1s 2 2s 2 2p 6 3s 2 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f Ni = 28 e

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PreAP Chemistry Chapter 4 Notes
Section 4.1 The Development of a New Atomic
Model
Previously, Rutherford reshaped our thoughts
of the atom by showing the protons were
located in the nucleus of the atom, but he could
not model for us where the electrons were,
other than outside the nucleus somewhere.
Fortunately, studies into the properties of light
and the effects of light on matter soon gave
clues to where electrons actually are.
Light is a small part of all the radiation
(something that spreads from a source) called
electromagnetic radiation. Electromagnetic
radiation is energy in the form of waves (of
electric and magnetic fields). Electromagnetic
radiation includes radio waves, microwaves,
infrared, visible light, X-rays, and Gamma
rays. All these together are considered the
Electromagnetic Spectrum.
As all the forms of electromagnetic radiation
are waves, they all have similar properties.
•All electromagnetic radiation travels at the
speed of light (c), 299,792,458 m/s (3 x 108) in a
vacuum
•The crest is the top of the waves, the trough is
the bottom of the waves, and the amplitude is a
measurement from the rest or zero line to a
crest or trough
•The wavelength (λ – lambda) is the distance
between successive crests/troughs and is
measured in meters (often nm = 10-9 m)
•The wavelength (λ – lambda) is the distance
between successive crests/troughs and is
measured in meters (often nm = 10-9 m)
•The frequency (ν – nu) is the number of waves
that pass a point in one second and is measured
in
1
(per second – can be written as s-1) or
s
Hz (Hertz)
Wavelength is the distance between two
crests/troughs
λ (lamda) is the symbol of wavelength and m is
the unit
Frequency is the number of crests passing
through a point per second
How many hertz is the first wave?
How many hertz is the second wave?
The speed of a wave is directly proportional to
the wavelength and the frequency; c = λν is the
formula
c
λ
ν
Example. A certain violet light has a
wavelength of 413 nm. What is the frequency
of the light?
Example. A certain violet light has a
wavelength of 413 nm. What is the frequency
of the light?
c
ν=
λ
Example. A certain violet light has a
wavelength of 413 nm. What is the frequency
of the light?
c
ν=
λ
ν=
3.00 × 108 m/s
413 × 10-9 m
Example. A certain violet light has a
wavelength of 413 nm. What is the frequency
of the light?
c
ν=
λ
ν=
3.00 × 108 m/s
413 × 10-9 m
ν = 7.26 × 1014 Hz
Unfortunately, thinking of light as waves lead
to a problem. It was noticed that if light strikes
a metal, then sometimes it could cause electrons
to be emitted (leave the atoms entirely – like in
a solar panel); called the photoelectric effect. If
light was a wave, then all amounts of light
energy should cause this to happen, but this
was not the case. It always took some
minimum amount of energy to get the electrons
to be emitted.
This lead Max Planck to theorize that light
must carry energy in basic minimum amounts
that he called quanta. Like a delivery person
cannot correctly deliver half a box, the
electrons in atoms cannot gain a fraction of a
quantum of energy (it has to be in whole
numbers).
He proposed that this energy was directly
proportional to the frequency of the
electromagnetic radiation and a constant, now
called Planck’s constant. E = h ν
E = energy in Joules (J)
h = Planck’s constant = 6.63 × 10-34 Js
ν = frequency in Hz or 1/s
E
h
ν
Example. What is the energy content of one
quantum of the light with a wavelength of 413
nm?
Example. What is the energy content of one
quantum of the light with a wavelength of 413
nm?
Note: wavelength is not in the energy equation,
but frequency is. So first, you must solve for
the frequency. As seen in the earlier example, a
wavelength of 413 nm gives a ν = 7.26 × 1014
Hz.
Example. What is the energy content of one
quantum of the light with a wavelength of 413
nm?
ν = 7.26 × 1014 Hz
E=h×ν
E = 6.63 × 10-34 Js × 7.26 × 1014 1/s
Example. What is the energy content of one
quantum of the light with a wavelength of 413
nm?
ν = 7.26 × 1014 Hz
E=h×ν
E = 6.63 × 10-34 Js × 7.26 × 1014 1/s
Example. What is the energy content of one
quantum of the light with a wavelength of 413
nm?
ν = 7.26 × 1014 Hz
E=h×ν
E = 6.63 × 10-34 Js × 7.26 × 1014 1/s
E = 4.81 × 10-19 J
In 1905 Einstein used Plancks work to propose
that electromagnetic radiation had a dual
wave-particle nature. As a particle,
electromagnetic radiation carries a quantum of
energy of energy, has no mass, and is called a
photon.
So to get an electron to emit from a metal, it
must be struck with a photon having quantum
energy big enough, or nothing will happen.
Each metal requires a different quantum
energy, thus each metal can be identified by the
frequency of light needed to emit electron.
This idea was expanded
upon to develop an idea
of where the electrons
were in an atom. It was
found that low pressure
gases could be trapped
in a tube and electrified,
and would then glow a
color particular to the
gas inside.
Furthermore this light could be passed into a
prism, and instead of getting the entire
spectrum (rainbow) of colors, only certain
wavelengths of light would be seen as small
bars of color, called a line-emission spectrum.
This would indicated that the electrons in an
atom were only absorbing specific amounts of
energy from the electricity, causing the
electrons to move from their ground state
(normal position close to the nucleus) to an
excited state (higher energy position further
away from the nucleus). The electrons do not
stay in the excited state for long and fall back
to their ground state, losing the energy equal to
what they gained.
Quanta Video
Niels Bohr used this to develop a model of the
atom where the electrons could only be in
certain, specific energy level (n) orbits around
the nucleus. Just as you cannot go up half a
rung on a ladder, the electron could not go up a
partial energy level. The electrons gained or
lost enough energy to move a whole number
amount of energy levels (n) away from or closer
to the nucleus, or it did not move.
He calculated the amount of energy needed for
an electron of hydrogen to move between each
energy level (n) (which was not constant) and
his calculations agreed with experimental
results.
The Balmer series of hydrogen spectral lines
refer to the four lines seen in the visible light
region (the four colored bars). If the electron
was excited to energy level (n) 6, 5, 4, or 3 and
fell to energy level (n) 2, the resulting energy
given off would have a frequency in the visible
region of electromagnetic radiation. (One line
for dropping from 6 to 2, one for 5 to 2, one for
4 to 2, and one for 3 to 2).
Line Spectra Video
However, there are other possibilities. If the
electrons drop from n=6, 5, or 4 to n=3, then
the energy given off is not big enough to be seen
as it is in the infrared region. These three lines
in the infrared region are referred to as the
Paschen series. If the electrons drop to n=1,
then the five lines given off are too high in
energy to be seen, as they are in the ultraviolet
region. These lines are referred to as the
Lyman series.
Model of Atom Review:
1. Thomson’s Plum Pudding Model – the atom
is a ball of evenly spread positive stuff with
random negative particles (electrons).
1. Thomson’s Plum Pudding Model – the atom
is a ball of evenly spread positive stuff with
random negative particles (electrons).
2. Rutherford’s Nuclear Model – the atom has
a central nucleus containing the positive
particles (protons) with the electrons outside.
1. Thomson’s Plum Pudding Model – the atom
is a ball of evenly spread positive stuff with
random negative particles (electrons).
2. Rutherford’s Nuclear Model – the atom has
a central nucleus containing the positive
particles (protons) with the electrons outside.
3. Bohr’s Orbital Model – The electrons circle
the nucleus in specific energy orbits, like the
planets orbit the sun. Unfortunately this
only works for atoms with one electron…
4. Quantum Mechanical Model – electrons are
found in specific regions around the nucleus,
but the exact location of the electrons inside
the regions cannot be determined
Atom Models Video
The quantum mechanical model of the atom is
built on the ideas and calculations of several
scientists.
•
Louis deBroglie suggested a way to show
that a particle could have wave like behavior
with the equation:
h
λ=
mv
h = Planck’s constant
m = mass of particle
v = velocity of particle
• The Heisenberg Uncertainty principle states
that it is impossible to know both the velocity
and the location of an electron at the same
time. If the position was known, then there
is no way to know where it has moved to,
and if the velocity is known, then there is no
way to know where it was.
• Schrödinger developed wave-based
equations that form the basis of the current
Quantum theory, which mathematically
describe the probably location of electrons,
often referred to as an electron cloud. The
electrons clouds describe areas around the
nucleus with a 90% chance of finding the
electron inside. Solving the equations has
lead to Quantum Numbers, which will be
studied later.
The quantum mechanical model starts with a
Principal Quantum Number (n), which is the
basic energy level of an electron, and often
matches the period number. Possible values
(currently) are 1-7.
The quantum mechanical model starts with a
Principal Quantum Number (n), which is the
basic energy level of an electron, and often
matches the period number. Possible values
(currently) are 1-7.
Inside the principal quantum energy level are
sublevels that correspond to different cloud
shapes. The sublevels are designated as s
(sharp), p (principal), d (diffuse), and f
(fundamental).
Inside the sublevels are orbitals, specific
regions with a 90% probability of finding
electrons.
• s –orbitals are spherically shaped clouds
around the nucleus
• p -orbitals are bar-bell shaped clouds with
the nucleus between the lobes
• d and f are much more complex in shape
Each sublevel has room for a different amount
of electrons, because an orbital can hold two
electrons, then each sublevel has a different
amount of orbitals
• s –sublevel can hold 2 electrons, so it has 1
orbital (shape)
• p –sublevel can hold 6 electrons, so it has 3
orbitals (shapes)
• d –sublevel can hold 10 electrons, so it has 5
orbitals (shapes)
• f –sublevel can hold 14 electrons, so it has 7
orbitals (shapes)
The s sublevel is simply a sphere centered on the nucleus.
The p sublevel has three orbitals. These are often referred to a dumbbell shape.
The d sublevel has five orbitals:
The f sublevel has seven orbitals:
To know the maximum amount of electrons
that could be in any principal quantum level
(and the number of elements that could be
represented) use the formula 2n2
if n=1, then
To know the maximum amount of electrons
that could be in any principal quantum level
(and the number of elements that could be
represented) use the formula 2n2
if n=1, then 2 electrons will fit
if n=4,
To know the maximum amount of electrons
that could be in any principal quantum level
(and the number of elements that could be
represented) use the formula 2n2
if n=1, then 2 electrons will fit
if n=4, then 32 electrons will fit
Section 4.3 Electron Configurations
In order to show on paper where electrons are
likely to be located in an atom, orbital filling
diagrams and electron configurations are
drawn or written. When this is done, three
rules must be followed:
1. Aufbau principle – electrons fill lower
energy levels first, thus 1 before 2 and s
before p, etc.
a. orbitals within a sublevel are equal in energy
(called degenerate)
b. the principal energy levels often overlap,
making them seem a little out of order
c. boxes are used to represent orbitals
Another way of writing the aufbau principle
diagram:
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
2. Pauli Exclusion principle – an orbital (box)
can hold a maximum of two electrons
(arrows)
a. for two electrons to fit, they have to have
opposite spins
b.
c.
for one electron in the orbital
for two electrons in the orbital (opposite
spins)
3. Hund’s Rule – when electrons occupy
degenerate orbitals, one electron is placed
into each orbital with parallel spins before
doubling up
Ex. _____ _____ _____
3p
NOT _____ _____ _____
3p
At this time show the orbital viewer at
http://intro.chem.okstate.edu/WorkshopFolder/
Electronconfnew.html
Have out periodic tables before
starting.
Orbital Notation shows the arrows in the boxes
to represent the electrons in an atom. To
shorten this process, an electron
configuration can be written. It leaves out
the information about the number of orbitals
in each sublevel, so it will be expect you
remember that information.
It has the general form nΔ°
n = principal quantum number (1-7…)
Δ = sublevel letter (s, p, d, or f)
° = number of e- in that orbital (1-14)
The sublevels can be listed in order of filling as
from the periodic table, but for the notes,
quizzes, and tests the sublevels will be
grouped together by principal quantum
number.
Ni = 28 e1s 2s
2p
3s
3p
3d
4s
4p
4d
4f 5s
5p
5d
5f
Ni = 28 e1s2 2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s
3p
3d
4s
4p
4d
4f 5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s2 3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s2 3p6 3d 4s
4p
4d
4f
5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s2 3p6 3d
4s2 4p
4d
4f 5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s2 3p6 3d8 4s2 4p
4d
4f
5s
5p
5d
5f
Ni = 28 e1s2 2s2 2p6 3s2 3p6 3d8 4s2 4p
2 + 2 + 6 + 2 + 6 + 8 + 2 = 28
4d
4f
5s
5p
5d
5f
Sn = 50 e-
Your Turn to Try
1s 2s
2p
3s
3p
3d
4s
4p
4d
4f 5s
5p
5d
5f
Sn = 50 e1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f 5s2 5p2
5d
5f
A few elements have electron configurations
that do not follow the normal pattern. These
two groups are represented by the first
elements in their group:
By normal configuration:
Cr -- 1s2 2s2 2p6 3s2 3p6 3d4 4s2 …
Cu -- 1s2 2s2 2p6 3s2 3p6 3d9 4s2
…
By actual (seen in nature) configuration:
Cr -- 1s2 2s2 2p6 3s2 3p6 3d5 4s1 …
Cu -- 1s2 2s2 2p6 3s2 3p6 3d10 4s1 …
The other elements in those groups work in
similar matters. Instead of writing d4, a half
filled sublevel seems to be more stable, so
make it d5. Instead of writing a d9, a
completely filled sublevel seems to be more
stable, so make it d10.
If writing out the entire electron configuration is too
much, we can use the previous (in the periodic
table) noble gas to take the place of part of the
electron configuration:
Polonium:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p4
Xenon:
1s22s22p63s23p64s23d104p65s24d105p6
Polonium: [Xe] 6s24f145d106p4
When the electron configuration is written for an
element using the noble gas configuration the
electrons written after the noble gas are the ones that
appear on the outside of the atom, called valence
electrons..
When elements bond to form compounds, it is these
electrons that are involved. The amount of valence
electrons makes a big difference in how the element
will bond, so to make it easy to predict, we draw
electron dot diagrams.
A) In an electron dot diagram, we use the symbol of
the element and dots to represent the number of
valence electrons.
B) Only s and p electrons with the highest quantum
number count for dot diagrams, even if there are d
and f electrons after the noble gas.
Lithium = [He]
1
2s
So
Li
Beryllium = [He]
So
Be
2
2s
Boron = [He]
So
2
1
2s 2p
B
Carbon = [He]
2
2
2s 2p
So
C
Nitrogen = [He]
So
2
3
2s 2p
N
Oxygen = [He]
So
2
4
2s 2p
O
Fluorine = [He]
So
2
5
2s 2p
F
Neon = [He]
So
2
6
2s 2p or
Ne
[Ne]

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