Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Topics to Cover • Electromagnetic radiation wavelength, frequency, speed, & energy • Quantum = packet of energy • Bohr's atom & line spectra • Matter waves/ uncertainty • Atomic orbitals • Quantum numbers • Electron configurations

Quantum Theory

Properties of Waves Electromagnetic radiation 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. 7.1

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 7.1

Longer wavelength Lower frequency Shorter wavelength Higher frequency The speed (u) of the wave = l x n

Speed of light (c) in vacuum = 3.00 x 108 m/s Maxwell (1873), proposed that visible light consists of electromagnetic waves.– provides a mathematical description of the general behavior of light. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves, which has an electrical field component and a magnetic field component . Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c 7.1

7.1

l x n = c n = c/l n = 3.00 x 108 m/s / 500 x 10 -9m n = 6.00 x 10 14 What is the frequency of light with wavelength of 500nm? l x n = c n = c/l n = 3.00 x 108 m/s / 500 x 10 -9m n = 6.00 x 10 14 Practice problem of chap7:Q1 7.1

Mystery #1, “Black Body Problem” Solved by Planck in 1900 When solids are heated, they emit electromagnetic radiation. The amount of radiation energy emitted by the object at a certain temperature depends on the wavelength. Classical physics assumed molecules could absorb/emit any arbitrary amount of radiation energy. Plank: Energy (light) is emitted or absorbed in discrete units (quantum), a packet of energy. Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy of a single quantum E = h x n= h x c/l Planck’s constant (h) h = 6.63 x 10-34 J•s Energy is always emitted in multiples of hn: 1 hn,2 hn,3hn..... 7.1

Photon is a “particle” of light Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn Einstein suggested that light is a stream of particles, which were named Photons. KE e- Light has both: wave nature 2. particle nature Photon is a “particle” of light hn = KE + BE KE = hn - BE How does light interact with atoms? • Photoelectric effect: light can cause some metals to emit electrons • Einstein said: energy of incoming light must be ≥ the energy needed for ionization 7.2

Energy of light: Quantum = packet of energy Photon = packet of light • E = h n Planck’s constant (h) h = 6.63 x 10-34 J•s

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 0.154 nm. E = h x n E = h x c / l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J Practice problem of chap7:Q2 7.2

Bohr’s theory of the Hydrogen Atom

Line Emission Spectrum of Hydrogen Atoms Emission spectra: either continuous or line spectra of radiation emitted by substances. Line Emission Spectrum of Hydrogen Atoms 7.3

7.3

Quantum theory and electronic structure How does emission occur that causes the line spectra? Bohr: The single electron in hydrogen atom could be located only in certain orbits, which have a particular energy associated with it. The energies associated with electron motion in particular orbits must be fixed in value, or quantized.

Bohr’s Model of the Atom (1913) 1. e- can only have specific (quantized) energy values • Atom absorbs energy – causes electron move to a higher energy level (the excited state) • When the electron falls back to a lower energy level, the excess energy is emitted as light of a specific wavelength. • Only certain energy levels are allowed. • Sometimes we can see the emitted light because the wavelength is in the visible region.

E = hn E = hn 7.3

n (principal quantum number) = 1,2,3,… Bohr’s Model of the Atom (1913) ---based on emission spectrum of the Hydrogen atom (or one electron species, He+) 2. 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 7.3

( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1 Transition between energy states Ephoton = DE = Ef - Ei 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 = h x n if ni > nf, then emission occurs 7.3

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 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm 7.3

The wave nature of particles De Broglie (1924) reasoned that e- is both particle and wave. If light can behave like a stream of particles (photons,) then perhaps particles such as electrons can behave as waves. Electron bound to nucleus behaves like a standing wave. Standing waves: the circumference of the orbit in an atom is equal to an integral number of wavelengths. 2pr = nl l = h mu Light can act like matter (photons) & like waves (energy) Matter Waves: λ = h/mv where h = Planck's constant m = mass v = velocity Because the energy of electrons depends on the size of orbit (the value of r), its value must be quantized. m = mass of e- u = velocity of e- 7.4

Integral number of standing waves The circumference of the orbit in an atom is equal to an integral number of wavelengths. Not integral; not possible for an orbit 7.4

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 10-34 / (2.5 x 10-3 x 15.6) l = 1.7 x 10-32 m = 1.7 x 10-23 nm The wave properties become observable only for submicroscopic objects. This distinction is due to the smallness of Plank’s constant. 7.4

Schrodinger Wave Equation The beginning of the Quantum Mechanical description of the Atom In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- Very complex mathematical treatment of the atom to calculate the wave function of an electron. Now we discuss the electron within an atom in terms of probabilities of finding the electron within a particular region of the atom - or electron density - called orbitals Probability is the wave function squared.

Quantum Mechanical Description of the distribution of electrons in the atom: • 4 quantum numbers: n (principal quantum number) l (angular momentum quantum number) ml (magnetic quantum number) ms (spin quantum number) Allowed values: • n = 1,2,3 …,n • l = 0,1,2,…,(n-1) • ml = (-l, …, 0. …+l) • ms = +1/2, -1/2

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 n=2 n=3 7.6

Atomic Orbitals---shape? An orbital does not have a defined shape since the wavefunction characterizing the atomic orbitals extends from nucleus to infinity. A boundary surface diagram encloses about 90% of the total electron density in an orbital. These boundary surfaces are complex shapes.

Boundary surface diagram Where 90% of the e- density is found for the 1s orbital 7.6

Schrodinger Wave Equation 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 = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 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 7.6

l = 0 (s orbitals) l = 1 (p orbitals) 7.6

l = 2 (d orbitals) 7.6

Schrodinger Wave Equation 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 7.6

ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2 7.6

Schrodinger Wave Equation Y = fn(n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½ Two possible spinning motions of an electron result in two different magnetic fields. 7.6

Schrodinger Wave Equation 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 7.6

Which of the following sets of quantum numbers in an atom is acceptable? (1, 0, +½, +½) (3, 2, -2, -½) (3, 3, 0, -½). (d) (4, 2, +3, +½) (e) (3, 0, +1, +½) Allowed values: • n = 1,2,3 …,n • l = 0,1,2,…,(n-1) • ml = (-l, …, 0. …+l) • ms = +1/2, -1/2 Group practice problem of chap7:Q6

Energy diagram 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 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

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 7.8

Electron Configurations Periods 1, 2, and 3 Three rules: 1.Electrons fill orbitals starting with lowest n and moving upwards (Aufbau principle: Fill up electrons in lowest energy orbitals ) 2. No more than two electrons can be placed in each orbital. No two electrons can fill one orbital with the same spin (Pauli exclusion principle: no two electrons in an atom can have the same four quantum numbers.) 3. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins ) Period 4 and Beyond the d orbitals begin to fill

Practice problem of chap7:Q12,13 Which of the following violates the Pauli Exclusion Principle? ↑ ↑__ ↑↑ (b) ↑ ↑↓ ↑_ (c) ↑ ↑↓ ↓_ (d) ↑↓_ __ ↑ _ ↑_ ↑_ (e) ↑_ ↑__ ↑ _ ↓_ ↑↓_ Pauli exclusion principle: no two electrons in an atom can have the same four quantum numbers. No two electrons can fill one orbital with the same spin. ) Which of the following violates the Hund’s rule? ↑ ↑__ ↑↑ (b) ↑ ↑↓ ↑_ (c) ↑ ↑↓ ↓_ (d) ↑↓_ __ ↑ _ ↑_ ↑_ (e) ↑_ ↑__ ↑ _ ↓_ ↑↓_ Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron. Practice problem of chap7:Q12,13

“Fill up” electrons in lowest energy orbitals (Aufbau principle) ? ? B 5 electrons B 1s22s22p1 Be 4 electrons Be 1s22s2 Li 3 electrons Li 1s22s1 He 2 electrons He 1s2 H 1s1 H 1 electron 7.9

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 N 1s22s22p3 N 7 electrons O 8 electrons O 1s22s22p4 C 1s22s22p2 C 6 electrons 7.7

Core electrons = electrons in inner shells; Valence electrons = electrons in outer shell

General rules for assigning electrons to atomic orbitals Each shell or principle level of quantum number n contains n subshells. E.g. if n=2, there are two subshells (two values of l) of angular momentum quantum number. Each subshell of quantum number l contains (2l+1) orbitals. E.g. if l=1, there are 3 p orbitals. No more than two electrons can be placed in each orbital. A quick way to determine the maximum number of electrons that an atom can have in principle level n is to use the formula of 2n2.

Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p (attracted by a magnetic field) (not drawn into a magnetic field) 7.8

What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 core electrons valence electrons What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½ 7.8

How many unpaired electrons are present in Al,N,Si,S? Al: 1s22s22p63s23p1 1 N: 1s22s22p3 3 Si: 1s22s22p63s23p2 2 S: 1s22s22p63s23p4 2 Practice problems of chap7:Q8,9

7.8

Outermost subshell being filled with electrons 7.8

Ground State Electron Configurations of the Elements ns2np6 Ground State Electron Configurations of the Elements ns1 ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2 d10 d1 d5 4f 5f 8.2

• Be able to write Electron Configurations of the first 25 elements (also using noble-gas configurations) • Electrons occupy lowest energy levels possible (called the ground state) • Core electrons = electrons in inner shells; Valence electrons = electrons in outer shell • Be able to relate Electron Configurations & the Periodic Table

Electron Configurations and the Periodic Table Shorthand way of writing electron configurations: Write the core electrons corresponding to the filled Noble gas in square brackets. Write the valence electrons explicitly. Example, P: 1s22s22p6 3s23p3 but Ne is 1s22s22p6 Therefore, P: [Ne] 3s23p3

The reason for this irregularity is that a slightly stability is associated with the half-filled (d5) and completely filled (d10) subshells.