Quantum Theory and the Electronic Structure of Atoms

Quantum Theory and the Electronic Structure of Atoms
Chapter 7

The Wave Nature of Light
A wave is a continuously repeating change or oscillation in matter or in a physical field. Light is also a wave. It consists of oscillations in electric and magnetic fields that travel through space. Visible light, X rays, and radio waves are all forms of electromagnetic radiation. 2

A wave can be characterized by its wavelength and frequency. The wavelength, l (lambda), is the distance between any two adjacent identical points of a wave. The frequency, n (nu), of a wave is the number of wavelengths that pass a fixed point in one second. 2

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.

The speed (u) of the wave = l x n
Properties of Waves Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n

The product of the frequency, n (waves/sec) and the wavelength, l (m/wave) would give the speed of the wave in m/s. In a vacuum, the speed of light, c, is 3.00 x 108 m/s. Therefore, So, given the frequency of light, its wavelength can be calculated, or vice versa. 2

Speed of light (c) in vacuum = 3.00 x 108 m/s
Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c

What is the wavelength of yellow light with a frequency of 5.09 x 1014 s-1? (Note: s-1, commonly referred to as Hertz (Hz) is defined as “cycles or waves per second”.) If c = nl, then rearranging, we obtain l = c/n 2

What is the frequency of violet light with a wavelength of 408 nm? If c = nl, then rearranging, we obtain n = c/l. 2

The range of frequencies or wavelengths of electromagnetic radiation is called the electromagnetic spectrum. Visible light extends from the violet end of the spectrum at about 400 nm to the red end with wavelengths about 800 nm. Beyond these extremes, electromagnetic radiation is not visible to the human eye. 2

high energy short low energy wavelength long wavelength
Electromagnetic Spectrum high energy short wavelength low energy long wavelength

l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m
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 n l x n = c 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

Mystery #1, “Black Body Problem” Solved by Planck in 1900
Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x J•s

Quantum Effects and Photons
Planck’s Quantization of Energy (1900) According to Max Planck, the atoms of a solid oscillate with a definite frequency, n. where h (Planck’s constant) is assigned a value of 6.63 x J. s and n must be an integer. He proposed that an atom could have only certain energies of vibration, E, those allowed by the formula 2

Planck’s Quantization of Energy. Thus, the only energies a vibrating atom can have are hn, 2hn, 3hn, and so forth. The numbers symbolized by n are quantum numbers. The vibrational energies of the atoms are said to be quantized. 2

By the early part of twentieth century, the wave theory of light seemed to be well entrenched. In 1905, Albert Einstein proposed that light had both wave and particle properties as observed in the photoelectric effect Einstein based this idea on the work of a German physicist, Max Planck. 2

Photoelectric Effect Einstein extended Planck’s work to include the structure of light itself. If a vibrating atom changed energy from 3hn to 2hn, it would decrease in energy by hn. He proposed that this energy would be emitted as a bit (or quantum) of light energy. Einstein postulated that light consists of quanta (now called photons), or particles of electromagnetic energy. 2

Photoelectric Effect The energy of the photons proposed by Einstein would be proportional to the observed frequency, and the proportionality constant would be Planck’s constant. In 1905, Einstein used this concept to explain the “photoelectric effect.” 2

Photoelectric Effect The photoelectric effect is the ejection of electrons from the surface of a metal when light shines on it. Electrons are ejected only if the light exceeds a certain “threshold” frequency. Violet light, for example, will cause potassium to eject electrons, but no amount of red light (which has a lower frequency) has any effect. 2

Photoelectric Effect Einstein’s assumption that an electron is ejected when struck by a single photon implies that it behaves like a particle. When the photon hits the metal, its energy, hn is taken up by the electron. The photon ceases to exist as a particle; it is said to be “absorbed.” 2

Photoelectric Effect The “wave” and “particle” pictures of light should be regarded as complementary views of the same physical entity. This is called the wave-particle duality of light. The equation E = hn displays this duality; E is the energy of the “particle” photon, and n is the frequency of the associated “wave.” 2

Photon is a “particle” of light
Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn Light has both: wave nature particle nature KE e- Photon is a “particle” of light hn = KE + BE KE = hn - BE

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
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 nm. E = h x n E = h x c / l E = 6.63 x (J•s) x 3.00 x 10 8 (m/s) / x 10-9 (m) E = 1.29 x J

Radio Wave Energy What is the energy of a photon corresponding to radio waves of frequency x 10 6 s-1? 11

Radio Wave Energy What is the energy of a photon corresponding to radio waves of frequency x 10 6 s-1? Solve for E, using E = hn, and four significant figures for h. 12

Radio Wave Energy (6.626 x 10-34 J.s) x (1.255 x 106 s-1) =
What is the energy of a photon corresponding to radio waves of frequency x 10 6 s-1? Solve for E, using E = hn, and four significant figures for h. (6.626 x J.s) x (1.255 x 106 s-1) = x = x J 13

The Bohr Theory of the Hydrogen Atom
Prior to the work of Niels Bohr, the stability of the atom could not be explained using the then-current theories. In 1913, using the work of Einstein and Planck, he applied a new theory to the simplest atom, hydrogen. Before looking at Bohr’s theory, we must first examine the “line spectra” of atoms. 2

Atomic Line Spectra When a heated metal filament emits light, we can use a prism to spread out the light to give a continuous spectrum-that is, a spectrum containing light of all wavelengths. The light emitted by a heated gas, such as hydrogen, results in a line spectrum-a spectrum showing only specific wavelengths of light. 2

Line Emission Spectrum of Hydrogen Atoms

The emission spectrum of hydrogen lead to the modern understanding of the electronic structure of the atom. emission spectrum - light emitted when a substance is excited by an energy source.

n (principal quantum number) = 1,2,3,…
Bohr’s Model of the Atom (1913) e- can only have specific (quantized) energy values light is emitted as e- moves from one energy level to a lower energy level En = -RH ( ) 1 n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J

The Bohr Atom 8 Initial understanding of the atom by Niels Bohr
Electrons exist in fixed energy levels surrounding the nucleus. Quantization of energy Promotion of electron occurs as it absorbs energy Excited State Energy is released as the electron travels back to lower levels. Relaxation

Orbit - what Bohr called the fixed energy levels.
Ground state - the lowest possible energy state.

n = 1, 2, 3, … The orbits are also identified using “quantum numbers”
When the electron relaxes (c) the energy released is observed as a single wavelength of light.

E = hn E = hn

Atomic Line Spectra In 1885, J. J. Balmer showed that the wavelengths, l, in the visible spectrum of hydrogen could be reproduced by a simple formula. The known wavelengths of the four visible lines for hydrogen correspond to values of n = 3, n = 4, n = 5, and n = 6. 2

Bohr’s Postulates Bohr set down postulates to account for (1) the stability of the hydrogen atom and (2) the line spectrum of the atom. 1. Energy level postulate An electron can have only specific energy levels in an atom. 2. Transitions between energy levels An electron in an atom can change energy levels by undergoing a “transition” from one energy level to another. 2

Bohr’s Postulates Bohr derived the following formula for the energy levels of the electron in the hydrogen atom. Rh is a constant (expressed in energy units) with a value of 2.18 x J. 2

Bohr’s Postulates When an electron undergoes a transition from a higher energy level to a lower one, the energy is emitted as a photon. From Postulate 1, 2

Bohr’s Postulates If we make a substitution into the previous equation that states the energy of the emitted photon, hn, equals Ei - Ef, Rearranging, we obtain 2

Bohr’s Postulates Bohr’s theory explains not only the emission of light, but also the absorbtion of light. When an electron falls from n = 3 to n = 2 energy level, a photon of red light (wavelength, 685 nm) is emitted. When red light of this same wavelength shines on a hydrogen atom in the n = 2 level, the energy is gained by the electron that undergoes a transition to n = 3. 2

A Problem to Consider Calculate the energy of a photon of light emitted from a hydrogen atom when an electron falls from level n = 3 to level n = 1. 2

( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1
nf = 1 ni = 3 nf = 2 ni = 3 Ef = -RH ( ) 1 n2 f nf = 1 ni = 2 Ei = -RH ( ) 1 n2 i i f DE = RH ( ) 1 n2

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. i f DE = RH ( ) 1 n2 Ephoton = Ephoton = 2.18 x J x (1/25 - 1/9) Ephoton = DE = x J Ephoton = h x c / l l = h x c / Ephoton l = 6.63 x (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm

Why is e- energy quantized?
De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nl l = h mu u = velocity of e- m = mass of e-

Quantum Mechanics Bohr’s theory established the concept of atomic energy levels but did not thoroughly explain the “wave-like” behavior of the electron. Current ideas about atomic structure depend on the principles of quantum mechanics, a theory that applies to subatomic particles such as electrons. 2

Quantum Mechanics The first clue in the development of quantum theory came with the discovery of the de Broglie relation. In 1923, Louis de Broglie reasoned that if light exhibits particle aspects, perhaps particles of matter show characteristics of waves. He postulated that a particle with mass m and a velocity v has an associated wavelength. The equation l = h/mv is called the de Broglie relation. 2

Quantum Mechanics If matter has wave properties, why are they not commonly observed? The de Broglie relation shows that a baseball (0.145 kg) moving at about 60 mph (27 m/s) has a wavelength of about 1.7 x m. This value is so incredibly small that such waves cannot be detected. 2

Quantum Mechanics If matter has wave properties, why are they not commonly observed? Electrons have wavelengths on the order of a few picometers (1 pm = m). Under the proper circumstances, the wave character of electrons should be observable. 2

What is the de Broglie wavelength (in nm) associated with a 2
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/mu h in J•s m in kg u in (m/s) l = 6.63 x / (2.5 x 10-3 x 15.6) l = 1.7 x m = 1.7 x nm

Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent

Quantum Mechanics If matter has wave properties, why are they not commonly observed? In 1927, it was demonstrated that a beam of electrons, just like X rays, could be diffracted by a crystal. The German physicist, Ernst Ruska, used this wave property to construct the first “electron microscope” in 1933. 2

Chemistry in Action: Electron Microscopy
le = nm STM image of iron atoms on copper surface

Scanning electron microscope. Photo courtesy of Carl Zeiss, Inc
Scanning electron microscope.Photo courtesy of Carl Zeiss, Inc., Thornwood, NY. .

Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- Wave function (Y) describes: . energy of e- with a given Y . probability of finding e- in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

Quantum Mechanics Quantum mechanics is the branch of physics that mathematically describes the wave properties of submicroscopic particles. We can no longer think of an electron as having a precise orbit in an atom. To describe such an orbit would require knowing its exact position and velocity. In 1927, Werner Heisenberg showed (from quantum mechanics) that it is impossible to know both simultaneously. 2

Quantum Mechanics Heisenberg’s uncertainty principle is a relation that states that the product of the uncertainty in position (Dx) and the uncertainty in momentum (mDvx) of a particle can be no larger than h/4p. When m is large (for example, a baseball) the uncertainties are small, but for electrons, high uncertainties disallow defining an exact orbit. 2

Quantum Mechanics Although we cannot precisely define an electron’s orbit, we can obtain the probability of finding an electron at a given point around the nucleus. Erwin Schrodinger defined this probability in a mathematical expression called a wave function, denoted y (psi). The probability of finding a particle in a region of space is defined by y2. 2

Quantum Numbers and Atomic Orbitals
According to quantum mechanics, each electron is described by four quantum numbers. Principal quantum number (n) Angular momentum quantum number (l) Magnetic quantum number (ml) Spin quantum number (ms) The first three define the wave function for a particular electron. The fourth quantum number refers to the magnetic property of electrons. 2

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

The principal quantum number(n) represents the “shell number” in which an electron “resides.” The smaller n is, the smaller the orbital. The smaller n is, the lower the energy of the electron. 2

Where 90% of the e- density is found for the 1s orbital

Y = fn(n, l, ml, ms) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 l = s orbital l = p orbital l = d orbital l = f orbital n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e- occupies

The angular momentum quantum number (l) distinguishes “sub shells” within a given shell that have different shapes. Each main “shell” is subdivided into “sub shells.” Within each shell of quantum number n, there are n sub shells, each with a distinctive shape. l can have any integer value from 0 to (n - 1) The different subshells are denoted by letters. Letter s p d f g … l …. 2

Using calculated probabilities of electron “position,” the shapes of the orbitals can be described. The s sub shell orbital (there is only one) is spherical The p sub shell orbitals (there are three) are dumbbell shape. The d sub shell orbitals (there are five ) are a mix of cloverleaf and dumbbell shapes. 2

l = 0 (s orbitals) l = 1 (p orbitals)

l = 2 (d orbitals)

The magnetic quantum number (ml) distinguishes orbitals within a given sub-shell that have different shapes and orientations in space. Each sub shell is subdivided into “orbitals,” each capable of holding a pair of electrons. ml can have any integer value from -l to +l. Each orbital within a given sub shell has the same energy. 2

Y = fn(n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 orientation of the orbital in space

ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2

The spin quantum number (ms) refers to the two possible spin orientations of the electrons residing within a given orbital. Each orbital can hold only two electrons whose spins must oppose one another. The possible values of ms are +1/2 and –1/2. 2

Y = fn(n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½

Y = fn(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Y. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time

Y = fn(n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and ml How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½ Y = (n, l, ml, ½) or Y = (n, l, ml, -½) An orbital can hold 2 electrons

How many 2p orbitals are there in an atom?
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 If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e- l = 2

Electron Configuration
An “electron configuration” of an atom is a particular distribution of electrons among available sub shells. The notation for a configuration lists the sub-shell symbols sequentially with a superscript indicating the number of electrons occupying that sub shell. For example, lithium (atomic number 3) has two electrons in the “1s” sub shell and one electron in the “2s” sub shell 1s2 2s1. 2

Electron Configuration
An orbital diagram is used to show how the orbitals of a sub shell are occupied by electrons. Each orbital is represented by a circle. Each group of orbitals is labeled by its sub shell notation. 1s s p Electrons are represented by arrows: up for ms = +1/2 and down for ms = -1/2 2

Orbital Diagrams Consider carbon (Z = 6) with the ground state configuration 1s22s22p2. Three possible arrangements are given in the following orbital diagrams. Diagram 1: Diagram 2: Diagram 3: 1s s p Each state has a different energy and different magnetic characteristics. 2

Orbital Diagrams Hund’s rule states that the lowest energy arrangement (the “ground state”) of electrons in a sub-shell is obtained by putting electrons into separate orbitals of the sub shell with the same spin before pairing electrons. Looking at carbon again, we see that the ground state configuration corresponds to diagram 1 when following Hund’s rule. 1s s p 2

Orbital Diagrams To apply Hund’s rule to oxygen, whose ground state configuration is 1s22s22p4, we place the first seven electrons as follows. 1s s p The last electron is paired with one of the 2p electrons to give a doubly occupied orbital. 1s s p 2

The Pauli Exclusion Principle
The Pauli exclusion principle, which summarizes experimental observations, states that no two electrons can have the same four quantum numbers. In other words, an orbital can hold at most two electrons, and then only if the electrons have opposite spins. 2

Magnetic Properties Although an electron behaves like a tiny magnet, two electrons that are opposite in spin cancel each other. Only atoms with unpaired electrons exhibit magnetic susceptibility. A paramagnetic substance is one that is weakly attracted by a magnetic field, usually the result of unpaired electrons. A diamagnetic substance is not attracted by a magnetic field generally because it has only paired electrons. 2

Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p

( ) Energy of orbitals in a single electron atom 1 En = -RH n2
Energy only depends on principal quantum number n n=3 n=2 En = -RH ( ) 1 n2 n=1

Energy of orbitals in a multi-electron atom
Energy depends on n and l n=3 l = 2 n=3 l = 1 n=3 l = 0 n=2 l = 1 n=2 l = 0 n=1 l = 0

The Pauli Exclusion Principle
The maximum number of electrons and their orbital diagrams are: Sub shell Number of Orbitals Maximum Number of Electrons s (l = 0) 1 2 p (l = 1) 3 6 d (l =2) 5 10 f (l =3) 7 14 2

Aufbau Principle Every atom has an infinite number of possible electron configurations. The configuration associated with the lowest energy level of the atom is called the “ground state.” Other configurations correspond to “excited states.” 2

Aufbau Principle The Aufbau principle is a scheme used to reproduce the ground state electron configurations of atoms by following the “building up” order. Listed below is the order in which all the possible sub-shells fill with electrons. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f You need not memorize this order. As you will see, it can be easily obtained. 2

“Fill up” electrons in lowest energy orbitals (Aufbau principle)
? ? C 6 electrons C 1s22s22p2 B 5 electrons B 1s22s22p1 Be 4 electrons Be 1s22s2 Li 3 electrons Li 1s22s1 He 2 electrons He 1s2 H 1 electron H 1s1

The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). Ne 10 electrons Ne 1s22s22p6 F 9 electrons F 1s22s22p5 O 8 electrons O 1s22s22p4 N 7 electrons N 1s22s22p3 C 6 electrons C 1s22s22p2

Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

Orbital Energy Levels in Multi-electron Systems
3d 4s 3p 3s Energy 2p 2s 1s 3

in the orbital or subshell
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 1s1 H

Aufbau Principle Here are a few examples.
Using the abbreviation [He] for 1s2, the configurations are Z=4 Beryllium 1s22s2 or [He]2s2 Z=3 Lithium 1s22s1 or [He]2s1 2

Aufbau Principle With boron (Z=5), the electrons begin filling the 2p subshell. Z=5 Boron 1s22s22p1 or [He]2s22p1 Z=6 Carbon 1s22s22p2 or [He]2s22p2 Z=7 Nitrogen 1s22s22p3 or [He]2s22p3 Z=8 Oxygen 1s22s22p4 or [He]2s22p4 Z=9 Fluorine 1s22s22p5 or [He]2s22p5 Z=10 Neon 1s22s22p6 or [He]2s62p6 2

Aufbau Principle With sodium (Z = 11), the 3s sub shell begins to fill. Z=11 Sodium 1s22s22p63s1 or [Ne]3s1 Z=12 Magnesium 1s22s22p23s2 or [Ne]3s2 Then the 3p sub shell begins to fill. Z=13 Aluminum 1s22s22p63s23p1 or [Ne]3s23p1 [Ne]3s23p6 or 1s22s22p63s23p6 Argon Z=18 2

Aufbau Principle The “building up” order corresponds for the most part to increasing energy of the subshells. By filling orbitals of the lowest energy first, you usually get the lowest total energy (“ground state”) of the atom. Now you can see how to reproduce the electron configurations of Table 8.1 using the Aufbau principle. Remember, the number of electrons in the neutral atom equals the atomic number, Z. 2

What is the electron configuration of Mg?
Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½

Outermost subshell being filled with electrons

n=1, 2(1)2=2 n=2, 2(2)2=8 n=3, 2(3)2=18 n=4, 2(4)2=32

Configurations and the Periodic Table
Note that elements within a given family have similar configurations. For instance, look at the noble gases. Helium 1s2 Neon 1s22s22p6 Argon 1s22s22p63s23p6 Krypton 1s22s22p63s23p63d104s24p6 2

Note that elements within a given family have similar configurations. The Group IIA elements are sometimes called the alkaline earth metals. Beryllium 1s22s2 Magnesium 1s22s22p63s2 Calcium 1s22s22p63s23p64s2 2

Electrons that reside in the outermost shell of an atom - or in other words, those electrons outside the “noble gas core”- are called valence electrons. These electrons are primarily involved in chemical reactions. Elements within a given group have the same “valence shell configuration.” This accounts for the similarity of the chemical properties among groups of elements. 2

The following slide illustrates how the periodic table provides a sound way to remember the Aufbau sequence. In many cases you need only the configuration of the outer elements. You can determine this from their position on the periodic table. The total number of valence electrons for an atom equals its group number. 2

Quantum Theory and the Electronic Structure of Atoms

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