Last Set of Semester Notes
Last Set of Semester Notes CHEM - 10060 - 001
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This 4 page Class Notes was uploaded by Nick Manning on Saturday May 7, 2016. The Class Notes belongs to CHEM - 10060 - 001 at Kent State University taught by TBA in Fall 2015. Since its upload, it has received 10 views. For similar materials see GENERAL CHEMISTRY I in Chemistry at Kent State University.
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Date Created: 05/07/16
I will stress going to class for this next week, as a lot of in class examples are being shown each day that will help so much more with concepts and the structure which the professor will ask questions. I’ll attempt to recreate an example right now for you: CH2O -2 C---H bonds -1 C===O double bond -C (sp2) and H(s) -one sigma and one pi -2 sigma bonds b/c -sp2 from C of single of bonds *All Single bonds = sigma bonds; all pi bonds only use p orbitals* ROTATION ABOUT BONDS - Sigma bonds rotate freely - Pi bonds cause restriction in the rotation due to pulling it to the bonded atom, solidifying it in place MOLECULAR ORBITAL THEORY While Valence Bond Theory is good at explaining observed geometry & equivalence of bonding orbitals, it fails to explain magnetic and spectral qualities of the atom and it leans more towards the particle view, while electrons contain both wave and particle qualities. So, we can also use the Molecular Orbital model. - Constructive Interference- when the troughs and crests of two waves are lined up in two waves being added together, resulting in an overall larger wave - Destructive Interference- when the troughs and crests of two waves are opposite (i.e. one crest lines up with one trough) and are added together, resulting in the destruction of the wave overall This is important for the Molecular Orbital (MO) model because these are also two forms of atomic bonding. CONSTRUCTIVE - Sigma bonds and LESS energy than destructive, due to two nuclei puling on two electrons DESTRUCTIVE - Called an Antibonding sigma bond (σ*), HIGHER energy, has space called a “node” between atoms. BOND ORDER Bond order is calculated by the equation: (bonding electrons - antibonding electrons)/2 If the bond order is GREATER THAN zero, the species IS ABLE to exist. IF NOT, the species CANNOT exist. Here we see the MO for two Hydrogen atoms. We fill the bonding atoms (bottom) first and then the antibonding (top) second. We can see there are two bonding electrons and 0 nonbonding, so (2-0)/2 = 1; the species can exist. We are responsible to know the ways that all of Period 2 elements react. This includes B, C, N, O, F, and Ne. The steps to make orbital diagrams for these are a little different, and are technically only partial orbital diagrams, because the 1s electrons are omitted. - Find the electron Configuration for the element - Make the shell the correct way (different for B, C, N and for O, F, and Ne; will be given to us) - Calculate the valence electrons, these will go in the middle boxes -Enter in the electrons (the arrows) on the side atoms -Enter in the electrons in the middle boxes, filling each box from the bottom up. You have to notice the 2 parts, the 2s and the 2p. The Electron Config. for Nitrogen is 2s2 2p3 so that is filed on either side; if it was a cation or anion, the # of arrows would change both in the middle and on one side according to how many electrons are gained/lost.
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