# Introduction to Inorganic Chemistry/Ionic and Covalent Solids - Energetics

## Contents

- 1
**Chapter 9: Ionic and Covalent Solids - Energetics** - 2 9.1 Ionic radii and radius ratios
- 3 9.2 Structure maps
- 4 9.3 Energetics of crystalline solids: the ionic model
- 5 9.4 Born-Haber cycles for NaCl and silver halides
- 6 9.5 Kapustinskii equation
- 7 9.6 Discovery of noble gas compounds
- 8 9.7 Stabilization of high and low oxidation states
- 9 9.8 Alkalides and electrides
- 10 9.9 Resonance energy of metals
- 11 9.10 Lattice energies and solubility
- 12 9.11 Discussion questions
- 13 9.12 Problems
- 14 9.13 References

## **Chapter 9: Ionic and Covalent Solids - Energetics**[edit]

In Chapter 8, we learned all about crystal structures of **ionic compounds**. A good question to ask is, *what makes a compound choose a particular structure?* In addressing this question, we will learn about the forces that hold crystals together and the relative energies of different structures. This will in turn help us understand in a more quantitative way some of the heuristic concepts we have learned about, such as hard-soft acid-base theory.

## 9.1 Ionic radii and radius ratios[edit]

## 9.2 Structure maps[edit]

## 9.3 Energetics of crystalline solids: the ionic model[edit]

Let's start looking at the lattice energies from the sense of chemical bonding. There are mainly two kinds of force that determine the energy of an ionic bond.

1) **Electrostatic Force** of attraction and repulsion (Coulomb's Law): Two ions with charges q1 and q2, separated by a distance r, experience and attractive force -F:

The Coulombic potential energy, V, is then given by

2) **Closed-shell repulsion**. When electrons in the closed shells of one atom overlap with those of another atom, there is a repulsive force comes from the Pauli exclusion principle. A third electron cannot enter an orbital that already contains two electrons. This force is short range, and is typically modeled as falling off exponentially or with a high power of the distance r between atoms. For example, in the Born approximation, B is a constant and n is a number ranging from 5~12

The energy of the ionic bond between two atoms is then calculated by the combination of net electrostatic attraction and the closed-shell repulsion energies, as shown in the figure at the right. In the crystal lattice, the equilibrium distance between ions is determined by the minimum in the total energy curve. At this distance, the net force on each ion is zero.

We can calculate the **electrostatic energy** of a crystal lattice using Coulomb's law as follows:

Consider a row of anions and cations in the NaCl structure.

We can see that the nearest neighbor interaction (+ -) is attractive, the next nearest neighbor interactions (- - and + +) are repulsive, and so on. In the NaCl structure, counting from the ion in the center of the unit cell, there are 6 nearest neighbors (on the faces of the cube), 12 next nearest neighbors (on the edges of the cube), 8 in the next shell (at the vertices of the cube), and so on:

The net attractive energy between the cation and anion in this infinite series will then result in a function:

Generalizing this formula for any three-dimensional ionic crystal we get a function:

where A is called the *Madelung constant*. The **Madelung constant** depends only on the geometrical arrangement of the point charges so it varies between different types of crystal structures, but within a structure type it does not change. Thus MgO and NaCl have the same Madelung constant because they both have the NaCl structure.

Having that formula in hand we can then combine the two forces into an equation which gives us the lattice energy.

Lattice Energy is equivalent to the equilibrium bond length so....

.

with .

and by expressing the repulsion term in exponential form we result in the Born-Mayer equation:

where p is the fudge factor~0.35Å

## 9.4 Born-Haber cycles for NaCl and silver halides[edit]

Which the equation and the derivation mentioned above we know that lattice energy is equivalent to its heat of formation. Lets consider the formation of table salt (NaCl)

Na^{+}(g)+Cl^{-}(g) -------> NaCl(s)

__Born-Harber cycle:__

S= Sublimation

IP= ionization of Na(g)

D= Dissocation energy

EA= Electron affinity

E_{L}=Lattice energy

R= Gas constant

t= Temperature

**From Hess' Law:**

dH_{f} = S + 1/2D + IP + EA + E_{L} + 2RT = -92.2

Acutal dH_{f}= -98.3

The error is only about 3% off. The result is promising because we neglected the van der Waals term.

__But....Why did we get away when Van der Waals term is neglected?__

This is because we used energy minimization during the calculation. This results in an overestimation of the repulsion force and underestimation of the attraction force. The two errors compensate each other out.

__Silver Halides__

Silver Halide | Calculated | Cycle | Difference |
---|---|---|---|

AgF | 220 | 228 | 8 |

AgCl | 199 | 217 | 18 |

AgBr | 105 | 215 | 20 |

AgI | 186 | 211 | 25 |

Looking at the table, we know that the ionic model didn't work for AgI from section 9.1 and 9.2 due to the electronegativity different dx=0.6. However we are still acquiring answers within ~10% error. How do we account for the partial covalent part of the E_{L}? The answers follows the same rule explained above. The ionic model over estimates the Madelung energy and this compensate for neglect of covalency.

## 9.5 Kapustinskii equation[edit]

The Lattice energy, E_{L}, for an ionic crystals are difficult to determine experimentally . In 1956 a Russian chemist, Anatoli Fedorovich Kapustinskii, published a formula that allow us to calculate E_{L} for any compound if we know the univalent radii. The Formula was later named The *Kapustinskii Formula.*

A. F. Kapustinskii noticed that the Madelung constant, A, is proportional to the number of ions in formula unit, n. showing:

Structure | A/n |
---|---|

NaCl | 0.874 |

CsCl | 0.882 |

Rutile | 0.803 |

Fluorite | 0.800 |

The difference in ionic radii between M^{+} and M^{2+} compensates for the difference in A/n for monovalent (NaCl, CsCl) and divalent (Rutile, CaF_{2}) structure.

This gives the Kapustinskii formula in () as:

The formula allows Neil Bartlett to fill out the fissing thems in the Born-Harder cycle, in this case EA, which is hard to measure and discovered the first Noble Gas.

## 9.6 Discovery of noble gas compounds[edit]

In 1962 at the University of British Columbia, Neil Bartlett synthesized PtF_{6} and accidentally noticed the compound’s reaction with O_{2} to generate O_{2}^{+}PtF_{6}^{-}.

EA can be calculated by following step. EA = -38 - 278 + 134 = -182 kcal/mol

Bartlett noticed Xe has ionization energy of +280 kcal, which is identical to the ionization energy of O_{2}, so he concluded that Xe^{+} should be about the same size as O_{2}^{+} and that Xe^{+}PtF_{6}^{-} should be a stable compound. The successful synthesis of Xe^{+}PtF_{6}^{-} demonstrated that similarities in lattice energies can predict the stability of unknown compounds. In the formula shown below detailing the reaction between Xe and PtF_{6}, Xenon prefers the +2 oxidation state.

For example, CuF and AuF are unknown compounds, but AgF is a known, stable compounds, so from the Born Haber cycle for CuF, the compound should be marginally stable with respect to the elements.

## 9.7 Stabilization of high and low oxidation states[edit]

To predict the stability of a certain oxidation state of a given ion, we can apply the lattice energy theories. Let's see two cycles showing the formation of CuF_{2} which is known to exist, and CuF which is unknown for its existence.

From the first cycle, CuF should be marginally stable with respect to the elements. The second cycle is for CuF_{2} which is known to exist.

Combining the two cycles shows that disproportionation is spontaneous (Same for 3AuF → AuF_{3} + 2Au). AgF is stable because the second IP is very high (495 kcal vs. 468 kcal for Cu, 473 for Au). Note that the difference in E_{L} (690-230=460 kcal) drives the disproportionation. This difference is bigger for smaller ions. For example, in fluorides, CuF is unstable but CuF_{2} is stable. However, in iodides, CuI is stable while CuI_{2} is unstable. As a conclusion: **small anions (O,F) stabilize high oxidation states, large anions (S, Br, I...) stabilize low oxidation states.**

## 9.8 Alkalides and electrides[edit]

Another interesting consequence of lattice energies involves the formation of certain salts containing Na^{-} + e^{-} as anions.

Complexing Na^{+} (K^{+}, Rb^{+}, Cs^{+}) with crown ethers stabilizes the M^{+} form of the metal ("salt" form) without generating a large E_{L} cost due to large atomic size of the anion and "particle in a box" effect.

## 9.9 Resonance energy of metals[edit]

Consider the stability of Na (metal) relative to Na+ e-. The Na (metal) is more stable relative to the Na+ e- "salt" due to the fact that valence electrons are shared by all atoms in the crystal. In the "salt" form, electrons are localized and the additional kinetic energy (ie. "particle in a box") adds to the total energy.

where:

h = Plank's constant; n = energy level, assume to be the lowest, n = 1 m = electron mass L = size of box

For a 3Å box:

So the Na^{+}e^{-} "salt" is unstable. The calculation is not very accurate because e- potential is not zero, and the "box" size is not so well defined.

## 9.10 Lattice energies and solubility[edit]

Lattice energies can also help predict compound solubilities.

For large anions, E_{L} doesn't change much with respect to r_{+}; E_{H} changes since E_{H} is proportional to . For example, with sulfate salts, MgSO_{4} (epsom salts) are soluble, while the larger BaSO_{4} (K_{sp} = 10^{-10})is insoluble.

For small anions, E_{L} is proportional to r^{+}, while E_{H} does not depend on r^{+} as strongly. For fluorides and hydroxides, LiF is slightly soluble while CsF is vastly soluble, and Mg(OH)_{2} is insoluble where Ba(OH)_{2} is very soluble. In both cases, the solubility depends on the size.

Combining the trends listed above:

1) Increasing size mismatch between the anion and cation leads to greater solubility, so CsF and LiI are the most soluble alkali halides.

2) Increasing covalency leads to lower solubility in the salts (due to larger E_{L}. For example, AgCl, AgBr, and AgI exhibit lower solubility as the atoms move down the row.

- AgCl(K
_{sp}= 10^{-10}) > AgBr> AgI(K_{sp}= 10^{-17})

3) Increasing charge on the anion generates lower solubility due to the increased effect on the E_{L} compared to the E_{H}.

4) Small, polyvalent cations (having E_{H}) generate soluble salts with large, univalent anions such as I^{-}, NO_{3}^{-}, ClO_{4}^{-}, PF_{6}^{-}, and acetate.

Examples:

- Lanthanides

- Ln
^{3+}: Nitrate salts are soluble, but oxides and hydroxides are insoluble.

- Fe
^{3+}: Perchlorate is soluble, but sulfate is insoluble.

5) Multiple charged anions such as O^{2-}, S^{2-}, PO_{4}^{3-}, and SO_{4}^{2-} generate insoluble salts with most M^{2+}, M^{3+}, and M^{4+} metals.

## 9.11 Discussion questions[edit]

- Explain why lattice energy calculations are very accurate for NaCl and CaCl
_{2}, but less accurate (by about 10%) for AgCl and PbCl_{2}. Does the Born-Mayer equation under- or overestimate the latter values? - Fluorine is more electronegative than oxygen. However, for many transition metals, we can make higher oxidation states in oxides than we can in fluorides. For example, Mn(IV) is stable in an oxide (MnO
_{2}), but MnF_{4}is unstable relative to MnF_{3}and fluorine.^{[1]}Can you explain this in terms of lattice energies?

## 9.12 Problems[edit]

1. Use lattice energies to explain why MgSO_{4} decomposes to magnesium oxide and SO_{3} at a much lower temperature than does BaSO_{4}.

2. Solid MgO might be formulated as Mg^{+}O^{-} or Mg^{2+}O^{2-}. Use the thermochemical data below (some of which are irrelevant) and Kapustinskii's formula to determine which is more stable. The lattice constant for MgO (NaCl structure) is 4.213 Å. While the idea of an O^{-} ion might seem strange, note that the second electron affinity of O and the second ionization potential of Mg (in the table below) are both quite endothermic.

Reaction | ∆H^{o}, kcal/mol |
---|---|

Mg(s) = Mg(g) | 35.3 |

Mg(g) = Mg^{+}(g) + e^{-} |
176.6 |

Mg^{+}(g) = Mg^{2+}(g) + e^{-} |
347.0 |

O_{2}(g) = 2 O(g) |
119.0 |

O(g) + e^{-} = O^{-}(g) |
-33.7 |

O^{-}(g) + e^{-} = O^{2-}(g) |
188.9 |

3. From the heat of formation of solid NH_{4}Cl (-75.2 kcal/mol) and gaseous NH_{3} (-11.0), the bond dissociation energies of H_{2} (104.2) and Cl_{2} (58.2), the ionization potential of atomic hydrogen (313.4), and the electron affinity of atomic chlorine (-83.4), calculate the gas-phase proton affinity of NH_{3}. The lattice energy of NH_{4}Cl may be estimated from Kapustinskii's formula using r_{N-Cl} = 3.50 Å.

4. Bottles of aqueous ammonia are often labeled “ammonium hydroxide.” We will test this idea by using a lattice energy calculation to determine whether the salt NH_{4}^{+}OH^{-} can exist.

The heats of formation of gaseous OH- and H2O are respectively -33.7 and -57.8 kcal/mol. Assuming that NH_{4}^{+} is about the same size as Rb^{+}, and OH^{-} about the same size as F^{-}, using Kapustinskii's formula, ionic radii, and the NH_{3} proton affinity calculated in problem 3, determine whether NH_{4}^{+}OH^{-} should be a stable salt relative to NH_{3} and H_{2}O. At what temperature should NH_{4}^{+}Cl^{-} be unstable relative to NH_{3} and HCl, if ΔH_{f}^{o} for HCl is -22.0 kcal/mol and ΔS^{o} (NH_{4}Cl --> NH_{3} + HCl) = 67 cal/mol K?

5. (a) Do you expect BaSO_{4} or MgSO_{4} to be more soluble in water? (b) Is LiF more soluble than LiClO_{4}? Explain.

6. Which polymorph of ZnS (zincblende or wurzite) would you expect to be more stable on the basis of electrostatic energy?

## 9.13 References[edit]

- ↑ K. O. Christe, "Chemical synthesis of elemental fluorine,"
*Inorg. Chem.***1986**,*25,*3721–3722. DOI: 10.1021/ic00241a001