Inorganic Chemistry Solutions Manual

231 Pages·2014·9.768 MB·English

by  Gary L. Miessler, Paul J. Fischer, Donald A. Tarr

Chapter 2 Atomic Structure 1
CHAPTER 2: ATOMIC STRUCTURE

h 6.6261034J s
2.1 a.  = = 2.4261011 m
mv (9.11010–31kg)(0.1)(2.998108m s–1)

h 6.6261034J s
b.  = = 61034m
mv 0.400kg(10km/hr103m/km1hr/3600s)

h 6.6261034J s
c.  = = 9.11035 m
mv 8.0 lb0.4536 kg/lb2.0 m s-1

h 6.6261034J s
d.  = =
mv 13.7 gkg/103g30.0 mi hr-11 hr/3600s1609.3 m/mi

 3.611033 m

1 1  hc 1
2.2 E R  – ; R 1.097107m–11.097105cm–1 ;    
H22 n2 H E __
h 
1 1  12
n 4    E  R –  R 20,570cm–1 4.08510–19J 
   
h H4 16 H64
4.86210–5cm=486.2 nm 

 1 1   21 
n 5    E  R  –   R   23,040cm–1 4.57710–19J ; 
h H4 25  H100
    4.34110–5cm=434.1 nm 

 1 1   8 
n 6    E  R  –   R   24,380cm–1 4.84110–19J ; 
h H4 36  H36
      4.10210–5cm=410.2 nm 

 
1 1 45 
2.3 ER  –   R   25,190cm–1 5.00210–19J 
H4 49  H196
 
1
       3.97010–5cm=397.0 nm
__


hc (6.6261034Js)(2.998108m/s)
2.4 383.65 nm E   5.1781019J
 (383.65nm)(m/109nm)
hc (6.6261034Js)(2.998108m/s)
379.90 nm E   5.2291019J
 (379.90nm)(m/109nm)
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2 Chapter 2 Atomic Structure
1
1 1  1 1 E 1 E 2
E R    ;   and n   
H22 n 2 n 2 4 R h 4 R 
h h H H
1

1 5.1781019J  2
For 383.65 nm: n    9
h 4 2.17871018J
1

1 5.2291019J  2
For 379.90 nm: n    10
h 4 2.17871018J

2.5 The least energy would be for electrons falling from the n = 4 to the n = 3 level:

1 1   7 
E R   2.17871018J 1.0591019J
H32 42 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102823.8530211cm197492.221701cm15331.6313201cm1
Ehc(6.6261034Js)(2.998108m s)(100cm m)(5331.6313201cm1)
E1.0591019J
This is the same difference found via the Balmer equation in Problem 2.5 for
a transition from n4ton3. The Balmer equation does work well for hydrogen.

2.7 a. We begin by symbolically determining the ratio of these Rydberg constants:
22 (Z )2e4 22 (Z )2e4
R  He+ He+ R  H H
He+ (4 )2h2 H (4)2h2
0 0

R  (Z )2 4
He+  He+ He+  He+ (since Z 2 for He)
RH H(ZH)2 H
The reduced masses are required (in terms of atomic mass units):
1 1 1 1 1
   
H me mproton 5.4857990946104mu 1.007276466812 mu
1
1823.88m1; 5.482813104m
 u H u
H
1 1 1 1 1
   
He me mHe2 5.4857990946104mu 4.001506179125mu
1
1823.13m1; 5.485047104m
 u He+ u
He+

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Chapter 2 Atomic Structure 3
The ratio of Rydberg constants can now be calculated:
R 4 4(5.485047104m )
He+  He+  u 4.0016
R  5.482813104m
H H u

b. R 4.0016R (4.0016)(2.17871018J)8.71841018J
He+ H

c. The energy difference is first converted to Joules:
Ehc(6.6261034Js)(2.998108m s)(100cm m)(329179.76197cm1)

E6.53911018J
The Rydberg equation is applied, affording nearly the identical Rydberg constant for He +:
1 1 1 1
E R (  )6.53911018J R (  )
He+ 2 2 He+ 1 4
n n
l h
R 8.71881018J
He+

h2 2
2.8 a. –  E ;  Asin rxBcossx  
82m x2

= Ar cos rx – Bs sin sx 
x

2
= –Ar2 sin rx – Bs2 cos sx  
x2
– h2 –Ar2 sin rx – Bs2 cos sx EA sin rx + B cos sx
82m

If this is true, then the coefficients of the sine and cosine terms must be independently
equal:

h2Ar2 h2Bs2
 EA ; = EB 
82m 82m

82mE 2
    r2 s2  ; rs 2mE  
h2 h

b.  A sin rx ; when x0,  Asin 00 
 when x a,  Asin ra0 
n
 ran; r   
a

r2h2 n22 h2 n2h2
c. E     
82m a2 82m 8ma2

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4 Chapter 2 Atomic Structure
a a a
nx a nx nx
d.  2dx  A2sin2  dx A2 sin2  d  
 a  n  a   a 
0 0 0
a
a 1nx 1 2nx
     A2   – sin  1 
n 2 a  4  a 
0
aA2 1 na 1 1 
  – sin2n–0 sin01 
n 2 a 4 4 
2
A  
a

2.9 a. 3p 4d
z xz

b. 3p 4d
z xz

c. 3p 4d
z xz

For contour map, see Figure 2.8.  

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Chapter 2 Atomic Structure 5

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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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 7

2.12 4f orbital
z(x2y2)
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. 

2.13 A 5f  orbital has the same general shape as the 4f , with the addition of a radial node
xyz z(x2–y2)
and rotation so the lobes are between the xy, xz, and yz planes.

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. 

2.14 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)
2 z2 = 3x2 + 3y2
z 2 = 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. n l m
l
5d 5 2 –2,–1,0,1,2

4f 4 3 –3,–2,–1,0,1,2,3
7g 7 4 –4,–3,–2,–1,0,1,2,3,4

b. n l m m
l s
3d 3 2 –2 ±2

3 2 –1 ±2
3 2 0 ±2
3 2 1 ±2
3 2 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!)

2.16 a. For a 5d electron, l = 2 and n = 5.

1
b. At most there can be ten 4d electrons, half of which can have ms =  .
2

c. f electrons have quantum number l = 3 and can have m = –3, –2, –1, 0, 1, 2, or 3.
l

d. For l = 4 (g electrons), m can be –4, –3, –2, –1, 0, 1, 2, 3, or 4.
l

2.17 a. The l quantum number limits the number of electrons. For l = 3, m can have
l
seven values (–3 to +3) each defining an atomic orbital, and for each value of m the
l
1 1
value of m can be  or  ; there can be two electrons in each orbital. At most,
s
2 2
therefore, there can be 14 electrons with n = 5 and l = 3.

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 m and therefore 9 orbitals.
l

2.18 a.     The first configuration is favored. 
  One exchange One pair in same orbital This configuration is stabilized by
E =  E =   (which is negative), and the second
e c e
is destabilized by  (which is positive).
c

b.   and    The first configuration is favored. It is
3 exchanges 1 exchange, 1 pair stabilized by 2  more than the second
e
(1–2, 1–3, 2–3) E =  +  configuration, and the second
e c
E = 3  configuration is also destabilized by  .
e c

2.19 W:      X:    
2 exchanges (1–2, 3–4) 3 exchanges (1–2, 1–3, 2–3)
E = 2 E = 3 , so this state has lower energy
e e
than W (because  is negative).
e
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Chapter 2 Atomic Structure 9

2.20 Y:      Z:     
  4 exchanges (1–2, 1–3, 6 exchanges (1–2, 1–3, 1–4, 2–3, 2–4, 3–4)
2–3, 4–5); E = 4 E = 6 , so this state has lower energy than Y.
e e
(because  is negative).
e

2.21 Three states are possible:     (three exchanges, E3 );
e
    (one exchange, E );     (one
e
exchange and one pair, E  ). Energy ranking:
e c

c
Energy
2
e

2.22 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.

2.23 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

2.24 a. Rb– [Kr] 5s2

b. Pt2– [Xe] 6s2 4f14 5d10

2.25 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
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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 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.

2.27 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  in comparison with the
c
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.

2.28 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
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