CHEM 31 SI
Wednesday November 4
Quantum Numbers and Electron Configurations Worksheet
(review)
(#70 from text) An electron in a hydrogen atom relaxes to the n = 4 level, emitting light of 114 THz. What is the
value of n for the level in which the electron originated? (Tera = 1012)
∆Eatom = Efinal – Einitial = hv = hc/λ = -2.18 x 10-18 J
c = 2.99792458 x 108 meters/second
h = planck’s constant = 6.626 x 10-34 J ∙ Second
1. Use the equation in the box. First let’s solve for the left half of the equation hv:
(6.626 x 10-34 J ∙ Second)(114 x 1012 s-1) = 7.55364 x 10-20 J
2. So the change in energy ∆E = 7.55 x 10-20 J. Now, is this number positive or negative?
Well, the atom loses / releases energy since the electron relaxes, so ∆E = - 7.55364 x 10-20 J.
3. Now solve for ninitial. (nfinal = 4) Note that the equation reads: (1/nfinal2 – 1/ninitial2) so:
- 7.55364 x 10-20 J = (-2.18x10-18)(1/42 – 1/ninitial2)
.03464 = (1/16 – 1/ninitial2)
.03464 + (1/ninitial2)= 1/16
(1/ninitial2) = (1/16) – (.03464)
(1/ninitial2) = .02785
1 = (.02785)(ninitial2)
35.9 = ninitial2
5.992 = ninitial
This value is very close to a whole number, so:
Ninitial = 6 !!
Note: the trick was to keep track of the signs. If you are solving for wavelength or
frequency, you do not need to worry about that…only if you are solving for an unknown n.
During the test, if you are solving for n and do not get a solution that is close to a whole
number, check your math closely (do not round between each step), and try flipping the sign
on the left side of the equation!
Answer: n = 6
(# 78 from text) The energy required to ionize sodium is 496 kJ / mol. What minimum frequency of light is required
to ionize sodium? (refer back to the photoelectric experiment, and refer ahead to ionization energy)
hv = φ
h = planck’s constant = 6.626 x 10-34 J ∙ Second
v = frequency of light
φ = the energy of a photon when threshold (dislodging of electron) is achieved
The trick is that we want to know how much energy to ionize a sodium atom. So we need to convert moles to
# of atoms using avogadro’s number. Since h is in terms of joules, we need to convert kJ to J. Then we can
solve for frequency.
1. (496 kJ / mol) * (1mol / 6.022 x 1023) * (1000 J / 1 kJ) = 8.236 x 10-19 Joules per atom = φ
2. v = φ/h
v = (8.236 x 10-19 Joules /atom) / (6.626 x 10-34 Joules ∙ Second)
v = 1.24 x 1015 cycles per second
Answer = 1.24 x 1015 Hz
CHEM 31 SI
Wednesday November 4
What are the possible values of n, l, ml, and ms? What does each of these determine?
n = the principal quantum number = specifies the size and energy of the orbital
n can equal any positive integer (1, 2, 3, 4, etc…)
l = the angular momentum quantum number = specifies the shape of the orbital
l is all whole numbers between zero and n-1…so if n = 3, l = 0,1, and 2
l=0=s
l=1=p
l=2=d
l=3=f
ml = magnetic quantum number = specifies the orientation of the orbital
ml = 0 and +/- l…..so if l = 3, ml = -3, -2, -1, 0, 1, 2, and 3
(this really means is that there are seven possible orientations of the f orbital…so don’t
get too hung up on -3, -2, etc…they are abstract…the total # of ml values is what matters)
ms = the spin quantum number
ms can either be ½ or – ½ AKA ↑ or ↓….these represent opposite spins
electrons with opposite spins can occupy the same orbital, so 2s can have two electrons
19. List all of the possible values of l (angular momentum), ml (magnetic), and ms (spin) quantum numbers for the
following values of n (the principal quantum number):
n= 2
n = 2, l = 0, ml = 0, ms = ½
n = 2, l = 0, ml = 0, ms = - ½
n = 2, l = 1, ml = -1, ms = ½
n = 2, l = 1, ml = -1, ms = - ½
n = 2, l = 1, ml = 0, ms = ½
n = 2, l = 1, ml = 0, ms = - ½
n = 2, l = 1, ml =1, ms = ½
n = 2, l = 1, ml = 1, ms = - ½
(8 total possibilities)
n=4
n = 4, l = 0, ml = 0, ms = ½
n = 4, l = 0, ml = 0, ms = - ½
n = 4, l = 1, ml = -1, ms = ½
n = 4, l = 1, ml = -1, ms = - ½
n = 4, l = 1, ml = 0, ms = ½
n = 4, l = 1, ml = 0, ms = - ½
n = 4, l = 1, ml =1, ms = ½
n = 4, l = 1, ml = 1, ms = - ½
n = 4, l = 2, ml = -2, ms = ½
n = 4, l = 2, ml = -2, ms = - ½
n = 4, l = 2, ml = -1, ms = ½
n = 4, l = 2, ml = -1, ms = - ½
n = 4, l = 2, ml = 0, ms = ½
n = 4, l = 2, ml = 0, ms = - ½
n = 4, l = 2, ml =1, ms = ½
n = 4, l = 2, ml = 1, ms = - ½
n = 4, l = 2, ml = 2, ms = ½
n = 4, l = 2, ml = 2, ms = - ½
n = 4, l = 3, ml = -3, ms = ½
n = 4, l = 3, ml = -3, ms = - ½
n = 4, l = 3, ml = -2, ms = ½
n = 4, l = 3, ml = -2, ms = - ½
(32 total possibilities)
n = 4, l = 3, ml = -1, ms = ½
n = 4, l = 3, ml = -1, ms = - ½
n = 4, l = 3, ml = 0, ms = ½
n = 4, l = 3, ml = 0, ms = - ½
n = 4, l = 3, ml =1, ms = ½
n = 4, l = 3, ml = 1, ms = - ½
n = 4, l = 3, ml = 2, ms = ½
n = 4, l = 3, ml = 2, ms = - ½
n = 4, l = 3, ml = 3, ms = ½
n = 4, l = 3, ml = 3, ms = - ½
20. Based on the above question, the total number of electrons possible for a given n-value is equal to:
A. 2n
B. 2n2
C. 2 (n-1)
D. 2(2n + 1)
18. Which of the following accurately describes l (the angular momentum quantum number)?
A. The value of l can be both positive and negative
B. The value of l specifies the orientation of the orbital
C. The value of l specifies the energy level of the orbital.
D. The number of possible l values equals the value of n.
CHEM 31 SI
Wednesday November 4
21. Supply the missing quantum numbers and sublevel names:
n
l
ml
Name
4
1
0
4p
2
1
0
2p
3
2
-2
3d
2
0
0
2s
Are the following sets of quantum numbers allowed (Yes or No)? If not, change the value of
only one of the quantum numbers to make the set allowed.
Allowed?
Changed Quantum Number if No:
n = 3, l = 0, ml = 0, ms = 1
Y/N
_________ms must equal ½ or – ½
n = 1, l = 1, ml = 0, ms = 1/2
Y/N
________l must equal 0
n = 8, l = 7, ml = -6, ms = -1/2
Y/N
____________________________
n = 2, l = -1, ml = 1, ms = 1/2
Y/N
__l must be 0 since n = 2 and ml = 0
What is the maximum number of electrons in an atom that can have these quantum numbers? (so
they are restricted to the following criteria)
n = 2, l = 0, ml = 0
______2_________
n = 2, l = 0, ml = 0, ms = ½
n = 2, l = 0, ml = 0, ms = - ½
(so there are two possible electrons in the 2s orbital)
n = 3, l = 1, ms = ½ _______3_______
n = 3, l = 1, ml = 1, ms = ½
n = 3, l = 1, ml = 0, ms = ½
n = 3, l = 1, ml = -1, ms = ½
(so there are three electrons possible in the 3p orbital with a spin ↑)
n = 5, ml = 1
_______8________
n = 5, l = 1, ml = 1, ms = ½
n = 5, l = 1, ml = 1, ms = - ½
n = 5, l = 2, ml = 1, ms = ½
n = 5, l = 2, ml = 1, ms = - ½
n = 5, l = 3, ml = 1, ms = ½
n = 5, l = 3, ml = 1, ms = - ½
n = 5, l = 4, ml = 1, ms = ½
n = 5, l = 4, ml = 1, ms = - ½
_______32_______
maximum number of electrons = 2n2 = 2(42) = 32
(think about it…if n=4, l = 0,1,2,and 3AKA l = s,p,d, and f…so, count across the 4s, 4p, 4d,
and 4f series on the periodic table…there are 32 elements)
n=4
CHEM 31 SI
Wednesday November 4
Write out the mnemonic for remembering the order to assigning electron configurations to
elements in their ground state orbitals:
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
6f
HAVING FUN YET?!
7s
(follow the arrows…so we fill the orbitals in the following sequence:
1s, 2s, 2p,3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, etc…….
What two transition metals do we need to know that are exceptions to the rules when writing
ground state electron configurations?
1. Chromium: Looks like [Ar] 2s1 3d5 instead of [Ar] 2s2 3d4
2. Copper: Looks like [Ar] 2s1 3d10 instead of [Ar] 2s2 3d9
Write the Ground State Electron Configuration and orbital diagram for the following atoms,
using the noble gas abbreviation:
(refer to the textbook and the solutions manual for these…
After you have done enough practice, these should be no sweat!)
N
Cu
Nd
Sr
I
Use a phrase to describe why the 2s orbital is more stable (lower energy) versus 2p.
When you superimpose the total radial probability of 2s and 2p onto the plot of 1s, you
notice that the 2s has a small peak that is inside the 1s shield, which causes them to have
more exposure to the full nuclear charge… so:
“2s electrons penetrate the 1s shield”
Explain briefly why the atomic radii increase moving down the periodic table, and decrease
moving to the right on the periodic table.
1. As we move down a column the n value (energy level) increases by 1. These
electrons are farther from the nucleus thus the radius is larger.
2. As you add electrons to an energy level, they do not shield each other very
extensively, and since you are adding protons as you move across, the valence
electrons experience successively greater nuclear charge…so they are drawn closer
to the nucleus and thus the radius decreases!
(see the solutions to the 11/8 worksheet for more details)