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

(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




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,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 +




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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 + (  )
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




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, 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.




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