# C arrange the following elements in order of

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(c) Arrange the following elements in order of increasing ionization potential: Li, Ne, Na Na < Li < Ne Li and Na each have a single outer shell electron, but IP decreases down a column. IP increases across a row and then drops in going to the next row, so Ne has an IP greater than either Li or Na. 10 . (10) For the reaction, 2A + B + 2C —> D + E , the following initial rate data were collected at constant temperature. Determine the correct rate law for this reaction. All units are arbitrary. Trial [A] [B] [C] 1 0.225 0.150 0.350 0.0217 2 0.320 0.150 0.350 0.0439 3 0.225 0.250 0.350 0.0362 4 0.225 0.150 0.600 0.0127 Rate Circle the correct answer: a. Rate = k [A] [B] [C] b. Rate = k [A] 2 [B] [C] c. Rate = k [A] 2 [B] [C] d. Rate = k [A] [B] 2 [C] e. None of these choices are correct –1 –1
11 . (5) The Schrödinger wave equation
FINAL EXAM - Karlin page 5 of 11 Chem 030.102 / Section 2 NAME____________________ May 17, 2011 The following molecular orbital diagram should be used for the problems on this page. For diatomic oxygen (O 2 ) and fluorine (F 2 ), the σ 2p orbital should be lower in energy than the π 2p . However, the diagram will still yield correct bond order and magnetic behavior for these molecules. Energy ________ σ * 2p ________ ________ π * 2p ________ σ 2p ________ ________ π 2p ________ σ * 2s ________ σ 2s ________ σ * 1s ________ σ 1s 12 . (5) According to molecular orbital theory, which of the following species is the most likely to exist? 2 2–
13 . (5) According to molecular orbital theory, which of the following species has the highest bond order?
14 . (5) Assuming that the molecular orbital energy diagram for a homonuclear diatomic molecule applies to a heteronuclear diatomic molecule, determine which of the following species has the highest bond order. a) NO b) OF c) CN d) O 2 e) NO Answer: c
FINAL EXAM - Karlin page 6 of 11 Chem 030.102 / Section 2 NAME____________________ May 17, 2011 15 . (12) (a) In the NO 2 (nitrite) ion, each atom can be viewed as sp 2 hybridized. Then, each atom has one remaining unhybridized p orbital. How many π 2p molecular orbitals (MO’s) (including the possibility of bonding, antibonding and non-bonding orbitals) are then formed using the un- hybridized p orbitals? i) 1 ii) 3 iii) 4 iv) 6 v) 12