General Chemistry I
Examnination II
July 3, 2008
1. (10 points) A green laser has a wave length or 500.0 nm. Determine the frequency and energy
of this laser.
" 10 -9 meters %
! = 500.0 nm $
= 5.000 X 10 -7 meters
# nanometer '&
c = !( or ( =
E = h( )
c
!
) (=
2.99792 X 10 8 m/sec
= 5.9958 X 10 +14 1/seconds = 5.996 X 1014 Hertz
-7
5.000 X 10 meters
E = (6.2608 X 10 -34 Joule-seconds)(5.9958 X 10 +14 1/seconds) = 3.973 X 10 -19 Joules
2. (10 points) The total energy from a beam of light is 957.0 kJ/mol. Calculate the wave length
and frequency of a single photon of this beam.
957.0 kJ/mol = 957.0 X 10 3 Joules/mole &
9.570 X 10 5 J/mol
= 1.59814 X 10 -18 Joules/photon
23
6.0221367 X 10 photons/mol
E
1.59814 X 10 -18 Joules
=
= 2.398 X 1015 Hz
-34
h
6.2608 X 10 Joule-seconds
c
2.99792 X 10 8 m/sec
c = #! or # =
" #=
= 1.250 X 10 -7 meters = 125.0 nm
!
2.398 X 1015 1/sec
E = h! " ! =
3. (10 points) Below is in energy level diagram for the electrons in an atom in an excited state.
This atom “de-excites.” Calculate the energy, wave length, and frequency of the photon associated
with this de-excitation.
E1 (initial energy) = -2.42 X 10 -19 J
E 2 (final energy) = -5.54 X 10 -19 J
!E = ( final energy ) - ( initial energy )
De-excitation means the electron
moves to a lower energy level
-1.36 X 10-19 Joules
n=3
-19
= E 2 - E1 = -3.03 X 10 Joules
(The negative means the energy is emitted and can be ignored)
n=2
-2.42 X 10-19 Joules
n=1
-5.45 X
10-19
Joules
E
3.03 X 10 -19 Joules
=
= 4.57 X 1014 Hz
h
6.2608 X 10 -34 Joule-seconds
c
2.99792 X 10 8 m/sec
c = #! or # =
" #=
= 6.56 X 10 -7 meters = 656. nm
14
!
4.57 X 10 1/sec
E = h! " ! =
4. (5 points) Explain why a green light can induce a photoelectric effect in a piece of metal but a
yellow light will not.
Green light is (closer to the blue end of the spectrum) higher in energy than yellow (closer to the red
end). The energy of the green light exceeds the “threshold” energy that will break the glue that
holds the electrons to the metal.
5. (20 points) Write out the electronic configurations for the following:
a) Cf (Z = 98)
[Rn] 7s2 6d1 5f9
b) W (Z = 74) = [Xe] 6s2 5d1 4f14 5d3
which becomes [Xe] 6s2 4f14 5d4
which becomes [Xe] 6s1 4f14 5d5
c) Bi+3 (Z = 83) = [Xe] 6s2 5d1 4f14 5d9 6p3 which becomes [Xe] 6s2 4f14 5d10 6p3 now
remove three electrons & you get [Xe] 6s2 4f14 5d10
d) Zr+2 (Z = 40) = [Kr] 5s2 4d2 remove s before d & you get [Kr] 4d2
6. (15 points) Construct an auf bau diagram and an energy level diagram for the last 5 electrons
added to a Niobium atom (Z = 41) = [Kr] 5s2 4d3
n
l
m
s
37
5
0
0
-1/2
38
5
0
0
+1/2
39
4
2
-2
-1/2
40
4
2
-1
-1/2
41
4
2
0
-1/2
4d
5s
8. (5 points) Which is the largest Br-1 , Kr or Rb+1 (and explain your choice)
All have the same number of electrons (36) but Br only has 35 protons. That means the electrons are
not held as tight and that makes Br- the largest.
9. (5 points) Which has the highest ionization potential: Na-1 , Na or Na+1 (and explain your choice)
All have the same number of protons (11) but Na+ has the fewest number of electrons (10). Than
means it takes more energy to remove and electron from Na+ and it has the highest ionization
potential.
10. (10 points) Aluminum only exists in the +3 oxidation state but gallium (right below Al) exists
in both the +3 and the +1. Explain why.
Ga has the stabilizing influence of the filled (d10) level
11. (10 points) Identify the oxidation state of all the elements in the following:
a) Cs2S2O8
Cs = +1, O = -2, S = +7
b) Ca3(PO3)2
Ca = +2, O = -2, P = +3