Chapter 7 Chem 109 Notes
Chapter 7 Chem 109 Notes Chem 109
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This 8 page Class Notes was uploaded by mkennedy24 on Tuesday March 15, 2016. The Class Notes belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 19 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.
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Date Created: 03/15/16
Chapter 7: Quantum Mechanical Model of Atom Quantum Mechanics Model: Explains strange behavior of electrons Classical Psychics o Unable to define quantum mechanics due to deterministic views o Deterministic environment is defined as a present set of conditions completely determining the future Quantum Mechanics o Not Deterministic o If quantum mechanics doesn’t confuse you, you haven’t understood quantum mechanics yet Section 7.2: Light Objective: Describe Characteristics of light o Wave: Disturbance in a medium o Electromagnetic Radiation: Type of energy embodied in oscillating electric and magnetic fields Magnetic Field: Region of space where a magnetic particles experiences a force (i.e. When holding magnetic material close to a magnet, you can feel a pull between the two, that is a force) Electric Field: Region of space where an electrically charged particle experiences a force (i.e. A proton has an electric field around it. If you bring another charged particle into that field, that particle will experience a force) o Amplitude: Height from the midline to the maximum/minimum of the wave Different amplitudes, different brightness (intensity) o Wavelength: Measured distance from crest to crest Different wavelengths, different colors in the spectrum o Frequency: Count of wavelengths passing a given point in a given period of time Inversely proportional to wavelength ( λ ) Directly proportional to speed Speed of Light (c) 8 o c= 3.0∙10 m/s o c=λυ Frequency (Hz) Wavelength (m) o Example: What is the frequency of 650nm of light? −9 Convert 650 nm to meters by just adding (x10 ) **Prefixes!** 8m −9 14 o Ex(mple: Cs)culate the wavelength (in nm) of the red light emitted by a barcode scanner that has a frequency of If you are given the frequency you can easily figure out the wavelength using this equation: c=λυ , by rearranging the equation to fit what we need we get: λ= c υ Convert wavelength from meters to nanometers by using the conversion factor between the two: (1nm= 10 - 9 m) 4.62∙10 s −1 Electromagnetic Spectrum o Interference: Two waves are in the same space Constructive: “In phase”; peak overlaps peak making the overall amplitude of the wave increase Destructive: “Out of phase”; Amplitude of the wave decreases because peaks overlap toughs, canceling out the wave (as seen above) Diffraction: Bend of wave passing through object Diffraction Pattern: Series of bright and dark spots past a double slit Photoelectric Effect: When a metal can emit electrons when light hits it o Example: Care head light turns on when its dim enough outside o Specific energy required to emit electron o Threshold Frequency: Minimum frequency to eject electrons No electrons are emitted from metal, no matter how long the light shines on the metal Low-frequency (long wavelength) light does not eject electrons from metal regardless of its intensity or its duration High-frequency (short-wavelength) light does eject electrons, even if its intensity is low Albert Einstein: “Light energy must come in packets” Photon: Quantized packet of energy E=hν o Example: A nitrogen gas lazer with a wavelength of 337 nm contains 3.83 mJ of energy. How many photons does it Frequency(Hz) Planck’s Constant: contain? Given: Epulse.83mJ λ=337nm Find: Number of Photons 10 m −7 λ=337nm∙ =3.37∙10 m 1nm −34 8m hc (6.626∙10 J ∙s)(3.0∙10s ) E Photon = =5.8985∙10 −1J v 3.37∙10−7m −3 3.83mJ ∙10 J =3.83∙10−3J 1mJ E −3 Numberof Photons= Pul=e 3.83∙10 J =6.49∙10 photons EPhoton5.8985∙10−19J Section 7.4: The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy de Broglie Relation: h λ= mv o **Notice: The velocity of a moving electron is related to its wavelength: Knowing one is equivalent to knowing the other Heinsenberg’s Uncertainty Principle o Conclusion: Best we can do is probability Section 7.5: Quantum Mechanical Model A constant Uncertainty in position Uncertainty in velocity Instead of orbits. . . o Orbitals: Probability distribution map showing where elctrongs most likely are found o Schrodingers Equation: Wave function( ψ ) describes wave behavior of electrons ( ψ ) = orbital o Quantum Numbers Objective: Describe what characteristic each quantum number represents Objective: Know which Quantum Numbers are allowed n principle number Overall size and energy that we actually have (Bohr put it as a distance away from the nucleus) n= 1,2,3,. . . ∞ (positive integers for n- values) l angular quantum number Shape of orbital “Subshell” If n=1, then the only possible number for l is 0 according to equation for l= n-1 so 1-1=0 If n=3, the only possible number for 1 is (n-1 3-1=2) 0,1,2 l= o 0 s orbital o 1 p orbital o 2 d orbital o 3 f orbital M lM sub l) Magnetic Quantum Number Orientation in space “orbital” Integers of M = -l, . . . -2, -1, 0, 1, 2, . . . +l l M sM sub s) Spin Quantum Number (not required by Schrodingers equation) Electron spin M s +1/2 , -1/2 o Objective: Know relationships between n, l, M l Subshells in a principal level (n) is always equal to the n-value Number of orbitals in a subshell is always equal to 2(l)+1 Number of orbitals in an energy level is always n 2 Energy level in atom and spectra Bohr’s equation Hydrogen atom o En=−2.18∙10 −18J/n2 Example: What is the wavelength of light emitted when electron move from n=2 to n=1 in a hydrogen atom? −18 1 1 −18 1 1 −18 Δ E=−2.18∙10 J ( )− 2 →=−2.18∙10 J ( )− 2→ΔE=−1.635∙10 J→ nF nI 1 2 −34 8m (6.626∙10 J∙s)(3.0∙10 ) −1.635∙10 −1= s →λ=1.216∙10 meters λ Section 7.6: Shapes of Atomic Orbitals Objective: Describe orbital shape given n and l o S orbital (n=1 & l=0) **For s orbitals, l is always 0** Probability Density: Probability (Volume @ shell=r) Unit volume Radical Distribution Probability o “Node”: Zero probability; at the nucleus, the volume is 0 therefore making the probability 0 of finding an electron in the nucleus 1s orbital Spherical 2s orbital Sphere inside a sphere with a node in the middle 3s orbital Sphere inside a sphere inside a sphere o P Orbital (n=2 & l=1) 2p orbital Above are all the orientations of a p orbital 3p orbitals are just like 2p but there is a lope inside each lope like such: And so on with increasing numbers o D orbitals Same with inside layers increasing with numbers The shape of d orbitals below: o F orbitals Same with inside layers increasing with numbers The shape of F orbitals below: