Unit 4 Review
Section I - Problems (Given h = 6.626 x 10-34 J.s)
The speed of light, c, equals 3.00 x 108 m/s.
What is the frequency of a wave with a wavelength of 3.5 x 108
3. What is the energy of a photon with a
frequency of 5.41 x 10-7 1/s?
E = h .
4. Substitute, rearrange, and write an equation
that relates E, h, c, and l.
c = l
l l l
Section II -
Label both ends of the spectrum with high/low frequency, high/low
energy, and long/short wavelength
Long wavelength radio
waves microwaves infrared light ROYGBIV ultraviolet
gamma rays short
Low frequency high frequency
Low energy high energy
2. Which has a higher energy, gamma or x-rays? gamma
3. Which has a shorter wavelength, radio or
4. Which has a lower frequency, yellow or green
5. In the equation E = hn, energy and frequency are directly proportional.
6. In the equation c = ln, wavelength and frequency are inversely proportional.
The symbol for wavelength is l .
the number of waves that pass a point per second.
Electrons give off energy in the form of a photon (or quantum) when returning to the ground state.
Which scientist proposed the idea that electrons travel around the
nucleus in fixed paths? Bohr
11. When an electron moves from the ground state to the excited state, energy is absorbed .
12. Bohr chose the element hydrogen to prove his theory.
13. The dual wave-particle
nature of electrons describes how the electrons in atoms can behave as
waves and particles .
Section III - Electrons
1. What is an electron cloud? Area around the nucleus where electrons are located
*2. Who proposed the uncertainty principle? Heisenberg
*3. Who is credited with the idea that electrons are placed in the lowest energy level first? Aufbau
*4. What rule requires that
each of the "p" orbitals (at a particular
energy level) receive one electron before any
of the orbitals can have two electrons? Hund’s Rule
5. What is the maximum number of electrons in any orbital? 2
6. The principal quantum number, n, indicates the energy level .
7. The maximum number of
electrons in an energy level can be determined by the equation
That means the maximum number of electrons in the 3rd energy level is 18 .
8. The number of sublevels in any energy level can be determined by the number of the energy level .
9. The number of orbitals in an energy level can be determined by the equation n2 .
So, the 3rd energy level has 9 orbitals. ( 1 is "s" orbital, 3 is/are "p" orbitals, and 5
is/are "d" orbitals)
10. List the four sublevels according to increasing energy. s à p à d à f
11. The "s" sublevel is shaped like a sphere and has 1 orbital.
12. A "p" sublevel is shaped like a dumbbell and has 3 orbitals.
13. The "d" sublevel has 5 orbitals and the "f" sublevel has 7 orbitals.
Section IV - Electron configuration, noble gas configuration, valence electrons, orbital notations
1. What is the electron configuration for phosphorus? 1s2 2s2 2p6 3s2 3p3
2. How many total electrons are in a neutral atom of phosphorus? 15
3. Write the noble gas configuration for phosphorus. [Ne] 3s2 3p3
4. What is the highest occupied energy level for phosphorus? 3
5. What is the atomic number of phosphorus? 15
6. Draw the orbital notation
for phosphorus. hi h h h
7. How many electrons are in the highest occupied energy level of phosphorus? 5
8. How many unpaired electrons does phosphorus have? 3
9. How many inner-shell electrons does phosphorus have? 10
10. In which orbitals are the inner-shell electrons located? 1s, 2s, 2p
11. Draw the electron dot
diagram for phosphorus. .
: P .
Section V - Quantum numbers (Honors level only)
1. How many electrons can be described by the quantum numbers n = 3 and l = 1? 6
2. How many electrons in an atom can have the quantum numbers n = 2 and l = 3? 0
3. How many "d" orbitals have the value n = 3? 5
4. How many electrons in an atom have the quantum numbers n = 4 and l = 2? 10
5. Which of the following
sets of quantum numbers does NOT represent a possible set of quantum numbers
hydrogen atom? (There may be more than one correct answer.)
n l m s
(A) 4 8 -4  p; 1/2 maximum value for l = n - 1
(B) 6 5 -5  p; 1/2
(C) 3 2 2 1/2
(D) 6 0 1 1/2 m = -l .. +l