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Electrons
Light and Quantized Energy
Electrons part 1
The spectrum and refraction
• White light (sunlight) is a blend of all colors (ROY G BIV)
combined together.
• The wavelength (λ) and frequency (υ) for each color are unique to
that color.
• As light passes through a prism…
- the different wavelengths of the colors are separated.
- individual colors can be detected by the eye.
- a rainbow appears.
EMS
• All substances (radioactive or not) emit electromagnetic radiation.
• Only part of the spectrum that human eyes can detect is visible
light: (ROY G BIV)
• All other radiations have wavelengths that are either too long or
too short for our eyes to detect. (Goldilocks!)
Cosmic
rays
Microwaves
too short to see
too long to see
7.4 x 10-6 m
longest
4.2 x 10-6 m
shortest
Increasing danger
Wavelength vs. Frequency
• Wavelength (λ): distance b/n crests of a wave.
long wavelength
crest
Wavelength
Wavelength
trough
short wavelength
Cont…
• Frequency (ν): # of wavelengths that pass a certain point in a
given amount of time.
- units are Hertz (Hz)
• These 2 waves are traveling at = speeds… which wave will have
more crests cross the ‘finish line’ in a matter of one min.?
high frequency = short wavelength
*** higher frequency = higher energy ***
FINISH
low frequency = long wavelength
Wave Calculations
• all waves on the EM spectrum travel at the speed of light (c).
• wavelength (λ) and frequency (υ) are inversely related.
c = λ(ν)
νλ
λν
• c = speed of light = 3.00 x 108 m/s.
• to solve for λ …
• to solve for υ …
c
λ= ν
c
ν=
λ
cont…
• Practice…
Calculate the υ of a wave that has a wavelength of 5.00 x 10-6 m.
ν= c
λ
3.00 x 108
ν =
5.00 x 10-6
ν = 6.00 x 1013 Hz
What is the λ of radiation with a frequency of 1.50 x 1013 Hz?
λ= c
ν
3.00 x 108
λ=
1.50 x 1013
λ=
2.00 x 10-5 m
Does this radiation have a shorter or longer λ than red light?
longer…red light ~ 7.4 x 10-6 m
Emissions of light by atoms
• all elements will emit light when excited (i.e. by electricity).
• atoms absorb energy and then emit an = amount of energy in the
form of light.
- atoms emit a characteristic color
- Ne = orange - red
- Na = bright yellow
• if we pass this light through a prism (separate the λ) we get an
atomic emission spectrum.
set-up
ex. of wavelengths emitted
Continuous vs. emission
• emission spectra are unique to particular elements.
• only show certain lines of the continuous spectrum (white light).
• have helped us gather a lot of info. about our universe!
• atomic absorption spectra shows colors missing from the
continuous spectrum (missing λ were absorbed by the element).
continuous
absorption
emission
Quantum of energy
• e- are found on certain energy levels (orbitals) around the atom.
- there is a maximum of seven energy levels in an atom.
- e- on the energy level closest to the nucleus have the lowest
energy. The 7th energy level has the highest energy.
- an e- requires one ‘quanta’ of energy to jump to the next
energy level.
• e- at their lowest energy level are considered to be at the
ground state (most stable).
• if e- absorb a quantum or more of energy (from electricity), they
can jump to higher energy levels (excited state).
• e- must lose energy in order to fall from the excited state back to
the ground state.
- this energy is emitted in the form of visible light!
The energy of each orbital can be calculated using Bohr’s equation.
En = -(2.18 x 10-18 J) / n2
Where n = (1,2,3,ect…)
Bohr’s model of the atom was able to correctly
explain why line spectra results when atoms are
heated.
Quantum Theory and the
Atom
Electrons part 2
Quantum Numbers
• the ‘address’ of an e-. Will look like this { 1, 1, 0, +1/2)
• there are four quantum numbers ...
1st = n = energy level of the e-. Possible values are (1,2,3,etc…)
- ‘the principle quantum number’
- represented by rows on the periodic table
- 1 = lowest, 7 = highest
2nd = l = shape of sub-orbital
- 4 different shapes…found on each energy level
3rd = ml = orientation of orbital in space.
- which axis the orbital lies on
4th = ms = spin of the e- on its axis.
- clockwise or counter-clockwise
Principle energy level: Maximum Number of
Electrons
- Corresponds to a ring on the Bohr Model. Each ring is principle energy.
• the formula 2n2 is used to find the max. # of e- on any principle
energy level.
- n = the energy level = (row on P.T.)
• Ex…
{ 1, 1, 0, +1/2}
• energy level 3… 2(3)2 = 18 e• energy level 6… 2(6)2 = 72 e-
First ring on Bohr
Model from our
first example.
The
nd
2
Quantum # (sublevels)
• l = shape of orbital. Possible values: 0,1,2,3. n = # of possible l
• there are four orbital shapes, represented by letters…
• each orbital can only hold 2 e-.
• (0) s = sphere= 1 orbital = 2 total e- lowest energy
• (1) p = dumbbell = 3 orbitals = 6 total e-
• (2) d = clover-leaf = 5 orbitals = 10 total e• (3) f = double clover-leaf
- highest energy
{ 1, 1, 0, +1/2)
Tells us that
the electron is
in the p orbital.
= 7 orbitals = 14 total e-
The 3rd Quantum number
Called the magnetic quantum # (ml)

Possible values: between (l to -l)
Ex. If 2nd quantum number is 1 then all possible
values for 3rd quantum number is: -1, 0, +1

This is the orientation of orbital around the
axis.
{ 1, 1, 0, +1/2)
Tells us the e- is in
the second of three
orbitals.
The 4th quantum number
The spin of the electron. (ms)



Can have two possible spins: clockwise,
counter clockwise.
We will use +½ for clockwise.
We will use -½ for counterclockwise.
{ 1, 1, 0, +1/2)
Tells us the e- has a
clockwise spin.
1st quantum
number tells how
large the orbital.
cont…
‘s’ orbital
2nd quantum
number
designates
shape of
orbitals.
And each
can have 2e-
3rd quantum
# is
orientation.
‘p’ orbitals
4th
quantum
number
tells the
spin of e-
How many e- can each
orientation have?
Answer: 2e-
cont…
‘d’ orbitals
How many total ecan the d orbitals
have?
‘f’ orbitals
10e-
Principle (n) 2nd Quant. #
(sublevel l)
# of orbitals of
sublevels.
(-l to +l )
Total # of
orbitals of
(n)
1
s = 0
1 (0)
1
2
s = 0
p = 1
1 (0)
3 (-1, 0, +1)
4
3
s = 0
p = 1
d= 2
1 (0)
3 (-1,0,+1)
5 (-2,-1,0,+1,+2)
9
4
s=0
p=1
d=2
f=3
1 (0)
16
3 (-1,0,+1)
5 (-2,-1,0,+1,+2)
7 (-3,-2,-1,0,+1,+2,+3)
Questions about address of
electron.
If n=3, can l = 1?

Yes
If n=2, can l = 2?
What are the possible values of the
sublevels in the 3 principle energy level?

0,1,2 ( n=3 so 3 possible and start with 0)
How many orbitals does the p sublevel
have?

3 (-1,0,+1)
Electron Configuration
Electrons part 3
Aufbau Principle
States that each electron occupies the
lowest energy orbital available.



All orbitals related to an energy sublevel are
of equal energy. Ex. All “p” orientations have
same energy.
The energy sublevels within a principle
energy level have different energies. Ex. The
2s orbitals have less energy than the 2p
orbitals.
The sequence of energy is: s (lowest),p,d,
then f (highest).
Pauli’s Exclusion Principle
• Pauli stated that no two e- in the same atom will have the same set
of quantum #s.
• only 2 e- can fill an orbital
- each e- has a different spin (‘s’ quantum #)
- even if first 3 quantum #s are the same, it is always the 4th that
will define an individual e-.
Hund’s Rule
• Hund stated that e- will fill empty orbitals 1st before they pair up.
• ‘elbow room’
The Diagonal Rule
s
p
d
f
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
cont…
• Pictorial configuration…
- draw a line to represent each orbital
- use arrows to represent e-...point up or down to represent spin.
- follow Aufbau, Pauli, & Hund’s Rule - fill orbitals in singly 1st
before you pair e-.
• Ex. Use diagonol rule: Give the pictorial notation for B (Z = 5)
1s
2s
1s2 2s2 2p1
2p
• Give the pictorial notation for S (Z = 16)
1s
2s
2p
1s2 2s2 2p6 3s2 3p4
3s
3p
What is the address of the last
electron added?
{2, 1, -1, +1/2}
1s
2s
1s2 2s2 2p1
2p
{3,1,-1,-1/2}
1s
2s
1s2 2s2 2p6 3s2 3p4
2p
3s
3p
Electron Configurations
• the periodic table is laid out in order of the quantum #s.
• in e- configurations, you must represent (in order), the:
energy level #, orbital (s,p,d or f) and e- # for each e- in the atom.
• must fill in lowest energy level (1) 1st, and lowest orbital (s) 1st on
each energy level.
• recall that # of p+ and e- = each other in a neutral atom!
Fe is
here.
Boron is
here.
Lets give the electronic configuration of Boron (B).
1s2 2s2 2p1
Lets give the electronic configuration for Fe
1s2 2s2 2p6 3s2 3p6 4s2 3d6
Look at the
diagonal rule; after
4s, then 3d.
cont…
• Practice…
• Write out the correct e- configuration for Mg (Z = 12)
1s2 2s2 2p6 3s2
• Write out the correct e- configuration for P (Z = 15)
1s2 2s2 2p6 3s2 3p3
• ‘d’ orbitals are so large that they reach into the next energy level.
• therefore, the 1st ‘d’ orbital belongs to the 3rd energy level…even
though we don’t see it until the 4th row of the P.T.!
• Write out the correct e- configuration for Br (Z = 35)
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
• ‘f’ orbitals are so large that they reach into the next 2 energy levels.
• therefore, the 1st ‘f’ orbital belongs to the 4th energy level…even
though we don’t see it until the 6th row of the P.T.!
Last Energy Level using Noble gas notation
• To give the noble gas notation, find the element of interest,
then find the previous noble gas and start from that point.
• Put the noble gas in brackets which will represent all of the
electrons up to the noble gas.
Ex…
Potassium (K) = [Ar] 4s1
[He]2s2 =
Beryllium (Be)
Iodine (I) = [Kr]5s2 4d10 5p5
[Ne]3s2 3p2 = Silicon (Si)
Nickel (Ni) = [Ar] 4s2 3d8
[Kr]5s2 4d9 = Silver (Ag)
Electron Dot Diagrams
• represents all of the valence e- in an atom.
• valence e- are all the e- on the outer-most energy level of an atom.
- only ‘s’ and ‘p’ orbitals on highest level!!!
- ‘f’ and ‘d’ orbitals are embedded in the other two.
- represented by the ‘A’ columns on the P.T.
• electron dot diagrams consist of the symbol of the element
surrounded by dots for each valence e-.
Ex…
Iodine (I):
[Kr]
5 s 2 5 p 5 Beryllium (Be):[He] 2s2
I
Be