Chs. 11/12 - Types of Solids and IAF: How do you know what type of a solid you have by looking at the formula? Two of the types are pretty easy. 1) Metals: metallic forces (sea of e- and band theory). You will simply see the symbol for the element a) Ex: Na, Fe, Cu, Bi, W, etc. Strength (hardness) and melting point varies widely. Depends strongly on the number of valence e- and thus the position in the periodic table. Generally, the hardness and m.p. increase from either end of the metals toward the middle of the transition metals. The transition metals in the middle of the P.T. tend to be the hardest and have the highest m.p. (Cr, Mn, Fe, Co, Ni, Mo, W). The metals on the "sides" of the P.T. (ends of the rows) are the softest and have lower m.p. (Na, K, Rb, Cs, Mg, Ca, Ga, In, Sn, Pb). Hardness and m.p. INC. in the P.T. from the outer edges toward the middle of the transition metals rep. metals ------> middle of transition metals <----- rep. metals 2) Covalent Network: One "giant molecule" (crystal) held together by strong covalent bonds and lots and lots of them. Among strongest and highest melting. a) C, Si, Ge, Sn(gray tin) - note these are all Group 4A elements These all have the diamond arrangement, each atom is tetrahedrally covalently bonded to 4 other atoms. Diamond are the hardest and highest melting substance. b) Just a few others: C(graphite - not as strong as diamond), SiC, BN, GeO2, SiO2 (quartz form). If I write SiO2(s) I mean quartz. NOTE: These are in group 4A or on either side of group 4A. Now for ionic and molecular solids. These aren't as hard as you may think. 3) Ionic: Ionic forces of attraction. Depends on the Lattice Energy, which depends directly on the charges of the ions and inversely on distance between them, (Q+)*(Q-) (Q+)*(Q-) F(ion-ion) => --------- LE => --------- d2 d Ionic cmpds. with smaller higher charged ions will have greater LE and be higher melting (and also tend to be less soluble). a) If the formula has "metal""nonmetal" it is ionic unless you're told otherwise Ex.: NaCl, CaO, Fe2O3, etc. 1) A few ionic substances when melted or boiled do form molecular compounds (MgCl2, AlCl3), but not in the solid under normal conditions. b) There are cases when a formula (compound) contains nothing but NONmetals and the substance is ionic (normally all nonmetals indicate a molecular compound - see #4 below). These special ionic compounds contain the Ammonium ION, NH4+. These are ionic substances. Ex.: NH4Cl, (NH4)2SO4, NH4NO3, etc. These are easy to recgonize. If you see NH4 in a formula (written like this of course) it is the ammonium ion and the substance is ionic. I don't know of a formula where NH4 is present and it is not the ammonium ion. 4) Molecular: IntERmolecular Att. Forces (IAF). Weaker than those above. London Forces: ALL molecules have these and NONpolar molecules have only LF Dipole-Dipole: must have a polar molecule H-bonds: molecule will have a H covalently (directly) bonded to a N, O or F atom and there will be a N, O or F with a lone-pair of e- somewhere in the molecule. Ex.: H2O, NH3, HF, CH3OH, CH3NH2 but not CH3-O-CH3, (CH3)3N, CH3F a) Formulas for these compounds will consist of atoms (noble gases - He, Ne, Ar, Kr, Xe) or several atoms (of the same or diff. type). They will be all NONmetals, or compounds consisting of NONmetals with SEMImetals. 1) Molecular elements: H2, N2, O2, F2, Cl2, Br2, I2, S8 (may see just S), and P4 (may see just P). 2) Molecular compounds: more than one element Ex.: H2O, CH4, CCl4, NH3, C6H6, C2H6, C2H5OH, SiH4, SiCl4, etc. NOTE: all nonmetals or nonmetals & semimetals b) Remember the exception above with NH4+. These compounds will be ionic. In the above discussion I did not mention LF between the particles for metals, covalent network or ionic compounds. Generally speaking the LF are so weak compared to the other forces holding the particles together in these compounds we ignore the LF in our discussions. For a covalent network or metals it doesn't even make much sense at all to speak of LF between the atoms in the solids. This of course would not be the case for graphite in which the LF are very important to the properties of graphite. Graphite consists of large sheets of benzene-like molecules held together with covalent bonds (electrons are delocalized and can "flow", accounting for graphite's electrical conductivity) and these sheets are held together by very strong LF. However, these forces are weak enough to allow the sheets to slide past one another and still retain its "shape" and properties (accounting for its lubricating properties).
Ch. 13 - Types of IAF and Solubility: Next subject deals with solubility and what happens to these various types of compounds in solution. In order for a solvent to dissolve something the solvent has to be able to effectively surround the solute particles. Also, remember the forces of att. formed between solute-solvent must be comparable to those which have to be broken between solute-solute and solvent-solvent particles (i.e. the solute and solvent must possess similar AF). Generally, remember "LIKE DISSOLVES LIKE": a) Polar and ionic solutes dissolve in polar solvents. Ex.: NaCl in H2O, NaCl in CH3OH, CH3OH in H2O, NH3 in H2O C6H6 (nonpolar) will not dissolve in H2O b) Nonpolar solutes dissolve in nonpolar solvents. Ex.: C6H6 in CCl4, SiCl4 in CCl4, I2 in C6H12 greases and oils (which are mostly nonpolar) in nonpolar solvents - Use Lestoil to get grease or oil out of clothing (test for color fastness first) Specifically, 1) Metals: tend to be fairly insoluble in most solvents. a) This does not mean those which REACT with the solvent. This is not considered dissolution in the true sense, but is due to a chemical reaction which produces products which dissolve (we can't get back what we started with simply by evaporating the solvent). Ex.: Group 1A and 2A metals react with H2O as do a few others Remember: You are breaking metallic forces (fairly strong) to get essentially individual atoms of the metal and trying to replace these with things like LF. Just doesn't tend to happen. 2) Covalent Network Solids: Insoluble in just about anything. In order for these to dissolve in either a polar or nonpolar solvent you would have to break thousands of COVALENT bonds and replace these att. forces in the solution with LF (the fragments of the solid would tend to be nonpolar and have only LF) between the fragments and the solvent. The LF would be so weak compared to the covalent bonding forces you would be breaking in the solid that you can't get enough energy back when you bring solute and solvent together to overcome the energy needed to pull things apart. 3) Ionic Solids: Soluble in POLAR solvents. You have to break ionic bonds (very strong) between the ions (see #3 above in Ch 11 summary). You replace these with ION-DIPOLE att. forces, which are almost as strong as ION-ION (and stronger than the other IAF - LF, DD, and even H-bonding). The ION-DIPOLE force and energy are given by the following (note the similarity to ion-ion att. forces expression, F(ion-ion) and LE for ionic cmpds), Q*u Q*u F(ion-dipole) => --------- E(i-d) => --------- d3 d2 where, Q = charge on ion u = dipole moment of polar molecule d = distance between ion and molecule (center to center) For ionic substances you are breaking ION-ION forces and replacing them with something almost as strong. You are also breaking the att. forces between the polar molecules in the solvent and replacing these with something which is as strong, if not stronger in most cases. The strength of the various AF is: LF < DD < H-bonding < Ion-Dipole < Ionic & Covalent a) Ionic solids are NOT soluble in NONpolar solvents. That is because you break strong ION-ION AF and you try to replace these with nothing but LF between the ion and the nonpolar molecule. This just won't happen. 4) Molecular Compounds: "Like Dissolves Like" It's more or less that simple. You tend to be breaking LF, DD and H-bonds and then reforming these. So you are either replacing one type of force with a similar force, or not. Sometimes you can actually form a stronger AF than the ones you break. A good example of this is when acids dissolve in polar solvents. When acids dissolve in a polar solvent they ionize. Thus, you are breaking LF, DD (and perhaps H-bonds depending on the acid) and you are forming ION-DIPOLE AF between the ions and the solvent molecules (as well as LF, DD and perhaps H-bonds). You get a lot of energy back compared to what has to be put into the system to separate the solute particles and separate the solvent particles. In fact you get so much energy back that the process is generally exothermic and a great deal of heat is generated. a) There are times when you do have to consider other things. 1) CH3OH is polar and is miscible in H2O (as are CH3CH2OH, CH3CH2CH2OH and CH3CH(OH)CH3 (isoproply alcohol, rubbing alcohol). However, the solubility in H2O falls off fairly rapidly as we start adding more CH2 units. By the time we are up to octanol, C8H17OH, this molecule is essentially insoluble in H2O and is actually more soluble in a nonpolar solvent (such as C6H14) or in a "mixed" solvent (such as another alcohol which has both polar and nonplar parts). The H2O interacts with the polar OH "head" but can not surround the large nonpolar "tail". The nonpolar solvent can surround most of the molecule. The "mixed" solvent is even better since it can do both. 2) Dioxane (often used as a solvent - shown below) is nonpolar but fairly soluble in H2O. Why? If you look at the structure below it is because dioxane has 2 O-atoms which can form H-bonds to the H-atoms in the H2O molecules. O / \ CH2 CH2 | | CH2 CH2 \ / O 3) For organic molecules containing mostly C and H atoms the more polar parts it has (such as O atoms, N atoms, S atoms) and the more possible hydrogen-bonding sites it has the more soluble it will be in a polar solvent such as H2O. Sugars have a lot of OH groups and O atoms and are quite polar and have lots of possible HB sites so they tend to be very soluble in water. Again, I did not mention breaking LF for metals, covalent network or ionic solids. This is because these are very weak compared to the other forces we are breaking to pull these solids apart.