Chemistry Homework #3
Answer Key
 

Problems from Kotz & Treichel

49    a) one of the 4d orbitals (exactly which one is not defined)

        b) 5 subshells exist with a total of 25 orbitals
                5s (1 orbital)
                5p (3 orbitals)
                5d (5 orbitals)
                5f (7 orbitals)
                5g (9 orbitals)

        c) 7 f orbitals are in a given principal quantum level (for n > 4); ml = 3, 2, 1, 0, -1, -2, -3
 

51) For an electron in a 5d orbital, the possible quantum numbers are:
 

n l ml
5 2 2
5 2 1
5
2
0
5
2
-1
5
2
-2

57)   a) There are 7 4f orbitals.
        b) For n = 5, there are a total of 25 orbitals (see # 49b, above).
        c) None, l cannot equal 2 when n = 2.
        d) These quantum numbers describe one 3p orbital.

 
59) The number of nodal planes (or angular nodes) for the orbitals listed is:
        a) 4f: 3 nodal planes (0 radial nodes)
        b) 2p: 1 nodal plane (0 radial nodes)
        c) 6s: 0 nodal planes (5 radial nodes)
 

61)   3p orbital exists
        4s orbital exists
        2f does not exist (l cannot equal 3 when n = 2)
        1p does not exist (l cannot equal 1 when n = 1)
 

65) These quantum numbers refer to a 4d orbital.
 

77)

designation
number of orbitals
3p
3
4p
3
4px
1
6d
5
5d
5
5f
7
n = 5
25
7s
1
 

79)   a) energy and size, shape
        b) 2, 1, 0
        c) f orbital
        d) 4, 2, 2 (the third quantum number can also be 1, 0, -1, -2)
        e)
 

   
 
letter
      
l value
    
nodal planes
  

Additional Problems:

1) The wavelength of an n = 6 ®n = 5 transition for He+ is:


An example of a transition that would yield a photon of visible light would be an n = 6 ® n = 4, which would emit 656 nm light (red).
 

2) The ionization energy of a hydrogen atom with a principal quantum number of five is: