Prof.P. Ravindran, - folk.uio.nofolk.uio.no/ravi/cutn/cmp/latticeEnergyf.pdf · Born-Mayer equation...

P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 16 July: Lattice Energy

http://folk.uio.no/ravi/CMP2013

Prof.P. Ravindran, Department of Physics, Central University of Tamil

Nadu, India

Lattice Energy

1


Introduction

Cohesive energy

energy required to

break up crystal into

neutral free atoms.

Lattice energy (ionic

crystals) energy

required to break up

crystal into free ions.


Kcal/mol = 0.0434 eV/molecule KJ/mol = 0.0104 eV/molecule




Interactions between Ions 6

Typically consider an ionic solid with many cations and many anions

All ions are interacting with each other: repulsion and attraction

Lattice energy of a solid – ΔE of ions in gas vs solid

High LE – strong interaction between ions, tightly bonded solid

Start with the CPE of 2 ions with charges z1 and z2:

Total PE of ionic solid is sum of CPE interactions between all ions

120

2

21

120

2112

r4

ezz

r4

ezezCPE


Electronegativity and Chemical Bonds

1. Ionization enthalpy

The enthalpy change when one mole of electrons are removed from one mole of atoms or positive ions in gaseous state.

X(g) X+

(g) + e- H 1

st I.E.

X+

(g) X2+

(g) + e- H 2

nd I.E.

Ionization enthalpies are always positive.


X(g) + e- X

-(g) H 1

st E.A.

X

(g) + e- X

2(g) H 2

nd E.A.


2. Electron affinity

The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state.

Electron affinities can be positive or negative.



I.E. and E.A. only show e- releasing/attracting power of

free, isolated atoms

However, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.



3. Electronegativity

The ability of an atom to attract electrons in a chemical bond.

Mulliken’s scale of electronegativity

Electronegativity(Mulliken) )(2

1EAIE

Nobel Laureate in Chemistry, 1966


Pauling’s scale of electronegativity

Nobel Laureate in Chemistry, 1954

Nobel Laureate in Peace, 1962



calculated from bond enthalpies

cannot be measured directly

having no unit

always non-zero

the most electronegative element, F, is arbitrarily assigned a value of 4.00


IA IIA IIIA IVA VA VIA VIIA VIIIA

H

2.1

He

-

Li

1.0

Be

1.5

B

2.0

C

2.5

N

3.0

O

3.5

F

4.0

Ne

-

Na

0.9

Mg

1.2

Al

1.5

Si

1.8

P

2.1

S

2.5

Cl

3.0

Ar

-

K

0.8

Ca

1.0

Ga

1.6

Ge

1.8

As

2.0

Se

2.4

Br

2.8

Kr

-

Rb

0.8

Sr

1.0

In

1.7

Sn

1.8

Sb

2.0

Te

2.1

I

2.5

Xe

-

Cs

0.7

Ba

0.9

Tl

1.8

Pb

1.8

Bi

1.9

Po

2.0

At

2.2

Rn

-



What trends do you notice about the EN values in the Periodic Table? Explain.

IA IIA IIIA IVA VA VIA VIIA VIIIA

H

2.1

He

-

Li

1.0

Be

1.5

B

2.0

C

2.5

N

3.0

O

3.5

F

4.0

Ne

-

Na

0.9

Mg

1.2

Al

1.5

Si

1.8

P

2.1

S

2.5

Cl

3.0

Ar

-

K

0.8

Ca

1.0

Ga

1.6

Ge

1.8

As

2.0

Se

2.4

Br

2.8

Kr

-

Rb

0.8

Sr

1.0

In

1.7

Sn

1.8

Sb

2.0

Te

2.1

I

2.5

Xe

-

Cs

0.7

Ba

0.9

Tl

1.8

Pb

1.8

Bi

1.9

Po

2.0

At

2.2

Rn

-


What trends do you notice about the EN values in the Periodic Table? Explain.

The EN values increase from left to right across a Period.

The atomic radius decreases from left to right across a Period.Thus,the nuclear attraction experienced by the bonding electrons increases accordingly.


The EN values decrease down a Group.

The atomic radius increases down a Group, thus weakening the forces of attraction between the nucleus and the bonding electrons.

What trends do you notice about the EN values in the Periodic

Table? Explain.


Born-Mayer Equation:

U0 = (e2 / 4 0) * (N zA zB / d0) * A * (1 – (d* / d0))

U0 = 1390 (zA zB / d0) * A * (1 – (d* / d0)) in kJ/mol

Kapustinskii equation :

U0 = (1210 kJ Å / mol) * (n zA zB / d0) * (1 – (d* / d0))

Where:

e is the charge of the electron, 0 is the permittivity of a vacuum

N is Avogadro’s number

zA is the charge on ion “A”, zB is the charge on ion “B”

d0 is the distance between the cations and anions (in Å) = r+ + r-

A is a Madelung constant

d* = exponential scaling factor for repulsive term = 0.345 Å

n = the number of ions in the formula unit

Ionic Bonding

The equations that we will use to predict lattice energies for crystalline solids are the

Born-Mayer equation and the Kapustinskii equation, which are very similar to one

another. These equations are simple models that calculate the attraction and

repulsion for a given arrangement of ions.


In this case the energy of coulombic forces (electrostatic attraction and repulsion) are:

Ecoul = (e2 / 4 0) * (zA zB / d) * [+2(1/1) - 2(1/2) + 2(1/3) - 2(1/4) + ....]

because for any given ion, the two adjacent ions are each a distance of d away, the next two

ions are 2d, then 3d, then 4d etc. The series in the square brackets can be summarized to

give the expression:

Ecoul = (e2 / 4 0) * (zA zB / d) * (2 ln 2)

where (2 ln 2) is a geometric factor that is adeqate for describing the 1-D nature of the infinite

alternating chain of cations and anions.

For an Infinite Chain of Alternating Cations and Anions:

Ionic Bonding

The origin of the equations for lattice energies.

The lattice energy U0 is composed of both coulombic (electrostatic) energies and an additional

close-range repulsion term - there is some repulsion even between cations and anions because of

the electrons on these ions. Let us first consider the coulombic energy term:

U0 = Ecoul + Erep


This means that the general equation of Coulombic energy for any 3-D ionic solids is:

Ecoul = (e2 / 4 0) * (zA zB / d) * A

Note that the value of Ecoul must be negative for a stable crystal lattice.

Ionic Bonding

For a 3-dimensional arrangement, the geometric factor will be different for each different

arrangement of ions. For example, in a NaCl-type structure:

Ecoul = (e2 / 4 0) * (zA zB / d) * [6(1/1) - 12(1/2) + 8(1/3) - 6(1/4) + 24(1/5) ....]

The geometric factor in the square brackets only works for the NaCl-type structure, but

people have calculated these series for a large number of different types of structures

and the value of the series for a given structural type is given by the Madelung constant,

A.


The numerical values of Madelung constants for a variety of different structures are

listed in the following table. CN is the coordination number (cation,anion) and n is the

total number of ions in the empirical formula e.g. in fluorite (CaF2) there is one cation

and two anions so n = 1 + 2 = 3.

Ionic Bonding

lattice A CN stoich A / n

CsCl 1.763 (8,8) AB 0.882

NaCl 1.748 (6,6) AB 0.874

Zinc blende 1.638 (4,4) AB 0.819

wurtzite 1.641 (4,4) AB 0.821

fluorite 2.519 (8,4) AB2 0.840

rutile 2.408 (6,3) AB2 0.803

CdI2 2.355 (6,3) AB2 0.785

Al2O3 4.172 (6,4) A2B3 0.834

Notice that the value of A is fairly constant for each given stoichiometry and that the

value of A/n is very similar regardless of the type of lattice.


Ionic Bonding

This is the Born-Mayer equation, when the constants are evaluated

we get the form of the equation that we will use:

U0 = 1390 (zA zB / d0) * A * (1 - (d* / d0)) in kJ/mol

Note: d* is the exponential scaling factor for the repulsive term and

a value that we will use for this is 0.345 Å.

If only the point charge model for Coulombic energy is used to estimate the lattice energy (i.e. if U0 = Ecoul) the

calculated values are much higher than the experimentally measured lattice energies.

E.g. for NaCl (rNa+ = 0.97Å, rCl- = 1.81Å):

U0 = 1390 (zA zB / d0) * A = 1390 ((1)(-1)/2.78) * (1.748) kJ/mol = - 874 kJ/mol

But the experimental energy is -788 kJ/mol. The difference in energy is caused by the repulsion between the

electron clouds on each ion as they are forced close together. A correction factor, Erep, was derived to

account for this.

Erep = - (e2 / 4 0) * (zA zB d*/ d2) * A and since

Ecoul = (e2 / 4 0) * (zA zB / d) * A the total is given by

U0 = (e2 / 4 0) * (zA zB / d0) * A * (1-(d*/d0))


Ionic Bonding

Using the Born-Mayer equation, for our example with NaCl.

U0 = 1390 (zA zB / d0) * A * (1 - (d* / d0))

= 1390 ((1)(-1)/2.78) * (1.748) * (1-(0.345/2.78) kJ/mol = - 765 kJ/mol

Which is much closer to the experimental energy of -788 kJ/mol.

Kapustinskii observed that A/n is relatively constant but increases slightly with

coordination number. Because coordination number also increases with d, the value of

A/nd should also be relatively constant. From these (and a few other) assumptions he

derived an equation that does not involve the Madelung constant:

Kapustinskii equation :

U0 = (1210 kJ Å / mol) * (n zA zB / d0) * (1 – (d* / d0))

One advantage of the Kapustinskii equation is that the type of crystal lattice is not

important. This means that the equation can be used to determine ionic radii for non-

spherical ions (e.g. BF4-, NO3-, OH-, SnCl6-2 etc.) from experimental lattice energies. The

self-consistent set of radii obtained in this way are called thermochemical radii.



Consider a line of alternating cations and anions:

CPE of an ion in center:

A = 2 ln 2

CPE is negative, net attraction between the ions

Now expand the model to 3D:

Coefficient A – Madelung constant – related to arrangement of ions

E A z2NAe

2

40d



As ions are separated, the attraction decreases

If ions are too close, past the point of contact, they repel each other

There is an ideal separation between ions:

Born-Meyer equation

d* = 34.5 x 10-12 m

PEmin NA z1z2 e

2

40d1d*

d

A


Ionic Crystals

ions: closed outermost shells

~ spherical charge

distribution

Cohesive/Binding energy

= 7.9+3.615.14 = 6.4 eV



Electrostatic (Madelung) Energy Interactions involving ith

ion: i i j

j i

U U

2/

2

. .R

i j

i j

qn ne

RU

qotherwise

p R

tot iU NUFor N pairs of

ions:

2/R q

N z eR

z ﹦number of n.n.

ρ ~ .1 R0

j i i jp

﹦Madelung

constant

At equilibrium: 0totdU

dR

2/

2

Rz qN e

R

→ 0

2/2

0

R qR e

z

2

0

0 0

1tot

N qU

R R

2

0

N qMadelung Energy

R


Evaluation of Madelung Constant

App. B: Ewald’s method

j i i jp

1 1 12 1

2 3 4

2ln 2

KCl

i fixed


Kcal/mol = 0.0434 eV/molecule


Neighbors Signs Numbers Distance

1st - 6 r

2nd + 12

3rd - 8

4th + 6 2r

31

31

Madelung Constant

r2

r3

Arrangement of ions in sodium chloride structures.


32 32

Total Coulombic potential:

r

e

r

e

r

e

r

e

r

e

r

eU

oA

0

2

0

2

2

0

2

0

2

0

2

4

748.1

4

...)2(4

634

824

124

6


The Madelung constants for other structures are listed

below:

NaCl structure 1.747565..

CsCl 1.762675

ZnS (Zinc Blende) 1.6381


34

34

Crystal Potential

The crystal potential is the summation of repulsive

energy and the Coulombic attraction

r

e

r

B

rUrUrV

on

AR

4

)()()(

2

Attractive energy EA

Net energy EN

Repulsive energy ER

Interatomic separation r

Ed


Cry

sta

l P

ote

ntial V

(r)

distance r

Coulombic attraction UA

Exclusion repulsive force UR

V(r)

equilibrium position

r = ro


Equilibrium position

The position at which the energy is minimum.

Balance between the repulsive force and attractive

force.

To evaluate, the first order derivative of V with

respect to r is zero, i.e.

02

2

1

o

norr r

e

r

nB

dr

dV

o


12

no

e

nBr

The equilibrium position is then

given by

And the minimum energy is

)1

1(4

)(2

minnr

erVV

oo

o


38

Example

Parameters for Sodium Chloride:

– B = 1.209x10-96 eV m9

– n = 9

– =1.7576

– Ionization energy for sodium = 5.14eV, and electron

affinity of chlorine = 3.61 eV

Find the cohesive energy Ecoh


39 39

Definitions

Ionization Energy

A loss in energy for the ionization

+

Sodium ion

Na+

Sodium atom

Eion

http://upload.wikimedia.org/wikipedia/commons/7/76/Electron_shell_011_Sodium.svg

http://upload.wikimedia.org/wikipedia/commons/7/76/Electron_shell_011_Sodium.svg


Electron affinity

Capture

electron

-

Chorine atom Cl Chorine ion Cl-

Energy released by capturing electron

http://upload.wikimedia.org/wikipedia/commons/5/56/Electron_shell_017_chlorine.png

http://upload.wikimedia.org/wikipedia/commons/5/56/Electron_shell_017_chlorine.png


NaCl molecule

dissociation

Sodium

atom

Chlorine

atom

Cohesive energy is the energy required to

dissociate the molecule into atoms..

Cohesive energy


42 42

Equilibrium position of Sodium Chloride Crystal

12

no

e

nBr

819

96

10609.17476.1

10209.19

xx

xx

Did I make mistake? It must

be e2


There is no mistake. Sine the unit of B is eV m9. It must be

converted into Joule by multiplying the electron charge e. The

electron charge cancels out one of the e factor in the

denominator.

Effectively, it can express B in eV m9, while eliminating one e

in the denominator.

The calculation result for ro becomes

r0 = 2.887x10-10 m


44 44

The energy required to dissociate a sodium

chloride molecule into separate ions:Ed

ClNaENaCl d

)1

1(4

)(2

nr

erVE

oo

od

9

8

1085.81416.44

10602.1747565.112

19

xxxx

xx

Why e instead of e2

?

= 7.96 eV


45

Ionization energy of Na

eNaENa ion

Electron affinity of Cl

affECleCl

affion EEClNaClNa

Eion = 5.14 eV

Eaff = 3.61 eV

recall

dENaClClNa


46 46

affiond EEENaClClNa

cohENaClClNaClNa

where

Ecoh = Ed -Eion + Eaff

In our present case Ecoh = 7.96 -5.14 +3.61 =

6.43 eV

Cohesive energy per ion = 6.43/2 = 3.21eV


Born-Haber Cycles

Calculating lattice enthalpies.

http://images.google.co.uk/imgres?imgurl=http://nobelprize.org/nobel_prizes/chemistry/laureates/1918/haber.jpg&imgrefurl=http://nobelprize.org/nobel_prizes/chemistry/laureates/1918/haber-bio.html&h=227&w=162&sz=12&hl=en&start=3&um=1&usg=__LUcInb-FG4aA4EapsSG61paDBkY=&tbnid=IC3qcrKZ3GaloM:&tbnh=108&tbnw=77&prev=/images?q=fritz+haber&um=1&hl=en&rls=GFRC,GFRC:2006-51,GFRC:en&sa=N


Lattice enthalpy terms

∆Hө L.E = Lattice enthalpy. The standard enthalpy change when one mole of

crystalline substance is formed from its constituent gaseous ions.

Na+(g) + Cl-(g) → NaCl(S)


Born-Haber cycles

Define and apply the terms enthalpy of formation,

ionisation enthalpy, enthalpy of atomisation of an

element and of a compound, bond dissociation

enthalpy, electron affinity, lattice enthalpy (defined

as either lattice dissociation or lattice formation),

enthalpy of hydration and enthalpy of solution.

Construct Born–Haber cycles to calculate lattice

enthalpies from experimental data.


For an ionic compound the lattice

enthalpy is the enthalpy change

when one mole of solid in its

standard state is formed from its

ions in the gaseous state.

The lattice enthalpy cannot be measured

directly and so we make use of other

known enthalpies and link them together

with an enthalpy cycle.

This enthalpy cycle is the Born-Haber cycle.

What do we mean by lattice enthalpy?

- -

-

- -

-

-

-

For an ionic compound the lattice

enthalpy is the enthalpy change

when one mole of solid in its

standard state is formed from its

ions in the gaseous state.


1 Sublimation of Sodium

Na(s) + 1/2 Cl2(g)

Na(g) + 1/2 Cl2(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

kJmol-1

-400

-300

-200

-100

H = +107kJmol-1 θ S

Born-Haber Cycle for Sodium Chloride


2 Bond Dissociation of Chlorine

Na(s) + 1/2 Cl2(g)

Na(g) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

H = +121kJmol-1 θ D

½

Born-Haber Cycle for Sodium Chloride kJmol-1


e-

e-

e-

e-

e-

Na(s) + 1/2 Cl2(g)

Na(g) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na+(g) + Cl(g)

H = +502kJmol-1 θ I


3 First Ionisation of Sodium

kJmol-1


4 Electron Affinity of Chlorine

e

-

Na(g) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(s) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na+(g) + Cl(g)

Na+(g) + Cl-(g)

H = -355kJmol-1 θ E

Born-Haber Cycle for Sodium Chloride kJmol-1

-


Na(s) + 1/2 Cl2(g)

Na(g) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

NaCl(s)


H = -411kJmol-1 θ F

Na+(g) + Cl(g)

kJmol-1

- -

-

- -

5 Formation of Sodium Chloride

Na+(g) + Cl-(g)


Na(s) + 1/2 Cl2(g)

Na(g) + 1/2 Cl2(g)

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

Na(g) + Cl(g)

0

+100

+200

+300

+400

+500

+600

+700

+800

-400

-300

-200

-100

NaCl(s)

Lattice Enthalpy for

Sodium Chloride

Na+(g) + Cl-(g)

H = -786 kJmol-1 θ L

Na+(g) + Cl(g)


- -

-

- -

kJmol-1


Born-Haber cycles

Explain and use the term: lattice enthalpy.

Use the lattice enthalpy of a simple ionic solid and

relevant energy terms to construct Born–Haber cycles.


Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy Values (kJ mol-1)

Smaller ions will have a greater attraction for each other

because of their higher charge density. They will have larger

Lattice Enthalpies and larger melting points because of the

extra energy which must be put in to separate the oppositely

charged ions.


Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their

higher charge density. They will have larger Lattice Enthalpies and larger

melting points because of the extra energy which must be put in to separate

the oppositely charged ions.

Cl¯ Na+ Cl¯

The sodium ion has the same charge as a potassium ion but is smaller. It has a higher

charge density so will have a more effective attraction for the chloride ion. More energy

will be released when they come together.

K+


Calculating Lattice Enthalpy

SPECIAL POINTS

One CANNOT MEASURE LATTICE ENTHALPY DIRECTLY

it is CALCULATED USING A BORN-HABER CYCLE

greater charge

densities of ions = greater attraction

= larger lattice enthalpy

Effects

Melting point the higher the lattice enthalpy,

the higher the melting point of an ionic compound

Solubility solubility of ionic compounds is affected by the

relative

values of Lattice and Hydration Enthalpies


NaCl2(s)

Na(s) Na(g) Na+2(g)

Cl2(g) 2 Cl(g) 2Cl-(g)

H°ea H°d

(H°ie1 + H°ie2)

H°sub

H°f

Lattice Energy, Uo

Ionic Bonding If we can predict the lattice energy, a Born-Haber cycle analysis can tell us why

certain compounds do not form. E.g. NaCl2

H°f = H°sub + H°ie1 + H°ie2 + H°d + H°ea + Uo

H°f = 109 + 496 + 4562 + 242 + 2*(-349) + -2180

H°f = +2531 kJ/mol

This shows us that the formation of NaCl2 would be highly endothermic and very

unfavourable. Being able to predict lattice energies can help us to solve many problems so we

must learn some simple ways to do this.


1

6

5

4

3

2

Mg(s) + Cl2(g)

MgCl2(s)

Mg(g) + Cl2(g)

Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–

(g)

7

Mg+(g) + 2Cl(g)

Mg2+(g) + 2Cl(g)

Enthalpy of formation of MgCl2

Mg(s) + Cl2(g) ——> MgCl2(s)

Enthalpy of sublimation of magnesium

Mg(s) ——> Mg(g)

Enthalpy of atomisation of chlorine

½Cl2(g) ——> Cl(g) x2

Ist Ionisation Energy of magnesium

Mg(g) ——> Mg+(g) + e¯

2nd Ionisation Energy of magnesium

Mg+(g) ——> Mg2+(g) + e¯

Electron Affinity of chlorine

Cl(g) + e¯ ——> Cl¯(g) x2

Lattice Enthalpy of MgCl2

Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)

1

2

3

4

5

7

6

Born-Haber Cycle - MgCl2


H

CaO (s)

Ca2+ (g) + O2- (g)

H lattice energy

of formation

Ca (s) + ½ O2 (g)

H formation

H atomisation(s)

Ca (g) + O (g)

Ca2+ (g) + 2 e- + O (g)

H ionisation energy/ies

H electron affinity/ies

CaO

193

248

590

1150

–142

+844

?

– 635 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice

Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844

= – 3518 kJ mol-1

–635

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Prof.P. Ravindran, - folk.uio.nofolk.uio.no/ravi/cutn/cmp/latticeEnergyf.pdf · Born-Mayer equation...

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