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[CHEM 2OA3] - Midterm Exam Guide - Everything you need to know! (74 pages long)
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74 Pages
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Fall 2014

Department
Chemistry
Course Code
CHEM 2OA3
Professor
Paul Harrison
Study Guide
Midterm

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McMaster
CHEM 2OA3
MIDTERM EXAM
STUDY GUIDE
CHEM, chapter one
1-47
We can construct all compounds by combining atomic molecules, forming degenerate
hybrid orbitals.
All the bonds in this molecule has an antibonding MO
Sigma bonds: low in energy, head-on overlap
Sigma star: high in energy, complimentary
Pi star: high in energy
Strongest to weakest: Sigma, Pi, Pi star, Sigma star
1-48
Pi is weaker than sigma if we start with an alkene, we destroy the pi bond during a
reaction, but the sigma bond remains intact.
Bond length: sigma bond between sp: s e closer to the nucleus.
More s character: sp bond is shorter than an sp3 bond.
1-49
Bond length contracts as hybridization becomes more s in character
Bond strength increases as the hybridization becomes more s in character
Double bond is stronger than single bond
1-50
Valence shell electron pair repulsion (VSEPR)
1. Determine the steric number (number of substituents around the atom)
Methane steric number: 4.
Lone pairs count as steric numbers it behaves as a substituent when finding
steric number.
When steric number is four, we need four orbitals (hybridize one s and 3 p in
methane). When steric number is three, sp^2. When steric number is two, sp.
1-52
For any sp^3 hybridized atom, the 4 valence e pairs will form a tetrahedral e group geo
Lone pairs repel bonded pairs more than bonded pairs repel bonded pairs.
Number of lone pairs impacts the bond angles.
1-53
The geometry of ammonia: trigonal pyramidal.
1-54
BF3: six e species, strong lewis acid. Steric number of 3 hybridization is sp2. Unlike
ammonia, because there is no lone pair, there is a trigonal molecular geometry.
1-55
Boro has a uhyridized p orital, ut it is epty, so it does’t out. It’s just a regio
of space where e could be if they wanted to. If we were to react BF3 with an e donor,
the p orbital tells us where the e could go if we were to add them.
1-56
Lone pair, sigma bond to C on left, sigma to C on right, pi bond
Sp^2 hybridized one of the two sp^2 hybrids will accommodate the lone pair, the
others will hang with C
find more resources at oneclass.com
find more resources at oneclass.com
When we have one pi bond to the atom = sp^2 hybridized
1-57
CO2 = sp hybridized to leave behind two p orbitals to make the two pi bonds.
1-59
Molecular polarity: can be predicted by a molecules structure.
Bonds between identical atoms are non-polar bonds.
Bonds between atoms of different EN can be polar partial pos and partial neg charges.
Dipole moment measures polarity, is the amount of charge (0 to 1) times the d the delta
plus and delta minus are separated. Most bonds are of the same length: 1 1.5 A.
Polarity is determined predominantly by the amount of delta in the delta +, delta
Delta units: 1 debye = 10^-18 esu * cm.
One e has a charge of 4.8*10^-10 esu.
1-60
A sigma bond between C and Cl: Cl grabs e in sigma bond, pulls them away from C.
1-61
Actual dipole moment is much less than a dipole moment for a fully ionic bond.
C-Cl is not fully covalent nor ionic. Predominantly covalent with a significant dipole.
What percentage is ionic? 1.8/8.5. Sufficient to determine all reactions of
choloromethane. Much more polar solvent than methane.
1-62
C-O: 10% ionic. Predominantly covalent.
O-H: 33% ionic. Quite substantially ionic large dipole. Alcohols are polar molecules
because of large dipole.
C=O: 41% ionic. Carbonyl group. Extremely polar bond. A lot more polar than C-O
because C-O has a strog siga od, C=O has a eaker pi od, so it’s easier for O to
pull e aay. The pi is ore easily roke tha siga. The e are’t shielded i the ulei.
1-64
Bod dipoles are etors. Whe e add etors, e do the etor su thigy, the et
molecular dipole is straight between the Cls in dicholomethane. Components cancel out.
We need to know the angle between the two bonds. Steric number = 4 = sp^3
hybridized. More polar than an alkane because it has a net dipole moment.
Vector sum: tip to tail.
Basicity of ammonia: dipole moment is pointing in the right direction to accept new e
1-65
Water would be a very polar molecule if it was linear but there are two lone pairs.
Steric number of 4, 4 substituents pointing to the corners of a tetrahedron, bent
molecular shape.
1-66
Red is negative, Blue is positive in electrostatic maps.
Taking e out of chloromethane, they would want to end up on Cl end.
1-67
Common solvents used for chromatography.
Mild solvent: pentane
Alkane, alkene, ether, amine, alcohol, water, alkyl halide, ketone
find more resources at oneclass.com
find more resources at oneclass.com

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[CHEM 2OA3] Comprehensive Midterm Exam guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com CHEM, chapter one 1-47 We can construct all compounds by combining atomic molecules, forming degenerate hybrid orbitals. All the bonds in this molecule has an antibonding MO Sigma bonds: low in energy, head-on overlap Sigma star: high in energy, complimentary Pi star: high in energy Strongest to weakest: Sigma, Pi, Pi star, Sigma star 1-48 Pi is weaker than sigma – if we start with an alkene, we destroy the pi bond during a reaction, but the sigma bond remains intact. Bond length: sigma bond between sp: s e closer to the nucleus. More s character: sp bond is shorter than an sp3 bond. 1-49 Bond length contracts as hybridization becomes more s in character Bond strength increases as the hybridization becomes more s in character Double bond is stronger than single bond 1-50 Valence shell electron pair repulsion (VSEPR) 1. Determine the steric number (number of substituents around the atom) Methane steric number: 4. Lone pairs count as steric numbers – it behaves as a substituent when finding steric number. When steric number is four, we need four orbitals (hybridize one s and 3 p in methane). When steric number is three, sp^2. When steric number is two, sp. 1-52 For any sp^3 hybridized atom, the 4 valence e pairs will form a tetrahedral e group geo Lone pairs repel bonded pairs more than bonded pairs repel bonded pairs. Number of lone pairs impacts the bond angles. 1-53 The geometry of ammonia: trigonal pyramidal. 1-54 BF3: six e species, strong lewis acid. Steric number of 3 – hybridization is sp2. Unlike ammonia, because there is no lone pair, there is a trigonal molecular geometry. 1-55 Boro▯ has a▯ u▯hy▯ridized p or▯ital, ▯ut it is e▯pty, so it does▯’t ▯ou▯t. It’s just a regio▯ of space where e could be if they wanted to. If we were to react BF3 with an e donor, the p orbital tells us where the e could go if we were to add them. 1-56 Lone pair, sigma bond to C on left, sigma to C on right, pi bond Sp^2 hybridized – one of the two sp^2 hybrids will accommodate the lone pair, the others will hang with C find more resources at oneclass.com
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