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Inorganic Chemistry 5th Edition By Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)

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Inorganic Chemistry 5e Gary Miessler, Paul Fischer

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Inorganic Chemistry 5e Gary Miessler, Paul Fischer

Inorganic Chemistry 5e Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual) Inorganic Chemistry 5e Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)

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Institution Inorganic Chemistry 5e Gary Miessler, Paul Fischer
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(Inorganic Chemistry 5e Gary Miessler, Paul Fischer, Donald Tarr)

(Solution Manual, There is no Solution for Chapter 1)

CHAPTER 2: ATOMIC STRUCTURE
h 6.626  10 34 J s
2.1 a. = =  2.426  10 11 m
mv (9.110  10 –31kg)  (0.1)  (2.998  108 m s –1 )

h 6.626  1034 J s
b.  = =  6  10 34 m
mv 0.400 kg  (10 km/hr  10 m/km  1hr/3600s)
3

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

h 6.626  10 34 J s
d. = =
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
 1 1  hc 1
2.2 E  RH  2 – 2 ; RH  1.097 10 7 m–1  1.097 10 5 cm–1 ;   
2 nh  E __

1 1   12 
nh  4 E  RH  –   RH    20,570cm –1  4.085  10 –19 J
 4 16   64 
  4.862 10 –5 cm = 486.2 nm

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

 1 1   8
nh  6 E  RH  –   RH    24,380cm –1  4.841 10 –19 J ;
 4 36   36 
  4.102 10 –5 cm = 410.2 nm
 1 1   45 
2.3 E  RH 
4 –   RH    25,190cm–1  5.002 10 –19 J
 49 
 196 
1
  __  3.970 10 –5 cm = 397.0 nm


hc (6.626  1034 J s)(2.998  108 m/s)
2.4 383.65 nm E   5.178  1019 J
 (383.65 nm)(m/10 nm)9

hc(6.626  10 34 J s)(2.998  108 m/s)
379.90 nm E   5.229  10 19 J
 (379.90 nm)(m/10 nm) 9

Copyright © 2014 Pearson Education, Inc.

,2 Chapter 2 Atomic Structure

1

 1 1  1 1 E 1 E  2
E  RH  2  2  ;   and nh    
2 nh  nh
2
4 RH 4 RH 
1

1 5.178 10 19 J  2
For 383.65 nm: nh    18 
9
4 2.1787 10 J 
1

1 5.229 10 19 J  2
For 379.90 nm: nh    18 
 10
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:
2 2  He + (Z He+ )2 e4 2 2  H (Z H )2 e4
RHe +  RH 
(4 0 )2 h2 (4 0 )2 h2
RHe +  He+ (Z He+ )2 4  He+
  (since Z  2 for He  )
RH  H (Z H ) 2 H
The reduced masses are required (in terms of atomic mass units):
1 1 1 1 1
   
H me m proton 5.4857990946  10 m 1.007276466812 mu
4
u
1
 1823.88 mu1;  H  5.482813  104 mu
H
1 1 1 1 1
   
He  me mHe2 5.4857990946  10 m 4.001506179125 mu
4
u
1
 1823.13 mu1;  He +  5.485047  10 4 mu
He +

Copyright © 2014 Pearson Education, Inc.

, 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 + (  )
n n 1 4
l h

RHe +  8.7188  1018 J

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

2 
= –Ar 2 sin rx – Bs 2 cos sx
x 2

h2
–
8 m
 – Ar
2
2

sin rx – Bs2 cos sx  E  A sin rx + B cos sx 

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

h 2 Ar 2 h 2 Bs 2
 EA ; = EB
8 2 m 8 2 m

8 2 mE 2
r 2  s2  ; r  s  2mE
h 2
h
b.   A sin rx ; when x  0,   A sin 0  0
when x  a,   A sin ra  0
n
 ra  n ; r  
a

r 2 h 2 n 2 2 h 2 n 2h 2
c. E  
8 2 m a 2 8 2 m 8ma 2

Copyright © 2014 Pearson Education, Inc.

, 4 Chapter 2 Atomic Structure

a a a
x 
2 n a nx  nx 
d.   dx   
2 2
A sin   dx  A
2
sin 2   d  
 a  n  a   a 
0 0 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
n 2 a 4 4 


2
A
a

2.9 a. 3pz 4dxz

b. 3pz 4dxz

c. 3pz 4dxz

For contour map, see Figure 2.8.

Copyright © 2014 Pearson Education, Inc.

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Exam (elaborations)

Inorganic Chemistry 5th Edition By Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)

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