Unformatted text preview: Chem 312 Winter 2010 "Chem–Talk" #1 Problems KEY
Wed, January 6, 2010 1. Balance the following redox equation, assuming it takes place in H2O under basic conditions. +5 2 NO3– +4 + 2 e- + 2 H2O N2O4 + 4 OH- a. What is the oxidation state of nitrogen in NO3–? +5 b. What is the oxidation state of nitrogen in N2O4? +4 c. How many electrons are involved in this reaction if you assume that one equivalent of N2O4 forms? 2e- 2a. Using the table below, explain why the electron affinity of phosphorus (P) is greater than that of nitrogen (N), and that of sulfur (S) is greater than that of oxygen (O). Is this what you would expect? Since the added electron must pair with another electron (in a porbital), the cost would be an increase in e-/e- repulsion, which, in some cases, does not offset the e-/Zeff attractive forces. This reduces the affinity for electrons. The magnitude of the e-/e- repulsive forces depends on the size of the orbital. Since the 3p orbital of P (and S) is larger than the 2p orbital of N (and O), e-/e- repulsion is less for these elements (P vs N (and S vs O)).
2b. Why is the atomic radius of O (0.73 Å) larger than N (0.70 Å)? Wouldn't you expect the opposite to be true? Oxygen has paired electrons in its highest energy 2p orbital, whereas nitrogen does not. In order to minimize the e-/e- repulsion caused by pairing electrons, the oxygen 2p orbital expands, resulting in a larger average distance from the nucleus, and thus larger atomic radius. ...
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- Winter '10
- Redox, Chemical element, Oxide, e-/e- repulsion