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Unformatted text preview: Chemistry 2090 October 10, 2008 Please hand in the following problems, at the end of class on October 17. In addition, attach the answers to the following questions found in our text: Chapter 9: 40, 58, 68 Chapter 10: 30, 32, 40, 52 1. (1 point) Atoms X, Y and Z are found in the same period of the periodic table, and have 2, 6 and 7 valence electrons, respectively. (a) Write the formula for the compound most probably formed between X and Z; is their bond primarily ionic, or covalent? We expect X −→ X 2 + + 2e − and Z + e − −→ Z − So the compound we expect to form between them is of the form XZ 2 , and that it would be largely ionic. (b) Write the formula for the compound most probably formed between Y and Z. Y brings 6 electrons to a bonding situation, and so is deficient by two with respect to an octet. Z brings 7 electrons, and so is deficient by one. So one likely candidate is YZ 2 , with two covalent bonds; an example would be OF 2 . If X is a larger atom–say S–then expanded shells are possible, and SCl 4 would also be a candidate. (1 point) 2. Classify as paramagnetic or diamagnetic each of the following species, as found in their ground state: He, Be, C, Na, Na + . For He: electron configuation is 1 s 2 ; all electrons paired, diamagnetic. For Be: electron configuation is 1 s 2 2 s 2 ; all electrons paired, diamagnetic. For C: electron configuation is 1 s 2 2 s 2 2 p 2 ; by Hund’s rule the two electrons in the 2 p orbitals are in different orbitals and spins are both “up” (or “down”) in the ground state; therefore, the C atom is paramagnetic. For Na: electron configuation is 1 s 2 2 s 2 2 p 6 3 s 1 ; with an odd number of electrons, (at least) one must be unpaired (here, exactly one is unpaired) and the atom is paramagnetic. For Na + : electron configuation is 1 s 2 2 s 2 2 p 6 ; the ion is diamagnetic. 1 3. (2 points) (a) Without consulting any reference materials, compare 4 Be and 3 Li. Estimate for each of the electrons found in the ground state of the neutral molecule the Z eff which is active; answers of the form “between 3.4 and 3.7” would provide the appropriate level of approximation. Suggest which electron(s) of the 7 listed would have lowest ionization energies; if you list more than 1, then make sure that they are ordered from lowest to highest. Li has in its ground state 3 electrons, in 1 s and 2 s orbitals; Be has the same orbitals occupied but the fourth electron is in the 2 s orbital. Each 1 s shields the other 1 s by about 1/3 of a charge unit, and the 2 s (and further away orbitals) by almost one full charge unit. The 2 s electrons shield the 1 s electrons by very little, and the other 2 s by about 1/3 of a charge unit. (For completeness, we might add that the 2 s shields a 2 p by about 1/3 of a charge unit, and the 2 p shields a 2 s by a little less than 1/3 of a charge. But electron-electron repulsion also plays a role as we go through half-filled shells, so it gets a bit more complicated as we fill the...
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This note was uploaded on 03/27/2009 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell.
- Fall '07