MASC 110 Solution HW2

# MASC 110 Solution HW2 - University of Southern California...

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University of Southern California MASC 110L Solution Set #2 E. Goo 1. Calculate the lattice energy of LiCl which has the same crystal structure as NaCl. You will need to use the information provided in the slides in lecture. From this slide the Li + ion radius is 76 picometers and the Cl - ion radius is 181 pm. The Li-Cl separation is therefore 76 + 181 = 257 pm Energy of attraction is () ( ) mol kJ pm mol pm kJ x V / 540 257 1 1 10 389 . 1 5 = +

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The electron ionization for Li is 500 kJ/mol from the above graph The electron affinity for Cl is around -335 kJ/mol from the above graph. The lattice energy is then -540+500-335 = -375 kJ/mol Finally you should correct the energy of attraction by multiplying by the Madelung constant which is 1.75 for this crystal structure -540 x 1.75 = -945 kJ/mol The lattice energy is -945+500-335 = -780 kJ/mol This is a negative quantity so the crystal should be stable. 2. Draw the electron dot notation for the following molecules. a) CH 4 b) CO 2 c) PCl 3
d) ClO 4 - (a) (b) (c) H Cl H H C O C O P Cl

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## This note was uploaded on 02/13/2009 for the course MASC 110L taught by Professor Goo during the Spring '07 term at USC.

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MASC 110 Solution HW2 - University of Southern California...

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