Friday, December 17, 2010
Fri-Day 1
Physics - took part 2 of our Work and Energy unit exam.
There was a correction to the last question on the multiple choice; a newborn human infant cannot way 900 Newtons. Think of it this way, 1 kg is about 10 N of weight. So, to convert from Newtons to kg, just divide by 10, and vice-versa. A 90 kg baby at about 2 lbs. per kg would way 180 pounds! ouch.
For another rough approximation, there are about 5 Newtons to every 1 lb. of weight. So, if you weigh 100 lbs., in metric units, you weigh 500 N.
We finished the write-up for our power/bavaa lab.
On Monday, we start the next major section of Physics: Electricity and Magnetism (mostly electricity in our course). In a college course, this part of physics is given in a different semester.
We will have to do a lot of drawing because the particles that we will deal with are invisible; however, with the correct picture, the forces and other factors are not difficult to deal with. You just have to remember that like charges repel, and that opposite charges attract!
AP Chem - did a couple of reinforcement bonding/molecular geometry/hybridization , and IMFA examples.
We will continue on with the application of intermolecular forces of attraction (IMFA) in discussing solids, liquids, and gases before our break.
The Bonding unit exam will be given next Wednesday.
I will post more practice questions on Blackboard.
The question types on the next test will be as follows:
1.Most of our time and therefore most of our test will be on all of the types of molecules or ions that we drew out in our grand chart. So anything and everything on that chart will be asked for such as:
- how to draw the correct Lewis dot structure including any and all reasonable (by the rules) resonance structures for any atom, molecule, or ion (given one central atom)
-Be able to show the orbital hybridization that occurs on a CENTRAL atom starting with the orbital diagram of the un-hybridized atom. Then show the electron shifts that occur in order to form the proper bonding arrangement in the molecule. For example, in CH4, carbon has a valence electron configuration of 2s2 2p2, which had only two unpaired electrons available for bonding. Not only that, the unpaired p electrons are in perpendicular p orbitals. So, in order to explain the four bonds that form in a tetrahedral arrangement, the C must hybridize its 2s and all three 2p orbitals to form FOUR equal energy, tetrahedrally oriented sp3 orbitals. As bonds form, these sp3 orbitals overlap with the Hydrogen 1s orbital (that is all that H has occupied with one electron; generally don't worry about hybridization of terminal atoms, unless there is only one type of hybridization that can apply) and the opposite spin electron pairs (one from each atom in this case) form a bond between each pair of atomic nuclei.
- know how to calculate and apply formal charge ( you must SHOW your calculation; the circle method is fine - bring a PENCIL for that part) in determining the most significant resonance structure for a molecule or ion.
- how to determine bond order and its relation to bond length AND bond strength (BDE).
- know the rules for electron deficient species, odd-electron molecules, and expanded octets.
- explain the dimerization of odd-electron species/ radicals, and how/why they are relatively so reactive.
2. Be sure to draw OXOacids correctly: for OXOACIDS, each "H" in the formula IS NOT BONDED TO THE CENTRAL ATOM, but rather to one of the Oxygen terminal atoms, which is why they are called oxoacids. So, in nitric acid, HNO3, the H is bonded to one of the terminal oxygens that is bonded to the N. Don't forget that. Also, keep in mind that Lewis structures for ANY ions, whether monatomic or polyatomic ions, ALWAYS get put in brackets with the magnitude of the charge FIRST and the sign of the charge NEXT outside the brackets; do not say that ions are polar or nonpolar - ions have full INTEGER charges of just ONE sign. Ions can only form multiple bonds to all surrounding oppositely charged ions! Ions can ATTRACT molecules via ion-molecule (duh!) attractions.
3. covalent vs. ionic bonds vs. metallic bonds
4. Molecular polarity and its relation to IMFA- intermolecular forces of attraction - types
The polarity of covalent bonds may result in the formation of permanent dipoles in a molecule if:
there is a net PARTIAL positive region of a molecule with a net partial negative region on the opposite region of the molecule
- this usually results from dipoles that are NOT equal and opposite OR dipoles that, due to lack of symmetrical distribution, do NOT cancel out.
5. Explain and show resonance stabilization in a given molecule via valence bond theory/orbital hybridization, and by showing overlapping p orbitals containing a pair of electrons; the electrons in these overlapping p orbitals are above and below the 3 or more bonded atoms' nuclear plane.
Relate the fractional bond order to integer bond order lengths and strengths.
6. Know the difference between and the REASON for the relative strengths of sigma bonds and pi bonds.
7. For intermolecular attractions, know the different types: induced dipole, dipole-dipole, and extreme dipole-dipole (hydrogen "bonding") attractions. Based on electronic and molecular geometry from your Lewis structure and VSEPR, predict the molecular polarity and the expected dominant intermolecular types!! of attraction. For example, CF4 has tetrahedral electronic and molecular geometry, which results in a symmetric distribution of charge/electrons. The molecule is therefore nonpolar because there is, on average, no net partial positive or partial negative "side" of the molecule so its intermolecular forces of attraction are induced dipole attractions.
REMEMBER, BOTH polar and non-polar molecules have induced dipole attractions BUT polar molecules ADDITIONALLY have dipole-dipole or, if there is an H-F, H-O or H-N within the molecule, there will be EXTREME dipole-extreme dipole attractions among/between the molecules.
There was a correction to the last question on the multiple choice; a newborn human infant cannot way 900 Newtons. Think of it this way, 1 kg is about 10 N of weight. So, to convert from Newtons to kg, just divide by 10, and vice-versa. A 90 kg baby at about 2 lbs. per kg would way 180 pounds! ouch.
For another rough approximation, there are about 5 Newtons to every 1 lb. of weight. So, if you weigh 100 lbs., in metric units, you weigh 500 N.
We finished the write-up for our power/bavaa lab.
On Monday, we start the next major section of Physics: Electricity and Magnetism (mostly electricity in our course). In a college course, this part of physics is given in a different semester.
We will have to do a lot of drawing because the particles that we will deal with are invisible; however, with the correct picture, the forces and other factors are not difficult to deal with. You just have to remember that like charges repel, and that opposite charges attract!
AP Chem - did a couple of reinforcement bonding/molecular geometry/hybridization , and IMFA examples.
We will continue on with the application of intermolecular forces of attraction (IMFA) in discussing solids, liquids, and gases before our break.
The Bonding unit exam will be given next Wednesday.
I will post more practice questions on Blackboard.
The question types on the next test will be as follows:
1.Most of our time and therefore most of our test will be on all of the types of molecules or ions that we drew out in our grand chart. So anything and everything on that chart will be asked for such as:
- how to draw the correct Lewis dot structure including any and all reasonable (by the rules) resonance structures for any atom, molecule, or ion (given one central atom)
-Be able to show the orbital hybridization that occurs on a CENTRAL atom starting with the orbital diagram of the un-hybridized atom. Then show the electron shifts that occur in order to form the proper bonding arrangement in the molecule. For example, in CH4, carbon has a valence electron configuration of 2s2 2p2, which had only two unpaired electrons available for bonding. Not only that, the unpaired p electrons are in perpendicular p orbitals. So, in order to explain the four bonds that form in a tetrahedral arrangement, the C must hybridize its 2s and all three 2p orbitals to form FOUR equal energy, tetrahedrally oriented sp3 orbitals. As bonds form, these sp3 orbitals overlap with the Hydrogen 1s orbital (that is all that H has occupied with one electron; generally don't worry about hybridization of terminal atoms, unless there is only one type of hybridization that can apply) and the opposite spin electron pairs (one from each atom in this case) form a bond between each pair of atomic nuclei.
- know how to calculate and apply formal charge ( you must SHOW your calculation; the circle method is fine - bring a PENCIL for that part) in determining the most significant resonance structure for a molecule or ion.
- how to determine bond order and its relation to bond length AND bond strength (BDE).
- know the rules for electron deficient species, odd-electron molecules, and expanded octets.
- explain the dimerization of odd-electron species/ radicals, and how/why they are relatively so reactive.
2. Be sure to draw OXOacids correctly: for OXOACIDS, each "H" in the formula IS NOT BONDED TO THE CENTRAL ATOM, but rather to one of the Oxygen terminal atoms, which is why they are called oxoacids. So, in nitric acid, HNO3, the H is bonded to one of the terminal oxygens that is bonded to the N. Don't forget that. Also, keep in mind that Lewis structures for ANY ions, whether monatomic or polyatomic ions, ALWAYS get put in brackets with the magnitude of the charge FIRST and the sign of the charge NEXT outside the brackets; do not say that ions are polar or nonpolar - ions have full INTEGER charges of just ONE sign. Ions can only form multiple bonds to all surrounding oppositely charged ions! Ions can ATTRACT molecules via ion-molecule (duh!) attractions.
3. covalent vs. ionic bonds vs. metallic bonds
4. Molecular polarity and its relation to IMFA- intermolecular forces of attraction - types
The polarity of covalent bonds may result in the formation of permanent dipoles in a molecule if:
there is a net PARTIAL positive region of a molecule with a net partial negative region on the opposite region of the molecule
- this usually results from dipoles that are NOT equal and opposite OR dipoles that, due to lack of symmetrical distribution, do NOT cancel out.
5. Explain and show resonance stabilization in a given molecule via valence bond theory/orbital hybridization, and by showing overlapping p orbitals containing a pair of electrons; the electrons in these overlapping p orbitals are above and below the 3 or more bonded atoms' nuclear plane.
Relate the fractional bond order to integer bond order lengths and strengths.
6. Know the difference between and the REASON for the relative strengths of sigma bonds and pi bonds.
7. For intermolecular attractions, know the different types: induced dipole, dipole-dipole, and extreme dipole-dipole (hydrogen "bonding") attractions. Based on electronic and molecular geometry from your Lewis structure and VSEPR, predict the molecular polarity and the expected dominant intermolecular types!! of attraction. For example, CF4 has tetrahedral electronic and molecular geometry, which results in a symmetric distribution of charge/electrons. The molecule is therefore nonpolar because there is, on average, no net partial positive or partial negative "side" of the molecule so its intermolecular forces of attraction are induced dipole attractions.
REMEMBER, BOTH polar and non-polar molecules have induced dipole attractions BUT polar molecules ADDITIONALLY have dipole-dipole or, if there is an H-F, H-O or H-N within the molecule, there will be EXTREME dipole-extreme dipole attractions among/between the molecules.
# posted by C @ 2:42 PM 
