CHEM 101 Midterm: Exam 3 Material
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Department
Chemistry
Course
CHEM 101
Professor
Ruth Topich
Semester
Spring

Description
Exam 3 Material Periodicity and the Electronic Structure of Atoms  Wavelength- the distance between successive wave peaks  Frequency- the number of wave peaks that pass a given point per unit time  Amplitude- the height of the wave maximum from the center  What we perceive as different kinds of EM energy are waves with different wavelengths and frequencies  Wavelength x frequency = speed of light (3.00 x 10^8 m/s) o Wavelength in meters (m) o Frequency in /s (1 /s = 1Hz)  Example: What is the wavelength associated with a frequency of 106.5 MHz? o 106.5 MHz= 106.5 x 10^6 Hz  =106.5 x 10^6 /s  Example: ?= v (frequency) of red light with a wavelength of 700nm (1nm= 1 x 10^-9 m) o V= 4.29 x 10^14 Hz  Particles of light = photons  E=hv per photon o h= Planck’s constant (6.626 x 10^-34 J•s) o Ex: ? = energy associated with a photon of light with a wavelength = 700 nm red  V= 4.29 x 10^14 Hz  E=hv: E= (6.626 x 10^-34 J•s)(4.29 x 10^14 /s)  E= 2.84 x 10^-19 J per photon  Note: you could also be asked in J/mol of photons or kJ/mol of phtons  1.71 x 10^5 J/mol x 1 kJ/ 1000 J = 171 kJ/mol of photons  Ex: ? = E associated with a photon of light with a wavelength = 2.625 x 10^3 nm: o Equation: E=hc/wavelength (in m) o C= 3.00 x 10^8 m/s (you must convert nm m)  C= 2.625 x 10^3 nm x 1x10^-9m/1 nm= 2.625 x 10^-6m  E= (6.626 x 10^-34 J•s)(3.00 x 10^8 m/s)/ 2.625 x 10^-6 m  E= 7.57 x 10^-20 J/photon  Bohr Model o When energy is absorbed by an electron it jumps up a level (or rung on the ladder) o Energy given off is a higher number to a lower number (largest is 21) o Lyman series (UV) has largest amount of energy  Balmer series (vis)  Paschen series (IR) o Shortest wavelength = furthest drop on ladder because of highest energy content  Quantum Mechanics o Wave function is characterized by 3 parameters  Quantum numbers n, l, m  Describes energy level and 3-dimensional shape of region  N=1 is closest to nucleus o Shape of orbitals:  S-orbital (theres one in every energy level) (spherical):  1s belongs to n=1 0 nodes  2s:n=2  1 node  3s: n=3 2 nodes  Note: node is area in space where you cannot find an electron (# nodes= n-1)  2p-orbitals (shaped like figure 8) (come in groups of 3):  have equal energy  the principle quantity: #s=# subshells  D-orbitals (come in groups of 5) (four leaf clover shape):  Dxy, dyz, and dxz have lobes between axes  Dx^2-y^2 has lobes on every axis  Dz^2 has figure 8 on z axis and a ring around the middle Principle Energy Level Orbitals N=1 1s (1) N=2 2s (1) 2p (3) N=3 3s (1) 3p (3) 3d (5) N=4 4s (1) 4p (3) 4d (5) 4f (7) N=5 5s (1) 5p (3) 5d (5) 5f (7) 5g (9)  N^2= total orbitals o Total number of electrons in a particular level= N^2 x 2 o E.g. (4)^2 x 2= 32  How are the orbitals in atoms occupied? o Aufbau principle o Pauli Exclusion Rule- an orbital can only hold 2 electrons o Hund’s Rule- if 2 or more degenerate orbitals, 1 electron goes into each until all are half full  2 different ways to express with Hund’s Rule:  electron configuration: 1s^1  orbital filling diagram o valence electrons are the only ones to be involved in chemical bonding (outermost shells)  inert gas: full outer shell (nonreactive)  26 Fe is an example of a paramagnetic element- having unpaired electrons o diamagnetic- all electrons are paired  Anomalous Electron Configuration o There are 19 but you only need to know these 5:  24 Cr (similar- Mo): 4s^1 3d^5  29 Cu (similar- Ag): 4s^1 3d^10  also the one to the right of Mo but not really  Quantum Num
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