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AP Chapter 6
Electronic Structure of AtomsHW: 5 7 29 35 63 67 69 71 73 75
-Electromagnetic Radiation = Emission and transmission of energy in the form of waves
-examples: visible light, infrared, UV, X-Rays-Electromagnetic wave = Travels at the speed of light = 3.0x108 m/s in a vacuum
6.1 – Wave Nature of Light
Electromagnetic Spectrum = (display of electromagnetic radiation by wavelength)
6.1 – Wave Nature of Light-Wavelength = l = Distance between identical points on successive
waves. Unit = Angstrom (A), nm, mm-Frequency = n = # of waves passing a point in a given unit of time
(typically 1 second). Unit = Hz = 1/s-Speed of the wave = c = ln
-Amplitude = Height-Node = Amplitude = 0
A high frequency wave must have a short wavelengthA low frequency wave must have a long wavelength
6.1 – Wave Nature of Light
Calculate:The yellow light given off by a sodium vapor lamp
used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?
(answer = 5.09 x 1014 s-1)
• Wave nature alone cannot explain all behaviors of light– Emission of light from hot objects
(blackbody radiation)
– Emission of electrons from metal surfaces on which light shines (photoelectric effect)
– Emission of light from electronically excited gas atoms (emission spectra)
6.2 – Quantized Energy and Photons
6.2 – Quantized Energy and Photons• He concluded that energy of a single quantum is equal
to a constant times the frequency of the radiation:E = h
where h is Planck’s constant, 6.63 10−34 Js.• Max Planck explained it by assuming that energy
comes in “chunks” called quanta.• Quantum = The smallest quantity of energy that can
be emitted or absorbed• Quantum Theory – Atoms and molecules emit and
absorb energy in discrete quantities only• Photon = A quantum of light energy
• Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light:
c = E = h
6.2 – Quantized Energy and Photonsa) A laser emits light with a frequency of 4.69 x 1014 s-1.
What is the energy of one photon of the radiation from this laser?
b) If the laser emits a pulse of energy containing 5.0 x 1017 photons, what is the total energy of that pulse?
c) If the laser emits 1.3 x 10-2 J of energy during a pulse, how many photons are emitted during the pulse?
Answers:d) 3.11 x 10-19 Je) 0.16 Jf) 4.2 x 1016 photons
6.3 – Line Spectra and the Bohr Model
When observing the emission spectra of atoms/molecules, only a line spectrum of discrete wavelengths is observed.
Emission Spectrum = Spectrum of radiation emitted by a substance/energy source – can be continuous or line spectrum depending on the substance
(continuous spectrum)
(line spectrum)
6.3 – Line Spectra and the Bohr ModelThe energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation:
E = -RH ( )1nf
2
1ni
2-
where RH is the Rydberg constant, 2.18 10−18 J, and ni and nf are the energy levels of the electron-If nf is smaller than ni, then the e- moves closer to the nucleus and DE is negative-If nf is larger than ni, then the e- moves farther from the nucleus and DE is positive-Each line on the line spectrum of Hydrogen can be calculated using this equation.
6.3 – Line Spectra and the Bohr Model
Ground State = Lowest energy state for the electron
Excited State = A higher energy state
• Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
1. Electrons in an atom can only occupy certain orbits (corresponding to certain energies).
2. Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
3. Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by
E = h
6.3 – Bohr’s Model of the Hydrogen Atom
Quantum Mechanics Preview:
6.4 - The Wave Behavior of Matter
• Louis de Broglie posited that if light can have material properties, matter (electrons in atoms) should exhibit wave properties.
• He demonstrated that the relationship between mass and wavelength was:
=hmn
This equation relates the wave ( ) l and particle (m) natures
6.4 – The Wave Behavior of Matter• Heisenberg Uncertainty Principle:
Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely its position known
It is impossible to know both the momentum and the position of an electron
6.5 - Quantum Mechanics
• Erwin Schrödinger developed a mathematical treatment (Schrodinger Wave Equation) into which both the wave and particle nature of matter could be incorporated.
• It is known as quantum mechanics.
6.5 -Quantum Mechanics• Uses advanced calculus• The wave equation is
designated with a lower case Greek psi ().
• The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time = electron density
Orbitals and Quantum Numbers
• Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies.
• Each orbital describes a spatial distribution of electron density.
• An electron is described by a set of four quantum numbers.
Principal Quantum Number, n
• The principal quantum number, n, describes the energy level on which the orbital resides.
• It is a measure of the distance from the nucleus.
• The values of n are integers ≥ 0. • Now 1-7
Angular Momentum Quantum Number, l
• This quantum number defines the shape of the orbital.
• Allowed values of l are integers ranging from 0 to (n − 1).
• We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.
Angular MomentumQuantum Number, l
Value of l 0 1 2 3
Type of orbital s p d f
Magnetic Quantum Number, m l
• Describes the three-dimensional orientation of the orbital.
• Values are integers ranging from - l to l :
− l ≤ m l ≤ l• Therefore, on any given energy level, there can
be up to– s = 0 = one s orbital– p = -1, 0, 1 = three p orbitals– d= -2,-1,0,1,2 = five d orbitals,– f=-3,-2,-1,0,1,2,3 = seven f orbitals
Magnetic Quantum Number, m l
• Orbitals with the same value of n form a shell.– Ex – The n=3 shell has an s orbital, three p orbitals and
five d orbitals
• The set of orbitals with the same shape within a shell form a subshells.– Ex = there are three orbitals in a p subshell)
Spin Quantum Number
• Each e- in one orbital must have opposite spins• Symbol – ms
• + ½ , - ½– Two “allowed” values and corresponds to
direction of spin
6.6 – Representation of OrbitalsThe s Orbital
• Value of l = 0.• Spherical in shape.• Radius of sphere
increases with increasing value of n.
• Radius of sphere increases with increasing energy of the electron(s).
p Orbitals
• Value of l = 1.• Have two lobes with a node (no probability of
finding electron) between them.
d Orbitals• Value of l is 2.• Four of the five
orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center.
f Orbitals
Higher than f not currently needed…
AP EXAM QUESTIONS:
List the four quantum numbers for the valence electrons in magnesium.
List in order – n, l, m l , ms
Valence electrons: 3s2
Electron 1 = 3,0,0,Electron 2 = 3,0,0,-
1
21
2
AP EXAM QUESTIONS:List the four quantum numbers for the valence
electrons in Nitrogen.
Energies of Orbitals
• For a one-electron hydrogen atom, orbitals on the same energy level have the same energy.
• That is, they are degenerate.
6.7 – Many Electron Atoms• As the number of electrons
increases, though, so does the repulsion between them.
• Therefore, in many-electron atoms, orbitals on the same energy level (principle quantum number) are no longer degenerate.
• This is the order we use from the periodic table to fill orbitals = Aufbau Principle = electrons fill from low to high energy
Pauli Exclusion Principle
• No two electrons in the same atom can have exactly the same energy.
• Also means that no two electrons in the same atom can have identical sets of quantum numbers.
6.8 - Electron Configurations
• Shows the distribution of all electrons in an atom
• Consist of – Number denoting the
energy level
Electron Configurations
• Distribution of all electrons in an atom
• Consist of – Number denoting the
energy level– Letter denoting the type of
orbital
Electron Configurations
• Distribution of all electrons in an atom.
• Consist of – Number denoting the
energy level.– Letter denoting the type of
orbital.– Superscript denoting the
number of electrons in those orbitals.
Practice: N, Zr, Bi
Orbital Diagrams
• Each box represents one orbital.
• Half-arrows represent the electrons.
• The direction of the arrow represents the spin of the electron.
Practice: B, Si
Hund’s Rule
“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
(In a p, d or f subshell, fill each orbital with ONE electron before pairing any)
• Diagmagnetism = Repelled by a magnet = Has all paired electrons with opposite spins
• Paramagnetism = Attracted to a magnet = has at least one unpaired electron
• Shielding = Inner electrons block outer electrons from the electrostatic force of the nucleus
• Valence electrons = Outer-shell electrons involved in bonding
Condensed Electron Configurations
• Begin at the NOBLE GAS (Has full valence shell) before the element. Write that symbol in [brackets]
• Continue on with the rest of the configuration
Practice: Na, As, Ag, At
Periodic Table
• Different blocks on the periodic table, then correspond to different types of orbitals.
Some Anomalies
Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.
Some AnomaliesGroup 6:
Ex - Chromium is[Ar] 4s1 3d5
rather than the expected[Ar] 4s2 3d4.Group 11:
Ex - Copper is[Ar] 4s1 3d10
rather than the expected[Ar] 4s2 3d9.
Some Anomalies
• This occurs because the 4s and 3d orbitals are very close in energy and a half-filled d is more stable than “missing” 1 e-.
• These anomalies occur in f-block atoms, as well (Sm and Pu and Tm and Md)