Name: KEY Chem 202 Section: Quiz 3 9/19/2008

Transcription

1 Name: KEY Chem 202 Section: Quiz 3 9/19/2008 a. 1) Draw the following for this quantum number: (n=3, l=1, m l =0, m s = +! ) a. Wave function b. Probability distribution c. Orbital lobes (draw all orientations) b. c.

2 2) I am a very important element for your body and usually am ionized. My first four ionization energy levels are 738 kj/mol, 1450 kj/mol, 7.7 x 10 3 kj/mol, and 1.1 x 10 4 kj/mol. I have a larger atomic radius than Beryllium and a smaller one than Barium. When I am ionized with my most common charge, I have two sets of p orbitals and three sets of s orbitals. All of my electrons are paired in this state and in my neutral state. Determine what I am, what my electron configuration is in my most common ionized state, and how you used what I told you about me to find your answer. Answer: Calcium. 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, or [Ar], or [Ne]3s^2, 3p^6. Students should explain that the jump in the ionization energy from the second to third state allows one to conclude the element must be in Group 2A. The atomic radii allow the students to eliminate barium and beryllium. The most useful bit of info is the number of orbitals, which eliminate strontium and magnesium as possibilities. 3) Determine the electron configurations for nitrogen and carbon. Then, draw atomic orbitals for each element and fill them with their electrons appropriately and according to Hund s rule (don t worry about energy differences). Use your drawings to explain why the electron affinity for nitrogen is approximately zero, while the electron affinitiy for carbon is substantially positive. An electron added to nitrogen to form an anion would have to occupy a 2p orbital that already contains one electron. The extra repulsion between the electrons in this already occupied orbital causes the anion to be unstable. When an electron is added to carbon, no such repulsions take place because there is an empty orbital present. Answer: Nitrogen - 1s^2, 2s^2, 2p^3 Carbon - 1s^2, 2s^2, 2p^2 E E Nitrogen Carbon

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5 3) Explain the reason why the ratio of Na second IP(Ionization potential) to it s first IP is 9.2 compared to Mg which has a ratio of second IP to first IP of 2.0(3) The first IP of Na is low, since it has [Ne] 3s 1 and by losing an e- it takes on [Ne] configuration. But once you move to the second e-, this is Na +! Na 2+ +e- and this is from the complete 2p 6 orbital. For Mg, it has [Ne]3s 2, and both e- come from the 3s orbital. By losing two e-, it takes on the [Ne] electron configuration.

6 Name: Chem 202 Section: Quiz 3 9/19/2008 1a) Determine the Z eff values for the elements aluminum (Z = 13) and sulfur (Z = 16). (1) b) Which element should have the higher ionization potential? Which should have the higher electron affinity? Explain in terms of the relation between Zeff and the ability to remove or add an electron. (2) c) Draw an atomic orbital diagram for sulfur. Make sure to add in your core electrons (1) 1a) ; 1b) As Zeff increases, ionization potential will on the whole increase, as each valence electron will experience a greater nuclear charge. Similarly, electron affinity will increase as it is electrostatically easier for an electron with a greater Zeff to accept another electron. Therefore, sulfur will have both the higher IP and the greater EA. 1c) The AO diagram should have the 10 core electrons (2 1s electrons, 2 2s electrons, and 6 2p electrons at progressively higher energy states), then a full 3s orbital, then (at a slightly higher energy) a set of 3p orbitals with 4 electrons in them; one p orbital should be completely filled, the other two should be half-filled 2) Draw the following: a) A graph of the probability of finding an electron in an orbital with a principle quantum number of 4. (1.5) b) A graph of! for an n=6 electron. (1.5) a b

7 3) Why is the electron affinity for Na -53 kj/mol while for F it is -328 kj/mol? Draw the electron configurations for both atoms with an added electron. (3) Answer: Fluorine gains the configuration of Ar by adding an e-, this is much more stabilizing than the configuration Na obtains by adding an e-, which is unfavorable due to e- repulsion.