1. Use a general Born-Haber approach to explain why only group 1 and 2 metals form metal hydrides.
Step
Symbol
Energy
(kj/mol)M (s) --- > M (g)
DHsubl (M)
M (g) --- > M+ (g) + e-
I (M)
1/2 H2 (g) ---> H (g)
B (H2) / 2
+ 218
e- + H (g) ---> H- (g)
-Ea (H)
- 77
M+ (g) + H- (g) --- > MH (s)
- DHL
M (s) + 1/2 H2 (g) ---> MH (s)
DHf (MH(s))
Since H has a relatively small electron affinity (77 kJ/mol) and H2 has a strong bond (436 kJ/mol), metal hydrides are generally less stable than metal compounds with other anions. In order to make stable compounds, we should look for low sublimation enthalpies, low ionization enthalpies, and form compounds with relatively high lattice enthalpies. The group 1 and 2 elements have the lowest ionization enthalpies, and low sublimation enthalpies.
2. When CsCl is dissolved and then recrystallized from conc. HCl at low temperature , a new stable solid compound is formed. The same reaction does not occur, however, with either LiCl or NaCl. What is the new compound, and why does it only form from CsCl?
This was a challenging question, I think, but the answer was not too complicated. First, it seems unlikely that any redox chemistry will occur, there are no strong oxidants or reductants that would be required to change the oxidation states of these stable ions. The stable solid that forms is CsHCl2, that is an ionic lattice containing Cs+ cations and HCl2- (bichloride) anions. The bichloride anion is not common, but we can predict it will have a linear structure similar to bifluoride (HF2-). In order to stabilize the bichloride anion, we need a large cation such as Cs+. With smaller cations, the anion will decompose to HCl and Cl-. For example, in the case of LiCl, the reaction:
LiHCl2 (s) ---> LiCl (s) + HCl (g)
is exothermic (and also entropically favored) and therefore prevents LiHCl2 from being stable. With larger cations, the lattice enthalpy of the chloride is not as large and the reaction above therefore less favorable.
3. Problem 13.7 in the text. -- read the article and describe
4. Problem 14.3 in the text
Pyrophyllite and talc are charge neutral sheets structures, with no strong interactions between sheets. They are soft, can be used as lubricants due to easy sheet sliding. Mica is a charged sheet structure, very much a layered habit, but stronger interactions between interlayer cations and negative sheets. It can cleave readily, but is a more brittle compound.
5. BCl3 can be prepared from B2O3 and carbon tetrachloride, but only at high temperature. Write a balanced equation for the reaction. What provides the driving force for this reaction?
2 B2O3 (s) + 3 CCl4 (g) --> 4 BCl3 (g) + 3 CO2 (g)
The reaction produces more moles of gas than it consumes, so the entropic term makes this is more favorable at high temperature.
compound
DHf (kJ/mol)
So (J/Kmol)
BCl3 (g)
-395
290
CO2 (g)
-394
214
B2O3 (s)
-1264
54
CCl4 (g)
-107
309
for above reaction
+87
727
So at 25 °C, DGrxn = 87 kJ/mol - (298 K)(727 J/Kmol) / 1000 = 87 - 217 = - 130 kJ/mol, slightly favorable, but at 1000 °C, DGrxn = - 840 kJ/mol.
6. Draw the structures of several repeat units of [SiO32-]n, [Si4O116-]n, and [Si4O104-]n.
[SiO32-]n SiO4 tetrahedra sharing 2 corners in a linear chain or ring
[Si4O116-]n a double chain comprised of 2 linear chain as above, with alternate SiO4 tetrahedra sharing corners between chains
[Si4O104-]n a double chain as above, but with all SiO4 sharing one corner with the other chain, notice this is equivalently written as [Si2O52-]n
7. SnF4 is difficult to prepare directly from the elements, due to the passivation of the Sn metal surface with metal fluoride. Suggest an alternate route to prepare stannic fluoride beginning with any needed elements in their standard states. Give all balanced reactions required, and indicate, where necessary, when the reagents or products are air or moisture sensitive.
(1) Sn (m) + 2 Br2 (g) ---> SnBr4 (liq)
(2) SnBr4 (liq) + 4 F2 (g) ---> SnF4 (s) + 4 BrF (g)
The first reaction must be carried out under anhydrous conditions, since SnBr4 can hydrolyze:
SnBr4 (liq) + 2 H2O ---> SnO2 (s) + 4 HBr
The second reaction must also be carried out under anhydrous conditions, for the above reason, and also because F2 will hydrolyze to make HF and O2. The reaction may also generate BrF3 or BrF5 byproducts. All of these BrFx byproducts can be evacuated off under vacuum, leaving SnF4 solid.
This sequence exploits a common trick in inorganic chemistry to prevent surface passivation, namely, get the reactant metal atoms (Sn in this case) separated into molecular units (as in SnBr4) first.
As an aside, the same reaction can be done with Cl2 in place of Br2. For step 2, using the bromide is more favorable.