Download 1 - Development of Atomic Models

Development of the AtomicTheory
Nearly 2500 years ago Greek
philosophers expressed a belief matter
is composed of tiny indivisible
particles called atoms (atomos is the
Greek word for “indivisible”)
These conclusions were not based on
any evidence; they were derived from
philosophical reasoning.
Experimentation by many scientists
during the 18th and 19th centuries led
to the development of 2 Laws
The Law of Conservation of Mass
During chemical change no loss or
gain of mass occurs.
and
The Law of Definite Proportions
Compounds contain elements in fixed
proportions by mass.
John Dalton, in the early 19th Century, made
sense of these Laws in an Atomic Theory
1. Matter consists of particles called atoms.
2. Atoms are indestructible. In chemical
reactions atoms rearrange but are not
broken apart.
3. Atoms in one particular element are
identical, but differ from atoms of other
elements.
4. Compounds are created when atoms of
different elements combine in definite
proportions.
Developments in technology led to
further refinement of the Atomic
Theory.
Evacuated gas discharge tubes
(like the modern TV tube) were
used to demonstrate the existence
of negatively charged particles
which were named electrons.
Here is an example of a cathode
ray tube.
air molecule
To vacuum pump
Cathode
Anode
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source of
electricity
air molecule
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electricity
air molecule
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electricity
air molecule
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electricity
air molecule
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electricity
air molecule
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electricity
air molecule
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source of
electricity
air molecule
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Anode
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source of
electricity
air molecule
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Anode
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source of
electricity
air molecule
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Anode
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source of
electricity
air molecule
To vacuum pump
Cathode
Anode
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source of
electricity
air molecule
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Anode
High voltage
source of
electricity
air molecule
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Cathode
Anode
High voltage
source of
electricity
air molecule
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Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
Affect of a
magnet
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
Experimenting with
different metals for
cathodes and anodes led
to identical results.
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
From that observation
scientists concluded that
all metals have identical
negative particles which they
called
electrons.
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
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Cathode
Anode
High voltage
source of
electricity
To vacuum pump
Cathode
Anode
High voltage
source of
electricity
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source of
electricity
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source of
electricity
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Cathode
Anode
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source of
electricity
Objects placed in these
tubes were observed to
rotate.
Conclusion:
electrons have mass
To vacuum pump
High voltage
source of
electricity
Anode
What conclusion can be drawn from this
observation?
Electrons are negatively charged
Negative plate
Cathode
Anode
High voltage Positive plate
source of
electricity
What conclusion can be drawn from this
observation?
Electrons are negatively charged
Negative plate
Cathode
Anode
High voltage Positive plate
source of
electricity
In 1897 J.J. Thomson
experimented with a special
cathode ray tube and made
some interesting observations.
When hydrogen gas is placed inside this
tube particles were observed to travel
towards the Cathode?
What possible explanation is there?
Cathode
Anode
High voltage
source of
electricity
Electrons collide with hydrogen atoms and
repel their electrons leaving behind a single
What is this particle called?
positively charged particle which is
A proton.
attracted to the negative cathode.
Anode
Cathode
High voltage
source of
electricity
Here is what Thomson inferred about
electrons and protons.
Atoms contain both positive protons
and negative electrons.
These charges are equal and opposite
Protons are much more massive than
electrons.
Thomson proposed an atomic model
which resembles a raisin bun.
His model suggested an atom is like a
positive bun with negative “raisins”.
J.J. Thomson’s Atomic Model
The
negative
electrons
are like
raisins
embedded
in a positive
bun
The discovery of radioactivity led
to the use of alpha particles as
probes to investigate the interior
structure of atoms.
Ernest Rutherford performed a
series of experiments where he
bombarded a thin sheet of gold
foil with positively charged alpha
particles.
Click here for a movie showing Rutherford’s
Experiment
Comparison of Subatomic Particles
Particle
Mass (g)
*
Mass (u)
electron
proton
neutron
9.1 x 10-28
-24
1.7 x 10
1.7 x 10-24
0.00055
1.007
1.009
Electrical
Symbol
Charge
11+
0
* u - this is the symbol for an atomic
mass unit which is based on the
mass of a carbon 12 atom
1 u = 1/12 the mass of a C-12 atom
e11+
p
no
Despite the evidence in support of
Rutherford’s model it failed to explain
why the negative electrons did not fall
into the positive nucleus.
Niels Bohr, in 1913, explained this by
suggesting electrons are found in
electron shells which surround the
nucleus. The energy electrons possess is
sufficient to allow them to move around
the nucleus, within these shells.
Electron Shells (Energy Levels)
1st shell
2nd shell
3rd shell
The interesting thing about Bohr’s atomic
model is electrons can absorb energy and
jump to higher energy levels, where they
remain for short periods of time before
“falling” back down.
1st shell
2nd shell
3rd shell
When they “fall” back down they emit
certain colours of light.
Each element, with its distinctive energy
levels, emits a unique colour of light.
1st shell
energy
2nd shell
3rd shell
Electrons are found in the energy
levels which surround the nucleus.
As the energy levels (shells) get
farther from the nucleus they are
capable of holding more electrons.
The number of electrons which fit in
each energy level can be determined
by looking at the changes in width of
the periodic table.
The 1st energy level holds 2e1-, the
2nd holds 8e1-, 3rd-18 e1-, 4th-32 e12e18e118e132e1-
Here are some examples
of Bohr-Rutherford
diagrams of atoms
where protons, neutrons
and electrons are all
shown.
H
1
2
1p1+ 1e11no
He
2
5
2p1+ 2e13no
Li
3
7
3p1+ 2e1-1e14no
Be
4
10
4p1+ 2e1-2e16no
Be-10 and Be-9 are called isotopes.
Atoms of the same element with
different masses are called isotopes.
When a number immediately follows
an element’s symbol, that number is
the atomic mass.
30Si
14
8e1-
14p1+ 2e116no
4e1-
Determine the isotope
notation (symbol, atomic #,
mass #) for each of the
following Bohr-Rutherford
Diagrams
33P
15
8e1-
15p1+ 2e118no
5e1-
O
8
19
8p1+ 2e1-6e111no
How does one know how many electrons
are found in each energy level for an atom
with lots of electrons?
Here are some simple steps to follow when
determining the number of electrons in
each energy level.
First locate where the element is found in
the Periodic Table.
To demonstrate here’s how to determine the
energy level arrangement for S.
Next assign group numbers to the periodic table.
8
1 2
3 4 5 6 7
2 2 2 2 2 2 2 2 2 2
These numbers represent the number
of electrons in the last energy level
Next assign row numbers.
1
2
3
8
1 2
3 4 5 6 7
2 2 2 2 2 2 2 2 2 2
+
4
5
6
7
These numbers represent the number
of energy levels atoms of this element
have.
S has 3 energy levels and 6 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
4
5
6
7
2 2 2 2 2 2 2 2 2 2
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a S
atom.
16 - (2+6) = 16-8 = 8
# of e1- in
each level
Energy
level
#
2
8
6
1
2
3
Here is a Bohr-Rutherford Diagram
S
16
6e18e1-
16 p1+ 2e1-
Now it’s your turn.
Determine the number
of electrons in each
energy level for an atom
of tin and then construct
a Bohr-Rutherford
Next assign row numbers.
1
2
3
4
5
8
1 2
3 4 5 6 7
2 2 2 2 2 2 2 2 2 2
+
6
7
These numbers represent the number
of energy levels atoms of this element
have.
Sn has 5 energy levelsand 4 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
4
5
6
7
2 2 2 2 2 2 2 2 2 2
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a Sn
atom.
50 - (2+8+18+4) = 50-32 = 18
# of e1- in
each level
Energy
level
#
2
8
18
18
4
1
2
3
4
5
Here is a Bohr-Rutherford Diagram
Sn
50
4e1-
18e118e18e1-
50 p1+ 2e1-
Now it’s your turn.
Determine the energy level
arrangements and draw the
Bohr Rutherford Diagrams
for each of the following:
Co,
Rh,
Ba,
V.
27
45
56
23
Co has 4 energy levels and 2 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
4
5
6
7
2 2 2 2 2 2 2 2 2 2
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a Co
atom (27 - (2+8+2) = 27-12 = 15
# of e1- in
each level
Energy
level
#
2
8
15
2
1
2
3
4
Here is a Bohr-Rutherford Diagram
27Co
2e115e18e1-
27 p1+ 2e1-
Rh has 5 energy levelsand 2 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
4
5
6
7
2 2 2 2 2 2 2 2 2 2
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a Rh
atom (45 - (2+8+18+2) = 45-30 = 15
# of e1- in
each level
Energy
level
#
2
8
18
15
2
1
2
3
4
5
Here is a Bohr-Rutherford Diagram
45Rh
2e1-
15e118e18e1-
45 p1+ 2e1-
Ba has 6 energy levels and 2 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
2 2 2 2 2 2 2 2 2 2
4
5
6
7
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a Ba
atom (56 - (2+8+18+18+2) = 56-48 = 8
# of e1- in
each level
2
Energy
level
1
#
8
18
18
8
2
2
3
4
5
6
Here is a Bohr-Rutherford Diagram
56Ba
8e1-
18e118e18e1-
56 p1+ 2e1-
2e1-
V has 4 energy levels and 2 e1- in the last
8
energy level
1 2
3 4 5 6 7
1
2
3
4
5
6
7
2 2 2 2 2 2 2 2 2 2
+
Draw blank lines for each of the energy levels
The number of e1- in the 2nd last energy level is
always determined by subtracting the number of
e1- already placed from the total number in a Va
atom (23 - (2+8+2) = 23-12 = 11
# of e1- in
each level
Energy
level
#
2
8
11
2
1
2
3
4
Here is a Bohr-Rutherford Diagram
23V
2e111e18e1-
27 p1+ 2e1-
This atomic model worked until a theoretical
physicist, Heisenburg, reasoned that electrons
are so small, whenever their presence is detected
the act of detection changes the direction they
are travelling.
This led to the development of a new atomic
model called the Quantum Mechanical Atomic
model. This model was developed using
sophisticated mathematical equations which
describe the motion of electrons within atoms.
This model assumes that the position and
direction of motion of an electron cannot be
simultaneously determined.
This means the location of an electron can only
be predicted and not know with absolute
certainty.
The modification required to the more simple
Bohr-Rutherford model is to separate the energy
levels in sublevels called orbitals. Each level is
subdivided into an equal number of sublevels.
The Quantum Theory of the Atom
http://www.shef.ac.uk/chemistry/orbitron/AOs/4f/index.html
It was proposed that energy levels
contained electrons with slight
differences in energy and it was proposed
that an energy level could be subdivided
into sublevels: (1)s; (2)s,p; (3)s,p,d; and
(4)s,p,d,f (orbitals).
Each sublevel (orbital) can hold a
different number of electrons.
2e1- in an s orbital
6 e1- , 2e1- in each of the 3 p orbitals
10e1-, 2e1- in each of the 5 d orbitals
14e1-, 2e1- in each of the 7 f orbitals
S - Orbitals are spherical regions around the
nucleus. An e1- can move anywhere within
this region of space. If an e1- is in an s orbital
in the first energy level it’s said to be a
1s electron.
Energy
level
1s
Orbital
shape
2s
3s
P - Orbitals - The 2nd energy level can hold a
maximum of (2n2) 8e1-. (n represents the
number of the energy level). Since the 2s can
hold 2e1- where do the other 6e1- go? They
are contained in the 3 “2p” orbitals.
D - Orbitals
There are 5 d-orbitals that can exist, each
holding 2e1- so that any energy level can hold
a maximum of 10 d orbital electrons.
F - Orbitals
There are 7 f -orbitals that can exist, each
holding 2e1- so that any energy level can hold
a maximum of 14 f orbital electrons.
7 different f orbitals
These energy levels are
superimposed on each other
in the region surrounding
the nucleus.
3d x2 - y2 2py
3dyz
1s
+
3dxz
2px
2s
3dxy
3s
3pz
3d z2
3py
4s
3px
Maximum # Maximum # of
Principle
Energy
Quantum Orbitals
of e 1- in
e 1- in the
Level
Number
each
energy level
1st
1
1s
2
2
2s
2
2nd
2
8
2p
6
3s
2
3rd
3
3p
6
18
3d
10
4th
4
4s
4p
4d
4f
2
6
10
14
32
Rules for Filling Orbitals
1. Electrons occupy the lowest energy
orbital of the lowest energy level
1st.
2. No electron pairing takes place in
the p, d or f orbitals until each
orbital contains 1 electron. This is
called Hund’s Rule.
3. No orbital can contain more than 2
electrons. (Pauli’s Exculsion Principle)
To determine the
order with which
the orbitals are
filled one can use
a periodic table.
s orbitals are filled here
p orbitals are filled here
d orbitals are filled here
f orbitals are filled here
s
p
d
f
The orbital order can be determined by counting
up the periodic table using the atomic number.
1s
2s
3s
4s
5s
6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
4f
5f
The orbital order can be determined by counting
up the periodic table using the atomic number.
Another way of
determining this
order is to write all
the orbitals first
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p
7s
Next
draw
arrows
like this
and
follow
them
from
start to
finish
Now let’s apply this concept to see
where all the electrons in individual
atoms are found.
An atom of Ruthenium has an
atomic number of
44
so the electrons are found in the
following orbitals:
2
1s
2
2s
3s2
2
4s
5s2
10
3d
4d6
2p6
3p6
6
4p
This electron configuration, for 44Ru is
usually written like this:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d6
notice the total number of electrons is
2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 6 = 44
Now it’s your turn to
try. Determine the
electronic
configuration for an
atom of titanium.
2
1s
2
2s
3s2
2
2
4s 3d
2p6
3p6
The electron configuration, for 22Ti is:
1s2 2s2 2p6 3s2 3p6 4s2 3d2
notice the total number of electrons is
2 + 2 + 6 + 2 + 6 + 2 + 2 = 22
Determine the electronic
configurations for
Ag- 47, As - 33, Ca - 20
2
1s
2
2s
3s2
2
4s
5s2
10
3d
4d9
2p6
3p6
6
4p
2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d9
1s
47Ag
2
1s
2
2s
3s2
2
4s
10
3d
2p6
3p6
3
4p
2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
1s
33As
2
1s
2
2s
3s2
2
4s
2 2s2 2p6 3s2 3p6 4s2
1s
20Ca
2p6
3p6
Energy Level Diagrams