CH511 Problem Set 1 - solutions
1. Determine the value of r for which the RDF of a 2pz orbital is a maximum.
RDF = 4 p r2 R2for 2p orbitals this is = ( p Z5 / 6 a05 ) r4 exp ( - Z r / a0 )
Rmax where d / dr = 0
d R(r) / dr = ( p Z5 / 6 a05 ) ( 4 r3 - Z r4 / a0 ) exp ( - Z r / a0 )
this is zero where 4 r3 = Z r4 / a0
r = 4 a0 / Z
as we might expect, the answer will depend inversely on Z
2. Prove that the electron probability function for a ground
state N atom has spherical symmetry, i.e. show that the sum of a px,
py and pz orbital has no angular dependence.
Y2pz2 = C cos2q
Y2px2 = C sin2q
cos2f
Y2py2 = C sin2q
sin2f
Ytot2 = Y2pz2 + Y2px2 + Y2py2
= C cos2q + C sin2q cos2f + C sin2q sin2f
Y2 = C [cos2q + sin2q (cos2f + sin2f)] = C [cos2q + sin2q] = C
so Y2 is a constant and not a function of q or f and is therefore spherically symmetric.
You can also do this with complex waveunctions (such as those in the textbook), if you do so remember that probabilility is associated with the product YY* (where Y* is the complex conjugate) rather than simply Y2
3. Madelung constants can be derived by calculating a summation of coulombic interactions, each term in the series indicates all the interactions for a specific ion-ion distance. For each term, the sign (from anion or cation), the total number of interactions, and interaction distance needs to be determined. The notes have a simple example for the infinite linear chain.
Derive the first 20 terms (arising from 20 shortest distances) for determining the Madelung constant of CsCl. For each term, indicate the number of interactions, the sign, and the distance in terms of the lattice parameter a. To accomplish this, apply symmetry and permutations to (xyz) coordinates, do not try to use drawings or models.
In the CsCl structure, if we put a Cl at the origin (000), then the lattice generation rules are:
Cl at all sites (n, n, n) and Cs at all sites (n/2, n/2, n/2); where n is any integer (positive or negative)
The distance from the origin is found from the coordinates by: distance = (x2 + y2 + z2)1/2
The number of ions at any given distance can be determined from all possible permutations with the same distance, interchanging the coordinates and switching positive and negative values. For example:
(½ ½ ½) has 8 permutations with the same distance: (½ ½ ½), (-½ ½ ½), (½ -½ ½), (½ ½ -½), (-½ -½ ½), (½ -½ -½), (-½ ½ -½), (-½ -½ -½)
(1 0 0) has 6: (1 0 0), (0 1 0), (0 0 1), (-1 0 0), (0 -1 0), (0 0 -1)
Any general coordinates (x y 0) will have 12 permutations if x = y, and 24 if x ≠ y
At most, the coordinates (x y z) where all are non-zero and x ≠ y ≠ z has 48 permutations.
The total energy contribution for the interactions at a particular distance will be proportional to the number of interactions / distance.
Finally, note the Madelung constant term is related to d (the closest approach of ions) rather than a (the unit cell edge length). In this cell, d = 0.866 a, so we multiply the total energy contributions by this factor to get the Madelung constant terms.
coordinates ion type number of ions distance in a units energy contribution ½ ½ ½ Cs 8 0.87 -9.1 1 0 0 Cl 6 1.00 +6.0 1 1 0 Cl 12 1.41 +8.5 3/2 ½ ½ Cs 24 1.66 -14.5 1 1 1 Cl 8 1.73 +4.6 2 0 0 Cl 6 2.00 -3.0 3/2 3/2 ½ Cs 24 2.18 -11.0 2 1 0 Cl 24 2.24 -10.7 2 1 1 Cl 24 2.45 +9.8 3/2 3/2 3/2 Cs 8 2.60 -3.1 5/2 ½ ½ Cs 24 2.60 -9.2 2 2 0 Cl 12 2.83 +4.2 5/2 3/2 ½ Cs 48 2.96 -16.2 3 0 0 Cl 6 3.00 +2.0 2 2 1 Cl 24 3.00 +8.0 3 1 0 Cl 24 3.16 +7.6 5/2 3/2 3/2 Cs 24 3.28 -7.3 3 1 1 Cl 24 3.32 +7.2 7/2 ½ ½ Cs 24 3.57 -6.7 3 2 0 Cl 24 3.60 +6.7 total of 20 terms 378 -26.2
4. After solving the above problem, what can you conclude about the convergence of this constant. Are 20 terms sufficient to determine which structures are most stable in terms of lattice enthalpy?
Notice the last few terms (approx. + or -7) change the value of the interaction energy (sum is -26.2 after 20 terms) significantly. We need more terms to obtain a Madelung constant with useful precision, where the truncation does not effect the result by an amount that is less than the difference in Madelung constants in different structures.
5. Ternary compounds (with 3 elements) often have structures closely related to the common structure types we will discuss for binary compounds. Describe the structure, and its relation to a simple binary structure type, for both K2PtCl6 and NaFeO2.
One good resource is Wells, Structural Inorganic Chemistry
K2PtCl6 has an antifluorite structure, where the PtCl62- octahedra form an fcc lattice, with K+ occupying all Td sites. It is also related to the perovskite structure.
http://www.chemistry.ohio-state.edu/~woodward/ch754/struct/K2PtCl6.htm
NaFeO2 is isostructural with LiCoO2, which has discussed in class. This is a superlattice of the NaCl structure type, with alternating layers of Na+ and Fe3+ in Oh sites (AcBaCb)n, where the uppercase letters are O2- and lowercase letters are alternately Na+ and Fe3+. This leads to an anisotropic, layered structure. The strongly bound FeO2 sheets contain edge-sharing FeO6 octahedra.
6. Show that the ionic compound CaCl (s), which does not exist, would be thermodynamically unstable with respect to disproportionation.
For the reaction: 2 CaCl (s) -> CaCl2 (s) + Ca (s)
ΔHrxn = 2 ΔHL (CaCl) - ΔHL (CaCl2) - ΔHat (Ca) + I2 (Ca) - I1 (Ca)
where I1 and I2 refer to the 1st and 2nd ionization energies for Ca
Estimate ΔHL using the Kapustinskii eqn and the following data:
r(Ca2+) = 110 pm, r(Cl-) = 167 pm
r(Ca+) is not known, however, it must be smaller than r(Ca) = 197 pm, and larger than r(Ca2+). Also, we can expect that r(Ca+) will be larger than r(K+) ≈ 150 pm, so let's estimate r(Ca+) ≈ 160 pm
then, ΔHL(CaCl) ≈ - [(2)(+1)(-1) / 3.27 Å] [1 - (0.345 Å / 3.27 Å)] x 1.21 MJÅ/mol ≈ 660 kJ/mol
and ΔHL(CaCl2) ≈ - [(3)(+2)(-1) / 2.77 Å] [1 - (0.345 Å / 2.77 Å)] x 1.21 MJÅ/mol ≈ 2290 kJ/mol
ΔHrxn ≈ 2 (660) - 2290 - 176 + 1145 - 590 ≈ - 600 kJ/mol
Even with the assumptions made, it is clear that this disproportionation reaction is exothermic. There should not be any large entropic term at moderate temperature, as all the phases are solids. ΔGrxn should thus be negative, the reaction is favorable and spontaneous. This suggests that it would be difficult to ever isolate CaCl(s). Looking at the energy terms, it is reasonable to state the "driving force" for the reaction is the large lattice enthalpy for CaCl2. If we consider K, where loss of the second electron involves removing a core electron, and I2 ≈3050 kJ/mol, it becomes clear why we don't see this reaction for group 1 elements.