Download CH 3 power point atomic structure

Atomic Structure
Dalton’s Atomic Theory (1808)
1. Elements are composed of extremely small
particles called atoms. All atoms of a given
element are identical, having the same size,
mass and chemical properties. The atoms of one
element are different from the atoms of all other
elements.
2. Compounds are composed of atoms of more
than one element. The relative number of atoms
of each element in a given compound is always
the same.
3. Chemical reactions only involve the rearrangement
of atoms. Atoms are not created or destroyed in
chemical reactions.
2.1
Dalton
16 X
+
8Y
8 X2Y
2.1
2
2.1
I.
Atomic Structure
A.
Three Fundamental Particles
1.
Electrons
a.
Work by Michael Faraday and George
Stoney lead to the discovery of electrons
b.
In 1897, J.J. Thompson determined the
charge to mass ratio of electrons
c.
In 1909, Robert Millikan determined the
mass of an electron
i.
9.1096 x 10-28 g
Measured mass of e(1923 Nobel Prize in Physics)
e- charge = -1.60 x 10-19 C
Thomson’s charge/mass of e- = -1.76 x 108 C/g
e- mass = 9.10 x 10-28 g
2.2
d.
The electron is assigned a charge of -1
e.
Cathode ray tubes - two electrodes are
sealed in a glass tube with low pressure
i.
when high voltage is applied and rays
are given off
ii.
these rays travel from the negative
electrode ( cathode) to the positive
electrode (anode)
2.
Protons
a.
component of all atoms
b.
mass of the proton is 1836 times bigger than
the electron
i.
c.
1.6726 x 10-24g
assigned a charge of +1
d. Canal rays - in 1886 Eugene Goldstein observed a
cathode ray tube also generates a stream of positively
charged particles
3.
Neutrons
a.
because atoms are electrically neutral - there
must be an equal number of protons and electrons
- but the sum of the protons and electrons
do not add up to equal the mass of an atom
b.
in 1920 Ernest Rutherford postulated the
existence of an uncharged particle
c. in 1932, James Chadwick is given credit for
discovering the neutron
d.
Heaviest of the three particles
i.
1.6749 x 10-24g
Rutherford’s Gold Experiment
Logical conclusion atoms are mostly empty space
“Honey I shrunk the KIds
The modern view of the atom was
developed by Ernest Rutherford (18711937).
Rutherford’s Model of
the Atom
atomic radius ~ 100 pm = 1 x 10-10 m
nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m
“If the atom is the Houston Astrodome, then
the nucleus is a marble on the 50-yard line.”
2.2
Atomic number (z) of an element is the number of
protons in the nucleus of each atom of that element
It is the number of protons in the nucleus that
determines what the element is
Mass number is the total number of protons and
neutrons in the nucleus of and atom
Isotopes – atoms of the same element that have
different mass numbers
Which means that it has different number
of neutrons
The Atom Hydrogen
Atomic number = 1 mass number = 1
Proton
Electron
Hydrogen has one proton, one electron and NO
neutrons
The Atom Hydrogen (isotope)
Atomic number = 1 mass number = 2
Proton
neutron
Electron
Hydrogen has one proton, one electron and NO
neutrons
The Atom Helium
Proton
Electron
Neutron
Helium has two electrons, two protons and two neutrons
The Atom Lithium
Protons
Neutrons
Electrons
The Atom Beryllium
Protons
Neutron
s
Beryllium has four electrons, four protons and five
neutrons.
Electrons
The Atom Boron
Protons
Electrons
Neutron
s
Boron has five electrons, five protons and six neutrons.
The Atom Carbon
Protons
Electrons
Neutron
s
Carbon has six electrons, six protons and six neutrons.
The Atom Nitrogen
Protons
Electrons
Neutron
s
Nitrogen has seven electrons, seven protons and seven
The Atom Oxygen
Protons
Electrons
Neutron
s
Oxygen has eight electrons, eight protons and eight
The Atom Fluorine
Protons
Electrons
Neutron
s
Fluorine has nine electrons, nine protons and ten neutrons.
The Atom Neon
Protons
Electrons
Neutron
s
Neon has ten electrons, ten protons and ten neutrons.
The Atom Sodium
Protons
Electrons
Neutrons
Sodium has eleven electrons, eleven protons and twelve
Average atomic mass is the weighted average of
the atomic mass’s of the naturally occurring
isotopes of an element
Name
Hydrogen
Symbol
Bromine
Oxygen
Br
atomic Number
1
35
O
Na
8
11
19
Phosphorus
K
P
15
39.10
30.98
39
31
19
15
19
15
20
16
Magnesium
Mg
12
12
12
Cl
17
24.
35
12
Chlorine
24.30
35.45
17
Carbon
Francium
C
Fr
6
12.01
12
6
17
6
18
6
87
Iron
Calcium
Silicon
Fe
Ca
Si
26
223
55.85
223
56
87
26
87
26
136
30
20
14
40.08
28.09
40
28
20
14
20
14
20
14
Tungsten
W
copper
Cu
74
29
183.85
63.55
184
64
74
29
74
29
110
35
82
207.2
126.90
207
127
82
82
53
238.03
14.01
238
Sodium
H
Potassium
Lead
Iodine
Uranium
Nitrogen
Tin
Silver
Manganese
Aluminum
Arsenic
Pb
I
53
U
92
N
Sn
Ag
Mn
Al
As
7
50
47
25
13
33
Atomic Mass Mass number # protons # electrons # neutrons
1.01
1
1
1
0
79.90
35
35
45
80
8
16.00
16
8
8
22.99
23
11
12
11
53
92
7
7
118.71
14
119
146
7
50
50
69
107.87
54.94
108
55
47
25
47
25
61
30
27
75
13
33
13
14
42
26.98
74.92
92
125
74
33
VII
Moles
A.
Avogadro’s number
1.
6.023 x 1023
2.
this is one mole - think as though it is a dozen
a.
3.
One mole of anything = 6.023 x 1023
One mole of an element = atomic mass
a.
1 mole of Na = 22.9898 g
this is the molar mass of Na
i. that is - if you take the atomic mass in grams of and
element, it will = 1 mole = 6.023 x 1023 atoms
How to use Avogadro’s number
e.g. How many atoms are there in 10.0 moles of Mg
(10.0 moles Mg) ( 6.023 x 1023atoms Mg)
(1 mole of Mg)
= 6.02 x 10 24 atoms of Mg
e.g
How many mole of Zn are in 1.2 x 1024 atoms
(1.2 x 1024 atoms of Zn) (
1mole of Zn
)
(6.023 x 1023 atoms Zn)
= 2.0 moles of Zn
You try two
How many atoms are in 3.0 moles of Cu
( 3.0 moles Cu ) (
6.023 x 10 23 Atoms Cu
1 mole Cu
)
= 1.8 x 1024 atoms of Cu
How many moles of W are in 2.50 x 1025 atoms of W
(
2.50 x
1025 atoms
W )(
1 mole W
)
6.023 x 10 23 Atoms of W
= 41.5 moles W
e.g
How many grams are in 3.0 moles of Fluorine
(3.0 moles F ) ( 19 g of F )
( 1 mole of F)
= 57 g of F
Changing grams to atoms
e.g How many atoms are in 10.0 g of Oxygen
23 Atoms of O)
(
1mole
of
O
)
(
6.023
x
10
(10.0 g of O )
( 16.0 g of O) ( 1mole of O)
= 3.76 x 1023 atoms of O
Changing atoms to grams
How many grams of potassium are in 2.30 x 1025 atoms
(2.30 x 1025 atoms of K)( 1 mole of K )
( 39.1 g of K )
( 6.023 x 1023 atoms ) ( 1 mole of K )
= 1490 g of K
Homework
(
(
25 g K
)(
2.50 moles Br
(
1 mole K
)
39.1 g K
)(
15.0 moles Ca
79.9 g Br
1 mole Br
= .64 moles K
)
= 200. g Br
) ( 6.023 x 10
atoms Ca
1 mole Ca
23
=
)
9.03 x 1024 atoms Ca
(
) ( 6.023 x 10
5.0 x 1024 atoms N
)
(
atoms N
1 mole N
23
14.0 g N
1 mole N
)
= 120 g N
(
2.0 g of Cu
)(
1 mole Cu
64 g Cu
)(
1.9 x 1022 atoms Cu
)
6.023 x 1023 atoms Cu
1 mole Cu
ATOMIC STRUCTURE
• properties of light
• spectroscopy
• quantum hypothesis
• hydrogen atom
• Heisenberg
Uncertainty Principle
• orbitals
ELECTROMAGNETIC RADIATION
• subatomic particles (electron, photon, etc)
have both PARTICLE and WAVE properties
• Light is electromagnetic radiation crossed electric and magnetic waves:
Properties :
Wavelength, l (nm)
Frequency, n (s-1, Hz)
Amplitude, A
constant speed. c
3.00 x 108 m.s-1
7.1
Properties of Waves
Wavelength (l) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
7.1
Electromagnetic Radiation (3)
• All waves have:
frequency
and
wavelength
• symbol: n (Greek letter “nu”)
l (Greek “lambda”)
• units:
“distance” (nm)
“cycles per sec” = Hertz
• All radiation:
l•n = c
where c = velocity of light = 3.00 x 10 8 m/sec
Note: Long wavelength
 small frequency
Short wavelength
 high frequency
increasing
frequency
increasing
wavelength
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
l
lxn=c
n
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
7.1
Electromagnetic Radiation (4)
Example: Red light has l = 700 nm.
Calculate the frequency,
n.
8
3.00
x10
m/sec
14
c
=
4.29 x 10 sec
=
n=
-7
l
7.00 x 10 m
• Wave nature of light is shown by classical
wave properties such as
• interference
• diffraction
Quantization of Energy (2)
Energy of radiation is proportional to frequency.
E = h•n
where h = Planck’s constant = 6.6262 x 10 -34 J•s
Light with large l (small n) has a small E.
Light with a short l (large n) has a large E.
When copper is bombarded with high-energy electrons,
X rays are emitted. Calculate the energy (in joules)
associated with the photons if the wavelength of the X
rays is 0.154 nm.
E=hxn
E=hxc/l
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10 -15 J
7.2
Atomic Line Spectra
• Bohr’s greatest contribution to
science was in building a
simple model of the atom.
• It was based on understanding
the SHARP LINE SPECTRA
of excited atoms.
Niels Bohr (1885-1962)
(Nobel Prize, 1922)
Line Spectra of Excited Atoms
• Excited atoms emit light of only certain wavelengths
• The wavelengths of emitted light depend on the
element.
H
Hg
Ne
7.3
Atomic Spectra and Bohr Model
One view of atomic structure in early 20th
century was that an electron (e-) traveled
about the nucleus in an orbit.
+
Electron
orbit
1. Classically any orbit should be
possible and so is any energy.
Atomic Spectra and Bohr Model (2)
• Bohr said classical view is wrong.
• Need a new theory — now called QUANTUM
or WAVE MECHANICS.
• e- can only exist in certain discrete orbits
— called stationary states.
• e- is restricted to QUANTIZED energy states.
Energy of state = - C/n2
where
C is a CONSTANT
n = QUANTUM NUMBER, n = 1, 2, 3, 4, ....
E = hn
E = hn
7.3
Chemistry in Action: Element from the Sun
In 1868, Pierre Janssen detected a new dark line in the solar
emission spectrum that did not match known emission lines
Mystery element was named Helium
In 1895, William Ramsey discovered helium in a mineral of
uranium (from alpha decay).
Atomic Spectra and Bohr Model (3)
Energy of quantized state = - C/n2
• Only orbits where n = integral
number are permitted.
• Radius of allowed orbitals
= n2 x (0.0529 nm)
• Results can be used to
explain atomic spectra.
Atomic Spectra and Bohr Model (4)
If e-’s are in quantized energy
states, then DE of states can
have only certain values. This
explains sharp line spectra.
E = -C
(1/22)
H atom
n=2
07m07an1.mov
E = -C (1/12)
n=1
Calculate DE for e- in H “falling” from
n = 2 to n = 1 (higher to lower energy) .
Energy
Atomic Spectra and Bohr Model (5)
n=2
n=1
DE = Efinal - Einitial = -C[(1/12) - (1/2 2)] = -(3/4)C
• (-ve sign for DE indicates emission (+ve for absorption)
• since energy (wavelength, frequency) of light can only be
+ve it is best to consider such calculations as DE = Eupper
- Elower
C has been found from experiment. It is now called R,
the Rydberg constant. R = 1312 kJ/mol or 3.29 x 1015 Hz
so, E of emitted light = (3/4)R = 2.47 x 1015 Hz
and l = c/n = 121.6 nm (in ULTRAVIOLET region)
This is exactly in agreement with experiment!
Hydrogen atom spectra
High E
Short l
High n
Low E
Long l
Low n
Visible lines in H atom
spectrum are called the
BALMER series.
6
5
4
Energy
3
2
1
En = -1312
n2
Ultra Violet
Lyman
Visible
Balmer
Infrared
Paschen
n
Calculate the wavelength (in nm) of a photon
emitted by a hydrogen atom when its electron
drops from the n = 5 state to the n = 3 state.
Ephoton = DE = RH(
1
n2i
1
n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = DE = -1.55 x 10-19 J
Ephoton = h x c / l
l = h x c / Ephoton
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
7.3
Because you have only one H atom, the R constant
R = 1312 kJ/mol
Must be changed to per atom.
That means that you must divide
R by 6.023 x 1023
This gives you 2.18 x 10-18 J
ni = 3
ni = 3
ni = 2
nf = 2
Ephoton = DE = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
DE = RH( 2
ni
)
)
1
n2f
)
nnf f==11
7.3
From Bohr model to Quantum mechanics
Bohr’s theory was a great accomplishment
and radically changed our view of matter.
But problems existed with Bohr theory —
– theory only successful for the H atom.
– introduced quantum idea artificially.
• So, we go on to QUANTUM or WAVE
MECHANICS
Quantum or Wave Mechanics
• Light has both wave & particle
properties
• de Broglie (1924) proposed that all
moving objects have wave
properties.
• For light: E = hn = hc / l
L. de Broglie
(1892-1987)
• For particles: E = mc2
Therefore, mc = h / l
(Einstein)
and for particles
(mass)x(velocity) = h / l
l for particles is called the de Broglie wavelength
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
l = h/mv
h in J•s m in kg v in (m/s)
l = 6.63 x 10-34 j sec / (2.5 x 10-3 kg x 15.6 m/sec)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
7.4
Quantum or Wave Mechanics
Schrodinger applied idea of ebehaving as a wave to the
problem of electrons in atoms.
E. Schrodinger
1887-1961
Uncertainty Principle
W. Heisenberg
1901-1976
Problem of defining nature of
electrons in atoms solved by
W. Heisenberg.
Cannot simultaneously define
the position and momentum
(= m•v) of an electron.
Dx. Dp = h
At best we can describe the
position and velocity of an
electron by a
PROBABILITY DISTRIBUTION,
QUANTUM NUMBERS
The shape, size, and energy of each orbital is a function
of 3 quantum numbers which describe the location of
an electron within an atom or ion
n (principal) ---> energy level
l (orbital) ---> shape of orbital
ml (magnetic) ---> designates a particular
suborbital
The fourth quantum number is not derived from the
wave function
s (spin) ---> spin of the electron
(clockwise or counterclockwise: ½ or – ½)
Orbital Quantum Numbers
An atomic orbital is defined by 3 quantum
numbers:
– n l ml
Electrons are arranged in shells and
subshells of ORBITALS .
n  shell
l
 subshell
ml  designates an orbital within a subshell
Where 90% of the
e- density is found
for the 1s orbital
e- density (1s orbital) falls off rapidly
as distance from nucleus increases
7.6
l = 0 (s orbitals)
l = 1 (p orbitals)
7.6
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3
7.6
How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d
subshell?
n=3
3d
l=2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e7.6
Quantum Numbers
Symbol
Values
Description
n (major)
1, 2, 3, ..
Orbital size and
energy = -R(1/n2)
l (angular)
0, 1, 2, .. n-1
Orbital shape
type (subshell)
ml (magnetic)
-l..0..+l
Orbital orientation
in space
Total # of orbitals in lth subshell = 2 l + 1
Shells and Subshells
For n = 1, l = 0 and ml = 0
There is only one subshell and that
subshell has a single orbital
(ml has a single value ---> 1 orbital)
This subshell is labeled s (“ess”) and
we call this orbital 1s
Each shell has 1 orbital labeled s.
It is SPHERICAL in shape.
Types of Orbitals (l)
s orbital
p orbital
d orbital
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
7.6
s Orbitals
All s orbitals are spherical in shape.
p Orbitals
For n = 2, l = 0 and 1
There are 2 types of
orbitals — 2 subshells
For l = 0 ml = 0
this is a s subshell
For l = 1 ml = -1, 0, +1
this is a p subshell
with 3 orbitals
Typical p orbital
planar node
When l = 1, there is
a PLANAR NODE
through the
nucleus.
p Orbitals
• The three p orbitals lie 90o apart in space
d Orbitals
For n = 3, what are the values of l?
l = 0, 1, 2
and so there are 3 subshells in the shell.
For l = 0, ml = 0
 s subshell with single orbital
For l = 1, ml = -1, 0, +1

p subshell with 3 orbitals
For l = 2, ml = -2, -1, 0, +1, +2
 d subshell with 5 orbitals
l = 2 (d orbitals)
7.6
Boundary surfaces for all orbitals of the
n = 1, n = 2 and n = 3 shells
n=
3d
3
2
There are
n2
orbitals in
the nth SHELL
1
f
Orbital
The last quantum number is the spin quantum num
Electrons act as though they spin about an axis through their ce
Because there are two directions in which they can spin,
the spin quantum number has 2 values +1/2,-1/2
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
7.8
Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
H
1s1
7.8
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
Be
Li
B5
C
3
64electrons
electrons
22s
222s
22p
12 1
BBe
Li1s1s
1s
2s
H
He12electron
electrons
He
H 1s
1s12
7.7
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
Ne97
C
N
O
F
6
810
electrons
electrons
electrons
22s
222p
22p
5
246
3
Ne
C
N
O
F 1s
1s222s
7.7
Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=3 l = 1
n=2 l = 1
n=1 l = 0
7.7
Exceptions to the Aufbau Principle
• Remember d and f orbitals require LARGE amounts of
energy
• If we can’t fill these sublevels, then the next best thing
is to be HALF full (one electron in each orbital in the
sublevel)
• There are many exceptions, but the most common
ones are
d4 and d9
For the purposes of this class, we are going to assume
that ALL atoms (or ions) that end in d4 or d9 are
exceptions to the rule. This may or may not be true, it
just depends on the atom.
Exceptions to the Aufbau Principle
• The next most common are f1 and f8
• The electron goes into the next d orbital
• Example:
– La [Xe]6s2 5d1
– Gd [Xe]6s2 4f7 5d1
Exceptions to the Aufbau Principle
d4 is one electron short of being HALF full
In order to become more stable (require less energy), one
of the closest s electrons will actually go into the d,
making it d5 instead of d4.
For example: Cr would be [Ar] 4s2 3d4, but since this
ends exactly with a d4 it is an exception to the rule.
Thus, Cr should be [Ar] 4s1 3d5.
Procedure: Find the closest s orbital. Steal one electron
from it, and add it to the d.
Why are d and f orbitals always in lower
energy levels?
• d and f orbitals require LARGE amounts of
energy
• It’s better (lower in energy) to skip a sublevel
that requires a large amount of energy (d and
f orbtials) for one in a higher level but lower
energy
This is the reason for the diagonal rule! BE
SURE TO FOLLOW THE ARROWS IN ORDER!
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons
1s22s22p63s23p5
1s < 2s < 2p < 3s < 3p < 4s
2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n=3
l=1
ml = -1, 0, or +1
ms = ½ or -½
7.8
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Isotopes are atoms of the same element (X) with different
numbers of neutrons in their nuclei
Mass Number
A
ZX
Atomic Number
1
1H
235
92
2
1H
U
Element Symbol
(D)
238
92
3
1H
(T)
U
2.3
2.3
Do You Understand Isotopes?
d.
Canal rays - in 1886 Eugen Goldstein observed a cathode ra
tube also generates a stream of positively charged particles
14
How many protons, neutrons, and electrons are in 6 C ?
6 protons, 8 (14 - 6) neutrons, 6 electrons
How many protons, neutrons, and electrons are
11
in 6 C ?
6 protons, 5 (11 - 6) neutrons, 6 electrons
2.3
Keep an Eye On Those Ions!
• Tin
Atom: [Kr] 5s2 4d10 5p2
Sn+4 ion: [Kr] 4d10
Sn+2 ion: [Kr] 5s2 4d10
Note that the electrons came out of
the highest energy level, not the
highest energy orbital!
Try These!
• Write the shorthand
notation for:
1 3d10
[Ar]
4s
Cu
1 4f14 5d5
[Xe]
6s
W
1 4f14 5d10
[Xe]
6s
Au
Odds and ends
VI.
Hund’s rule - electrons must be placed in each
orbital before pairing them up
VII.
Valance electrons - those in the outer most energy level
- gives elements its chemical properties
VIII.
orbital notation - lines
IX.
Electron configuration -
1s2 2s2 2p6
X.
Nobel gas notation - {Ar} 4s1
XI.
Electron dot notation - only use the valance electron
ELEMENTS THAT EXIST AS
DIATOMIC MOLECULES
Remember:
BrINClHOF
These elements
only exist as
PAIRS. Note that
when they
combine to make
compounds, they
are no longer
elements so they
are no longer in
pairs!
On page 87
do questions
1-11
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