Quiz 5_Sol

Quiz 5_Sol - Ch1a Qui z 5 Solutions 1. (40 points) Consider...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ch1a Qui z 5 Solutions 1. (40 points) Consider the heteronuclear diato mic free-radical molecule NO. a. (5 points) Draw a Lewis dot structure for NO and calcu late the formal charge on each ato m. N O The formal charge on both ato ms is zero. b. (20 points) Using the 2s and 2p atomi c orbitals on each atom, construct a molecular orbital diagram for NO and include the electrons. Be sure that the relative energies o f the various ato mic and molecular orbitals are clear and label each molecular orbital with an appropriate identifier (2sσb, etc.). Fo r NO, the 2px,yπb molecular orbitals are l ower in energy than the 2pzσb. 2 p z! * 2 p x ,y " * 2p 2p 2 p z! b 2 p x ,y " b 2s 2 s! * 2s 2 s! b N NO O c. (6 points) Based on the molecular orbital diagram, calculate the bond order of NO and rank NO, NO+, and NO- in order of increasing stability. Specify whether NO is paramagnetic. There are 8 bonding electrons and 3 antibonding electrons, so the bond o rder is (8-3)/2 = 2 .5. Since NO+ has one fewer antibonding electron than NO and NO- has one more antibonding electron than NO, NO- is the least stable, follo wed by NO and then NO+. Since there is an unpaired electron, NO is paramagnetic. d. (9 points) Sketch the 2sσ*, 2pzσb, and 2pxπ* molecular o rbitals and indicate the phase o f each lobe and all nodal planes perpendicular to the bond axis. Be sure your diagrams also indicate the relative electron density around each atom. For much more accurate sketches than the ones belo w, see Gray 97. For the bonding MO, there is more electron density around the oxygen since it is more electronegative. For the t wo antibonding MOs, there is more electron d ensity around the nitrogen. The nodes are the dashed lines. + N - O 2 s! * - N + O - 2 p z! b + O N 2 p x "* - + 2. (20 points) Cyclooctatetraene (COT) is an eight-membered ring and has the fo rmula C8 H8. COT is not a planar molecule, but both COT2+ and COT2- are planar molecules. a. (10 points) Us e the eight unhybridized 2p orbitals on th e carbon ato ms of COT2+ to create eight π molecular orbitals and arrang e these orbitals in a molecular orbital diagram with the highestenergy orbital at the top and the lo west-energy o rbital at the bottom. Sketch each orbital as viewed fro m above the plane o f the molecule and indicate the positions of all nodes . !* !* n .b . !b !b It would be more accurate to draw adjacent p o rbitals of the same phase as one large seg ment of electron density, but either way is acceptable. There are also other ways to draw so me o f the molecular orbitals, but it is import ant that each MO con tain the correct nu mber of nodes . The nodes are indicated by d ashed lines. b. (5 points) Speci fy whether each π molecular orbital abo ve is bonding, nonbonding, or antibonding and indicate which orbitals contain electro ns. See the above diagram. c. (5 points) Co mpare the bond o rders o f the carbon atoms in COT2+ and COT2-. Since the four additional electrons in COT2- go into the nonbonding MOs , the bond orders are the same in both COT2+ and COT2-. 3. (40 points) When looking at its resonance structures, o ne can easily see ho w electrons in NO2- must be delocalized . Using hybridization and molecular orbit al theory we can understand this a bit mo re deeply. N O O O N O a. (10 points) Assu me that each O atom is sp2 hybridized. Ho w many total electrons occupy the hybridized orbitals? Ho w many are involved in π-bondi ng? These should not exceed the nu mber you start out with! Since the geo metry around the N ato m is planar, and its SN = 3, it is sp2 hybridized. This means that each atom in the molecule has three sp2 orbitals as shown belo w (18e- total): This makes 14 e- in the hybridized o rbitals, which leaves 4e- for the π-system. For full credit, only the number of electrons in hybridized orbitals and unhy bridized p orbitals need to be indicated. It is not necessary to draw the diagram above or sho w an y additional work . b. (20 points) Draw the molecular orbitals in the π-system of NO2-, clearly indicating the relative energies, electron densities, and phases around each atom and the locations of any nodes . Hint: The oxygen ato ms are too far awa y to interact with each other. The MO diagram is belo w: !" !n b 2p x 3 !b c. (10 points) Fill in your MO diagram with the appropriate number of electrons. Are the double bond and the neg ative charge really sep arate, as implied in each individual resonance structure? Is it an average of the two ? Explain. The filled MO diagram is belo w: !" !n b 2p x 3 !b No, both the double bond and the negative charge (lone pair) are simultaneously delocalized in the π-system. There is al ways a partial double bond and a p artial negative charg e on each oxygen, which is consistent with the “av erage” of the t wo reson ance structures. ...
View Full Document

This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.

Ask a homework question - tutors are online