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[CHEM 231] - Final Exam Guide - Comprehensive Notes fot the exam (38 pages long!)


Department
Chemistry and Biochemistry
Course Code
CHEM 231
Professor
Lee Friedman
Study Guide
Final

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UMD
CHEM 231
FINAL EXAM
STUDY GUIDE

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Lecture 1 1/26/17
Bonding
o Ionic-formal electron transfer, electrostatic attraction
o Covalent-electron sharing, bonding pair
o Polar covalent-unequal bonding
o Determine type of bond by change in electronegativity
o EN less than or equal to 0.4: covalent
o EN greater than or equal to 2.0: ionic
o EN greater than 0.4, less than 2.0: polar covalent
Ions
o Determine negative charge by dipole moment or formal charge
o Ex. Hydroxide- ide indicates negative chage, dipole points to O
o Formal charge=group number-electrons
Lecture 2 1/31/17
Formal Charge
o Carbon monoxide acts neutral even though it is polar because of formal charge
o Formal charge opposes dipole in this case
o In BF4 formal charge does not make sense with electro-negativities
o Formal charge is not always correct, must look at electron distribution
VSEPR
o Electron group-bonded atom or lone pair
o Shape-only atom positions
o 2 electron groups- angle:180 degrees linear
o 3 electron groups- angle:120 degrees trigonal planar or triangular
o 4 electron groups- angle:109.5 tetrahedral
o 4 electron groups- 1 un-bonded trigonal pyramidal
o 4 electron groups 2 un-bonded bent
Electrons
o Electrons occupy orbitals in atoms
o The sharing/transfer of electronsbonds
o 1s orbital: n=1, l=0, ml=0, spherical shaped
o Schrodinge’s euation- gives equation for 1s orbital
o Cartesian coordinates: defined by x, y, z
o Polar coordinates: r=radius, theta=angle from z, phi=angle from x axis
o 1s orbital: n=1, l=0, m=0, Phi1s is proportional to e^-r
o Electron more likely to be closer to the nucleus
o Phi is proportional to the probability of an electron being somewhere
o 2s orbital: n=2, l=0, m=0, Phi2s is proportional to (2-r)e^-r
o When r is less than 2, Phi is greater than 0
o When r is equal to 2, Phi is equal to 0
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o When r is greater than 2, Phi is less than 0
o 2p orbital: l=1, m=0, z configuration
o l=1, m=+/-1, x/y configuration
o electron is equally likely to be in both lobes because the sign is squared and
therefore the number is the same
Bonding
o Ex. H+HH-H
o Overlap of 2x 1s orbitals
o Constructive interference
o Two Phi1s orbitals combine
o Overlap constructively1new orbital-bonding orbital
o Must conserve orbital quantitiesorbital subtraction
o Phi1s-Phi1s=Phi1s+(-Phi1s)
o Wave function from positivenegative-destructive interference
o Results in antibonding orbital
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