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Chem 351 G. E. Hofmeister

Spring 2005

Problem Set 2--Answer Key 1. a. Prepare a molecular orbital energy level diagram for the cyanide ion (CN-). Use the MO diagram of CO as a model for this, but start with atomic orbitals of appropriate relative energies, based on the table of orbital energies that I gave you in class on Wednesday.

E

2p 2p

2s 2s

CN C N

b. What is the bond order in CN-, and how many unpaired electrons does cyanide have? bond order = 3; 0 unpaired electrons c. Which molecular orbital(s) do you predict would interact most strongly with a hydrogen 1s orbital to form an H-C bond? Explain. The HOMO is the s2p orbital, which can interact with the 1s of the H+, as in the diagram below. The bonding orbital has an energy near that of the p orbitals; the antibonding orbital becomes the highest energy orbital.

E

1s

s2p

H

CN

2.

Prepare a molecular orbital energy diagram for SH-, including sketches of the orbital shapes and the number of electrons in each of the orbitals. Calculate the MO diagram using Spartan (see directions on the course common folder) and compare the Spartan output with your predictions. The molecular orbitals for SH- are given below. Net one bond

Chem 351 G. E. Hofmeister

Spring 2005

S orbital energies are ­12.0 eV (3s) and ­20.7 eV (3p). the 1s orbital of H has an energy of ­13.6 eV.

3s

p

p

3p

1s

E 2s

1s H S

3s

I didn't connect the 3s with the 3s orbital, but that would be okay, to show mixing. There will be less mixing between these two, since the energy difference is so great. The Spartan output (attached) indicates the same relative energy levels as shown in the diagram above, although the absolute energies in Spartan are not comparable to the atomic orbital energies listed above. Because the S 3s orbital is so much lower than the 3p, I predict little mixing between the two, which is also indicated by Spartan. 3. Prepare a MO energy level diagram for the ozone molecule, O3, a. without mixing of the s & p orbitals

"degenerate" (all non-bonding, but not same symmetry) 2p 5s E 1p 1p 4s 3s 2p

2p

2s 1s O2 O3 O

2s

Chem 351 G. E. Hofmeister

Spring 2005

The 5s and 2p orbitals are formally non-bonding. Net of 3 bonds. b. indicate the energy changes that you would expect on mixing s & p orbitals. Mixing s and p character would lower the energy of 3s (and perhaps 2s) and raise the energy of 5s (and perhaps 4s). 4. Using the MO diagram for O2 (in the book), compare the bonding in O22-, O2-, and O2 (calculate the bond order) in each, and indicate which species is (are) paramagnetic. Which species has the shortest bond? O22- Bond order = 1 & diamagnetic O2Bond order = 1.5 & paramagnetic O2 Bond order = 2 & paramagnetic Use a process similar to how we constructed the MO diagram of BH3 in class on Friday to construct the MO diagram for linear BeH2, which is a transient species in the gas phase. You may want to look at section 3.13 and figure 3.39 in S & A to help you with this. 1s is bonding, 2s is non-bonding, and 3s is antibonding:

3s

5.

2s E

2s

1s H-H Be

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