Download Atomic structure - ISA DP Chemistry with Ms Tsui

Atomic structure
Part 1
2.1 The nuclear atom
• Essential idea: The mass of an atom is
concentrated in its minute, positively charged
nucleus
Understandings
• Atoms contain a positively charged dense
nucleus composed of protons and neutrons
(nucleons).
• Negatively charged electrons occupy the space
outside the nucleus.
• The mass spectrometer is used to determine
the relative atomic mass of an element from
its isotopic composition.
Application and skills
• Use of the nuclear symbol notation to
deduce the number of protons, neutrons and
electrons in atoms and ions.
• Calculations involving non-integer relative
atomic masses and abundance of isotopes
from given data, including mass spectra.
John Dalton (1766 – 1844)
John Dalton (1766 – 1844)
• Dalton’s atomic theory
1. All matter is made of atoms. Atoms are
indivisible and indestructible.
2. All atoms of a given element are identical in
mass and properties. Atoms of different
elements are different
3. Compounds are formed by a combination of two
or more different kinds of atoms in specific
whole number ratios.
4. A chemical reaction is a rearrangement of
atoms, not a change in the atoms themselves.
Sir Joseph John “J. J.” Thomson (1856 – 1940)
Sir Joseph John “J. J.” Thomson (1856 – 1940)
• Challenged Dalton’s
notion that atoms
are indestructible
• First suggestion of
electrons
• “Negatively charged
electrons scattered
in a positively
charged sponge-like
substance”
Deflecting “cathode ray”, now known
as electron beam
When a high voltage was applied to the metal plate, a
“cathode ray” is released.
The cathode ray’s path can be deflected towards a
positively charged metal plate, suggesting that the
cathode ray is negatively charged.
Sir Joseph John “J. J.” Thomson (1856 – 1940)
Ernest Rutherford (1871 – 1937)
Gold foil experiment
Discovery of subatomic particles
• If Thomson’s “plum pudding model” was
correct, the alpha particles (positively
charged) should either pass straight through
or get stuck in the positive “sponge”
Discovery of subatomic particles
• Results:
– Most alpha particles passed straight through and
get detected at the expected location on the
screen
– A very small number were repelled and bounced
back
Discovery of subatomic particles
• Conclusion
– The atom is mainly
empty space
– Deflection occurred
when the alpha
particles hit and
were repelled by a
dense, positively
charged center
called the “nucleus”
Subatomic particles
Niels Bohr (1885 – 1962)
Bohr model of the hydrogen atom
• A small solar system with an electron moving
in an orbit
• Most of the atom is empty space
• Protons?
• Neutrons?
• Electrons?
Ions
40Ca2+
Number of subatomic particles
• Protons?
• Neutrons?
• Electrons?
127I-
Number of subatomic particles
• Protons?
• Neutrons?
• Electrons?
How should these three symbols be
interpreted?
Isotopes
Isotopes: Atoms of the same element
with different number of neutrons
Isotopes and Atomic mass
• https://phet.colorado.edu/en/simulation/isot
opes-and-atomic-mass
Take a look at the data booklet
6. Periodic table with relative atomic mass
• The values are expressed to 2 decimal places
• Calculated as the weighed, relative average
mass based on the percentage abundance of
different isotopes
Question type 1. Calculate the relative atomic
mass from the information provided below
Relative atomic mass (Ar)
= 34.97 x 75.77% + 36.97 x 24.23%
= 35.45
Question type 2:
Mass spectra interpretation
Question type 3:
• Relative atomic mass of Chlorine is 35.5
• The two isotopes are 35Cl and 37Cl
• Calculate the relative abundance of each
isotope
Question type 3:
• Relative atomic mass of Chlorine is 35.5
• The two isotopes are 35Cl and 37Cl
• Calculate the relative abundance of each isotope
• Let
– x = relative abundance of 35Cl
– y = relative abundance of 37Cl
x+y=1
y=1–x
35x + 37y = 35.5
x = relative abundance of 35Cl
y = relative abundance of 37Cl
x+y=1
y=1–x
35x + 37y = 35.5
35x + 37 (1-x) = 35.5
35x + 37 – 37x = 35.5
-2x = -1.5
x = 0.75
y = 1 – x = 0.25
x = relative abundance of 35Cl = 0.75 or 75%
y = relative abundance of 37Cl = 0.25 or 25%
x+y=1
y=1–x
35x + 37y = 35.5
35x + 37 (1-x) = 35.5
35x + 37 – 37x = 35.5
-2x = -1.5
x = 0.75
y = 1 – x = 0.25
Practice: Magnesium has three stable isotopes
– 24Mg, 25Mg, and 26Mg.
• The lightest isotope has an abundance of
78.90%. Calculate the percentage abundance
of the other isotopes.
The Quantum Mechanical model of the atom
… continued in Part 2