chapter 13 electrons in atoms milbank high school
TRANSCRIPT
Section 13.1Models of the Atom
OBJECTIVES: Explain the significance of
quantized energies of electrons as they relate to the quantum mechanical model of the atom.
Ernest Rutherford’s Model Discovered dense
positive piece at the center of the atom- nucleus
“Nuclear model”
Niels Bohr’s Model Move like planets around the sun. In circular orbits at different levels. Amounts of energy separate one
level from another. “Planetary model”
Bohr’s planetary model electron cannot exist between energy
levels, just like you can’t stand between rungs on ladder
Quantum of energy required to move to the next highest level
The Quantum Mechanical Model
Since the energy of an atom is never “in between” there must be a quantum leap in energy.
Erwin Schrodinger Mathematical solution
Has energy levels for electrons.
Orbits are not circular. It can only tell us the
probability of finding an electron a certain distance from the nucleus.
The Quantum Mechanical Model
The atom is found inside a blurry “electron cloud”
An area where there is a chance of finding an electron.
Think of fan blades
The Quantum Mechanical Model
Atomic Orbitals Principal Quantum Number (n) = the
energy level of the electron. Within each energy level, the complex
math of Schrodinger’s equation describes several shapes.
Atomic orbitals - regions where there is a high probability of finding an electron.
Sublevels
By Energy Level First Energy Level only s orbital only 2 electrons 1s2
Second Energy Level
s and p orbitals are available
2 in s, 6 in p 2s22p6
8 total electrons
By Energy Level Third energy level s, p, and d
orbitals 2 in s, 6 in p, and
10 in d 3s23p63d10
18 total electrons
Fourth energy level
s,p,d, and f orbitals
2 in s, 6 in p, 10 in d, ahd 14 in f
4s24p64d104f14
32 total electrons
Section 13.2Electron Arrangement in Atoms
OBJECTIVES: Apply the aufbau principle, the
Pauli exclusion principle, and Hund’s rule in writing the electron configurations of elements.
Section 13.2Electron Arrangement in Atoms
OBJECTIVES: Explain why the electron
configurations for some elements differ from those assigned using the aufbau principle.
Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
Aufbau diagram - page 367
Electron Configurations Aufbau principle- electrons enter the
lowest energy first. This causes difficulties because of the
overlap of orbitals of different energies. Pauli Exclusion Principle- at most 2
electrons per orbital - different spins
Electron Configuration Hund’s Rule- When electrons
occupy orbitals of equal energy they don’t pair up until they have to.
Phosphorus (15 electrons)
The first two electrons go into the 1s orbital
Notice the opposite spins
only 13 more to go...
Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
The next electrons go into the 2s orbital
only 11 more...Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
• The next electrons go into the 2p orbital
• only 5 more...Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
• The next electrons go into the 3s orbital
• only 3 more...Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
Incr
easi
ng e
nerg
y
1s
2s
3s
4s
5s6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
7p 6d
4f
5f
• The last three electrons go into the 3p orbitals.
• They each go into separate shapes
• 3 unpaired electrons
• = 1s22s22p63s23p3
The easy way to remember
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2
• 2 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2
• 4 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
• 12 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
3p6 4s2
• 20 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
3p6 4s2 3d10 4p6
5s2
• 38 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
3p6 4s2 3d10 4p6
5s2 4d10 5p6 6s2
• 56 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
3p6 4s2 3d10 4p6
5s2 4d10 5p6 6s2
4f14 5d10 6p6 7s2
• 88 electrons
Fill from the bottom up following the arrows
1s2s 2p3s 3p 3d4s 4p 4d 4f
5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f
• 1s2 2s2 2p6 3s2
3p6 4s2 3d10 4p6
5s2 4d10 5p6 6s2
4f14 5d10 6p6 7s2
5f14 6d10 7p6 • 108 electrons
Orbitals fill in order Lowest energy to higher energy. Adding electrons can change the
energy of the orbital. Half filled orbitals have a lower
energy. Makes them more stable. Changes the filling order
Write these electron configurations
Titanium - 22 electrons 1s22s22p63s23p64s23d2
Vanadium - 23 electrons 1s22s22p63s23p64s23d3
Chromium - 24 electrons 1s22s22p63s23p64s23d4 expected But this is wrong!!
Chromium is actually: 1s22s22p63s23p64s13d5
Why? This gives us two half filled orbitals. Slightly lower in energy. The same principal applies to
copper.
Copper’s electron configuration
Copper has 29 electrons so we expect: 1s22s22p63s23p64s23d9
But the actual configuration is: 1s22s22p63s23p64s13d10
This gives one filled orbital and one half filled orbital.
Remember these exceptions: d4, d9
Section 13.3Physics and the Quantum
Mechanical Model OBJECTIVES:
Calculate the wavelength, frequency, or energy of light, given two of these values.
Section 13.3Physics and the Quantum
Mechanical Model OBJECTIVES:
Explain the origin of the atomic emission spectrum of an element.
Light The study of light led to the
development of the quantum mechanical model.
Light is a kind of electromagnetic radiation.
Electromagnetic radiation includes many kinds of waves
All move at 3.00 x 108 m/s = c
Parts of Wave - p.372 Origin - the base line of the energy. Crest - high point on a wave Trough - Low point on a wave Amplitude - distance from origin to crest Wavelength - distance from crest to
crest Wavelength is abbreviated by the Greek
letter lambda =
Frequency The number of waves that pass a
given point per second. Units: cycles/sec or hertz (hz or sec-1) Abbreviated by Greek letter nu =
c =
Frequency and wavelength Are inversely related As one goes up the other goes down. Different frequencies of light are
different colors of light. There is a wide variety of frequencies The whole range is called a spectrum,
Fig. 13.10, page 373
Radiowaves
Microwaves
Infrared .
Ultra-violet
X-Rays
GammaRays
Low energy
High energy
Low Frequency
High Frequency
Long Wavelength
Short WavelengthVisible Light
Atomic Spectrum Each element
gives off its own characteristic colors.
Can be used to identify the atom.
How we know what stars are made of.
• These are called discontinuous spectra, or line spectra
• unique to each element.
• These are emission spectra
• The light is emitted given off
• Sample 13-2 p.375
Light is a Particle Energy is quantized. Light is energy Light must be quantized These smallest pieces of light are
called photons. Photoelectric effect? Energy & frequency: directly related.
Energy and frequency E = h x E is the energy of the photon is the frequency h is Planck’s constant h = 6.6262 x 10 -34 Joules x sec. joule is the metric unit of Energy
Examples What is the wavelength of blue light
with a frequency of 8.3 x 1015 hz? What is the frequency of red light
with a wavelength of 4.2 x 10-5 m? What is the energy of a photon of
each of the above?
Explanation of atomic spectra When we write electron
configurations, we are writing the lowest energy.
The energy level, and where the electron starts from, is called it’s ground state- the lowest energy level.
Further they fall, more energy, higher frequency.
This is simplified the orbitals also have different energies
inside energy levels All the electrons can move around.
Ultraviolet Visible Infrared
What is light? Light is a particle - it comes in chunks. Light is a wave- we can measure its
wavelength and it behaves as a wave If we combine E=mc2 , c=, E = 1/2 mv2
and E = h We can get: = h/mv called de Broglie’s equation Calculates the wavelength of a particle.
Sample problem What is the approximate mass of a
particle having a wavelength of 10-7 meters, and a speed of 1 m/s? Use = h/mv
= 6.6 x 10-27
(Note: 1 J = N x m; 1 N = 1 kg x m/s2