Chem 312, Au13 Problem Set 8 Due in class, Friday, November 22 1. Our focus so far in class has been solids held together by ionic or metallic bonding. But many solids are held together by covalent bonding, and then the rules about molecular structures apply, such as the octet rule. For instance, in diamond, each carbon atom makes four bonds, in a tetrahedral array, just as you would expect for an sp3 hybridized carbon atom. (a) Based on the octet rule, how many bonds are made by each atom in the structure of solid: aluminum, silicon, phosphorus, sulfur, and chlorine? (Chlorine is a solid below -101 ˚C.) Based on the octet rule, each silicon atom in solid silicon should make four bonds (c.f., SiH4), each phosphorus atom in solid phosphorus should make three bonds, each sulfur atom in solid sulfur should make two bonds, and each chlorine atom in solid chlorine should make one bond. Using normal two electron-two center bonding, aluminum cannot attain an octet, but should make three bonds. This should tip you off that a simple octet rule picture will not apply to aluminum. Aluminum is, in fact, a metal -- not a covalent solid. Boron, directly above aluminum, is a non-metal but adopts strange structures to get around this non-octet problem (please look in Wulfsberg, p. 188). (b) Which of those elements is most likely to be molecular in the solid state (composed of molecules)? Which is most likely to have a linear chain structure (-E-E-E-E- etc.)? Which is most likely to be similar to diamond? Chlorine is molecular, Cl2, since each atom only makes one bond. Sulfur has a chain structure, -S-S-S-. The most stable form consists of eight-membered rings, S8, and is therefore molecular also. But most samples of sulfur also contain rings of other sizes, and sulfur can be polymeric under some conditions as well. The structure of silicon is the same as the structure of diamond, except that the Si-Si bonds are significantly longer than the C-C bonds. 2. Now consider the heat of atomization of solid aluminum, silicon, phosphorus, sulfur and chlorine. As discussed in Wulfsberg chapter 6 (especially sections 6.1, 6.2, and 6.3), the heat of atomization of an element is the enthalpy required to take a mole of an element from its standard state at 25 ˚C (~room temperature) to individual gaseous atoms. For instance, the enthalpy of conversion of diamond to gas phase individual carbon atoms is quite unfavorable: C (diamond) → C (atomic, g) ΔH˚ = +717 kJ mol-1. (a) From the Tables in chapter 6, how does the heat of atomization vary as one traverses the periodic table from Al to Si to P to S to Cl? The heats of atomization, from Table 6.1, are (in kJ/mol): Al: 326; Si: 456; P: 315; S: 279; Cl: 122. This is shown graphically below. Heat of Atomization 500 Heat of Atomization per "bond" 400 300 kJ/mol 200 100 0 Al Si P S Cl Chem 312, Au13; PS 8, p. 2 (b) Explain why the heat of atomization has this specific trend (use your answer to question 1a). The heat of atomization parallels the number of bonds made by the element, according to the simple rules used in question 8. Thus Al could make three bonds, silicon four, phosphorus three, sulfur two, and chlorine one. The heat of atomization divided by the formal "number of bonds" is pretty constant for these elements, as shown by the solid squares above. (c) Formulate a general rule for the heat of atomization across a row of the periodic table: The most favorable heat of atomization is found when the number of electrons is in what ratio to the number of valence orbitals? The maximum number of bonds occurs, for these elements, when the atom has four valence electrons. These atoms have four valence orbitals (3s, 3p). The general rule is that maximum bonding (and maximum heat of atomization) occurs when the number of electrons equals the number of orbitals, in other words when the orbitals are half full (since each orbital can hold two electrons). (d) Why is the heat of atomization of argon zero? The heat of atomization is defined as the heat required to take an element from its standard state at room temperature to gas phase isolated atoms. Argon at room temperature is already isolated (monatomic) gas phase atoms, so the heat of atomization is zero. (e) In what way is the definition of the heat of atomization of a solid similar to the lattice energy of an ionic solid? In what way is it different? The heat of atomization and the lattice energy are both enthalpies. They both give the heat required to break up a material (usually a solid) into independent (non-interacting) particles in the gas phase. These two quantities are also different. In the heat of atomization, one is forming gas phase atoms, while the lattice energy refers to the formation of gas phase ions. (Also, the lattice energy specifically refers to a solid while the heat of atomization can apply to solid, liquid, or gas-phase materials.) (f) How is the heat of atomization of a pure element E related to the heat of formation (ΔHf˚) of gas phase atoms of that element? [Don’t overthink this, the answer is very simple.] They are the same. 3. Now consider the series N2, O2, and F2. (a) Draw Lewis dot structures for each and predict which has the strongest bond. How many valence orbitals are present in each case? How many electrons are present in each case? :N≡N: eight valence orbitals, ten valence electrons. :O=O: eight valence orbitals, twelve valence electrons. :F–F: eight valence orbitals, fourteen valence electrons. N2 has the strongest bond, because it is a triple bond. (b) The bond strength in a molecule X–Y is defined as: ΔH˚ for X–Y (g) → X (g) + Y (g) Calculate the bond strengths in N2, O2, and F2 from their heats of atomization. The heat of atomization is defined as the enthalpy of taking one mole of an element at its normal state (25 ˚C, 1 atm pressure) and converting it to gas phase atoms. For an element that exists gas phase diatomic element, X2 (g), the heat of atomization is: Chem 312, Au13; PS 8, p. 3 1/2 X2 (g) → X (g) ΔH˚atm(X) The bond strength is defined as the heat required to cleave the bond: X2 (g) → 2 X (g) ΔH˚(Bond Dissociation Energy of X2) So the bond strength is simply double the heat of atomization: N2, 946 kJ/mol; O2, 498 kJ/mol; F2, 158 kJ/mol. (c) Do your answers fit with the general rule from band filling in metals, relating the best bonding to the number of electrons and the number of orbitals? Yes, the data from the questions above are consistent with the general rule that the best bonding occurs when the number of electrons is equal to the number of orbitals. [This breaks down if you consider the molecule C2, which is formed in flames but is not a “stable” molecular species. Because of the shapes of the s and p orbitals, C2 can make only a triple bond, so it cannot use two of its valence electrons to form chemical bonds. This is one reason why elemental carbon is not diatomic but rather an extended structure, graphite or diamond.] 4. (a) List in order of decreasing covalent bond strength: I–I, C–C, P–P, Ge–Ge. In order of decreasing covalent bond strength: C–C, P–P, Ge–Ge, I–I. Covalent bonds get weaker as you go down the periodic table in the p block. (b) List in order of decreasing covalent bond strength: N–O, C–O, C–F, C–N. Explain your answers in terms of the factors that influence covalent bond strengths. In order of decreasing covalent bond strength: C–F, C–O, C–N, N–O. Covalent bonds are stronger when there is a larger difference in electronegativity. Note that the comparison in part (a) is made between bonds of equal polarity (all are non-polar covalent bonds) while the comparison in (b) is between bonds all involving atoms of the second row. Thus the two trends can be seen separately. 5. (a) Draw a thermochemical cycle for the enthalpy of MgCl2 dissolving in water. Clearly label the lattice energy, heat of solution of each ion, and whatever other terms appear in the cycle. Calculate the lattice energy from the equation with the Madelung constant – based on the radius ratio, which structure does MgCl2 adopt? Use this lattice energy to calculate the ΔH˚ of solution of MgCl2. (b) Draw a similar thermochemical cycle for the enthalpy of reaction of Mg(s) plus Cl2 (g) to make MgCl2 (s). Again, label all the necessary terms. Calculate the ΔH˚ of reaction. (c) Draw a similar thermochemical cycle for the enthalpy of reaction of Mg(s) plus Cl2 (g) to make MgCl2 (aq) -- that is, MgCl2 dissolved in water. Again, label all the necessary terms. Calculate the ΔH˚ of reaction. (d) Two of these values should add to give the third. Which two need to be added? Explain why this is true, using the chemical equations that describe parts (a), (b), and (c). Chem 312, Au13; PS 8, p. 4 Mg 2+ (g) + 2 -Cl (g) Lattice !H hyd 2x !H hyd Heats of Energy 2+ - hydration (Mg ) (Cl) 2+ - MgCl 2 (s) Mg (aq) + 2 Cl (aq) Heat of Solution (b) IP(1) + IP(2) Mg (g) + 2 Cl (g) Mg 2+ (g) + 2 -Cl (g) Electron Affinity !H atom 2x !H atom (Cl) Lattice (Mg) Energy Mg (s) + Cl2 (g) MgCl 2 (s) Heat of Reaction (c) IP(1) + IP(2) Mg (g) + 2 Cl (g) Mg 2+ (g) + 2 -Cl (g) Electron Affinity - !H hyd 2x !H hyd (Cl ) !H atom 2x !H atom (Cl) 2+ (Mg) (Mg ) 2+ - Mg (s) + Cl2 (g) Mg (aq) + 2 Cl (aq) Heat of Reaction Data: ΔHatm (Mg) = 146 kJ/mol; ΔHatm (Cl) = 122 kJ/mol.