III-1 1

PART THREE: Ionic Bonding In

. Molecules And Solids

So far we have defined several energies of interaction between charged particles.

- Ionization Energy - Electron Affinity - Coulombic (Electrostatic) Attraction & Repulsion.

We now use these to look at further bonding properties: Mostly solids.

Why are solids important? Catalysts Ad & Ab – Sorbents Lasers Fibre Optics Magnetic Memories Optical Switching (Computers) Batteries Fluorescent Lights Superconductors LED’s ………. III-2 2 We Need To Start At The Beginning

One of the simplest ionic solids is sodium chloride (NaCl).

Various depictions of the rocksalt structure. Not all will be met in this course.

III-3 3

Discrete NaCl molecules can exist in the gas phase.

NaCl(g) → Na(g) + Cl(g)

∆H for this reaction is called the Bond Dissociation Energy (409 kJ mol-1)

• We can judge the stability of solids whether they form or not by looking at the free energy change for:

M+(g) + X-(g) → MX(s)

∆G = ∆H - T∆S

• If ∆G is –ve then spontaneous.

• Note: Process of Lattice formation (solid ) is very exothermic at room temperature (∆S may be neglected). III-4 4

• We will use ∆H’s (enthalpy).

Born Haber Cycle

+ - Na (g) + Cl (g)

IP EA Electrostatic (U) _1 Attraction Na(g) + 2 Cl2(g) NaCl(g) ∆H f ( NaCl )

We can calculate U from this cycle

Can You Do It ?: watch the signs

However: NaCl(g) is NOT NaCl(s) III-5 5

- We want to get an idea of how stable different solids are.

- Then U is more complicated.

WHY?

Sodium Chloride Lattice

Cubic Na Cl + Cl Each Na has 2r - 3r Na 6 near neighbour Cl Cl Na Each Cl- has Na + r Cl 6 near neighbour Na • The attractive energy between a Na+ and its 6 near neighbour Cl- is offset by repulsion from 12 next near neighbour Na+.

• We must sum these up. III-6 6

First 6 nearest neighbour Na-Cl (Attractive)

-q q Na ⋅ Cl E = r × 6

Second 12 next nearest neighbor Na-Na (Repulsive) +q ⋅q E = Na Na × 12 2r

Third 8 next nearest neighbour Na-Cl

-q q E = Na Cl ×8 3r

(and so on … for ever) This is a convergent series. The sum of all the attractive and repulsive terms can be lumped together. -q2 E = r ⋅ A A is called the Madelung Constant. For NaCl and other cubic structures A = 1.74756. III-7 7

• NOTE: In this case the coulombic term is

overall attractive. Where is the repulsion to stop

the solid collapsing?

Born-Meyer Repulsion (DeKock & Gray, p. 457)

-ar ER = be (See problem set 4; Q. 4)

(remember Van der Waals repulsion?)

• Again due to overlap of electron clouds. b is a constant related to compressibility of solid. overall we have

III-8 8 E

Born Meyer Repulsion be-ar

dE = 0 r(dr DIST

Coulombic Attraction -Aq q 1 2 ( r

(a common value for a is 2.899) • We are almost ready to determine whether bonding in NaCl is Ionic (that is can it be described by Madelung, Coulomb Born and Meyer).

• How? â We can Calculate the Lattice Energy. III-9 9 ã We can Measure the Lattice Energy.

â Is from Ionic Bonding only.

ã Is real value.

Lattice Energies (Enthalpies) (p. 455, DeKock & Gray)

Definition: lattice engergy is the energy released when 1 mole of a substance is formed from its gas phase ions.

e.g. Na+(g) + Cl-(g) → NaCl(s) ∆G = U • first we will determine the lattice energy from experimental values.

III-10 10

• Born Haber Cycle

+ - N a ( g) + e + Cl(g) _ EA + T E N a - N a ( g) + C l( g) C l ( g) + (Na (g) _1 2 B E C l 2 _1

N a ( g) + 2 C l2 ( g)

U ( L A T T I C E E N E R G Y ) ∆Ha t _1 +

N a ( s ) N a ( s ) + 2 C l2 ( g)

∆H f N a C l ( s )

If we go round this cycle we must expend no energy (overall).

III-11 11 So, (Follow the Arrows)

-∆H +∆H (Na)+1BE(Cl )+IE(Na) f at 2 2 - EA(Cl) + U = 0

U = ∆H −∆H (Na) - 1BE(Cl )−IE+EA f at 2 2

U = -411 – 108 –121 –502 + 354

-1 UNaCl = -788 kJ mol

Now Let’s Calculate U (p. 456, DeKock & Gray)

• Earlier We wrote down the contributions to Ionic bonding as the sum of Attractive & Repulsive terms.

−Aq q 1 2 −ar E= r +be III-12 12 We Now Use the More General Form Given in DeKock and Gray

E = EC + ER (Coulombic and Repulsive)

For Avogadros number of units

+ − 2 AN(z )(z )⋅e −ar E = r +Nbe

(Z is charge: 1st term is –ve Attractive when charges are of different sign.

Now see DeKock and Gray p. 457-459.

We Follow Them

N ≡ AVOGADRO’S NUMBER

III-13 13

ANz+z−e2 −ar E = r + Nbe â E In The Energy Curve r dE = 0 at r = r dr eq

So r = r dE__ = 0 eq dr dE + − 2 = 0 = -ANz z e − Nabe−ar 2 dr req

⎛ ⎞ d ⎜1⎟ -1 ⎜ ⎟ = ⎜ r ⎟ 2 dr ⎝ ⎠ r + − 2 ar b = -Az z e e ã ar2 d ear = aear eq dr

PLEASE: NOTE MISTAKE IN DEKOCK & GRAY, P. 458, EQUATION 7-8.

III-14 14

SUB ã IN â

⎧ ⎫ ⎪ + - 2 ar ⎪ + 2 -Az z−e ⎪-Az z e e ⎪ -ar E = + N⎨ ⎬ e r ⎪ 2 ⎪ eq ⎪ r a ⎪ ⎩⎪ eq ⎭⎪

+ 2 + − 2 - CONSTANT E = -Az z e - ANz z e 2 8 -1 req a = 2.9×10 cm areq

LATTICE ⎧ ⎫ ANz+z-e2 ⎪ 1 ⎪ Er = - ⎨1 - ⎬ ENERGY eq req ⎪ areq ⎪ ⎩ ⎭ Couloubic + Repulsive BM Term. For NaCl req = 2.814D (See Later)

E = -759 kJ mol-1 (Substituting in the Values)

-1 Compare to EEXPTL = -788 kJ mol

III-15 15 • Good Agreement (within 5%) – we have ignored VDW but this is small.

• CONCLUSION: Bonding in NaCl is Substantially Ionic.

- Extra Lattice Energy is From Covalent Bonding (Part Five) ************************************* The following may not be covered Kapustinskii Equation

• If Madelung Constant for a number of structures is divided by the number of ins/formula unit: approximately same value results.

⎧ ⎫ ⎧ ⎫ ⎪ 1 ⎪ ⎪ 0.345⎪ • Replace ⎨1− ⎬ with ⎨1- ⎬ ⎪ r a ⎪ ⎪ r +r ⎪ ⎩ eq ⎭ ⎩ c a ⎭

⎧ ⎫ ANz+z-e2 ⎪ 1 ⎪ IN Er = ⋅ ⎨1 - ⎬ eq r ⎪ ar ⎪ eq ⎩ eq ⎭ III-16 16 • Also replace A with 1.21 MJ D mol-1

We Get

⎛ ⎞ -nz+z- ⎜ 0.345⎟ E = ⎜1− ⎟ ⎜ ⎟ d ⎝ rc +ra ⎠ r : cation radius⎫ c ⎪ ⎬ d = rc+ ra ⎪ ra : anion radius ⎭⎪

Example KNO3 n = 2 (2 Ions/formula unit) z+ = z- = 1

+ ⎫ r =1.38⎪ ⎬ rc + ra =3.27D - ⎪ r = 1.89⎭⎪

⎛ ⎞ 2 ⎜ 0.345⎟ -1 E = ⎜1 − ⎟ × 1.21 MJ mol ⎜ ⎟ 3.27 ⎝ 3.27 ⎠ = -622 kJ mol-1 III-17 17 Chemical Significance of Lattice Energies

• Lattice Energy (and EA) are principal (only)

reasons why ∆Hf is –ve. These alone make the solid stable.

• Look Again at the values used in the Born- Haber Calculation for NaCl.

• Kapustinskii Equation.

III-18 18 • Lattice Energy is a good measure of stability. Increases with Ion Charge with decrease in r.

Thermal Stabilities of Ionic Solids (Not in DeKock and Gray)

EXAMPLE: Carbonates

MCO3(s) → MO(s) + CO2(g)

Decomposition Temp: Is where ∆G for above react goes –ve

nt Observed to increase as M gets bigger. CaCO3 more stable than MgCO3 . i.e. Ionic Radii Ca2+ Mg2+ 1.00D 0.49D

WHY?

III-19 19

NOTE: Thermodynamics:

∆G = ∆H - T∆S

Decomp occurs when T = ∆H (after ∆G is ∆S –ve) For MCO3 decomp entropy is const – CO2 formation. ∴Enthalpy Change is good Guide of Stability

Large Cations Stabilize Large Anions

MCO3(s) → MO(s) + CO2(g)

∆H = ∆HL(MO) - ∆HL(MCO3)

i.e. Thermal Stability is measured by difference in Lattice Energies.

III-20 20 IF MO has much bigger lattice energy than MCO3.

THEN MCO3 will be very unstable wrt to MO.

Large Cation Small Cation

Cation

Lattice Small % scale } Large % change }change

Cation

(From: Inorganic Chemistry, D. Shriver, P. Atkins, C. Langford, p. 132, Pub. W.H. Freeman.) III-21 21 In Kapustinskii Equation

⎛ ⎞ −nz+z− ⎜ 0.345⎟ E = ⎜1 − ⎟ ⎜ ⎟ d ⎝ d ⎠ change in E ∝ 1 - 1 + 2- + 2- rM +r0 rM + rCO3

(all else is same)

If M+ is large diff is not large

(i.e. If M+ → ∞ we can ignore anion radii)

Greatest Change Occurs when r+M is small.

III-22 22 CHARGE

• Which are more stable M+ carbonates or M2+ carbonates (To thermal Decomp to Oxide)

KAPUSTINSKII

Difference in Latice Energy between M+ and M2+

⎧ ⎫ ⎪ 1 1 ⎪ ∆E∝zM+ ⎨ - ⎬ ⎪rM+ +rO2 rM+ +rCO2− ⎪ ⎩⎪ 3 ⎭⎪

DOMINATES

∴ M2+ change is largest

2+ + M carbonates less stable than M

*********************************** III-23 23 Structure of Ionic Solids (DeKock & Gray p. 427- )(Shriver 115-119)

IONIC MATERIALS: Properties

+ - • Contain Ions e.g. Na Cl , LiCl, CaF2 • Ionic Bonding Predominates • Low Electrical Conductivity • High Melting Points • Hard and Brittle • Soluble in Polar solvents

Structure

• Anions and Cations pack together in solid. - Electrostatic Attraction maximized - Electrostatic Repulsion Minimized

• Leads to different Geometric Arrangements of Ions

III-24 24 Unit Cell

• A simple arrangement of atoms which when repeated in 3 Dimensions produces the crystal lattice.

COMMON TYPES OF LATTICE

III-25 25

MORE ABOUT THESE LATER III-26 26

AB IONIC SOLIDS Madelung Const.

(A) CN (B) NaCl 6 6 1.7475 CsCl 8 8 1.7626 ZnS (Zincblende) 4 4 1.6380 ZnS(wurtzite) 4 4 1.6413 CaF2 8 4 2.5193 TiO2 (rutile) 6 3 2.408

Crystal Packing

• Can we make some simple rules about crystal packing that enable us to predict these structures?

â Ions are essentially spherical ã Pack together by size.

III-27 27

Radius Ratio Rules

• In ideal ionic crystals, coordination numbers are determined largely by electrostatic considerations.

• Cations surround themselves with as many anions as possible and vice-versa.

• This can be related to the relative sizes of the ions. ⇒ radius ratio rules.

as r+ ↑ the more anions of a particular size can pack around it.

III-28 28

III-29 29

Coordination r+/R- 8 0.732 6 0.414 4 0.225 3 0.155 Examples

BeS rBe2+/rsS2- = 0.59/1.7 = 0.35 ∴ CN = 4

NaCl rNa+/rCl- = 1.16/1.67 = 0.69 ∴ CN = 6 (anions do not touch) CsCl rCs+rCl- = 1.81/1.67 = 1.08 ∴ CN = 8

RR rule doesn’t always work ZnS: r+/r- = 0.52 ∴ CN = 6 III-30 30 (Actually 4)

Graph compares structures (CsCl / NaCl) with predictions by radius ratio rules from r+/r- {r-/r+ if cation is larger} For Li+ and Na+ salts, ratios calculated from both r6 and r4 are indicated .Radius ratios suggest adoption of CsCl structure more than is observed in reality NaCl structure is observed more than is predicted Radius ratios are only correct ca. 50% of the time, not very good for a family of ionic solids

III-31 31 Close Packing

• An alternative way of looking at Ionic Solids. • Anions are often larger than cations and therefore touch. • Small cations then fit in the “holes” between anions. • Think of packing basketballs and baseballs in the most efficient way.

1926 Goldschmidt proposed atoms could be considered as packing in solids as hard spheres

This reduces the problem of examining the packing of like atoms to that of examining the most efficient packing of any spherical object - e.g. have you noticed how oranges are most effectively packed in displays at your local shop? III-32 32

CLOSE-PACKING OF SPHERES

A single layer of spheres is closest-packed with a HEXAGONAL coordination of each sphere

III-33 33 A second layer of spheres is placed in the indentations left by the first layer

space is trapped between the layers that is not filled by the spheres TWO different types of HOLES (so-called INTERSTITIAL sites) are left OCTAHEDRAL (O) holes with 6 nearest sphere neighbours TETRAHEDRAL (T±) holes with 4 nearest sphere neighbours

III-34 34

When a third layer of spheres is placed in the indentations of the second layer there are TWO choices

The third layer lies in indentations directly in line (eclipsed) with the 1st layer Layer ordering may be described as ABA The third layer lies in the alternative indentations leaving it staggered with respect to both previous layers Layer ordering may be described as ABC

III-35 35

Close-Packed Structures

The most efficient way to fill space with spheres

Is there another way of packing spheres that is more space-efficient?

In 1611 Johannes Kepler asserted that there was no way of packing equivalent spheres at a greater density than that of a face-centred cubic arrangement. This is now known as the Kepler Conjecture.

This assertion has long remained without rigorous proof, but in August 1998 Prof. Thomas Hales of the University of Michigan announced a computer-based solution. This proof is contained in over 250 manuscript pages and relies on over 3 gigabytes of computer files and so it will be some time before it has been checked rigorously by the scientific community to ensure that the Kepler Conjecture is indeed proven!

An article by Dr. Simon Singh © Daily Telegraph, 13th August 1998 http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Lecture1/oranges.htmlT

III-36 36

Features of Close-Packing

Coordination Number = 12 74% of space is occupied Simplest Close-Packing Structures

ABABAB.... repeat gives Hexagonal Close-Packing (HCP) Unit cell showing the full symmetry of the arrangement is Hexagonal

III-37 37

ABCABC.... repeat gives Cubic Close-Packing (CCP) Unit cell showing the full symmetry of the arrangement is Face-Centred Cubic

III-38 38 The most common close-packed structures are METALS

A NON-CLOSE-PACKED structure adopted by some metals is:-

(Like CsCl)

III-39 39

III-40 40

Holes in the FCC Structure

• Anions pack close – cations fit in “holes”.

The locations of (a) octahedral holes and (b) tetrahedral holes in the fcc structure.

(a) Octahedral hole in cleft between six spheres (b) tetrahedral hole in the cleft between four spheres III-41 41 Hexagonal Close Paking (ABABAB)

Recall:

III-42 42 • Coordination # of HCP or FCC is 12.

• OCT Holes (6) TET Holes (4)

Some examples:

ZINC BLENDE or Sphalerite CCP S2- with Zn2+ in half Td holes Lattice: FCC 4 ZnS in unit cell

Coordination: 4:4 (tetrahedral)

III-43 43

Wurtzite

HCP S2- with Zn2+ in half Tetrahedral holes

III-44 44

Fluorite CCP Ca2+ with F- in all Tetrahedral holes

Lattice: fcc Coordination: Ca2+ 8 (cubic) : F- 4 (tetrahedral) In the related Anti-Fluorite structure (e.g., Li2O Cation and Anion positions are reversed

III-45 45

CARBON

III-46 46 Crystal Planes

• In looking at ionic structures in class we have seen that anions and cations tend to lie in “sheets” or planes.

Back to NaCl

Close packed layers III-47 47 • Planes of Atoms or groups are labeled using Miller Indices.

Miller Indices

DEFINITION: Smallest Integers which are reciprocals of the intercepts of the plane on the a, b & c axes of the tl • These are given symbols h, k, l and are written {1, 1, 1} or {2, 1, 0} etc.

Examples

{1, 0, 0} Planes in cubic system

c

b

a

CUTS a axis at 1 ∴ 1 Doesn’t cut c or b axes ∴ 0, 0 III-48 48 {1, 1, 1} Planes in cubic system. c

b a Cuts each axis at 1 ∴ {1, 1, 1}

{0, 1, 1} Plane

c

b a

Cuts b axis at 1 c axis at 1 a axis – doesn’t cut

⎛ ⎞ ⎜ 1 1 1⎟ ∴ {0, 1, 1} ⎜ , , ⎟ ⎜ ∞ ⎟ ⎝ 1 1⎠ Can you draw the arrangement of atoms on these Planes? See also problem sets. III-49 49 An Introduction to Diffraction Methods

• So far we have accepted these structures as given.

• They were determined using diffraction techniques.

• X-Ray Diffraction - tells us the size of the unit cell, where the atoms are.

X-RAYS are (partially) Reflected by planes of Atoms.

III-50 50

Bragg's Law wave in wave out

2 d 2

When λ = 2d sin θ, reflected waves have a phase difference of 2π, and interfere constructively.

What would happen if we squeezed in another plane of atoms at d/2 III-51 51 Rotating Crystal Powder (Debeye Scherer) Specim Reflect Incident ed Incident

Fil Cylindrical X-ray

constructive Interference Occurs when extra distance Beam 2 travels is a whole number of wavelengths nλ = 2R

but R = dsinθ Bragg Diffraction Law

∴ nλ =2dsinθ III-52 52 • nλ = 2dsinθ

λ = wavelength of x-rays θ = angle d = distance between crystal planes.

We know λ and θ therefore we can determine d for all the planes in the crystal.

• X-Ray diffraction patterns nowadays are produced on a computer screen from an X- ray detector. • NOTE d spacing is not normally the interatomic distance. e.g. (100) in fcc

Interatomic Distance

d100 III-53 53

Bragg Reflection - On-line Exercise

http://www.eserc.stonybrook.edu/ProjectJava/Br agg/ Try it for fun.