# CHM 501

### Homework 3

#### Due April 4, 2016

1. The repulsive term in the lattice energy equation can be assigned to be an exponential function of the form Erep = Ce–r/d* (this is the Mayer repulsion). Use this function and derive a lattice energy equation similar to the Born-Lande equation used in class.

Elatt = N°MZ+Ze2/(4πε°r) + Ce–r/d*

At r = r° dElatt/dr = 0

dElatt/dr = –N°MZ+Ze2/(4πε°r°2) – (C/d*)e–r°/d* = 0

C = [–d*N°MZ+Ze2/(4πε°r°2)]er°/d*

Elatt = N°MZ+Ze2/(4πε°r°) – [d*N°MZ+Ze2/(4πε°r°2)]er°/d*e–r°/d*

Elatt = [N°MZ+Ze2/(4πε°r°)](1 – d*/r°)

2. Estimate ΔH°f for the hypothetical compound KBr2. Cite your sources for data you use in your calculation. Use the lattice energy equation you derived in question 1 with d* = 34.5 pm. Assume that KBr2 has the same structure as CaBr2 and that the ionic radii are similar for two cations.

Use a Born-Haber cycle but rather than calculating Elatt, estimate Elatt from the Born-Mayer equation. Assume a CaCl2 lattice. All data is found in Inorganic Chemistry A Unified Approach, W. W. Porterfield, 2nd edition.

 K(s) → K(g) S = 89 kJ/mol K(g) → K+(g) IE1 = 4.339 eV = 428.7 kJ/mol K+(g) → K2+(g) IE2 = 31.81 eV = 3069 kJ/mol Br2(l) → Br2(g) ΔHvap = 53.4 kJ/mol Br2(g) → 2 Br(g) ΔHdiss = 190.2 kJ/mol 2 Br(g) → 2 Br– 2EA = 2(324.6) = 649.2 kJ/mol Elatt = [N°MZ+Z–e2/(4πε°r°)](1 – d*/r°) Elatt = [(6.02×1023)(2.365)(2)(–1)(1.602×10–19)2/((1.113×10–10)(2.96×10–10))](1 – 34.5/(296)) = 1959 kJ/mol N° = 6.02×1023 M = 2.365 Z+ = 2 Z– = –1 e = 1.602×10–19 4πε° = 1.113×10–10 r+ = 114 pm r– = 182 pm r° = r+ + r– = 114 + 182 = 296 pm

ΔH°f = S + IE1 + IE2 + ΔHvap + ΔHdiss – 2EA – Elatt = 89 + 428.7 + 3069 + 53.4 + 190.2 – 649.2 – 1959 = 1222 kJ/mol

The heat of formation is large and positive, which is why the compound is hypothetical. The increased lattice energy is not sufficient to overcome the cost of the second ionization of K.

3. The group 14 atoms (C (diamond), Si, Ge, and α-Sn) have the following band gaps: 5.4 eV, 1.1 eV, 0.67 eV, and 0.08 eV, respectively, and the following room temperature electrical conductivities, 10–6 S/cm, 10–5 S/cm, 10–2 S/cm, and 100 S/cm, respectively. Are these two data sets consistent? Why or why not? What underlying periodic property (or properties) drives these trends?

The band gaps suggest a much larger spread in electrical conductivities: using e–Eg/2RT gives the Boltzmann term values of ~10–45 for C(diamond), ~10–9 for Si, ~10–6 for Ge, and ~10–1 for α-Sn. The mobility and the number of carriers also change considerably for these examples, which must account for the smaller variation in conductivities.

The increasing atomic size going down the periodic table accounts for the trend. As the atoms become larger, the electron cloud becomes more polarizable, which allows for better mobility of electrons in the bands.