2 has the shorter bond.
b.
F2 has a bond order of 1.0 (see Figure 5.7). F2+ has one less antibonding (π*)
electron and a higher bond order, 1.5. F2+ would be expected to have the shorter bond.
c.
Expected bond orders (see Figure 5.1):
He2+
Bonding electrons
2
Antibonding electrons
1
HHe+
2
0
+
1
0
H2
Bond order
1
(2 – 1) = 0.5
2
1
(2 – 0) = 1
2
1
(2 – 1) = 0.5
2
Both He2+ and H2+ have bond orders of 0.5. HHe+ would therefore be expected to have
the shortest bond because it has a bond order of 1.
5.3
a.
These diatomic molecules should have similar bond orders to the analogous
diatomics from the row directly above them in the periodic table:
P2
S2
Cl2
b.
Cl2 has the weakest bond.
The bond orders match those of the analogous oxygen species (Section 5.2.3):
S2 +
S2
S2 –
c.
bond order = 3 (like N2)
bond order = 2 (like O2)
bond order = 1 (like F2)
bond order = 2.5
bond order = 2
bond order = 1.5
S2– has the weakest bond.
Bond orders:
NO+
NO
NO–
bond order = 3 (isoelectronic with CO, Figure 5.13)
bond order = 2.5 (one more (antibonding) electron than CO)
bond order = 2 (two more (antibonding) electrons than CO)
NO– has the lowest bond order and therefore the weakest bond.
Copyright © 2014 Pearson Education, Inc.
54
Chapter 5 Molecular Orbitals
5.4
O22– has a single bond, with four
electrons in the π* orbitals canceling
those in the π orbitals.
–
O2 has three electrons in the π* orbitals, and
a bond order of 1.5. The Lewis structures
have an unpaired electron and an average
bond order of 1.5.
O2 has two unpaired electrons in its π*
orbitals, and a bond order of 2. The simple
Lewis structure has all electrons paired,
which does not match the paramagnetism
observed experimentally.
Bond lengths are therefore in the order
–
O22– > O2 > O2, and bond strengths are
the reverse of this order.
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
5.5
Bond Order
(Figures 5.5 and 5.7)
C22–
N22–
O22–
O2
Bond Distance (pm)
3
2
1
2
55
Unpaired Electrons
119
122.4
149 (very long)
120.7
0
2
0
2
The bond distance in N22– is very close to the expected bond distance for a diatomic with 12
valence electrons, as shown in Figure 5.8.
5.6
The energy level pattern would be similar to the one shown in Figure 5.5, with the interacting
orbitals the 3s and 3p rather than 2s and 2p. All molecular orbitals except the highest would be
occupied by electron pairs, and the highest orbital (u*) would be singly occupied, giving a bond
order of 0.5. Because the bond in Ar2+ would be weaker than in Cl2, the Ar–Ar distance would be
expected to be longer (calculated to be > 300 pm; see the reference).
5.7
a.
The energy level diagram for NO is on
the right. The odd electron is in a π2p*
orbital.
 2p 
 2p 
b.
c.
d.
O is more electronegative than N, so
its orbitals are slightly lower in energy.
The bonding orbitals are slightly more
concentrated on O.
The bond order is 2.5, with one unpaired
electron.
NO+
NO
–
NO
5.8
2p
2p
2s
2s
Bond order = 3
shortest bond (106 pm)
 2s
Bond order = 2.5
N
NO
intermediate (115 pm)
Bond order = 2
longest bond (127 pm), two electrons in antibonding orbitals.
–
2p
2p
2s
O
a.
The CN energy level diagram is similar to that of NO (Problem 5.7) without the
antibonding π* electron.
b.
The bond order is three, with no unpaired electrons.
c.
The HOMO is the 2p orbital, which can interact
with the 1s of the H+, as in the diagram at right.
The bonding orbital has an energy near that of the
π orbitals; the antibonding orbital becomes the
highest energy orbital.
Copyright © 2014 Pearson Education, Inc.
56
5.9
5.10
Chapter 5 Molecular Orbitals
a.
A diagram is sketched at the right. Since the difference in valence orbital potential energy
between the 2s of N (-25.56 eV) and the 2p of F (-18.65 eV) is 6.91 eV, the 2p orbital is
expected to be higher in energy
2p*
relative to the degenerate 2p set.
b.
NF is isoelectronic (has the same
number of valence electrons) with
O2. Therefore, NF is predicted to
be paramagnetic with a bond
order of 2. The populations of the
bonding (8 electrons) and
antibonding (4 electrons)
molecular orbitals in the diagram
suggest a double bond.
2p
2p*
2p
2p
2p
2s
2s*
2s
c.
The 2s, 2s*, 2p, and 2p* orbitals
2s
exhibit Cv symmetry, with the
NF
NF bond axis the infinite-fold
N
F
rotation axis. The 2p and 2p*
orbitals exhibit Cs symmetry. The latter do not possess C2 rotation axes coincident to the
infinite-fold rotation axis of the  orbitals on the basis of the change in wave function
sign upon crossing the nodes on the bond axis.
a.
OF– has 14 valence electrons, four in the π2p* orbitals (see the diagram in the answer to
Problem 5.9).
b.
The net result is a single bond between two very electronegative atoms, and no unpaired
electrons.
c.
The concentration of electrons in the π* orbital is more on the O, so combination with
the positive proton at that end is more likely. In fact, H+ bonds to the oxygen atom, at
an angle of 97°, as if the bonding were through a p orbital on O.
5.11
The molecular orbital description of KrF+ would predict that this ion, which has the same number
of valence electrons as F2, would have a single bond. KrF2 would also be expected, on the basis
of the VSEPR approach, to have single Kr–F bonds, in addition to three lone pairs on Kr.
Reported Kr–F distances: KrF+: 176.5-178.3 pm; KrF2: 186.8-188.9 pm. The presence of lone
pairs in KrF2 may account for the longer bond distances in this compound.
5.12
a.

The KrBr+ energy level diagram is at the right.
b.
The HOMO is polarized toward Br, since its
energy is closer to that of the Br 4p orbital.
c.
Bond order = 1
d.
Kr is more electronegative. Its greater
nuclear charge exerts a stronger pull on
the shared electrons.

HOMO
4p
4p



4s
4s

Kr
Copyright © 2014 Pearson Education, Inc.
KrBr+
Br
Chapter 5 Molecular Orbitals
5.13
57
The energy level diagram for SH– is shown below. A bond order of 1 is predicted.
The S orbital energies are –22.7 eV (3s) and –11.6 eV (3p); the 1s of H has an energy of –13.6
eV. Because of the difference in their atomic orbital energies, the 1s orbital of hydrogen and the
3s orbital of sulfur interact only weakly; this is shown in the diagram by a slight stabilization of
the lowest energy molecular orbital with respect to the 3s orbital of sulfur. This lowest energy
orbital is essentially nonbonding. These orbitals are similar in appearance to those of HF in
Example 5.3, with more balanced contribution of the hydrogen 1s and sulfur valence orbitals
since the valence orbitals of sulfur are closer to the energy of the hydrogen 1s orbital than the
valence orbitals of fluorine.

3p
1s

3s

H
5.14
a.
SH–
S
The group orbitals on the
hydrogen atoms are

2p
and
1s
The first group orbital interacts
with the 2s orbital on carbon:
2s

And the second group orbital
interacts with a 2p orbital on
carbon:
Carbon’s remaining 2p orbitals are
nonbonding.
b.

C H C H
H H
Linear CH2 is a paramagnetic diradical, with one electron in each of the px and py orbitals
of carbon. (A bent singlet state, with all electrons paired, is also known, with a calculated
bond angle of approximately 130°.)
Copyright © 2014 Pearson Education, Inc.
58
Chapter 5 Molecular Orbitals
H
5.15
a.
BeH2
Be
H
H
Be
H
Group Orbitals:
Be Orbitals with
Matching Symmetry:
MO Diagram:
b.
5.16
2s
2pz
The energy level diagrams for CH2 and BeH2 feature the same orbital interactions. One
difference is that the different number of valence electrons renders the linear species
BeH2 diamagnetic and CH2 paramagnetic. The energy difference between the Be and H
valence orbitals is larger than that between the valence orbitals of C and H, and both the
2s and 2p orbitals of Be are higher in energy than the 1s orbital of H. The result is
greater bond polarity in BeH2.
BeF2 uses s and p orbitals on all three atoms, and is isoelectronic with CO2. The energy level
diagram for CO2 in Figure 5.25 can be used as a guide, with the orbitals of Be higher in energy
than those of C and the orbitals of F lower in energy than those of O. Calculated molecular
orbital shapes are below, for comparison for those of CO2 in Figure 5.25.
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
4b1u
4ag
LUMO (2)
2b2u, 2b3u
HOMO (2)
1b2g, 1b3g
Bonding (2)
1b2u, 1b3u
3b1u Nonbonding
3ag, Bonding
2b1u
2ag
Of the occupied orbitals, there are three bonding (two π and one ) and five nonbonding
(two π and three ). (References: W. R. Wadt, W. A. Goddard III, J. Am. Chem. Soc.,
1974, 96, 5996; R. Gleiter, R. Hoffmann, J. Am. Chem. Soc., 1968, 90, 5457; C. W.
Bauschlicher, Jr., I. Shavitt, J. Am. Chem. Soc., 1978, 100, 739.)
Copyright © 2014 Pearson Education, Inc.
59
60
5.17
Chapter 5 Molecular Orbitals
a.
The group orbitals of the
fluorines are:
x
F
b.
Xe
F
z
F
The matching orbitals on
xenon are:
F
F
F
pz
s, dz2
pz
s, dz2
dxz
px
dyz
py
F
F
Xe
F
F
Xe
Xe
F
F
Xe
F
H
5.18
The point group of TaH5 is C4v.
1.
H
H
Ta
H
H
z
2. Axes can be assigned as shown:
Ta
y
x
3. Construction of reducible representation:
C4v

A1
A2
B1
B2
E
E
5
1
1
1
1
2
2C4
1
1
1
–1
–1
0
C2
1
1
1
1
1
–2
2v
3
1
–1
1
–1
0
2d
1
1
–1
–1
1
0
z
Rz
(x, y), (Rx, Ry)
Copyright © 2014 Pearson Education, Inc.
z2
x2–y2
xy
(xz, yz)
F
Chapter 5 Molecular Orbitals
 reduces to 2 A1 + B1 + E
5, 6. Two group orbitals, shown at right, have A1
symmetry. These may interact with the pz, dz2,
and s orbitals of Ta.
Ta
Ta
One group orbital has B1 symmetry. It
can interact with the dx2–y2 orbital of Ta.
A degenerate pair of group orbitals has
E symmetry. It may interact with the
(px, py) and (dxz, dyz) pairs of Ta.
5.19
61
Ta
Ta
Ta
The energy level diagram for O3 with the simple combinations of s and p orbitals is shown below.
Mixing of s and p orbitals is fairly small, showing mostly in the four lowest orbitals. The order of
orbitals may vary depending on the calculation method (for example, PM3 and AM1 methods
reverse the orders of HOMO and HOMO –1).
12
11
10
9
8
7
6
5
4
3
2
1
O
O
O
O
O
Copyright © 2014 Pearson Education, Inc.
O
62
5.20
Chapter 5 Molecular Orbitals
SO3 has molecular orbitals similar to those of BF3 (Section 5.4.6). The irreducible
representations below are labeled for the oxygen orbitals.
D3h
 (s)
 (py)
 (px) 
 (pz)
A1
A2
A2 
E
E

5.21


π

E
3
3
3
3
1
1
1
2
2
2C3
0
0
0
0
1
1
1
–1
–1
 (s) = A1 + E
 (py) = A1 + E
 (px) = A2 + E
 (pz) = A2 + E
3C2
1
1
–1
–1
1
–1
–1
0
0
h
3
3
3
–3
1
1
–1
2
–2
2S3
0
0
0
0
1
1
–1
–1
1
3v
1
1
–1
1
1
–1
1
0
0
z
(x, y)
Sulfur s, px, and py
Sulfur s, px, and py
Sulfur px and py
Sulfur pz
As a cyclic (triangular) ion, H3+ has a pair of electrons in a bonding orbital and two vacant
orbitals that are slightly antibonding:
H
H
+
E'
H
A 1'
H3 +
5.22
–
The thiocyanate ion, SCN , has molecular orbitals similar to those of CO2, but
with more mixing between the s orbital of C and the s and p orbitals of S. The
valence orbital potential energies of S are very close to those of C, and those of
N are only slightly lower. There is significant double bonding in thiocyanate
on the basis of this excellent orbital energy compatibility. This is consistent
with the resonance structures shown at right, with the top structure favored.
1–
S
C
N
1–
S
C
1+
S
N
2–
C
–
N
For cyanate, OCN , the s and p orbitals of carbon effectively interact with valence orbitals on the
N side, but less on the O side because the oxygen orbital energies are much more negative. The
structures described in Section 3.1.3 (a mix of two double bonds and O–C and CN) fit this ion
also.
–
For fulminate, CNO , the large differences between the C and O orbital energies render the
contributions of the terminal atom orbitals to the group orbitals relatively uneven. A practical
result of this imbalance is less delocalization of electron density within this anion. This is
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
5.23
63
suggested by the formal charge effects described in Example 3.3. As a result, the bonding in this
ion is weak. Fulminate is stable only when complexed with a metal ion.
–
–
The highest occupied orbitals in SCN or OCN are π nonbonding orbitals (see Figure
5.25 for the similar CO2 orbitals). Combination with H+ or with metal ions depends on the energy
match of these orbitals with those of the positive ion. The preference for attack can be examined
by comparing the energies of the valence orbitals of the terminal atoms that contribute to these
nonbonding molecular orbitals. For example, the H 1s orbital energy matches the energy of the N
valence orbitals orbital better than either the S or O valence orbitals, and the nitrogen atom is the
site of protonation in both anions. The energies of metal ion valence orbitals vary from element to
element (and with oxidation state). Some metal ions will be more compatible energetically with
the S orbitals while others will be more compatible with the N orbitals. These electronic effects
contribute to which site of thiocyanate is appropriate for bonding to metals. The S can also use
the empty 3d orbitals to accept electron density via π bonding from some metal ions.
–
5.24
The CN molecular orbitals are similar to those of CO (Figure 5.13), but with less difference
between the C and N atomic orbital energies than between C and O orbitals. As a result, the
HOMO should be more evenly balanced between the two atoms, and bonding at both ends seems
–
more likely with CN relative CO. The Prussian blue structures (Fe4[Fe(CN)6]3 or KFe(Fe(CN)6)
–
have iron and CN in arrangements that have both Fe–C and Fe–N bonds.
5.25
a.
The resonance structures were considered in Problem 3.3, showing bent structures
with primarily double bond character in S=N and single bonding in N–O or S–O.
–
SNO is more stable on the basis of formal charges.
b.
The molecular orbitals should be similar to those of O3, with more mixing of s and p
orbitals because of the difference between atomic orbital energies of S and O as
terminal atoms. The π bonding, nonbonding, and antibonding orbitals are numbers 6, 9,
and 10 in the ozone diagram in the Problem 5.19 answer.
The relative contributions of the valence orbitals of each atom to the π molecular orbitals
–
of SNO can be partially rationalized on the basis of electronegativity. In the π bonding
orbital, the electron density is highest on the most electronegative O atom, a reasonable
expectation for the most stabilized π interaction. In the π antibonding orbital, the electron
density is highest on the least electronegative S atom, a feature that contributes to the
–
destabilization of this orbital. The π molecular orbitals of NSO are not as clearly
explained by these electronegativity arguments, possibly due to the ability of S to expand
its valence shell to increase its participation in bonding as a central atom.
Copyright © 2014 Pearson Education, Inc.
64
Chapter 5 Molecular Orbitals
c.
π
–
S—N—O
πn
–
S—N—O
S—N—O
π
–
N—S—O
πn
–
N—S—O
π*
–
N—S—O
π*
–
The calculated bond distances for these ions are:
ion
N–S
S–O
N–O
–
171 pm
120 pm
SNO
–
146 pm
149 pm
NSO
–
NSO has the shorter N–S bond and the higher energy N–S stretching vibration.
5.26
H2O has C2v symmetry. Figure 5.26 defines the coordinate system to be used. The representation
that describes the symmetry of the group orbitals (Section 5.4.3) is A1  B1 . The next step is to
track the fate of one of the hydrogen 1s orbitals as the symmetry operations are carried out:
Original Orbital
E
C2
 v( xz)
 v(
 yz)
H1s(a) becomes….
H1s(a)
H1s(b)
H1s(a)
H1s(b)
H1s(b)
H1s(a)
Now we multiply these outcomes by the characters associated with
each operation for the A1 and B1 representations, and then add the results to obtain the linear
combinations of the H 1s atomic orbitals that define the group orbitals.
 v( xz)
C2
E
 v(
 yz)
SALCs
A1
H1s(a)
+
H1s(b)
+
H1s(a)
+
H1s(b)
 2( H1s(a) )  2(H1s(b) )
B1
H1s(a)
–
H1s(b)
+
H1s(a) –
H1s(b)
 2( H1s(a) )  2(H1s(b) )
Each group orbital equation must be normalized, so that the sum of the squares of the coefficients
1
within each equation equals 1. The normalization factors, N   (ca2  cb2 ) , where ca and cb


are the lowest common integer coefficients for the hydrogen 1s orbital wave functions in each
SALC, are:
A1 : N   (1) 2  (1) 2 


1

1
2
B1 : N   (1) 2  (1) 2 )


1

This results in the normalized SALC equations for the two group orbitals:
1
1
 (H a )  ( H b )  B1 :
 (H a )  (H b )  .
A1 :
2
2
Copyright © 2014 Pearson Education, Inc.
1 .
2
Chapter 5 Molecular Orbitals
65
SALC Coefficients and Evidence of Normalization:
Squares of SALC
Coefficients
Coefficients in Normalized SALCs
A1
ca
cb
ca2
cb2
1
1
2
1
2
1
1
2
1
2
1
2
1
2
1
1

2
2
Sum of the squares for each 1s wavefunction
must total 1 for an identical contribution of
each atomic orbital to the group orbitals
B1
5.27
Sum of Squares =1
for Normalization
1
1
b
The irreducible representation associated with the 2s set is   A1  E  . The
atoms will be labeled as shown. The next step is to track the fate of one of the
fluorine 2s orbitals as the symmetry operations are carried out:
a
c
Original Orbital
E
C3
C32
C2(a)
C2(b)
C2(c)
h
S3
S32
 v(a)
 v(b)
 v(c)
2sa becomes…
2sa
2sb
2sc
2sa
2sc
2sb
2sa
2sb
2sc
2sa
2sc
2sb
Now we multiply these outcomes by the characters associated with each operation for A1 in the
D3h character table to obtain the linear combination. All of the characters are 1 for the totally
symmetric A1 irreducible representation. Therefore:
A1 : 2sa  2sb  2sc  2sa  2sc  2sb  2sa  2sb  2sc  2sa  2sc  2sb  4(2sa )  4(2sb )  4(2sc )
The lowest common integer coefficient is 1, and N   (1) 2  (1) 2  (1)2 


normalized A1 SALC is
1

1
3
. The
1
 (2sa )  (2sb )  (2sc )  . Next we multiply the terms in the
3
table above by the characters of the E  irreducible representation. The characters for C2 and  v
are 0, so multiplication by these characters leads to no contribution to the SALC:
E  : 2(2s )  2s  2s  2(2s )  2s  2s  4(2s )  2(2s )  2(2s ) .
a
b
c
a
b
c
a
b
c
Reduction to the lowest common integer coefficient affords 2(2sa )  (2sb )  (2sc ) , and
N   (2) 2  (1) 2  (1)2 


1

1
6
. The normalized E  SALC is therefore
1
 2((2sa )  (2sb )  (2 pc )  .
6
The remaining E  SALC can be deduced by examination of the coefficients and the sums of their
squares.
Copyright © 2014 Pearson Education, Inc.
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Chapter 5 Molecular Orbitals
Coefficients of Normalized SALCs
A1
E
Squares of SALC Coefficients
ca
cb
cc
ca2
cb2
cc2
1
1
1
3
2
3
1
3
1
1
3
2
3
1
3
1
6
1
2
1
3
1
6
1
2
1
1
6

6

1
6
1

2
2
Sum of the squares for each 2s wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
E
0
Sum of the
Squares = 1
for
Normalization
0
1
1
1
1
Since the sum of the squares of the coefficients for 2sa equals 1 without any contribution from
the second E  SALC, this SALC must have ca  0 . The squares of cb and cc must equal
1
2
to
satisfy the normalization and identical contribution from each orbital requirements. Since E 
matches the symmetry of the x and y axes, and the origin is the center of the group orbital (the
boron atom), one of the coefficients must be positive and the other negative. The second E 
1
 (2sb )  (2sc )  .
SALC is therefore
2
b
The irreducible representation associated with the 2 p y set is also   A1  E  .
The atoms are labeled as shown. The fate of one of the fluorine 2 p y orbitals as the
symmetry operations are carried out is:
E
2 p y(a)
2 p y(a)
C3
C32
C2(a)
C2(b)
c
C2(c)
h
S3
S32
a
 v(a)  v(b)  v(c)
2 p y(b) 2 p y(c) 2 p y(a) 2 p y(c) 2 p y(b) 2 p y(a) 2 p y(b) 2 p y(c) 2 p y(a) 2 p y(c) 2 p y(b)
becomes…
Now we multiply these outcomes by the characters associated with each operation for A1 in the
D3h character table. All of the characters are 1 for the A1 irreducible representation. Therefore:
A1  : 2 p y ( a )  2 p y (b)  2 p y (c )  2 p y ( a )  2 p y ( c )  2 p y (b )  2 p y ( a )  2 p y (b)  2 p y ( c )  2 p y ( a )  2 p y ( c )  2 p y (b)
 4(2 p y ( a ) )  4(2 p y (b) )  4(2 p y (c ) ).
The lowest common integer coefficient is 1, and N   (1) 2  (1) 2  (1)2 


normalized A1 SALC is therefore
1

1 
(2 p y(a) )  (2 p y(b) )  (2 p y(c) )  .

3
Copyright © 2014 Pearson Education, Inc.
1
3
. The
Chapter 5 Molecular Orbitals
67
Next we multiply the terms in the table above by the characters of the E  irreducible
representation. The characters for C2 and  v are 0, so multiplication by these characters leads to
no contribution to the SALC:
E  : 2(2 p y(a) )  2 p y(b)  2 p y(c)  2(2 p y(a) )  2 p y(b)  2 p y(c)  4(2 p y(a) )  2(2 p y(b) )  2(2 p y(c) ) .
Reduction to the lowest common integer coefficient affords 2(2 p y(a) )  (2 p y(b) )  (2 p y(c) ) , and
N   (2) 2  (1) 2  (1)2 


1

1
6
. The normalized E  SALC is
1 
2((2 p y(a) )  (2 p y(b) )  (2 p y(c) )  .


6
The remaining E  SALC can be deduced by examination of the coefficients and the sums of their
squares.
Coefficients of Normalized SALCs
A1
E
Squares of SALC Coefficients
ca
cb
cc
ca2
cb2
cc2
1
1
1
3
2
3
1
3
1
1
3
1
3
1
3
1
2
3
1
6
1
6
1
0
1
2
1
2
1
1
1
1
6
E
Sum of the
Squares = 1
for
Normalization
0

6
1


6
1
2
2
Sum of the squares for each 2py wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
Since the sum of the squares of the coefficients for 2 p y(a) equals 1 without any contribution
from the second E  SALC, this SALC must have ca  0 . The squares of cb and cc must equal
1
2
to satisfy the normalization and identical contribution from each orbital requirements. Since
E  matches the symmetry of the x and y axes, and the origin is the center of the group orbital (the
boron atom), one of the coefficients must be positive and the other negative. The second E 
1 
(2 p y(b) )  (2 p y(c) )  .
SALC is

2
The irreducible representation associated with the 2 px set is   A2  E  . The
b
atoms are labeled as shown. The fate of one of the fluorine 2 px orbitals as the
symmetry operations are carried out is:
c
Copyright © 2014 Pearson Education, Inc.
a
68
Chapter 5 Molecular Orbitals
2 px(a)
becomes…
E
C3
C32
C2(a)
C2(b)
C2(c)
h
S3
S32
 v(a)  v(b)
 v(c)
2 px(a)
2 px(b)
2 px(c)
2 px(a)
2 px(c)
2 px(b)
2 px(a)
2 px(b)
2 px(c)
2 px(a)
2 px(b)
2 px(c)
Now we multiply these outcomes by the characters associated with each operation for A2 in the
D3h character table:
A2 : 2 px(a)  2 px(b)  2 px(c)  2 px(a)  2 px(c)  2 px(b)  2 px(a)  2 px(b)  2 px(c)  2 px(a)  2 p y(c)  2 px(b)
 4(2 px(a) )  4(2 px(b) )  4(2 px(c) )
The lowest common integer coefficient is 1, and N   (1) 2  (1) 2  (1)2 


normalized A2 SALC is
1

1
3
. The
1 
(2 px(a) )  (2 px(b) )  (2 px(c) )  .

3
Next we multiply the terms in the table by the characters of the E  irreducible representation.
E  : 2(2 px(a) )  2 px(b)  2 px(c)  2(2 p y(a) )  2 px(b)  2 px(c)  4(2 px(a) )  2(2 px(b) )  2(2 px(c) )
Reduction to the lowest common integer coefficient affords 2(2 px(a) )  (2 px(b) )  (2 px(c) ) , and
N   (2) 2  (1) 2  (1)2 


1

1
6
. The normalized E  SALC is
1 
2((2 px(a) )  (2 px(b) )  (2 px(c) )  .

6
The remaining E  SALC can be deduced by examination of the coefficients and the sums of their
squares.
Coefficients of Normalized SALCs
A2
E
E
Squares of SALC Coefficients
Sum of the
Squares = 1
for
Normalization
Requirement
ca
cb
cc
ca2
cb2
cc2
1
1
1
3
2
3
1
3
1
1
3
1
3
1
3
1
2
3
1
6
1
6
1
0
1
2
1
2
1
1
1
1
6
0

6
1


6
1
2
2
Sum of the squares for each 2px wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
69
Since the sum of the squares of the coefficients for 2 px(a) equals 1 without any contribution
from the second E  SALC, this SALC must have ca  0 . The squares of cb and cc must equal
1
2
to satisfy the normalization and identical contribution from each orbital requirements. Because
E  matches the symmetry of the x and y axes, and the origin is the center of the group orbital
(the boron atom), one of the coefficients must be positive and the other negative. The second E 
1 
(2 px(b) )  (2 px(c) )  .
SALC is

2
The irreducible representation associated with the 2 pz set is
b
  A2 E  . The atoms are labeled as shown. The fate of one of the
fluorine 2 pz orbitals as the symmetry operations are carried out is:
2 pz(a)
becomes…
a
c
E
C3
C32
C2(a)
C2(b)
C2(c)
h
S3
S32
 v(a)  v(b)  v(c)
2 pz(a)
2 pz(b)
2 pz(c)
2 pz(a)
2 pz(c)
2 pz(b)
2 pz(a)
2 pz(b)
2 pz(c)
2 pz(a)
2 pz(c)
2 pz(b)
Now we multiply these outcomes by the characters associated with each operation for A2 in the
D3h character table:
A2: 2 pz(a)  2 pz(b)  2 pz(c)  2 pz(a)  2 pz(c)  2 pz(b)  2 pz(a)  2 pz(b)  2 pz(c)  2 px(a)  2 pz(c)  2 pz(b)
 4(2 pz(a) )  4(2 pz(b) )  4(2 pz(c) )
The lowest common integer coefficient is 1, and N   (1) 2  (1) 2  (1)2 


normalized A2 SALC is
1

1
3
. The
1 
(2 pz(a) )  (2 pz(b) )  (2 pz(c) )  .

3
Next we multiply the terms in the table by the characters of the E  irreducible representation:
E  : 2(2 pz(a) )  2 pz(b)  2 pz(c)  2(2 pz(a) )  2 pz(b)  2 pz(c)  4(2 pz(a) )  2(2 pz(b) )  2(2 pz(c) ).
Reduction to the lowest common integer coefficient affords 2(2 pz(a) )  (2 pz(b) )  (2 pz(c) ) , and
N   (2) 2  (1) 2  (1)2 


1

1
6
. The normalized E  SALC is
1 
2((2 pz(a) )  (2 pz(b) )  (2 pz(c) )  .

6
The remaining E  SALC can be deduced by examination of the coefficients and the sums of
their squares.
Copyright © 2014 Pearson Education, Inc.
70
Chapter 5 Molecular Orbitals
Coefficients of Normalized SALCs
A2
E 
ca
cb
cc
ca2
cb2
cc2
1
1
1
3
2
3
1
3
1
1
3
1
3
1
3
1
2
3
1
6
1
6
1
0
1
2
1
2
1
1
1
1

6
E 
Squares of SALC Coefficients
Sum of the
Squares = 1
for
Normalization
Requirement

6
1
0

6
1
2
2
Sum of the squares for each 2pz wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
Since the sum of the squares of the coefficients for 2 pz(a) equals 1 without any contribution from
the second E  SALC, this SALC must have ca  0 . The squares of cb and cc must equal
1
2
to
satisfy the normalization and identical contribution from each orbital requirements. Since E  has
the same symmetry as the xz and yz orbitals that have nodes defined by the yz and xz planes,
respectively, one of the coefficients must be positive and the other negative. The second E 
1 
(2 pz(b) )  (2 pz(c) )  .
SALC is

2
5.28
The point group is D4h . The reducible representation that describes the symmetries of the group
orbitals is:
D4h
E
2C4
C2
4
0
0
2C2
2
2C2
0
i
2S 4
h
2 v
2 d
0
0
4
2
0
Reduction affords   A1g  B1g  Eu . To deduce the SALCs, we need to track the fate of one of
the 3s orbitals through each symmetry operation of the character table.
3s( A)
becomes…
E
C4
C43
C2
C2 (x)
C2 ( y)
C2 (1)
C2 (2)
3s( A)
3s(B)
3s(D)
3s(C)
3s(C)
3s( A)
3s(D)
3s(B)
i
S4
h
 v (x)
 v ( y)
 d (1)
 d (2)
3s(C)
3s(B)
S43
3s(D)
3s(D)
3s(B)
3s( A)
3s(C)
Since all of the characters for A1g are 1, the SALC for A1g is:
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3s( A)
Chapter 5 Molecular Orbitals
71
A1g : 3s( A)  3s( B)  3s(D)  3s(C)  3s(C)  3s( A)  3s(D)  3s(B)  3s(C)  3s( B)  3s(D)  3s( A) 
3s(C)  3s( A)  3s( D)  3s( B)  4(3s( A))  4(3s(B))  4(3s(C))  4(3s(D))
This simplifies to 3s( A)  3s(B)  3s(C)  3s(D) , which provides a normalization constant of
1
1
1
and the SALC  (3s( A))  (3s(B))  (3s(C))  (3s(D))  .
2
2
Multiplication of the orbitals in the above table by the B1g characters gives:
N   (1) 2  (1) 2  (1)2  (1)2 



B1g : 3s( A)  3s( B)  3s(D)  3s(C)  3s(C)  3s( A)  3s(D)  3s(B)  3s(C)  3s( B)  3s(D)  3s( A) 
3s(C)  3s( A)  3s( D)  3s( B)  4(3s( A))  4(3s(B))  4(3s(C))  4(3s(D))
This simplifies to 3s( A)  3s(B)  3s(C)  3s(D) , which provides a normalization constant of
N   (1) 2  (1) 2  (1)2  (1)2 


1

1
1
and  (3s( A))  (3s(B))  (3s(C))  (3s(D)) 
2
2
as the B1g normalized SALC.
Multiplication of the orbitals in the above table by the Eu characters affords:
Eu : 2(3s( A))  2(3s(C))  2(3s(C))  2(3s( A))  4(3s( A))  4(3s(C))
This simplifies to 3s ( A)  3s (C ), which provides a normalization constant of
N   (1) 2  (1) 2 


1

1
2
and
1
 (3s( A))  (3s(C))  as one of the normalized Eu
2
SALCs.
The equation for the other Eu SALC can be deduced by consideration of the normalized SALC
coefficients and their squares.
Coefficients of Normalized SALCs
A1g
B1g
Eu
Eu
Squares of SALC Coefficients
cA
cB
cC
cD
c A2
cB2
cC2
cD2
1
2
1
2
1
1
2
1

2
1
2
1
2
1
1
2
1

2
1
4
1
4
1
2
1
4
1
4
1
4
1
4
1
2
1
4
1
4
2
0
0
1

0
2
0

1
2
2
Sum of the squares for each 3s wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
0
1
1
0
1
1
0
1
2
0
1
2
1
1
1
1
Copyright © 2014 Pearson Education, Inc.
Sum of the
Squares = 1 for
Normalization
Requirement
72
Chapter 5 Molecular Orbitals
The sum of c A2 and cC2 equals 1 without any contribution from the second Eu SALC; c A  cC  0 for
this second Eu SALC. A magnitude of
1
is required for both c B and cD of the second Eu SALC on
2
the basis of the normalization requirement and equal contribution of each atomic orbital to the group
orbital’s requirement. The alternate signs are required since Eu has the symmetry of the x and y axes;
these Eu group orbitals need to match the symmetry of the valence p orbitals of the central atom. The
1
 (3s(B))  (3s( D))  .
2
Sketches of these group orbitals are below, using the coordinate system specified in this problem.
Note the scaling of the orbitals to reflect the larger contribution of the 3s orbitals in the Eu
normalized equation for the second Eu SALC is
SALCs compared to that in A1g and B1g SALCs.
A
A
B
B
D
A1g
D
B1g
C
C
A
D
B
Eu
5.29
Eu
C
For this question, we will label the 2 pz orbitals simply by their letters (A–F) for clarity. The first
task is to track the fate of A through all of the D6 h symmetry operations:
A
becomes…
E
C6
C3
B
C65
F
A
i
S3
S32
-D
-C
-E
h
C2 (1)
-A
 d (1)
C2 (2)
-C
 d (2)
C2 (3)
-E
 d (3)
C2 (1)
-B
 v (1)
C2 (2)
-D
 v (2)
C2 (3)
-F
 v (3)
-A
B
D
F
A
C
E
C2
C
C32
E
S6
S65
-B
-F
D
Multiplication by the characters of each irreducible representation of the D6 h character table
(and for each symmetry operation) is a tedious, but effective, method to deduce the SALCs.
A1g : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
A2g : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
73
B1g : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
B2g : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E
 4 A  4B  4C  4D  4E  4F
E1g : 2 A  B  F  C  E  2D  2D  C  E  B  F  2 A  4 A  2B  2C  4D  2E  2F
E2g : 2 A  B  F  C  E  2D  2D  C  E  B  F  2 A  0
A1u : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
A2u : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E
 4 A  4B  4C  4D  4E  4F
B1u : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
B2u : A  B  F  C  E  D  A  C  E  B  D  F  D  C  E  B  F  A  B  D  F  A  C  E  0
E1u : 2 A  B  F  C  E  2D  2D  C  E  B  F  2 A  0
E2u : 2 A  B  F  C  E  2D  2D  C  E  B  F  2 A  4 A  2B  2C  4D  2E  2F
The six group orbitals have the symmetries B2g , A2u , E1g , and E2u , expressed in simplified
form below, with normalization constants shown at right.
1
B2g : A  B  C  D  E  F
B2g : N   (1) 2  (1) 2  (1)2  (1)2  (1)2  (1)2 


A2u : A  B  C  D  E  F
A2u : N   (1) 2  (1) 2  (1)2  (1)2  (1)2  (1)2 


E1g : 2 A  B  C  2D  E  F
E1g : N   (2) 2  (1) 2  (1)2  (2)2  (1)2  (1)2 


E2u : 2 A  B  C  2D  E  F
E2u : N   (2) 2  (1) 2  (1)2  (2)2  (1)2  (1)2 


The first four SALC equations are:
B2g :
1 
(2 pz( A) )  (2 pz( B) )  (2 pz(C ) )  (2 pz( D) )  (2 pz( E ) )  (2 pz( F ) ) 

6
A2u :
1 
(2 pz( A) )  (2 pz( B) )  (2 pz(C) )  (2 pz( D) )  (2 pz( E) )  (2 pz( F ) ) 

6
1 
2(2 pz( A) )  (2 pz( B) )  (2 pz(C ) )  2(2 pz( D) )  (2 pz( E) )  (2 pz( F ) ) 


12
1 
E2u :
2(2 pz( A) )  (2 pz( B) )  (2 pz(C) )  2(2 pz( D) )  (2 pz( E) )  (2 pz( F ) ) 

12 
E1g :
Copyright © 2014 Pearson Education, Inc.
1

1
6
1

6
1
1


1
12
1
12
74
Chapter 5 Molecular Orbitals
The remaining two SALCs ( E1g and E2u ) can be deduced by examination of the coefficients of the
normalized equations.
Coefficients of Normalized SALCs
cA
B2g
A2u
E1g
1
6
1

6
2
cB
cC
1
1
6
1
6
1
6
1
6
1

cE
1
1
6
1
6
1
6
2
6
1


12
1
2
1
12

cF
c A2 cB2 cC2 cD2
cE2
cF2
1
1
6
1
6
1
3
1
6
1
6
1
12
1
4
1
12
1
4
1
6
1
6
1
12
1
4
1
12
1
4
1
1

6
1
6
1
12
12
1
1
E1g
0
0


2
2
1
1
2
2




E2u
12
12
12
12
12
12
1
1
1
1
E2u
0
0


2
2
2
2
Sum of the squares for each 2pz wavefunction must
total 1 for an identical contribution of each atomic
orbital to the group orbitals
12
12
1
2
1
cD
Squares of SALC Coefficients
0
1
3
0
1
1 1
6 6
1 1
6 6
1 1
12 12
1 1
4 4
1 1
12 12
1 1
4 4
1
6
1
6
1
3
1
1
1
0
1
3
0
Sum of the
Squares = 1 for
Normalization
Requirement
1
1
1
1
1
1
The sum of the squares of the SALC coefficients for A and D equal 0 without any contribution
from the second E1g and E2u SALCs; therefore, c A and cD are zero for these two SALCs. The
1
without contributions from the
2
1
second E1g and E2u SALCs. This suggests that c B2 , cC2 , c D2 , and c E2 equal for these two
4
1
SALCs, and that cC , c D , c E , and c F equal  . The choice of signs in the table above are
2
those required for the SALCs to satisfy the symmetry requirements of the functions associated
with the E1g and E2u representations, and to obtain the number of nodes expected (see sketches
sum of the squares of the coefficients for B, C, E, and F equal
below).
The last two normalized SALCs are:
1
E1g :  (2 pz( B) )  (2 pz(C ) )  (2 pz( E ) )  (2 pz( F ) ) 

2
1
E2u :  (2 pz( B) )  (2 pz(C ) )  (2 pz( E ) )  (2 pz( F ) ) 

2
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
75
The six group orbitals are sketched below, ranked by their relative energy. The number of nodes
increases from zero to three with orbital energy. These orbitals, and a discussion of their energies,
are in Section 13.4.4.
B2g
E2u
E2u
E1g
E1g
Energy
A2u
5.30
5.31
a.
Cl2+ has one fewer electron than Cl2, so the π* levels have three, rather than four,
electrons. As a result, Cl2+ has a bond order of 1.5, and the bond is shorter and stronger
than that of of Cl2 (189 pm, compared with 199 pm for Cl2).
b.
Cl4+ has such an elongated rectangular shape (194 pm by 294 pm) that it must be
essentially a Cl2 and a Cl2+ side by side, with only a weak attraction between them
through the π* orbitals. The Cl–Cl bond in Cl2 is 199 pm long; apparently, the weak
side-to-side bond draws off some of the antibonding electron density, strengthening
and shortening the other two shorter Cl–Cl bonds.
1+
a.
F
5.32
F
F
B
B
F
F
F
F
1–
B
F
1–
1+ F
B
F
1–
F
F 1+
b.
The 1a2 orbital near the middle of the figure is the π-bonding orbital.
c.
The LUMO, 2a2, is the best orbital for accepting a lone pair.
d.
The 1a2 orbital is formed by adding all the pz orbitals together. The 2a2 orbital is
formed by adding the B pz orbital and subtracting the three F pz orbitals.
SF4 has C2v symmetry. Treating the four F atoms as simple spherical
orbitals, the reducible representation  can be found and reduced to
F
S
Copyright © 2014 Pearson Education, Inc.
F
F
F
76
Chapter 5 Molecular Orbitals
 = 2A1 + B1 + B2. Overall, the bonding orbitals can be dsp2 or d2sp,
with the s and pz or dz2 orbital with A1 symmetry, px or dxy with
B1 symmetry, and py or dyz with B2 symmetry. The pz or dz2 orbital
remaining accounts for the lone pair. (The use of the trigonal
bipyramidal hybrids dsp3 or d3sp include the lone pair as one of the five locations.)
C2v

A1
A2
B1
B2
5.33
C2
0
1
1
–1
–1
E
4
1
1
1
1
(xz)
2
1
–1
1
–1
(yz)
2
1
–1
–1
1
z, z2
x, xz
y, yz
A square pyramidal molecule has the reducible representation  = E + 2A1 + B1.
C4v

E
A1
B1
E
5
2
1
1
C2
1
–2
1
1
2C4
1
0
1
–1
2v
3
0
1
1
2d
1
0
1
–1
(x, y) (xz, yz)
z, z2, x2+y2
x2–y2
There appear to be three possibilities for combining orbitals, depending on the details of their
relative energies: dsp3 (px and py for E, s and pz for A1, dx2–y2 for B1), d2sp2 (substituting dz2
for pz), and d3sp (substituting dxz and dyz for px and py). Although dxz and dyz appear to work,
they actually have their electron concentration between the B atoms, and therefore do not
participate in  bonding, so d3sp or d2sp2 fit better.
5.34
Square planar compounds have D4h symmetry.
D4h

Eu
A1g
B1g
E
4
2
1
1
2C4
0
0
1
–1
C2
0
–2
1
1
C2
2
0
1
1
C2
0
0
1
–1
i
0
–2
1
1
2S4
0
0
1
–1
h
4
2
1
1
2v
2
0
1
1
2d
0
0
1
–1
 = A1g + B1g + Eu



2
2
2
s, dz dx –y px, py
dsp2 hybrids are the usual ones used for square planar compounds, although d2p2 is also
possible. Since the dz2 orbital does not extend far in the xy plane, it is less likely to
participate in  bonding.
Copyright © 2014 Pearson Education, Inc.
(x,y)
z2
2 2
x –y
Chapter 5 Molecular Orbitals
5.35
a.
77
PCl5 has D3h symmetry.
D3h

E
A1
A2
E
5
2
1
1
2C3
2
–1
1
1
3C2
1
0
1
–1
h
3
2
1
–1
2S3
0
–1
1
–1
3v
3
0
1
1
(x, y) (x2–y2, xy)
z2
z
 = E + 2A1 + A2, so the hybrids are dsp3 or d3sp.
b.
This could also be analyzed separately for the axial and the equatorial positions. The
pz and dz2 orbitals can bond to the axial chlorines (A1 + A2) and the s, px, and py
orbitals or the s, dx2–y2, and dxy orbitals can bond to the equatorial chlorines (E).
c.
The dz2 orbital extends farther than the p orbitals, making the axial bonds a bit longer.
5.36
Ignoring
Orbital Lobe Signs
Including
Orbital Lobe Signs
D3h
D3h
D3h
C2v
C3v
C3v
C3h
C1
1a2
2a2
1a2
1e
Results should be similar to Figure 5.32. The energies of some of the orbitals in the
middle of the diagram are similar, and the order may vary with different calculation
methods. In addition, the locations of the nodes in degenerate orbitals (e and e) may
vary depending on how the software assigns orientations of atomic orbitals. If nodes cut
through atomic nuclei, 1e orbitals may have C2 symmetry, matching the symmetry of
the first E group orbital shown in Figure 5.31. The table of orbital contributions for each
of the orbitals should show the same orbitals as in Figure 5.32. There may be some
differences in contributions with different calculation methods, but they should be minor.
Assignments to px, py, and pz will also differ, depending on how the software defines
orientations of orbitals. Semi-empirical calculation AM1 gives these as the major
contributors to the specified orbitals:
B
F
3a1
4a1
1a2
1a2
2a2
2s
2s
2s
2s, 2py
2pz
2pz
2px
2pz
2pz
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78
5.37
Chapter 5 Molecular Orbitals
a.
The shapes of the orbitals, generated using one of the simplest computational
methods, Extended Hückel Theory, are shown below, with the most
electronegative element shown at right in the heteronuclear cases.
1π
3
1π *
N2
NO+
–
CN
CO
b.
In the 1π orbitals (bonding), the lobes are increasingly concentrated on the more
electronegative atom as the difference in electronegativity between the two atoms
increases. This effect is seen most significantly in CO, where the difference in
electronegativity is the greatest.
In the 1π* orbitals, the antibonding partners of the 1π orbitals, the effect is
reversed, with the largest lobes now concentrated on the less electronegative
atoms. The greatest effect is again shown in CO, with the lobes on carbon much
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
79
larger than those on oxygen.
The 3 orbitals also show the influence of electronegativity, this time with the lobe
extending to the left from the less electronegative atom being the largest, with CO once
more showing the greatest effect. This can be viewed in part as a consequence of the 3
orbital being a better match for the energy of the less electronegative atom’s 2s orbital
which, together with the 2pz orbital of the same atom, interacts with the 2pz orbital of the
more electronegative atom (the atom shown on the right).
c.
The results vary greatly depending on the software used. The results using one approach,
AM1, are shown below (numerical values are energies in electron volts).
*
π*

LUMO
HOMO
π
*

CN
–
14.7
0.13
–3.13
–5.10
–9.37
–28.0
CO
5.28
0.94
–13.31
–16.30
–22.00
–41.2
6.03
1.00
–14.32
–16.19
–21.43
–41.39
–4.42
–9.62
–26.13
–28.80
–35.80
–56.89
N2
+
NO
In this table, the energies decrease as the atomic numbers increase (with CO and N2
giving mixed results). There is considerable mixing of the  orbitals, a phenomenon
that may raise the energy of the  (HOMO) orbital above the energy of the π orbitals–
as is the case in each of these examples.
5.38
Among the trends that should be observed is the effect on the shapes of the π and π* orbitals
(see orbitals of CO labeled as 1π and 1π* in Figure 5.13) as the difference in electronegativity
between the atoms increases (this trend is also observed in Problem 37). For BF and BeNe, the
lobes of the π orbitals should become increasingly concentrated on the more electronegative
atoms, and the lobes of the π* orbitals should become increasingly concentrated on the less
electronegative atoms (a pattern that has begun with CO, if the orbital shapes for CO are
compared with those of the isoelectronic N2).
An additional effect is that the size of the protruding orbital lobe of the less electronegative atom
should increase as the difference in electronegativity between the atoms increases; this can be see
in the 3 orbital of CO in Figure 5.13. Additional trends in the other molecular orbitals can also
be noted.
5.39
In one bonding orbital, the H s orbitals have the same sign and add to the Be s orbital in the
HOMO–1 orbital. Subtracting the Be s orbital results in the antibonding LUMO. The difference
between the two H s orbitals added to the Be pz orbital results in the HOMO; subtracting the Be
pz results in the LUMO+3 orbital. LUMO+1 and LUMO+2 are the Be px and py orbitals and are
nonbonding (and degenerate) in BeH2. For an energy level diagram, see the solution to Exercise
5.8 in Appendix A.
5.40
BeF2 is similar to BeH2, with the addition of π and π* orbitals from the px and py orbitals,
extending over all three atoms. The F px orbitals with opposite signs do not combine with the Be
orbitals, and neither do the py orbitals; the px and py orbitals form the HOMO and HOMO+1 pair.
Copyright © 2014 Pearson Education, Inc.
80
Chapter 5 Molecular Orbitals
The answer to Problem 5.16 shows more details.
5.41
The azide orbitals are similar to the CO2 orbitals, with some differences in contributions
from the atomic orbitals because the CO2 atomic orbitals do not have the identical energies
–
as the nitrogen atoms do. The two highest occupied orbitals of CO2, BeF2, and N3 all consist
of px or py orbitals of the outer atoms with opposite signs, essentially nonbonding orbitals. The
third orbital down has more s orbital contribution from the outer atoms than either of the other
two; in those cases, the lower orbital energies of the atoms reduce that contribution. See also the
solution to Exercise 5.7 in Appendix A.
5.42
One aspect of ozone’s molecular orbitals that should be noted is its π system. For reference, it is
useful to compare the bonding π orbital that extends over all three atoms (the atomic orbitals that
are involved are shown as molecular orbital 6 in the solution to Problem 5.19); this orbital is the
lowest in energy of the 3-orbital bonding/nonbonding/antibonding set (orbitals 6, 9, and 10 in
Problem 5.19) involving the 2p orbitals that are not involved in  bonding. Another
bonding/nonbonding/antibonding set can be seen in the molecular orbitals derived from 2s
orbitals (orbitals 1, 2, and 3 in Problem 5.19).
5.43
a.
Linear
H
H
Cyclic
H
H
H
H
In the linear arrangement, the molecular orbitals shown, from bottom to top, are bonding,
nonbonding, and antibonding, with only the bonding orbital occupied. In the cyclic
geometry, the lowest energy orbital is bonding, and the other two orbitals are degenerate,
each with a node slicing through the center; again, only the lowest energy orbital is
occupied.
5.44
b.
Cyclic H3+ is slightly more stable than linear H3+, based on the energy of the lowest
orbital in an AM1 calculation (–28.4 eV versus –26.7 eV).
a.
The full group theory treatment (D2h symmetry), shown in Section 8.5.1, uses the two
bridging hydrogens as one set for group orbitals and the four terminal hydrogens as
another set; these sets are shown in Figure 8.11. The representations for these sets can be
Copyright © 2014 Pearson Education, Inc.
Chapter 5 Molecular Orbitals
81
reduced as follows:
The bridging hydrogens have  = Ag + B3u.
The boron s orbitals have  = Ag + B1u.
The px orbitals (in the plane of the bridging hydrogens) have  = B2g + B3u.
The pz orbitals (perpendicular to the plane of the bridging hydrogens) have
 = Ag + B1u.
The boron Ag and B3u orbitals combine with the bridging hydrogen orbitals,
resulting in two bonding and two antibonding orbitals. Electron pairs in each of the
bonding orbitals result in two bonds holding the molecule together through hydrogen
bridges.
b.
Examples of diborane molecular orbitals are in Figure 8.14.
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82
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
CHAPTER 6: ACID-BASE AND DONOR-ACCEPTOR CHEMISTRY
6.1
Acid
Base
a.
AlBr3
Br
–
b.
HClO 4
CH 3CN
Lewis, Brønsted-Lowry
c.
Ni2+
NH 3
Lewis
d.
ClF
NH 3
Lewis
e.
SO 2
ClO3 –
Lewis
f.
HF
C3H 7COOH
Lewis, Brønsted-Lowry
Acid
Base
Definition
XeO3
OH –
Lewis
6.2
a.
Definition
Lewis
This is not a Brønsted-Lowry reaction since the product connectivity is [XeO3 (OH)]– .
b.
Pt
XeF4
Lewis
c.
H 2SeO 4
C2 H 5OH
Lewis, Brønsted-Lowry
d.
Lewis
[CH 3Hg(H 2O)]+
SH –
This reaction likely occurs via the following steps:
[CH 3Hg(H 2O)]+  SH –  [CH 3Hg(SH)] + H 2O
[CH 3Hg(SH)] + H 2O  [CH 3HgS]– + H 3O +
While a Brønsted-Lowry reaction occurs in the second step, the species listed in Problem
6.2d are involved in the initial Lewis acid-base reaction.
6.3
e.
CH 3COOH
(benzyl)3N
Lewis, Brønsted-Lowry
f.
HCl
SO 2
Lewis
Al3+ is acidic: [Al(H2O)6]3+ + H2O
[Al(H2O)5(OH)]2+ + H3O+
The hydronium ions react with the basic bicarbonate to form CO2:
–
H3O+ (aq) + HCO3 (aq) 2 H2O (l) + CO2 (g)
–
With pKa values of 5.0 for [Al(H2O)6]3+, 6.4 for H2CO3, and 2.0 for HSO4 , the pH is about 3, low
enough to convert the bicarbonate to CO2.
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Chapter 6 Acid-Base and Donor-Acceptor Chemistry
83
–
6.4
An increase in conductivity suggests that ions are formed: BrF3 + KF
6.5
a.
The ions are [BrF6] and [BrF4]+: 2 Cs[BrF6] + [BrF4][Sb2F11]
b.
[BrF6] : Oh (this complex features a stereochemically inactive nonbonding pair on the
central Br atom. See A. R. Mahjoub, X. Zhang, K. Seppelt, Chem. Eur. J. 1995, 1, 261.)
BrF4 + K+
–
3 BrF5 + 2 CsSbF6
–
F
[BrF4]+: C2v
F
c.
–
+
F
[BrF4] acts as a Lewis acid, accepting F .
+
Br
F
6.6
2 H2SO4
H3SO4+ + HSO4– and 2 H3PO4
H4PO4+ + H2PO4–
form enough ions to allow conductance in the pure acids.
6.7
Gas-phase basicity is defined as G for BH + (g)  B(g)  H + (g) , while proton affinity is
H for the same reaction. Since G  H  T S , and S is undoubtedly positive for these
reactions (where one mole of gaseous reactant is converted to two moles of gaseous products), it
makes sense that the gas-phase basicities in Table 6.6 are less positive than the corresponding
proton affinities.
6.8
These data suggest the following basicity ranking for these ketones:
O
O
O
<
<
H 3C
CH3
H3CH2C
CH2CH3
Ph
Ph
A convenient way to rationalize this basicity ranking is to examine the conjugate acids via
resonance arguments. One resonance form features a positive charge on the carbonyl carbon
(structure B, right). Since benzophenone can further delocalize this positive charge into its phenyl
groups, the conjugate acid of benzophenone is the most stabilized of the three acids, leading to
benzophenone being the strongest base. Since an ethyl group is
H
H
slightly more electron-releasing than a methyl group, the
O
O
conjugate acid of diethylketone is slightly more stabilized
relative to the conjugate acid of acetone (structure B is more
effectively stabilized in the conjugate acid of diethylketone
R
R
relative to A, the conjugate acid of acetone). Acetone is
R
R
B
A
consequently the weakest base among these ketones.
6.9
These data indicate that triphenylphosphine ( PPh 3) is more basic than triphenylamine ( NPh3) in
the gas phase. On the sole basis of electronics, triphenylamine would be expected to be more
basic by virtue of the higher electronegativity of N relative to P, leading to the N center being
more electron rich than the P center. However, the origin of the observed gas-phase basicity
ranking must be the varying abilities of these atoms to accommodate the tetrahedral geometries of
the conjugate acids. The larger covalent radius of P relative to N results in longer P—C(phenyl)
bonds than N—C(phenyl) bonds, resulting in less steric strain between the phenyl rings in
Copyright © 2014 Pearson Education, Inc.
84
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
[HPPh3 ] relative to [HNPh3 ] . The conjugate acid [HPPh3 ] has less steric hindrance than
[HNPh3 ] , and PPh 3 is more basic.
810
The data are graphed below.
750
Dimethylether
730
Ethanol
670
690
710
Methanol
Water
Gas-Phase Basicity (kJ mol-1)
770
790
Diethylether
650
6.10
‐2.7
‐2.5
‐2.3
‐2.1
‐1.9
‐1.7
‐1.5
pKa of Conjugate Acid in Water
a.
The gas-phase and aqueous basicity data correlate poorly. The strongest base in aqueous
solution ( H 2O ) is the weakest base in the gas phase. While basicity in water increases as
Me 2O < Et 2O < MeOH < EtOH < H 2O , the basicity in the gas-phase increases as
H 2O < MeOH < EtOH < Me 2O < Et 2O .
b.
The ethers are the strongest gas-phase bases within this series on the basis of the
electronic impact of two electron-releasing alkyl groups bound to the oxygen atom
compared to one group (ROH) or no alkyl groups ( H 2O ). The oxygen atoms of these
ethers are relatively electron rich as a result. However, the ethers are the weakest bases in
water of this series; their conjugate acids have fewer sites for hydrogen bonding with
H 2O relative to the conjugate acids of alcohols and water. The poorer ability of the
conjugate acids of the ethers to be solvated by water renders these ethers very poor bases
in aqueous solution.
c.
As shown in the graph and the basicity rankings above, the “ethyl” molecule is more
basic than the “methyl” molecule in both the gas-phase and in aqueous solution. This is
undoubtedly an inductive effect; the more electron-releasing ethyl group renders Et 2O
and EtOH more basic than Me 2O and MeOH , respectively, in both phases.
d.
H 2O is the strongest base in water of this series on the basis of the excellent ability of its
conjugate acid ( H 3O + ) to be solvated by water via hydrogen bonding. It is the weakest
gas-phase base within this series as a consequence of the relatively poor inductive effect
of H compared to methyl and ethyl in increasing electron density at the oxygen atom.
Copyright © 2014 Pearson Education, Inc.
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.11
85
This BF3 affinity trend is strongly correlated to the inductive ability of the groups bound to
sulfur. These data suggest that the electron-releasing ability of substituents in sulfoxides ( R 2SO )
increases as R  Ph  Me  n Bu  cyclo-(CH 2 ) . A resonance argument can be employed to
further rationalize the relatively low BF3 affinities of Ph 2SO and PhSOMe . The phenyl group
permits delocalization of the formal positive charge at the oxygen in structure C (below),
rendering this oxygen atom less Lewis basic than when alkyl groups—which cannot enable this
attenuation of the positive formal charge—are present instead of phenyl groups.
BF3
BF3
6.12
BF3
O
O
O
S
S
S
A
B
C
Other resonance forms with
further delocalization of
positive charge into phenyl
rings
a.
According the the authors, a reference acid should be a strong enough Lewis acid to react
with most common bases; form 1:1 acid-base adducts; have spectroscopic characteristics
that can be monitored to observe variations in the strength of Lewis bases when reactions
are conducted; and should not undergo side reactions while acting as a Lewis acid.
b.
Lewis basicity towards the zinc(II) reference is governed significantly by the steric
hindrance created at the nitrogen upon complexation. The bases quinuclidine and
pyridine, both of which feature insignificant geometric changes at nitrogen upon binding,
were found more Lewis basic than all primary, secondary, and tertiary amines examined
with the zinc(II) reference. This is not the case when assessing Lewis basicity via BF3
affinities. For example, the BF3 affinity for pyridine (128.08 kJ mol ) is less than for
some tertiary amines (for example, the BF3 affinity of Me3N is 139.53 kJ mol ).
6.13
c.
The importance of steric hindrance is reflected in the trends observed. When acyclic
amines are considered, the less hindered primary amines were generally the strongest
Lewis bases. A clear trend was not observed with secondary and tertiary amines, but
secondary amines were found stronger bases than tertiary amines when bases with the
same alkyl group were examined. Among alicyclic amines, the trend is opposite in that
quinuclidine (tertiary) was found stronger than the secondary amine piperidine. The
authors state the relative order of the Lewis basicity for acyclic amines as primary >
secondary > tertiary, but an inverted order for alicyclic amines
( tertiary > secondary ≈ primary (acyclic) ).
a.
The frustrated Lewis pair of the sterically encumbered
P(t-C4 H 9 )3 , in combination with the highly Lewis acidic
B(C6 F5 )3 , binds N 2O to give a PNNOB linkage.
b.
This complex has been characterized by single crystal X-ray
crystallography (figure at right).
Copyright © 2014 Pearson Education, Inc.
tBu
tBu
tBu
P
N
N
O
B
C6F5
C6F5
C6 F5
86
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.14
a.
The reaction coordinate diagram is below. The van der Waals complex is hypothesized to
be stabilized via significant pi-stacking between the aromatic rings of the borane and the
secondary amine. In the transition state, the B—H bond has fully formed, and the proton
is beginning to form a C—H bond with the previously aromatic ring of the amine. The
formal positive charge of the hydrogenated intermediate (not shown on the ispo carbon of
the amine) is stabilized by resonance.
F
F
C 6F 5
F
B
C 6F 5
F
F
C6F5
H
H
F
N
H
F
B
C 6F 5 H
F
tBu
H
H F
F
tBu
H
F
B
TS
N
F
C6 F5
F
15.7 kcal/mol
C 6F 5
van der
Waals
complex
(12.3
kcal/mol)
H
F
F
H
N
H
tBu
H
Hydrogenated Intermediate
(4.8 kcal/mol)
N
H
tBu
Ph
C 6F 5
B
C6 F5
C 6F 5
H2
Frustrated Lewis Pair and Hydrogen (0.0 kcal/mol)
[Ph(tBu)NH2][HB(C6F5)3]
(-10.0 kcal/mol)
[(C5H11)(tBu)NH2][HB(C6F5)3]
(Relative energy unspecified)
b.
As proposed in the reference, the phenyl ring of the amine in the van der Waals complex
rotates towards the boron atom. The hydrogen molecule is then “split” in this cavity
formed by the borane and the phenyl ring of the secondary amine. The hydride bonds to
the boron, and the proton binds to the para carbon of the amine phenyl ring.
c.
The activation barrier for hydrogenation of the phenyl ring of the secondary amine is
higher than that for formation of [Ph(tBu)NH 2 ][HB(C6 F5 )3 ] , which is considered the
resting state for this reaction. Therefore the rate of the hydrogenation reaction is
exceedingly low in pentane at ambient temperature, but viable in refluxing toluene.
The salt [i Pr2 NHPh][HB(C6 F5 )3 ] forms when the more basic i Pr2 NPh is employed. It is
d.
hypothesized that i Pr2 NPh is sufficiently basic to not allow formation of the areneborane van der Waals complex necessary to permit hydrogenation of the aromatic ring of
the amine.
6.15
The N—O distance in the complex 2b is 129.6 pm, longer than the corresponding bond in free
NO (115.1 pm). Bond lengthening is expected since NO likely uses its  * orbital (its LUMO) to
accept electron density from the phosphine. Population of an orbital that is antibonding with
respect to the N—O bond will result in a longer bond distance. It is also noteworthy that the
LUMO of NO features greater orbital contribution from the nitrogen atom relative to the oxygen
atom, and the nitrogen atom binds to the phosphorous atom in 2b.
a.
Complex 2b has strong radical character at oxygen, and reacts with 1,4-cyclohexadiene
via a hydrogen-atom abstraction pathway to afford benzene.
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Chapter 6 Acid-Base and Donor-Acceptor Chemistry
(C6F5)2
B
Resonance
stabilzed radical
(C6F5)2
B
H
H
O
N
H
H
N
P
(Mes)2
P
(Mes)2
(C6F5)2
B
(C6F5)2
B
H
H
O
N
N
H
87
OH
+
OH
+
H
H
H
P
(Mes)2
P
(Mes)2
Complex 2b also reacts with the stronger C—H bond of toluene, with hydrogen-atom
abstraction followed by radical coupling to form an O—C bond.
(C6F5)2
B
O
N
(C6F5)2
B
H
H2 C
P
(Mes)2
N
(C6F5)2
B
O
O
N
H2 C
P
(Mes)2
P
(Mes)2
While the radical formed upon hydrogen-atom abstraction from the CH 3 group of
toluene is resonance stabilized, C—O coupling involving a carbon atom in the ring would
lead to a nonaromatic product (one such product is shown at right). Coupling at the
primary carbon results in the retention of aromaticity in the product (left).
N
CH2
(C6F5)2
B
(C6F5)2
B
O
vs.
P
(Mes)2
6.16
H2 C
P
(Mes)2
(C6F5)2
B
b.
+
OH
N
N
O
P
(Mes)2
a.
The pK a for the water/HSCN pair is much smaller than for the water/HCN pair. The
hydrogen bonding between water and HSCN is predicted as much stronger.
b.
The hydrogen bond donors that have the smallest pK a with nitriles are HClO 4 and HI.
c.
(NC)3CH has the smallest pK a with organic sulfides among the organic acids.
d.
Classification on the basis of estimated pK a values from the Slide Rule and the criteria
in Section 6.5.1: Amines (medium strong), triphosphines (medium strong), sulfoxides
(medium), ketones (medium/medium weak), and nitro compounds (medium weak).
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Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.17
a.
The structure has the Br atoms in a staggered structure, resulting in S6 symmetry.
b.
It is convenient to visualize this using tetrahedral, sp3
hybridized As. An sp3 orbital on each As points
inward toward the benzene ring. If the hybrid lobes
have opposite signs of their wave functions, they fit
the symmetry of the π orbitals of the benzene ring to
form bonding and antibonding molecular orbitals.
The bonding interaction is shown.
6.18
As
As
a.
The very high electronegativity of O in comparison with Al pulls the bonding pair very
close to O. This increases the repulsion between the bonding pairs and causes the large
angle. The steric bulk of the AlCl3 and PCl3 is not considered an important factor in
dictating the large angle.
b.
The dative bond between OPCl3 and AlCl3 likely employs a nonbonding donor orbital
of OPCl3 . The donation of electron density from a nonbonding orbital of OPCl3 would
not be expected to significantly impact the P—O bond order.
6.19
6.20
a.
The methyl groups in
(CH3)3N—SO3 donate electrons
to the nitrogen, making (CH3)3N
a stronger Lewis base and
strengthening and shortening the
N—S bond.
(CH3)3N—SO3
H3N—SO3
N—S
191.2 pm
195.7 pm
N—S—O
100.1°
97.6°
b.
The greater concentration of electrons in the N—S bond of (CH3)3N—SO3 increases
electron-electron (bp-bp) repulsions, opening up the N—S—O bond in comparison
with H3N—SO3.
a.
The polarity of the Xe–F bonds concentrates electrons on the fluorine atoms, which act as
the centers of Lewis basicity. As shown in the reference, which provides the structure of
[Cd(XeF2)](BF4)2, the Xe–F–Cd bond is strongly bent at the fluorine and the geometry
around xenon is nearly linear, with an F–Xe–F bond angle of 179.1°.
b.
The BF4– ion is smaller than AsF6–, and the charge per fluorine is also greater in BF4–,
making BF4– the stronger Lewis base. In addition, the higher oxidation state of
cadmium(II) in [Cd(XeF2)](BF4)2 enables stronger interaction with the fluorines in
BF4– than occurs between the silver(I) ion and AsF6– in [Ag(XeF2)]AsF6.
Copyright © 2014 Pearson Education, Inc.
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.21
89
An energy level diagram for NO- is below. The HOMO is the π* orbital. Bonding with
H+ depends on which end of the π* orbital carries more electron density. Calculation shows
slightly more electron density on N, making HNO the more likely (bent) molecule. This is
consistent with the energy of the nitrogen 2p subshell (-13.18 eV) relative that of the oxygen 2p
subshell (-15.85 eV); the nitrogen 2p orbitals should contribute more to these π* orbitals relative
to the oxygen 2p orbitals. Problem 6.43 asks for calculation of the molecular orbitals of both
HNO and HON to address this question from an alternate perspective.
 2p 
 2p
2p
2p
2p
 2p
2s
2s
2s
 2s
NO–
N
6.22
O
a.
This is similar to the effects described in Section 6.4.2 for I2. Br2 forms charge-transfer
complexes with donor solvents such as methanol.
b.
The 500 nm band (π* *) should shift to shorter wavelength (higher energy)
because the difference in energy between the π* and * orbitals is greater in
Br2•CH3OH than in Br2.
–
–
–
6.23
AlF3 + F  AlF4 . The Na+, AlF4 salt can dissolve in HF. When BF3 is added, it has a stronger
–
–
attraction for the fluoride ions, with the reaction AlF4 + BF3  AlF3  + BF4 .
6.24
Soft metal ions do not combine with oxygen as strongly as hard metal ions, so reactions such as
2 HgO

2 Hg + O2(g)
2 Ag2O

2 CuO


2 Cu + O2(g)
4 Ag + O2(g)

CuO + C
Cu + CO(g)
2 CuO + C
2 Cu + CO2(g)
are more easily carried out. Reduction of some of the softer metals can be carried out with
relatively low temperatures (campfires); some think use of rocks containing ores could have
led to accidental reduction to the metals and discovery of the smelting process. Harder metals
such as iron require much higher temperatures for the reduction process.
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Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.25
Hg is a much softer metal, and combines with the soft sulfide ion more strongly. Zinc and
cadmium are more borderline metals, and combine with all the anions with more nearly equal
facility.
6.26
When any of these salts vaporize, the vapor phase consists of molecules. When they are in
the solid state, they are ionic, with some covalent properties. The liquid state is between the
two, and can be made up of ions, covalent molecules, or something between these two
extremes. If the liquid is molecular, vaporization should be easier (molecules in the liquid
phase being converted to molecules in the vapor phase). If the liquid is mostly ionic,
vaporization is more difficult. Using this criterion, the most ionic liquids should be ZnF2 and
CdF2, and the most molecular liquids should be HgF2 and HgCl2. On a more general view,
mercury, as the softest metal in the series, forms the more molecular compounds and zinc, as
the hardest, forms the more ionic compounds. Cadmium forms the most molecular
compound with the borderline bromide.
6.27
a.
pyridine + BF3:

H =  E py E BF3  Cpy CBF3

= –[(1.17)(9.88) + (6.40)(1.62)] = –21.9 kcal/mol or –91.6 kJ/mol. This predicted value is
roughly 13% less exothermic than the experimental value of -105 kJ/mol.
pyridine + B(CH3)3:

H =  E py E B(CH 3 )3  Cpy CB(CH 3 )3

= –[(1.17)(6.14) + (6.40)(1.70)] = –18.1 kcal/mol, or –75.7 kJ/mol. This predicted value
is more exothermic than the experimental values of -64 and -71.1 kJ/mol, respectively.
6.28
b.
Fluorine is electron-withdrawing, and CH3 electron-releasing. Therefore, B in BF3
carries a greater positive charge and interacts more strongly with Lewis bases such as
pyridine.
c.
The harder acid BF3 interacts more strongly with the borderline base pyridine.
NH3 + BF3:

H =  E NH 3 E BF3  CNH 3 CBF3

= –[(1.36)(9.88) + (3.46)(1.62)] = –19.0 kcal/mol, or –79.5 kJ/mol
NH3 + B(CH3)3:

H =  E NH 3 EB(CH 3 )3  CNH 3 CB(CH 3 )3

= [(1.36)(6.14) + (3.46)(1.70)] = –14.2 kcal/mol, or –59.4 kJ/mol
The order is pyridine + BF3 > NH3 + BF3 > pyridine + B(CH3)3 > NH3 + B(CH3)3. The change
from BF3 to B(CH3)3 is larger than the change from pyridine to NH3.
Copyright © 2014 Pearson Education, Inc.
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.29
91
Absolute hardness parameters:
I
15.81
10.7
9.3
7.8
BF3
NH3
C5H5N
N(CH3)3

A
–3.5
–5.6
–0.6
–4.8
9.7
8.2
5.0
6.3
The difference between the HOMO of pyridine (9.3) and the LUMO of BF3 (–3.5) is smaller
than the other possible interaction, so this pair has the largest –H, in spite of the difference
in hardness. By comparison with the nitrogen compounds, B(CH3)3 is expected to have an
absolute hardness lower than that of BF3 , approximately 7.5—8.
6.30
CsI and LiF fit Basalo’s principle that ions of similar size and equal (but opposite) charge form
the least soluble salts. CsF and LiI have ionic sizes that are very different, and they do not fit
as well into an ionic lattice. In addition, CsI and LiF are soft—soft and hard—hard combinations,
which combine better than the hard—soft LiI and soft—hard CsF.
6.31
The following reaction is unlikely since C is soft and O is hard.
H3C
CH3
H3C + CH3
O
H
H
H
O
H
The second reaction is is more likely. Adding the carbonyl oxygen makes C harder, and C in CH3
and the H atom are soft.
O
O
H 3C
C
CH3
O
H
H
6.32
a.
H 3C
C
+
O
CH3
H
H
Solubilities:
MgSO4 > CaSO4 > SrSO4 > BaSO4
Electrostatic forces predict the reverse order due to cation sizes (the charge density of
Mg 2+ is greater than that of the larger Ba 2+ ), but the larger cations fit better with the
large sulfate anion in the crystals. Hydration of the cations is the strongest for Mg2+,
weakest for Ba2+, agreeing with the solubility order.
b.
6.33
Solubilities:
PbCl2 > PbBr2 > PbI2 > PbS
As a moderately soft cation, Pb2+ has stronger interactions with the softer anions
–
–
–
(hardness order: Cl > Br > I > S2–). In addition, hydration of the anions is largest for
the chloride, smallest for sulfide, based on size.
The order is Al–OC–W (see Figure 13.17). The oxygen of CO is the “harder” end of this
molecule and bonds with the harder Al atom; carbon and W engage in a soft–soft interaction.
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92
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.34
a.
TeH2 is the strongest acid, because Te is more electronegative than either Sn or Sb.
Therefore, the hydrogen is more positive and acidic.
b.
NH3 is the strongest base because N is more electronegative than either P or Sb.
c.
(CH3)3N is the strongest base in the gas phase because the methyl groups contribute
electron density to the nitrogen. In solution, the order is scrambled, probably due to
solvation (Section 6.3.6).
d.
4-Methylpyridine > pyridine > 2-methylpyridine. Again, the methyl group adds electron
density to the N. However, with methyl in the 2 position, steric hindrance makes bonding
to BMe3 more difficult.
–
6.35
In general, oxide ion reacts with water to form hydroxide: O2– + H2O  2 OH . In B2O3, the
small, hard B3+ holds on to the oxide ions strongly. As a result, B2O3 + 3 H2O  2B(OH)3
–
H+ + H2BO3 , and the solution is very weakly acidic (pKa = 9.25). In Al2O3, the Al3+ ion
–
is larger and softer. It can form either [Al(OH)4] (acting as an acid) or [Al(H2O)6]3+ (acting as a
base), depending on the other species in solution. Sc3+ is still larger and softer, so it combines
better with water than with hydroxide ion. The result is the reaction Sc2O3 + 15 H2O 
–
2[Sc(H2O)6]3+ + 6 OH .
6.36
a.
CaH2 + 2 H2O  Ca2+ + 2 H2 + 2 OH Calcium has a lower electronegativity than
hydrogen, so CaH2 is Ca2+(H–)2 and the hydride ions react with water.
b.
HBr + H2O  H3O+ + Br Bromine is more electronegativite than hydrogen,
so the hydrogen is strongly positive and is readily transferred to the lone pair of water.
c.
H2S + H2O
–
–
–
+
H3O + HS Sulfur is slightly more electronegative than hydrogen,
and the positive hydrogen in H2S can dissociate to a small extent.
d.
CH4 + H2O no reaction. The C–H bond is almost nonpolar; the interaction between
the oxygen atom of water and this C–H bond required for hydrogen ion transfer is
exceedingly weak.
6.37
BF3 > B(CH3)3 > B(C2H5)3 > B(C6H2(CH3)3)3 Alkyl groups are electron-donating and increase
the electron density on B and reduce the attraction for the lone pair of NH3. The bulky
mesityl groups render B(C6H2(CH3)3)3 less Lewis acidic on the basis of the steric hindrance
introduced when these boron centers adopt the required tetrahedral geometry upon complexation
of NH 3 .
6.38
a.
CH3NH2 is a stronger base on the basis of the electron-releasing ability of the
methyl group.
b.
Although 2-methylpyridine is the stronger base with smaller acid molecules, the
methyl group interferes with adduct formation with trimethylboron (F-strain) and
the pyridine-trimethylboron dative bond is stronger.
Copyright © 2014 Pearson Education, Inc.
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.39
c.
Trimethylboron forms a stronger adduct with ammonia because the three phenyl
rings of triphenylboron cannot bend back readily to allow the boron to become
tetrahedral (B-strain).
a.
With the acids listed in order of increasing acidity:
H3AsO4
H2SO3
H2SO4
pKa (9-7n)
2
2
–5
pKa (8-5n)
3
3
–2
pKa (exp)
9.2
2.2
1.8
b.
With the acids listed in order of increasing acidity:
pKa (9-7n)
pKa (8-5n)
pKa (exp)
6.40
HClO
9
8
7.4
HClO2
2
3
2
HClO3
–5
–2
–1
93
HMnO4
–12
–7
–11
HClO4
–12
–7
–10
Dimethylamine acts as weak base in water, with a very small amount of OH– provided by
–
the reaction (CH3)2NH + H2O
(CH3)2NH2+ + OH . Acetic acid is a stronger acid than
water, so dimethylamine acts as a stronger base, and the reaction
–
(CH3)2NH + HOAc (CH3)2NH2+ + OAc
goes to completion. 2-Butanone is a neutral solvent; there is no significant acid-base reaction
with dimethylamine.
6.41
SbF5 in HF reacts to increase the H+ concentration and decrease H0:
SbF5 + HF
6.42
–
H+ + SbF6 . These ions then can react with alkenes.
a.
As the Lewis acids BF3 and BCl3 interact with NH3, the geometry around boron changes
from planar to trigonal pyramidal; however, in accord with the LCP model the
nonbonded F…F and Cl…Cl distances are nearly constant, suggesting that these atoms
remain in contact with each other. Because these nonbonded distances remain essentially
constant, the boron–halogen distance must increase as the distortion from trigonal
geometry occurs. Because of the strength of the B–F bond, a consequence of the large
electronegativity difference between B and F and the small size of the fluorine atom,
more energy is required to distort BF3 from planarity than to similarly distort BCl3.
(Calculations in support of this argument are presented in the reference.) The
consequence of this energy requirement is that BF3 is a weaker Lewis base than BCl3
toward NH3. The article does not address the relative Lewis basicity of BBr3, but a
similar argument could apply for this compound.
b.
This article does not consider the LCP model but focuses on ab initio calculations on the
adducts X3B–NH3. These calculations show a higher B–N bond dissociation energy in
the BCl3 adduct than in the BF3 adduct. Although many factors are involved, an
important issue is that BCl3 has a lower energy LUMO that is able to interact more
strongly with the donor orbital of NH3, giving a stronger covalent interaction (and
stronger B–N bond) in Cl3B–NH3.
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94
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
6.43
a.
The energy diagram and the orbitals are shown below.
 2p
 2p 
2p
2p
2p
2p
2s
2s
2s
 2s
N
NO–
O
The HOMO is the π*, with greater concentration of electron density on N. This orbital
can overlap with the empty H+ 1s orbital, forming HNO, a bent molecule.
b.
Calculations predict that HNO is more stable than NOH, with the energy of the HNO
HOMO lower than the energy of the NOH HOMO. The HOMO and HOMO-1 of HNO
and NOH, respectively, two of the nine molecular orbitals of these species arising from
the valence orbitals, are shown below.
HNO
NOH
HOMO
HOMO–1
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Chapter 6 Acid-Base and Donor-Acceptor Chemistry
95
6.44
LUMO
HOMO
HOMO –1
HOMO –2
Br2
Br2–HOCH3
HOCH3
The interaction of the HOMO of the methanol and the LUMO of the Br2 results in the adduct
LUMO and HOMO –2 orbitals, respectively. The geometry shown features the Br2 at
approximately a trigonal angle (H—O—Br is 113° and C—O—Br is 106° in a PM3 calculation).
6.45
6.46
a.
For orbitals of BF3 and NH3, see Figures 5.32 and 5.30, respectively.
b.
The B–N bonding molecular orbital, shown below, is polarized toward the more
electronegative nitrogen; the matching antibonding orbital, which has a node between the
boron and nitrogen atoms, is polarized toward the boron.
a.
The sketch below is a first approximation of the postulated interaction responsible for the
halogen bond with acetylene.
Copyright © 2014 Pearson Education, Inc.
96
Chapter 6 Acid-Base and Donor-Acceptor Chemistry
The donor orbital of acetylene
is a pi-bonding orbital
b.
The acceptor orbital of Br 2
is its * orbital
A key orbital to look for is one in which the adjacent lobes of the pi-bonding orbital of
acetylene, shown an a, and the * orbital of Br2 have merged into a larger lobe
(occupied by an electron pair) that links the two molecules. This bonding orbital
should have a matching antibonding orbital (empty) at higher energy.
Copyright © 2014 Pearson Education, Inc.
Chapter 7 The Crystalline Solid State
97
CHAPTER 7: THE CRYSTALLINE SOLID STATE
7.1
a.
Oh
b.
e.
The image in Figure 7.10, which has D3d symmetry, actually consists of three unit cells.
For an image of a single unit cell, which has point group C2h, see page 556 in the
Greenwood and Earnshaw reference in the “General References” section.
D4h
c.
Oh
d.
Td
7.2
The unit-cell dimension is 2r, the volume is 8r3. Since this cell contains one molecule
4 / 3r 3
whose volume is 4/3 π r3, the fraction occupied is
 0.524  52.4%.
8r 3
7.3
The unit-cell length for a primitive cubic cell is 2r. Using the Pythagorean theorem, we
can calculate the face diagonal as
2r 2  2r  2  2.828r and the body diagonal as
2.828r  2  2r  2  3.464r . Each corner atom contributes r to this distance, so the
diameter of the body center is 1.464 r and the radius is 0.732 r, 73.2% of the corner atom
size.
7.4
a.
A face-centered cubic cell is shown preceding Exercise 7.1 in Section 7.1.1. The cell
1
contains 6 × 1 = 3 atoms at the centers of the faces and 8 × 8 = 1 atom at the corners, a
2
total of 4 atoms. The total volume of these 4 atoms  4 
radius of each atom (treated as a sphere).
4 3 16 3
r  r , where r is the
3
3
As can be seen in the diagram, the diagonal of a face of the cube
= 4r. By the Pythagorean theorem, the dimension (length of a
diagonal 4r
side) of the cube =
.

2
2
3
 4r  32r 3
.
The volume of the cube, therefore, =   
 2 
2
16 3 
 r 

volume of spheres  3
 0.740 .
The ratio
=
3
32r 
volume of cube


 2 
Therefore, 74.0% of the volume of the cube is occupied by the spheres.
b.
In a body-centered cube, the unit cell contains 1 atom in the center of
1
the cube and 8 × 8 = 1 atom at the corners, a total of 2 atoms. The total volume of these
4
8
2 atoms  2  r 3  r 3 , where r is the radius of each atom.
3
3
Because atoms touch along the diagonal, the diagonal of the cube = 4r. Using the
diagonal 4r
Pythagorean theorem, it can be shown that the dimension of the cube =
.

3
3
Copyright © 2014 Pearson Education, Inc.
98
Chapter 7 The Crystalline Solid State
3
 4r  64r 3
Therefore, the volume of the cube =   
 3 
3 3
8 3 
 r 

volume of spheres 3
 0.680 or 68.0%
The ratio
=
3
64r 
volume of cube


 3 3 
7.5
In the table below, CaF2 is considered to have a fluoride ion in the body center of the overall
unit cell and calcium ions in the body centers of the subunits (labeled “Internal”).
Compound
Corners
NaCl cations
NaCl anions
8 × 1/8
CsCl cations
CsCl anions
8 × 1/8
CaF2 cations
CaF2 anions
7.6
Edges
Face
Centers
6 × 1/2
12 × 1/4
Body
Centers
Internal
Total
Type
1×1
4
4
MX
MX
1×1
1
1
MX
MX
4
8
MX2
MX2
4×1
8 × 1/8
12 × 1/4
6 × 1/2
1×1
LiBr has a formula weight of 86.845, and the unit cell contains four cations and four anions
(or four formula units per molecular unit cell).
10 –6 m3
86.845 g mol–1
3
–1
 25.07 cm mol 
 2.507 10 –5 m3mol –1
–3
3
3.464 g cm
cm
2.507 10 –5 m3mol –1
4 units 1.665 10 –28 m3


unit cell
6.022 10 23 units mol–1 unit cell
3
1.665 10 –28 m3  5.502 10 –10 m = unit cell length
2(r++ r–) = 5.502 × 10–10 m; r++ r– = 2.751 × 10–10 m = 275.1 pm
The sum of the ionic radii from Appendix B.1 is 261 pm.
7.7
CsCl has 8 Cl– at the corners of the unit cell cube, with the Cs+ at the center.
r+/ r– = 173/167 =1.04.
CaF2 has the same structure in a single cube of F– ions, but only half the cubes contain Ca2+.
r+/ r– = 126/119 = 1.06.
Both should have coordination number = 12 based on the radius ratios.
Copyright © 2014 Pearson Education, Inc.
Chapter 7 The Crystalline Solid State
7.8
Figure 7.8 shows the zinc blende unit cell, which contains four S atoms (net) in an fcc lattice
and four Zn atoms in the body centers of the alternate smaller cubes. The diagrams below
are each a view of two layers of such a cell, with • indicating a Zn atom in this layer and a 
indicating a S atom in the layer below. The next pair of layers either above or below these
has the opposite pattern, and the third repeats the original. The fcc lattice pattern can be
seen for the Zn atoms (four corners and face center on the top face, four face centers in the
middle later, and four corners and face center on the bottom face). Each S atom (and each
Zn atom) has two nearest neighbors in the layer above and two in the layer below, in the
arrangement for a tetrahedral hole. Extending the patterns below shows the S fcc lattice.

•
•
•

•

•
•

•
7.9
99
•

•
•

•

•
The graphite layers have essentially the same energy levels as
benzene, but each level becomes a wider band because of the large
number of atoms. This leads to the energy levels shown at right,
with the bands coming from the lowest energy π orbitals filled and
those coming from the highest energy π orbitals empty. The
difference between the highest occupied and lowest unoccupied
bands is small enough to allow conduction electrons to make the
jump and the electrons and holes to move within the bands.
Conduction perpendicular to the layers is smaller, because there are
no direct orbitals connecting them. In polycrystalline graphite, the
overall conductance is an average of the two. (The π orbitals of
benzene are shown in Figure 13.22.)
•
•
*

C6H6
Graphite
In diamond, each carbon atom has four  bonds to its nearest neighbors.
The gap between these filled orbitals and the corresponding antibonding orbitals is larger,
effectively limiting conductance. The bonding in carbon nanotubes resembles that of graphite,
and conductivity is expected. The conductivity of carbon nanotubes is a function of the diameter
of the tubes, and spans the range associated with semiconductors to metallic conductors, up to
approximately 1000 times the conductivity of copper (Section 8.6.1).
7.10
Solutions of alkali halides in water conduct electricity. This does not prove that they are
ionic as solids, but is suggestive of ions in the solid state. Their high melting points are also
suggestive of ionic structures, and the molten salts also conduct electricity. Perhaps the
most conclusive evidence is from X-ray diffraction studies, in which these compounds show
uniform cation—anion distances. If they were molecular species, the interatomic distances
within a molecule should be smaller than the interatomic distances between molecules.
7.11
Hg(I) appears in compounds such as Hg22+ units. The 6s1 4f 14 5d 10 structure of Hg+ forms 
and * molecular orbitals from the s atomic orbitals, allowing the two s electrons to pair in the
 orbital for a diamagnetic unit.
Copyright © 2014 Pearson Education, Inc.
100
7.12
Chapter 7 The Crystalline Solid State
a.
Forming anions from neutral atoms results in the addition of an electron. More electrons
means a larger size, due to increasing electron—electron repulsion. By the same
argument,
forming cations from neutral atoms results in removal of an electron and smaller size.
The cations have fewer electrons and less electron-electron repulsions.
b.
7.13
The oxide ion is larger than the fluoride ion because its nuclear charge is smaller; the
oxygen nucleus has less attraction for the electrons.
Radius ratios for the alkali halides, using ionic radii for CN = 6 (Appendix B.1):
Ions
Radii (pm)
F–
Cl–
Br–
I–
Li+
79
0.66
0.47
0.43
0.38
Radii (pm)
119
167
182
206
Na+
107
0.90
0.64
0.59
0.52
K+
138
1.16
0.83
0.76
0.67
Rb+
166
1.39
0.99
0.91
0.81
Cs+
181
1.52
1.08
0.99
0.88
These salts all actually exhibit the sodium chloride lattice (CN = 6), and radius ratios between
0.414 and 0.732 are expected (Table 7.1). The seven boldface entries are the only structures
correctly predicted on the basis of these data. These data incorrectly predict LiI to have CN = 4,
NaF, KCl, KBr, RbCl, RbBr, RbI, CsBr, and CsI to have CN = 8, and KF, RbF, CsF, and CsCl to
have CN = 12.
7.14
The interionic distances for salts of the same cation increase as Br – > Cl– > F – as expected on
the basis of the increasing sizes of these monoanions with increasing atomic number.
A comparison of these interionic distances with the sum of the corresponding ionic radii is
interesting. Ionic radii for CN = 6 are used (Appendix B.1) in determining the sums (pm) below.
Salt
Interionic
Distance
LiF
LiCl
LiBr
201
257
275
Sum of
Ionic
Radii
198
246
261
Salt
Interionic
Distance
NaF
NaCl
NaBr
231
281
298
Sum of
Ionic
Radii
226
274
289
Salt
Interionic
Distance
AgF
AgCl
AgBr
246
277
288
Sum of
Ionic
Radii
248
296
311
It is interesting that only the silver salts exhibit shorter interionic distances than predicted by the
sum of the ionic radii. As the softness of the anion increases, the absolute contraction of the
interionic distance relative to the sum of the ionic radii increases (from only 2 pm (AgF) to 23 pm
(AgBr)). This suggests an increasing covalent contribution in the interaction with Ag + as the
halide softness increases from F – to Br – . Estimation of the interionic distance using ionic radii
clearly introduces more error as the softness of the interaction towards Ag + increases.
For the lithium and sodium salts, the interionic distances are all longer than the sum of the ionic
radii. The agreement between the distances on a percentage basis is best for LiF (the interionic
distance is 1.5% greater than the sum of the ionic radii) and worst for LiBr (the interionic distance
is 5.4% greater than the sum of the ionic radii). The agreement between these values tracks
qualitatively with the hard-soft compatibility of these ions. The relative differences in these
Copyright © 2014 Pearson Education, Inc.
Chapter 7 The Crystalline Solid State
101
distances changes more dramatically with decreasing halide softness for the harder Li+ (from
1.5% to 5.4%) than with the less hard Na + (2.2% to 3.1%).
7.15
½ Cl2 (g)  Cl (g)
239/2 = 119.5 (Energies in kJ/mol)
–
Cl (g) + e  Cl (g)
–EA
Na (s)  Na (g)
109
Na (g)  Na+ (g) + e–
496 (=5.14 eV × 96.4853 kJ/mol•eV)
–
–
Na+ (g) + Cl (g)  NaCl (s)
Total: Na (s) + ½ Cl2 (g)  NaCl (s)
–789 (Lattice energy calculated below)
–413
EA = 348.5 kJ/mol (The value in Appendix B.3 is 349 kJ/mol.)
U
 
23
NMZ  Z –  e2  

30 
  6.022  10  1.74756  1 –1
 2.3071 10 –28 J m   1
  1  

–12
r0
 274 
274  10 m
 4 0   r0 
 – 789 kJ/mol
7.16
CaO has charges of 2+ and 2–, and radii of 114 and 126 pm, total distance of 240 pm.
KF has charges of 1+ and 1–, and radii of 138 and 119 pm, total distance of 257 pm.
The distance in CaO is 7% smaller and the charge factor is four times as large, both
leading to stronger interionic attraction and contributing to the hardness of the crystal.
MgO has charges of 2+ and 2–, and radii of 86 and 126 pm, total distance of 212 pm, with
NaCl structure and Madelung constant = 1.75.
CaF2 has charges of 2+ and 1–, and radii of 126 and 119 pm, total distance of 245 pm with
the fluorite structure and Madelung constant = 2.52.
The size difference and charges favor stronger MgO interionic attraction, enough to overcome the
Madelung constant difference.
7.17
MgCl2 has a rutile structure, so it will be used for the NaCl2 calculation.
The lattice energy for a rutile structure with combined radii of 253 pm (either NaCl2 or
MgCl2) is:
U
U
NMZ  Z –  e 2   

1 
r0
4 0  r0 
   2.307  10–28 Jm   1 30 
6.022  1023 mol –1  2.385  2  –1
253 10
–12
m

253 
 –2,310 kJ mol –1
The lattice energy of MgCl2 (and the hypothetical NaCl2) is negative. If the Born—Haber cycles
of MgCl2 and NaCl2, respectively, were compared, important contributing terms that would differ
Copyright © 2014 Pearson Education, Inc.
102
Chapter 7 The Crystalline Solid State
would be the H sub and ionization energy values. The magnitudes of the second ionization
energies vary tremendously. Na+  Na2+ + e–, removing an electron from a core 2p orbital,
requires much more energy than the corresponding Mg+  Mg2+ + e– reaction that removes the
second 3s valence electron.
The two metals differ by about 2,800 kJ/mol in these steps (all in kJ/mol):
Na (s)  Na(g)
Na (g)  Na+(g) + e–
Na+ (g)  Na2+(g) + e–
Totals: Na (s)  Na2+(g) + 2e–
Mg (s)  Mg (g)
Mg (g)  Mg+(g) + e–
Mg+ (g)  Mg2+(g) + e–
Mg (s)  Mg2+(g) + 2e–
107
495
4562
5164
147
738
1451
2336
Therefore, it is extremely unlikely that NaCl2 can be made. The large second ionization energy of
Na does not permit the H f for hypothetical NaCl2 to be negative despite the predicted large and
negative lattice energy of this salt.
Repeating the process for a sodium chloride lattice (sum of radii = 274 pm) to compare
The lattice energy and the energies necessary to form Na+ and Mg+:
U
NMZ  Z –  e 2   

1 
r0
4 0  r0 
U
   2.307  10–28 Jm   1 30 
6.022  1023 mol–1  1.748  1 –1
274  10
–12
m

274 
 –789 kJ mol–1
Na (s)  Na(g)
Na (g)  Na+(g) + e–
Na (s)  Na+(g) + e–
107
495
602
Mg (s)  Mg (g)
Mg (g)  Mg+(g) + e–
Mg (s)  Mg+(g) + e–
147
738
885
Unlike with NaCl2, the lattice energy associated with forming MgCl nearly compensates for the
energy required to generate Mg + . From this perspective, the formation of MgCl is more realistic.
However, the ionic bonding in MgCl2 is so much stronger (nearly triple the lattice energy) that
MgCl is a very unlikely product of the reaction of Mg and Cl2 . Indeed, MgCl2 is the
thermodynamic sink in this system.
7.18
½ Br2 (l)  ½ Br2 (g)
½ Br2 (g)  Br (g)
Br (g) + e–  Br–
K (s)  K (g)
K (g)  K++ e–
K+ (g) + Br– (g)  KBr (s)
Total
14.9 (Energies in kJ/mol)
95.1
–324.7
79.0
418.8
–687.7
–404.6 kJ mol–1 for K (s) + ½ Br2 (l)  KBr (s)
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Chapter 7 The Crystalline Solid State
103
For a sodium chloride lattice with total radii of 320 pm:
U
U
NMZ  Z –  e 2   

1 
r0
4 0  r0 
   2.307  10–28 Jm   1 30 
6.022  1023 mol–1  1.748  1 –1
320  10 –12 m

320 
 –687.7 kJ mol–1
7.19
½ O2 (g)  O (g)
O (g) + 2 e–  O2–
Mg (s)  Mg (g)
Mg (g)  Mg2++ 2e–
Mg2++ O2– MgO (s)
Total
247
(Energies in kJ/mol)
603
37
2188
–3934
–859 kJ mol–1 for Mg (s) + ½ O2 (g)  MgO (s)
For a sodium chloride lattice with total radii of 212 pm:
7.20
U
NMZ  Z –  e 2   

1 
r0
4 0  r0 
U
6.022 10 23 mol –1 1.748  2   –2 
 30 
–1
 2.307 10 –28 Jm 1
  –3934 kJ mol
–12


212
212 10 m
PbS has total radii of 303 pm. For the NaCl lattice:
U
NMZ  Z –  e 2   

1 
r0
4 0  r0 
U
6.022 10 23 mol –1 1.748  2   –2 
 30 
 2.307 10 –28 Jm 1

–12
 303 
30310 m
 –2888 kJ mol–1
1/8 S8 (s)  S2– (g)
Pb (s)  Pb (g)
Pb (g)  Pb+ (g) + e–
Pb+ (g)  Pb2+ (g) + e–
Pb2+ (g) + S2– (g)  PbS (s)
535
196
716
1450
U
Total
–98 kJ mol–1 for Pb (s) + 1/8 S8 (s) PbS (s)
(Energies in kJ/mol)
U = –2995 kJ mol–1, roughly 3.5% more exothermic than -2888 kJ mol–1.
Copyright © 2014 Pearson Education, Inc.
104
7.21
Chapter 7 The Crystalline Solid State
In ZnO or TiO, additional Zn or Ti would have two more electrons than the metallic ions.
As a result, any nonstoichiometry in the direction of excess Zn or Ti would supply extra
electrons, making an n-type semiconductor.
In Cu2S, CuI, or ZnO, excess S, I, or O would have fewer electrons than the corresponding
ions. Therefore, the result of excess nonmetals in the lattice would be a p-type semiconductor.
7.22
Vibrational motions of the atoms in the lattice become at least partly synchronized, with positive
centers moving closer together. This concentration of positive charge can attract electrons,
allowing two electrons to be closer to each other than would usually be the case. When the
vibrations are synchronized, this attraction can ripple through the material, helping the electrons
move. Apparently the whole system acts as if it is at ground state energies, so no net change in
energy is needed to keep the process going indefinitely.
7.23
The general reaction is Na2Z + Ca2+ (aq)  CaZ + 2Na+ (aq), where Na2Z is the original
zeolite with sodium ions providing the positive charge. When hard water, containing Ca2+
or Mg2+, passes through the zeolite, the ions exchange, leaving only Na+ in the
softened water. The zeolite can be regenerated by flushing with concentrated brine. The
large Na+ concentration reverses the reaction above.
7.24
The ion C24– should have the following molecular orbitals (see Figure 5.7):
 2p
2p
 2p
distortion
 2p
 2s
 2s
Distortion could result in removal of the degeneracy of the πu and πg* orbitals, giving a
diamagnetic ion.
7.25
Gallium nitride has a larger band gap than gallium arsenide (continuing the trend in which
gallium phosphide has a larger band gap than gallium arsenide; see Table 7.3) and emits higher
energy light. Gallium nitride has grown rapidly in importance; it is used in high-energy lasers,
LEDs that emit light in the high-energy part of the visible spectrum as well as at lower energies,
and a variety of other electronics applications.
Copyright © 2014 Pearson Education, Inc.
Chapter 7 The Crystalline Solid State
105
7.26
The smaller the size of the quantum dots, the greater the separation between energy levels within
the dots, and the higher the energy the photoluminescence. Consequently, the largest dots would
produce the lowest energy emission bands.
7.27
First appearing in the literature in 2001, articles on medical applications of quantum dots and
related topics have been more than doubling in number every two years.
7.28
The spectral effect of the L-cysteine/ Cd 2+ ratio on the absorbance of CdS quantum dots is shown
in Figure 1 of this reference. The molar ratio 4:2 causes the highest blue shift of the absorption
edge (363 nm) and affords the smallest CdS quantum dots. These QDs absorb UV-Vis radiation
of highest energy among those studied in this paper.
7.29
These quantum dots were prepared via first generating a solution of CdCl2  2.5 H 2O and
thiourea in a small volume of ultrapure water. An aqueous solution of 3-mercaptopropionic acid
was added, followed by 1 M NaOH, until the pH = 10. The molar ratio of Cd 2+ /thiourea/3mercaptopropionic acid was typically 1/1.7/2.3. The solution was saturated with N 2 and
transferred to an autoclave. The mixture was subjected to 100 ° C for specified times before being
lowered to ambient temperature. The pressure above the sealed solution increased during the
reaction. A reaction scheme is provided here:
NaO2C
CdCl2
HSCH2CH2CO2H
NaO2C
S
NaO2C
H2N
S
S
S
H2N
(Chemicals in autoclave)
S
CO2Na
S
CO2Na
O
in H2O at pH=10
100oC
CdS
S
NH2
CO2Na
S
CO2Na
NH2
(Species in autoclave after the reaction)
The diameters of these QDs depend on the reaction time, with longer times leading to larger QDs.
Diameters were calculated via two mathematical approaches using band gap and UV-Vis data,
respectively.
Reaction Time
(min)
45
60
90
120
180
Diameters (nm) from
Band Gap Data
2.5(4)
2.7(4)
3.1(4)
3.5(4)
3.9(4)
Diameters (nm)
from UV-Vis Data
2.3(4)
2.5(4)
2.9(4)
3.1(4)
3.7(4)
As impressively shown in Figures 2b and 2c of the reference, the smaller the QDs (the shorter the
reaction time), the higher the energy of the emitted radiation when these QDs are excited with
UV radiation. The emission peak maxima vary from 510 nm (smallest QDs, 45 min reaction
time) to 650 mn (largest QDs, 180 min reaction time).
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106
Chapter 7 The Crystalline Solid State
7.30
These authors employed two spectroscopic methods to assess CdSe and CdTe quantum dot size.
UV-Vis spectra were obtained throughout the digestive ripening stage to monitor particle size.
Fluorescence spectra were subsequently acquired to confirm the UV-Vis results. The UV-Vis
spectra acquired after 4 and 8 hours of digestive ripening are in Figure 3 of the reference. The
corresponding fluorescence spectra are in Figure 4. The lower energy fluorescence  max of the
CdTe QDs (572 nm) relative to that of the CdSe QDs (530 nm) indicates the larger size of the
CdTe QDs (Figure 4). A fundamental concept for QDs, that decreasing size leads to higherenergy emitted radiation, is applied to make the size assessment. The relative sizes were further
examined by transmission electron microscopy (TEM) experiments that indicated the average
particle sizes of the CdSe and CdTe QDs to be 4.0 and 4.5 nm, respectively.
7.31
The metallic nature of BaGe3 was probed by measuring the temperature dependence of its
electrical resistivity; the metallic nature was established since the resistivity decreases with
decreasing temperature. Zero resistivity was observed below 4.0 K, suggesting that BaGe3 is a
superconductor with this critical temperature. The Meissner effect was also observed at 4.0 K,
providing further support of this critical temperature.
The conduction bands of BaGe3 are mainly composed of Ge 4p orbitals with some contribution
from Ba orbitals. The states near the Fermi level are primarily made up of Ge 4p and Ba 5d
orbitals. It is interesting that the Ge 4 px , 4 p y , and 4 pz orbitals contribute to different relative
extents to the conduction bands near the Fermi level, with large contributions of Ge 4 px and
4 p y orbitals near the Fermi level of BaGe3 .
7.32
Figure 3 of the reference provides a plot of Tc versus a-axis lattice constants. For the LnFeAsO-
based materials, the highest Tc is associated with an a-axis lattice constant of approximately 3.91
Å. The Tc values reach a maximum when the lanthanide element lies roughly in the middle of the
lanthanide series (Sm, Gd), and decreases as the masses of the lanthanide atoms increase and
decrease, respectively, from these elements, with La, Ce, (the lightest lanthanides) and Er (a
heavier lanthanide) lying at the extremes of the Tc curve for LnFeAsO-based materials. In this
regard, a periodic trend is roughly observed.
The authors suggest an upper limit on the a-axis constant (~4.0 Å) for the possibility of
superconductivity. The lack of superconductivity of the two perovskite-based As materials
( (Sr3Sc 2O5 )(Fe 2As2 ) and (Ba 3Sc 2O5 )(Fe 2As2 ) ) is attributed to their very long a-axis constants
of roughly 4.07 and 4.13 Å, respectively.
7.33
a.
Si4O128–
b.
Si8O2416–
c.
[Si6O1710–]n
Copyright © 2014 Pearson Education, Inc.
Chapter 8 Chemistry of the Main Group Elements
107
CHAPTER 8: CHEMISTRY OF THE MAIN GROUP ELEMENTS
8.1
a.
H2
H2+
74.2 pm
106 pm
436 kJ/mol
255 kJ/mol
These values are consistent with the molecular orbital descriptions. H2 has two
electrons in the bonding  orbital, H2+ has only one. The net attraction between the
bonding electron and the nuclei are weaker in H2+ relative to that between the two
bonding electrons and the nuclei in H2.Therefore, the bond distance is longer in H2+.
b.
8.2
H3+ was described in Problems 5.21 and 5.43. It has a single pair of electrons in a
2-electron, 3-center bond, with bond orders of 1/3.
He2+ has 2 electrons in the bonding  orbital and 1 in the antibonding * orbital, with a
bond order of ½. HeH+ has a bond order of one, but with a poorer match between the
energy levels of the two atoms (–13.6 eV for H, –24.5 for He).
1s*
1s
1s*
1s
1s
1s
1s
1s
He
He2+
He+
HHe+
He
H+
8.3
If the isoelectronic series could be continued, the ion CsF4+ would be expected to have a similar
–
structure to XeF4 and IF4 , with two lone pairs in trans positions. However, a variety of factors
–
make such a species truly unlikely. For example, while the bonds in XeF4 and IF4 are best
described as polar covalent, the tremendous difference in electronegativity between Cs and F
suggest that ionic bonding would be necessary in CsF4+. However, complexes containing Cs(V)
are unknown.
8.4
The equilibrium constants for formation of alkali metal ion complexes with cryptands are listed in
Figure 8.7, with the formation constant for the sodium complex with cryptand [2.2.1] larger than
for either the lithium or potassium complexes. A similar trend is found for the alkaline earths,
with the maximum at strontium. Apparently the optimum size for an ion fitting in the cryptand is
larger than sodium (atomic radius: 107 pm), but smaller than barium (149 pm) or potassium (138
pm), leaving strontium (132 pm) as the closest fit.
8.5
The diagram at the right shows
the primary interactions forming
2p
molecular orbitals. The other
orbitals on the fluorine atoms
2s
form lone-pair orbitals and π
orbitals. The interactions are similar to those in carbon
dioxide (Figure 5.25), but the poorer energy match
between the valence orbitals leads to significantly
less covalent character in the resulting bonds of BeF2.
Be
Copyright © 2014 Pearson Education, Inc.
2p
F Be F
F
F
108
8.6
Chapter 8 Chemistry of the Main Group Elements
A combination of the in-plane p orbitals of Cl and the s and in-plane p orbitals of Be form
orbitals that link the Be and Cl atoms in 3-atom, 2-electron bonds. The atomic orbital interactions
that result in bonding molecular orbitals involving these bridging chlorine atoms are shown
below. The antibonding orbitals also formed by these interactions, along with other orbitals with
significant contribution from the terminal chlorine atoms, are not shown.
pz
pz
px
Cl
Be
Be
Cl
pz
px
s
x
z
pz
px
px
s
The same orbitals, shown as calculated using extended Hückel theory software, are below.
Orbitals from the terminal chlorine atoms are included in these molecular orbitals.
Copyright © 2014 Pearson Education, Inc.
Chapter 8 Chemistry of the Main Group Elements
8.7
The much greater difference in
orbital energies of B (–8.3 and –14.0
eV) and F (–18.7 and –40.2 eV)
makes the BF bond weaker than that
of CO. The s orbital of B and the p
orbital of F can interact, but the p-p
combination cannot interact significantly
because of the large difference in p orbital
energies. This mismatch results
in weak single bonding, rather than
the triple bond of CO.
109
2p

2s
2p


2s
B
BF
F
R
8.8
a.
Evidence cited for a double bond
includes an X-ray crystal structure that shows
a very short boron–boron distance (156 pm)
that is considerably shorter than compounds
having boron—boron single bonds and
comparable to the bond distance in
compounds with double bonds between boron
atoms. Density functional theory
computations also showed a HOMO with
boron—boron π bonding.
N
R
H
R
B
R
R
N
B
N
H
R
R
N
R = isopropyl
R
R
b.
The product in this case was a silicon compound
having a structure similar to the boron compound
shown above, but lacking hydrogen atoms;
nevertheless, it has bent geometry at the silicon
atoms, leading to the zigzag structure shown at
right. An X-ray crystal structure similarly shows a
silicon–silicon distance consistent with a double
bond, and density functional theory yields a
HOMO with π bonding between the silicon atoms.
N
N
R
R
Si
R
Si
R
N
R
N
R
8.9
The combination of orbitals forming the bridging orbitals in Al2(CH3)6 is similar to those of
diborane (Section 8.5.1). The difference is that the CH3 group has a p orbital (or sp3 hybrid
orbital) available for bonding to the aluminum atoms, with the same symmetry possibilities
as those of H atoms in B2H6.
Copyright © 2014 Pearson Education, Inc.
R
110
8.10
Chapter 8 Chemistry of the Main Group Elements
D2h
(pz)
(px)
(1s)
Ag
B2g
B1u
B3u
E
2
2
2
1
1
1
1
C2(z)
2
–2
0
1
–1
1
–1
C2(y)
0
0
0
1
1
–1
–1
C2(x)
0
0
2
1
–1
–1
1
(xy)
0
0
2
1
–1
–1
1
i
0
0
0
1
1
–1
–1
(xz)
2
2
2
1
1
1
1
(yz)
2
–2
0
1
–1
1
–1
z2
xz
xy
x
Reduction of the representations  gives the following:
Boron orbitals:
a.
(pz) = Ag + B1u
b.
(px) = B2g + B3u
Hydrogen orbitals:
8.11
c.
(1s) = Ag + B3u
d.
Treating each group orbital as a single orbital, the orbitals below have the indicated
symmetries.
D2h
E
C2(z)
C2(z)
C2(z)
i
(xy)
(xz)
(yz)
Ag (pz)
1
1
1
1
1
1
1
1
B1u (pz)
1
1
–1
–1
–1
–1
1
1
B2g(px)
1
–1
1
–1
1
–1
1
–1
B3u (px)
1
–1
–1
1
–1
1
1
–1
Ag (s)
1
1
1
1
1
1
1
1
B3u
1
–1
–1
1
–1
1
1
–1
Carbon has two lone pairs in C(PPh3)2, with VSEPR predicting a tetrahedral electron group
geometry. The bulky phenyl groups limit the impact of the nonbonding pairs and force a larger
angle.
Copyright © 2014 Pearson Education, Inc.
Chapter 8 Chemistry of the Main Group Elements
8.12
111
The cations of CaC2, CeC2, and YC2 are not isoelectronic. While Ca 2+ has no valence electrons,
the unusual 2+ group 3 ion Y 2+ and the lanthanide Ce 2+ contain 1 and 2 valence electrons,
respectively. It has been suggested (Greenwood and Earnshaw, Chemistry of the Elements,
2nd ed., p. 299) that there is some transfer of this valence electron density to the π* orbitals of the
dicarbide (or acetylide) ion, resulting in a longer bond.
8.13
Radioactive decay obeys a first order kinetic equation:
 x 
dx
 –kx; ln  –kt
dy
x 0 
The relationship of the half-life (the time at which x = ½ xo) and the rate constant is
 x  1 
ln  ln  –kt1/ 2  –0.693
x 0  2 
k
0.693
t1/ 2
0.693
 1.2110 –4 y –1
t1/ 2
 x 
ln  ln.56  –1.21 10 –4 t
x 0 
k
t  4.8 10 3 y
8.14
Ih symmetry includes 12 C5 axes, 12 C52 axes, 20 C3 axes, 15 C2 axes, an inversion center, 12 S10
axes, 20 S6 axes, and 15 mirror planes. One of the C5 axes and an S10 axis can be seen in the
middle of the second fullerene figure (end view) in Figure 8.18. There are six of each of these,
each representing two rotation axes, C5 and C54, for a total of 12. The same axes fit the 12 S52
axes. If the structure is rotated to line up two hexagons surrounded by alternating pentagons and
hexagons, the 20 C3 and 20 S6 axes can be seen (one in the center, three in the first group around
that hexagon, and three adjacent pairs in the next group, doubled because of C3 and C32). The
C2 axes pass through bonds shared by two hexagons, with pentagons at each end. These original
hexagons each have two more, making five, the next ring out has eight, and the perimeter (seen
edge on) has two for a total of 20 C2. There are five mirror planes through the center of the
pentagons surrounded by five hexagons (see the end view of Figure 8.18 again), and there are
three sets of these for a total of 15. (All this is best seen using a model!)
8.15
a.
Td
b.
Ih
c.
D5h
d.
zigzag: C2h
armchair: D2h
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112
Chapter 8 Chemistry of the Main Group Elements
8.16
Graphane, obtained from the hydrogenation of graphene, was first reported in 2009 (see J.
Agbenyega, Mater. Today, 2009, 12, 13), and research interest has grown rapidly.
8.17
Printing on transparencies is suggested for this exercise. Rolling up of the diagram should
show that more than one chiral form is possible.
8.18
Readers are encouraged to search for the most recent references in this promising area. At this
writing (February 2013) the reference has been cited in 176 other articles in scientific journals.
8.19
The reference provides absorption spectra and colors of solutions containing mixtures
of armchair-enriched single-walled carbon nanotubes with different diameters, ranging from
0.83 nm to 1.5 nm. The strongest absorption band for the nanotubes shifted to longer wavelength
(lower energy) as the diameter increased, with the smallest absorbing at 407 nm and the largest at
785 nm, and the observed colors (complementary to the colors of the strongest absorption bands)
changed accordingly. The observations were consistent with the energy levels of the nanotubes
becoming closer as the diameters of the tubes increased. A similar size effect is observed for
quantum dots (Section 7.3.2).
8.20
The increased stability of 2+ oxidation states as compared to 4+ is an example of the “inert
pair” effect (see Greenwood and Earnshaw, Chemistry of the Elements, 2nd ed., pp. 226, 227,
374). In general, the ionization energy decreases going down a column of the periodic table,
because of greater shielding by the inner electrons. In this family, removal of the second
electron is fairly easy, as it is the first in the higher-energy p orbitals. However, the next two
electrons to be removed are the s electrons, and they are not as thoroughly shielded in the ions.
The effect is larger for the three lower members of the group because the d10 electrons are added
between the s and p electrons. The lower electronegativity of C and Si make them more likely to
form covalent bonds than ions.
8.21
a.
H2ISiSiH2I has a C2 axis perpendicular to the plane of the diagram as drawn, a
perpendicular mirror plane (the plane of the diagram), and an inversion center, C2h.
b.
The Si—H stretches have the reducible representation shown below, which reduces to
 = Ag +Bg +Au+Bu. Au and Bu are IR-active.
C2
C2h
E
i
h
4
0
0
0

1
1
1
1
Ag
Rz
1
–1
1
–1
R x, R y
Bg
1
1
–1
–1
Au
z
1
–1
–1
1
x, y
Bu
8.22
P4 (g)
2 P2 (g), H = 217 kJ mol–1
P4 has six P–P bonds, so six bonds are broken and two triple bonds are formed.
H = bond dissociation energy (reactant) – bond dissociation energy (product)
217 = 6 × 200 – 2(bond dissociation energy of PP)
Bond dissociation energy of PP = 492 kJ mol–1
The pπ orbitals in P2 do not overlap as effectively as those in N2 on the basis of the larger atomic
radius of P relative to N, resulting in weaker π bonds in P2.
Copyright © 2014 Pearson Education, Inc.
Chapter 8 Chemistry of the Main Group Elements
8.23
a.
b.
c.
8.24
–
N3 has molecular orbitals similar to those of CO2 (Section
5.4.2), with two occupied  orbitals and two
occupied π orbitals for a total of four bonds. In contrast
to the situation in CO2 , the atomic orbitals involved in the
–
bonding in N3 have equal energies.
113
*
n
–
Because the HOMOs of N3 (at right) are primarily
composed of p orbitals of the terminal nitrogen atoms, H+
bonds at an angle to the N=N=N axis. The angle is 114°,
larger than the 90° predicted by simple bonding to a p
orbital.
The H–N–N angle is 114º. This observation suggests that
the first resonance structure (with an N(sp)—H bond)
contributes less to the electronic ground state of HN3
relative to the second structure (with an N(sp3)—H bond.
A decreased terminal N—N distance relative to the central
N—N distance is consistent with these contributions.

H
1+
1+
N
N
1–
H
N
1+
N
2–
N
N
The electronegativity of H (2.300) is less than that of N (3.066). Hydrazine has two hydrogen
atoms and an equally electronegative NH2 group bound to each nitrogen atom, while ammonia
has three hydrogens bound to N. In this way, ammonia features three polar covalent bonds
enriching the electron density at nitrogen, while hydrazine features two polar covalent N—H
bonds and one nonpolar covalent bond (to the NH2 group). The N of ammonia is therefore more
electron-rich, and hydrazine is the weaker aqueous base.
Another factor is the potential for hydrogen bonding with water in the resulting conjugate acids.
An ammonium ion (NH4+) features four excellent hydrogen bond donors per N atom, while
hydrazinium ion features just three hydrogen atoms bound to a positively charged N atom. The
enhanced opportunities for NH4+ solvation relative to N2H5+ contribute to NH3 being a stronger
base.
8.25
The larger central atoms feature longer bond lengths, leading to lower concentrations of bonding
electron density close to the Group 15 atom, reducing bonding pair-bonding pair repulsion. The
lone pair subsequently exerts increasing influence down the column, forcing smaller angles as the
atomic radius of the central atom increases. This decrease in bonding pair-bonding pair repulsion
can also be explained using electronegativity arguments, with the most electronegative nitrogen
atom drawing a larger share of the bonding electron density towards it relative to the other group
15 elements that progressively become less electronegative down the column.
8.26
One p orbital of each oxygen is used for the  bond to nitrogen, and the p orbital
perpendicular to the plane of the molecule is used for a π orbital. The remaining p orbital
is in the plane of the molecule, so addition of a proton to this lone pair leaves the entire
molecule planar.
8.27
From Table 8.10
N2O
Cv
N2O4 D2h
–
NO2
C2v
HNO2 Cs
NO
N2O5
–
NO3
HNO3
Cv
D2h
D3h
Cs
NO2
C2v
NO+ Cv
N2O22– C2h
Copyright © 2014 Pearson Education, Inc.
N2O3 Cs
NO2+ Dh
NO43– Td
114
8.28
Chapter 8 Chemistry of the Main Group Elements
a.
b.
Computational work suggests that the electronic ground state of cis- N 2 F4 is 1.4 kcal/mol
lower than that of trans- N 2 F4 at 298 K.
Two trans-cis isomerization mechanisms are proposed. The simpler mechanistic
possibility involves rotation about the N=N bond. A lower limit of 59.6 kcal/mol was
calculated for this isomerization mechanism. The second possibility proceeds via a planar
transition state that resembles [N 2 F]F , with relatively short and long N—F bonds, and a
nitrogen–nitrogen triple bond, respectively. The transformation of nitrogen–nitrogen 
bonding electron density to nonbonding electron density at nitrogen completes the
isomerization. An activation barrier of 68.7 kcal/mol was determined for this mechanism.
The activation barrier for isomerization via N=N rotation is 9.1 kcal/mol less than the
proposed barrier for the [N 2 F]F mechanism.
c.
As a strong Lewis acid, SbF5 can engage in a donor-acceptor interaction with a
nonbonding pair on a fluorine atom (left). The structure on the right, where the F atom
has been completely transferred, resulting in [N 2 F][SbF6 ] , is the limiting result of this
interaction.
F
F
F
N
Sb
F
F
F
F
F
N
N
F
Sb
F
N
F
F
F
F
While fluoride transfer is not envisioned to occur when SbF5 assists in the isomerization,
this possibility is instructive since it suggests that the interaction on the left is expected to
lengthen one N—F bond while shortening the other N—F bond and the N—N bond.
These changes are predicted to lower the activation barrier of the donor-acceptor complex
(left) towards formation of the planar isomerization transition state that resembles
[N 2 F]F , with a relatively short and long N—F bonds, and a nitrogen–nitrogen triple
bond, respectively.
8.29
The balanced reduction half-reactions, for acidic and basic conditions, respectively, follow.
Acidic:
H 3PO4 (aq) + 2 H + (aq) + 2e – 
 H 3PO3 (aq) + H 2O(l)
H 3PO3 (aq) + 2 H + (aq) + 2e– 
 H 3PO2 (aq) + H 2O(l)
4 H 3PO 2 (aq) + 4 H + (aq) + 4e– 
 P4 + 8 H 2O(l)
P4 + 8 H + (aq) + 8e– 
 2 P2 H 4
P2 H 4  2 H +(aq)  2e– 
 2 PH 3
Basic:
PO43– (aq) + 2 H 2O(l) + 2e – 
 HPO32– (aq) + 3 OH – (aq)
HPO32– (aq) + 2 H 2O(l) + 2e – 
 H 2 PO2– (aq) + 3 OH – (aq)
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Chapter 8 Chemistry of the Main Group Elements
115
4 H 2 PO 2– (aq) + 4e – 
 P4 + 8 OH – (aq)
P4 + 12 H 2O(l) + 12e– 
 4 PH 3 + 12 OH – (aq)
The Frost diagram for phosphorus in acidic solution features points at (5, –1.915; H 3PO4 ),
(3, –1.363; H 3PO3 ), (1, –0.365; H 3PO2 ), (0, 0; P4 ), (–2, 0.200; P2 H 4 ), and (–3, 0.206; PH 3).
The Frost diagram for phosphorus in basic solution features points at (5, –7.43; PO 43– ),
(3, –5.19; HPO32– ), (1, –2.05; H 2 PO 2– ), (0, 0; P4 ), (–3, 2.67; PH 3).
8.30
The bonding between O2 units is proposed to
involve interactions between the singly occupied
π* orbitals of neighboring molecules. Such
interactions would generate both bonding and
antibonding molecular orbitals, with the
π*
Copyright © 2014 Pearson Education, Inc.
π*
116
Chapter 8 Chemistry of the Main Group Elements
originally unpaired π* electrons stabilized (i.e., lowered in energy) as they occupy the bonding
orbitals in O8. One such interaction is shown at right. Diagrams of the four highest occupied
molecular orbitals of O8 (which may be viewed as (O2) 4) are in the reference.
8.31
S2 is similar to O2, with a double bond. As a result, the bond is shorter than single bonds in S8.
8.32
MnF62– acts as a Lewis base, transferring two F ions to two SbF5 molecules:
–
–
MnF62– + 2 SbF5  2 SbF6 + MnF4
The MnF4 intermediate can then lose an F atom, with two of these radicals coupling to form F2:
2 MnF4 F2 + 2 MnF3
8.33
a.
Although the ClO3 groups in Cl2O7 are highly electronegative, their size is apparently
responsible for the larger angle in Cl2O7 (118.6°) than in Cl2O (110.9°).
O
Cl
O3Cl
Cl
110.9°
b.
118.6°
2-
O
O
ClO3
O3Cr
126°
CrO3
In the dichromate ion the highly electropositive Cr atoms allow the central oxygen
to attract electrons more strongly than is possible in Cl2O7, resulting in stronger bonding
pair-bonding pair repulsions around the central oxygen in [Cr2O7]2– and a larger bond
angle (126°).
–
8.34
I3 is linear because there are three lone pairs in a trigonal geometry on the central I. I3+ is
bent because there are only two lone pairs on the central I.
8.35
B has only three valence electrons. Adding six from the hydrogen atoms gives B2H6 a total of 12
electrons. The 3-center 2-electron bonds result in four pairs around each boron atom and
nearly tetrahedral symmetry at each boron atom. Iodine has
seven valence electrons initially. In I2Cl6, five more are added
Cl
Cl
Cl
to each iodine, resulting in 12 electrons and octahedral
I
I
Cl
Cl
Cl
geometry around each iodine.
8.36
F + BrF3  BrF4
–
–
–
KF acts as a base; BrF4 is the solvent anion.
SbF5 + BrF3  BrF2+ + SbF6
8.37
a.
BrF2+ is the cation of the solvent; SbF5 acts as the acid.
Br2+ and I2+ have one less antibonding electron than Br2 and I2, so the cations have bond
orders of 1.5 and shorter bonds than the neutral molecules:
Br2
Br2+
b.
–
Bond Order
1.0
1.5
Bond Distance (pm)
228
213
Bond Order
I2
1.0
I2+
1.5
Bond distance (pm)
267
256
The most likely transition is from the πg* (HOMO) to u* (LUMO) (see Figure 5.7 for
comparable energy levels of F2). Because the observed colors are complementary to the
colors absorbed, Br2+ is expected to absorb primarily green light, and I2+ is expected to
absorb primarily orange light. Because orange light is less energetic than green, I2+ has
Copyright © 2014 Pearson Education, Inc.
Chapter 8 Chemistry of the Main Group Elements
117
the more closely spaced HOMO and LUMO.
8.38
c.
In both I2 and I2+ the HOMO and LUMO are the πg* and u*, respectively. I2 absorbs
primarily yellow light (to give the observed violet color) and I2+ absorbs primarily orange
(to give blue). Because yellow light is more energetic than orange, I2 has the more
widely spaced HOMO and LUMO.
a.
In I2+ there are three
electrons in the π* orbitals that
result from interactions between 5p

orbitals (these are the orbitals
I
I
labeled 4πg* in Figure 6.10).
I
I
Interaction between the singly
I2+
I42+
occupied π* orbitals of two
I2+ ions can lead to stabilization of electrons by formation of a bonding
molecular orbital that can connect the I2+ units when I42+ is formed.
b.
I2+
Higher temperature would provide the energy to break the bonds between
the I2+ units and favor the monomer. In addition, formation of two I2+ ions from I42+
would be accompanied by an increase in entropy, also favoring the monomer at higher
temperatures.
F
O
8.39

There are three possibilities:
F
I
F
I
O
F
O
F
F
F
The third structure, with the lone
pair and double bonds in a facial
F
O
arrangement, is least likely because it
C2v
Cs
would have the greatest degree of
electron-electron repulsions involving these regions of high electron concentrations.
I
O
O
F
Cs
Infrared data can help distinguish between the other two structures. In the first structure, only the
antisymmetric stretch would be IR-active, giving a single absorption band. In the second
structure, both the symmetric and antisymmetric stretches would be IR-active, giving two bands.
Observation of two I–O bands is therefore consistent with the second structure. This structure,
which has fewer 90° lone pair–double bond repulsions than the first structure, is also most likely
by VSEPR considerations. Other experimental data are also consistent with the second structure.
8.40
Superhalogens are species which, like the halogens, have very high electron affinities. Classic
superhalogens contain a central atom surrounded by highly electronegative atoms or groups of
–
–
atoms. Examples include MnO4, AlCl4 , and BO2 . See K. Pradhan, G. L. Gutsev, P, Jena, J.
Chem. Phys., 2010, 133, 144301; C. Sikorska, P. Skurski, Inorg. Chem., 2011, 50, 6384; and
references cited therein.
8.41
These reactions take place in the gas phase. The initial product of Xe with PtF6 is believed
–
to be Xe+ PtF6 ; however, when these two ions are in close proximity, they may react further
–
to give [XeF]+, [Pt2F11] , and other products. SF6, if present in large excess, prevents the
formation of these secondary products, apparently by acting as an inert diluent and preventing
effective collisions between the desired products.
Copyright © 2014 Pearson Education, Inc.
118
Chapter 8 Chemistry of the Main Group Elements
8.42
F
F
Xe
O
F
F
O
O
Xe
O
Xe
O
O
O
C4v
F
C2v
8.43
F
F
Xe
F
F
C2v
F
D3h
Two electrons are in the bonding orbital and two are in the nonbonding orbital:
5p
2p
Xe
8.44
F Xe F
F
F
Xe(OTeF5)4: The OTeF5 group has one electron available for bonding on the O, so the four
groups form a square planar structure around Xe, with lone pairs in the axial positions.
O=Xe(OTeF5)4: A square pyramidal structure, with O in one of the axial positions and a
lone pair on the other, and OTeF5 groups in the square base.
8.45
Half reactions:
–
Mn2+ + 4 H2O  MnO4 + 8 H+ + 5e–
XeO64– + 12 H+ + 8e–  Xe + 6 H2O
Overall reaction:
–
8 Mn2+ + 5 XeO64– + 2 H2O  8 MnO4 + 4 H+ + 5 Xe
8.46
–
XeF5 has D5h symmetry. The reducible representation for Xe–F stretching is
D5h

E
5
2C5
0
2C52
0
5C2
1
h
5
2S5
0
2S53
0
which reduces to A1+ E1+ E2, with only E1 IR-active.
O
F
Copyright © 2014 Pearson Education, Inc.
5v
1
F
Xe
F
F
Chapter 8 Chemistry of the Main Group Elements
8.47
a.
119
Point group: C4v
C4v

A1
A2
B1
B2
E
E
18
1
1
1
1
2
2C4
2
1
1
–1
–1
0
C2
–2
1
1
1
1
–2
2v
4
1
–1
1
–1
0
b.
 = 4 A1 + A2 + 2B1 + B2 + 5E
c.
Translation: A1 + E (match x, y, and z)
Rotation: A2 + E (match Rx, Ry, and Rz)
Vibration: all that remain: 3 A1 + 2B1 + B2 + 3E
2d
2
1
–1
–1
1
0
z
Rz
(x, y), (Rx, Ry)
8.48
By the VSEPR approach, XeF22– would be
F
expected to have a steric number of 6, with
four lone pairs on xenon. Two structures,
F
Xe
Xe
cis and trans, need to be considered. In
F
the cis structure there would be five lone
pair–lone pair interactions at 90°, and in the
F
trans structure there would be only four such
cis
trans
interactions. The trans structure, therefore,
would be expected to be more likely. However,
the number of lone pair–lone pair repulsions would still be high, likely making this a difficult ion
to prepare.
8.49
The energy necessary for vibrational excitation decreases with reduced mass; consequently, the
lowest energy band, at 1003.3 cm-1, is assigned to the Xe–D stretch of DXeOXeD, the deuterium
analogue of HXeOXeH. The remaining unassigned bands, at 1432.7 and 1034.7 cm-1, are
therefore due to the Xe–H and Xe–D stretches, respectively, of HXeOXeD, which is also formed.
8.50
a.
HXeO63  2 H 2O(l)  2e 
 HXeO4  4 OH   o  0.94 V
HXeO 4  3 H 2O(l)  6e 
 Xe + 7 OH   o  1.24 V
b.
The disproportionation reaction of interest is:
4 HXeO4  5 OH  
 Xe + 3 HXeO63  3 H 2O(l)  o  0.30 V
G o  nFE o  (6 mol)(96485
8.51
a.
C
J
)(0.30 )  1.7 *105 J (170 kJ ) .
C
mol
The synthesis of XeO 2 is very challenging to carry out. A special reactor composed of
a perfluoro-ethylene/propylene copolymer must be rendered completely moisture free by
extended treatment under vacuum. The reactivity of the container surface is then further
reduced by treatment with F2 gas. Once the container is prepared, a small volume of
water is added in an Ar gas atmosphere, and cooled to 0 °C. The reaction is initiated by
Copyright © 2014 Pearson Education, Inc.
120
Chapter 8 Chemistry of the Main Group Elements
slowly adding XeF4 crystals. The authors emphasize the importance of both the order of
the addition (water first, then XeF4 ) and the speed of addition. The large and negative
reaction H , coupled with the relatively small volume of solvent to dissipate the
heat, can result in complications, including decomposition of the product; the mixture
must be kept at 0 °C. Within 30 seconds after mixing, yellow XeO 2 precipitates from the
solution. The XeO 2 solid is then separated from the reaction mixture via centrifugation
of the mini-reactor itself at 0 °C. The resulting solid is thermally unstable at ambient
temperature and must be kept cold (-78 °C) for long-term storage.
b.
VSEPR theory predicts monomeric XeO 2 to be a bent molecule with a persistent dipole
moment that should permit solubility in water. The insolubility of obtained XeO 2
samples suggests an extended XeO 2 structure; individual monomeric units apparently do
not persist in water.
c.
Application of isotopically labeled water in the synthesis facilitated XeO 2
characterization. The presence of exclusively Xe  O bonding was inferred by preparing
XeO 2 as described in (a) using H 216O and H 218O , respectively. The Raman spectrum of
the H 216O -derived XeO 2 was identical to that of the H 218O -derived XeO 2 sample,
except that all of the bands of the H 216O -derived XeO 2 sample were shifted to slightly
higher energies. The identical spectral patterns, coupled with an isotopic shift in each
band, suggested that only Xe  O bonding was present. Identical Raman spectra were
obtained for XeO 2 samples prepared with either H 216O or D 216O , suggesting no
hydrogen in the samples; no isotopic shifts of any Raman bands were observed. These
chemists were particularly interested in the possibility of Xe  OH units, which was
ruled out by the previous observations.
8.52
a.
The [XeF]+ ion, with a total of 14 valence s and p electrons, would be expected to have a
bond order of 1 (see Section 5.2.1).
b.
Using the VSEPR approach, the following structures can be drawn for the cations of
compounds 1-4:
F
F
F
S
N
Xe
F
F
F
S
F
F
N
S
1
F
2
Xe
F
N
F
F
F F
H
H
F
3
S
F
F
N
4
H
Xe
The [F3NXeF]+ ion, 1, with a triple bond between sulfur and nitrogen, has the shortest
N–S distance (139.7 pm), and [F5SN(H)Xe]+, with a single bond, has the longest distance
(176.1 pm). The other ions have bond distances between these extremes that are
consistent with double bonds between sulfur and nitrogen.
c.
Compound 1 is expected to have a cation with linear bonding around nitrogen, as shown.
However, its crystal structure shows significant bending, with an S–N–Xe angle of
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Chapter 8 Chemistry of the Main Group Elements
121
142.6°. This bending is attributed to close N…F contacts within the crystalline lattice.
d.
VSEPR would predict linear bonding with three lone pairs on xenon as, for example, in
XeF2. The angle measured by X-ray crystallography is 179.6°.
e.
With significant S–N double bonding, these groups are equatorial.
f.
The S–Faxial bonds are longer, as in many other VSEPR examples (see Figures 3.17
through 3.19). Average distances for these ions are: (2: S–F (axial), 158.2 pm; S–F
(equatorial), 152.4 pm) and (3: S–F (axial), 156.1 pm; S–F (equatorial), 151.8 pm).
8.53
a.
b.
D2h
D4d
8.54
[FBeNe]+ The relative order of molecular orbitals may vary depending on the software used. A
simple approach, using an extended Hückel calculation, gave the following results:
F–Be bonding
c.
d.
C1
C2v
e.
f.
Cs
C2
Several orbitals involved:
HOMO: degenerate pair
π bonding between Be and F
Next orbital below HOMO
 bonding by the F pz and the Be s
Be–Ne bonding
One molecular orbital contributing significantly:
Four orbitals below HOMO
 bonding by the Ne pz and the Be s and pz
Overall, the interactions between Be and F are stronger than between Be and Ne.
8.55
In Xe2+, there are a total of 54 orbitals. We will classify only the higher-energy orbitals
here, starting with the highest energy (as calculated using the relatively simple extended Hückel
approach). The orbitals used are the 5s, 5p, and 4d.
Molecular
Orbital
HOMO
Type of
Interaction
*
Atomic
Orbitals
pz
HOMO – 1
*
px
HOMO – 2
HOMO – 3

HOMO – 4

py
pz
px
py
Molecular
Orbital
HOMO – 8
HOMO – 9
HOMO – 10
HOMO – 11
Type of
Interaction
*
Atomic
Orbitals
d 2 2 , d xy

d x 2  y 2 , d xy
HOMO – 12
HOMO – 13
HOMO – 14
HOMO – 15
*
d xz , d yz

d xz , d yz
dz2
HOMO – 5
HOMO – 6
*
s
HOMO – 16
*
HOMO – 7

s
HOMO – 17

Copyright © 2014 Pearson Education, Inc.
x y
dz2
122
Chapter 8 Chemistry of the Main Group Elements
If this were neutral Xe2, there would be no bond, because every occupied bonding orbital
would be offset by an occupied antibonding orbital. In Xe2+, one electron has been removed, so
the bond order is
1
2
, making for a long bond. The large atomic radius of Xe further predisposes
Xe2+ to having a long bond. It is noteworthy that  bonding is exceedingly weak in this species
since the overlap leading to formation of the  and * orbitals is very small.
8.56
The reference by Steudel and Wong provides illustrations for the four highest occupied molecular
orbitals. In addition to the interesting symmetry of these orbitals, the way in which the * orbitals
of O2 are imbedded in the molecular orbitals of O8 should be noted.
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Chapter 9 Coordination Chemistry I: Structures and Isomers
123
CHAPTER 9: COORDINATION CHEMISTRY I: STRUCTURES AND
ISOMERS
9.1
Hexagonal:
C2v
C2v
D2h
Hexagonal pyramidal:
Cs
Cs
C2v
Trigonal prismatic:
Cs
C2v
C2
Trigonal antiprismatic:
Cs
C2
C2h
The structures with C2 symmetry would be optically active.
9.2
a.
dicyanotetrakis(methylisocyano)iron(II) or dicyanotetrakis(methylisocyano)iron(0)
b.
rubidium tetrafluoroargentate(III) or rubidium tetrafluoroargentate(1–)
c.
cis- and trans-carbonylchlorobis(triphenylphosphine)iridium(I) or cis- and transcarbon
ylchlorobis(triphenylphospine)iridium(0)
9.3
d.
pentaammineazidocobalt(III) sulfate or pentaammineazidocobalt(2+) sulfate
e.
diamminesilver(I) tetrafluoroborate(III) or diamminesilver(1+) tetrafluoroborate(1–)
–
(The BF4 ion is commonly called “tetrafluoroborate.”)
a.
tris(oxalato)vanadate(III) or tris(oxalato)vanadate(3–)
b.
sodium tetrachloroaluminate(III) or sodium tetrachloroaluminate(1–)
c.
carbonatobis(ethylenediamine)cobalt(III) chloride or
carbonatobis(ethy
lenediamine)cobalt(1+) chloride
9.4
d.
tris(2,2-bipyridine)nickel(II) nitrate or tris(2,2-bipyridine)nickel(2+) nitrate (The
IUPAC name of the bidentate ligand, 2,2-bipyridyl may also be used; this ligand is most
familiarly called “bipy.”)
e.
hexacarbonylmolybdenum(0) (also commonly called “molybdenum hexacarbonyl”).
The (0) is often omitted.
a.
tetraamminecopper(II) or tetraamminecopper(2+)
b.
tetrachloroplatinate(II) or tetrachloroplatinate(2–)
c.
tris(dimethyldithiocarbamato)iron(III) or tris(dimethyldithiocarbamato)iron(0)
d.
hexacyanomanganate(II) or hexacyanomanganate(4–)
e.
nonahydridorhenate(VII) or nonahydridorhenate(2–) (This ion is commonly called
“enneahydridorhenate.”)
Copyright © 2014 Pearson Education, Inc.
124
9.5
Chapter 9 Coordination Chemistry I: Structures and Isomers
a.
triamminetrichloroplatinum(IV) or triamminetrichloroplatinum(1+)
b.
diamminediaquadichlorocobalt(III) or diamminediaquadichlorocobalt(1+)
c.
diamminediaquabromochlorocobalt(III) or diamminediaquabromochlorocobalt(1+)
d.
triaquabromochloroiodochromium(III) or triaquabromochloroiodochromium(0)
e.
or
dichlorobis(ethylenediamine)platinum(IV) or dichlorobis(ethylenediamine)platinum(2+)
dichlorobis(1,2-ethanediamine)platinum(IV) or dichlorobis(1,2ethanediamine)platinum(2+)
f.
diamminedichloro(o-phenanthroline)chromium(III) or diamminedichloro(ophenanthroline)chro
mium(1+) or diamminedichloro(1,10-phenanthroline)chromium(III)
or
diamminedichloro(1,10-phenanthroline)chromium(1+)
g.
or
bis(2,2-bipyridine)bromochloroplatinum(IV) or bis(2,2bypyridine)bromochloroplatinum(2+)
bis(2,2-bipyridyl)bromochloroplatinum(IV) or bis(2,2bipyridyl)bromochloroplatinum(2+)
h.
dibromo[o-phenylene(dimethylarsine)(dimethylphosphine)]rhenium(II) or
dibrom
o[o-phenylene(dimethylarsine)(dimethylphosphine)]rhenium(0) or
dibrom
o[1,2-phenylene(dimethylarsine)(dimethylphosphine)]rhenium(II) or
dibrom
o[1,2-phenylene(dimethylarsine)(dimethylphosphine)]rhenium(0)
i.
dibromochlorodiethylenetriaminerhenium(III) or
dibrom
ochlorodiethylenetriaminerhenium(0) or dibromochloro(2,2diam
inodiethylamine)rhenium(III) or dibromochloro(2,2diam
inodiethylamine)rhenium(0)
9.6
9.7
a.
dicarbonylbis(dimethyldithiocarbamato)ruthenium(III) or
dicarbonylbis(dimethyldithiocarbamato)ruthenium(1+)
b.
trisoxalatocobaltate(III) or trisoxalatocobaltate(3–)
c.
tris(ethylenediamine)ruthenium(II) or tris(ethylenediamine)ruthenium(2+)
d.
bis(2,2-bipyridine)dichloronickel(II) or bis(2,2-bipyridine)dichloronickel(2+)
a.
Bis(en)Co(III)-µ-amido-µ-hydroxobis(en)Co(III)
N
N
Co
N
4+
N
N
H2
N
O
H
Co
N
N
N
Copyright © 2014 Pearson Education, Inc.
Chapter 9 Coordination Chemistry I: Structures and Isomers
b.
125
Diaquadiiododinitritopalladium(IV)
I
ONO
H2O
Pd
H2O
ONO
H 2O
ONO
H 2O
I
I
OH2
ONO
OH2
H 2O
ONO
I
I
I
OH2
I
OH2
Pd
ONO
ONO
ONO
Pd
I
I
I
Pd
I
Pd
ONO
ONO
I
Pd
ONO
OH2
ONO
OH2
H2O
enantiomers
c.
Fe(dtc)3
S
S
S
S
Fe
S
S
S
S
Fe
S
S
S
S
=
C
–
N
S
S
S
CH3
H
S
At low temperature, restricted rotation about the C—N bond can lead to additional
isomers as a consequence of the different substituents on the nitrogen. These isomers can
be observed by NMR.
Copyright © 2014 Pearson Education, Inc.
126
9.8
Chapter 9 Coordination Chemistry I: Structures and Isomers
a.
triammineaquadichlorocobalt(III) chloride
+
H2O
H3N
H 3N
Co
Isomers are of the cation:
+
Cl
Cl
H3N
NH3
H3N
Co
OH2
H2O
NH3
Cl
Co
Cl
Cl
Cl
cis
trans
mer
b.
H3N
O
Cr
NH3
H3N
4+
NH3
H3N H3N
H3N
c.
fac
-oxo-bis(pentammine-chromium(III)) ion
Cr
NH3
NH3
NH3
potassium diaquabis(oxalato)manganate(III)
–
O
O
H2O
Mn
O
O
O
O
H2O
Mn
Isomers are of the anion:
–
O
–
H2O
O
O
OH2
O
OH2
O
H2O
trans
Cl
a.
O
Mn
cis enantiomers
9.9
NH3
cis-diamminebromochloroplatinum(II)
Pt
Br
NH3
I
b.
diaquadiiododinitritopalladium(IV)
H2O
Pt
ONO
ONO
OH2
I
c.
tri--carbonylbis(tricarbonyliron(0))
O
C
O O
C C
Fe
OC
+
NH3
O
C CO
Fe
CC
O O
Copyright © 2014 Pearson Education, Inc.
CO
NH3
NH3
Chapter 9 Coordination Chemistry I: Structures and Isomers
O
9.10
C
CH2
O –
H2 N
O
N
= N
O
O
M
O
N
N
O
O
N
M
N
N
O
O
N
N
M
N
N
N
O
O
N
O
mer
M(AB)3
B
A
B
M
B
A
A
B
B
A
M
A
A
B
B
A
A
M
A
A
A
B
B
A
B
mer
[Pt(NH3)3Cl3]+
a.
B
M
B
fac
9.12
O
M
O
fac
9.11
127
+
NH 3
Cl
Pt
Cl
+
Cl
NH3
H3N
NH3
Cl
NH3
Pt
NH3
Cl
Cl
fac
mer
[Co(NH3)2(H2O)2Cl2]+
b.
+
NH3
H 2O
Cl
Co
+
Cl
NH3
H 2O
OH2
H 2O
Cl
Co
NH3
Cl
NH3
Cl
+
OH2
Cl
Co
Co
Cl
OH2
H2 O
NH3
H 3N
NH3
H3N
Cl
Co
Cl
enantiomers
Copyright © 2014 Pearson Education, Inc.
Co
NH3
+
H 2O
OH2
Cl
+
NH3
OH2
NH3
Cl
H 2O
+
NH3
OH2
Cl
128
Chapter 9 Coordination Chemistry I: Structures and Isomers
[Co(NH3)2(H2O)2BrCl]+
c.
+
OH2
H 3N
H 3N
Co
Br
H 2O
Cl
H 2O
Co
OH2
+
Br
Co
NH3
Br
NH3
Cl
Co
+
H 3N
NH3
H 3N
Co
Cl
OH2
Cl
OH2
H2 O
H2 O
Br
Cl
+
Co
NH3
H3 N
NH3
H3N
+
OH2
Co
Cl
Br
enantiomers
d.
Br
H2O
Br
Cl
OH2
Co
NH3
OH2
OH2
+
NH3
NH3
H2O
NH3
+
NH3
Br
OH2
H 2O
+
Cl
enantiomers
Cr(H2O)3BrClI
OH2
H2 O
H2O
Cr
Cl
OH2
H2 O
I
I
OH2
Cr
Br
Cl
Br
enantiomers
Cl
H2 O
Cr
OH2
I
Br
H 2O
Cr
OH2
Cl
Br
H2 O
OH2
Cr
I
Br
Cl
I
e.
OH2
OH2
OH2
[Pt(en)2Cl2]2+
2+
N
N
Cl
Pt
2+
N
N
N
N
N
Cl
Pt
N
N
Cl
N
Cl
cis enantiomers
Copyright © 2014 Pearson Education, Inc.
2+
Cl
Pt
Cl
trans
N
N
Chapter 9 Coordination Chemistry I: Structures and Isomers
[Cr(o-phen)(NH3)2Cl2]+
f.
+
N
Cl
NH3
H 3N
NH3
H3N
N
N
Cr
enantiomers
N
NH3
H3N
2+
N
N
N
trans NH3 ligands
2+
N
Pt
Cl
Cl
trans Cl ligands
N
N
NH3
Cr
Cl
2+
Br
Pt
Cl
N
NH3
[Pt(bipy)2BrCl]2+
N
Cr
Cl
Cl
Cl
Cl
N
N
N
Cr
+
+
+
N
g.
129
Br
Br
Cl
Cl
N
Pt
N
N
N
enantiomers
h.
Re(arphos)2Br2
Abbreviating the bidentate ligands As P :
Br
P
As
Re
Br
P
P
As
As
Re
Br
Br
Re
As
Br
P
Br
Br
P
P
P
P
Re
Re
P
As
As
As
As
P
As
As
Br
P
Br
Br
Br
Br

P

As
As
Re
As
As
P



Copyright © 2014 Pearson Education, Inc.
Re
P
P
As
As
Re
P

Br
Br
Br
Br
130
Chapter 9 Coordination Chemistry I: Structures and Isomers
i.
Re(dien)Br2Cl
Cl
N
Re
N
Br
Cl
Br
Br
Br
Br
Re
N
N
N
N
N
Br
N
Cl
Re
Cl
N
N
N
Br
Re
Br
9.13
Br
N
N
N
Cl
Re
N
Br
a. M(ABA)(CDC)
C
A
A
M
B
D
C
A
D
B
M
C
B
C
A
D
C
M
B
A
A
A
A
C
C
b. M(ABA)(CDE)
C
A
B
M
A
A
D
C
A
D
E
C
D
C
D
M
B
B
A
A
M
E
E
B
B
M
A
A
A
A
M
E
E
A
A
M
A
A
B
B
E
Copyright © 2014 Pearson Education, Inc.
M
M
E
C
D
C
D
C
D
C
D
Chapter 9 Coordination Chemistry I: Structures and Isomers
131
9.14
A
A
C
B
M
B
A
C
B
M
B
A
B
B
B
C
M
C
C
A
B
C
C
M
A
A
B
B
C
M
A
A
B
B
M
C
C
C
C
B
B
B
B
M
C
C
A
B
M
A
A
B
B
M
C
C
9.15
B
C
A
A
M
B
C
A
A
A
A
A
C
C
M
C
A
B
A
B
C
a. The “softer” phosphorus atom bonds preferentially to the soft metal Pd (see Section 6.6.1).
b, c. Abbreviating the bidentate ligands N P :
Cl
P
N
Ni
Cl
P
P
N
N
Ni
Cl
Cl
Ni
N
Cl
P
Cl
Cl
N
Cl
P
P
P
P
Ni
N
N
N
N
P
N
P
Ni
P
Cl
Cl
Cl
Cl
Ni

P

N
N
Ni
N
N
P



Copyright © 2014 Pearson Education, Inc.
P
P
N
N
Ni
P

Cl
Cl
Cl
Cl
132
9.16
Chapter 9 Coordination Chemistry I: Structures and Isomers
a, b.
Abbreviating the bidentate ligands N P and O S :
Cl
S
O
N
Cl
Cl
M
P
P
O
O
Cl
P
S
N
O
M
Cl
Cl
N
P
M
Cl
Cl
Cl
Cl
M
N
P
P
N
N
O
O
M

S


S

N
N
P
P
S
Cl
M
Cl
M
P
P
S
S
M
S
Cl
Cl
Cl
Cl
M
N
N
S
S
M
O
O
O
O




Cl
Cl
Cl
Cl
–
N
9.17
The single C–N stretching frequency indicates a trans
structure for the cyanides (the symmetric stretch of the
C—N bonds is not IR active), while the two C–O bonds
indicate a cis structure for the carbonyls (both the
symmetric and antisymmetric C–O stretches are IR
active). As a result, the bromo ligands are also cis.
C
O
C
Br
Co
C
N
Copyright © 2014 Pearson Education, Inc.
O
C
Br
Chapter 9 Coordination Chemistry I: Structures and Isomers
9.18
133
There are 18 isomers overall, six with the chelating ligand in a mer geometry and 12 with
the chelating ligand in a fac geometry. All are enantiomers. They are all shown below, with
dashed lines separating the enantiomers.
N
P
M
As
P
P
Br
H2O
M
P
P
As
As
M
Br
Br
NH3
H 3N
N
M
P
P
As
As
N
M
NH 3
H3N
OH2
H2 O
M
NH3
OH2
OH2
Br
Br
N
N
N
N
N
N
OH2
H2O
Br
Br
M
P
P
As
As
M
NH3
H3 N
Br
Br
M
P
P
As
As
M
OH2
H 2O
NH3
H3N
M
NH3
NH3
OH2
OH2
Br
Br
NH3
NH3
OH2
OH2
NH3
NH3
M
As
Br
OH2
N
N
NH3
M
As
N
N
N
OH2
H2O
M
Br
P
P
As
As
M
Br
N
N
NH3
H3N
M
b.
9.20
All are chiral if the ring in b does not switch conformations.
9.21
20b  20c top ring: , bottom ring: 
9.22
a, b.
1.
O
Mo
2.
S
Cs
Mo
Mo
Cs
3.
Mo
C3v
O
Mo
Cr
O
S
O
O
Mo
Mo
Se
Se
Cr
W
Mo
Mo
O
Cr
Se
O
C1
S
O
S
O
Mo
S
S
Mo
O
W
Cr
W
Se
C1
O
Mo
Cr
W
Mo
C1
O
O
W
Mo
Cr
C1
Se
C1
W
Cs
C1
S
O
O
W
Mo
Mo
Mo
O
Se

S
Mo
Mo
Mo
Se
Cr
O
Se
O
W
C1
Copyright © 2014 Pearson Education, Inc.
O
Mo
Cr
W
C1
M
OH2
W
S
S
S
N
Br
O
Mo
W
N
Br
Mo
Mo
O
M
O
Mo
O
O
S
S
Mo
W
d.

Mo
O
Mo
O
c.
As
OH2
a.

P
As
Br
Br
9.19

P
S
P
As
P
As
P
As
134
Chapter 9 Coordination Chemistry I: Structures and Isomers
S
Mo
Mo
Mo
Cr
O
c.
9.23
S
Se
Cs
O
Mo
Se
W
W
O
Cs
O
Cr
Yes, provided the structure has no symmetry or only Cn axes. Examples are the
structures with C1 symmetry in part a.
The 19F doublet is from the two axial fluorines (split by the equatorial fluorine).
The 19F triplet is from the equatorial fluorine (split by the two axial fluorines).
The two doubly bonded oxygens are equatorial, as expected from VSEPR
considerations. Point group: C2v
9.24
+
F
O
O
Os
F
F
Examples include both cations and anions:
–
–
–
–
–
[Cu(CN)2] , [Cu2(CN)3] , [Cu3(CN)4] , [Cu4(CN)5] , [Cu5(CN)6]
[Cu2(CN)]+, [Cu3(CN)2]+, [Cu4(CN)3]+, [Cu5(CN)4]+, [Cu6(CN)5]+
Based primarily on calculations (rather than experimental data), Dance et al. proposed
linear structures such as the following:
–
NC—Cu—CN
–
NC—Cu—CN—Cu—CN
–
NC—Cu—CN—Cu—CN—Cu—CN
+
Cu—CN—Cu
+
Cu—CN—Cu—CN—Cu
+
Cu—CN—Cu—NC—Cu—CN—Cu
[Cu(CN)2] :
[Cu2(CN)3] :
[Cu3(CN)4] :
[Cu2(CN)] :
[Cu3(CN)2] :
[Cu4(CN)3] :
Where 2-coordinate copper appears in these ions, the geometry around the Cu is linear, as
expected from VSEPR.
9.25
The bulky mesityl groups cause sufficient crowding that the phosphine ligands can show
HC
chirality (C3 symmetry) and can be considered as similar to left-handed (PL) and
right-handed (PR) propellers. If two P(mesityl)3 phosphines are attached in a
linear arrangement to a gold atom, three isomers are possible:
3
PL—Au—PL
PR—Au—PR
PL—Au—PR
H 3C
mesityl
(PR—Au—PL is equivalent to PL—Au—PR, as can best be seen with models.) NMR data
at low temperature support the presence of these isomers, which interconvert at higher
temperatures.
Copyright © 2014 Pearson Education, Inc.
CH3
Chapter 9 Coordination Chemistry I: Structures and Isomers
9.26
The point group is D3h. A representation  based on the nine 1s orbitals
of the hydride ligands is:
D3h

A1
E
A2
E
E
9
1
2
1
2
2C3
0
1
–1
1
–1
3C2
1
1
0
–1
0
2S3
0
1
–1
–1
1
h
3
1
2
–1
–2
3v
3
0
0
1
0



Re

z2
(x, y), (x2–y2, xy)
z
(xz, yz)
135





= H
The representation  reduces to 2 A1 + 2 E + A2 + ECollectively these representations
match all the functions for s (totally symmetric, matching A1), p, and d orbitals of Re, so all the s,
p, and d orbitals of the metal have suitable symmetry for interaction. (The strength of these
interactions will also depend on the match in energies between the rhenium orbitals and the 1s
orbital of hydrogen.)
9.27
NN
NN
NN
O Cr O
O Cr O
O Cr O
O
OO
OO
OO
Mn O
O
O
OO
O Mn O
O
O
O
O
OO
OO
O Mn O
O
O
O
O
OO
O
O Mn
O
OO
O
O Cr O
O Cr O
O Cr O
NN
NN
NN
9.28
N
N
N
N
N
O
O
O
O
N
N
N
N
O
N
N
N
Co
O O
N
Co
N
O O
N
N
N
N
Co
O
O
O
N
N
N
N
O
O
N
Co
N
O
O
O
N
N
O
O O
N
N
N
Co
O O
N
N
N
Co
O
O
N
N
N
N
Copyright © 2014 Pearson Education, Inc.
136
9.29
9.30
Chapter 9 Coordination Chemistry I: Structures and Isomers
a.
Cu(acacCN)2: D2h
b.
C6
tpt: C2v
All four metal-organic frameworks studied (MOF-177, Co(BDP), Cu-BTTri, Mg 2 (dobdc) ) are
significantly more effective at adsorbing carbon dioxide relative to adsorbing hydrogen. This is
attributed, in part, to the higher polarizability of CO 2 relative to that of H 2 . The formation of an
induced dipole in these gases by exposed cations within MOFs is an important prerequisite for
adsorption. The two MOF properties that most strongly correlate with CO 2 adsorption capacity
are MOF surface area and MOF accessible pore volume. As these values (tabulated below)
increase, the CO 2 adsorption capacity increases.
MOF
Surface Area ( m2 g )
Accessible Pore Volume ( cm3 g )
MOF-177
4690
1.59
Co(BDP)
2030
0.93
Cu-BTTri
1750
0.713
Mg 2 (dobdc)
1800
0.573
The graphs in Figure 1 of the reference clearly indicate that Mg 2 (dobdc) adsorbs the most CO 2
at 5 bar. The arrangement and concentration of open Mg 2+ cation sites on the Mg 2 (dobdc)
surface is hypothesized to render this MOF more susceptible to CO 2 adsorption. This MOF,
along with Cu-BTTri, which also features exposed metal sites, are identified as the best prospects
for CO 2 H 2 separation.
9.31
The synthesis and application of amine-functionalized MOFs for CO 2 adsorption is the general
topic of the reference. While the M 2 (dobdc) series of MOFs were proposed as excellent
candidates for this functionalization (on the basis of their relatively large concentration of
exposed metal cation sites), their amine-functionalization proved difficult. This was attributed to
the relatively narrow MOF channels that may hinder amine diffusion into M 2 (dobdc) .
One hypothesized solution was to prepare a MOF with the M 2 (dobdc) structure-type, but with
larger pores. The wider linker dobpdc (below, along with dobdc for comparison) was used in the
hope of obtaining MOFs with larger pores.
O
O
O
O
O
O
O
O
O
O
O
O
dobdc
dobpdc
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Chapter 9 Coordination Chemistry I: Structures and Isomers
137
Amine-functionalized Mg 2 (dobpdc) was prepared by mixing H 4 (dobpdc) , magnesium bromide,
and a small solvent volume (a mixture of N,N’-diethylformamide and ethanol) in a Pyrex
container. The mixture was heated in a microwave reactor, and the M 2 (dobpdc) collected by
filtration after cooling. Dried samples of Mg 2 (dobpdc) were then heated for roughly one hour at
420 °C under dynamic vacuum. After this “activation” step, Mg 2 (dobpdc) was stirred with an
excess of N,N’-dimethylethylenediamine (mmen) in hexanes for one day. Subsequent heating
under vacuum resulted in removal of residual solvents to afford mmen-functionalized
Mg 2 (dobpdc) . The “activation” step was found necessary to completely remove residual N,N’diethylformamide from the Mg 2+ coordination sites.
9.32
This reference discusses application of porphyrin-containing MOFs where the porphyrin provides
a binding site for Fe(III) and Cu(II). The precursor to the porphyrin linker (TCPP) is provided
below; the resulting carboxylates of this linker permit its incorporation into the MOF.
HOOC
COOH
N
H
N
N
H
N
HOOC
COOH
The metallation options include premetallation and postmetallation. In premetallation,
H 4 -TCPP-Cu and H 4 -TCPP-FeCl , respectively, are used as reactants for the MOF synthesis. In
this case, the porphyrin linker and its bound metal ion are installed simultaneously into the MOF.
This general approach afforded MOF-525-Cu, MOF-545-Fe, and MOF-545-Cu. MOF-525-Fe
could not be obtained via this strategy. For this MOF, postmetallation was employed, via the
reaction of MOF-525 with Fe(III) chloride; Fe(III) ions were introduced into the MOF-525
porphyrin linkers via this method.
In terms of similarities and differences, MOF-545 can be metallated with both Fe(III) and Cu(II)
via a premetallation strategy, while MOF-525 requires alternate procedures for incorporation of
Cu(II) (premetallation) and Fe(III) (postmetallation), respectively.
Copyright © 2014 Pearson Education, Inc.
138
Chapter 10 Coordination Chemistry II: Bonding
CHAPTER 10: COORDINATION CHEMISTRY II: BONDING
10.1
10.2
a.
Tetrahedral d 6, 4 unpaired electrons
b.
[Co(H2O)6]2+, high spin octahedral d 7, 3 unpaired electrons
c.
[Cr(H2O)6]3+, octahedral d 3, 3 unpaired electrons
d.
square planar d 7, 1 unpaired electron
e.
5.1 BM =  =
a.
[M(H2O)6]3+ with 1 unpaired electron.
b.
[MBr4]– with the maximum number of unpaired electrons (5).
n  n  2  ; n  4.2  4
M = Ti
d1:
M3+ with 5 d electrons: M = Fe
c.
Diamagnetic [M(CN)6]3–. The strong field cyano ligand favors
low spin. M3+ with 6 d electrons: M = Co
d.
[M(H2O)6]2+ having LFSE =   o . Both high spin d4 and d9 have the correct  o.
3
5
0.6 
High spin d4: M = Cr
0.6 
d9: M = Cu
– 0.4 
10.3
a.
K3[M(CN)6] M is first row transition metal, 3 unpaired electrons.
M3+ with 3 d electrons: M = Cr
Copyright © 2014 Pearson Education, Inc.
– 0.4 
Chapter 10 Coordination Chemistry II: Bonding
b.
139
[M(H2O)6]3+ M is second row transition metal, LFSE = 2.4  o. This can be achieved
with a low spin d6 configuration. M3+ with 6 d electrons: M = Rh
0.6 
– 0.4 
c.
[MCl4]– M is first row transition metal, 5 unpaired electrons.
M3+ with 5 d electrons: Fe
d.
MCl2(NH3)2 M is third row d8 transition metal; two M–Cl stretching bands in IR.
Second and third row d8 complexes are often square planar. The presence of two
M–Cl stretching bands implies cis geometry of the chloro ligands.
Cl
M with 8 d electrons: M = Pt
2+
NH3
Pt
Cl
NH3
10.4
Angular overlap calculations for d 8 and d 9 ions show no energy difference between
D4h and Oh when exclusively considering  interactions. Both d 8 geometries have energies of
-3e; both d 9 geometries have energies of -6e. In general, stability constants decrease as more
ligands are added; the sequence for nickel is the common one. The huge drop in stability constant
between the second and third ethylenediamine on Cu2+ is a result of the d 9 Jahn-Teller effect. The
first two en ligands add in a square-planar geometry, with water molecules in the axial positions,
and the coordination of these two mondentate ligands allows for the Jahn-Teller distortion.
Adding a third en ligand requires a geometry change and with a preference for uniform M—N
bond distances towards the six nitrogen atoms. This is counter to the Jahn-Teller distortion, and
the third addition is much less favorable than the first two.
10.5
[M(H2O)6]2+ M is first row transition metal, µ = 3.9 BM. The magnetic moment implies 3
unpaired electrons, which would give rise to   3 3 5  = 3.9 BM. There are two possibilities,
d3 and d7:
M2+ , d3 : M = V
10.6
a.
[Cr(H2O)6]2+
b.
4–
M2+ , d7: M = Co
  4(6)  4.9  B
[Cr(CN)6]
n=2
  2(4)  2.8  B
–
c.
[FeCl4]
d.
[Fe(CN)6]3–
  1(3)  1.7  B
n=4
e.
[Ni(H2O)6]2+
f.
[Cu(en)2(H2O)2]2+ n = 1
n=5
  5(7)  5.9  B
n=1
Copyright © 2014 Pearson Education, Inc.
n=2
  2(4)  2.8  B
  1(3)  1.7  B
140
10.7
Chapter 10 Coordination Chemistry II: Bonding
Fe(H2O)4(CN)2 is really [Fe(H2O)6]2[Fe(CN)6], all containing Fe(II). [Fe(H2O)6]2+ is high spin
d 6, with  = 4.9 µB; [Fe(CN)6]4– is low spin d 6 , with  = 0 µB. The average value is then
2 × 4.9/3 = 3.3 µB.
2.67 unpaired electrons gives   2.67  4.67  3.53 µB.
10.8
Co(II) is d 7. In tetrahedral complexes, it is generally high spin and has 3 unpaired electrons; in
octahedral complexes, it is also typically high spin and also has 3 unpaired electrons; in square
planar complexes, it has 1 unpaired electron. The magnetic moments can be calculated as
  n(n  2)  3.9, 3.9, and 1.7  B , respectively.
10.9
For the red compounds (Me and Et at high temperatures, Pr, pip, and pyr at all temperatures), the
larger magnetic moment indicates approximately 5 unpaired electrons, appropriate for high-spin
Fe(III) species. At low temperatures for the Me and Et compounds, the magnetic moment
indicates 3 to 4 unpaired electrons, an average value indicating an equilibrium mixture of high
and low spin species. The low spin octahedral complexes have 1 unpaired electron. Increasing
the size of the R groups changes the structure enough that it is locked into high-spin species at all
temperatures.
10.10
Both [M(H2O)6]2+ and [M(NH3)6]2+ should show the double-humped curve of Figure 10.12,
with larger values for the NH3 compounds. Therefore, the difference between these curves
shows the same general features as in Figure 10.12.
10.11
These can be verified using the approach introduced in Chapters 4 and 5, in which a character
of +1 is counted for each operation that leaves a vector unchanged, a character of –1 for each
operation that leaves a vector in its position, but with the direction reversed. The general
transformation matrix for rotation is provided in Figure 4.16. Application is of this matrix is
necessary to deduce the characters involving rotations.
10.12
The e column gives d orbital energies for complexes involving  donor ligands only; the Total
column gives energies for complexes of ligands that act as both  donors and  acceptors:
a.
ML2, using positions 1 and 6:
z2
x2–y2
xy
xz
yz
b.
eπ
0
0
0
–2
–2
e
2
0
0
0
0
Total
2e
0
0
–2 eπ
–2 eπ
ML3, using positions 2, 11, 12:
z2
x2–y2
xy
xz
yz
e
0.75
1.125
1.125
0
0
eπ
0
–1.5
–1.5
–1.5
–1.5
Total
0.75e
1.125 e – 1.5 eπ
1.125 e – 1.5 eπ
–1.5 eπ
–1.5 eπ
Copyright © 2014 Pearson Education, Inc.
Chapter 10 Coordination Chemistry II: Bonding
c.
ML5, C4v, using positions 1, 2, 3, 4, 5:
z
x2–y2
xy
xz
yz
eπ
0
–1.5
–1.5
–3.5
–3.5
e
2.75
1.125
1.125
0
0
z
x2–y2
xy
xz
yz
Total
2.75e
1.125 e – 1.5 eπ
1.125 e – 1.5 eπ
–3.5 eπ
–3.5 eπ
ML8, cube, positions 7, 8, 9, 10, doubled for the other four corners:
e
0
0
2.67
2.67
2.67
2
z
x2–y2
xy
xz
yz
10.13
Total
2e
3e
–4eπ
–3eπ
–3eπ
ML5, D3h, using positions 1, 2, 6, 11, 12:
2
e.
eπ
0
0
–4
–3
–3
e
2
3
0
0
0
2
d.
141
eπ
–5.33
–5.33
–1.78
–1.78
–1.78
Total
–5.33 eπ
–5.33 eπ
2.67 e – 1.78 eπ
2.67 e – 1.78 eπ
2.67 e – 1.78 eπ
–
Metal d orbitals, NH3 influence
eπ
Total
e
2
1
0
z
1e
2 2
3
0
x –y
3e
0
0
0
xy
0
0
0
xz
0
0
0
yz
Metal d orbitals, Cl influence
eπ
Total
e
2
2
0
z
2e
2 2
0
0
0
x –y
0
0
0
xy
0
2
2 eπ
xz
0
2
2 eπ
yz
Ligand NH3:
Ligand Cl :
1
2
3
4
5
6
–
e
0
–1
–1
–1
–1
0
Total
0
–1e
–1e
–1e
–1e
0
1
2
3
4
5
6
e
–1
0
0
0
0
–1
eπ
–2
0
0
0
0
–2
Total
–1e – 2 eπ
0
0
0
0
–1e – 2 eπ
Overall energy = –8 e(NH3) – 4 e(Cl) – 8 eπ(Cl) + 4 eπ(Cl) = –8 e(NH3) – 4 e(Cl) – 4 eπ(Cl)
The metal electrons are still unpaired, one in the dxy orbital and one each in the dxz and dyz, raised
by π interaction with Cl–. Four of the ligand orbitals are lowered by e(NH3) and two are lowered
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Chapter 10 Coordination Chemistry II: Bonding
by e(Cl) + 2 eπ(Cl); each contains a pair of ligand electrons. The precise orbital energies cannot
be calculated since the angular overlap parameters for NH3 and Cl towards Cr(III) are not
provided. Table 10.13 does provide the necessary parameters for trans-[Cr(NH 3 )4 F2 ]+ .
10.14
a.
If we assume these ligands have similar donor abilities, the energies below are obtained
for the molecular orbitals with high d orbital character.
z2
metal
d orbitals
b.
x2-y2
xy
xz
yz
2.75e
1.125e 
For consideration of L as a -acceptor in the axial position, the identical energy level
diagram is obtained regardless of whether L is assigned to position 1 or 6. The xz and yz
orbitals are each stabilized by e.
z2
x2-y2
2.75e 
xy
1.125e 
metal
d orbitals
e
xz
yz
For consideration of L in an equatorial position, the angular overlap result is different
depending on whether the -acceptor is placed in position 2 or position 11/12. This is
problematic since all three equatorial positions are indistinguishable.
For L in position 2:
z2
2.75e 
x2-y2
xy
metal 1.125e e
d orbitals
1.125e 
yz
xz
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e
Chapter 10 Coordination Chemistry II: Bonding
143
For L in position 11 or 12:
z2
metal
d orbitals
2.75e 
xy
2
x -y2
b
xz
c
yz
a: 1.125e e
b: 1.125e e
a
c: 0.25e
0.75e
Since all three equatorial sites are equivalent, the optimum approach is to consider the
weighted average of these two equatorial diagrams, with the second one contributing
twice as much as the first since the second diagram is obtained with L in both position
11 and position 12. This average diagram is below:
z2
2.75e
xy
x2-y2
metal
d orbitals
yz
c.
xz
1.125e e
0.5e
The axial/equatorial preference for the -acceptor ligand depends on the number of metal
valence electrons. The table below assumes low-spin configurations, reasonable on the
basis of the -acceptor ligand. Alternate configuration energies would be obtained for
high-spin cases. This analysis predicts that the axial position would be preferred for the
-acceptor ligand in d 1 to d 7 cases. There is no difference between the configuration
energies for the d 8 to d 10 cases. This analysis includes neither the exchange energy nor
the coulombic energy of repulsion contributions. Ligand positions are strongly influenced
by steric considerations, also not accounted for in this generic case.
Equatorial L
Configuration
Energy
0.5e
Preference: Axial or
Equatorial L ?
d1
Axial L
Configuration
Energy
e
d2
2e
e
Axial
d3
3e
1.5e
Axial
d4
4e
2e
Axial
Metal Valence
Electron Count
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Axial
144
10.15
Chapter 10 Coordination Chemistry II: Bonding
d5
1.125e  4e
1.125e  2.5e
Axial
d6
2.25e  4e
2.25e  3e
Axial
d7
3.375e  4e
3.375e  3.5e
Axial
d8
4.5e  4e
4.5e  4e
None
d9
7.25e  4e
7.25e  4e
None
d 10
10e  4e
10e  4e
None
The angular overlap diagrams for the molecular orbitals with high d orbital character in square
pyramidal (left) and trigonal bipyramidal (right) geometries with -donor ligands are below. In
each of these geometries, the -donor ligands pairs are stabilized by 1e; these levels are omitted
here.
x2-y2
z2
z2
3e
2e
metal
d orbitals
xy
xz
yz
x2-y2
xy
xz
yz
2.75e
1.125e 
The energies of the resulting configurations, and the preference between these coordination
geometries, depend on electron count and whether the complex is low spin or high spin. The low
spin configuration energies (ignoring both the exchange energy and the coulombic energy of
repulsion contributions) and preferences are:
Low Spin
Preference
0e
Trigonal
Bipyramidal
Configuration
Energy
0e
d2
0e
0e
None
d3
0e
0e
None
d4
0e
0e
None
d5
0e
1.125e
Square Pyramidal
d6
0e
2.25e
Square Pyramidal
d7
2e
3.375e
Square Pyramidal
d8
4e
4.5e
Square Pyramidal
d9
7e
7.25e
Square Pyramidal
d 10
10e
10e
None
Metal Valence
Electron Count
Square Pyramidal
Configuration
Energy
d1
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None
Chapter 10 Coordination Chemistry II: Bonding
145
The corresponding high spin preferences (assuming that configurations with five unpaired
electrons are lower in energy than five-electron configurations with electron pairs) are:
Metal Valence
Electron Count
Square Pyramidal
Configuration
Energy
High Spin
Preference
0e
Trigonal
Bipyramidal
Configuration
Energy
0e
d1
d2
0e
0e
None
d3
0e
1.125e
Square pyramidal
d4
2e
2.25e
Square pyramidal
d5
5e
5e
None
d6
5e
5e
None
d7
5e
5e
None
d8
5e
6.125e
Square pyramidal
d9
7e
7.25e
Square Pyramidal
d 10
10e
10e
None
None
Although steric configurations are not part of this angular overlap analysis, the trigonal
bipyramidal geometry is never preferred relative to square pyramidal in these cases with five donor ligands.
For a set of five -donor/-acceptor ligands, the energy levels are as follows for square pyramidal
(left) and trigonal bipyramidal (right).
x2-y2
z2
z2
3e
2.75e
2e
metal
d orbitals
x2-y2
xy
1.125e e
3.5e
4e
xy
xz
yz
xz
yz
The energies of the resulting configurations, and the preference between these coordination
geometries, depend on electron count and whether the complex is low spin or high spin. We will
assume only low spin configurations here since ligands with combined -donor/-acceptor
capabilities tend to stabilize low spin complexes. The low spin configuration energies (ignoring
both the exchange energy and the coulombic energy of repulsion contributions) and preferences
are:
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Chapter 10 Coordination Chemistry II: Bonding
Low Spin
Preference
Metal Valence
Electron Count
Square Pyramidal
Configuration
Energy
d1
4e
Trigonal
Bipyramidal
Configuration
Energy
3.5e
d2
8e
7e
Square Pyramidal
d3
12e
10.5e
Square Pyramidal
d4
16e
14e
Square Pyramidal
d5
20e
1.125e  15.5e
Square Pyramidal
d6
24e
2.25e  17e
Square Pyramidal
d7
2e  24e
3.375e  18.5e
Square Pyramidal
d8
4e  24e
4.5e  20e
d9
7e  24e
7.25e  20e
TBP>SP by only
0.5e  4e
Square Pyramidal
d 10
10e  24e
10e  20e
Square Pyramidal
Square Pyramidal
Although steric configurations are not part of this analysis, the square pyramidal geometry is
clearly preferred relative to trigonal pyramidal in all these cases with five -donor/-acceptor
ligands, except with a d 8 metal where the preference is slight.
10.16
a.
Seesaw, using positions 1, 6, 11, and 12 ( donor only)
Position
z2
x2 – y2
xy
xz
yz
1
6
11
12
Total
1
1
1/4
1/4
2.5
0
0
3/16
3/16
0.375
0
0
9/16
9/16
1.125
0
0
0
0
0
0
0
0
0
0
Trigonal pyramidal, using positions 1, 2, 11, and 12 ( donor only)
Position
z2
x2 – y2
xy
xz
yz
1
2
11
12
Total
1
1/4
1/4
1/4
1.75
0
3/4
3/16
3/16
1.125
0
0
9/16
9/16
1.125
0
0
0
0
0
0
0
0
0
0
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Chapter 10 Coordination Chemistry II: Bonding
b.
Number of d
Electrons
1
2
3
4
5
6
7
8
9
10
10.17
Energies of d Configurations (in units of e)
Seesaw
Trigonal Pyramidal
Low Spin
High Spin
Low Spin
High Spin
0
0
0
0
0.375
0.750
1.875
3
5.5
8
0
0
0.375
1.5
4
4
4
4.375
5.5
8
0
0
0
0
1.125
2.25
3.375
4.5
6.25
8
0
0
1.125
2.25
4
4
4
5.125
6.25
8
c.
The seesaw geometry is favored for d 5—d 9 low spin and d 3, d 4, d 8, and d 9 high
spin configurations. For other configurations, both geometries have identical energies
according to this approach. (The values shown are for the d orbitals only; each
interaction also stabilizes a ligand orbital by e.)
a.
The energies of the molecular orbitals with high d orbital character are as follows:
z2
3.5e
xy
x2-y2
2.25e 
metal
d orbitals
yz
xz
The new positions (13 and 14 below) are opposite 11 and 12, and have the same
values in the table. z2 is affected most strongly, since two ligands are along the z
axis; x2 – y2 and xy are also strongly influenced, since they are in the plane of the six
ligands. xz and yz are not changed, since they miss the ligands in all directions.
Position
z2
x2 – y2
Xy
Xz
yz
1
2
6
11
12
4
13
14
Total
1
1/4
1
1/4
1/4
1/4
1/4
1/4
3.5
0
3/4
0
3/16
3/16
3/4
3/16
3/16
2.25
0
0
0
9/16
9/16
0
9/16
9/16
2.25
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
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148
Chapter 10 Coordination Chemistry II: Bonding
b.
The D6h character table indicates that z2 has A1g symmetry, x2 – y2 and xy have E2g symmetry,
and xz and yz exhibit E1g symmetry.
c.
The energies of the molecular orbitals with high d orbital character are as follows:
z2
3.5e
xy
x2-y2
2.25ee
metal
d orbitals
5e
yz
z2
0
0
0
0
0
0
0
0
0
3.5 e
Position
1
2
6
11
12
4
13
14
Total
Overall
d.
10.18
x2–y2
0
0
0
3/4
3/4
0
3/4
3/4
3 eπ
2.25 e - 3 eπ
xz
xy
0
1
0
1/4
1/4
1
1/4
1/4
3 eπ
2.25 e - 3 eπ
xz
1
1
1
1/4
1/4
1
1/4
1/4
5 eπ
-5 eπ
yz
1
0
1
3/4
3/4
0
3/4
3/4
5 eπ
-5 eπ
If we assume that each low spin configuration will have the maximum number of paired
electrons, then the configurations with asymmetrically occupied degenerate orbitals are
those for d1, d3, d5, and d7 metals. These are expected to give rise to Jahn-Teller
distortion.
The ammine Co(III) complex is considerably more stable and is less easily reduced, with the
difference primarily in the 3+ species. In addition, the metal in [Co(NH3)6]3+ is surrounded
by ammonia molecules, which are more difficult to oxidize than water. This makes transfer of
electrons through the ligand more difficult for the ammine complexes.
[Co(H2O)6]
[Co(H2O)6]2+
d (1s)
d 7 (hs)
–2.4o
–0.8o
o
16,750
8,400
[Co(NH3)6]3+
d 6 (1s)
–2.4o
24,000
–57,600
[Co(NH3)6]2+
d 7 (hs)
–0.8o
10,200
–8,160
LFSE
3+
6
LFSE
–40,200
–6,720
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Difference
33,480
49,440
Chapter 10 Coordination Chemistry II: Bonding
–
149
–
10.19
Cl has the lowest o value and fairly good π donor properties that reduce o. F is next,
with less π donor ability. Water has very small π donor ability (only one lone pair not
involved in  bonding), and ammonia and en have nearly insignificant π donor nor acceptor
ability (no lone pairs, antibonding orbitals with the wrong shapes and energies for π bonding).
–
CN has good π acceptor properties, making o largest for this ligand.
10.20
Ammonia is a stronger field ligand than water. It is a stronger Lewis base ( donor) than water.
Water also has a lone pair that can act as a π donor (which leads to a reduction in the ligand field
splitting). These factors result in the less electronegative nitrogen on ammonia being a better
donor atom than the oxygen atom of water. In the halide ions, all have the same valence
electronic structure, so the electronegativity is the determining factor in ligand field strength.
Fluoride is also a stronger Brønsted base than the other halides.
10.21
a.
Compression moves dz2 up in energy and
lowers the ligand energies of positions 1 and 6.
dz2
eg
d x2–y2
dxz, dyz
t2g
b.
Stretching reverses the changes. In the limit
of a square planar structure, dz2 is affected
only through interactions with the ring
in the xy plane.
dxy
dx2–y2
eg
d z2
dxy
t2g
dxz, dyz
Oh
D4h
10.22
Cr3+ has three singly occupied t2g orbitals and two empty eg orbitals. As a result, its complexes
exhibit no Jahn-Teller distortion. Mn3+ has one electron in the eg orbitals; its complexes show
Jahn-Teller distortion.
10.23
a.
4–
3+
4–
3+
[Co(CO)4]
[Cr(CN)6]
[Fe(H2O)6]
[Co(NO2)6]
[Co(NH3)6]
–
[MnO4]
2+
[Cu(H2O)6]
b.
LFSE (in o)
0

0
2
2.8
–1.6
5
5.9
0
3
3.9
–0.8
0
0
–2.4
0
0
0
1
1.7
–0.6
n
–
0
The two tetrahedral ions ([Co(CO)4]– and [MnO4]–) have zero LFSE (10 or 0 d
electrons, respectively) and have ligand equipped to participate in significant π bonding,
with CO as an acceptor and O2– as donor. With the exception of [Fe(H2O)6]3+, the others
have LFSE values that favor octahedral structures. [Fe(H2O)6]3+ has LFSE = 0 for either
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Chapter 10 Coordination Chemistry II: Bonding
octrahedral or tetrahedral shapes, but water is a slight π donor and Fe(III) is only a
moderately good π acceptor. As a result, electrostatics favors six ligands.
c.
The difference in the LFSEs can be used to assess the relative stabilities of these
geometries for Co(II) and Ni(II):
For Co2+:
High-spin octahedral d 7 has LFSE = –0.8o
Tetrahedral d 7 has LFSE = –1.2t = –0.53o
For Ni2+:
High-spin octahedral d 8 has LFSE = –1.2o
Tetrahedral d 8 has LFSE = –0.6t = –0.27o
Co(II) (d 7) has only –0.27o favoring the octahedral shape, while Ni(II) (d 8) has
–0.93o. Therefore, Co(II) compounds are more likely to be tetrahedral than are
Ni(II) compounds.
10.24
a.
The ligand field stabilization energies and the LFSEO  LFSET differences are below.
h
d
The LFSE is more negative for an octahedral field than the corresponding LFSE for a
tetrahedral field, regardless of the ligand field strength. The magnitude of the difference
( LFSEO  LFSET ) provides some insight into preferences between these geometries for
h
d
these ions. The likelihood of tetrahedral geometry increases as the energy difference
between the octahedral and tetrahedral LFSE decreases. Note that this solution considers
both high and low spin tetrahedral complexes even though high spin configurations are
more common in tetrahedral complexes. The rightmost column provides the difference
between the strong field octahedral LFSE and the weak field tetrahedral configurations.
Fe 2+ (d 6 )
Co2+ (d 7 )
Ni2+ (d 8 )
Octahedral Field
LFSE
Weak
Strong
Tetrahedral Field
LFSE
Weak
Strong
2
 o
5
4
 o
5
6
 o
5
8

45 o
16
 o
45
24
 o
45
12

5 o
9
 o
5
6
 o
5

LFSEO  LFSET
h
48

45 o
36
 o
45
24
 o
45


Weak —
Weak
Strong —
Strong
0.222 o
1.333 o
0.444 o
 o
0.667 o
0.667 o
d
Strong (L.S.)—
Weak (H.S.)
2.22 o
1.44 o
0.667 o
The magnitudes of the LFSE O  LFSET differences predict the preference for
h
d
tetrahedral geometry with weak field ligands to be Fe 2+  Co2+  Ni2+ . The
corresponding preference for tetrahedral geometries with a strong field ligand set is
predicted as Ni2+  Co2+  Fe 2+ . Note that this ranking is the same with strong field
ligands whether the comparison is made to the corresponding high or low spin tetrahedral
configuration (the “strong—strong” and “strong—weak” differences offer the same
preference rankings). These calculations suggest that the octahedral preference for Fe 2+
and Co2+ is even stronger with a strong field ligand set than with a weak field ligand set.
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Chapter 10 Coordination Chemistry II: Bonding
151
This approach predicts that the Ni2+ preference for octahedral geometry is independent
of the ligand field strength.
This LFSE analysis does not match the observation that the number of tetrahedral
complexes for these ions ranks as Co2+  Fe 2+  Ni2+ , but does suggest that Co2+ is
more likely to afford tetrahedral complexes relative to Ni2+ when the ligand field is
weak.
b.
The energies of these electronic configurations on the basis of the angular overlap model
are below. Note that this solution also considers both high and low spin tetrahedral
complexes even though high spin configurations are more common in tetrahedral
complexes. The rightmost column provides the difference between the strong field
octahedral and the weak field tetrahedral electronic configurations, respectively. As in a,
the electronic stabilization afforded by the octahedral field is greater than that provided
by a tetrahedral field, with the preference for an octahedral geometry over tetrahedral
geometry generally more pronounced with strong field ligands (with Ni2+ being an
exception with preferences for octahedral geometry that are independent of the ligand
field strength).
Octahedral
Weak
Strong
Tetrahedral
Weak
Strong
Weak
Oh  Td
Strong
Fe 2+ (d 6 )
6e
12e
4e
5.33e
2e
6.67e
Strong (L.S.)
—Weak (H.S.)
8e
Co 2+ (d 7 )
6e
9e
4e
4e
2e
5e
5e
Ni2+ (d 8 )
6e
6e
2.67e
2.67e
3.33e
3.33e
3.33e
The magnitudes of the Oh  Td angular overlap differences predict the preference for
. The
tetrahedral geometry with weak field ligands to be
corresponding preference for tetrahedral geometries with a strong field ligand set is
predicted as Ni2+  Co2+  Fe 2+ regardless of whether low spin/low-spin (strong—
strong) or low-spin/high spin (strong—weak) configurations are compared.
This angular overlap analysis does not match the observation that the number of
tetrahedral complexes for these ions ranks as Co2+  Fe 2+  Ni2+ , but does suggest that
the tetrahedral preference is
when the ligands are weak.
Neither of these approaches is highly predictive in these cases. The complexity of the
effects that govern geometric preferences cannot be encompassed in any single approach.
10.25
The energy levels of the square planar d orbitals are shown in Figure 10.14; those of the
octahedral orbitals are shown in Figure 10.5. The seventh, eighth, and ninth d electrons in an
octahedral complex go into the highest orbitals, raising the total energy of the complex. In a
square planar complex, the seventh and eighth metal d electrons occupy a relatively nonbonding
orbital. When strong field -acceptor ligands are employed, the  o between the t2g and eg
orbitals increases in octahedral complexes, further stabilizing configurations with 6 or less metal
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152
Chapter 10 Coordination Chemistry II: Bonding
d electrons. In these cases, the energy of the square planar complex may be more favorable even
though the total ligand electron energy is less than in octahedral complexes because there are only
four ligands rather than six.
10.26
a.
Square pyramidal complexes have C4v symmetry.
C4v

A1
B1
E
2C4
1
1
–1
0
E
5
1
1
2
The z2 and x2–y2 orbitals are
the major d orbitals used in
the bonding, so the d orbital
energies are as shown at right.
The z2 orbital is less involved
because it is directed
primarily toward one ligand,
and the x2–y2 orbital is
directed toward four ligands.
C2
1
1
1
–2
2v
3
1
1
0
2 d
1
1
–1
0
dx2–y2
d z2
dxz, dyz
d
dxy

M
b.
D5h

A1
E1
E2
A2
z, z2
x – y2
(x, y), (xz, yz)
2
ML5
5L
Pentagonal bipyramidal complexes have D5h symmetry.
E
7
1
2
2
1
2C5
2
1
2 cos 72°
2 cos 144°
1
2C52
2
1
2 cos 144°
2 cos 72°
1
5C2
1
1
0
0
–1
h
5
1
2
2
–1
2S5
0
1
2 cos 144°
2 cos 72°
–1
2S53
0
1
2 cos 72°
2 cos 144°
–1
5v
3
1
0
0
1
z2
(x, y)
2
(x – y2, xy)
z
The representation  reduces to 2 A1+ E1 + E2 + A2The d orbitals are in three sets,
including a nonbonding pair (dxz and dyz) in addition to the A1 and degenerate E2.
dxy, dx2–y2
dz2
d
dxz, dyz

M
ML7
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7L
Chapter 10 Coordination Chemistry II: Bonding
10.27
153
The angular overlap parameters are (Cl), 5430 cm 1 ; e (Cl), 1380 cm 1 ;
e (PPh3 ), 3340 cm 1 ; e (PPh3 ),  310 cm 1 . Application of these values using Tables 10.10
and 10.11 with the ligand positions arbitrarily defined as shown, provides these results:
Stabilization of Donor Pairs
(7)
1e (Cl)  2e (Cl)  8190 cm1
(8)
1e (PPh3 )  3340 cm1
(9)
1e (PPh3 )  3340 cm
1
1e (PPh3 )  3340 cm
1
(10)
Cl (7)
Co
Ph3P
(8)
PPh3 (10)
PPh3
(9)
The three  acceptor orbitals associated with the PPh 3 ligands are destabilized by
2
e (PPh3 )  207 cm 1 .
3 
The energies of the molecular orbitals with high metal d orbital character are shown below. The
triphenylphosphine and chloride  interactions and the chloride  interactions contribute to the
destabilization of these orbitals, while the  -acceptor capability of triphenylphosphine slightly
lowers the energies of the xy, xz, and yz orbitals.
z2
x2  y2
xy
xz
yz
2
e (Cl)  2e (PPh3 )  300 cm1
3
2
e (Cl)  2e (PPh3 )  300 cm1
3 
1
2
2
1e (PPh3 )  e (Cl)  e (Cl)  e (PPh 3 )  5250 cm1
3
9
3
1
2
2
1e (PPh3 )  e (Cl)  e (Cl)  e (PPh 3 )  5250 cm1
3
9
3
1
2
2
1e (PPh3 )  e (Cl)  e (Cl)  e (PPh 3 )  5250 cm1
3
9
3
(Continued)
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154
Chapter 10 Coordination Chemistry II: Bonding
The energy level diagram is:
207 cm-1
8
9 10
-acceptor orbitals of PPh3
xy xz yz
5250 cm-1
xy xz yz
z2
x2-y2
z2 x2-y2
300 cm-1
7 8
3340 cm-1
8190 cm-1
8
9 10
9 10
Cl- PPh3 PPh3 PPh3
7
The large degree of stabilization of the chloride donor pair relative to the PPh3 pairs is
surprising; triphenylphosphine is generally considered a stronger field ligand relative to chloride.
The steric bulk associated with the coordination of three PPh 3 ligands to the Co(I) center may
attenuate the donor ability (and acceptor ability) of these ligands in CoCl(PPh3 )3 . The increased
-donation of PPh 3 and increased -donation of Cl in NiCl2 (PPh 3 )2 relative to CoCl(PPh3 )3 ,
respectively, is likely linked to the higher oxidation state of Ni(II) relative to the Co(I). The
reduced steric bulk about the Ni(II) center in NiCl2 (PPh 3 )2 relative to around the Co(I) center in
CoCl(PPh3 )3 likely plays a role in increasing the overlap necessary for the -interactions.
10.28
a.
The MO diagram is similar to that for CO, shown in Figure 5.13, with the atomic orbitals
of fluorine significantly lower than the matching orbitals of nitrogen. The NF molecule
has 2 more electrons than CO; these occupy π* orbitals and have parallel spins.
b.
Because its highest occupied sigma orbital (labeled 5a1 for CO) is significantly
concentrated on the nitrogen, NF should act as a strong donor. In addition, because the
π* orbitals are also strongly concentrated on nitrogen, NF should act as a π acceptor at
nitrogen. Because the π* orbitals are singly occupied, NF is likely to be a weaker π
acceptor than CO, whose π* orbitals are empty. Overall, NF is likely to be relatively
high in the spectrochemical series, but lower than CO.
10.29
a.
When these compounds are oxidized, in the positively charged products there is less 
acceptance by the CO ligands than in the neutral compounds. Because CO is acting
less as a  acceptor in the products, the products have less donation into their π*
orbitals, which are antibonding with respect to the C–O bonds. This means that the
oxidation products have stronger and shorter C–O bonds than the reactants. In the
reference cited, the C–O bonds are calculated to be shortened by 0.014 to 0.018 Å in the
PH3 complex and by 0.015 to 0.020 Å in the NH3 complex.
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Chapter 10 Coordination Chemistry II: Bonding
b.
155
In the phosphine compound, oxidation reduces the  acceptance by the phosphine (as it
reduces the  acceptance by the carbonyls), and there is weaker Cr–P bonding.
Therefore, the Cr–P bond is longer in the product (by 0.094 Å).
In the ammine compound, the NH3 is not a  acceptor; it is a  donor only.
Because oxidation increases the positive charge on the metal, there is stronger
attraction between the metal and the sigma-donating NH3, shortening the Cr–N bond
(by 0.050 Å).
2–
10.30
The structure of [ReH9] (Figure 9.35) has a D3h point group.
D3h

A1 
E
A2
E
2C3
0
1
–1
1
–1
E
9
1
2
1
2
h
3
1
2
–1
–2
3C2
1
1
0
–1
0
2S3
0
1
–1
–1
1
3v
3
1
0
1
0
x2 + y2, z2
(x, y) (x2 – y2, xy)
z
(xz, yz)
The representation reduces to 2A1 + 2E + A2 + EThese representations
match atomic orbitals on Re as follows:
A1 
E
(px, py), (dx2– y2, dxy)
A2
pz
E
10.31
a.
s, dz2
(dxz, dyz)
Group orbitals:
H
b.
H
Group orbital–central atom interactions:
d z2
H
s
H
c.
Fe
Fe
pz
H
H
The hydrogen orbitals (potential energy = –13.6 eV) are likely to interact more strongly
with iron orbitals that have a better energy match, the 3d (–11.7 eV) and 4s (–7.9 eV).
The 3p orbitals of iron are likely to have very low energy (ca. –30 eV) and not to interact
significantly with the hydrogen orbitals. (Potential energies for transition metal orbitals
can be found in J. B. Mann, T. L. Meek, E. T. Knight, J. F. Capitani, and L. C. Allen, J.
Am. Chem. Soc., 2000, 122, 5132.)
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156
Chapter 10 Coordination Chemistry II: Bonding
10.32
a. In octahedral geometry the s orbitals on the hydride ligands would generate the
representation:
Oh

A1g
Eg
T1u
E
6
1
2
3
8C3
0
1
–1
0
6C2
0
1
0
–1
3C2(=C42)
2
1
2
–1
6C4
2
1
0
1
i
0
1
2
–3
6S4
0
1
0
–1
8S6
0
1
–1
0
3h
4
1
2
1
6d
2
1
0
1
(2z2–x2–y2, x2–y2)
(x, y, z)
b.
The representation reduces to A1g + Eg + T1u, as shown above.
c.
The matching orbitals on Cr are: A1g: s ; Eg: dx2-y2, dz2 ; T1u: px, py, pz. The orbital
potential energies for the Cr 3d orbitals and H 1s orbitals are -10.75 eV and -13.61 eV,
respectively. The energy compatibility of these orbitals suggests the possibility of a
relatively robust interaction between the Cr dx2-y2, dz2 orbitals and the group orbitals
comprised of hydrogen 1s orbitals.
dx2–y2, dz2
o
3d
dxy, dxz, dyz
s
Eg
Cr
10.33
a.
D4h
b.
1.
Other group orbitals
(A1g, T1u)
CrH6
6H
Because a deuterium atom has only its 1s electron to participate in bonding, there
are only three platinum group orbitals that can potentially interact with
deuterium:
Group Orbitals:
Group Orbital–D Interactions
d z2
pz
s
Pt
D
Pt
Pt
D
Pt
2.
In each case, sigma interactions could occur, as shown above.
3.
The strongest interactions are most likely between the dz2 orbitals of Pt and the 1s
orbital of D. Lobes of the 5dz2 orbitals point toward the D, and their energy
(–10.37 eV) is a good match for the valence orbital potential energy of hydrogen
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Chapter 10 Coordination Chemistry II: Bonding
157
(–13.61 eV). Both bonding and antibonding molecular orbitals would be formed:
D
Pt
D
Pt
Pt
Pt
10.34
MnO4– has no d electrons, while MnO42– has 1. The slight antibonding effect of this electron
is enough to lengthen the bonds. In addition, there is less electrostatic attraction by Mn(VI) in
–
MnO42– than by Mn(VII) in MnO4 .
10.35
The oxidation product is sketched at right, with the Pd—N distances (pm) provided. The longer
Pd—N axial distances (243 pm) relative to the equatorial Pd—N distances (202 and 210 pm)
illustrate the Jahn-Teller distortion. The reference indicates that the HOMO has extensive
character of the Pd 4d 2 orbital. This orbital also features significant sp 3 character at the axial
z
nitrogen atoms.
Using a ligand with more electron-rich axial nitrogen atoms could
possibly increase the energy of the HOMO. One proposal would be the
introduction of electron-releasing methyl substituents at the (now)
methylene carbons  the tertiary amine nitrogen atoms. One ligand
candidate is below.
One challenge in “ligand design” is that modifications of a molecule
will sometimes prevent its subsequent metal binding (for example, if
steric bulk near the donor atoms increases too much). And the
synthesis of new ligands often presents significant research challenges,
even when the desired modification appears minor.
H 3C
CH3
N
N
N
N
H 3C
CH3
Copyright © 2014 Pearson Education, Inc.
N
243
N
210
N
202
Pd
Cl
CH3
243
N
158
10.36
Chapter 10 Coordination Chemistry II: Bonding
The dimeric structure of [MnL1(N 3 )(OCH 3 )]2 is below, with an abbreviated sketch of the Mn(III)
coordinate sphere with bond lengths (pm) at right. The reference states that the three donor atoms
of the meridionally coordinated Schiff base ligand and the bridging oxygen atom 189 pm from
Mn(III) form an equatorial plane in this distorted octahedral geometry. The significantly longer
bond length to the other bridging oxygen atom (223 pm) suggests the presence of Jahn-Teller
distortion, but the possibility of a trans effect of the azide ligand (Section 12.7) must also be
considered.
CH3
N
O
CH3
H
O
N3
N
Mn
N
O
CH3
O
H
O
CH3
Mn
189
O 223
N3
N
189
H3C
199
Mn
N
217
O
210
CH3
N
N3
10.37. The structure of [Pd(tacn)(Htacn)]3+ is surprising since both non-coordinated nitrogen atoms are
positioned on the same side of the plane formed by the four nitrogen atoms bound to the Pd(II)
center (syn). This creates more steric hindrance than would be provided by an anti structure. The
authors suggest that the orientation of these nitrogen atoms maximizes hydrogen bonding with the
nitrate counterions and water in the solid state. An alternate explanation not explicitly stated is an
intramolecular hydrogen bond involving both N(1) and N(1) .
The palladium-nitrogen core of [Pd(tacn)2 ]3+ is sketched at right (bond
lengths in Å with estimated standard deviations), with the slight elongation
in the Pd—N(axial) bonds indicating Jahn-Teller distortion. The three
unlabeled bonds are equivalent to those trans to them by symmetry; they
have the identical bond distances.
10.38
a.
Oh

A1g
Eg
T1u
E
6
1
2
3
N
2.180(9)
N 2.118(9)
2.111(9)
N
Pd
N
N
N
8C3
0
1
–1
0
6C2
0
1
0
–1
6S4
2
1
0
1
3C2(=C42)
2
1
2
–1
i
0
1
2
–3
6S4
0
1
0
–1
8S6
0
1
–1
0
3h
4
1
2
1
6d
2
1
0
1
b.
The representation reduces to A1g + Eg + T1u, as shown above.
c.
The matching orbitals on Ti are: A1g: s ; Eg: dx2-y2, dz2 ; T1u: px, py, pz
Copyright © 2014 Pearson Education, Inc.
(2z2–x2–y2, x2–y2)
(x, y, z)
Chapter 10 Coordination Chemistry II: Bonding
159
d.
dx2–y2, dz2
o
3d
dxy, dxz, dyz
s
Eg
Ti
10.39
10.40
10.41
[TiH6]2–
Other group orbitals
(A1g, T1u)
6H
a.
The t2g molecular orbitals should involve titanium’s dxy, dxz, and dyz orbitals, the the eg
orbitals should involve the metal’s dx2-y2, dz2 orbitals.
b.
A difference from Figure 10.5 that should be noted is that the t2g orbitals of the metal are
no longer nonbonding in TiF63– but are involved in the formation of bonding and
antibonding molecular orbitals. In comparison with Figure 10.7, the t2g orbitals with
greatest metal character should form molecular orbitals that are higher rather than lower
in energy in comparison with the dxy, dxz, and dyz orbitals of the free metal ion.
a.
The iron orbitals matching the representation e are the dx2‐y2 and dz2 ; the orbitals matching
t2 are the dxy, dxz, and dyz. The d orbital–ligand orbital interactions should appear more
indirect than in the case of an octahedral transition metal complex because the lobes of
these orbitals do not point directly toward the ligands.
b.
The results for FeCl4– should be simpler than shown in Figure 10.19 because the chloro
ligand does not have π* orbitals to be taken into consideration (the LUMO orbitals of
CO).
Results may vary considerably, depending on the software used. It is probably more likely
that selecting the π-acceptor ligand (CN–), a -donor, and a π-donor will show a trend matching
that in Table 10.12 than if a set of three of  -donor or π-donor ligands is used—but examining
each type can be instructive.
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160
Chapter 11 Coordination Chemistry III: Electronic Spectra
CHAPTER 11: COORDINATION CHEMISTRY III: ELECTRONIC
SPECTRA
11.1
a.
p3
There are (6!)/(3!3!) = 20 microstates:
MS
–3/2
+2
+1
ML
0
1
–
–1
–2
–
0 –1
–
–1/2
–
–
+
1 1 0
–
–
1+ 1 –1
–
–
0+
1 0
–
–
+
1 0 –1
–
–
1 0+ –1
–
–
+
1 0 –1
–
–
–1+ –1 1
–
–
–1 0 0+
–
–
–1+ –1 0
1/2
–
1 1 0+
–
1+ 1 –1+
–
1+ 0 0+
–
+
+
1 0 –1
–
1+ 0 –1+
–
1 0+ –1+
–
–1+ –1 1+
–
+
–1 0 0+
–
–1+ –1 0+
3/2
+
1+ 0+ –1+
Terms: L = 0, S = 3/2: 4S (ground state)
L = 2, S = 1/2: 2D
L = 1, S = 1/2: 2P
b.
p1d1 There are
6!

10!
1!5! 1!9!
–1
–
–
1 2
–
–
1 1
–
–
0 2
–
–
1 0
–
–
0 1
–
–
–1 2
–
–
1 –1
–
–
0 0
–
–
–1 1
–
–
–1 0
–
–
0 –1
–
–
1 –2
–
–
–1 –1
–
–
0 –2
–
–
–1 –2
3
2
1
ML
0
–1
–2
–3
Terms: L = 3, S = 1
L = 3, S = 0
L = 2, S = 1
 60 microstates:
3
F (ground state)
F
3
D
1
MS
0
– +
–
1 2 , 1+ 2
–
–
1+ 1 , 1 1+
– +
–
0 2 , 0+ 2
–
–
1+ 0 , 0+ 1
– +
– +
1 0, 0 1
–
–
–1 2+, –1+ 2
–
–
1+ –1 , 1 –1+
–
–
0 0+, 0+ 0
–
– +
+
–1 1 , –1 1
–
–
–1+ 0 , 0+ –1
– +
–
–1 0 , 0 –1+
–
–
1 –2+, 1+ –2
+
–
–
–1 –1 , –1 –1+
–
–
0+ –2 , 0 –2+
–
–
–1 –2+, –1+ –2
L = 2, S = 0
L = 1, S = 1
L = 1, S = 0
1
1 2+
1+ 1+
0+ 2+
1+ 0+
0+ 1+
–1+ 2+
+
1+ –1+
0+ 0+
–1+ 1+
–1+ 0+
0+ –1+
1+ –2+
–1– –1+
0+ –2+
–1+ –2+
1
D
P
1
P
3
The two electrons have quantum numbers that are independent of each other, because
the electrons are in different orbitals. Because they have different l values, the electrons
can have the same ml and ms values.
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Chapter 11 Coordination Chemistry III: Electronic Spectra
11.2
For p3: L = 0, S = 3/2, the term 4S has J = 3/2 only (|L+S| = |L–S|). Therefore, the ground
state is 4S3/2.
For p1d1: L = 3, S = 1, the term 3F has J = 4, 3, 2. Since both levels are less than half filled,
the state having lowest J has lowest energy, and the ground state is 3F2.
2! 10!

 20 microstates:
11.3
a.
s1d1
There are
1!1! 1!9!
MS
No index entries found.
mmmm
M
–1
0
+1
+2
0– 2–
0 – 2 +, 0 + 2 –
0+ 2+
+1
0– 1–
0 – 1 +, 0 + 1 –
0+ 1+
–
–
–
+
+
–
0
0 0
0 0, 0 0
0+ 0+
ML
–1
0– –1–
0– –1+, 0+ –1–
0+ –1+
–
–
–
+
+
–
–2
0 –2
0 –2 , 0 –2
0+ –2+
b.
Terms: L = 2, S = 1: 3D; L = 2, S = 0: 1D
The 3D, with the higher spin multiplicity, is the lower energy term.
10! 14!

 140 microstates:
11.4
a.
d1f1
There are
1!9! 1!13!
MS
No index entries found.
mmmm
M
–1
0
+5
2– 3–
2– 3+, 2+ 3–
2– 2–
2– 2+, 2+ 2–
+4
1– 3–
1– 3+, 1+ 3–
– –
2 1
2– 1+, 2+ 1–
– –
+3
1 2
1– 2+, 1+ 2–
– –
0 3
0– 3+, 0+ 3–
– –
2 0
2– 0+, 2+ 0–
1– 1–
1– 1+, 1+ 1–
+2
– –
0 2
0– 2+, 0+ 2–
– –
–1 3
–1– 3+, –1+ 3–
–
–
2– –1+, 2+ –1–
2 –1
–
–
1 0
1– 0+, 1+ 0–
+1
0– 1–
0– 1+, 0+ 1–
–
–
–1 2
–1– 2+, –1+ 2–
–
–
–2 3
–2– 3+, –2– 3–
–
–
2 –2
2– –2+, 2+ –2–
ML
–
–
1 –1
1– –1+, 1+ –1–
–
–
0
0 0
0 – 0 +, 0 + 0 –
–1– 1–
–1– 1+, –1+ 1–
–
–
–2 2
–2– 2+ , –2+ 2–
–
–
–2 1
–2– 1+, –2+ 1–
–
–
–1 0
–1– 0+, –1+ 0–
–
–
–1
0 –1
0– –1+, 0+ –1–
1– –2–
1– –2+, 1+ –2–
–
–
2 –3
2– –3+, 2– –3–
c.
continued
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+1
2+ 3+
2+ 2+
1+ 3+
2+ 1+
1+ 2+
0+ 3+
2+ 0+
1+ 1+
0+ 2+
–1+ 3+
2+ –1+
1+ 0+
0+ 1+
–1+ 2+
–2– 3+
2+ –2+
1+ –1+
0+ 0+
–1+ 1+
–2+ 2+
–2+ 1+
–1+ 0+
0+ –1+
1+ –2+
2– –3+
161
162
Chapter 11 Coordination Chemistry III: Electronic Spectra
–2
–3
–4
–5
11.5
–2– 0–
–1– –1–
0– –2–
1– –3–
–2– –1–
–1– –2–
0– –3–
–2– –2–
–1– –3–
–2– –3–
–2– 0+ , –2+ 0–
–1– –1+, –1+ –1–
0– –2+, 0+ –2–
1– –3+, 1+ –3–
–2– –1+, –2+ –1–
–1– –2+, –1+ –2–
0– –3+, 0+ –3–
–2– –2+, –2+ –2–
–1– –3+, –1+ –3–
–2– –3+, –2+ –3–
–2+ 0+
–1+ –1+
0+ –2+
1+ –3+
–2+ –1+
–1+ –2+
0+ –3+
–2+ –2+
–1+ –3+
–2+ –3+
b.
Terms: L = 5, S = 1: 3H; L = 5, S = 0: 1H; L = 4, S = 1: 3G; L = 4, S = 0: 1G;
L = 3, S = 1: 3F; L = 3, S = 0: 1F ; L = 2, S = 1: 3D; L = 2, S = 0: 1D;
L = 1, S = 1: 3P; L = 1, S = 0: 1P
c.
The lowest energy term is the 3H. For this term, J has the values 6, 5, and 4. Because
the subshells are less than half full, the lowest value of J provides the lowest energy:
3
H4.
a.
From Problem 11.1a, for a p3 configuration there are three terms: 4S, 2D, and 2P.
The J values for each of these are determined below.
2P
For 4S: L = 0, S = 3/2 ;
Because J = L + S, L + S – 1,
…|L – S|, the quantum number J
can only be 3/2 and there is a
single state for 4S: 4S3/2
For 2D: L = 2, S = 1/2.
28839.31
28838.92
c
2D
19233.18
19224.46
 e
Possible J values are 5/2 and
3/2, and the two possible states
are 2D5/2 and 2D3/2.
4S
For 2P: L = 1, S = 1/2. Possible J values are 3/2 and 1/2, with states 2P3/2 and 2P1/2.
The lowest energy state is 4S3/2 (highest multiplicity). 2D5/2 and 2D3/2 are next, at
19233.18 and 19224.46 cm–1, and 2P3/2 and 2P1/2 are the highest energy at 28838.92
and 28839.31 cm–1.
b.
The difference in energy between the 4S and 2D states is 2e. From the average of
the two nearly degenerate 2D states, e = 9614.41 cm–1.
The difference in energy between the averages of the 2D and 2P states is
c = 28839.12 – 19228.82 = 9610.30 cm–1.
Copyright © 2014 Pearson Education, Inc.
0
Chapter 11 Coordination Chemistry III: Electronic Spectra
11.6
a.
s1f1 There are
2!

14!
1!1! 1!13!
 28 microstates:
MS
0
–1
11.8
+
–
2
–
1
–
0
–
0 3,
0 3
0
–
0 2+,
–
0+ 2
–
0+ 2+
–
0 1+,
–
0+ 1
–
0+ 1+
–
0
–
0 +, 0 + 0
–
0+ 0+
–
0+ –1
–
0+ –1+
+
+
3+
0
2
0
1
0
0
0
–1
0 –1
–
–
0 –1+,
–2
0 –2
–
–
0 –2+, 0+ –2
–
–
0+ –2+
–3
0 –3
–
–
0 –3+, 0+ –3
–
–
0+ –3+
3
F (ground state)
F
b.
Terms: L = 3, S = 1
L = 3, S = 0
c.
The 3F term, with the higher spin multiplicity, has the lower energy. This term has
J = 2, 3, 4; the lowest energy term, including J, is 3F2.
a.
2
D has L = 2 and S = 1/2, so ML = –2, –1, 0, 1, 2 and MS = –1/2, 1/2
b.
3
G has L = 4 and S = 1, so ML = –4, –3, –2, –1, 0, 1, 2, 3, 4 and MS = –1, 0, 1
c.
4
F has L = 3 and S = 3/2, so ML = –3, –2, –1, 0, 1, 2, 3 and MS = –3/2, –1/2, 1/2, 3/2
a.
2
D with J = 5/2, 3/2 fits an excited state of d3, 2D3/2
b.
3
G with J = 5, 4, 3 fits an excited state of d4, 3G3
c.
4
F with J = 9/2, 7/2, 5/2, 3/2 fits the ground state of d7, 4F9/2
1
L
A  0.10 l  1.00cm
mol cm
A
mol
0.10
A  lc; c  
 2.6
L 
L
l 
0.038
1.00cm

mol cm 
  0.038
11.9
11.10
3
–
1
–
3
ML
11.7
–
a.
__
  24,900 cm–1

1


1
24,900 cm
–1
 4.02  10 –5 cm = 402 nm
Copyright © 2014 Pearson Education, Inc.
163
164
Chapter 11 Coordination Chemistry III: Electronic Spectra
 2.998  10 m s 100 cm m   7.46  10 s
 
c

b.
8
–1
–1
14 –1
–5
4.02  10 cm
  366 nm
2.998 10 m s 10 nm m   8.19 10 s
 
c
8
–1

366 nm
9
–1
14 –1
E  h  (6.626 10 –34 Js)(8.19 1014 s –1 )  5.4310 –19 J
11.11
J values are included in these answers:
a.
d 8 Oh MS = 1 = S Spin multiplicity = 2 + 1 = 3
Max ML = 2+2+1+1+0+0–1–2 = 3 = L, so F term.
b.
d 5 Oh high spin MS = 5/2 = S Spin multiplicity = 5 + 1 = 6
Max ML = 2+1+0–1–2 = 0 = L, so S term.
J = 5/2
3
J = 4,3,2
6
F4
S5/2
d 5 Oh low spin MS = 1/2 = S Spin multiplicity = 1 + 1 = 2
Max ML = 2+2+1+1+0 = 6 = L, so I term.
J = |L ± S| = 11/2, 13/2 2I11/2, 2I13/2
(J is uncertain in this case; the usual rule does not apply because the level is exactly
half full)
11.12
c.
d 4 Td MS = 2 = S Spin multiplicity = 4 + 1=5
Max ML = 2+1+0–1 = 2 = L, so D term.
d.
d 9 D4h MS = 1/2 = S Spin multiplicity = 1+1=2
Max ML = 2+2+1+1+0+0–1–1–2 = 2 = L, so D term.
a.
[M(H2O)6]2+ There are two possibilities, M = Ni (d8) and M = Ti (d2). The complex
[Ni(H2O)6]2+ is well known. However, titanium strongly prefers the 3+ oxidation state,
and [Ti(H2O)6]2+ has not been well characterized. A third possibility would be low spin
[Cr(H2O)6]2+ —if this complex were low spin. However, stronger field ligands than H2O
are necessary for complexes of Cr(II) to be low spin.
b.
[M(NH3)6]3+ There are several possibilities, for M = Ti (d1), V (d2), Cr (d3), and Co
(d6), among first row transition metals that commonly exhibit a 3+ oxidation state.
J = 4, 3, 2, 1, 0
5
D0
J = |L ± S| = 5/2, 3/2
2
D5/2
Excitation of the single d electron from the t2g to eg levels in [Ti(NH3)6]3+ leads to
asymmetric occupation of the eg, a configuration susceptible to Jahn-Teller distortion
(Section 10.5). Like [Ti(H2O)6]3+ (Figure 11.8), [Ti(NH3)6]3+ shows splitting of its
absorption band (excitation from 2T2g to 2Eg (Figure 11.11)).
Similarly, in [V(NH3)6]3+, excitation of an electron from t2g to eg would give asymmetric
occupation of the eg, a configuration that could potentially give rise to distortion and
splitting of absorption bands. However, spectra of d3 octahedral complexes typically do
not show such splitting (see spectrum of [V(H2O)6]3+ in Figure 11.8), but have broad
overlapping bands that dominate the visible spectra.
Copyright © 2014 Pearson Education, Inc.
Chapter 11 Coordination Chemistry III: Electronic Spectra
165
Excitation of a t2g electron to an eg level in the d3 complex [Cr(NH3)6]3+ leads to
asymmetric occupation of the eg, potentially giving rise to band splitting in the absorption
spectrum.
The low-spin d6 complex [Co(NH3)6]3+ does show splitting consistent with distortion of
the excited state resulting from excitation of a t2g electron to the previously empty eg
level. Such splitting is also observed in the d6 iron (II) complex [Fe(H2O)6]2+ (Figure
11.8).
The last three transition metals in this row, Ni, Cu, and Zn, do not exhibit stable 3+
complexes of formula [M(NH3)6]3+. Comparable complexes for Mn and Fe have not been
well characterized.
c.
[M(H2O)6]2+ If M = Zn, all t2g and eg orbitals are filled, and consequently no d–d
transitions are possible. As a consequence, the d10 complex [Zn(H2O)6]2+ is colorless.
The d5 complex [Mn(H2O)6]2+ is nearly colorless (very pale pink) because it has
no excited state of the same spin multiplicity (6) as the ground state (see Tanabe-Sugano
diagram in Figure 11.7). The electronic spectrum (Figure 11.8) shows absorption bands
approximately two orders of magnitude smaller than for other first row [M(H2O)6]2+
complexes.
11.13
[Ni(H2O)6]2+ For d 8 ions, the energy of the lowest energy band is o, so o = 8,700 cm–1.
The bands are split due to Jahn-Teller distortion in the excited state.
11.14
a.
[Cr(C2O4)3]3– is Cr(III), d 3. o is equal to the lowest energy band, so o = 17,400 cm–1.
b.
[Ti(NCS)6]3– is Ti(III), d 1. o is the energy of the single band; o = 18,400 cm–1.
The band is split due to Jahn-Teller distortion of the excited state.
c.
[Ni(en)3]2+ is Ni(II), d 8. The lowest energy band corresponds to o; o = 11,200 cm–1.
d.
[VF6]3– is V(III), d 2. Following the example on pp. 427 - 428, we find the ratio 2/1
and then o /B: 2/1 = 1.57 at o /B = 26. From the Tanabe-Sugano diagram at
o /B = 26,
1: E/B = 24.1
E = 24.1 B = 14,800 cm–1
B = 614 cm–1
–1
2: E/B = 37.0
E = 37.0 B = 23,250 cm
B = 628 cm–1
–1
–1
Average B = 621 cm , o = 26 B = 16,100 cm (On the basis of the data in this and
following problems, the values of B and o should be rounded to two significant digits;
additional digits are shown here to assist in checking calculations.)
e.
V(III) is a d 2 ion. Again following the example on pp. 427 - 428, 2 = 21,413 cm–1
and 1 = 14,409 cm–1, 2/1 = 1.49. From the Tanabe-Sugano diagram at o/B = 34.5,
1: E/B = 29
E = 29 B =14,409 cm–1
B = 497 cm–1
–1
2: E/B = 44
E = 44 B = 21,413 cm
B = 487 cm–1
Average B = 492 cm–1, o = 34.5 B = 16,970 cm–1
11.15
[Co(NH3)6]2+, d 7. As in Problem 11.14d:
2/1 = 2.34 at o/B = 11. From the Tanabe-Sugano diagram at o = 11,
1: E/B = 10
E= 10 B = 9,000 cm–1
B = 900 cm–1
–1
2: E/B = 22.5
E= 22.5 B = 21,100 cm
B = 938 cm–1
Average B = 919 cm–1, o = 11 B = 10,100 cm–1
Copyright © 2014 Pearson Education, Inc.
166
Chapter 11 Coordination Chemistry III: Electronic Spectra
11.16
a.
t2g4eg2 The t2g level is a triply degenerate asymmetrically occupied state, so it is T.
b.
t2g6
c.
t2g3eg3 This is an excited state, with the t2g level uniformly occupied and the eg
level doubly degenerate, so it is E.
t2g5
This is a triply degenerate state, T. (Vacancies in the orbitals can be treated
similarly to electrons.)
d.
e.
eg
This is a nondegenerate state, completely occupied, so it is A.
Another excited state, this is doubly degenerate, E.
11.17
The complexes with potential degeneracies are those with d 1, d 2, d 4, d 7 and d 9, low-spin d 5,
and high-spin d 6 configurations. The strongest effects are with high-spin d 4, low-spin d 7, and
d 9 complexes, corresponding to [Mn(NH3)6]3+, [Ni(NH3)6]3+, and the unknown [Zn(NH3)6]3+.
Weaker effects might be seen with [Ti(NH3)6]3+, [V(NH3)6]3+, low-spin [Fe(NH3)6]3+, and highspin [Co(NH3)6]3+, all of which are unknown or unstable. Other ligands are needed to stabilize
ions containing these 3+ metal ions.
11.18
The 5d orbitals of Re are higher in energy than the 3d orbitals of Mn, so an LMCT excitation
–
requires more energy for ReO4 . In addition, since the molecular orbitals derived primarily from
–
–
–
the 3d orbitals of MnO4 are lower in energy than the corresponding MO’s of ReO4 , MnO4 is
better able to accept electrons; it is a better oxidizing agent.
11.19
The order of energy of the charge transfer bands is I < Br < Cl. LMCT in low-spin d 6
[Co(NH 3 )5X]2+ can be approximated (since this complex does not have Oh symmetry) as
excitation into the LUMO (empty eg orbitals of high z 2 and x 2  y 2 character) from lower energy,
relatively nonbonding (or weakly bonding -donor) orbitals with high halide valence orbital
character. It is reasonable to approximate the LUMO energies as very similar for this series of
Co(III) cations. However, the lower-energy orbitals with high halide valence orbital character will
be lowest for the most electronegative chloride, and highest for the least electronegative iodide. This
difference results in [Co(NH 3 )5I]2+ having the lowest-energy predicted LMCT band.
From an HSAB perspective, iodide is softer and can lose an electron most easily in an LMCT process.
11.20
Comparing [Fe(CN)6]3– (low spin d 5) and [Fe(CN)6]4– (low spin d 6), where CN– is a  donor
and a π acceptor: the t2g orbitals of [Fe(CN)6]3– contain 5 electrons, allowing LMCT from
the ligand orbitals to either t2g or eg levels. The t2g levels of [Fe(CN)6]4– are full, so only the
higher eg levels are available for LMCT. MLCT transitions (t2g  π*) are also
possible for either complex.
11.21
a.
O2– and Cl– are both  and π donors, and the metal ions are Cr(V) and Mo(V) (the
ligands are Cl– and O2–). Metal d orbitals, influence of Cl– ligands:
z2
2
x – y2
xy
xz
yz
e
2
3
0
0
0
eπ
0
0
4
3
3
Copyright © 2014 Pearson Education, Inc.
Total
2e
3e
4eπ
3eπ
3eπ
Chapter 11 Coordination Chemistry III: Electronic Spectra
167
Metal d orbitals, influence of O2– ligands:
z2
x2 – y2
xy
xz
yz
e
1
0
0
0
0
E
1
1
1
2
2C4
1
–1
–1
0
C4v
A1
B1
B2
E
eπ
0
0
0
1
1
C2
1
1
1
–2
Total
e
0
0
eπ
eπ
2v
1
1
–1
0
2d
1
–1
1
0
z2
x – y2
xy
(xz, yz)
2
Overall, from lower energy (top) to higher energy (bottom), the d orbital energies are:
xy
xz and yz
z2
x2 – y2
4eπ(Cl)
3eπ(Cl) + eπ(O)
2e(Cl) + e(O)
3e(Cl)
b.
The symmetry labels of the orbitals are given in the C4v character table.
c.
The lowest d orbital has an energy of 2eπ(Cl); the next have energies of 2eπ(Cl) +
eπ(O). Because these are d 1 complexes, the transitions given provide the HOMO/LUMO
gaps in these complexes. The interactions between metal and ligand are generally
stronger for a second row transition metal than for the first row, raising the LUMO
energy in the Mo case. It is reasonable to expect the larger Mo(V) to offer better 
overlap with the ligand donor orbitals relative to the smaller Cr(V).
11.22
[V(CO)6]– < Cr(CO)6 < [Mn(CO)6]+ As the nuclear charge on the metal increases, the metal
orbitals are drawn to lower energies. Consequently, the MLCT bands should increase
in energy.
11.23
a.
 s  0.65  n n  2 
At 80K:
 s  5.2  n n  2 
At 300K:
b.
11.24
a.
n = 0.19
n = 4.3
The complex is near the low-spin – high-spin boundary of the d 6 Tanabe-Sugano
diagram. High spin becomes increasingly favored as the temperature increases.
ML2, using positions 1 and 6, with O2– both a  and π donor:
2
z
x2 – y2
xy
xz
yz
e
2
0
0
0
0
eπ
0
0
0
2
2
Copyright © 2014 Pearson Education, Inc.
Total
2e
0
0
2eπ
2eπ
168
Chapter 11 Coordination Chemistry III: Electronic Spectra
b.
c.
11.25
If this is a high-spin complex, there are 4 electrons in the lowest
levels (xy, x2 – y2), 3 in the next two (xz, yz), and 1 in the highest
(z2). Electronic transitions can
be either from the middle levels to the top, and from the bottom
levels to the middle and the top—three possibilities in all.
According to the reference, the transitions seen are from the middle
and the bottom levels to the top level.
2e
1
2e
2
Assigning the transitions as in part b:
E = 2e = 16,000 cm–1 E = 2eπ = 9,000 cm–1
so e = 8,000 cm–1 and eπ = 4,500 cm–1
The reference provides a much more detailed analysis, including a discussion of the
paramagnetism of this complex and other factors related to its electronic spectrum.
Re(CO)3(P(OPh3))(DBSQ)
Re(CO)3(PPh3)(DBSQ)
Re(CO)3(NEt3)(DBSQ)
18,250 cm–1
17,300 cm–1
16,670 cm–1
NEt3 is the strongest donor ligand in this series. Therefore, the metal in the complex
Re(CO)3(NEt3)(DBSQ) has the greatest concentration of electrons and the greatest tendency for
electron transfer to acceptor orbitals. Since this complex also has the lowest-energy charge
transfer band, we may assign this as MLCT.
11.26
11.27
a.
RuO42– has the highest value of t. t increases with the oxidation state of the metal
and in general is greater for second row than for first row metals. The overall trend
is RuO42– > FeO42– > MnO43– > CrO44–.
b.
The nuclear charge of iron is greatest in this isoelectronic series and exerts the
strongest attraction for bonding electrons. As a result, FeO42– has the shortest
metal-oxygen distance, 165 pm, in comparison with 170 pm for MnO43– and 176 pm
for CrO44– and RuO42–.
c.
As the nuclear charge of the metal increases, the metal orbitals’ energies are decreased
(further stabilized). Consequently, less energy is needed to excite electrons from ligand
orbitals to these metal orbitals. These are LMCT absorptions.
Aqueous solutions of Ni(NO3)2 contain the green [Ni(H2O)6]2+ ion; nitrate is the counterion.
Addition of aqueous NH3 replaces the H2O ligands in [Ni(H2O)6]2+ to give blue [Ni(NH3)6]2+.
As a bidentate ligand, en can replace two NH3 ligands; three en ligands can therefore replace
all six NH3 ligands to form violet [Ni(en)3]2+:
2+
observed color:
complementary color:
[Ni(H2O)6]
green
red
NH3
2+
[Ni(NH3)6]
blue
orange
en
[Ni(en)3]2+
violet
yellow
The complementary colors in this series have increasing energies, indicating that en has the
strongest effect on o, and H2O has the weakest effect. This is consistent with the positions of
these ligands in the spectrochemical series.
Copyright © 2014 Pearson Education, Inc.
Chapter 11 Coordination Chemistry III: Electronic Spectra
11.28
a.
169
These colors are most likely the consequence of LMCT transitions, from orbitals
that are primarily from the oxide ligands to orbitals that are primarily from the metal:
t2
e
d
LMCT
MO4–
M7+
11.29
–
b.
In TcO4 , the separation between the donor orbitals of the O2– ligands and the
–
–
acceptor orbitals is greater than in MnO4 . As a consequence, TcO4 absorbs light
–
of higher energy (green) than MnO4 (yellow). Actually, most of the absorption by
–
TcO4 is in the ultraviolet, with the pale red color a result of a tail of the absorption band
extending into the visible.
c.
The metal-ligand interactions in MnO42– (Mn(VI)) are weaker than in MnO4
(Mn(VII)), and the separation of donor and acceptor orbitals in MnO42– is smaller,
–
meaning that less energy (red light) is necessary for excitation than in MnO4 (yellow).
2–
Also worth noting: the Mn—O bond distance is longer in MnO4 (165.9 pm) than in
–
MnO4 (162.9 pm), an indication of weaker bonding in the former.
a.
At 350 nm:
A
2.34
 
lc 1.00cm 2.00 10 –4 M
–

–1

 2,660 L mol –1 cm –1
At 590 nm:
A
0.370
 
lc 1.00cm 2.00 10 –4 M
At 1540 nm:
A
0.0016
 
lc 1.00cm 2.00 10 –4 M

–1
 1,850 L mol cm
b.

–1
At 514 nm:
A
0.532
 
lc 1.00cm 2.00 10 –4 M
 11,700 L mol cm
–1
11.30
4O2–


–1


–1
 8.0 L mol cm
Because of their high intensity, the bands at 350, 514, and 590 nm are probably
charge transfer bands. However, the low molar absorptivity of the band at 1540 nm
indicates that it is probably a d–d transition (see examples in Figure 11.8).
These are all d 8 complexes, with three excited states of the same spin multiplicity as the ground
state. For d 8, o = energy of the lowest energy band. B can be calculated by using the method of
Problem 11.14. o is the difference between the transitions 3A2  3T1 and 3A2  3T2; a graph of
2/1 versus o/B needs to be prepared for d 8 Ni2+ in order to calculate B, as described on pages
427 - 428. A plot of the ratio of the highest energy band to the lowest energy band is shown on
the next page.
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170
Chapter 11 Coordination Chemistry III: Electronic Spectra
Species
[Ni(H2O)6]2+
[Ni(NH3)6]2+
[Ni(OS(CH3)2)6]2+
[Ni(dma)6]2+
11.31
o/B
8.8
12.5
8.4
8.0
Ratio
3.06
2.62
3.11
3.14
o(cm–1)
8500
10500
7728
7576
B(cm–1)
970
840
920
950
a.
These are both high-spin d 7 complexes, for which the ground-state term symbol is 4F
(see Figure 11.7)
b.
There are three possible transitions, all originating from the 4T1(4F) and going to the
4
T2, 4A2, and 4T1(4P) levels.
c.
For [Co(bipy)3]2+, the ratio
comes at
 2 22,000 cm–1

 1.95. From Figure 11.14, this
1 11,300 cm–1
o
 17, at which:
B
for the 22,000 cm–1 band:
E
22,000 cm1
 28 and B 
 786 cm1
B
28
for the 11,300 cm–1 band:
E
11,300 cm1
 15 and B 
 754 cm1
B
15
Average B = 770 cm–1;
o
 17; o = 13,100 cm–1
B
4
For a d 7 complex, LFSE = –  o  –10,500 cm–1
5
o for [Co(NH3)6]2+ was calculated in Problem 11.15; o = 10,100 cm–1.
4
For this d 7 complex, LFSE = –  o  –8,080 cm–1
5
d.
These bands should be broad (see Figure 11.8).
e.
The molecular orbital energy level diagrams should be similar to Figure 10.5, with 6
electrons in t2g orbitals and 1 electron in an eg level in each case. The separation
between t2g and eg orbitals should be larger in the bipy complex.
Copyright © 2014 Pearson Education, Inc.
Chapter 11 Coordination Chemistry III: Electronic Spectra
11.32
These absorption bands correspond to LMCT transitions. As the nuclear charge increases
(Fe
Ni), the acceptor (largely d) orbitals decrease in energy, enabling the charge transfer
transitions to occur at lower energy in the nickel complex.
The target complex structure is at right. The conjugated
phenylacetlyene linker is first and foremost a conjugated
system to allow electronic communication between the
separated metal fragments; the linker is to permit the
OC
CO
OC
11.33
171
Fe
CO
Fe
CO
OC
S
S
N
delocalization of the [Ru(terpy)2 ]2+ MLCT excited state
towards the iron atoms. The reference mentions
literature precedence for the effectiveness of
phenylacetlyene linkers in increasing covalently attached
[Ru(terpy)2 ]2+ excited state lifetimes, advantageous to
facilitate effective transfer to the iron-sulfur fragment.
The rigidity of the linker permits precise control over the
distance between the Fe and Ru centers, another way to
modulate the effectiveness of the electronic
communication between these metal centers. Finally,
from a practical standpoint, the acetylene substituent in
an intermediate serves as a convenient functional group
in the multi-step synthesis of the complex sketched.
The most intense bands in the provided spectrum are
2+
1-
F
F
F
P
F
F
F
1-
F
F
F
P
N
F
F
N
F
N
Ru
N
N
N
assigned to    * intraligand transitions of terpyridine
(310 nm) and MLCT of the covalently attached [Ru(terpy)2 ]2+ fragment (500 nm). The complex
above is unable to induce proton reduction since the intramolecular electron transfer from the
[Ru(terpy)2 ]2+ excited state (via MLCT) to the iron-sulfur fragment is not thermodynamically
favorable. The authors speculate on various strategies to render this electron transfer favorable,
including substitution of the carbon monoxide molecules with different ligands.
11.34
The excellent electron withdrawing ability of the imide function is hypothesized to stabilize the
resulting reduced dithiolate diiron complex upon electron transfer from the photogenerated zinc
porphyrin excited state. This proposed stabilization is expected to render the electron transfer
from the zinc photosensitizer more thermodynamically favorable. The well-established utility of
the imide functional group for the coupling reactions necessary to synthesize the complex is also
advantageous.
The lack of ground state electronic interaction between the naphthalene monoimide (NMI)
dithiolate-tethered zinc porphyrin and the iron centers was inferred by infrared spectroscopy. The
infrared spectrum in the carbonyl region of the NMI ditholate diiron complex is essentially
unchanged upon covalent attachment of the zinc porphyrin. The presence of this zinc substituent
has an insignificant impact on the electron density at the iron centers.
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172
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
CHAPTER 12: COORDINATION CHEMISTRY IV: REACTIONS AND
MECHANISMS
12.1
[Cr(H2O)6]2+ is labile, and has 4 unpaired electrons, with 1 in the anti-bonding eg orbital.
Occupation of this orbital renders substitution easier by leading to relatively weak chromium(II)–
aqua ligand bonds. [Cr(CN)6]4– is inert. It has all 4 metal valence electrons in the bonding t2g
levels. These orbitals are rendered bonding in character due to π-backbonding with the cyanide
ligands.
12.2
If the rate constants for different entering ligands are significantly different, it suggests that
increasing the coordination number is important in the rate-determining step. A necessary
assumption to make this assertion is that the mechanism via which the complex undergoes
substitution does not vary with the entering ligand.
12.3
Pentachlorooxochromate(V) is a d 1 complex; it should be labile, with vacancies in the t2g levels.
Hexaiodomanganate(IV) is a d 3 complex; it should be inert.
Hexacyanoferrate(III) is a low-spin d 5 complex; it should be inert (vacant eg levels).
Hexammineiron(III) is a d 6 high-spin complex; with partly occupied orbitals in both levels; it
should be labile.
12.4
The [Fe(CN)6]4– ion is a low-spin d6 complex, with a maximum LFSE of –2.4 o. It is a notably
kinetically inert complex, hence its low reactivity towards ligand substitution that would release
the potentially toxic cyanide.
12.5
[Fe(H2O)6]3+ and [Co(H2O)6]2+ are high-spin species; the electrons in the upper eg levels render
them labile. [Cr(CN)6]4– is a d 4 low-spin species. The t2g levels are unequally occupied
and the eg are vacant, which makes it a borderline complex in terms of substitution rate.
[Cr(CN)6]3–, [Fe(CN)6]4–, and [Co(NH3)5(H2O)]3+ are low-spin species with the t2g levels either
half filled or completely filled. This, combined with empty eg levels, indicates inert species.
LFAE approximations suggest that the activation energies for substitution reactions with these
ions are relatively large.
12.6
The general rate law for square planar substitution is: Rate = (k1 + k2[Cl–]) [[Pt(NH3)4]2+].
The general procedure is to measure either the disappearance of [Pt(NH3)4]2+ or the formation
of [Pt(NH3)3Cl]+ to find the reaction rate. The most convenient method is by UV or visible
absorbance spectra, using a wavelength where there is a large difference in absorbance between
the two species. One experimental strategy would be to attempt to establish pseudo-first order
conditions, by using a significantly large chloride concentration, so that its concentration can be
assumed to not change appreciably while the reaction rate is measured. This may make the
reaction appear first order in Pt reactant. Measure the change in reactant and product
concentrations using spectroscopy. A plot ln([Pt(NH3)42+]) versus time will afford a straight line
with slope kobs if the pseudo-first order approximation is valid when excess chloride is used.
Assuming this approximation holds, repeat several times with different (but all sufficiently high)
chloride concentrations. A graph kobs vs. [Cl–] will result in a straight line with intercept = k1 and
slope = k2.
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Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
12.7
173
a.
DMSO exchange is extremely fast since this is present in large excess as the solvent.
b.
Saturation kinetics occurs with these entering ligands, as increasing [X  ] results in
apparent plateaus of the rate constant. The largest of these rate constants (for nitrate) is
slightly more than an order of magnitude larger than that of thiocyanate, suggesting the
incoming ligand may play a significant role in influencing the rate. The limiting rate
constant at high [X  ] suggests that either a D mechanism, I d mechanism (if we assume
that these substitution reactions are irreversible), or one involving a preassociation
complex are viable options.
c.
Application of a steady-state approximation for a possible D mechanism provides a rate
2+
law of Rate = kobs   Co(en)2 (NO 2 )(DMSO)   where kobs  k1 when the concentration


of entering ligand is high. A vital issue is whether these k1 values are sufficiently similar
when the chloride, nitrate, and thiocyanate trials are compared to justify assignment of a
D mechanism. Ideally, k1 would be expected to be independent of the incoming ligand if
a D mechanism were operative. For the possibility of assessing whether an I d
mechanism is operative with these data, it is necessary to assume that the substitution is
irreversible. In this case, when the concentration of entering ligand is high, pseudo-first
order conditions can be assumed, and the rate law simplifies to
2+
Rate = kobs   Co(en)2 (NO 2 )(DMSO)   where kobs  k1[X  ] .


d.
The similarity in the k1 values when those of chloride, nitrate, and thiocyanate are
compared for both D (where kobs  k1 ) and I d possibilities (where kobs  k1[X  ]
assuming the substitution reaction is irreversible) suggests that these mechanisms are
both viable possibilities. However, the variation in these rate constants with different
entering ligands indicates that the entering ligand likely plays some role in the rate
determining step, rendering an I d mechanism as perhaps a better hypothesis. The
possibility of a preassociation complex mechanism could also be argued, but more data is
needed to properly evaluate this possibility.
12.8
a.
Because the rate is independent of the concentration of 13CO, the rate determining step is
most likely:
Cr(12CO)6
Cr(12CO)5 + 12CO
Cr(12CO)5 then reacts rapidly with 13CO.
b.
These terms describe two pathways to product, a dissociative pathway (as in part a) and
an associative pathway:
Dissociative:
Cr(CO)6
Cr(CO)5 + CO
Cr(CO)5 + PR3
Cr(CO)5(PR3)
Associative:
Cr(CO)6 + PR3
(slow)
(fast)
Cr(CO)5(PR3) + CO
Copyright © 2014 Pearson Education, Inc.
rate = k1[Cr(CO)6]
rate = k2[Cr(CO)6][PR3]
174
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
With two mechanistic pathways leading to the same product, the overall rate is the sum of
the rates of both.
c.
Bulky ligands will tend to favor the first order (dissociative pathway) because the
crowding around the metal will favor dissociation and hinder association with incoming
ligands. (This effect is discussed further in Section 14.1.1)
12.9
[Cu(H2O)6]2+ is a d 9 complex, subject to Jahn-Teller distortion. Therefore, different rates
are observed for exchange of axial and equatorial water molecules.
12.10
These exchange reactions are interesting since the reactants and products have the same structures
and energies. The hypothesized mechanism for [Li(H 2O)4 ]+ water exchange is shown below.
In A, a water molecule in the second coordination sphere interacts weakly with the lithium center
and one aqua ligand via hydrogen bonding. This solvated [Li(H 2O)4 (H 2O)]+ species is more
thermodynamically stable than separated [Li(H 2O)4 ]+ and H2O. In the transition state B, the
incoming water molecule approaches the metal center and pushes two aqua ligands towards the
axial positions of the approximately trigonal bipyramidal intermediate C, with the incoming H2O
in an equatorial site. Calculations suggest that the molecular volume of intermediate C is less than
that of B, supporting a limiting associative mechanism (A).
H 2O
H 2O
Li
OH2
OH2
OH2
O
H
H 2O
H
H2O
Li
OH2
OH2
A
B
O
H 2O
H
H
Li
H 2O
O
H
H
OH2
C
+
The mechanism for ammonia ligand exchange for [Li(NH 3 )4 ] is proposed to follow an alternate
mechanism that also begins with a solvated [Li(NH 3 )4 (NH 3 )]+ that is more thermodynamically
stable than separated [Li(NH 3 )4 ]+ and NH 3 . In the lowest energy solvated [Li(NH 3 )4 ]+ species
(D), the NH 3 molecule is proposed to interact simultaneously with N—H bonds of two ammine
ligands. Transition state E is approximately trigonal bipyramidal with the entering and leaving
ammine ligands positioned in axial sites. An associative interchange mechanism ( I a ) is inferred
on the basis of molecular volume calculations that indicate that the volume of E is nearly the
same as that of D (the bonds in E are long).
NH3
NH3
H3 N
Li
Li
NH3
NH3
H 3N
NH3
D
H3 N
N
H
H
H
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NH3
E
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
12.11
175
A plot of kobs versus [As(C6H5)3] exhibits a linear relationship, with intercept k1 = 2.3 × 10–5 s–1
and slope k2 = 2.06 × 10–5 M–1s–1.
Rate = (k1 + k2[As(C6H5)3])[Co(NO)(CO)3]
This reaction shows both first and second order kinetics; the ligand substitution likely occurs via
two mechanisms. The first order reaction appears to be a dissociative reaction or a solventassisted dissociation of CO, followed by a fast addition of As(C6H5)3. The other path shows first
order dependence on As(C6H5)3, perhaps caused by an associative reaction.
12.12
HNP is a measure of basicity, with more basic molecules having a smaller HNP. The
reactions appear to be associative, with the rate increasing with the basicity of the incoming
ligand. The two lines are a consequence of the different natures of the ligands; numbers 111 are all phosphorus ligands, while 12-14 are organic nitrogen compounds. The half
neutralization potential provides a relative measure of basicity within compounds of similar
structure, but is not an absolute measure of the reactivity of the compounds.
12.13
a.
G° = H° – TS° = 10300 – 298 × 55.6 = –6300 J/mol = –6.3 kJ/mol
G° = –RT lnK; ln K= – G°/RT = 6300/(8.3145 × 298.15) = 2.54; K = 12.7
b.
The cis isomer has the higher bond energy (actually, the lower overall energy and
collectively stronger bonding), since rearrangement to the trans isomer is endothermic.
Since the phosphines are better π acceptors, the cis isomer should be the more stable. If
the phosphines are mutually trans, they compete for overlap with the same d orbitals,
resulting in weaker bonds to Pt(II). When these phosphines are mutually cis, each can use
one of the pair of dxz and dyz orbitals and avoid competition.
c.
The free phosphine must aid the isomerization via an associative mechanism. Since
benzene is the solvent, and it is very nonpolar, it is not likely to assist the reaction.
Only the second term of the rate equations of Problem 12.6 is significant here, with
phosphine playing the role of the entering ligand.
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176
12.14
12.15
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
Two factors need to be considered:
1.
The best π acceptors (CO, PPh3) slow the dissociation of CO in the cis position; π donors
(halides) speed the dissociation.
2.
Halides are weaker  donors relative to phosphines and carbon monoxide. As a result,
there is less electron density available for -back-bonding in the complexes with halide
ligands. This results in weaker M—CO bonds in these complexes, which lowers the
activation energy for CO dissociation by raising the electronic ground state of the starting
complex. The -donor ability of these halides also assists in stabilizing the squarepyramidal transition state, further contributing to a lowering of the activation energy for
CO dissociation.
Trans-Pt(NH3)2Cl2 reacts with thiourea (tu) to form trans-[Pt(NH3)2(tu)2]2+; the first Cl– is
displaced and the strong trans effect of tu leads to preferential replacement of the second Cl–.
Both chlorides of cis-Pt(NH3)2Cl2 are initially replaced. The tu’s are then trans to the NH3’s,
which are replaced because of the strong trans effect of tu, resulting in [Pt(tu)4]2+.
12.16
12.17
a.
[Pt(CO)Cl3]– + NH3  trans-[Pt(CO)(NH3)Cl2]
CO is the stronger trans director
b.
[Pt(NH3)Br3]– + NH3  cis-[Pt(NH3)2Br2]
Br is the stronger trans director
c.
[Pt(C2H4)Cl3]– + NH3  trans-[Pt(C2H4)(NH3)Cl2] C2H4 is the stronger trans director.
a.
Two sets of reactions, with examples from Figure 12.13 identified:
[PtCl4]2– + 2NH3  cis-[PtCl2(NH3)2] + 2 Cl–
(b)
cis-[PtCl2(NH3)2] + 2 py  cis-[Pt(py)2(NH3)2]2+ + 2 Cl–
(h)
cis-[Pt(py)2(NH3)2]2+ + 2 NO2–  trans-[Pt(NO2)2(NH3)(py)] + NH3 + py
(e)
trans-[Pt(NO2)2(NH3)(py)] + CH3NH2 
py
NH2CH3
Pt
O 2N
+
+
NO2-
NH3
(g)
(Continued on next page)
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Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
177
[PtCl4]2– + 2 py  cis-[PtCl2(py)2] + 2 Cl–
(b)
cis-[PtCl2(py)2] + 2 CH3NH2  cis-[Pt(CH3NH2)2(py)2]2+ + 2 Cl–
(h)
cis-[Pt(CH3NH2)2(py)2]2+ + 2 NO2– 
trans-[Pt(NO2)2(CH3NH2)(py)] + py + CH3NH2
(c)
trans-[Pt(NO2)2(CH3NH2)(py)] + NH3 
O2 N
NH2CH3
+
Pt
+
py
NO2-
NH3
(a)
b.
Reaction with Cl2 puts Cl both above and below the plane, with the products
+
Cl
H 3N
H3 N
NH2CH3
and
Pt
py
+
Cl
NH2CH3
Pt
py
O2N
NO2
Cl
Cl
Reaction with one mole of Br– replaces one Cl–, with the products
+
Br
H 3N
NH2CH3
Pt
py
NO2
Cl
12.18
12.19
+
Br
H3 N
and
NH2CH3
Pt
py
O2N
Cl
a.
The large negative entropy of activation implies that the activated complex is much more
ordered than the reactants. This suggests an associative pathway.
b.
The iodo ligand leads to the fastest rate, involving bond breaking trans to the halogen,
and therefore has the strongest trans effect.
V2+ is d 3, and likely to be inert. V3+ is d 2, and labile. At low [H+] (slightly basic conditions), an
equilibrium mixture of [V(H 2O)6 ]3+ and [V(H 2O)5OH]2+ may persist. The hydroxo ligand of
[V(H 2O)5OH]2+ may permit a bridging interaction with labile [V(H 2O)6 ]2+ to form
[V(H 2O)5 (  -OH)V(H 2O)5 ]4+ to facilitate an inner sphere electron transfer mechanism. A
proposed rate law is below, with a and b determining the relative rates of the two paths, where
term a is associated with an outer sphere electron transfer and term b is associated with an inner
sphere pathway. In path a, the bimolecular rate-determining step is the outer sphere electron
Copyright © 2014 Pearson Education, Inc.
178
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
transfer while in path b the rate-determining step is the bimolecular formation of the bridging
hydroxo complex.
Rate = (k1 + k2/[H+])[V2+][V3+], so a = k1 [V2+][V3+] and b = k2[V2+][V3+]
12.20
[Cr(H2O)6]2+ is labile, but [Co(NH3)6]3+ is inert, and the NH3 ligands are not well equipped to
bridge two metals. Therefore, the reaction is likely to be an outer-sphere reaction.
12.21
This reaction is inner sphere, so the X– serves as a bridging ligand and a conduit for the electron
–
–
–
–
–
transfer. The order of rates is Br > Cl > N3 > F > NCS . The difference in reaction rate is
most strongly correlated to the varying abilities of these bridging ligands to accommodate
–
electron transfer between the metal centers. Br works best, and this could be rationalized by
–
estimating that the hard-soft compatibility of Cr2+ and Br is optimum among these ligands,
leading to a relatively robust bridging interaction. Chloride is slightly harder, leading to a slightly
–
lower electron transfer rate. F is apparently too hard; the high effective nuclear charge its
–
valence electrons experience render it a poor conduit for electron transfer to Cr2+. The NCS
ligand offers a very slow electron transfer, the Cr—NCS—Cr bridge features one interaction with
better HSAB compatibility than the other; the resulting bridge must not be ideal for electron
transfer despite the presence of a conjugated -system that can lower the activation barrier for
electron transfer. Azide also has a conjugated system, but the Cr-NNN-Cr interactions apparently
do not have an optimal HSAB match (nitrogen is too hard) to permit a suitable interaction. It is
somewhat surprising that azide results in a faster electron transfer rate relative to thiocyanate on
the basis of solely HASB arguments. The data in this problem are from D.L. Ball and E. L. King,
J. Am. Chem. Soc., 1958, 80, 1091.
12.22
a.
NSe has the stronger trans effect. The longer Os—N1 distance is consistent with
weakening of this bond by the ligand trans to it, the NSe ligand.
b.
The short bond distance and large Os—N—Se angle (164.7°) suggest that the ligand has
NSe+ character; the cation NSe+ would have a bond order of 3 and a very short bondbond distance. It is also worth noting that the N–Se stretching vibration in this complex
is much higher (by 211 cm-1) than in gas phase NSe.
a.
The more rapid reactivity of the Mn complex is consistent with the general observation
that first row transition metal complexes are generally more substitutionally labile than
second and third row complexes.
b.
The negative volume of activation is consistent with an A (or Ia) mechanism.
c.
Occurrence of two infrared bands is more consistent with a fac isomer; a mer isomer is
predicted by symmetry to have three IR-active bands:
12.23
mer isomer (C2v):
C2v

A1
B1
E
3
1
1
C2
1
1
–1
 (xz)
3
1
1
 (yz)
1
1
–1
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z
x
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
179
The representation  reduces to 2 A1 + B1, all IR-active. Three IR-active carbonyl bands
are expected.
fac isomer (C3v):
C3v

A1
E
E
3
1
2
2 C3
0
1
–1
3v
1
1
0
z
(x, y)
The representation  reduces to A1 + E, both IR-active. Two IR-active carbonyl bands
are expected.
12.24
These data support ring-closing reactions with dissociative activation. The uniformly positive
S act values suggest transition states in the rate determining steps where cyanide ligand
dissociation is more prominent than coordination of the pyridine-carboxylate nitrogen atom to the
Mn(V) center. The similarity in the rate constants (the average k1 is 1.03 103 M 1s1 with a
standard deviation of only 1.3 104 M 1s1 ) is remarkable. These rate constants appear
independent of the different nitrogen atom basicities and the charges of these pyridinecarboxylates. Varying these properties of the entering ligand has very little impact on the speed of
cyanide substitution. This is also consistent with dissociative activation.
12.25
a.
The consumption rates of U 4+ and [PtCl6 ]2- were monitored with UV-Vis spectroscopy,
by monitoring the absorptions of U 4+ ( max  648 nm) and [PtCl6 ]2- ( max  300 nm)
with time. Reaction orders with respect to both of these ions were determined to be 1 on
the basis of the linear relationships of the plots of ln( A  At ) versus time, where A
and At are the absorbances at time t and infinite time for U 4+ and [PtCl6 ]2- ,
respectively. These plots exhibited linear relationships for more than three half-lives.
b.
One hypothesis is that [U(H 2O) n ]4+ initially forms a
H
2-
complex with [PtCl6 ] that subsequently loses two
protons to yield a neutral intermediate where
O
PtCl6
(H2O)n-2U
O
[PtCl6 ]2- is coordinated to [U(H 2O) n-2 (OH)2 ]2+ via
H
hydroxide bridges (shown at right). The inner-sphere
electron transfer of two electrons from U(IV) to Pt(IV) is proposed to proceed via this
complex. The activation barrier for electron transfer via this intermediate may be reduced
(leading to a faster reaction) since the U fragment closely resembles the ultimate
uranium(VI)-containing product [(H 2O) n-2 UO 2 ]2+ .
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180
12.26
Chapter 12 Coordination Chemistry IV: Reactions and Mechanisms
a.
These complexes have been established as outer-sphere reactants in previous work.
b.
Using the reaction of [Co(edta)] and [Fe(CN)6 ]4 as an example, the rate law is
proposed as Rate  k1 [Fe(CN)6 ]4  [Co(edta)]  , first order with respect to each



complex. To create a graph such as that shown in Figure 3 of the reference, a series of
reactions were conducted with varying, but always excess [Fe(CN)6 ]4 and a smaller
fixed concentration of [Co(edta)] . The concentration of [Fe(CN)6 ]4 is assumed to not
change significantly during these trials, allowing the rate law to be expressed as
Rate  kobs [Co(edta)]  where kobs  k1 [Fe(CN)6 ]4  . The kobs rate constants were




determined and then plotted versus the [Fe(CN)6 ]4 concentration for each trial. The
linear relationship and the zero y-intercept support the validity of the pseudo-first order
conditions, and that the overall bimolecular rate law is operative in these electron transfer
reactions. The relationship between kobs and k1 is kobs  k1 [Fe(CN)6 ]4  , and the


slope of the plot of kobs versus [Fe(CN)6 ]4  provides the bimolecular rate constant


.
This
is
a
beautiful
example
of
the
application
of pseudo-first order conditions.
k1
c.
The hypothesis is that the reactants become compartmentalized, hindering their ability to
encounter each other for electron transfer, in the presence of reverse micelle (RM)
microemulsions. [Co(edta)] is believed to penetrate into the interfacial region of the
RM, while [Fe(CN)6 ]4 access to this region is limited, confining [Fe(CN)6 ]4 to the
aqueous solution. The physical separation of these complexes may be the dominant factor
in hindering the electron transfer reaction, but the possibility that the reduction potentials
of these complexes may be changed (for example, that [Co(edta)] may be rendered a
weaker oxidizing agent when embedded in the interfacial region of the RM) is also
mentioned.
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Chapter 13 Organometallic Chemistry
181
CHAPTER 13: ORGANOMETALLIC CHEMISTRY
13.1
Method A: = Donor-Pair Method, Method B: = Neutral-Ligand Method
a.
13.2
Fe(CO)5
A: 8 + 5×2 = 18
+
B: 8 + 5×2 = 18
b.
[Rh(bipy)2Cl]
A: 7 + 2×4 + 2 = 17
B: 9 + 2×4 + 1 – 1 = 17
c.
(5-Cp*)Re(=O)3
A: 6 + 0 + 3×4 = 18
B: 5 + 7 + 3×2 = 18
d.
Re(PPh3)2Cl2N
A: 2 + 2×2 + 2×2 + 6 = 16
B: 7 + 2×2 + 2×1 + 3 = 16
e.
Os(CO)(CPh)(PPh3)2Cl A: 7 + 2 + 3 + 2×2 + 2 = 18 B: 8 + 2 + 3 + 2×2 + 1 = 18
f.
Ru(CE)Cl2 (NC5H 4 NMe 2 )2 (C3H 4 N2 (C8 H9 )2 )
A: 6 2 2  2 2  2  2  18 B: 8  2  2 1 2 2  2  18
All of these compounds have 16-electron valence configurations.
a.
Ir(CO)Cl(PPh3)2
A: 8 + 2 + 2 + 2×2 = 16
B: 9 + 2 + 1 + 2×2 =
b.
RhCl(PPh3)3
A: 8 + 2 + 3×2 = 16
B: 9 + 1 + 3×2 = 16
c.
[Ni(CN)4]
2–
A: 8 + 4×2 = 16
B: 10 + 4×1 + 2 = 16
d.
cis-PtCl2(NH3)2
A: 8 + 2×2 + 2×2 = 16
a.
[M(CO)7]+
A: 18 – 7×2 = 4 = M+, V
b.
H3CM(CO)5
A: 18 – 2 – 5×2 = 6 = M+, Mn B: 18 – 1 – 5×2 = 7 = M, Mn
c.
M(CO)2(CS)(PPh3)Br
A: 18 – 2×2 – 2 – 2 – 2 = 8 = M+, Co
B: 18 – 2×2 – 2 – 2 – 1 = 9 = M, Co
d.
[3-C3H3)(5-C5H5)M(CO)]–
A: 18 – 4 – 6 – 2 = 6 = M+, Mn
–
B: 18 – 3 – 5 – 2 = 8 = M , Mn
e.
(OC)5M=C(OCH3)C6H5
A and B: 18 – 5×2 – 2 = 6 = M, Cr
f.
[4-C4H4)(5-C5H5)M]+
g.
(3-C3H5)(5-C5H5)M(CH3)(NO)
A: 18 – 2 – 6 – 2 – 2 = 6 = M, Cr
B: 18 – 3 – 5 – 1 – 3 = 6 = M, Cr
16
B: 10 + 2×2 + 2×1 =
16
13.3
A: 18 – 4 – 6 = 8 = M2+, Ni
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B: 18 – 7×2 = 4 = M+, V
B: 18 – 4 – 5 = 9 = M+, Ni
Chapter 13 Organometallic Chemistry
182
13.4
13.5
h.
[M(CO)4I(diphos)]–
a.
[Fe(CO)2(5-C5H5)]2
A: 7 + 2×2 + 6 = 17, single Fe–Fe
B: 8 + 2×2 + 5 = 17, single Fe–Fe
b.
[Mo(CO)2(5-Cp)]22–
A: 5 + 2×2 + 6 + 1 = 16, double Mo=Mo
B: 6 + 2×2 + 5 + 1 = 16, double Mo=Mo
a.
[M(CO)3(NO)]–
linear M–N–O:
A: 18 – 4×2 – 2 – 4 = 4 = M, Ti
B: 18 – 4×2 –1 – 4 = 5 = M–, Ti
Calculating for each metal atom:
bent M–N–O:
13.6
b.
[M(PF3)2(NO)2]+
linear NO:
c.
[M(CO)4(2–H)]3
As a triangular structure with three M–M bonds:
A: 18 – 4×2 – 2 – 2 = 6 = M+, Tc
B: 18 – 4×2 – 1 – 2 = 7 = M, Tc
d.
M(CO)(PMe3)2Cl
A: 16 – 2 – 2×2 – 2 = 8 = M+, Rh
B: 16 – 2 – 2×2 – 1 = 9 = M, Rh
A: 18 – 2×2 – 2×2 = 10 = M–, Rh
B: 18 – 2×2 – 2×3 = 8 = M+, Rh
Method B works better for calculating overall charge.
a.
13.7
A: 18 – 3×2 – 2 – 2 = 8 = M, Ru
B: 18 – 3×2 – 3 = 9 = M–, Ru
A: 18 – 3×2 – 2 = 10 = M, Pd
B: 18 – 3×2 – 1 = 11 = M–, Pd
[Co(CO)3]z
9 + 3×2 = 15, z = 3–
z
b.
[Ni(CO)3(NO)]
c.
[Ru(CO)4(GeMe3)]z
10 + 3×2 + 3 = 19, z = 1+
d.
[(3-C3H5)V(CNCH3)5]z
3 + 5 + 5×2 = 18, z = 0
e.
[(5-C5H5)Fe(CO)3]z
5 + 8 + 3×2 = 19, z = 1+
f.
[(5-C5H5)3Ni3(3–CO)2]z
a.
[(5-C5H5)W(CO)x]2, assuming a single W–W: A: 6 + 5 + 1 + x × 2 = 18, x = 3
B: 5 + 6 + 1 + x × 2 = 18, x = 3
b.
ReBr(CO)x(CO2C2H4)
c.
[(CO)3Ni–Co(CO)3]z
A and B: 3 × 2 + 10 + 2 + 9 + 3×2 – z = 36, z = 3–
d.
[Ni(NO)3(SiMe3)]z
B: 10 + 3×3 + 1 – z = 18, z = 2+
8 + 4×2 + 1 = 17, z = 1–
3 × 5 + 3×10 + 2×2 = 49, z = 1+, assuming three
Ni–Ni bonds; calculating for each Ni: 5 + 10 + 2(2/3)
+ 2 = 18 1/3; charge per Ni = 1/3+, overall charge = 1+
(Each triply bridging CO can be considered to donate 2 electrons overall, 2/3
electron to each metal.)
A: 6 + 2 + x × 2 + 2 = 18, x = 4
B: 7 + 1 + x × 2 + 2 = 18, x = 4
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Chapter 13 Organometallic Chemistry
13.8
e.
a.
[(5-C5H5)Mn(CO)x]2
Upper ring: 5-Cp
Co
b.
M:
183
B: 5 + 7 + 2 + x × 2 = 18, x = 2
5 electrons
9 electrons
14 electrons; 4 more needed, so lower ring must be 4
1-C7 ring
2 PEt3
GeR3
1 electron
4 electrons
1 electron
M:
6 electrons
7-C7 ring
5-C5 ring
7 electrons
5 electrons
12 electrons
4 more needed to reach 16: Ti
10 more needed to reach 16: Pt
O
C
13.9
Co (group 9):
Ni (group 10):
Co
CO
CO
Co
O
C
O
C
Ni
Co
C
O
Ni
Cu (group 11):
Cu
CO
Other compounds with different hapticities for cyclopentadienyl ligands may also be possible.
13.10
Figure 10.19 gives an MO diagram for Ni(CO)4 and cites additional references. The
HOMOs (t2) are strongly bonding, and there is a large energy gap between the HOMOs and the
LUMOs in this 18-electron molecule.
13.11
The energy of stretching vibrations depends on the square root of the force constant divided
by the reduced mass (Section 13.4.1).
The reduced masses are 14.73 for 16O and 16.41 for 18O, so (14.73/16.41)1/2 = 0.947, and the
18
O complex has vibrational energy of 0.947 ×975 cm–1 = 924 cm–1. The value given in
the reference is 926 cm–1.
13.12
Sulfur is less electronegative than oxygen. Therefore, the tungsten in W(S)Cl2(CO)(PMePh2)2
has greater electron density and a greater tendency to participate in -backbonding with CO; a
lower energy (CO) is expected (actual value: 1986 cm–1).
13.13
Adding electrons to a carbonyl complex puts more electrons into the back-bonding t2g orbitals.
As a result, the V–C bonds in [V(CO)6]– are strengthened and the distance shortened (but
the C–O bonds are weakened by having more electrons in the t2g orbitals, that are antibonding
with respect to the C–O bonds.).
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Chapter 13 Organometallic Chemistry
184
Sigma Donor
13.14
Pi Acceptor
a.
M
C N
b.
M
C N
R
R
R
R
R
R
M
PR3
M
M
S C N
PR3
c.
13.15
13.16

M
S C N
a.
If NO is counted as a linear donor, each of these has 18 electrons. The increasing
nuclear charge (from formally Cr(0) to Mn(I) to Fe(II)) results in progressively less
backbonding to the ligands and an increase in N–O bond order, leading to the higher NO
stretching frequencies. The Cr species has an exceptionally low NO stretching
frequency. It may have bent NO coordination, rendering it a 16 electron ion formally
containing Cr(II). The Fe complex has a rather high NO stretching frequency, suggesting
that π-backbonding is not a very large contribution to the electronic ground state of this
complex.
b.
The low energy band indicates bent NO coordination; the higher energy band is from
the linear ligand. This complex has one of each, with angles of 138° and 178°
(Greenwood and Earnshaw, Chemistry of the Elements, 2nd ed., pp. 450-52).
a.
The CO2 molecular orbitals are shown in Figure 5.25.
b.
The π orbitals of 1,3,5-hexatriene are shown in Figure 13.21.
c.
cyclo-C4H4:




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Chapter 13 Organometallic Chemistry
13.17
d.
cyclo-C7H7:
a.
Using the group theoretical method described in Chapter 4: The four CO ligands are in a
square planar arrangement, but the Mo atom is above them. The reducible representation
, shown below, can be derived for the C–O stretching vibrations.  reduces to A1 + B1 +
E, with A1 and E infrared active, so there are two bands visible (the E bands are
degenerate).
C4v

A1
B1
E
2C4
0
1
–1
0
E
4
1
1
2
C2
0
1
1
–2
2v
2
1
1
0
b.
x
d xy
p x, d xz
y
py , dyz
s, p z, dz2
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185
2d
0
1
–1
0
z
(x, y)
Chapter 13 Organometallic Chemistry
186
13.18
13.19
a.
From bottom to top:
0-node, left, is A1, 0 node, right is A2
1-node, far left and third from left are E1
1-node, second from left and far right are E1
2-node, far left and third far left are E2
2-node, second from left and far right are E2
b.
s, dz2
A 1
c.
The matching orbitals are shown in part b; there is no match in the Fe orbitals for the E2
ligand orbitals.
a.
The π orbitals of benzene are shown in Figure 13.22.
b.
Group orbitals
(dxy, dx2–y2)
E2 
(dxy, dyz)
E1 
pz
A2 
(px, py)
E 1
3-Node
Cr
11
12
7
8
9
10
3
4
5
6
1
2
2-Node
1-Node
0-Node
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Chapter 13 Organometallic Chemistry
c.
Matching Cr orbitals:
1
2
3
4
5
6
d.
7
8
9
10
11
12
s, dz2
pz
py
dyz
px
dxz
dxy
none
d x2 – y2
none
none
none
Energy level diagram
3-Node
4p
2-Node
4s
1-Node
3d
0-Node
Cr
13.20
a.
Cr(C6H6 ) 2
Group Orbitals
Point group: C2v (in orientation shown)
Fe
b.
The reducible representation is :
C2v

A1
A2
B1
B2
E
10
1
1
1
1
C2
0
1
1
–1
–1
v (xz)
2
1
–1
1
–1
v (yz)
0
1
–1
–1
1
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
z
x
y
x 2 , y 2, z 2
xy
xz
yz
187
188
c.
Chapter 13 Organometallic Chemistry
 = 3A1 + 2A2 + 3B1 + 2B2
Representations
A1
A2
B1
B2
Matching orbitals of Fe
s, pz, dz2
dxy
px, dxz
py, dyz
13.21
2C3
0
1
1
–1
E
3
1
1
2
C3v

A1
A2
E
3v
1
1
–1
0
z
(x, y)
 = A1 + E, both IR active. There are three vibrations, but two are degenerate, so two
C–O stretching bands are expected in the IR spectrum.
13.22
Ni(CO)4
 = A1 + T2
8C3
1
1
0
E
4
1
3
Td

A1
T2
3C2
0
1
–1
6S4
0
1
–1
6d
2
1
1
(x, y, z)
A1 is IR inactive, T2 is IR active: 1 band.
Cr(CO)6
Oh
E
8C3
6C2
6 C4
3 C2
i
6S4
8S6

A1g
Eg
T1u
6
1
2
3
0
1
–1
0
0
1
0
–1
2
1
0
1
2
1
2
–1
0
1
2
–3
0
1
0
–1
0
1
–1
0
= A1g + Eg + T1u Only T1u is IR active: 1 band
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3h
4
1
2
1
6d
2
1
0
1
(x, y, z)
Chapter 13 Organometallic Chemistry
C
O
O
C
C
O
a.
O
C
M
C
C
13.24
O
O
13.23
M
C
O
C
O
4C2
0
1
–1
0
0
0
C
O
C
O
D4 d
E
2S8
2C4
2S83
C2

A1
B2
E1
E2
E3
10
1
1
2
2
2
0
1
–1
2
1
1
0
–2
0
0
1
–1
2
1
1
–2
2
–2
2
0
– 2
189
– 2
0
2
4d
4
1
1
0
0
0
z
(x, y)
(Rx, Ry)
= 2A1 + 2B2 + E1 + E2 + E3
2B2 and E1 are IR active: 3 bands
W(CO)5(NCC3H7)
W(CO)4(NCC3H7)2, cis
W(CO)3(NCC3H7)3, probably fac
2077, 1975, 1938 cm–1
2107, 1898, 1842 cm–1
1910, 1792 cm–1
Pentacarbonyls have three active IR stretches, tetracarbonyls have one (other ligands
trans) or four (other ligands cis), and tricarbonyls have two (fac) or three (mer) (see
Table 13.10). It appears that two of the tetracarbonyl bands have similar enough energies
for their bands to overlap, so only three appear in the spectrum (or one band may be so
weak in intensity that it does not appear), but that the nitriles must be cis. The evidence
for the fac isomer of the tricarbonyl is not conclusive based on the IR spectrum alone,
since again two bands of the mer isomer might overlap.
13.25
b.
The stretching energy of the CO trans to the nitrile ligand is lower than that for CO
cis due to reduced competition for the π backbonding electrons (CO is a better πacceptor). In the second complex, two are cis and two are trans. In the third complex,
all three are trans, so the band energy is lower. In general, the trend is to lower energies
as CO is replaced by butyronitrile; butyronitrile is not as effective a π acceptor as CO,
so the π-acceptor nature of CO is enhanced as more nitrile ligands are added.
c.
As more CO ligands are replaced by nitriles, which are stronger at donating electrons to
Mo, the remaining CO ligands become stronger π acceptors, and the Mo–C bonds
become stronger. In Mo(CO)3(NCC3H7)3 the Mo–C bond is so strong that the complex
cannot react further with butyronitrile.
a.
The bands at lower energy are for the CO ligands. The C–O stretches, which involve
a greater change in dipole moment, are more intense than the C–N stretches. Because
CO has a slightly greater reduced mass than CN, C–O stretches should occur at lower
energies than C–N stretches; the energy of vibrational levels is inversely proportional to
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190
Chapter 13 Organometallic Chemistry
reduced mass . Also, the greater π-acceptor ability of the CO ligand is a contributing
factor to the lower energy stretches for this ligand.
b.
2–
The trans-[Fe(CO)2(CN)4] complex should have a single IR-active C–O stretch
(only the antisymmetric stretch is IR active) and a single IR-active C–N stretch (see
Table 13.7). The cis complex should have two IR-active C–O stretches (both
symmetric and antisymmetric) and four IR-active C–N stretches. In addition to its
3–
single carbonyl stretch, [Fe(CO)(CN)5] would be expected to have three IR-active
C–N stretches. The correct identifications are therefore:
Complex
2–
A: cis-[Fe(CO)2(CN)4]
2–
B: trans-[Fe(CO)2(CN)4]
3–
C: [Fe(CO)(CN)5]
Predicted on basis of symmetry
C–O stretches
C–N stretches
2
1
1
4
1
3
In complexes A and C, some of the C–N stretches are too weak to be seen or the
bands overlap; otherwise, the expected numbers of bands and the observed spectra
match.
13.26
a.
The CO ligand can be in either the axial or the equatorial position on the trigonal
bipyramidal complex. These environments are not equivalent, so the two
isomers should have different carbonyl stretching bands.
b.
The CO stretch absorbs at a higher energy in Fe(CO)(PF3)4, implying that the
C–O bond is stronger in that compound. If the CO bond is stronger, the Fe–C bond
is correspondingly weaker, indicating that the PF3 ligands are better π acceptors
and therefore higher in the spectrochemical series than CO.
O
13.27
C
Et3P
Ru
PEt3
PEt3
PEt3
C
Et3P
Ru
C
C
O
O
PEt3
O
One C–O stretch (antisymmetric)
(minor isomer)
Two C–O stretches (symmetric
and antisymmetric) (major isomer)
13.28
From Table 13.10, [Co(CO)3(PPh3)2]+ must be trigonal bipyramidal with all three CO
ligands in the equatorial plane, giving rise to a single C–O absorption band.
13.29
By extrapolation of the positions of the C–O bands for [Mn(CO)6]+ (2100 cm–1) and
[Fe(CO)6]2+ (2204 cm–1), one might predict a comparable band for [Ir(CO)6]3+ near 2300 cm–1.
Actual value: 2254 cm–1.
13.30
a.
In these three cations, the high metal oxidation states significantly reduce backbonding.
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Chapter 13 Organometallic Chemistry
191
This effect is greatest for [Hg(CO)2]2+, with a charge of 1+ per CO, and weakest for
[Os(CO)6]2+, with a charge of 1/3+ per CO. Therefore, [Os(CO)6]2+ should have the
strongest backbonding, weakest C–O bond, and lowest energy carbon-oxygen stretching
vibration. The actual energies are listed on the next page.
(CO), cm–1
[Hg(CO)2]2+
2278
[Pt(CO) 4]2+
2244
[Os(CO) 6]2+
2190
b.
A reducible representation for this ion based on C–O stretching vibrations would
have the following characters:
D2d
E
2S4
C2

2 A1
2 B2
E
6
2
2
2
0
2
–2
0
2
2
2
–2
2C2
0
2
–2
0
2d
4
2
2
0
z
(x, y)
There should be three carbon–oxygen stretching bands, two of B2 symmetry and one
of E symmetry (a degenerate pair). With a charge 2+, this complex should exhibit
infrared bands near those of [Fe(CO)6]2+ (2204 cm–1). Actual bands are observed at
2173 cm–1 (E) and at 2187 and 2218 cm–1 (B2).
13.31
a.
The representation based on the set of six C–O vibrations in Oh symmetry is:
Oh

E
6
8C3
0
6C2
0
6C4
2
3C2(=C42)
2
I
0
6S4
0
8S6
0
3h
4
6d
2
This representation reduces to A1g + Eg + T1u (see Exercise 10.4). Of these, A1g and Eg
match squared functions and are therefore Raman active. (A1g matches the band at 2015
cm–1 and Eg matches the band at 2119 cm–1.)
b.
13.32
J, K, and L are, respectively, the products of substitution of CO by py: Mo(CO)5py,
cis-Mo(CO)4py2, and fac-Mo(CO)3py3. As the number of pyridine ligands increases, the
strong  donation by this ligand increases the backbonding by the CO ligands, resulting
in successive lowering in energy of the C–O vibrations. Symmetry analysis of each of
these complexes shows more expected Raman-active bands than reported in the article.
Additional weak bands can be seen in the Raman spectra (shown in the reference).
(5-C5H5)Cr(CO)2(NS) 1962, 2033 cm–1
(5-C5H5)Cr(CO)2(NO) 1955, 2028 cm–1
NS is a stronger π acceptor, so the CO backbonding is reduced in (5-C5H5)Cr(CO)2(NS)
and the CO bond is strengthened and has a higher stretching energy.
13.33
The energy for a stretching vibration is proportional to
k

, where k is the force constant and
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Chapter 13 Organometallic Chemistry
192
µ is the reduced mass (Section 13.4.1). In this case, the reduced masses for the N–S stretches
are:
14
N–S:  
14.00  32.06
 9.745
14.00  32.06
15
N–S:  
15.00  32.06
 10.22 .
15.00  32.06
Because the energy is inversely proportional to the square root of the reduced mass, we can write:
E( 15 N–S)
E( 14 N–S)

 (14 NS)
(15 NS)

9.745
 0.9765 The expected position of the N–S stretch
10.22
in the 15NS complex is therefore 0.9765 1284 cm–1 = 1254 cm–1. The reported value is
1248 cm–1.
13.34
Because the reduced mass of 13CO is greater than the reduced mass of 12CO, the separation
between vibrational energy levels should be less for 13CO, and the 13CO complex should
therefore show an infrared band at lower energy than 2199 cm–1. Actual value: 2149 cm–1.
13.35
a.
Mo(CO)6 + Ph2 PCH 2PPh 2
b.
(5-C5H5)(1-C3H5)Fe(CO)2
H2 C
h
O
C
Ph 2
P
Mo
P
Ph 2
C
O
CO
+ 2 CO
CO
(5-C5H5)(3-C3H5)Fe(CO) + CO
The allyl ligand can bond in either 1 or 3 fashion. Loss of CO converts the
reactant from 18 electrons to 16 electrons; rearrangement of allyl from 1 to 3
returns it to 18 electrons.
c.
(5-C5Me5)Rh(CO)2
[(5-C5Me5)Rh(CO)]2 + 2 CO
A double Rh–Rh bond is needed; CO ligands may be bridging or terminal (the electron
count is the same for both modes).
d.
V(CO)6 + NO
V(CO)5(NO) + CO
The compound changes from 17 to 18 electrons.
e.
–
[(CO)5WCC6H5]+ + F + F2BOC2H5
W(CO)5=C(C6H5)(OC2H5) + BF3
AlR 3
O
O
f.
C
[(5-C5H5)Fe(CO)2]2 + 2 Al(C2H5)3
Fe
(See Figure 13.17)
C
Fe
C
O
O
v(CO), cm
13.36
P(t-C4H9)3
–1
1923 (best  donor, poorest π acceptor)
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R3 Al
C
Chapter 13 Organometallic Chemistry
P(p-C6H4Me)3
P(p-C6H4F)3
P(C6F5)3
13.37
13.38
13.39
a.
Fe(CO)4(PF3)
b.
[Re(CO)6]+
193
1965
1984
2004 (best π acceptor)
(PF3 is a strong π acceptor)
c.
Mo(CO)3(PCl3)3 (PCl3 is the best π acceptor among the phosphines)
a.
Oxidation of these low-spin d6 octahedral complexes causes loss of 1 t2g electron,
resulting in a change in LFSE of 0.4  (from –2.4 to –2.0 ).
b.
The C—O distance decreases. As the oxidation state of Cr increases from 0 to 1+,
the CO ligands become weaker pi acceptors, and there is less occupation of π* orbitals in
the ligands; consequently, the bonding in these ligands is stronger in the cations than in
the neutral complexes (in the reference, the calculated decrease in C—O distances is in
the range of 0.015 - 0.020 Å).
c.
The Cr—P distance increases (calculated increase: 0.094Å) as the phosphine ligand
becomes a weaker pi acceptor; there is less backbonding in the cation than in the neutral
complex.
d.
The Cr—N distance decreases.
In order from highest to lowest :
Mo(CO)4 (F2PCH2CH2PF2)
Mo(CO)4 ((C2F5)2PCH2CH2P(C2F5)2)
Mo(CO)4 ((C6F5)2PCH2CH2P(C6F5)2)
Mo(CO)4 (Ph2PCH2CH2PPh2)
Mo(CO)4 (Et2PCH2CH2PEt2)
has strongest acceptor diphosphine
has strongest donor diphosphine
13.40
Coordinated N2 has a lower stretching energy than free N2. N2 can act as a π acceptor,
weakening the N–N bond and lowering the energy of the stretching vibration. The
stretching vibration of free N2 is IR inactive; there is no change in dipole moment on stretching.
13.41
At the higher temperature, the –O–CH3 group should rotate rapidly enough to show only a
single, average environment, so each CH3 has a single peak. At low temperatures,
this rotation can be restricted, and cis and trans isomers result (see Figure 13.45). The larger
peaks represent the more prevalent isomer.
13.42
At high temperature, the C5H5 rings undergo rapid 1,2 shifts to give
a single average proton signal on the NMR time scale. At sufficiently
low temperature, the different environments of the 1 and 5 rings can
be seen. The relative intensities are a = 1, b = 2, c = 2, and d = 5.
O
C
d
d
O
C
Fe
a
d
b
d
d
c
b
c
13.43
PF3 is a stronger π acceptor than PCl3. As a result, the chromium in Cr(CO)5(PCl3) has a greater
electron density, and CO acts as a stronger π acceptor in this complex. As a result:
a.
Cr(CO)5(PF3) has the stronger and shorter C–O bonds.
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Chapter 13 Organometallic Chemistry
194
13.44
b.
Cr(CO)5(PCl3) has the higher energy Cr–C bands; since CO acts as a better π
acceptor in this complex, the Cr–C bond is strengthened.
a.
[Fe(NO)(mnt)2]– has less electron density on Fe, less backbonding to NO, and a
stronger N–O bond with higher stretching frequency.
b.
(CO)5Cr:NN:Cr(CO)5 N2 acts as a π acceptor toward both metals, significantly
weakening the N–N bond.
c.
Ta=CH2 has a shorter Ta–C bond because it is a double bond.
d.
CrCCH3 has a triple bond, which is shorter than either of the Cr–C bonds to CO ligands.
e.
[Fe(CO)4]2- has more π backbonding because Fe has the lowest nuclear charge of
the metals in this isoelectronic series; this reduces the CO bonding and the energy
of the C–O vibration.
Figure 13.51 shows the highest occupied orbital in the four-atom pi system of the
ruthenium analogue to the dimanganese complexes in this problem. This orbital, with
two nodes between the metals, is shown below.
13.45
Mn
C C
Mn
A similar orbital is involved in the dimanganese complexes. Oxidation of the neutral
complex, by removing electrons from this orbital, should weaken and lengthen the
carbon–carbon bond (removing electrons from an orbital that is C–C bonding) and
strengthen and shorten the manganese–carbon bonds (removing electrons from an orbital
that is Mn–C antibonding).
13.46
13.47
a.
MX 6
b.
 ML3X 6
ML 4 X5 
c.
ML 4 2–  
 ML 4 X 2
d.
ML5 X2
e.
ML4 X2
f.
ML 4 X2+ 
 ML 3X3
a.
ML3X 3 3+ 
 MX6
[FeCl3(PPh3)3 ]3+ belongs to an unknown class of iron complexes, and this cation has not
been reported.
b.
ML 2 X3+ 
 MX4L
[FeCl3(PPh3)2 ]+ belongs to an unknown class of iron complexes, and this cation has not
been reported.
Copyright © 2014 Pearson Education, Inc.

Chapter 13 Organometallic Chemistry
195
c.
MX 2L2+ 
 MX3L
d.
[FeCl2(PPh 3 )2]+ belongs to an unknown class of iron complexes, and this cation has not
been reported.
MX 2L2
FeCl 2 (PPh3)2 is a member of a class with less than 1% of known Fe complexes. The
crystal structure of FeCl2 (PPh3)2 was reported in 2005 (O. Seewald, U. Flörke, G.
Henkel, Acta Cryst., 2005, E61, m1829).
13.48
13.49
13.50
One CO is replaced by 2-butyne. The NMR peaks are due to ethyl
CH3 ( = 0.90), ethyl CH2 ( = 1.63), and butyne CH3 ( = 3.16).
The  = 3.16 peak splits at low temperatures because the two ends
of the butyne are not identical; at higher temperatures, they become
identical on NMR time scale, perhaps through rotation about the
Mo–butyne bond. The single 31P peak indicates identical PEt3
groups, suggesting the isomer shown. The IR indicates that a CO
ligand remains on the compound; the molecular weight of the
compound shown is 574.2, well within the limits given.
O
CH3
C
Br
Mo
C
Et3 P Br
CH3
The analysis fits [(5-C5H5)Fe(–CO)3Fe(5-C5H5)].
The single CO band is consistent with this structure.
Using D3h symmetry for the central part of the molecule, a
representation based on the bridging carbonyls reduces to
A1 (IR-inactive) plus E (IR-active); the observed absorption
is for the E vibration.
D3h
E
2C3
3 C2

A 1
E
3
1
2
0
1
–1
1
1
0
h
3
1
2
2S3
0
1
–1
(5-C5H5) (3-C5H7)Ni There are 5 5 protons,
4 on the two carbons of the second Cp ring that are
not bonded to Ni, 2 on the first and third carbons
bonded to Ni, and 1 on the center C bonded to Ni.
3v
1
1
0
(x, y)
(5)
(2)
Ni H
(1) H
H
H
H
2
H
H (4)
2
13.51
Is/Ia = cot (/2) = cot 38° = 1.64
13.52
a.
In the 6-C6H6 complex, CO is acting as a stronger π acceptor, as shown in its lower
Copyright © 2014 Pearson Education, Inc.

PEt 3 C
Chapter 13 Organometallic Chemistry
196
energy C–O stretching vibrations. Because the concentration of electrons on Cr is
therefore greater in the 6-C6H6 complex, 6-C6H6 is donating more strongly to Cr
than is the 6-C4BNH6 ligand, so 6-C4BNH6 is the stronger acceptor.
13.53
b.
The longest C–C distance is the bond opposite the B–N bond,
as in the resonance structure shown. The nitrogen attracts
electrons from its neighboring carbon, which in turn attracts
electrons from the next carbon, enhancing the bond strength
between these two carbons and resulting in the shortest C–C bond,
1.374 Å. In the π orbitals (see Figure 13.22 for benzene), one of
the occupied π orbitals is antibonding with respect to the C–C
bond opposite the B–N bond, consistent with this C–C
bond being the longest in the molecule.
a.
Mass of C60:
720
Ir:
193 (most abundant isotope)
CO:
28
C9H7: 115
1056
C 60
b.
Because the carbonyl stretch decreases by 44 cm–1, there must be a significant decrease
in the electron density at Ir.
c.
The C60 is replaced by PPh3, giving (5-C9H7)Ir(CO)(PPh3).
OC
C
O
Cr
Spectral data are interpreted in the reference.
13.54
13.55
Ir
C
O
NMe 2
Cu
C
O
PPh3
If CO is liberated from (5-C5H5)Mn(CO)3, it must be replaced by some other ligand if the
18-electron rule is to be maintained. In this case, the new ligand is tetrahydrofuran (THF), a
cyclic ether that can act as a sigma donor, and compound Q is (5-C5H5)Mn(CO)2(THF):
(5-C5H5)Mn(CO)3 + THF
(5-C5H5)Mn(CO)2(THF) + CO
Q
The NC groups on carbon in H2C(NC)2 can act as donors to transition metals; each has a
lone pair on carbon. In the formation of R, the weakly bound THF is replaced by a H2C(NC)2
ligand:
H
H
C
H
N
C
C
N
C
H
+ CpMn(CO)2(THF)
Mn
C
C O
O
R
There are 0.00300 mmol of dimers in the solution. Of the metals in these dimers, one third
are Mo and two thirds are W.
Copyright © 2014 Pearson Education, Inc.

C
C

13.56
N
N
Chapter 13 Organometallic Chemistry
197
The probability that a particular molecule has the formula [CpMo(CO)3]2 = 1/3 × 1/3 = 1/9.
Therefore, 1/9 × total moles = 1/9 × 0.00300 mmol = 0.00033 mmol [CpMo(CO)3]2.
The probability that a molecule has the formula [CpW(CO)3]2 = 2/3 × 2/3 = 4/9, and
4/9 × 0.00300 mmol = 0.00133 mmol [CpW(CO)3]2.
The probability that a molecule has the formula Cp(CO)3Mo–W(CO)3Cp = 2 × 1/3 × 2/3 = 4/9,
and 4/9 × 0.00300 mmol = 0.00133 mmol Cp(CO)3Mo–W(CO)3Cp.
13.57
H
The IR bands at 1945 and 1811 cm–1 are similar to those of
[Ti(CO)4(5-C5H5)]–, suggesting similarites in structure and charge
B
between this Ti complex and Z. If the other IR bands are for B–H
H
H
stretches and the 1H NMR shows two signals of relative area 3:1, it
H
is reasonable to suggest that the boron might be present in the BH4–
Ti
ion, with three of the hydrogens in one environment, the fourth in
CO
OC
another. The peak at 2495 cm–1 is distinctly different in energy
C
from the peaks at 2132 and 2058 cm–1 and might correspond to a stretch
O C
O
involving the less abundant of the protons. All is consistent with a
structure having four CO ligands and one BH4– ligand (which has replaced
Z
the cyclopentadienyl ligand), and a negative charge.
–
Symmetry analysis indicates that this complex should have two C–O stretching bands in
the infrared (A1 and E in local C4v symmetry) and three B–H bands (two A1 and one E in
local C3v symmetry). The higher energy B–H stretch is for the terminal H; the lower
energy bands are for the bridging H atoms.
13.58
Elemental analysis: calculated: 36.4% C, 6.10% H by mass.
1
H NMR: peak at  2.02 corresponds to methyl groups on Cp ring
(15 protons), peak at  –11.00 corresponds to hydrides (which
typically have negative chemical shifts) (5 protons); protons are in
desired 3:1 ratio. IR: If the environment around the osmium is
assigned C4v symmetry, using the method of Chapter 4 gives the
representation:
C4v
E
2 C4
C2

5
1
1
2 v
3
2 d
1
CH3
CH3
CH3
CH3
H
H
CH 3
Os
H
H
H
This reduces to 2A1 + B1 + E. The A1 and E representations are IR-active, for a total of three IRactive vibrational modes, matching the three bands in the spectrum.
Copyright © 2014 Pearson Education, Inc.
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Chapter 13 Organometallic Chemistry
198

 2p

2p
13.59
a.
2s

2p
2 p


2s

B
13.60
13.61
13.62
BF
F
b.
Because the π* orbitals of BF are concentrated on the boron to an even greater
extent than π* orbitals of CO are concentrated on the carbon (because the difference
in electronegativity between the atoms in BF is greater than the difference in
electronegativity between the atoms in CO), BF would be expected to be a stronger πacceptor ligand. (For a computational comparison of BF, CO, and other diatomic
ligands, see U. Radius, F. M. Bickelhaupt, A. W. Ehlers, N. Goldberg, and R. Hoffmann,
Inorg. Chem. 1998, 37, 1080.)
c.
The stronger π-acceptor nature of BF should reduce the concentration of electrons on the
Ru atoms in comparison with [(5-C5H5)Ru(CO)2]2. Consequently, the carbonyl ligands
in the BF complex would not act as strongly as π-acceptors, and the C–O stretching
vibrations should be at higher energies than 1939 and 1971 cm-1 (reported values are
1960 and 2012 cm-1).
The product is hexaferrocenylbenzene (Figure 13.32)! In addition to being inherently interesting
because of its structure and symmetry, it has potential as a precursor to a variety of derivatives
whose properties could be tunable for electronic, catalytic, and other applications. The article
provides references to these and other potential applications.
OC
OC
O
C
O
C
Fe
Fe
C
O
C
O
CO
CO
Spectral data are interpreted in the reference.
Different software and different parameter settings will generate orbitals with slightly
different shapes and energies than those shown in the text, although the results should be
similar. The relative energies in Part c in particular may differ from those in the text,
because this calculation is sensitive to the methods used. In Part d the dz2 conical nodal
surface of the orbital is close to the p orbitals of the C5H5 rings (the dz2 lobes point toward the
centers of the rings), so the interaction between this orbital and the rings is weak.
Copyright © 2014 Pearson Education, Inc.

Chapter 13 Organometallic Chemistry
13.63
Two orbitals that show ethylene
as both a donor and acceptor are
shown below.
H
Cl
Cl
C
C
H
Donor interaction (from π orbital
of ethylene):
13.65
H
Pt
Cl
13.64
199
H
Acceptor interaction (involving
π* orbital of ethylene):
The shapes of the orbitals should be similar to those in Figure 13.8. The eg* orbitals have
different shapes because the d orbitals involved (dz2 and dx2–y2) have different shapes.
a.
See orbitals in Section 13.4.4, page 499.
b.
px , dxz
py , dyz
s, pz , dz 2
c.
Depending on the parameters used, the upper and lower lobes of the p orbitals should
merge in the lowest energy π orbital (to give clouds above and below the plane of the
nuclei), and the lobes of two p orbitals in one of the upper π orbitals (derived from the p
orbitals in front in the diagram at upper right, above) may also merge.
d.
The analysis can proceed similarly to the discussion of ferrocene in Section 13.5.2.
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Chapter 13 Organometallic Chemistry
200
13.66
a.
2-Node Group Orbitals
z
y
x
1-Node Group Orbitals
0-Node Group Orbitals
2-Node Group Orbitals:
b.
1-Node Group Orbitals:
px
0-Node Group Orbitals:
py
dxz
s, d
z2
dyz
pz
The representation , shown below, reduces to the irreducible representations listed.
These match the group orbitals and their matching metal orbital assignments in Parts a
and b. The valence s orbital matches the A1g representation, so both the s and dz2 orbitals
of nickel can interact with the zero-node group orbital on the left.
c.
D4h

A1g
B2g
A2u
B1u
Eg
Eu
d.
none
dxy
E
8
1
1
1
1
2
2
2C4
0
1
–1
1
–1
0
0
C2
0
1
1
1
1
–2
–2
2C2
0
1
–1
–1
1
0
0
2C2
0
1
1
–1
–1
0
0
i
0
1
1
–1
–1
2
–2
2S4
0
1
–1
–1
1
0
0
h
0
1
1
–1
–1
–2
2
2d
4
1
1
1
1
0
0
z2
xy
z
(xz, yz)
(x, y)
Comparisons can be made similarly to those for ferrocene in Section 13.5.2.
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
2v
0
1
–1
1
–1
0
0
Chapter 14 Organometallic Reactions and Catalysts
201
CHAPTER 14: ORGANOMETALLIC REACTIONS AND CATALYSIS
14.1
–
a.
[Mn(CO)5] + H2C=CH–CH2–Cl
b.
Oxidative addition
–
H2C=CH–CH2–Mn(CO)5 + Cl
CH3
Cl
PPh3 + CH3I
Ir
Ph3P
Ir
Ph3P
C
O
C
O
c.
Ir
Ph3P
PPh3
I

Cl
Ph3P
Ir
Ph2P
H
PPh3
PPh3
Two possibilities: methyl migration and addition, or ligand substitution
O
O
C
C
+ PPh3
Fe
CH3
PPh3
or
Fe
CH3
PPh3
C
O
14.2
PPh3
Cyclometallation
Cl
d.
Cl
C
O
+ CO
Fe
CH3
C
O
e.
(5-C5H5)Mn(CO)3[C(=O)CH3]
(dissociation and methyl migration)
f.
H3C–Mn(CO)5 + SO2
Mn bonds)
a.
H3C–Mn(CO)5 + P(CH3)(C6H5)2
H3C(C=O)–Mn(CO)4[P(CH3)(C6H5)2]
(methyl migration and phosphine addition)
b.
[Mn(CO)5] + (5-C5H5)Fe(CO)2Br
(nucleophilic displacement)
–
(5-C5H5)Mn(CO)3(CH3) + CO
H3C(SO2)–Mn(CO)5
(1,1 insertion, with C—S and S—
(CO)5Mn–Fe(5-C5H5)(CO)2 + Br
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–
202
Chapter 14 Organometallic Reactions and Catalysts
H
c.
OC
Ph3P
trans-Ir(CO)Cl(PPh3)2 + H2
(oxidative addition)
Ir
PPh3
H
Cl
14.3
d.
W(CO)6 + C6H5Li
e.
Alkyl migration of CH3 to one of the adjacent CO ligands, followed by addition of 13CO.
Isomeric products: 1/3 each of the fac enantiomers, 1/3 mer isomer. There should be
no 13C in the acyl group.
f.
Alkyl migration of CH3 to one of the adjacent CO ligands, followed by addition of 13CO.
Isomeric products: two mer species, enantiomers if the 13CO is taken into account,
identical otherwise.
a.
cis-(13CO)(CH3CO)Mn(CO)4
Mn(CH3)(CO)5, 25% with no 13CO, 25% with 13CO
13
trans to CH3, 50% with CO cis to CH3.
b.
–
[C6H5COW(CO)5] + Li+ (nucleophilic attack on carbonyl C)
C6H5–CH2Mn(CO)5
O O
C C
h
CO +
C
C O
O
c.
[V(CO)6] + NO
d.
Cr(CO)6 + 2 Na/NH3
e.
Fe(CO)5 + NaC5H5
Na+ + [(5-C5H5)Fe(CO)2]– + 3 CO
f.
[Fe(CO)4]2– + CH3I
[(CH3)Fe(CO)4]– + I–
g.
[V(CO)5(NO)] + CO
CH3
Ph3P
Rh
PPh3
2 Na+ + [Cr(CO)5]2– + CO

CH4 +
Ph3P
14.4
Mn
Ph2P
Rh
Ph3P
[(C5H5)Fe(CO)3]+ + NaH A (C7H6O2Fe)
B (colorless gas) + C (C7H5O2Fe) (purple-brown)
A
PPh3
A = (5-C5H5)Fe(CO)2H
B = H2
(See Figure 13.35, p. 509 for structure of C.)
C= [(5-C5H5)Fe(CO)2]2
D (C7H5O2FeI) (brown)
D = (5-C5H5)Fe(CO)2I
C + I2
D + TlC5H5
E (C12H10O2Fe) + TlI
(See Figure 13.35, p. 509 for structure of E.)
E = (C5H5)2Fe(CO)2
(one Cp 1 and one Cp 5)
E
F = (5-C5H5)2Fe, ferrocene
F (C10H10Fe) + colorless gas (CO)
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Chapter 14 Organometallic Reactions and Catalysts
14.5
A (C10H12FeO2S)
[(5-C5H5)Fe(CO)2]– + ClCH2CH2SCH3
A + heat
203
B
A has bands at 1980 and 1940 cm–1, B at 1920 and 1630 cm–1.
The sulfur reagent loses Cl– and bonds to the Fe as an alkyl ligand to form A, with two carbonyls.
This is an example of nucleophilic displacement of chloride by an organometallic anion. A then
rearranges, with the S becoming attached to Fe, and the alkyl migrates to a carbonyl carbon. B
contains an ordinary carbonyl and an acyl C=O bond, for the two quite different C–O stretching
energies.
O
C
S
Fe
CH2
C
O
CH3
O
C
CH3
Fe
CH2
S
C
O
A
14.6
a.
CH2
CH2
B
Two term rate laws like this could be the result of two parallel associative reactions—
the first by solvent, the second by the phosphate—or could result from a dissociative
reaction for the first term and an associative reaction for the second.
First term, dissociative, k1 [V(CO)5(NO)]:
V(CO)4(NO) + CO
V(CO)5(NO)
PR3
V(CO)4(NO)(PR3)
Second term, associative, k2[PR3][V(CO)5(NO)]:
PR3
V(CO)5(NO)
b.
14.7
V(CO)5(NO)(PR3)
V(CO)4(NO)(PR3) + CO
V(CO)5[P(OCH3)3](NO) If the NO is a bent, 1-electron donor, it can be an 18electron species ( 5 + 5×2 + 2 + 1 = 18). If the NO is a linear, 3-electron donor, the
total is 20 electrons.
Co2(CO)8
Co2(CO)7 + CO K1 (fast equilibrium) [Co2(CO)7] = K1[Co2(CO)8]/[CO]
Co2(CO)7 + H2
Co2(CO)7H2 k2 (slow)
Co2(CO)7H2 + CO
2 HCo(CO)4 (fast)
Rate = k2[Co2(CO)7][H2] = k2K1 [Co2(CO)8][H2]/[CO]
14.8
This depends on the cone angle of the phosphine ligands, with the order PPh3 > PBu3
(estimated) > P(OPh)3 > P(OMe)3 from Table 14.1 (p. 543). The PPh3 should dissociate
most rapidly and the P(OMe)3 should dissociate least rapidly.
14.9
K for the dissociation reaction is in the order PPh3 > PMePh2 > PEt3 > PMe3, as a result of a
combination of decreasing cone angle and increasing negative charge on the phosphorus.
Alkyls push more electron density onto P than phenyl rings.
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204
14.10
Chapter 14 Organometallic Reactions and Catalysts
a.
Tolman employed Ni(0) complexes of general formula Ni(CO)3 (PX1X 2 X 3 ) where
the substituents of the monodentate phosphine were varied.
b.
The parameter  was defined as follows to determine the energy of the
Ni(CO)3 (PX1X 2 X 3 ) A1 carbonyl stretching mode, with each phosphine substituent
contributing a  value. The 2056.1 cm -1 relatively low energy  CO absorption is the
A1 mode of Ni(CO)3(P t Bu 3 ) , containing the strong  -donor P t Bu 3 .
3
 CO ( A1 )  2056.1   cm -1
c.
i1
-1
These  values range from 0 to 19.6 cm , with the highly electronegative Cl (14.8), F
(18.2) and CF3 (19.6) substituents contributing the most towards shifting the A1 carbonyl
stretching mode to higher energy.
14.11
d.
The debate over the classification of phosphine ligands as  -donors and  -acceptors is
discussed in this reference. Tolman states that while highly electronegative substituents
have higher  values (and cause higher energy  CO absorptions) and more electrondonating substituents have lower  values, the relative importance of  donation and 
acceptance for these phosphine-nickel bonds is difficult to assess.
a.
The conversion of the Ni(II) complex to a Ni(0) complex is shown here:
PPh2
PPh2
N
Ni
H
CO
CO
NH
Ni
CO
PPh2
PPh2
Upfield-shifted metal-hydride 1H resonances are common. In this reaction, the Ni—H
resonance (-18 ppm) is replaced by one at 8.62 ppm, consistent with formation of the
secondary amine. Two carbonyl absorptions are consistent with the cis carbonyl product
shown. The authors speculate that the first step in this reaction is CO coordination at
Ni(II) to afford a five coordinate intermediate that presumably brings the amido nitrogen
and the hydride into cis positions. Reductive elimination results in N—H bond formation.
Initial CO binding renders the Ni(II) center more electrophilic (via -backbonding) and a
better oxidizing agent. CO binding also generates a five-coordinate complex with the
amido nitrogen and the hydride in closer proximity to facilitate the orbital overlap
necessary for reductive elimination.
Copyright © 2014 Pearson Education, Inc.
Chapter 14 Organometallic Reactions and Catalysts
b.
205
The alternate reaction pathway exhibited by the Ni—CH3 complex is:
PPh2
N
Ni
H 3C
PPh2
CH3
CO
C
Ni
N
PPh2
O
PPh2
CO
N
CH3
PPh2
O
CO
Ni
PPh2
CO
The first reaction is proposed as a migratory insertion of CO into the Ni—CH3 bond,
affording an acyl complex with a diagnostic 1621 cm -1 carbonyl absorption. Further
reaction with CO results in reductive elimination of a N—C 2 bond and formation of a
sp
neutral amide-functionalized bidentate phosphine ligand. As in Part a, the first step in the
reductive elimination reaction is hypothesized as CO coordination at Ni(II) of the
intermediate, resulting in a five-coordinate species with the amido nitrogen atom and the
acyl ligand carbon atom in close proximity.
c.
The interesting feature of the Part b scheme is that no N—C
sp3
bond formation occurs
when CO is added to the Ni—CH3 complex. The authors of the reference primarily
attribute this to an electronic effect. The more electron-rich Ni(II) center in the Ni—CH3
complex is less reactive towards reductive elimination compared to the less electron-rich
Ni(II) center in the Ni—H complex. Note that these reductive eliminations are proposed to
begin via initial coordination of CO, rendering the metal center more electrophilic for
subsequent reductive elimination by virtue of -backbonding. In these reductive
elimination reactions, Ni(II) formally serves as the oxidizing agent (it goes from Ni(II) to
Ni(0)); a less electron-rich Ni(II) would be expected to be a better oxidizing agent. An
idea not mentioned in the reference is that activation barriers for reductive eliminations
involving H atoms are generally lower than those involving CH 3 groups (leading to faster
reductive elimination rates with H atoms). The spherical symmetry of the 1s H orbital
better facilitates effective overlap throughout reductive elimination pathways when
compared to the sp3 hybrid orbital employed by a methyl group.
14.12 a.
This double-cyclometallation very likely occurs via  -bond metathesis. A sketch of a
possible four-centered transition state (compare to the general  -bond metathesis
transition state in Figure 14.8) is shown below. This structure presumably leads to
toluene expulsion with concomitant orthometallation of the aromatic ring.
Ph
Ph H C
2
Ph
H 2C
H2C
N
b.
H
Zr
N
CH2
N
N
Since Zr(IV) is in its highest oxidation state, and is not electronically predisposed to
participate in sequential oxidative addition/reductive elimination steps to affect
orthometallation via metal hydride intermediates, a  -bond metathesis pathway seems
like the best choice. The rate of this intramolecular process would be expected
Copyright © 2014 Pearson Education, Inc.
206
Chapter 14 Organometallic Reactions and Catalysts
independent of H2 pressure. This pathway includes neither the release of hydrogen gas
nor hydride intermediates that could arise from hydrogen gas.
The reference does not speculate as to why the rate with R = CH 2 Ph is the slowest. It is
possible that the steric bulk of the four aromatic rings in the proposed transition state
structure may raise the activation barrier for formation of the four-centered transition
state, decreasing the reaction rate.
14.13
If it loses CO followed by migration of CH3 to an adjacent position, all the CO lost should
be 12CO, because the CO ligands cis to the CH3C=O will be the ones lost.
14.14
a.
CH3
There are two possible products,
O
both a result of methyl migration
C
1 13 O
followed by carbonyl addition. The
C
methyl group can move to either
Mn
PMe3
CO
the 1 position or the 2 position.
C
The resulting products are different
O
PMe3
only in the location of the 13CO.
b.
14.15
[(C5H5)2Fe2(CO)4] + Na/Hg
2
C
O
H3C
O
C
A
Mn
PMe3
PMe3
C
C + PhNa
A + D
A:
B:
C:
D:
H
Br
Fe
B
B + LiAlH4
a.
C
The new IR band is from the acyl carbonyl, and should be near 1630 cm–1 (see
Problem 14.5).
–
A + Br2
14.16
O
O
C 13 O
C
Fe
Fe
CO
C
O
CO
C
O
C
O
A
B
C
Sodium acts as reducing agent. Anion A has lower energy C–O stretches, as
expected (see p. 488).
Br2 acts as oxidizing agent and is the source of the bromo ligand.
The NMR peak at –12 ppm is caused by the hydride ligand.
benzene, C6H6
The bromoethoxide anion adds to a carbonyl; the hard oxygen of the anion adds to the
hard carbon. Then Br– is lost, leaving a positive carbon. The alkyl tail can then bend
around and react with the carbonyl oxygen, giving the compound shown here:
O
C Br
Re(CO)5Br + BrCH2CH2O–
OC
Re
–
C C
O O
O
C Br
O
OC
C
OCH2CH2Br
Re
O
C
OCH2CH2Br
C C
O O
O
C Br
OC
Re
O
+ Br–
C
C C
O O
Y
Copyright © 2014 Pearson Education, Inc.
–
O
Chapter 14 Organometallic Reactions and Catalysts
b.
14.17
207
There are 2 electrons donated from each of the six ligand positions, and Re+ has 6
electrons for a total of 18. In the isomer shown, there are three different carbonyls,
and the carbene ligand has two identical carbons, so there are five different magnetic
environments for carbon. In the other possible isomer, with Br trans to the alkoxide
ligand, there are not five different carbon environments. Finally, Ag+ can remove
Br– from the complex.
Nitrenium ligands are similar to N-heterocyclic carbene (NHC) ligands (see Figure 14.33(a))
except that they have a nitrogen atom in place of the coordinating carbon; three nitrogens are in a
row. R groups that have been studied include those with phosphine arms that can also coordinate
to metals.
R2
R2
M
P
M
P
R
N
N
N
+
R
N
N
+
N
Nitrenium ligands have been demonstrated to be weaker sigma donors than NHC ligands on the
basis of infrared and bond length evidence, supported by density functional theory calculations;
see the reference for details. The phosphine arms that also coordinate to metals give nitrenium
ligands pincer characteristics comparable to the ligands discussed in Section 14.1.5
14.18
a.
This is the example described in Example 13.4, p. 531.
I + PPh3
II
II:
IR: 2038, 1958, 1906 cm–1
NMR: 7.62, 7.41 muliplets (15), 4.19 multiplet (4)
II + PPh3
III
III:
IR: 1944, 1860 cm–1
NMR: 7.70, 7.32 multiplets (15), 3.39 singlet (2)
II has 3 CO ligands in a fac geometry. The NMR shows one PPh3 (15) and the ethylene
hydrogens (4).
III has only 2 CO ligands in a cis configuration. The ratio of NMR integrated peaks is
now 15:2 because there are two PPh3 groups.
b.
I + Ph2PCH2PPh2
IV
IV
2036, 1959, 1914 cm–1, 35.8% C, 2.73% H
Replacing two CO’s with the phosphine gives a compound with 25.7% C, 3.1% H.
Replacing one CO and the Br with the phosphine gives a compound with 28.9% C,
3.3% H.
The only way to get the analysis to work out is to have a single diphos ligand bridging
two Re atoms after loss of CO from each:
[(Re(CO)3Br(C(O2C2H4))]2{(PPh2)2CH2}, IV, has 36.14% C, 2.46% H.
Copyright © 2014 Pearson Education, Inc.
208
Chapter 14 Organometallic Reactions and Catalysts
c.
Re(CO)5Br + V
I + S2CN(CH3)2–
V has no metal, has no IR bands between 1700 and 2300 cm–1, has NMR bands at
3.91 (triplet), 3.60 (triplet), 3.57 (singlet), and 3.41 (singlet).
Acting as a nucleophile, the dithiocarbamate ion attacks the carbene to form
(CH3)2NC(=S)SC2H4O–, which can pick up a proton from trace amounts
of water to make the hydroxy compound (CH3)2NC(=S)SC2H4OH, V. The 1500 and
977 cm–1 bands are the N–C and C–S bands, and the NMR shows triplets for the two CH2
units, singlets for the CH3’s.
14.19
The mechanism is the one described in problem 16, with formation of the alkoxide ion by the
reaction:
O + Br–
14.20
BrCH2CH2O–
The reaction proceeds by substitution of Mn(CO)5– for Br–, followed by alkyl migration,
addition of Mn(CO)5–, and finally cyclization:
Mn(CO)5– + BrCH2CH2CH2Br
(CO)5MnCH2CH2CH2Br + Br–
Mn(CO)5–
O O
C C
–
Mn
(CO)5Mn
C C
O O
O O
C C
O
C
(CO)5Mn
CH2CH2CH2Br
Mn
O
C
CH2CH2CH2Br
C C
O O
O O
C C
(CO)5Mn
Mn
–
O
C
+ Br–
C C
O O
14.21
The acyl metal carbonyl has a resonance structure with a negative charge on oxygen, shown
below, where the proton can add. Protonation is then similar to the alkylation reaction described
in Section 13.6.2.
–
O
O
C
R
M(CO)x
C
+
M(CO)x
R
Copyright © 2014 Pearson Education, Inc.
Chapter 14 Organometallic Reactions and Catalysts
14.22
a.
Acetaldehyde from ethylene:
C2H4 + CO + H2
[PdCl4]2–
209
O
H3C
C
+ H2O
H
The Wacker process (Figure 14.21) with ethylene as the starting alkene will
result in acetaldehyde.
b.
Ethyl propionate from chloroethane:
[Co(CO)4]–
14.23
CH2CH2Cl
CH3CH2Co(CO)4
CO
(CH3CH2CO)Co(CO)4
C2H5OH
CH3CH2COOC2H5 + HCo(CO)4
c.
Pentanal from 1-butene: The hydroformylation process (Figure 14.18) does
this.
d.
4-phenylbutanal from an alkene: Again, the hydroformylation process should do
this, starting with 4-phenyl-1-propene.
e.
Wilkinson’s catalyst [ClRh(PPh3)3] (Figure 14.22) should do this. D2 would be added
across the least hindered double bond.
f.
Catalytic deuteration with H3TaCp2 (Figure 14.17) should deuterate the
phenyl ring without affecting the methyl hydrogens.
Figure 14.19 shows this process of hydroformylation. n-pentanal results if
R = C2 H5. Identification of the steps:
1. Dissociation of one CO
2. Addition of the alkene
3. 1,2 insertion
4. Addition of CO
5. Alkyl migration
6. Oxidative addition
7. Reductive elimination of the aldehyde
14.24
Hydroformylation (Figure 14.19) with Rh(CO)2(PPh3)2 as the catalytic species
will work, starting with 2-methyl-1-butene: H3C–CH2–C(CH3)=CH2.
Copyright © 2014 Pearson Education, Inc.
210
14.25
Chapter 14 Organometallic Reactions and Catalysts
a.
Direct metathesis would occur as follows:
In addition, reactants could undergo self-metathesis, for example
+
and the products of the direct metathesis could undergo further metathesis to give a
variety of products. One of these could be formed as shown here:
+
b.
c.
d.
This is Hérisson and Chauvin’s classic experiment, with products shown in Figure 14.27.
As in Part a, self metathesis can also occur, and the products of direct metathesis can also
undergo further metathesis to give a variety of products.
+
+
Copyright © 2014 Pearson Education, Inc.
Chapter 14 Organometallic Reactions and Catalysts
14.26
211
The catalyst is an 18-electron species with M = W. The first steps in the catalytic cycle are
diagrammed here:
W
W = (CO)5W
R = C6H5
CR2
W
W
CR2
CR2
+
=
CR2
W
+
CR2 =
W
CR2
W
CR2
W
...
Polymer
14.27
a.
Me
Pr
+
+
Me
Pr
+
Me
Me
Me
Pr
Me
+
Pr
Me = methyl
Pr = n-propyl
b.
+
Pr
“Double cross” product
+
Later
metathesis
products
Pr
The pairwise mechanism would call for the dimethyl and dipropyl products to be
formed first, followed by the “double cross” product. In the non-pairwise mechanism,
the double cross product should form simultaneously with the dimethyl and the
dipropyl products. Experimental results showed that the double cross product formed
(along with the dimethyl and dipropyl products) at the beginning of the reaction,
consistent with the non-pairwise mechanism.
Copyright © 2014 Pearson Education, Inc.
212
Chapter 14 Organometallic Reactions and Catalysts
H H
14.28
a.
C
(C6F5)2B
B(C6F5)2
H
b.
Zr
(Cp)2
H
One possible route:
H
Cp2Zr
+ H2 C
CH2
Cp2Zr
H2C
empty site
1, 2-insertion
H
Cp2Zr
CH2CH3
CH2
C2H4
etc.
Cp2Zr
1, 2-insertion
CH2CH2CH2CH3
Cp2Zr
H2C
14.29
a.
CH2
Details of the scheme, together with schematic diagrams of the proposed catalytic cycle
and related cycles, are presented in the reference. The catalyst precursor is the 1,2dimethoxyethane (DME) adduct of N  W(OC(CF3 )2 Me)3 , dubbed 1-DME (the crystal
structure is provided in the reference). Proposed steps include:
Reaction of 1-DME with 3-hexyne to form metallacycle
Dissociation of nitrile, leaving tungsten carbyne complex
Reaction with p-methoxybenzonitrile to form second metallacycle
Dissociation of product to regenerate 1-DME
1.
2.
3.
4.
b.
CH2CH3
The first product was considered to be 1-(4-methoxyphenyl)-1-butyne, formed as shown
below. However, this compound accounted for only 15 mole percent of the reaction
products. The major product, bis(4-methoxyphenyl)acetylene, accounting for 77 mole
percent of the products, was attributed to secondary metathesis, also involving
metallacycle intermediates. The article discusses in detail alternative cycles for forming
bis(4-methoxyphenyl)acetylene.
Et
N
W
X3
1-DME
+
Et
C
N
C
C
W
X3
C
Et
N
C
+
X3W C
Et
Et
Et
+ N
C
OCH3
=
N
W
X3
+ Et
C
C
OCH3
1-(4-methoxyphenyl)-1-butyne
H3CO
C
N
C
X3W
C
C
bis-(4-methoxyphenyl)acetylene
Copyright © 2014 Pearson Education, Inc.
OCH3
Et
OCH3
Chapter 14 Organometallic Reactions and Catalysts
213
14.30
CH3
Fe
CH3
CH3
C 2 H4
H2C
Fe
CH2
H3 C
CH2
P(OCH3)3
Fe
(H3CO)3P
P(OCH3)3
H2C
X
14.31
RhCl3  3 H2O + P(o-MePh)3
Rh–Cl  = 351 cm–1
I (blue-green) (C42H42P2Cl2Rh) eff = 2.3 B.M.,
I is trans-RhCl2(PR3)2 The IR is the Rh–Cl asymmetric stretch. The cis isomer would give
two IR bands. One unpaired electron (15 electron species, square planar).
I + heat
II (yellow, diamagnetic) Rh:Cl = 1,  = 920 cm–1
II Loss of one Cl and combination of two CH3’s to form a single tridentate phosphine ligand
with a π bond. Sixteen electrons around Rh. The double bond in the ligand is perpendicular
to the Rh–Cl–P plane because of the size and geometry of the benzene rings.
II + SCN–
NMR: 
6.9–7.5
3.50
2.84
2.40
III Rh(SCN)
Area
12
1
3
3
II:
Type
aromatic
doublet of 1:2:1 triplets
singlet
singlet
R2P
Rh
PR2
Cl
III: Same overall structure as II, with SCN substituted for Cl. The singlets are the methyl
protons (two separate environments—not obvious from the drawing), the doublet is from the
vinylic protons, and the aromatic multiplets are the sum of all the phenyl protons.
II + NaCN
NMR: 
7.64
6.9 - 7.5
2.37
IV (C21H19P, mw = 604)  = 965 cm–1
Area
1
12
6
Type
singlet
aromatic
singlet
IV is the diphosphine ligand, C42H38P2, shown above in compound II. (CN– substitutes for all the
ligands in this reaction.) The IR band is characteristic of trans vinylic hydrogens. All
methyl protons are equivalent in the free ligand (singlet), the vinylic protons are a singlet at
7.64, and the phenyl protons are a multiplet at 6.9–7.5. The change in the vinyl hydrogen IR
band shows that the ethylene does coordinate to Rh. Coordination reduces the C=C bonding,
which also reduces the C–H bending energy (electrons are drawn away from C–H, toward
C–Rh).
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214
Chapter 14 Organometallic Reactions and Catalysts
14.32
a.
[MeC
C(CH2)2OOC(CH2)]2 = H3CC
CCH2CH2OOCCH2CH2COOCH2CH2 C
O
H 3C
C
C
C
C
O
C
O
C
O
C
C
MeC
C(CH2)8COO(CH2)9 C
O
C
C
C
C
CH3
C
C
CH3
O
O
b.
CCH3
O
CMe
O
H 3C
C
O
C
O
C
O
14.33
Evidence in support of the intermediate shown at the right includes:
(1)
(2)
An infrared band at 2104 cm–1 has been observed. This
I
band is similar to the higher energy band in the Ir
I
analogue, [CH3Ir(CO)2I3]–, which has carbonyl bands at
2102 and 2049 cm–1. A second band, expected for
[CH3Ir(CO)2I3]–, would be hidden under strong bands of
the reactants and products of steps 2 and 3 of the mechanism.
–
CH3
Rh
CO
CO
I
The ratio of the absorbance of the 2104 cm–1 band to the absorbance of a
band at 1985 cm–1 of [Rh(CO)2I2]–, the reactant in step 2, is proportional to
the concentration of CH3I. This is consistent with what would be expected
Copyright © 2014 Pearson Education, Inc.
Chapter 14 Organometallic Reactions and Catalysts
215
from the equilibrium constant expression for formation of the intermediate,
K
[Rh(CO) 2 I2 ] – [CH 3 I]
[CH 3 Rh(CO) 2 I3 ] –
and is consistent with the steady state approximation for the mechanism.
(3)
(4)
14.34
The maximum intensity of the 2104 cm–1 band occurs when the product of
step 3 is formed most rapidly.
When 13CH3I is used, a doublet is observed in the NMR consistent with
C–103Rh coupling. Other NMR data also support the proposed structure.
13
–
A: [Fe(CO)3(CN)3]
B: cis-[Fe(CO)2(CN)4]2
–
–
The cyanide ion has the capacity to replace ligands such as I and CO. The C–N stretch
involves a large change in dipole moment, so this vibration, like that of CO, can be useful
in characterizing cyano complexes. Because CN has a slightly smaller reduced mass than CO,
C–N stretches typically occur at slightly higher energies than C–O stretches (see Section 13.4.1).
In this situation, both A and B have two sets of C–N and C–O stretches, indicating that both
complexes have at least two of both ligands. As more cyano ligands are added, the concentration
of electrons on Fe increases, and both types of ligands become stronger π acceptors, reducing the
energy of the stretching vibrations.
The formula of B suggests the possibility of both cis and trans isomers. The observation of two
bands in the carbonyl region is consistent with the cis isomer. (See also Problem 13.25 and its
reference.)
14.35
2:
Mo
P
Mo
C
O
H
CO
H
The two IR bands are as expected for this dicarbonyl complex. NMR peaks can be assigned as
follows: chemical shift 5.28 (relative area 5): Cp; 1.31 (27): nine methyl groups on t-butyl
groups on benzene ring; 5.15 (3): protons on benzene ring; 5.46 (2), 4.22 (2) and hidden small
peak: 4-C5H6. In the 13C NMR, the resonance at 236.9 ppm can be assigned to the carbonyl
carbons. This product is formed via an unusual attack of a hydride on a cyclopentadienyl
carbon; for an additional example of this type of attack, see footnote 17 in the reference.
Copyright © 2014 Pearson Education, Inc.
216
14.36
Chapter 14 Organometallic Reactions and Catalysts
X:
B
N
H
H
Y:
B
N
H
Z:
B
N
TBS
Cr
OC
CO
C
O
H
H
In heterocycle Z coupling of the N–H proton with the 14N nucleus (which has S = 1) results in
a triplet, and coupling for the B–H proton with the 11B nucleus (which has S = 3/2) results in a
quartet. (The original spectrum is shown in the reference.)
14.37
See Figure 13.32 for a diagram of this molecular “ferrous wheel”!
Copyright © 2014 Pearson Education, Inc.
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
CHAPTER 15: PARALLELS BETWEEN MAIN GROUP AND
ORGANOMETALLIC CHEMISTRY
15.1
a.
Mn2(CO)10 + Br2
2 Mn(CO)5Br
H
C
b.
–
HCCl3 + excess[Co(CO)4]
Co(CO)3
(CO)3Co
Co
(CO)3
c.
Co2(CO)8 + (SCN)2
2 NCSCo(CO)4
R
C
O
C
d.
Co2(CO)8 + C6H5–CC–C6H5
Co
O
15.2
15.3
e.
Mn2(CO)10 + [(5-C5H5)Fe(CO)2]2
a.
Tc(CO)5
b.
[Re(CO)4]–
CH2
c.
[Co(CN)5]3–
CH3
d.
[CpFe(C6H5)]+
e.
[Mn(CO)5]+
CH3+
f.
Os2(CO)8
H2C=CH2
C
C
C
O
CH4
Many examples are possible. Two for each:
CH3
CH3
(6-C6H6)Mn(PPh3)2
[Ni(CS)(PMe3)2Cl2]+
b.
CH
CH
(4-C4H4)Co(CS)
(6-C6H6)Rh
O
C
Co
C
O
C
O
2 (CO)5Mn–Fe(5-C5H5)(CO)2
CH3
a.
R
Copyright © 2014 Pearson Education, Inc.
217
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
218
15.4
15.5
c.
CH3+
CH3+
d.
CH3
–
CH3
e.
(5-C5H5)Fe(CO)2
(5-C5H5)Fe(CO)2
f.
Sn(CH3)2
Sn(CH3)2
Mo(borazine)(PMe3)(CS)
–
[V(CO)3(en)]
–
(8-C8H8)Ru(PEt3)
[Co(N2)(CO)2(bipy)]+
Co(CO)3Cl2
–
[(6-C6H6)Cr(CO)2]
(5-C5H5)Ir(CO)
–
[Tc(CO)4]
Many possibilities exist; some of those given here may not be synthetically accessible.
a.
C2H4
b.
P4
c.
cyclo-C4H8
d.
S8
a.
All three complexes are 18 electron species, with the benzene ring of
[(C5Me5)Fe(C6H6)]+ replaced by three carbonyl groups or two carbonyls and a
phosphine. All have a formal coordination number of six.
b.
The experimental results are more complex than might be expected.
(CO)4Fe=Fe(CO)4
[Ir(CO)3]4
(CO)4Fe
Ru(CO)4
(CO)4Ru
Fe(CO)4
cyclo-[Fe(CO)4]8
[(C5Me5)Fe(C6H5)]++ H–
Fe
H
H
[(5-C5H5)Fe(CO)3]+ + H–
5


+
(5-C5H5)Fe(CO)2H
–
[( -C5H5)Fe(CO)2PPh3] + H


OC
[(5-C5H5)Fe(CO)2]2 + H2
PPh3
Fe
CO
(See A. Davison, M. L. H. Green,
and G. Wilkinson, J. Chem. Soc.,
1961, 3172.)
H
H
Copyright © 2014 Pearson Education, Inc.
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
15.6
a.
CH2, Fe(CO)4, [Mn(CO)4]–, and PR2 each has two frontier orbitals, each with one
electron. Each fragment is two ligands short of the parent polyhedron (octahedron
or tetrahedron).
b.
Fe(CO)4 and CpRh(CO) each has two frontier orbitals, each with one electron.
c.
[Re(CO)4]– has two frontier orbitals, with one electron in each. R2C is isolobal with
two frontier orbitals, each with one electron.
219
15.7
The W(Cp)(CO)2 fragments are isolobal with CR and CPh, each with three orbitals containing
one electron each; PtR2 and Cu(C5Me5) are isolobal, each with two orbitals containing one
electron each.
15.8
a.
If Mn has the lower energy orbital, then the bonding molecular orbital has a greater
contribution by the Mn orbital, and the electrons in the orbital are polarized toward Mn.
b.
The gold orbitals are higher in energy. Rather than matching energies with the lowest
of the π orbitals of the Cp ring, the Au orbitals will match better with the higher π
orbitals, which have a nodal plane cutting across the ring.
a.
CH2 and Fe(CO)4 are isolobal, each with two orbitals containing one electron apiece.
b.
Mn(CO)2(C5Me5) is isolobal with [Mn(CO)5]+, a 16 electron species, which is in turn
isolobal with CH2. The Mn–Sn–Mo fragment is similar to allene, C=C=C, in which the
double bonds force a linear geometry.
15.9
15.10
[C(AuPPh3)5]+ Structure: slightly distorted trigonal bipyramid, as expected from VSEPR.
[C(AuPPh3)6]2+ Structure: slightly distorted octahedron, as expected from VSEPR.
The structure of the 7-coordinate [C(AuPPh3)7]3+ has not been reported, but interest in the realm
of highly coordinated carbon compounds continues; for a recent update, see H. G. Raubenheimer
and H. Schmidbaur, Organometallics, 2012, 31, 2507.
In these complexes, each of the AuPPh3 groups can be viewed as having an sp hybrid orbital
pointing in toward the carbon. Interactions between these hybrids with the s and p orbitals
of carbon give rise to four bonding orbitals in each case. The eight valence electrons available
fill these orbitals. The result is the equivalent of four bonds spread over the complex (bond
order of 4/5 in [C(AuPPh3)5]+ and 4/6 in [C(AuPPh3)6]2+. Metal-metal bonding is also likely
to contribute to the stability of these structures.
For more details, see H. Schmidbaur, et. al., Angew. Chem. Int. Ed. Engl., 1989, 28, 463
and Angew. Chem. Int. Ed. Engl., 1988, 27, 1544.
15.11
Carbonyl ligands are significantly stronger π acceptors than cyclopentadienyl. Replacing three
CO ligands with Cp on each metal atom, therefore, leaves the metals with greater electron density
for involvement in metal–metal bonding, leading to shorter metal–metal bonds in the cobalt
complexes. In addition, there is a general decrease in atomic radii from left to right in this row of
transition metals, so Co atoms are slightly smaller than Fe atoms (see Table 2.8).
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
220
15.12
15.13
a.
The orbital interactions are similar in the two cases, except that the π orbitals of the
P5 ring are lower in energy than those of the C5H5 ring, resulting in a stronger ability
of the P5 ring to accept electrons from the metal (see also reference 11 in this chapter).
b.
Because the P5 ring acts as a strong acceptor, electrons are accepted into antibonding
–
orbitals in the titanium complexes, leading to longer P–P distances than in P5 (calculated
at 2.12 Å). The Ti–P distances are calculated to be shorter in the theoretical structure
–
2–
[Ti(P5)] than in [Ti(P5)2] , in which ligands compete for electrons in the same d orbitals
on the metal. The consequence is that the predicted P–P distance is longer (2.24 Å) in
–
[Ti(P5)] , where the ligand can act as a stronger acceptor because it is closer to the metal,
2–
than in [Ti(P5)2] (calculated: 2.175 Å; experimental in reference 11: 2.154 Å).
a.
The Mn(CO)5 fragments have a single electron in their HOMO, largely derived from
the dz2 (hybridized with the pz orbital) of Mn. Lobes of the HOMOs of two Mn(CO)5
fragments interact in a sigma fashion with the g orbital of C2 (see Figure 5.5 for the
approximate shape) derived primarily from the pz orbitals of the C atoms:
O
C O
C
OC
Mn
O
C O
C
C
C
C
O C
O
b.
The empty π* orbitals of C2 can interact with occupied d orbitals of Mn(CO)5 fragments:
OC
Mn
C
O C
O
15.14
CO
C
O C
O
O
C O
C
c.
Mn
O
C O
C
C
C
Mn
CO
C
O C
O
The reference points out other interactions, such as between π (bonding) orbitals of
C2 and Mn(CO)5, and discusses the relative energies of the molecular orbitals of this
molecule and other molecules having bridging C2 ligands.
A staggered configuration is more likely, as predicted by VSEPR. The bonding is similar to
that in the triply bonded [Os2Cl8]2–.
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
15.15
221
End view:
matches dx2–y2
15.16
Cotton’s explanation is that removal of an electron effectively changes the oxidation number
of Tc, causing the d orbitals in [TcCl8]2– to be smaller than those of [TcCl8]3–. This reduces
the orbital overlap, weakening the bonding in spite of the higher formal bond order. The change
is small—about 3 pm. (See F. A. Cotton and G. Wilkinson, Advanced Inorganic Chemistry,
5th Ed., Wiley, 1988. p. 1090.)
15.17
As the reference describes, these ions are isostructural and are based on octahedra fused at
one mutual face (occupied by three bridging chlorine atoms). The shortening of the Re–Re
distance upon reduction is attributed chiefly to two factors. As the complex is reduced, the net
positive charge on each metal is reduced (the oxidation state changes from 4 to 3.5), which
reduces the metal-metal repulsion. In addition, with the reduction of the metals, the metal d
orbitals expand, enabling more effective overlap and stronger bonding.
15.18
This example is similar to the one presented in Figure 15.10. In compound 1 there are six
DTolF ligands, each with a charge of 1–. Adding the charges of the bridging hydroxides, the
total charge of the ligands in this complex is 8–. Consequently, the total charge on the Mo atoms
must be 8+, or a charge of 4+ per Mo2 unit. Each Mo2+ has 4 d electrons, giving 8 d electrons
per Mo2. Mo24+, with its 8 d electrons, has an Mo–Mo bond order of 4 (Figure 15.10).
In compound 2 there are also six DTolF ligands. The bridging O2– ligands result in a total
ligand charge of 10–; consequently, the four molybdenums must carry an average charge
of 2.5+, or 5+ per Mo2 unit. There are now just 7 d electrons per Mo2. From Figure 15.10 we
can see that Mo25+, with 7 d electrons, corresponds to a bond order of 3.5. The lower Mo–Mo
bond order in 2 results in a longer bond.
Copyright © 2014 Pearson Education, Inc.
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
222
15.19
a.
Because the diphenylforamidinate ion has a charge of 1–, the iron atoms have an
average oxidation state of 1.5, alternatively viewed as mixed-valent Fe(I) and Fe(II).
b.
The iron atoms have a total of 7 [Fe(I)] + 6 [Fe(II)] = 13 d electrons. The 3d orbitals
interact in sigma (dz2), pi (dxz and dyz), and delta (dxy and dx2–y2) fashions to generate
bonding and antibonding orbitals of these types. The calculated order of energy of these
orbitals is shown below. Experimental data and supporting calculations suggest that the
singly occupied π*, *, , and * orbitals are very close in energy, favoring a high-spin
arrangement that, in accordance with Hund’s rule, results in seven electrons with parallel
spin.







15.20
The clamping of the nitrogen atoms that bridge the chromium atoms is postulated to enforce short
chromium—chromium bonds. The ligand, in effect, assists in pushing the chromium atoms
together. Ligand design is critical to minimize inter-ligand substituent repulsion to maximize
ligand chelation ability. While Figure 1 of the reference emphasizes steric effects, it is stated that
repulsion between filled nitrogen and chromium nonbonding orbitals is maximized (resulting in
further metal—metal bond contraction) when chelation is more robust. The aminopyridinates are
shown in the same figure to chelate less effectively to the chromium atoms; the nitrogen
substituents of the two ligands are directed slightly towards each other, reducing the efficacy of
the chelate, and leading to longer chromium—chromium distances. The amidate substituents are
directed away from the chromium—chromium bond, resulting in less inter-ligand repulsion, and
slightly shorter chromium—chromium bonds. The guanidinates tune the clamping effect further
by introducing steric pressure provided by the substituents emerging from the central carbon atom
of the chelating ligand. This steric hindrance pushes the substituents of the chelating nitrogen
atoms slightly towards the chromium atoms, resulting in the chromium atoms being pushed closer
together.
The magnetic susceptibility results indicate that the dichromium complex 3 (see reference)
features a singlet ground state, with no unpaired electrons. If the chromium atoms are both d5
centers, then these 10 metal valence electrons will fill all five bonding molecular orbitals of the
energy level diagram in Figure 15.9 to afford a quintuple bond.
15.21
It is proposed that CH 3C  CCH 3 reacts with the quintuply bonded dichromium complex in a
manner consistent with 2+2 cycloaddition resulting in a dichromium metallacycle with a
carbon—carbon double bond and a chromium—chromium quadruple bond. The crystallographic
data support this hypothesis on the basis of a carbon—carbon distance (132.6 pm) similar to that
of ethylene (133.9 pm), and a chromium—chromium distance (192.5 pm) longer than that in the
starting quintuply bonded complex (180.3 pm), consistent with a fourfold chromium—chromium
bond order. Calculated bond orders (via DFT calculations) for the dichromium metallacyle
product are 3.43 and 1.98 for the chromium—chromium and carbon—carbon bonds, respectively.
Copyright © 2014 Pearson Education, Inc.
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
15.22
a.
Oh
E
6
1
3
2
 (s, pz)
A1g
T1u
Eg
8C3
0
1
0
–1
6C2
0
1
–1
0
6C4
2
1
1
0
3C2
2
1
–1
2
i
0
1
–3
2
6S4
0
1
–1
0
6S6
0
1
0
–1
3h
4
1
1
2
223
6d
2
1
1
0
 fits either the s or the pz orbitals.
b.
It can be seen from the table above that  = A1g + T1u + Eg. This can also be worked out
by the more elaborate methods used in Chapter 4.
c.
The reducible representation for the px and py orbitals and its components are shown in
the following table. Figure 15.13 shows the three T1u and the three T2g group orbitals
required (which also include some pz contribution from two of the boron atoms). The
T1g and T2u representations are for antibonding orbitals.
Oh
x,y
T1g
T1u
T2g
T2u
15.23
E
12
3
3
3
3
8C3
0
0
0
0
0
6C2
0
–1
–1
1
1
6C4
0
1
1
–1
–1
3C2
–4
–1
–1
–1
–1
i
0
3
–3
3
–3
6S4
0
1
–1
–1
1
6S6
0
0
0
0
0
3h
0
–1
1
–1
1
6d
0
–1
1
1
–1
(Rx, Ry, Rz)
(x, y, z)
The number of orbitals of each type can be obtained by analogy with the results for B6H62–
(Section 15.4.1):
2 valence atomic orbitals of B combine to form:
15 bonding orbitals (2n + 1) consisting of:
8 framework MOs (n + 1)
1 bonding orbital from overlap of sp orbitals
7 bonding orbitals from overlap of p orbitals of B with sp hybrid
orbitals or p orbitals of other B atoms
7 B–H bonding orbitals (n)
13 nonbonding or antibonding orbitals
15.24
a.
C2B3H7  B5H9  B5H54–
nido
b.
B6H12  B6H66–
arachno
c.
B11H112–
closo
d.
C3B5H7  B8H10  B8H82–
closo
e.
CB10H13  B11H14  B11H114–
nido
f.
B10H142–  B10H106–
arachno
–
–
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
224
15.25
15.26
15.27
15.28
a.
SB10H102–  B11H132–  B11H114–
nido
b.
NCB10H11  B12H14  B12H122–
closo
c.
SiC2B4H10  B7H13  B7H76–
arachno
d.
As2C2B7H9  B11H15  B11H114–
nido
e.
PCB9H11  B11H14  B11H114–
nido
a.
B3H8(Mn(CO)3)  B4H8  B4H44–
nido
b.
B4H6(CoCp)2  B6H8  B6H62–
closo
c.
C2B7H11CoCp  B9H13CoCp  B10H14  B10H104–
d.
B5H10FeCp  B6H10  B6H64–
e.
C2B9H11Ru(CO)3  B11H13Ru(CO)3  B12H14  B12H122–
a.
Ni(CO)4: CH4
Ni(CO)3: CH3+
b.
[Bi3Ni4(CO)6]x
The cluster needs 4n + 2 = 4(7) + 2 = 30 cluster valence
electrons for a closo classification (there are 7 Bi and Ni atoms in the core of the
cluster). Each Bi atom contributes its 5 valence electrons, and each CO ligand
contributes 2, for a total of 3(5) + 6(2) =27. Three additional electrons are
needed for a total of 30; consequently, the charge must be x = 3–; the formula is
[Bi3Ni4(CO)6]3–. (See reference and citations therein for additional details on
electron counting.)
c.
The cluster valence count starts with 6 CO ligands (12 electrons)
[BixNi4(CO)6]2–
and a charge of 2– , a total of 14 electrons. If x = 0, the value of 4n + 2 would be 18.
Each Bi atom counts an additional 5 electrons, and each additional core atom increases
4n + 2 by 4. At x = 4, the total cluster valence electron count (34) equals 4n + 2 (4 Bi
atoms and 4 Ni atoms in core, so n = 8). Consequently, the formula is [Bi4Ni4(CO)6]2– .
–
–
nido
nido
closo
Ni(CO)2: CH22+
Ni(CO): CH3+
B2 H 6 (Cp*RuCO)2 : Cp*RuCO is a 15 electron fragment, 3 electrons short of 18. This fragment
is isolobal to the 5-electron fragment BH 2 , 3 short of 8 electrons. The metallaborane
B2 H 6 (Cp*RuCO)2 is therefore isolobal with B4 H10 .
B4 H10  6H + 
 B4 H 46- (arachno)
(Continued on next page)
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
225
B3H 7 (Cp*RuH)2 : Cp*RuH is a 14 electron fragment, 4 electrons short of 18. This fragment is
isolobal to the 4-electron fragment BH, 4 short of 8 electrons. The metallaborane
B3H 7 (Cp*RuH)2 is therefore isolobal with B5H 9 .
B5H 9  4H + 
 B5H 54- (nido)
The reactions of B2 H 6 (Cp*RuCO)2 with Mn 2 (CO)10 and Re 2 (CO)10 , respectively, afford
products with dramatically different structures. The Mn 2 (CO)10 reaction yields
arachno  [(Cp*RuCO)2 B3H 7 ] (left) and the novel bridged borylene complex
[3 -BH)(Cp*RuCO)2 ( -H)( -CO){Mn(CO)3}] (right, perspective from above the triply bridging
borylene ligand).
H
H
B
H
H3C
B
B
OC
(CO)3
Mn
H
H
H3C
H
CH3
H3C
CH3 Ru
CH3 Ru
C
O
H
OC
CH3
H3C
CH3
H3C
CH3
H3C
CH3
CH3
Ru
Ru
H3C
H3C
HB
H
OC
OC
CH3
H3C
H3C
CH3
The reaction of B2 H 6 (Cp*RuCO)2 and Re 2 (CO)10 provides [Cp*Ru(CO)( -H)]2 (below) as the
only characterized product.
H3C
CH3
H 3C
OC
H3 C
CH3
H
Ru
Ru
H
CO H C
3
H3C
CH3
CH3
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CH3
226
15.29
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
The cluster is below, with each Mo an abbreviation for ( 5 -Cp*Mo) . The presence of
molybdenum—molybdenum bonds is postulated to prevent BH capping of some of the cubane
faces. The two boron atoms labeled as “1” are five-coordinate and are assigned a 11B{1H} NMR
chemical shift of 131.5 ppm. The boron atoms labeled “2” are four-coordinate with bonds to three
Mo atoms and one boron atom; the corresponding chemical shift for these two atoms is 94.4 ppm.
The three boron atoms labeled “3” are four-coordinate with two bonds each to boron and
molybdenum; the chemical shift for these three boron atoms is 58.8 ppm.
H
3 B
H 3
B
Mo
1
HB
BH
1
Mo
HB
3
2
Mo
B
H
Mo
B
H
2
This cluster features four  5 -Cp* ligands, and pairs of these are chemically equivalent on the
basis of the 1H NMR spectrum. The “top” two Mo atoms feature bonds to five boron atoms and
one Mo atom. The “bottom” two Mo atoms feature bonds to four boron atoms and two Mo
atoms. These provide two unique chemical environments for the  5 -Cp* ligands.
15.30
Product yields for insertion of the dicarbollide cage into [( 6 -arene)Fe][PF6 ]2 were deemed lower
with the pentamethylbenzene and hexamethylbenzene iron complexes as reactants (15-20%,
compared to 48-70% with the less substituted benzenes) on the basis of stronger Fe—arene bonds
with these more highly methylated cations. A variety of approaches were used to show
correlations to the degree of methylation of the benzene. Figure 1 of the reference shows that the
diamagnetic shielding of the CH cage hydrogen atoms increases with increasing number of arene
methyl groups. The resonances for these CH hydrogen atoms shift upfield as the number of
methyl substituents changes from 1 to 6. Similar linear correlations were observed between the
11
B chemical shift of the antipodal dicarbollide cage boron atom (labeled as 12 in Scheme 1) and
the 1H chemical shift of the hydrogen atom on this same boron atom; as the degree of methyl
substitution increases at the arene ligand, these resonances progressively shift upfield. It is
remarkable how sensitive the latter chemical shifts are to modulation of the donor ability of the
arene ring since these atoms are separated from the iron atom by a relatively large number of
bonds. The Fe2+/Fe3+ redox potential decreases linearly as the number of methyl substituents
increases; the complexes become progressively easier to oxidize (the complexes become better
reducing agents) as the number of methyl groups increases.
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
15.31
15.32
227
a.
Ge94– has 40 valence electrons = 4n + 4. Its classification is nido.
b.
InBi32– has 3 + 3(5) + 2 = 20 valence electrons = 4n + 4. Its classification is also nido.
c.
Bi82+ has 8(5) – 2 = 38 valence electrons = 4n + 6. It is an arachno cluster.
The reference notes that significant ion-pairing exists between the potassium ions and the highly
negatively charged Zintl ions in K 3E7 (E = P, As, Sb). Addition of cryptand[2.2.2] results in
potassium ion complexation, attenuates the ion-pairing interaction, and renders the Zintl ions
more effective nucleophiles. The IR spectroscopic data in the carbonyl region for
[K(2.2.2)]3[E7Cr(CO)3 ] (below) indicate extensive -backbonding in each case. On the basis of
the lowest energy CO stretching frequency, one would tentatively rank the E73 donor ability to
the Cr(CO)3 fragment as increasing as Sb < P < As, but these frequencies are sufficiently similar
that it is difficult to clearly differentiate the donor ability of these Zintl ions.
Salt
IR  (CO) (cm -1 )
[K(2.2.2)]3[P7Cr(CO)3]
1829, 1738, 1716
[K(2.2.2)]3[As7Cr(CO)3 ]
1824, 1741, 1708
[K(2.2.2)]3[Sb7Cr(CO)3 ]
1823, 1748, 1718
The electronic spectra of [K(2.2.2)]3[E7Cr(CO)3 ] are provided in Figure 5a of the reference. The
tails of the absorptions assigned to intra-ligand E73 charge transfer bands decrease in energy
from PCr > AsCr > SbCr. Note that the max of these charge transfer bands (Table 4) are nearly
identical, although they do exhibit differences in their molar absorptivity constants,  , in
L
L
); AsCr: 365 nm, (11400
); SbCr: 365 nm,
parentheses: PCr: 363 nm, (6600
mol cm
mol cm
L
(16000
). The assignment of these intra-ligand charge transfer bands in
mol cm
[K(2.2.2)]3[E7Cr(CO)3 ] is reasonable since the identical trends are exhibited by the Zintl ions
E73 themselves; these absorptions are essentially unchanged upon coordination to the Cr(CO)3
fragment.
A sketch of [Cr(CO)3 (HSb7 )]2- is below. The hydrogen ion binds to the Sb atom furthest from
the Cr atom.
2OC CO CO
Cr
Sb
Sb
Sb
Sb
Sb
Sb
Sb
H
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
228
15.33
The Lewis structure of [HP7 ]2- is below. It is interesting that this Zintl ion exhibits fluxional
behavior in solution that renders all the phosphorus atoms equivalent on the NMR time scale.
P
P
P
P
H
P
P
P
The most likely carbodiimide hydrophosphination mechanism is as follows, with nucleophilic
attack at the carbodiimide carbon affording an intermediate amidinate that rapidly abstracts a
proton from the Zintl ion cage. Evidence for intramolecular proton transfer is provided by
isotopic labeling studies. When [HP7 ]2- is used as a reactant towards carbodiimides in
deuterated-solvents, the only amidine-functionalized products obtained feature N—H bonds.
When the corresponding reactions with [DP7 ]2- are conducted in protio-solvents, only amidinefunctionalized products with N—D bonds are obtained. These experiments definitively prove the
role of [HP7 ]2- as the proton source, and strongly suggest the mechanism below.
NR
P
P
P
P
P
P
P
RN
H
C
NR
P
P
P
N
H
P
P
nucelophilic attack
C
R
P
P
intramolecular proton transfer
NR
P
P
P
P
P
P
15.34
a.
m = 1 (a single polyhedron)
n = 5 (each B atom counts)
o = 0 (no bridging atoms)
p = 2 (2 missing vertices)
8 skeletal electron pairs
b.
m = 2 (2 polyhedra)
n = 10 (each B, C, and Co atom counts)
Copyright © 2014 Pearson Education, Inc.
P
C
HN
R
Chapter 15 Parallels Between Main Group and Organometallic Chemistry
229
o = 1 (1 bridging atom, the cobalt)
p = 2 (2 missing vertices: the top part of the molecule is considered as a pentagonal
bipyramid with the top vertex missing, and the bottom as an octahedron with
the bottom vertex missing)
15 skeletal electron pairs
c.
m = 2 (2 polyhedra)
n = 17 (each B, C, and Fe atom counts)
o = 1 (1 bridging atom, Fe)
p = 1 (1 missing vertex, the top atom of the incomplete pentagonal bipyramid)
21 skeletal electron pairs
15.35
15.36
15.37
a.
C2v
b.
C5v, D5h
c.
[Re2Cl8] : D4h; [Os2Cl8] : D4d
d.
D2h
e.
C5v
2–
2–
f.
Cs, C2h
g.
Te62+: D3h
Ge94–: C4v
a.
Td
b.
Ih
a.
The group orbitals are derived primarily from the 3p orbitals of phosphorus, which
collectively resemble the five π orbitals for C5H5 in Figure 13.22. Diagrams of these
orbitals can be found in the second reference. (See also Z-Z. Liu, W-Q. Tian, J-K. Feng,
G. Zhang, and W-Q. Li, J. Phys. Chem. A, 2005, 109, 5645.)
Atomic orbitals on transition metals suitable for interaction (assuming metal
centered below P5 plane)
lowest energy group orbital: s, pz, dz2
1-node degenerate pair: px, dxz, py, dyz
2-node degenerate pair: dxy, dx2–y2
b.
The molecular orbitals of P5– are lower in energy than the similar orbitals of C5H5–,
giving rise to a generally stronger ability of P5– to π accept. For example, the energy
match between 2-node orbitals and metal d orbitals may be closer in P5– complexes than
in the case of ferrocene (Figure 13.28), enabling stronger interaction. An example of
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
230
such an interaction would be between an empty 2-node orbital of P5– and a dx2–y2 orbital
of a metal:
dx2–y2
c.
The reference provides an energy level diagram of the molecular orbitals of
[(5-C5H5)2Ti]2–. In analyzing the orbitals it is important to see how the π orbitals of P5–
ligands match up with d orbitals on Ti and how the resulting shapes illustrate how the
lobes of these interacting orbitals merge (in the bonding orbitals) and how nodes are
formed between them (in the antibonding orbitals).
15.38 The three types of interaction, , π, and , should be evident in the orbitals generated. In
addition to the bonding interactions (see Figure 15.8) there should be matching antibonding
interactions, with a nodal plane bisecting the metal–metal bond. The relative energies of the
molecular orbitals should be similar to those on the right side of Figure 15.9, depending on the
level of sophistication of the calculations used.
15.39
In addition to the , π, and  interactions between d orbitals, shown in Figure 15.8, a  interaction
could occur between s orbitals of the two atoms:
(In a chromium atom the valence 4p orbitals are empty, so interactions between them need not be
considered.) The extent of the calculated interactions between the atomic orbitals is likely to
differ significantly. Consequently, even though there may be six possible orbital interactions, the
strength of such bonding is calculated to be significantly weaker than would be expected in a true
“sextuple” bond. It is suggested that, for this exercise, the Cr–Cr distance be set at 166 pm, the
equilibrium distance reported in the references.
15.40
a.
The A1g orbital should have a very large, nearly spherical
lobe in the center of the cluster and six smaller, also nearly
spherical lobes, on the outside of the borons (centered on
the hydrogens).
b.
The T1u orbitals should each have two regions where lobes of p
orbitals on four B atoms merge, plus additional lobes, nearly
spherical, centered on opposite hydrogens.
c.
The T2g orbitals should show how the lobes of
adjacent p orbitals merge. The result should have four
lobes, somewhat similar in appearance overall to a d
orbital.
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Chapter 15 Parallels Between Main Group and Organometallic Chemistry
15.41
15.42
231
The orbitals are naturally much more complex, with 167 orbitals and 270 electrons. The T1u
and T2g orbitals shown are from a Scigress Extended Hückel calculation.
a.
The A1g orbital should be similar to that for B6H62– in
Problem 15.31, with one large lobe in the center and smaller
d orbital lobes on each of the Ru atoms. It is the HOMO,
antibonding in symmetry, and the only orbital near this
energy that involves carbon orbitals.
b.
The T1u orbitals should also be similar to the T1u orbitals of
B6H62–. There should be two large lobes, each derived primarily
from d orbitals on four Ru atoms and a p orbital of the central
carbon, plus lobes on opposite Ru atoms (the other Ru atoms).
There may also be fragment lobes of d orbitals on the Ru atoms.
c.
The T2g orbitals are not directly involved in bonding with the
carbon, but they do strengthen the cluster framework. The
atomic orbitals should interact in sets of four, similarly to the
T2g orbitals in B6H62–, but some of the distinctive features of d
orbitals should also be observable in the ruthenium cluster.
a.
As in ferrocene, important interactions should be observed between 1-node group orbitals
on the (in this case, Ge5) ligands and dxz and dyz orbitals on cobalt (see example for
ferrocene above Figure 13.27). With the Ge5 ligands held in closer proximity by bonds
between the top and bottom rings, interactions should also be found between 2-node
group orbitals and the dxy and dx2–y2 orbitals of Co and between 1-node group orbitals and
px and py orbitals of Co.
b.
In contrast to the cyclopentadienyl rings in ferrocene, primarily sigma bonding should be
observed within the Ge5 rings; the reported Ge–Ge distance in the rings is within the
range of single bonds.
c.
In addition to the comparisons mentioned above, it would be informative to see if in the
[CoGe10]3– cluster the interactions between the rings and the dz2 and pz orbitals of Co are
weak (because the lobes of these orbitals point toward the center of the rings) and to look
for interactions involving the “doughnut” of the dz2. If the software generates energies of
the molecular orbitals, additional comparisons can be made between the energies of the
cluster orbitals and those of ferrocene shown in Figure 13.28.
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