electron structure of the atom
DESCRIPTION
Electron Structure of the Atom. Chapter 7. 7.1 Electromagnetic Radiation and Energy. Electromagnetic Radiation. EM Radiation travels through space as an oscillating waveform . EM Radiation travels through a vacuum at a constant speed of 3.00×10 8 m/s. Properties of EM Radiation. - PowerPoint PPT PresentationTRANSCRIPT
Electron Structure of the Atom
Chapter 7
7.1 Electromagnetic Radiation and Energy
Electromagnetic Radiation• EM Radiation travels through space
as an oscillating waveform.
• EM Radiation travels through a vacuum at a constant speed of 3.00×108 m/s
Properties of EM Radiation• Wavelength (λ, measured in nm)• Frequency (υ, measured in Hertz, Hz)
Relationship between λ and υ
Electromagnetic Spectrum
Mathematical Relationships
υλ = cυ = Frequency of the light (1/s, or Hz)λ = Wavelength of light (nm or m)c = CONSTANT, Speed of light (3.00×108 m/s)
Mathematical Relationships
Ephoton=hυ Ephoton=(hc)/λυ = Frequency of the light (1/s, or Hz)λ = Wavelength of light (nm or m)c = CONSTANT, Speed of light (3.00×108 m/s)h = Planck’s Constant (6.626×10-34 J×s)Ephoton = Energy of a single photon (J)
Example• Assume we want to determine the
frequency of orange light and the energy of a single photon of this light.
• Orange light = 600 nm = 6.00×10-7 m• υλ = c, therefore υ = c/λ• = 5.00×1014 Hz• Ephoton=hυ=(6.626×10-34 J×s)(5.00×1014
Hz)• Ephoton=3.31×10-19 J
PROBLEM• Calculate the frequency and photon
energy for an X-ray of wavelength 1.00 nm.
• X-Ray= 1.00 nm = 1.00×10-9 m• υλ = c, therefore υ = c/λ• = 3.00×1017 Hz• Ephoton=hυ=(6.626×10-34 J×s)(3.00×1017
Hz)• Ephoton=1.99×10-16 J
PROBLEM• What color is laser with a frequency
of 6.0×1014 Hz?• therefore • = 5.00×10-7 m = 500 nm• 500 nm = Green Light
Continuous vs. Line Spectra
7.2 The Bohr Model of the Hydrogen Atom
Bohr Model of the Atom• Propsed by Niels Bohr• Explains the Emission
Spectrum of Hydrogen• Relies of quantitized
energy levels.• Does not work for
atoms with more than one electron.
7.3 The Modern Model of the Atom
Orbitals and Orbits• Bohr’s model had electrons orbit in
tight paths, but this only worked for Hydrogen.
• Schrödinger expanded the model by using 3 dimensional orbitals
Energy Levels and Orbital Shape
• Electrons are still in quantitized energy levels.
• Orbitals of roughly the same size are in the same overarching, or principal, energy level.
• There are four ground state orbital geometries: s, p, d and f.
Naming Orbitals• Orbitals are named for their principal
energy level and their orbital geometry.
• The n=1 principal energy level has only one geometry, s.
• The n=2 principal energy level has two geometries, s and p.
• n=3 is composed of s, p, and d• n=4 is composed of s, p, d and f.
Orbital Geometries
Orbital Diagrams
Rules for Filling in Orbitals• Ground State Atoms have the same number
of electrons as protons.• Aufbau Principle – Start with the lowest
energy level.• Pauli Exclusion Principle – Max of two
electrons in each orbital with opposite spins• Hund’s Rule – Electrons are distributed in
orbitals of the same energy as to maximize the number of unpaired electrons.
Example
Sodiump= 11e= 11
PROBLEM
Carbon
PROBLEM
Titanium
Electron Configurations• Orbital diagrams are informative but
take a lot of space.• Electron Configurations are a
shorthand for these diagrams.• Though they convey the same
information, they do not show sublevel organization.
Example
Sodiump= 11e= 11
Na 1s2 2s2 2p6 3s1
PROBLEM
Nitrogen
PROBLEM
Iron
7.4 Periodicity of Electron Configuration
Periodic Table
Another Way to Look at It
7.5 Valance Electrons in the Main Group Elements
Main Group Elements
Valance Electrons• Valance Electrons are those electrons in
the last filled principal energy level.• Core Electrons are those below the
valance level.• Valance Electrons for Main Group
Elements are those in the highest s and p orbitals.
• Main Elements in the same group have the same number of valance electrons.
7.6 Electron Configurations for Ions
ExampleSodium ionp= 11e= 10
Na 1s2 2s2 2p6
Ion Electron Configurations• Ion charges are as they are due to
the role of orbitals.• Ions are stable at 1+, 2+, or such
because that gets the electron configuration to a completed principal energy shell (for main group elements).
• Na (1+) is isoelectronic with Neon (a completed n=2)
7.7 Periodic Properties of Atoms
Valance Electrons and Chemistry
• Valance electrons are the ones participating in chemical reactions.
• Compounds are stabilized by reaching a filled principal energy level.
• We will return to this next chapter.
Ionization Energy• Ionization Energy, the amount of
energy required to remove en electron from an gaseous atom (kJ/mol)
• The lower the ionization energy the more reactive a compound is.