Download Chapter 7 The Quantum- Mechanical Model of the Atom - NTOU-Chem

Chapter 7 Lecture
Lecture Presentation
Chapter 7
The QuantumMechanical Model
of the Atom
Sherril Soman
Grand Valley State University
© 2014 Pearson Education, Inc.
授課教師：林秀美
辦公室：綜合二館302B
TEL: 5562 email:[email protected]
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評量方式：
(一) 課堂互動討論（占總成績比例10%）：由授課教師決定。
(二) 會考（占總成績比例90%）：
額外加分：維基百科3000字元一分，最多5分；可以問星期三的課輔助教，以潔與嘉珊；或
是參加維基台北寫作聚 每月第二個禮拜六。。
http://zh.wikipedia.org/wiki/Wikipedia:WPTP-W
1. 日期：第一次會考：10月23日(三)第二次會考：11月13日(三)
第三次會考：12月18日(三): Ch7,8,9
第四次會考：01月15日(三): Ch10,11,12
2. 時間：18:00~19:30 PM
3. 地點：延平技術大樓（考試當週於網站公布試場及座位）
4. 注意事項：如普通化學課程會考施行細則（CHEM003）。
5. 會考成績：會考結束七日內公告（http://academic.ntou.edu.tw/exam）
(三) 積極性補強教學出席率、化學課後作業（10%加分制）：由授課教師決定。
詳參1021海洋大學整合型普通化學課程積極性補強教學施行細則。（CHEM004）。
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7.1 Schrodinger’s Cat
薛丁格的貓
http://www.youtube.com/watch?v=BHEEdp79rIY
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The Beginnings of Quantum Mechanics量子
力學的開端
• Until the beginning of the twentieth century it was
believed that all physical phenomena were
deterministic.直到二十世紀開始，所有的物理現象都
具有確定性。
• Work done at that time by many famous physicists
discovered that for sub-atomic particles, the present
condition does not determine the future condition.
– Albert Einstein, Neils Bohr, Louis de Broglie, Max Planck, Werner
Heisenberg, P. A. M. Dirac, and Erwin Schrödinger
• 當時許多著名的物理學家發現了次原子粒子，目前
的狀況無法確定未來的狀況。
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The Beginnings of Quantum Mechanics
• Quantum mechanics forms the foundation
of chemistry量子力學奠定了基礎化學
– Explaining the periodic table
– The behavior of the elements in chemical
bonding
– Provides the practical basis for lasers,
computers, and countless other applications
解釋週期表
的行為的元素的化學鍵
雷射，計算機，以及其他無數的應用提供了實踐基
礎
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The Behavior of the Very Small非常小的
行為
• Electrons are incredibly small.電子是小得令人難以置信。
– A single speck of dust has more electrons than the
number of people who have ever lived on Earth.
– 一個點塵埃的電子數比曾經生活在地球上的人的數量都還多。
• Electron behavior determines much of the
behavior of atoms. 電子行為決定大部分原子的行為。
• Directly observing electrons in the atom is
impossible; the electron is so small that observing
it changes its behavior.
– Even shining a light on the electron would affect it.
直接觀察的原子中的電子是不可能的，電子是如此之小，觀察其行為改
變。
即使對電子發光光會影響它。
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A Theory That Explains Electron Behavior
一個理論來解釋的電子行為
• The quantum-mechanical model explains the
manner in which electrons exist and behave
in atoms.量子力學模型解釋原子中電子的存在和行為？
• It helps us understand and predict the properties
of atoms that are directly related to the behavior
of the electrons:它可以幫助我們了解和預測的原子的電子的行為有直接關係的屬性：
– Why some elements are metals and others are nonmetals
– Why some elements gain one electron when forming an
anion, whereas others gain two
– Why some elements are very reactive, while others are
practically inert
– Why in other periodic patterns we see in the properties of
the elements
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7.2 The Nature of Light: Its Wave Nature
• Light: a form of electromagnetic radiation
– Composed of perpendicular oscillating waves,
one for the electric field and one for the
magnetic field
• An electric field is a region where an electrically charged
particle experiences a force.
• A magnetic field is a region where a magnetized particle
experiences a force.
• All electromagnetic waves move through
space at the same, constant speed.
– 3.00 × 108 m/s = the speed of light
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7.2 The Nature of Light: Its Wave Nature
Electromagnetic Radiation
Composed of perpendicular oscillating waves, one for the electric field and one for the magnetic field
An electric field is a region where an electrically charged particle experiences a force.
A magnetic field is a region where a magnetized particle experiences a force.
• All electromagnetic waves move through
space at the same, constant speed.
– 3.00 × 108 m/s = the speed of light
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Speed of Energy Transmission
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Wave Characteristics
•
•
The amplitude is the height of the wave.
– The distance from node to crest or node to trough
– The amplitude is a measure of light intensity—the larger the amplitude, the brighter
the light.
The wavelength (l) is a measure of the distance covered by the wave.
– The distance from one crest to the next
• The distance from one trough to the next, or the distance between alternate
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nodes
Characterizing Waves
•
The frequency (n) is the number of waves that pass a point in a given period of time.
– The number of waves = the number of cycles.
– Units are hertz (Hz) or cycles/s = s−1 (1 Hz = 1 s−1).
•
The total energy is proportional to the amplitude of the waves and the frequency.
– The larger the amplitude, the more force it has.
– The more frequently the waves strike, the more total force there is.
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**Frequency, Wavelength and Speed of
Electromagnetic Radiation
• Frequency (n) in Hertz—Hz or s-1.
• Wavelength (λ) in meters—m.
• mm
m
nm
Å
pm
(10-3 m) (10-6 m) (10-9 m) (10-10 m) (10-12 m)
• Velocity (c)—2.997925  108 m s-1.
c = λn
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λ = c/n
n = c/λ
Color xx
• The color of light is determined by its
wavelength or frequency.
• White light is a mixture of all the colors
of visible light.
– A spectrum
– Red Orange Yellow Green Blue Indigo
Violet
• When an object absorbs some of the
wavelengths of white light and reflects
others, it appears colored; the observed
color is predominantly the colors reflected.
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Example 7.1 Wavelength and Frequency
Calculate the wavelength (in nm) of the red light emitted by a barcode scanner that has a frequency of
4.62 × 1014 s–1.
Solution
You are given the frequency of the light and asked to find its wavelength. Use Equation 7.1, which relates
frequency to wavelength. You can convert the wavelength from meters to nanometers by using the conversion
factor between the two (1 nm = 10–9 m).
For Practice 7.1
A laser dazzles the audience in a rock concert by emitting green light with a wavelength of 515 nm. Calculate the
frequency of the light.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
**Electromagnetic Spectrum
•
Visible light comprises only a small fraction of all the wavelengths of light, called the electromagnetic spectrum.
•
Shorter wavelength (high-frequency) light has higher energy.
– Radio wave light has the lowest energy.
– Gamma ray light has the highest energy.
•
High-energy electromagnetic radiation can potentially damage biological molecules.
– Ionizing radiation
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Thermal Imaging using Infrared Light （IR）
(X)
To produce a
medical X-ray, the
patient is exposed
to shortwavelength
Using High-Energy Radiation
to Kill Cancer Cells
electromagnetic
radiation that can
pass
through the skin to
create an image of
bones and internal
organs.
During radiation therapy, a tumor is
targeted from multiple directions in order to
minimize the exposure of healthy cells, while
maximizing the exposure of cancerous cells.
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Interference(X)
建設性干涉
破壞性干涉
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Diffraction(X)
http://www.dcfever.com/learn/viewglossary.php?glossary_id=19
•
•
•
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When traveling waves
encounter an obstacle or
opening in a barrier that is
about the same size as the
wavelength, they bend
around it; this is called
diffraction.
– Traveling particles do
not diffract.
The diffraction of light
through two slits separated
by a distance comparable to
the wavelength results in an
interference pattern of the
diffracted waves.
An interference pattern is a
characteristic of all light
waves.
Two-Slit Interference(X)
http://www.youtube.com/watch?v=ZImlf6S9WnE
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**pg.302 The Particle Nature of Light:
The Photoelectric Effect
http://www.youtube.com/watch?v=ubkNGwu_66s
•
•
•
It was observed that many metals emit electrons when a light shines on their surface.
– This is called the photoelectric effect.
Classic wave theory attributed this effect to the light energy being transferred to the
electron.
According to this theory, if the wavelength of light is made shorter, or the light wave’s intensity made brighter, more
electrons should be ejected.
– Remember that the energy of a wave is directly proportional to its amplitude and its frequency.
– This idea predicts if a dim light were used there would be a lag time before electrons were emitted.
• To give the electrons time to absorb enough energy
The Particle Nature of Light:
The Photoelectric Effect
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The Photoelectric Effect: The Problem
• Experimental observations indicate the
following:
– A minimum frequency was needed before
electrons would be emitted regardless of the
intensity called the threshold frequency.臨界頻率
– High-frequency light from a dim source caused
electron emission without any lag time.
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**Einstein’s Explanation
• Einstein proposed that the light energy was
delivered to the atoms in packets, called quanta
or photons.量子或光子。
• The energy of a photon of light is directly
proportional to its frequency.
– Inversely proportional to its wavelength
– The proportionality constant is called Planck’s
Constant, (h), and has the value 6.626 × 10−34 J ∙ s.
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Ejected Electrons
• One photon at the threshold frequency gives
the electron just enough energy for it to
escape the atom.克服原子核對電子的吸引力
– Binding energy, f
• When irradiated with a shorter wavelength
photon, the electron absorbs more energy
than is necessary to escape.
• This excess energy becomes kinetic energy
of the ejected electron.
Kinetic Energy = Ephoton – Ebinding
KE = hn − f
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Example 7.2 Photon Energy
A nitrogen gas laser pulse with a wavelength of 337 nm contains 3.83 mJ of energy. How many photons does it
contain?
Sort
You are given the wavelength and total energy of a light pulse and asked to find the number of photons it contains.
Given:
Find: number of photons
Strategize
In the first part of the conceptual plan, calculate the energy of an individual photon from its wavelength.
In the second part, divide the total energy of the pulse by the energy of a photon to get the number of photons in
the pulse.
Conceptual Plan
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.2 Photon Energy
Continued
Relationships Used
E = hc/λ (Equation 7.3)
Solve
To execute the first part of the conceptual plan, convert the wavelength to meters and substitute it into the equation
to calculate the energy of a 337 nm photon.
To execute the second part of the conceptual plan, convert the energy of the pulse from mJ to J. Then divide the
energy of the pulse by the energy of a photon to obtain the number of photons.
Solution
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.2 Photon Energy
Continued
Check
The units of the answer, photons, are correct. The magnitude of the answer (10 15) is reasonable. Photons are small
particles and any macroscopic collection should contain a large number of them.
For Practice 7.2
A 100-watt lightbulb radiates energy at a rate of 100 J/s (The watt, a unit of power, or energy over time, is defined as
1 J/s.) If all of the light emitted has a wavelength of 525 nm, how many photons are emitted per second? (Assume
three significant figures in this calculation.)
For More Practice 7.2
The energy required to dislodge electrons from sodium metal via the photoelectric effect is 275 kJ/mol. What
wavelength in nm of light has sufficient energy per photon to dislodge an electron from the surface of sodium?
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.3 Wavelength, Energy, and Frequency
Arrange the three types of electromagnetic radiation—visible light, X-rays, and microwaves—in order of increasing
a. wavelength.
b. frequency.
c. energy per photon.
Solution
Examine Figure 7.5 and note that X-rays have
the shortest wavelength, followed by visible
light and then microwaves.
a. wavelength
X rays < visible < microwaves
Since frequency and wavelength are inversely
proportional—the longer the wavelength the
shorter the frequency—the ordering with respect
to frequency is the reverse of the ordering with
respect to wavelength.
b. frequency
microwaves < visible < X rays
FIGURE 7.5 The Electromagnetic Spectrum The right side of the spectrum
consists of high-energy, high-frequency, short-wavelength radiation. The left
side consists of low-energy, low-frequency, long-wavelength radiation. Visible
light constitutes a small segment in the middle.
Energy per photon decreases with increasing wavelength but increases with increasing frequency; therefore, the
ordering with respect to energy per photon is the same as for frequency.
c. energy per photon
microwaves < visible < X rays
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.3 Wavelength, Energy, and Frequency
Continued
For Practice 7.3
Arrange these three colors of visible light—green, red, and blue—in order of increasing
a. wavelength.
b. frequency.
c. energy per photon.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
7.3 Atomic Spectroscopy and the Bohr Model
Spectra
Exciting Gas Atoms to Emit Light with Electrical Energy
Na
K
Li
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Ba
When atoms or molecules absorb energy, that
energy is often released as light energy,
Fireworks, neon lights, etc.
When that emitted light is passed through a
prism, a pattern of particular wavelengths of
light is seen that is unique to that type of atom
or molecule – the pattern is called an emission
spectrum.
Noncontinuous
Can be used to identify the material
Flame tests
Continuous Spectrum
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古典 Continuous Spectrum
量子 Emission Spectra
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Examples of Spectra
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Rydberg’s Spectrum Analysis
Johannes Rydberg (1854–1919)
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Bohr Model of H Atoms
The Bohr Model of the Atom
•
Bohr’s major idea was that the energy of the
atom was quantized, and that the amount of
energy in the atom was related to the electron’s
position in the atom.
– Quantized means that the atom could only
have very specific amounts of energy.
•
The electrons travel in orbits that are at a fixed
distance from the nucleus.
– Stationary states
– Therefore, the energy of the electron was
proportional to the distance the orbit was
from the nucleus.
Electrons emit radiation when they “jump” from
an orbit with higher energy down to an orbit with
lower energy.
– The emitted radiation was a photon of light.
– The distance between the orbits
determined the energy of the photon of light
produced.
•
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Neils Bohr (1885–1962
101
Fireworks typically
contain the salts of
such metals as
sodium, calcium,
strontium, barium,
and copper.
Emissions from
these elements
produce the
brilliant colors of
pyrotechnic
displays.
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Emission versus Absorption Spectra
Emission
Absorption
Spectra of Mercury
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7.4 The Wave Nature of Matter: The de Broglie
Wavelength, the Uncertainty Principle, and
Indeterminacy
Wave Behavior of Electrons, m = 9.1 x 10-31
Louis de Broglie (1892–1987)
• de Broglie predicted that the wavelength of a
particle was inversely proportional to its
momentum.
• Because it is so small, the wave character of
electrons is significant.
https://www.youtube.com/watch?v=4dHu9aJuDCs
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xxElectron Diffraction
• Proof that the electron had wave nature came a few
years later with the demonstration that a beam of
electrons would produce an interference pattern the
same as waves do.
However, electrons
actually present an
interference pattern,
demonstrating they
behave like waves.
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xxElectron Diffraction
• Proof that the electron had wave nature came a few
years later with the demonstration that a beam of
electrons would produce an interference pattern the
same as waves do.
If electrons behave only like
particles, there should only
be two bright spots on the
target.
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Complementary Properties互補屬性
粒子與波不會同時存在
• When you try to observe the wave nature of
the electron, you cannot observe its particle
nature, and vice versa.
– Wave nature = interference pattern
– Particle nature = position, which slit it is passing
through
• The wave and particle nature of the electron
are complementary properties.
– As you know more about one you know less
about the other.
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Example 7.4 De Broglie Wavelength
Calculate the wavelength of an electron traveling with a speed of 2.65 × 106 m/s.
Sort
You are given the speed of an electron and asked to calculate its wavelength.
Given: v = 2.65 × 106 m/s
Find: λ
Strategize
The conceptual plan shows how the de Broglie relation relates the wavelength of an electron to its mass and velocity.
Conceptual Plan
Relationships Used
λ = h/mv (de Broglie relation, Equation 7.4)
Solve
Substitute the velocity, Planck’s constant, and the mass of an electron to calculate the electron’s wavelength. To
correctly cancel the units, break down the J in Planck’s constant into its SI base units (1 J = 1 kg · m 2/s2).
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.4 De Broglie Wavelength
Continued
Solution
Check
The units of the answer (m) are correct. The magnitude of the answer is very small, as expected for the wavelength of
an electron.
For Practice 7.4
What is the velocity of an electron that has a de Broglie wavelength approximately the length of a chemical bond?
Assume this length to be 1.2 × 10–10 m.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Uncertainty Principle
• Heisenberg stated that the product of the
uncertainties in both the position and speed of a
particle was inversely proportional to its mass.
– x = position, Dx = uncertainty in position
– v = velocity, Dv = uncertainty in velocity
– m = mass
• This means that the more accurately you know
the position of a small particle, such as an
electron, the less you know about its speed, and
vice versa.
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xxUncertainty Principle Demonstration
Any experiment designed to
observe the electron results
in detection of a single
electron particle and no
interference pattern.
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Xx Determinacy versus Indeterminacy
• According to classical physics, particles move in a
path determined by the particle’s velocity,
position, and forces acting on it.
– Determinacy = definite, predictable future
• Because we cannot know both the position and
velocity of an electron, we cannot predict the path
it will follow.
– Indeterminacy = indefinite future, can only predict
probability
• The best we can do is to describe the probability
an electron will be found in a particular region
using statistical functions.
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xxTrajectory versus Probability
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xxElectron Energy
• Electron energy and position are
complementary.
– KE = ½ mv2
• For an electron with a given energy, the best
we can do is describe a region in the atom of
high probability of finding it.
• Many of the properties of atoms are related to
the energies of the electrons.
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xx7.5 Quantum Mechanics and the Atom
Schrö dinger’s Equation
• Schrödinger’s equation allows us to
calculate the probability of finding an
electron with a particular amount of energy
at a particular location in the atom.
• Solutions to Schrödinger’s equation
produce many wave functions, Y.
• A plot of distance versus Y2 represents an
orbital, a probability distribution map of a
region where the electron is likely to be found.
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53
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
Prentice Hall © 2005
n=2
n=3
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
7.6
Chapter Seven
Solutions to the Wave Function, Y
• These integers are called quantum
numbers.
– Principal quantum number, n
energy
– Angular momentum quantum number, l
size, shape
– Magnetic quantum number, ml
orientation
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Principal Quantum Number, n :Energy
En= (-13.6 eV) / n2
Bohr’s energy level
• n  1 integer.
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Angular Momentum Quantum Number, l
shape of the orbital
• l ＝ 0 to (n – 1).
• Each value of l is called by a particular letter
that designates the shape of the orbital.
– l=0, s orbitals are spherical.
– l=1, p orbitals are like two balloons tied at the
knots.
– l=2, d orbitals are mainly like four balloons tied at
the knots.
– l=3, f orbitals are mainly like eight balloons tied
at the knots.
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Magnetic Quantum Number, ml
orientation of the orbital
• ml = −l ~ +l
– Including 0
– Gives the number of orbitals of a particular shape
• When l = 2, ml = −2, −1, 0, +1, +2, 5 orbitals
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Describing an Orbital
• Each set of n, l, and ml describes one
orbital.
• n : principal energy level.
– Also called the principal shell
• l , ml : sublevel.
– Also called a subshell
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Energy Levels and Sublevels
The n = 2 principal energy
level contains two sublevels:
a.The l = 0: 2s sublevel with
one orbital with ml = 0
b.The l = 1: 2p sublevel with
three p orbitals with
ml = -1, 0, +1
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Energy Levels and Sublevels
• In general,
– the number of sublevels within a level = n.
– the number of orbitals within a sublevel = 2l + 1.
– the number of orbitals in a level n2.
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Example 7.5 Quantum Numbers I
What are the quantum numbers and names (for example, 2s, 2p) of the orbitals in the n = 4 principal level? How
many n = 4 orbitals exist?
Solution
First determine the possible values of l (from the given value of n). For a given value of n, the possible values of l
are 0, 1, 2, . . . , (n – 1).
n = 4; therefore, l = 0, 1, 2, and 3
Next, determine the the possible values of ml for each value of l. For a given value of l, the possible values of ml are
the integer values including zero ranging from –l to +l . The name of an orbital is its principal quantum number (n)
followed by the letter corresponding to the value l. The total number of orbitals is given by n2.
For Practice 7.5
List the quantum numbers associated with all of the 5d orbitals. How many 5d orbitals exist?
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.6 Quantum Numbers II
These sets of quantum numbers are each supposed to specify an orbital. One set, however, is erroneous. Which one
and why?
a. n = 3; l = 0; ml = 0
b. n = 2; l = 1; ml = –1
c. n = 1; l = 0; ml = 0
d. n = 4; l = 1; ml = –2
Solution
Choice (d) is erroneous because for l = 1, the possible values of ml are only –1, 0, and +1.
For Practice 7.6
Each set of quantum numbers is supposed to specify an orbital. However, each set contains one quantum number
that is not allowed. Replace the quantum number that is not allowed with one that is allowed.
a. n = 3; l = 3; ml = +2
b. n = 2; l = 1; ml = –2
c. n = 1; l = 1; ml = 0
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Quantum Leaps
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Hydrogen Energy Transitions and Radiation
DEelectron = Efinal state − Einitial state
Eemitted photon = −DEelectron
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n = 32 656nm, red light
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© 2014 Pearson Education, Inc.
Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom
Determine the wavelength of light emitted when an electron in a hydrogen atom makes a transition from an orbital
in n = 6 to an orbital in n = 5.
Sort
You are given the energy levels of an atomic transition and asked to find the wavelength of emitted light.
Given:
Find: λ
Strategize
In the first part of the conceptual plan, calculate the energy of the electron in the n = 6 and n = 5 orbitals using
Equation 7.7 and subtract to find ΔEatom.
In the second part, find Ephoton by taking the negative of ΔEatom and then calculate the wavelength corresponding to a
photon of this energy using Equation 7.3. (The difference in sign between Ephoton and ΔEatom applies only to emission.
The energy of a photon must always be positive.)
Conceptual Plan
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom
Continued
Relationships Used
En = –2.18 × 10–18 J(1/n2)
E = hc/λ
Solve
Follow the conceptual plan. Begin by calculating ΔEatom.
Calculate Ephoton by changing the sign of ΔEatom.
Solve the equation relating the energy of a photon to its wavelength for λ. Substitute the energy of the photon and
calculate λ.
Solution
Ephoton = –ΔEatom = +2.6644 × 10–20 J
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 7.7 Wavelength of Light for a Transition in the Hydrogen Atom
Continued
Check
The units of the answer (m) are correct for wavelength. The magnitude is reasonable because 10 –6 m is in the infrared
region of the electromagnetic spectrum. We know that transitions from n = 3 or n = 4 to n = 2 lie in the visible region,
so it makes sense that a transition between levels of higher n value (which are energetically closer to one another)
would result in light of longer wavelength.
For Practice 7.7
Determine the wavelength of the light absorbed when an electron in a hydrogen atom makes a transition from an
orbital in which n = 2 to an orbital in which n = 7.
For More Practice 7.7
An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 93.8 nm.
Find the principal level to which the electron relaxed.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
7.6 The Shapes of Atomic Orbitals, l
Probability and Radial Distribution Functions
• y 2 is the probability density.
– The probability of finding an electron at a particular point
in space
– For s orbital maximum at the nucleus?
– Decreases as you move away from the nucleus
• The radial distribution function represents the total
probability at a certain distance from the nucleus.
– Maximum at most probable radius
• Nodes in the functions are where the probability
drops to 0.
© 2014 Pearson Education, Inc.
The Shapes of Atomic Orbitals
• The l quantum number primarily determines
the shape of the orbital.
• l can have integer values from 0 to (n – 1).
• Each value of l is called by a particular letter
that designates the shape of the orbital.
– s orbitals are spherical.
– p orbitals are like two balloons tied at the knots.
– d orbitals are mainly like four balloons tied at
the knots.
– f orbitals are mainly like eight balloons tied at
the knots.
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72
The 1s Orbital
• The 1s orbital (n = 1, l = 0, ml = 0) has spherical symmetry.
• An electron in this orbital spends most of its time near the
nucleus.
Spherical symmetry;
probability of finding
the electron is the same
in each direction.
Prentice Hall © 2005
The electron
cloud doesn’t
“end” here …
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
… the electron just
spends very little
time farther out.
Chapter Seven
73
The 2s Orbital, l=0
• The 2s orbital has two concentric, spherical regions of
high electron probability.
• The region near the nucleus is separated from the outer
region by a node—a region (a spherical shell in this case)
in which the electron probability is zero.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Seven
74
The Three p Orbitals, l=1
Three values of ml (-1,
0, 1)gives three p
orbitals in the p
subshell.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Seven
75
The Five d Orbitals, l=2
Five values of ml (–
2, –1, 0, 1, 2) gives five
d orbitals in the d
subshell.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Seven
The seven f Orbitals, l=3
seven values of ml (–3,
-2, –1, 0, 1, 2, 3) gives
seven f orbitals in the
d subshell.
© 2014 Pearson Education, Inc.
Phases
Orbitals are determined from mathematical wave
functions.
A wave function can have positive or negative
values.
As well as nodes where the wave function
=0
The sign of the wave function is called its
phase.
When orbitals interact, their wave functions may
be in phase (same sign) or out of phase
(opposite signs).
This is important in bonding as will be
examined in a later chapter.
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Why Are Atoms Spherical?
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.