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Atomic Structure
Introduction
• All substances are made up of matter and
the fundamental unit of matter is the atom.
• The atom constitutes the smallest particle of
an element which can take part in chemical
reactions and may or may not exist
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independently.
Introduction
• Most of what is known about atomic structure
is based on basically 2 types of research:– Electrical nature of matter
– Interaction of matter with light energy
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Dalton’s Atomic Theory
• The Beginning of modern atomic theory is
credited to John Dalton.
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Dalton’s Atomic Theory
• Dalton’s model was formulated on a number
“laws” about how matter behaves in a
chemical reaction.
– These laws were based on experimental
evidence
– Observations on many chemical substances &
their reactions
– These laws are the foundation on which the
modern atomic theory is based
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Dalton’s Atomic Theory
• Dalton formulated his theory:
– All matter is made up of atoms (small, indivisible,
indestructible, fundamental particles)
– Atoms can neither be created or destroyed (they
persist unchanged for all eternity)
– Atoms of a particular element are all alike (in size,
mass & properties)
– Atoms of different elements are different from one
another (different sizes, masses & properties)
– A chemical reaction involves either the union or
the separation of individual atoms
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Dalton’s Atomic Theory
• We know now that Dalton’s theory is not entirely
true, for example:
– Atoms are not the most fundamental particles –
they are composed of smaller particles
– Atoms can be created or destroyed but a nuclear
process is needed to do so
• Nonetheless, Dalton’s model was superb for his
time and it laid the foundation for further
developments in atomic theory.
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Fundamental Particles
• Three fundamental particles make up atoms.
• The following table lists these particles together
with their masses and their charges.
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The Discovery of Electrons
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The Discovery of Electrons
• Earliest evidence for atomic structure was
supplied in the early 1800’s by the English
chemist, Humphrey Davy.
• Davy passed electricity
through compounds
and noted:
– that the compounds
decomposed into elements.
– concluded that compounds
are held together by
electrical forces.
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The Discovery of Electrons
• Most convincing evidence came from Cathode Ray
Tubes experiments performed in the late 1800’s &
early 1900’s.
– Consist of two electrodes sealed in a glass tube
containing a gas at very low pressure.
– When a voltage is applied to the cathodes a glow
discharge is emitted.
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The Discovery of Electrons
• These “rays” are emitted from cathode (- ve
end) and travel to anode (+ve end).
– Cathode Rays must be negatively charged!
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The Discovery of Electrons
• J.J. Thomson modified
the cathode ray tube
experiments in 1897 by
adding two adjustable
voltage electrodes.
– Studied the amount that
the cathode ray beam
was deflected by
additional electric field.
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The Discovery of Electrons
• Modifications to the basic cathode ray tube
experiment show the nature of cathode rays
(a) A cathode ray discharge tube,
showing the production of a beam
of electrons (cathode rays). The
beam is detected by observing the
glow on a flourescent screen.
(b) A small object placed in front of
the beam, casts a shadow
indicating that cathode rays
travel in straight lines.
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(c) Cathode rays have a negative
electrical charge, as demonstrated
by their deflection in an electrical
field.
(d) Interaction with a magnetic field
also consistent with negative
charge.
(e) Cathode rays have mass, as
shown by their ability to turn a
small paddle wheel in their path.
The Discovery of Electrons
• Thomson used his modification to measure the
charge to mass ratio of electrons.
Charge to mass ratio
e/m = -1.75881 x 108 coulomb/g of e-
• Thomson named the cathode rays electrons.
• Thomson is considered to be the “discoverer of
electrons”.
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Canal Rays and Protons
• In 1886 Eugene Goldstein noted that cathode ray tube
also generated streams of positively charged particles
that moved toward the cathode.
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Canal Rays and Protons
– Particles move in opposite direction of cathode rays.
– Called “Canal Rays” because they passed through
holes (channels or canals) drilled through the negative
electrode.
- Canal rays must be positive.
– Goldstein postulated the existence of a positive
fundamental particle called the “proton”.
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Model for Atomic Structure
• By early 1900s it was clear that atoms contained
regions of +ve and -ve charge.
• But how these charges were distributed was still
unclear.
• 1st model for the structure of the atom was
proposed by Thompson based on the following:
• Atoms contain small –ve charged
particles (electrons)
• Atoms of an element behave as if
they had no electrical charge
• So there must be something in
the atom to neutralize the –ve
electrons (protons not yet
discovered)
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Rutherford and the Nuclear Atom
• Further insight into atomic structure was provided by Ernest
Rutherford.
• He has established that - particles were +ve charged particles
– They are emitted by some radioactive atoms (when they
disintegrate spontaneously)
• Bombarded thin Au foils with - particles from a radioactive
source
– Gave us the basic picture of the atom’s structure.
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Rutherford and the Nuclear Atom
• If
Thompson’s model was
correct then any - particles
passing through the foil would
be deflected by small angles.
• Unexpectedly most of the particles passed through the foil
with little or no deflections
(shown in black).
• Many were deflected through
moderate angles (shown in
red).
• These deflections were
surprising, but the 0.001% of
the total that were reflected at
acute angles (shown in blue)
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were totally unexpected!
Rutherford and the Nuclear Atom
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Rutherford and the Nuclear Atom
•
Rutherford’s major conclusions from the particle scattering experiment:
1. The atom is mostly empty space.
2. It contains a very small, dense center called the
nucleus.
3. Nearly all of the atom’s mass is in the nucleus.
4. The nuclear diameter is 1/10,000 to 1/100,000
times less than atom’s radius.
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Neutrons
• James Chadwick in 1932
analyzed the results of particle scattering on thin
Be films.
• Chadwick recognized
existence of massive
neutral particles which he
called neutrons.
– Chadwick discovered
the neutron.
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Atomic Number
• The atomic number = # of protons in the nucleus.
– Sometimes given the symbol Z.
– On the periodic table Z is the uppermost number in
each element’s box.
Charge of
particle
Mass
number
(= p + n)
A
Z
Atomic
number
(= # of p )
X
C
Symbol of
the atom
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Atomic Number
• In 1913 H.G.J. Moseley
realized that the atomic
number determines the
element.
– The elements differ from
each other by the number of
protons in the nucleus.
• So… it is the number of protons
that determine the identity of an
element
– The number of electrons in a
neutral atom is also equal to
the atomic number.
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Nucleon Number and Isotopes
Charge of
particle
Mass
number
(= p + n)
Atomic
number
A
Z
(= # of p)
X
C
Symbol of
the atom
• Nucleon number (formerly Mass number) is given
the symbol A.
• A = # of protons + # of neutrons.
– If Z = proton number
– Then A = Z + N
and N = neutron number
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• So, the Standard Notation used to show mass
and proton numbers is:
Charge of
particle
Mass
number
(= p + n)
A
Z
X
C
Atomic
number
Symbol of
the atom
(= # of p )
A
Z
E for example
12
6
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20
C, Ca,
197
79
• Can be shortened to this symbolism.
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63
N, Cu,
107
Ag, etc.
Au
Mass Number and Isotopes
• Isotopes are atoms of the same element but with
different neutron numbers.
– Isotopes have different masses and A values but are the
same element.
Isotopes of Hydrogen
Protium
(Hydrogen - 1)
Deuterium
(Hydrogen - 2)
Tritium
(Hydrogen - 3)
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Mass Number and Isotopes
• The stable oxygen isotopes provide another
example.
1. 16O is the most abundant stable O isotope. How
many protons and neutrons are in 16O?
8 protons and 8 neutrons
2. 17O is the least abundant stable O isotope. How
many protons and neutrons are in 17O?
8 protons and 9 neutrons
3. 18O is the second most abundant stable O isotope. How
many protons and neutrons in 18O?
8 protons and 10 neutrons
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Mass Spectrometry & Isotopic Abundances
• Identifies chemical composition of a compound or sample on
the basis of the mass-to-charge ratio of charged particles.
• A gas sample at low pressure is bombarded with high-energy
electrons.
– This causes electrons to be ejected from some of the gas
molecules  creating +ve ions.
• Positive ions then focused into a very narrow beam and
accelerated by an electric field.
• Then passes through a magnetic field which deflects the ions
from their straight path.
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Mass Spectrometry
• There are four factors which determine the extent
of deflection:
1 Accelerating voltage
• Higher voltages  beams move more rapidly and deflected less
than slower moving beams produced by lower voltages.
2 Magnetic field strength
• Stronger fields give more deflection
3 Masses of particles
• Heavier particles deflected less than lighter ones
4 Charge on particles
• Particles with higher charges interact more strongly with
magnetic fields and are thus deflected more than particles of
equal mass with small charge.
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Mass Spectrometry
A modern mass spectrometer
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Fig. 5-10a, p. 176
Mass Spectrometry & Isotopic Abundances
• Mass spectrum of Ne+ ions shown below.
– How do scientists determine the masses and
abundances of the isotopes of an element?
• Neon consists of 3
isotopes, of which Neon20 is the most abundant
(90.48%).
• The number by each peak
corresponds to the fraction
of all the Ne+ ions
represented by the isotope
with that mass.
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Mass Spectrometry & Isotopic Abundances
• The mass of an atom is measured relative to the C-12 atom
– Its’ mass is defined as exactly 12 atomic mass units (amu)
• Therefore the amu is 1/12 the mass of a C-12 atom
• Example: What is the mass in amu of a 28Si atom?
– The spectrometer will measure the ratio of the mass of an 28Si atom
to 12C:
» Mass of 28Si atom
Mass of 12C atom
= 2.331411
– From this mass ratio, the isotopic mass of the 28Si can be found:
» 2.331411 x 12 amu = 27.97693 amu
– The mass of the isotope relative to the mass of the C-12 isotope
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Table 5-3, p. 178
Isotopes
• Small differences in physical
properties.
• Similar chemical properties
because isotopes have same
number of p and e.
• Some isotopes are radioactive.
– nuclear behavior of isotopes is
unique
– Radioactive isotopes are
biologically useful
– Example: radioactive I-131 to
study thyroid gland
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Atomic Weight
• The relative atomic weight (also called relative
atomic mass) of an element is the weighted average
of the masses of its stable isotopes.
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Atomic Weight
• Atoms are amazingly small.
• Their masses are compared with the mass of
an atom of the carbon-12 isotope, as the
standard.
– One atom of the C-12 iostope weight exactly 12
units (Atomic mass units, amu).
– E.g. an atom of the most common isotope of Mg
weighs twice as much as one atom of C-12, its
relative isotopic mass is 24.
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Atomic Weight
• “weighted average” e.g. Chlorine
– Cl- 35 75%
– Cl-37  25%
– If you had 100 atoms, 75 would be Cl-35 and 25
would be Cl-37.
– The weighted average is closer to 35 than 37
because there are more Cl-35 than Cl-37 atoms.
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Atomic Weight
• Example: Naturally occurring Cu consists of
2 isotopes.
– It is 69.1% 63Cu with a mass of 62.9 amu,
– and 30.9% 65Cu, which has a mass of 64.9 amu.
– Calculate the atomic weight of Cu to one decimal
place.
atomic weight  (0.691)(62 .9 amu)  (0.309)(64 .9 amu)


 


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Cu isotope
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Cu isotope
atomic weight  63.5 amu for copper
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Atomic Weight
• Example: The relative atomic mass of boron
is 10.811 amu. The masses of the two
naturally occurring isotopes are 510B and
11B, are 10.013 and 11.009 amu,
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respectively. Calculate the fraction and
percentage of each isotope.
You do it!
• This problem requires a little algebra.
– A hint for this problem is x + (1-x) = 1
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Atomic Weight
10.811 amu  x(10.013 amu)  1  x (11.009 amu)

 

10
B isotope
11
B isotope
 10.013 x  11.009 - 11.009 x  amu
10.811 - 11.009 amu  10.013 x - 11.009 x  amu
- 0.198  -0.996 x
0.199  x
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Atomic Weight
• Note that because x is the multiplier for the
10B isotope, our solution gives us the fraction
of natural B that is 10B.
• Fraction of 10B = 0.199 and % abundance of
10B = 19.9%.
• The multiplier for 11B is (1-x) thus the fraction
of 11B is 1-0.199 = 0.801 and the %
abundance of 11B is 80.1%.
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Home Work
1. Calculations: Chemistry 9th Edition, Chapter 4,
Exercises 28 – 38.
2. What are the main points in Dalton’s atomic theory?
3. Briefly outline how the mass spectrometer works to
help determine the isotopic abundance and isotopic
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mass. Include a diagram in your answer.