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Midterm 2 Review

Midterm 2 Review. Line Structures. Problem 1: Draw the line structure of HOCH 2 CH 2 CCHCOOH. What is the hybridization of each atom. We will do this on the worksheet. .

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Midterm 2 Review

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  1. Midterm 2 Review

  2. Line Structures • Problem 1: Draw the line structure of HOCH2CH2CCHCOOH. What is the hybridization of each atom. We will do this on the worksheet.

  3. What is the empirical formula of the structures shown below. What is the hybridization on the nitrogen? What is the molecular and electron geometry on the oxygen in the OH? How many s bonds are there? How many p bonds are there? We will do this on the worksheet

  4. Problem 3: Draw the Lewis structure of O3. Give the hybridization, electron geometry, molecular geometry and decide if its polar. What orbitals is each bond formed. What is the formal charge on each oxygen, what is the bond order. (Good practice tip: you can answer these questions for any molecule you are asked to draw) We will do this on the worksheet.

  5. Question 1: • Question 1: Describe the difference between bonding and anti-bonding orbitals. How do pi and sigma orbitals compare. For each: In general, how can you recognize these with pictures? How are they made? How are their energies related? • Bonding vs. Antibonding: • Bonding: Lower in energy, formed by constructive interference, recognizable by most of the electron density being inbetween the nuclei. • Antibonding: higher in energy, formed by destructive interference, recognizable by most of the electron density not being inbetween the nuclei. 1s-1s Anti-bonding MO, σ*1s E isolated H atoms 1s+1s Bonding MO, σ1s

  6. Question 1:Cont…. • Question 1: Describe the difference between bonding and anti-bonding orbitals. How do pi and sigma orbitals compare. For each: In general, how can you recognize these with pictures? How are they made? How are their energies related? • s vs p • s : symmetrical around bond axis, “end-on-end” interference • p : nodal plane, side on side interference • Energy levels are dependent on the atoms involved and which s vs p you are talking about. 1s-1s Anti-bonding MO, σ*1s E isolated H atoms 1s+1s Bonding MO, σ1s

  7. Summary:

  8. Question 2: • If given two resonance structures of equal stability, how is the sigma and pi bonding electrons shared amongst the atoms in the molecule? How does this affect bond order? How does this affect the location of formal charges? Use ozone from above and benzene as an examples to help explain this. • s bonds are shared between bonding atoms. p orbitals “merge” to for delocalized electron cloud. • The bond order is an average of all the structures since the electrons in the p bonds are shared over the delocalized electron cloud. • The formal charges are an average of the structures since the electron density is shared over the delocalized electron cloud.

  9. Ozone Showing s bonds Showing bonds as drawn in lewis structures, aka not how they actually are! p s s Showing delocalized electrons

  10. Benzene

  11. Question 3: • What makes a bond more ionic than others? Be specific. Describe the behavior of both species in the bond and how each contributes to the ionic or covalent character. Key words to use, electronegativity, polarizable, polarizability, cation, anion, highly positively charged, highly negatively charged, atomic radius. • The larger the difference in electronegativity the more ionic it is. If they are too close in electronegativity they are considered polar covalently bound. If there is no difference in electronegativity, it is a covalent bond. • Ions that are highly positively charged and very small have high polarizing power. The higher the polarizing power, the more covalent the bond. • Ions that are highly negatively charged and very large have high polarizablity. The higher the polarizability the more covalent character there is. http://www.chem.uky.edu/courses/che514/Fajan.pdf NaCl and NaI are both ionic, but which has more covalent character? NaI Highly polarized anion, covalent bond Idealized ionic bond Polarized anion, some covalent character

  12. Question 4: • Why do electron affinity and electronegativity share the same trend? Why does electron affinity have p-block exceptions but electronegativity does not even though they share the same trend (be specific and give one example in your answer)? You are on cash cab and you are asked where the d-block exceptions to electron affinity are. First you curse me for not having that be required exam knowledge, but then you think back on the p block exceptions and your logic used above. What do you guess for the d-block exceptions in order to claim your prize? • Example: P, Si, S is the order of the third period electron affinity. The trend would follow that it should be Si, P, S. • Electron affinity is based on both effective nuclear charge AND electron configuration. Since P is [Ne]3s23p3 it is half filled and less likely to gain another electron. • Half and fully filled areas should be where the exceptions occur. So nd5<nd4 and nd10<nd9

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