Chapter 6: Ionic Compounds andThe Lattice Energy

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Review of the Previous Lecture

1. Solid state lattices vary in terms of packing efficiency Packing: Simple Cubic < Body-centered < Close-packed layers

2. Atom contribution differs in the unit cell depending on location Corner: 1/8 Edge: 1/4 Face: 1/2 Inside the unit cell: 1

3. Lattice types 1:1 Cation:Anion 1:2 Cation:Anion

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Let’s consider how some of these lattices are related byions filling in the holes of a common lattice type.

Ca2+ Zn2+

Na+

Bi3-

4CCP: Cubic close-packing.

A. Calculation of UO may appear straightforward, equal to the potential energy of a cation and anion interacting:

Use Coulomb’s Law

E = in kJ mol-1

1. Lattice Energy (U)

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Energy released when ions come together from infinite separation to form a crystal.

M+(g) + X- (g) MX (s)Uo

4πϵoro

Z+Z-e2Z+ , Z- = Charges on the ionsϵo = Permittivity of vacuum

= 8.854185 x 10-12 C2J-1m-1

ro = Distance between the ions = r++ r- (pm)e = Charge on electron = 1.6210x10-19 C

Plot of potential energy of ions coming together.

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E = 4πϵoro

Z+Z-e2

+

+

‐

‐

ro

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The problem with this model is that it assumes thateach cation only experiences attraction from one anionwithin the crystalline lattice.

1. Lattice Energy (U)

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B. Must account for close and long-range interactions

Na+

Nearest neighbors to Na+:6 attractions to Cl- at distance r

1. Lattice Energy (U)

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B. Must account for close and long-range interactions

Na+

Nearest neighbors to Na+:6 attractions to Cl- at distance r

2nd Nearest neighbors to Na+:12 repulsions from Na+

at distance √2 r

1. Lattice Energy (U)

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B. Must account for close and long-range interactions

Na+

Nearest neighbors to Na+:6 attractions to Cl- at distance r

2nd nearest neighbors to Na+:12 repulsions from Na+

at distance √2 r

3rd nearest neighbors to Na+:8 attractions to Cl-

at distance √3 r

1. Lattice Energy (U)

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B. I. The Madelung Constant (M or in some textbooks A)

Geometric factor that sums all the attractions and repulsions in a crystal form.

Madelung Constant

Nearest neighbors to Na+:6 attractions to Cl- at distance r

2nd nearest neighbors to Na+:12 repulsions from Na+

at distance √2 r

3rd nearest neighbors to Na+:8 attractions to Cl-

at distance √3 r

1. Lattice Energy (U)

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C. Lattice Energy for known lattices

Born-Mayer equation:

4πϵo ro ro

- NM│Z+ │ │ Z- │ e2 1 - ρΔU = N = Avogadro’s number = 6.022x1023 mol-1

M = Madelung Constant (Look it up)ρ = A constant dependent on the

compressibility of the crystal

in kJ mol-1

1. Lattice Energy (U)

13Copyright © 2014 Pearson Education, Inc.

1. Lattice Energy (U)

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D. Lattice Energy for unknown lattices

Kapustinskii equation:

ro

- 107,000 υ │Z+ │ │ Z- │ΔU = υ= Number of ions in the ionic compoundin kJ mol-1

2. Born-Haber Cycle

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Recall Hess’s Law: Enthalpy is independent of the path.

Let’s consider the enthalpy of formation of an ionic lattice:

M(s) + ½ X2 (s) MX(s)ΔHfEnthalpy of formation

2. Born-Haber Cycle

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Recall Hess’s Law: Enthalpy is independent of the path.

Let’s consider the enthalpy of formation of an ionic lattice:

M(s) + ½ X2 (s) MX(s)

M(g) M+(g)

X (g) X-(g)

ΔHsubsublimation

ΔHIEIonization Energy

ΔHAatomization

ΔHEAElectron Affinity

ΔHfEnthalpy of formation

ΔULattice Energy

+

A Born-Haber thermochemicalcycle of an ionic compoundgives the enthalpy changeassociated with the formation ofthe ionic lattice.

2. Born-Haber Cycle

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Recall Hess’s Law: Enthalpy is independent of the path.

Let’s consider the enthalpy of formation of an ionic lattice:

M(s) + ½ X2 (s) MX(s)

M(g) M+(g)

X (g) X-(g)

ΔHsubsublimation

ΔHIEIonization Energy

ΔHAatomization

ΔHEAElectron Affinity

ΔHfEnthalpy of formation

ΔULattice Energy

+ ΔHf = ΔHsub + ΔHA + ΔHIE + ΔHEA + ΔU

2. Born-Haber Cycle

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ΔHsub , ΔHA , ΔHIE , ΔHEA : Terms that can be looked up in tables.

ΔHEA , ΔU : Terms that are negative values. Recall that ΔHEA = - Electron Affinity

ΔHf = ΔHsub + ΔHA + ΔHIE + ΔHEA + ΔU

Let’s consider the thermodynamic drive for the formation of a solid state lattice:

Gibbs free energy ΔG = ΔH - TΔS

Where ΔG = Negative values for a thermodynamically favorable process

ΔH = ΔHf

2. Born-Haber Cycle

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ΔHf = ΔHsub + ΔHA + ΔHIE + ΔHEA + ΔU

ΔG = ΔH - TΔS

The formation of a solid state lattice is a process that results in a more orderly state for atoms in the compound. Therefore TΔS is a negative value ΔHf must be very negative to make ΔG negative ΔU must be very negative to make ΔHf very negative

Why can’t we have NaCl2?

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ΔHf = ΔHsub + ΔHA + ΔHIE + ΔHEA + ΔU

Na(s) + Cl2 (s) NaCl2(s)ΔHfEnthalpy of formation

Na(s) + Cl2 (s) NaCl2(s)

Na2+(g)

2Cl-(g)

ΔHfEnthalpy of formation

ΔULattice Energy

+

Remove two electrons from Narequires high second ΔHIE Value of ΔU will not be

negative enough to overcomethe value of ΔHIE

3. Size Effects

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Factors that affect ionic radii

I. Left to right in Periodic Table, ↓ Ionic radius of cation

II. ↑ Ionic charge of cation, ↓ Ionic radius Greater pull on e-

III. As coordination number of cation ↑ , ↑ Ionic radius

IV. The number of unpaired electrons affects the ionic radius of transition metals

3. Size Effects

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V. The radius ratio of cation/anion (r+/r-) can dictate coordination number and geometry of the cation.

r+/r- Coordination Number Geometry<0.15 2 Linear

0.15‐0.22 3 Trigonal Planar0.22‐0.41 4 Tetrahedral

0.41 4 Square Planar0.41‐0.73 6 Octahedral>0.73 8 Cubic

4. Covalent character in ionic bonds

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Must consider a certain amount of covalency in ionic bonds via polarization.

Polarization induced by one ion to the oppositely charged ion ↑ by

1. Small, highly charged cations2. Large, highly charged anions3. The number of unpaired electrons of cation

Results of polarization

1. High melting points2. Solubility in polar solvents3. Higher covalent character in a bond, shorter bond lengths