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Problem Set 12(2)

Problem Set 12(2) - Problem Set 12 Br 1 a Br Br Br b Br Br...

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Problem Set 12 1. a) b) Br Br Br Br Br Br .. x 3 (one for each axis) Te Br x 6 x 3 (one for each axis) x 3 (one for each axis) 2 non bonding orbitals per Br atom x 7 x 7 See next page for the final orbitals 1 non bonding orbitals per axis
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TeBr 6 2- has 50 e - If we fill the above diagram with them, then all of the orbitals will be filled except for the antibonded p’s, thus proving the said geometry as the correct one. 2. There are many aspects of this graph that are notable: a) The slopes of the metal oxides are negative: This is clear if one remembers the formula for the free energy: ∆G = ∆H - T∆S Since the slope equals -∆S, and the slope here is positive it means that the ∆S of these reactions is negative. This makes sense, since the reactions for these compounds are of some form like: M + O 2 = MO This has a negative change in entropy since there is at least one mole of gas on the reactants side, a highly entropic state, and none on the products side.
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