ANSWERS TO PROBLEM SET 1 (2012)-Some Principles of Bonding




1. Lithium is the first element in the second row of the Periodic Table.

a. How many electron does it have?

ANS: 3 (3 points)

b. Construct an energy diagram that predicts whether or not dilithium (Li­2) forms.


ANS: The atomic orbitals of lithium hold 3 electrons. These electrons are distributed in the 1s and 2s orbitals according to the Aufbau Principle as shown in A (5 points).





These atomic orbitals can be combined from two lithium atoms to form six molecular orbitals. The six electrons are then distributed into the MO’s according to the Aufbau Principle as shown in  Β. The electrons in the antibonding orbital at the first quantum level cancels the bonding  properties of the electrons in the bonding MO. These are the kernal electrons. The two remaining electrons are in a bonding  orbital at the second quantum level. The antibonding  orbital at this level is empty. We conclude the Li2 is a stable molecule. (5 points)











2. Describe the MOs of acetamide.




ANS: The methyl carbon is attached to 4 atoms so it hybridizes sp3 (3 points). Three of the 4 lobes of the sp‑3 hybrid overlap with a  1s orbital of a hydrogen to form 3 s bonds.  (3 points). The carbonyl carbon is attached to 3 atoms so it will hybridize sp2. (3 points)   One of the sp2 lobes will overlap the remaining sp3 lobe of the methyl carbon to form a s bond . The nitrogen makes 3 bonds so it will hybridize sp2 (3 points). One of its lobes will overlap an sp2 lobe of the carbonyl carbon. The other two lobes will overlap 1s orbitals of hydrogen. The s bond framework looks like this:



(5 points)


We have used all the lobes of the sp3 orbital at the methyl carbon. We have used all 3 of the sp2 lobes of the carbonyl carbon. We have used 1 of oxygen’s 6 valance electrons. Two lone pairs will occupy the other lobes of the sp2 hybrid leaving an electron in an atomic p orbital (3 points). We have used 3 of nitrogen’s 5 valance electron leaving a lone pair in an atomic p orbital (3 points).

The p orbitals look like this:


(3 points)


If we add the remaining electrons to these atomic orbitals, we get:


(3 points)


The electrons are localized in the atomic p orbitals. This correponds to:




where ΦA, ΦB and ΦC correspond to the atomic orbitals for the three atoms we are considering.

However, the p orbitals can overlap. For example:








Bonus points (5 points)



There are three possible MOs and only 4 electrons to place. Without calculations, it is difficult to decide which two MOs are used, but it is clear that the nitrogen has some double bond character ( 2 points) and is part of the  resonance hybrid (2 points):





5.  Draw five 3d orbitals.

ANS: See Fig 1.42 of CBM-2 (2 points for each = 10 points)

6. Explain why the double bond that we write for phosphates is different from the one we write for carbonyl compounds.

ANS: See section 1.16 of CBM-2. (10 points)

7. Is the following compound cis or trans?



It is trans (fumaric acid). (4 points; name not necessary)

a. The other isomer is:





b. They are geometrical isomers (4 points)

c. The p-atomic orbitals overlap to form a stable, p bond. It takes significant energy to break this interaction and rotate about the s bond (4 points). Therefore, the molecule exists as two stable isomers, cis and trans.

d. The s bond framework requires three MOs. These can be formed by mixing a 2s and two 2p to give three sp2 hybrid orbitals. When these three hybrid orbitals are combined, they point towards the apes of an equilateral  triangle:




The angles of an equilateral triangle are 120º. Therefore, the s bond angles should be close to 120º. ( 4 points)

8. Define tautomerization

ANS: For our purposes, tautomerization involves migration of a double bond produced by an acidic  hydrogen. This kind of tautomerization is sometimes called prototropy. Tautomers equilibrate rapidly.  (4 points)

9.  Write the tautomers of acetone.





(4 points)


Grade Yourself: >100 = A+, 90-100 = A, 75-89 = B, 50-74 = C, 25-49 = D, <25 = F


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