Solutions to Midterm #1

 
1. Because electronegativity increases as you go down a bond, the bond will become less polar as you descend a group.  Therefore, the strength of hydrogen bonds will likewise decrease.
 
2. There are actually 5 different substituents on each metal, so they should be stereochemically active, but the presence of internal symmetry in the form of an S2 axis (i.e., the inversion centre) causes the molecule to be achiral.
 
3. As demonstrated in class, there are 8 nearest neighbours to the central atom - they are of opposite charge at a distance d, so the first term is -8

The next-nearest neighbours are the like-charged ions in immediately adjacent units cells, and there are 6 of those.  Since the distance of the body diagonal is 2d, the edge distance is 2/31/2 so the second term is +12 / 31/2
 
4.
(a)
 
(b)
 
5.
(a) Using the Born-Meyer equation, remember that d = 0.50 + 1.96 = 2.46 A = 2.46 x 10-10 m.
Also remember that the charges on the ions are -1 and +2.

You should get a theoretical DHLo = 2340 kJ/mol.
 
(b) The two main factors affecting the validity of the Born-Meyer equation are the estimate of the intermolecular distances (rMg2+ and rBr-) and the fact that the equation assumes hard spheres (i.e., 100% ionic bonding).  Therefore, because the solution to (a) is close to the experimental value, it means that both the radii are close and that MgBr2 is essentially an ionic lattice.
 
6. Since H2 and PdO2 do not overlap, must find species that do by moving "up" on the H2 diagram and "down" on the Pd diagram.  There is overlap between H+ and Pd, so they are the products.  (Because there is an excess of H2, all Pd2+ will react, so it is not a product.)

H2 ---> H+   Eo = 0.00 V
PdO2  --->   Pd   Eo = (2*1.19 + 2*0.92)/4 = +1.06 V

Therefore the voltage for the overall reaction is 1.06 V
 
7.
(a) K is a little smaller than Cl-, so it can only fit in the octahedral holes

Zn2+ is a lot smaller than S2-, so it is too big to fit in the octahedral holes (it would "rattle"), so it is a better fit in the smaller tetrahedral holes.  Because there are twice as many tetrahedral holes, Zn2+ will only fill half of them.
 
(b) Zn2+ should be smaller than K+ because it is further to the right on the periodic table (and will suffer a lanthanide contraction), but more importantly, Zn has 2+ charge and K is only 1+.

Likewise, S should be bigger than Cl, and the 2- charge will make it even larger.
 
(c) If there is a ccp form, there is often an hcp form of similar energy, because hcp packing is very energetically similar to ccp packing.  Therefore, a polymorphic form is not surprising in this case.
 
Bonus There are 3 isomeric structures:

 


 

Last updated, October 28, 2003
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