• Allotropes: Section 6.2
• Conductors, semiconductors, and ferromagnetism: Section 6.6
• Metal strength: Section 6.6. Hardness and softness are related to the numbers of bonding electrons available in the metal
• Catenation: Section 6.7
• pi-bonding: Section 6.5. (pi-bonding is common when pi-bonds are stronger than sigma bonds)
• Metal-metal bonding (a form of catenation): Section 6.7
• Liquids at or near room temperature: there are only 4 nonradioactive elements that are liquids, 3 metals and one nonmetal. (Br₂, Hg, Cs, Ga)
• Gases at room temperature: 11 of them, all nonmetals.
• Only known substance with 2 liquid forms: helium (p. 184)

van der Waals radius and anionic radius are about the same. They are the largest radii.

Covalent radius and single-bonded metallic radius are about the same, and are about 60 pm smaller than the van der Waals/anionic radii.

Cation radius is the smallest, and is about 60 pm less than the covalent/single-bonded metallic radii.
 
 

(a) Draw Lewis structures for the product and the reactant. Answer:

(b) Calculate the enthalpy change for this reaction from the bond energies for products and reactants. The bond energy for C-O sigma bonds is 358 kJ/mol; for C-O pi bonds it is 374 kJ/mol. [Note: sigma and pi bond energies between two atoms of the same kind are in Table 6.5 in Wulfsberg.] Answer:

The bond energy is the enthalpy change for breaking a bond in a molecule with the reactants and products in the gas phase.

One mole of C₂O₄ has 6 C-O sigma bonds and 2 C-O pi bonds. Thus 6 x (358 kJ/mol to break sigma bonds) + 2 x (374 kJ/mol to break pi bonds) = 2896 kJ to break the bonds and form 2 moles of C atoms and 4 moles of O atoms.

C₂O₄ ---> 2 C (g) + 4 O (g); Delta H = +2896 kJ

Now let the C and O atoms combine to form two moles of CO₂, in which there are two sigma bonds and two pi bonds in each molecule. Thus four moles of C-O sigma bonds and two moles of pi-bonds will be formed, for a total energy gain of 4 x (-358 kJ/mol to form sigma bonds) + 4 x (-374 kJ/mol to form pi bonds) = -2928 kJ

2 C (g) + 4 O (g) ---> 2 O=C=O; Delta H = -2928 kJ

Hess's Law states that when you add two (or more) reactions to give you an overall reaction, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions. Thus,

C2O4 ---> 2 C (g) + 4 O (g)	Delta H = +2896 kJ
2 C (g) + 4 O (g) ---> 2 O=C=O	Delta H = -2928 kJ
Overall: C2O4 ---> 2 O=C=O	Delta H = -2928 kJ + 2896 kJ = -32 kJ

(c) Which is more stable? Answer CO₂ is more stable because the enthalpy change for the overall rreaction is -32 kJ.

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For problems 4, 5, and 6 you will need to make use of most or all of the following tables:

Table 2.1: Hydration Energies

Table 6.1: Atomization Energies

Tables 6.6-6.8: 1st, 2nd, and 3rd Ionization Energies

Table 6.9: Electron Affinities

Solution. A thermochemical cycle may be shown as a diagram, or as a set of equations (see following the figure). In the figure, the overall reaction is represented on the top line (step 1). This reaction defines the enthalpy of formation for Al₂O₃.

A thermochemical cycle shows that reactants can form a product in a single step (step 1 below), or in a series of steps (Steps 2, 3, 4, 5, and 6). Whichever way the reaction occurs, the enthalpy change is the same. This is what Hess's Law says: the enthalpy change associated with a reaction is the same whether the reaction proceeds in one step or many steps.

Hess's Law: If a reaction can be written as the sum of two or more constituent reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the constituent reactions.

Thus the enthalpy associated with step 1 is the sum of the enthalpy changes of steps 2, 3, 4, 5, and 6.

• In steps 2 and 3, we imagine atomizing the elements (atomization energies). Table 6.1
• In steps 4 and 5, we let them lose or gain electrons to form the ions needed in the final compound (ionization energies and electron affinities). Tables 6.6 - 6.9
• In step 6 we let the gaseous ions combine to form the crystalline compound (lattice energy).

Chemical equations and their associated enthalpies are shown below. Keep in mind that the nth electron affinity is the ionization energy of the -n ion:

Definition	Equation	Delta H, kJ/mol
1st electron affinity of O	O- (g) ---> O (g) + e-	141
2nd electron affinity of O	O2- (g) ---> O- (g) + e-	-780

(Note that the positive sign for the first electron affinity means that the O^- ion is stable with respect to losing an electron. However, the O^2- ion is NOT stable in the gas phase. The oxide ion only exists in ionic compounds where the cation attraction to an O^2- ion is high enough to outweigh the unfavourable electron affinity of forming the ion. Coulomb's law at work again!)

Thus the lattice energy is -1676 kJ - 13596 kJ = -15272 kJ

(a) Set up a thermochemical cycle and calculate the enthalpy of reaction.

(b) Ignoring entropy effects, calculate E^o for this reaction.

(c) Does this reaction go as written to produce products?

(b) Ignoring entropy effects means that we will assume that the entropy change is zero. In this reaction, two electrons are involved as Zn is oxidized to the +2 ion and the Mn²⁺ ion is reduced to manganese. Thus n = 2 in the equation below.

(c) The reaction isn't expected to go to any appreciable extent since the enthalpy change is positive (leading to a negative value of E^o for the reaction.)
 
 

6. Hints. This question works the same way as the previous two problems. You'll be using atomization energies, ionization energies (or their reverse) for metals and hydrogen, electron affinities for halogens, and hydration energies for all ions involved. Just proceed in an orderly manner as we have done in the above examples.
 
 

Last revised: March 27 2001 by M. F. Richardson © Brock University, 2001