Dr. Murad A. AlDamen

Inorganic Chemistry 221

1

Inorganic Chemistry

Chemistry is like the story about the group of blind men who encountered an elephant? Each one moved his hands over a different part of the elephant's body— the trunk, an ear, or a leg— and came up with an entirely different description of the beast. In this elephant (chemistry), the part we see is Inorganic Chemistry.

Symmetry-adapted orbitals in several MLn complexes

Figure 6.5. Construction of the MOs forH2O from AO on oxygen and thesymmetry-adapted orbitals on thehydrogen atoms.

a1

a1

a1

b2

b1

b2

1b2

2b2

1a1

2a1

3a1

1b1

symmetry, three molecular orbitals are formed: 1a1 (bonding), 2a1 (non-bonding), and 3a1 (antibonding). The atomic orbital with B1 symmetry(2px) cannot, by symmetry, interact with any other, and so it staysunchanged in shape and in energy (1b1, nonbonding).

6.6. Symmetry-adapted orbitals in severalMLn complexes

The aim of this section is to construct the symmetry-adapted orbitalsfor the principal ligand fields, making use of the reduction (6.5) andprojection (6.10) formulae. In what follows, with the exception of §6.6.6, we shall only consider a single orbital on each ligand, the onethat is used to create the σ bond with the metal (Chapter 1, § 1.5.1).The orbital on ligand Li will be written σi. We shall suppose that all theligands, and thus all the orbitals σi, are identical.

M

!1

!2!3

!4

6-23

C4,C2, S4z

x

y

C2a!

C2b!

C2b"

C2a"

ML4 L1

L2L3

6-24

6.6.1. Square-planar ML4 complexes

Consider a complex in which the metallic atom is surrounded by fourligands that are placed at the corners of a square (6-23). The sym-metry elements of this system are characteristic of the D4h point group.The axes are shown in 6-24. The planes of symmetry are xy (σh), xz

(σda), and yz (σdb), respectively, together with the planes that bisect xz

and yz and each contain two M−−L bonds (σva and σvb, respectively).The inversion centre is of course at the origin, coincident with thecentral atom.

C4

2

Fundamental particles of an atom

Atomic number, mass number and isotopes

Successes in early quantum theory

An introduction to wave mechanics

Atomic orbitals

Many-electron atoms

The periodic table

The aufbau principle

Ionization energies and electron affinities

Chapter 1: Basic Concepts: atoms

3

Chapter 1: Basic Concepts: atoms

An atom: the smallest unit quantity of an element that capable of existence

fm=femtometer

4

Fundamental Particles of an Atom

ProtonsFound in nucleusRelative charge of +1Relative mass of 1.0073 amu

NeutronsFound in nucleusNeutral chargeRelative mass of 1.0087 amu

ElectronsFound in electron cloudRelative charge of -1Relative mass of 0.00055 amu

Relative atomic mass

The mass of a single atom is very small ! non integral number ! we define the

atomic mass 1/12 th of a 126C atom, the atomic mass of 12

6C=1.660x10-27 kg,

1 amu (atomic mass unit)=1.660x10 -27 kg,

The mass on an atom=(the summation of protons)/(Avogadro’s number)

5

Atomic Number, Mass Number and Isotopes

Atomic number, ZThe number of protons in nucleusThe number of electrons in a neutral atomThe integer on the periodic table of each element

Mass Number, AInteger representing the approximate mass of an atom equal to the sum of the number of protons and neutrons in the nucleus=A-Z

IsotopesAtoms of the same element which differ in the number of neutrons in the nucleus designated by mass number

Some isotopes occur naturally others produce artificially,

Naturally elements (e.g. 3115P, 12

9F) are monotopic ! naturally occurs with one isotope only. With more than one isotope e.g. 12C, 13C and 14C.

Isotopes can be separated by mass spectrometry.

Allotropes: Each of two or more different physical forms in which an element can exist. Graphite, charcoal, diamond and fullerenes... are all allotropes of carbon (see page 384 for more information).

Caution! Do not confuse isotope and allotrope

EA

Z

Element symbol

6

Successes in early quantum theory

Quantum theory: An introduction

• Electrons in an atom occupy a region of space around the nucleus.

• Understanding of the electronic structures leads to understand the properties of atoms, ions and molecules including bonding and further interactions.

• A way of understanding the electronic structures is through quantum theory and wave mechanics.

• The quantum theory is necessary to describe electrons. It predicts discrete allowed energy levels and wavefunctions, which give probability distributions for electrons. Wavefunctions for electrons in atoms are called atomic orbitals.

• Although the electron is a wave and a particle at the same time, it easier and more convenient to consider the electron as a particle rather than a wave.

We with Quantum Chemistry is like Alice in marvel land.

Source:http://philosophyofscienceportal.blogspot.com/2008/04/lewis-carroll-fun.html

Source: http://philosophyofscienceportal.blogspot.com/2008/04/theoretical-physicsphilosophysci-fi.html

7

Classical Quantum Mechanics

Observation!At low temperatures, the radiation emitted by a hot body is mainly of low energy and occurs in the

infrared, but as the temperature increases, the radiation becomes successively dull red and white.

Explanation!

In the end of nineteenth century, much experimental work had been done on analysis this discrete spectrum

of radiation. attempts to explain this observation failed until, Planck suggested that energy could be

observed or emitted only in quanta of magnitude !E related to the frequency of the radiation, ".

!E ! "

!E =constant*"

!E =h"

c =#" "=c/#

Wavelength (#)

Speed of light (c)=2.998x108 m s-1

Energy (E)

Frequency (v)

Planck’s constant (h)=6.626x10-34 J s

!E =hc/" Planck relationship

wavenumber (v )=1

wavelength(v)!!cm

-1

Planck

8

Very precise spectroscopic measurements showed that hydrogen had the simplest spectrum of all the

elements. It was found that various lines in the optical and nonoptical regions where systematically spaced

in various series. Amazingly, it turned out that all the wavelengths of atomic hydrogen were given by a

single empirical relation Rydyberg formula.

where n1 =1 and n2=2,3,4... gives the Lyman series (ultraviolet region)where n1 =2 and n2=3,4,5... gives the Balmer series (visible region)where n1 =3 and n2=4,5,6... gives the Paschen series (infrared region)where n1 =4 and n2=5,6,7... gives the Brackett series (far infrared region)

RH=1.097x107 m-1

The existence of discrete spectroscopic frequencies is evidence that

the energy of the electron in the hydrogen atom is quantized. A

photon of light is emitted when the electron moves from a higher to

a lower energy level separated by energy difference, !E=hv.

v =1

!= R

H(1

n1

2"1

n2

2)

! =364.7n

2

n2" 4

!(n = 3, 4,5...) Balmer’s equation

The Emission Spectra of Atoms 25

Table 2.2 The first eight Balmer lines (2s → np; n = 3, 4, . . . , 10) of hydro-gen Balmer’s formula: λ(calc.) = 364.7n2/(n2 − 4)

state energy‡

n ω/cm−1 #ω/cm−1 λ(obs.)/nm λ(calc.)/nm

2 82 258.93 97 492.3 15 233 656.5 656.54 102 823.8 20 565 486.3 486.35 105 291.5 23 033 434.2 434.26 106 632.2 24 373 410.3 410.37 107 440.4 25 182 397.1 397.18 107 965.0 25 706 389.0 389.09 108 324.7 26 066 383.6 383.6

10 108 582.0 26 323 379.9 379.9∞ 109 678.8 27 420 364.7 364.7

‡Charlotte E. Moore, Atomic Energy Levels, Vol I, US National Bureau of Standards,Circular 467, Washington DC, 1949.

momentum. However, following the discussion of Chapter 1, it should be clear that, aswith the gas laws of Boyle and Gay-Lussac, there can be only one test of Bohr’s lawsand that is whether they reproduce the experimental findings.

Table 2.2 lists the wavelengths of a series of emission lines of the hydrogen atomwhich, in 1913, had been known for many years. They are called the Balmer lines becauseJohann Jakob Balmer (1825–1898), a Swiss school teacher, had shown in 1885 that thewavelengths of these lines (in nm = 10−9 m) fitted the simple formula:

λ = 364.7n2

(n2 − 4)(n = 3, 4, 5, . . .) (2.7.1)

Balmer’s equation, which predicts the wavelengths very accurately and was deducedfrom the experimental data alone, is illustrated with a graph in Figure 2.8. Thus, the test

350 400 450 500 550 600 650 700

0.5

1.0

1.5

2.0

∞ 8 6 5 4 3n =

n2 (n2

− 4)

Wavelength, l (nm)

Figure 2.8 A graph of Balmer’s equation for the hydrogen atom spectrum

Experimentally!

26 From Classical to Quantum Mechanics

of Bohr’s laws is their ability to interpret Equation (2.7.1). The application of the Bohrmodel to Balmer’s equation is described in detail in Box 2.1. In the next section theresults derived there are used to test the Bohr theory against experiment.

2.7.2 Comparison of Bohr’s model with experiment

As shown in Box 2.1, the Bohr model leads to Equation (B2.1.13) for the wavelengthsof the atomic spectral lines of the hydrogen atom:

λ = 32ε20ch

3

e4me

·(

n2

n2 − 4

)(B2.1.13)

On evaluating the factor 32ε20ch

3/e4me using the values for the fundamental constantslisted in Appendix 1, we obtain:

32ε20ch

3

e4me

= 32 × (8.85419)2 × 2.99792 × (6.62608)3 × 10118

(1.60218)4 × 9.10939 × 10107= 364.507 × 10−9 m

This is in excellent agreement with Balmer, i.e. with experiment, and confirms thevalidity of Bohr’s model for the hydrogen atom spectrum.

2.7.3 Further development of Bohr’s theory

The excellent agreement of Bohr’s theoretical result with the experimental data embodiedin Balmer’s equation was most gratifying and, at the time, it marked a very important stepforward. But law 2, the arbitrary imposition of quantisation upon what was essentially aclassical solution of the problem was to no one’s liking, least of all to Bohr’s. Furthermore,strenuous efforts by Bohr and others, notably Arnold Johannes Wilhelm Sommerfeld(1868–1951), to apply a similar, semi-classical approach to atoms with more than oneelectron failed to give results in agreement with experiment and it was clear that a fullyacceptable theory of the structure of the atom was yet to be found. The search for aform of mechanics applicable to atoms and molecules was to last for a further 12 years.Amazingly, when a viable theory was found, two apparently very different theories wereannounced at almost the same time. Karl Werner Heisenberg (1901–1975) describedhis matrix mechanics in 1925 and Erwin Schrodinger (1887–1961) his wave mechanicsin 1926. But the two approaches were soon shown to be different formulations of thesame theory (compare the Newtonian, Lagrangian and Hamiltonian forms of classicalmechanics), and we shall deal only with Schrodinger’s version here.

But before we take our final step to quantum mechanics we must note a suggestion madein his PhD thesis by de Broglie which, though it was not substantiated by experiment untilafter the advent of quantum mechanics, showed very clearly that a profound differencewas to be expected between the description of very small particles by classical and byquantum mechanics. It also had a great influence on Schrodinger who, in his famouspaper of 1926, described de Broglie’s thesis as ‘inspired’.

2.8 de BROGLIE’S PROPOSAL

Louis Victor Pierre Raymond Prince de Broglie (1892–1987) descended from a noblefamily who had served many French kings. He studied history before taking up physicsand in his doctoral dissertation (1924) he reasoned as follows:

Bohr’s equation

R. Grinter, THE QUANTUM IN CHEMISTRY: An Experimentalist’s View, , John Wiley &

Sons Ltd, 2005.

9

1. The electron in the hydrogen atom orbits the nucleus in a circular path, like a planet around the sun.

2. Of the infinite number of possible orbits, only those are allowed for which the orbital angular

momentum (mvr) of the electron is an integral multiple of Planck’s constant divided by 2! .

3. Contrary to classical electromagnetic theory, electrons in these allowed orbits do not radiate energy.

4. When an electron changes its orbit a quantum of energy (photon) is emitted or absorbed in accordance

with the equation h", where is the difference in the energy of the two orbits.

Bohr’s Laws of the Hydrogen Atom Structure

10

v

+e

M

-e me

from eq.2 and eq.3 we obtain

E=potential energy+kinetic energy

E= e2/(4#$0v2)+mev2/2 by eq.1 and eq.2 and after some algebraic rearrangements:

=-e2/(4#$0) % (e2#me)/($0n2h2) +me/2(e2/(2$0nh))2 =(-e4me)/(8$02n2h2)

E(n)=-1/n2 %(e4m)/(8$02h2)

Bohr’s 1st law: The electron in the hydrogen atom orbits the nucleus in a circular path, like a

planet around the sun.

The electron does not fall into the nucleus because of its velocity at right-angles to the line

between the two particles. the radial vector.

The coulombic force does cause an acceleration of the electron towards the nucleus equal to -v2/r.

According to Newton, force=mass x acceleration so: e2/(4$%0r2)=mev2/r " r= e2/(4#$0v2) ....eq.1

Bohr’s 2nd law: This law introduces quantization with the postulate that only those orbits for

which the angular momentum is an integer multiple of h/2$ are allowed,

i.e. the angular momentum=nh/2$ or mevr=nh/2$ where n=1,2,3...,# ...eq.2

to eliminate the v we use from eq.1. r= e2/(4#$0v2)= or r=e2/(4#$0) % 1/v2 and from eq 2. v=nh/

(2$mer) " 1/v2=(2#mer)2/(nh)2 then mer=($0n2h2)/(e2#) .....eq.3

v=nh/(2#mer)=nh/2# % e2#/($0n2h2)= e2/(2$0nh)

!E(n)=hv=hc/"

26 From Classical to Quantum Mechanics

of Bohr’s laws is their ability to interpret Equation (2.7.1). The application of the Bohrmodel to Balmer’s equation is described in detail in Box 2.1. In the next section theresults derived there are used to test the Bohr theory against experiment.

2.7.2 Comparison of Bohr’s model with experiment

As shown in Box 2.1, the Bohr model leads to Equation (B2.1.13) for the wavelengthsof the atomic spectral lines of the hydrogen atom:

λ = 32ε20ch

3

e4me

·(

n2

n2 − 4

)(B2.1.13)

On evaluating the factor 32ε20ch

3/e4me using the values for the fundamental constantslisted in Appendix 1, we obtain:

32ε20ch

3

e4me

= 32 × (8.85419)2 × 2.99792 × (6.62608)3 × 10118

(1.60218)4 × 9.10939 × 10107= 364.507 × 10−9 m

This is in excellent agreement with Balmer, i.e. with experiment, and confirms thevalidity of Bohr’s model for the hydrogen atom spectrum.

2.7.3 Further development of Bohr’s theory

The excellent agreement of Bohr’s theoretical result with the experimental data embodiedin Balmer’s equation was most gratifying and, at the time, it marked a very important stepforward. But law 2, the arbitrary imposition of quantisation upon what was essentially aclassical solution of the problem was to no one’s liking, least of all to Bohr’s. Furthermore,strenuous efforts by Bohr and others, notably Arnold Johannes Wilhelm Sommerfeld(1868–1951), to apply a similar, semi-classical approach to atoms with more than oneelectron failed to give results in agreement with experiment and it was clear that a fullyacceptable theory of the structure of the atom was yet to be found. The search for aform of mechanics applicable to atoms and molecules was to last for a further 12 years.Amazingly, when a viable theory was found, two apparently very different theories wereannounced at almost the same time. Karl Werner Heisenberg (1901–1975) describedhis matrix mechanics in 1925 and Erwin Schrodinger (1887–1961) his wave mechanicsin 1926. But the two approaches were soon shown to be different formulations of thesame theory (compare the Newtonian, Lagrangian and Hamiltonian forms of classicalmechanics), and we shall deal only with Schrodinger’s version here.

But before we take our final step to quantum mechanics we must note a suggestion madein his PhD thesis by de Broglie which, though it was not substantiated by experiment untilafter the advent of quantum mechanics, showed very clearly that a profound differencewas to be expected between the description of very small particles by classical and byquantum mechanics. It also had a great influence on Schrodinger who, in his famouspaper of 1926, described de Broglie’s thesis as ‘inspired’.

2.8 de BROGLIE’S PROPOSAL

Louis Victor Pierre Raymond Prince de Broglie (1892–1987) descended from a noblefamily who had served many French kings. He studied history before taking up physicsand in his doctoral dissertation (1924) he reasoned as follows:

Bohr’s equation

11

Amazingly, when a viable theory was found, two apparently very different theories were announced

at almost the same time. Karl Werner Heisenberg described his matrix mechanics in 1925 and

Erwin Schrödinger his wave mechanics in 1926. But the two approaches were soon shown to be

different formulations of the same theory (compare the Newtonian, Lagrangian and Hamiltonian

forms of classical mechanics), and we shall deal only with Schrödinger’s version here.

de Broglie (1892 – 1987) descended from a noble family who had served many French kings.

He studied history before taking up physics and in his doctoral dissertation (1924) he reasoned as follows:

According to Einstein’s theory of relativity:

E=mc2

but Einstein, following Planck also suggested that:

E=hv=hc/"

therefore, if the momentum (p) of the photon as a particle is mc, then:

p = mc = E/c = h/"

and if this result applies to other particles as well as to photons, then all moving particles

should be associated with a wavelength (#) given by the equation:

"= h/p = h/mv

where m and v are the mass and velocity of the particle respectively.

The wave-nature of electrons

Ein

stei

n l

ived

her

e

12

The uncertainty principle

Electrons moving in circles around the nucleus, as in Bohr's theory, can

be thought of as forming standing waves that can be described by the

de Broglie equation. However, we no longer believe that it is possible to

describe the motion of an electron in an atom so precisely. This is a

consequence of another fundamental principle of modern physics,

Heisenberg's uncertainty principle, which states that there is a

relationship

!p!x "h

4#

!x = uncertainty in the position of the electron !px = uncertainty in the momentum of the electron

This concept means that we cannot treat electrons as simple particles

with their motion described precisely, but we must instead consider the

wave properties of electrons, characterized by a degree of uncertainty

in their location. In other words, instead of being able to describe

precise orbits of electrons, as in the Bohr theory, we can only describe

orbitals, regions that describe the probable location of electrons. The

probability of finding the electron at a particular point in space (also

called the electron density) can be calculated, at least in principle.

The probability of finding an electron at a given point in space is

determined from the function $2 where $ is the wavefunction

Imagine a car in a circle, when this ball has zero velocity then its momentum will be zero, and so we can determine where is the ball exactly. If this ball moves by very high speed, then we will see the car in all the circle, it means that our accuracy in determining the position of the car will decreases and the energy increases.

v=0p=0

v=v0 p=mv0

i / Werner Heisenberg.

German military decided that it was too ambitious a project to undertake

in wartime, and too unlikely to produce results.

Some historians believe that Heisenberg intentionally delayed and

obstructed the project because he secretly did not want the Nazis to

get the bomb. Heisenberg’s apologists point out that he never joined

the Nazi party, and was not anti-Semitic. He actively resisted the gov-

ernment’s Deutsche-Physik policy of eliminating supposed Jewish in-

fluences from physics, and as a result was denounced by the S.S. as a

traitor, escaping punishment only because Himmler personally declared

him innocent. One strong piece of evidence is a secret message carried

to the U.S. in 1941, by one of the last Jews to escape from Berlin, and

eventually delivered to the chairman of the Uranium Committee, which

was then studying the feasibility of a bomb. The message stated “...that

a large number of German physicists are working intensively on the

problem of the uranium bomb under the direction of Heisenberg, [and]

that Heisenberg himself tries to delay the work as much as possible,

fearing the catastrophic results of success. But he cannot help fulfill-

ing the orders given to him, and if the problem can be solved, it will be

solved probably in the near future. So he gave the advice to us to hurry

up if U.S.A. will not come too late.” The message supports the view that

Heisenberg intentionally misled his government about the bomb’s tech-

nical feasibility; German Minister of Armaments Albert Speer wrote that

he was convinced to drop the project after a 1942 meeting with Heisen-

berg because “the physicists themselves didn’t want to put too much

into it.” Heisenberg also may have warned Danish physicist Niels Bohr

personally in September 1941 about the existence of the Nazi bomb

effort.

On the other side of the debate, critics of Heisenberg say that he

clearly wanted Germany to win the war, that he visited German-occupied

territories in a semi-official role, and that he simply may not have been

very good at his job directing the bomb project. On a visit to the oc-

cupied Netherlands in 1943, he told a colleague, “Democracy cannot

develop sufficient energy to rule Europe. There are, therefore, only two

alternatives: Germany and Russia. And then a Europe under German

leadership would be the lesser evil.” Some historians2 argue that the

real point of Heisenberg’s meeting with Bohr was to try to convince the

U.S. not to try to build a bomb, so that Germany, possessing a nuclear

monopoly, would defeat the Soviets — this was after the June 1941 en-

try of the U.S.S.R. into the war, but before the December 1941 Pearl

Harbor attack brought the U.S. in. Bohr apparently considered Heisen-

berg’s account of the meeting, published after the war was over, to be

inaccurate.3 The secret 1941 message also has a curious moral pas-

sivity to it, as if Heisenberg was saying “I hope you stop me before I do

something bad,” but we should also consider the great risk Heisenberg

would have been running if he actually originated the message.

2A Historical Perspective on Copenhagen, David C. Cassidy, Physics Today,July 2000, p. 28, http://www.aip.org/pt/vol-53/iss-7/p28.html

3Bohr drafted several replies, but never published them for fear ofhurting Heisenberg and his family. Bohr’s papers were to be sealed for50 years after his death, but his family released them early, in Febru-ary 2002. The texts, including English translations, are available athttp://www.nbi.dk/NBA/papers/docs/cover.html.

Section 4.4 The Uncertainty Principle 99

13

The Schrödinger equation describes the wave properties of an electron in terms of its position, mass, total

energy, and potential energy. The equation is based on the wave function, &, which describes an electron

wave in space; in other words, it describes an atomic orbital. In its simplest notation, the equation is:

The Schrödinger wave equation

H!=E!H=The Hamiltonian operator

!= The wave function

E= The energy of the electron

H =!h

2

8" 2m(#2

#x2)+VIn the form used for calculating energy levels, the Hamiltonian operator is

con lo que podemos comprobar que se cumple [X, Px] = ih. Y en general :

pm =h

i

∂

∂qm

Tenemos ya una regla para construir los operadores:

a) Expresar cualquier variable dinamica en funcion de las coordenadas y los mo-mentos.

b) Sustituir los momentos por el operador asociado al momento.

Para acabar con postulados, nos queda el ultimo, que nos relaciona la funcion deestado con el tiempo.

6) La funcion de estado viene ligada con el tiempo por la relacion siguiente:

− h

i

∂Ψ

∂t= HΨ ihΨ = HΨ

∂Ψ

∂t=

1

ihHΨ = − i

hHΨ

Es la ecuacion de Schrodinger dependiente del tiempo, que nos da la evolucion delestado de un sistema con el tiempo.

Erwin Rudolf Josef Alexander Schrodinger.Erdberg, Vienna, Austria(12-VII-1887) -Vienna, Austria(4-I-1961)

Es el primer postulado en que aparece un operador concreto, el operador H.

Existen ciertos sistemas en los que H no depende del tiempo, estos sistemas se llamanestacionarios.

Dado que la energıa cinetica no depende del t, el potencial tampoco debe depender det, por lo que solo sera funcion de las coordenadas, y las funciones de onda de cualquierestado de un sistema estacionario Ψ(q, t) se puede escribir como el producto de unafuncion dependiente de las coordenadas y otra que dependa del tiempo.

Ψ(q, t) = Ψ0(q)ϕ(t) ihΨ = HΨ

HΨ = HΨ0(q)ϕ(t) = ϕ(t)HΨ0(q)

ihΨ = ih∂Ψ∂t = ihΨ0(q)

∂ϕ(t)∂t

ih

ϕ(t)

∂ϕ(t)

∂t=

1

Ψ0(q)HΨ0(q) = W

3

The image above is not of an atom, but shows an alternative

electron corral pattern, predicted by the Schrödinger wave

equation and created by electrons in experiment

This equation can exactly be solved only for hydrogen-like atom: An

atom contains a nucleus and only one electron

Courtesy D. Eigler, IBM research division

14

Particle in a Box

1. The wave function $ is a solution of the Schrödinger equation and describes the

behavior of an electron in a region of space called atomic orbital.

2. We can find the energy values that are associated with particular wave functions.

3. The quantization of energy levels arises naturally from the Schrödinger equation.

!2"

!x2+8# 2m

h2(E $V )" = 0

$=Asin(rx)+Bcos(sx) substitution into Schrödinger equation leads: r=s=%(2mE) &2$/h

ra=±n$ or r=±n$/a=%(2mE) &2$/hE=n2h2/(8ma2)

The wave function "=#(2/a) sin (n#x/a)

where n: principal quantum number

A number of conditions required for a physically realistic solution for & in atoms:

1. The wave function $ must be single valued.

2. The wave function $ and its derivatives must be continuous.

3. The wave function $ must be quadratically integrable. This means that '$$*.d(=1. This

is called normalizing the wave function.

4. Any two solutions to the wave function must be orthogonal, which means '$A$B.d(=0

see Box 1.3 (Housecroft and Sharpe, Inorganic Chemistry 3rd)

V="

V=0

V="

Potential Energy Node Node

15

28 Chapter 2 Atomic Structure

I . r sin 8

FIGURE 2-5 Spherical Coordi- / nates and Volume Element for a Spherical Shell in Spherical Coordinates. Spherical Coordinates

X

Volume Element

The same procedure used on the d orbital functions for ml = f 1 and 1 2 gives the

functions in the column headed 0 @ ( 0 , +) in Table 2-3, which are the familiar d orbitals.

The d,z orbital (ml = 0) actually uses the function 2z2 - x 2 - y2, which we shorten to z2

for convenience. These functions are now real functions, so T = T* and TT* = T2. A more detailed look at the Schrodinger equation shows the mathematical origin

of atomic orbitals. In three dimensions, T may be expressed in terms of Cartesian co-

ordinates ( x , y, z ) or in terms of spherical coordinates ( r , 0 , +). Spherical coordinates,

as shown in Figure 2-5, are especially useful in that r represents the distance from the

nucleus. The spherical coordinate 0 is the angle from the z axis, varying from 0 to n ,

and 4 is the angle from the x axis, varying from 0 to 2 n . It is possible to convert be-

tween Cartesian and spherical coordinates using the following expressions:

x = r sin 0 cos + y = r sin 0 sin + z = r cos 0

In spherical coordinates, the three sides of the volume element are

r d0, r sin 0 d+, and dr. The product of the three sides is r2 sin 0 d0 d+ dr, equivalent

to dx dy dz. The volume of the thin shell between r and r + dr is 4 n r 2 dr, which is the

integral over + from 0 to n , and over 0 from 0 to 2 n . This integral is useful in describ-

ing the electron density as a function of distance from the nucleus.

T can be factored into a radial component and two angular components. The

radial function, R, describes electron density at different distances from the nucleus;

the angular functions, 8 and cP, describe the shape of the orbital and its orientation in

space. The two angular factors are sometimes combined into one factor, called Y:

K is a function only of r; Y is a function of 0 and 4, and gives the distinctivc shapes of

s, p, d, and other orbitals. R, 8, and @ are shown separately in Tables 2-3 and 2-4.

x=rsin !

y=rsin ! sin "

z=rcos !

dx.dy.dz=r2sin(!).d!.d".dr

$(r,!,")=R(r))(!)*(")

radial factor angular factors

Atomic Orbitals

The wavefunctions of the hydrogen atom are not easy to describe in a Cartesian co- ordinate system of three, mutually

perpendicular axes. The polar co-ordinate system with the nucleus at the origin is much more suitable. The mathematical functions

for atomic orbitals may be written as a product of two factors:

The radial wavefunction describes the behavior of the electron as a function of distance from the nucleus; the angular wavefunction

shows how it varies with the direction in space. (Describe the electron density)

Angular wavefunctions do not depend on n and are characteristic features of s, p, d,...orbitals. (Describe the shape and orientation)

16

Quantum numbers and their properties

Symbol Name Values Role

n Principal 1,2,3,... Determine the size and so the major part of energy

l Angular momentum 0,1,2,3,...n-1 Describes angular dependence (shape) and contributes to the energy

ml Magnetic 0,±1,±2,±3,...±l Describes orientation in space

ms Spin ±+ Describes orientation of the electron spin in space

Orbitals with different l values are usually known by the following labels, which are derived from early terms for different

families of spectroscopic lines:

l=

label

0 1 2 3 4 5, ...

s, p, d, f, g, continuing alphabetically

For a given value of n (n,1) there is one s atomic orbital

For a given value of n (n,2) there are three degenerate p atomic orbitals

For a given value of n (n,3) there are five degenerate d atomic orbitals

For a given value of n (n,4) there are seven degenerate f atomic orbitals

s sharp, p principal, d diffuse, f fundamental

17

+

-

21 3p

4pnodesnode

+-

n-1 Total

n-l-1 Radial

l Angular

Radial wavefunctions depend on n and l but not on m; thus each of the three 2p orbitals has the same radial form. The

wavefunctions may have positive or negative regions, but it is more instructive to look at how the radial probability distributions for

the electron depend on the distance from the nucleus.

• Radial distributions may have several peaks, the number being equal to n-1.

• The outermost peak is by far the largest, showing where the electron is most likely to be found.

The distance of this peak from the nucleus is a measure of the radius of the orbital, and is roughly

proportional to n2 (although it depends slightly on l also).

Radial distributions determine the energy of an electron in an atom. As the average distance from the

nucleus increases, an electron becomes less tightly bound. The subsidiary maxima at smaller distances

are not significant in hydrogen, but are important in understanding the energies in many-electron

atoms.

Radial Part

18

Radial Distribution Function

The Wave Functions of the Hydrogen Atom 99

0 5 10 15 20

0.0

0.5

1.0

1.5

2.0

Rn,

l(r)×

102

Rn,

l(r)

Rn,

l(r)

s-orbitals: l = 0n = 1, 2, 3

r (atomic units)

0 5 10 15 20

r (atomic units)

0.0

2.5

5.0d-orbital: l = 2n = 3

p-orbitals: l = 1n = 2, 3

−0.1

0.0

0.1

0.2

Figure 5.1 Hydrogen-atom radial functions

0 1 2 3 40

1

2

3

4

5

6

7

8

1s

4pr2 R

2

r (atomic units)

0.0

0.5

1.0

1.5

2.0

2.5

2p

2s

0 5 10 15 20 250.0

0.5

1.03p

3d3s

Figure 5.2 Hydrogen-atom radial density functions

enclosed between two spheres of radii r and r + δr is 4πr2δr . Therefore, the proportionof the hydrogen-atom electron which lies between two such spheres, i.e. at distance rfrom the nucleus, is 4πr2 · R2(r)δr . So a graph of 4πr2 · R2(r) against r is a measureof the probability of finding an electron at a distance r from the nucleus (Figure 5.2) butit can easily be misleading. Because the function r2 increases so rapidly with increasingr , the function 4πr2 · R2(r) has one or more maxima which must be interpreted withcare. For example, if we were able to stand at the hydrogen nucleus and experience theelectron density of the 1s orbital there in the way in which we experience a fog we wouldfind that the fog was very dense at the nucleus. As we walked away from the nucleus thethickness of the fog would decrease exponentially and we would not see any increase inthe region of r = a0 where the graph of 4πr2 · R2(r) against r shows a maximum. Themaximum is a consequence of the factor r2. Of course, where there are radial nodes theelectron density must fall to zero at these values of r and increase again at greater r .

5.2.2 The angular functions, !l,m(θ) and #m(φ)

We might well have expected that an electron circulating around a nucleus would havesome orbital angular momentum associated with it, and this is indeed the case. Further, we

1 atomic unit=Bohr’s radius=5.293pm

Probability of finding the electron at r = 4!r2R2

19

32 Chapter 2 Atomic Structure

TABLE 2-5 Nodal Surfaces

Spherical nodes [ R ( r ) = 01

Exanzples (number of spherical nodes) z Y

Is 0 2~ 0 3d 0

2s 1 3~ 1 4d 1

3s 2 4~ 2 5d 2

Angular nodes [Y (8, +) = 01 x

Examples (number of angular nodes)

s orbitals 0

p orbitals 1 plane for each orbital

d orbitals 2 planes for each orbital except dz2

1 conical aurfilcz Tor 4 2

Cartesian (x, y, 2) coordinates (see Table 2-3). In addition, the regions where the wave

function is positive and where it is negative can be found. This information will be use-

ful in working with molecular orbitals in later chapters. There are 1 angular nodes in any

orbital, with the conical surface in the d,2 and similar orbitals counted as two nodes.

Radial nodes, or spherical nodes, result when R = 0, and give the atom a lay-

ered appearance, shown in Figure 2-8 for the 3s and 3p, orbitals. These nodes occur

when the radial function changes sign; they are depicted in the radial function graphs

by R ( r ) = 0 and in the radial probability graphs by ~ I T ~ ~ R ~ = 0. The Is, 2p, and 3d

orbitals (the lowest energy orbitals of each shape) have no radial nodes and the number

of nodes increases as n increases. The number of radial nodes for a given orbital is al-

ways equal to n - l - l .

Nodal surfaces can be puzzling. For example, a p orbital has a nodal plane

through the nucleus. How can an electron be on both sides of a node at the same time

without ever having been at the node (at which the probability is zero)? One explanation

is that the probability does not go quite to zero.20

Another explanation is that such a question really has no meaning for an electron

thought of as a wave. Recall the particle in a box example. Figure 2-4 shows nodes at

x/a = 0.5 for n = 2 and at x/a = 0.33 and 0.67 for n = 3. The same diagrams could

represent the amplitudes of the motion of vibrating strings at the fundamental frequen-

cy ( n = 1 ) and multiples of 2 and 3. A plucked violin string vibrates at a specific fre-

quency, and nodes at which the amplitude of vibration is zero are a natural result. Zero

amplitude does not mean that the string does not exist at these points, but simply that

the magnitude of the vibration is zero. An electron wave exists at the node as well as on

both sides of a nodal surface, just as a violin string exists at the nodes and on both sides

of points having zero amplitude.

Still another explanation, in a lighter vein, was suggested by R. M. Fuoss to one

of the authors (DAT) in a class on bonding. Paraphrased from St. Thomas Aquinas,

"Angels are not material beings. Therefore, they can be first in one place and later in

another, without ever having been in between." If the word "electrons" replaces the

word "angels," a semitheological interpretation of nodes could result.

''A. Szabo, J. Chern. Educ., 1969, 46, 678, uses relativistic arguments to explain that the electron

probablity at a nodal surface has a very small, but finite, value.

34 Chapter 2 Atomic Structure

Here, the expression x2 - y2 appears in the equation, so the designation is d,2-y2. Because there are two solutions to the equation Y = O (or x2 - Y 2 = 0), x = y and x = - y , the planes defined by these equations are the angular nodal surfaces. They are planes containing the z axis and making 45" angles with the x and y axes (see Table 2-5). The function is positive where x > y and negative where x < y . Itl addition, a 3dx2-,2 orbital has no spherical nodes, a 4d&2 has one spherical node, and so on.

I EXERCISE 2-2 Describe the angular nodal surfaces for a d,2 orbital, whose angular wave function is

I EXERCISE 2-3

Describe the angular nodal surfaces for a d,, orbital, whose angular wave function is

The result of the calculations is the set of atomic orbitals familiar to all chemists.

Figure 2-7 shows diagrams of s, p, and d orbit& and Figure 2-8 shows lines of constant

electron density in several orbitals. The different signs on the wave functions are shown

by dil'ferent shadings of the orbital lobes in Figure 2-7, and the outer surfaces shown en-

close 90% of the total electron density of the orbitals. The orbitals we use are the com-

mon ones used by chemists; others that are also solutions of the Schrodinger equation

can be chosen for special purposes.21

2-2-3 THE AUFBAU PRINCIPLE

Limitations on the values of the quantum numbers lead to the familiar aufbau (German,

Auflau, building up) principle, where the buildup of electrons in atoms results from

continually increasing the quantum numbers. Any combination of the quantum numbers

presented so far correctly describes electron behavior in a hydrogen atom, where there

is only one electron. However, interactions between electrons in polyelectronic atoms

require that the order of filling of orbitals be specified when more than one electron is

in the same atom. In this process, we start with the lowest n, 1, and ml, values (1, 0, and 1

0, respectively) and either of the rn, values (we will arbitrarily use - 3 first). Three rules

will then give us the proper order for the remaining electrons as we increase the quan-

tum numbers in the order ml, m,, I , and n.

1. Electrons are placed in orbitals to give the lowest total energy to the atom. This

means that the lowest values of n and I are filled first. Because the orbitals within

each set (p, d, etc.) have the same energy, the orders for values of ml and m , are

indeterminate.

"R. E. Powell, .I. Chem. Educ., 1968,45,45.

Example

Nodal Planes

dxy

Polar Diagrams of the Angular Functions 103

Y1,c and Y1,s turn out to be exactly like Y1,0 consisting of two spheres in contact butaligned along the x- and y-axes respectively.

Accordingly, these three angular functions are known as the px , py and pz functions.It is important to note that a sum of the squares of the x-, y- and z-functions has nodependence upon θ or φ and is therefore spherically symmetrical. Consequently, if each p-function is occupied by one or by two electrons the total electron distribution is sphericallysymmetrical. This is clearly the case for the single s-function and it is also true of thefive d-functions, the seven f-functions and so on.

A further point should be made with regard to the forms of the squared functions. Ifwe make a polar plot of Y 2

1,0, Y2

1,c or Y 21,s we obtain results similar to those obtained

above but with an important difference; the two spheres become two pear-shaped lobes.But the directional properties of the functions remain quite unchanged and either form ofplotting is suitable for a discussion of the spatial characteristics of p-functions.

5.3.3 The d-functions

Polar diagrams of the five d-functions can be obtained in the manner described above.They are shown in Figure 5.5. Again, for purposes of illustration the real sine and cosinefunctions have been chosen. Reflecting their orientation with respect to the Cartesian axes,the d-functions are named, dxy , dxz and dyz for the three functions which are directedbetween the axes indicated, and dx2−y2 for the function with its lobes lying along the xand y axes. The function dz

2 ≡ Y2,0, is clearly different and the reason for this can be asource of difficulty. In actual fact, it is not as different from its companions as it appearssince it can be written as a combination of two functions, dz2−x2 and dz2−y2 , which haveexactly the same form as the other four d-orbitals, i.e. four lobes with their axes at rightangles and lying in a plane. We have to write the dz

2 function in the form given because

Y

ZX

dxy

−

− −

+

+

+

+ −

Z

Y

dyz

− +

+ −

Z

X

dxz

− +

+ −

Y

X

dx2 − y2

dz2

−

−

+ +

Figure 5.5 d-orbital angular functions

Polar Diagrams of the Angular Functions 103

Y1,c and Y1,s turn out to be exactly like Y1,0 consisting of two spheres in contact butaligned along the x- and y-axes respectively.

Accordingly, these three angular functions are known as the px , py and pz functions.It is important to note that a sum of the squares of the x-, y- and z-functions has nodependence upon θ or φ and is therefore spherically symmetrical. Consequently, if each p-function is occupied by one or by two electrons the total electron distribution is sphericallysymmetrical. This is clearly the case for the single s-function and it is also true of thefive d-functions, the seven f-functions and so on.

A further point should be made with regard to the forms of the squared functions. Ifwe make a polar plot of Y 2

1,0, Y2

1,c or Y 21,s we obtain results similar to those obtained

above but with an important difference; the two spheres become two pear-shaped lobes.But the directional properties of the functions remain quite unchanged and either form ofplotting is suitable for a discussion of the spatial characteristics of p-functions.

5.3.3 The d-functions

Polar diagrams of the five d-functions can be obtained in the manner described above.They are shown in Figure 5.5. Again, for purposes of illustration the real sine and cosinefunctions have been chosen. Reflecting their orientation with respect to the Cartesian axes,the d-functions are named, dxy , dxz and dyz for the three functions which are directedbetween the axes indicated, and dx2−y2 for the function with its lobes lying along the xand y axes. The function dz

2 ≡ Y2,0, is clearly different and the reason for this can be asource of difficulty. In actual fact, it is not as different from its companions as it appearssince it can be written as a combination of two functions, dz2−x2 and dz2−y2 , which haveexactly the same form as the other four d-orbitals, i.e. four lobes with their axes at rightangles and lying in a plane. We have to write the dz

2 function in the form given because

Y

ZX

dxy

−

− −

+

+

+

+ −

Z

Y

dyz

− +

+ −

Z

X

dxz

− +

+ −

Y

X

dx2 − y2

dz2

−

−

+ +

Figure 5.5 d-orbital angular functions

The five degenerate d orbitals

Angular Part

20

Orbital sign key

0 1 2 3 4 5 6

1 2 3 4 5 6

Total number of nodes=n-1

s

pz

2 3 4 5 6

2 3 4 5 6

dz2

dxz

nodes

nodes

nodes

nodes

n=1

l=0m=0

n=2 n=3 n=4 n=5 n=6 n=7

l=1m=0

l=2m=0

l=2m=1

21

Orbital angular momentum=h

2!l(l +1)

Spin angular momentum=h

2!s(s +1)

Example: l=2, ml=-2,-1,0,+1,+2

Orbital angular momentum=

Actual magnitude of z component of l= ml

h

2!

h

2!6

Total angular momentum=h

2!j( j +1)

Actual magnitude of z component of s= ms

h

2!

j = l +1

2!or ! l !

1

2

l=0 no angular momentum

Total angular momentum=h

2!8.75 !or !

h

2!3.75

j possible orientations

Angular momentum, the inner quantum number, j, and spin-orbit coupling

22

R = Rydberg!constant=1.097 !107m-1

h = Planck constant=6.626 !10-34 J!s

c=!Speed!of!light=3.000 !108m!s-1

Z !is!the!atomic!number

Hydrogen-like

Hydrogen"1s"spherical"only depends on n (l doesn't affect the energy)

En= !

RhcZ2

n2

r =n2a0

Z

3s

2s

1s

2p

3p 3d 3d 3d 3d3s

3s

3s

2s2s

2s

1s1s

1s

3p

3p

3p

2p

2p

2p

En

ergy (

not

to s

cale

$

Three multielectron atomsHydrogen atom

Li (Z=3) Na (Z=11) K (Z=19)

23

The aufbau Principle

1- Orbitals are filled in order of energy

2- Hund’s rule of maximum multiplicity: electrons be placed in orbitals so as to give the maximum total spin possible.

3- Pauli exclusion principle: each electron in an atom have a unique set of quantum numbers.

degenerate for Hydrogen atoms

degenerate for Hydrogen atoms

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

7p

3d

4d

5d

6d

4f

5f

First Rule

3s 3p 3d

2s 2p

n=2

n=3

n=%0

1s

n=1

24

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

adding to 3d (4s13dn)

adding to 4f (4fn6d2)

adding to 4f (4fn6s25d1)

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

adding to 6d (6dn7s2)

Black plate (18,1)

Table 1.3 Ground state electronic configurations of the elements up to Z ¼ 103.

Atomic Element Ground state Atomic Element Ground statenumber electronic configuration number electronic configuration

1 H 1s1

2 He 1s2 ¼ [He]

3 Li [He]2s1

4 Be [He]2s2

5 B [He]2s22p1

6 C [He]2s22p2

7 N [He]2s22p3

8 O [He]2s22p4

9 F [He]2s22p5

10 Ne [He]2s22p6 ¼ [Ne]

11 Na [Ne]3s1

12 Mg [Ne]3s2

13 Al [Ne]3s23p1

14 Si [Ne]3s23p2

15 P [Ne]3s23p3

16 S [Ne]3s23p4

17 Cl [Ne]3s23p5

18 Ar [Ne]3s23p6 ¼ [Ar]

19 K [Ar]4s1

20 Ca [Ar]4s2

21 Sc [Ar]4s23d 1

22 Ti [Ar]4s23d 2

23 V [Ar]4s23d 3

24 Cr [Ar]4s13d 5

25 Mn [Ar]4s23d 5

26 Fe [Ar]4s23d 6

27 Co [Ar]4s23d 7

28 Ni [Ar]4s23d 8

29 Cu [Ar]4s13d 10

30 Zn [Ar]4s23d 10

31 Ga [Ar]4s23d 104p1

32 Ge [Ar]4s23d 104p2

33 As [Ar]4s23d 104p3

34 Se [Ar]4s23d 104p4

35 Br [Ar]4s23d 104p5

36 Kr [Ar]4s23d 104p6 = [Kr]

37 Rb [Kr]5s1

38 Sr [Kr]5s2

39 Y [Kr]5s24d 1

40 Zr [Kr]5s24d 2

41 Nb [Kr]5s14d 4

42 Mo [Kr]5s14d 5

43 Tc [Kr]5s24d 5

44 Ru [Kr]5s14d 7

45 Rh [Kr]5s14d 8

46 Pd [Kr]5s04d 10

47 Ag [Kr]5s14d 10

48 Cd [Kr]5s24d 10

49 In [Kr]5s24d 105p1

50 Sn [Kr]5s24d 105p2

51 Sb [Kr]5s24d 105p3

52 Te [Kr]5s24d 105p4

53 I [Kr]5s24d 105p5

54 Xe [Kr]5s24d 105p6 ¼ [Xe]

55 Cs [Xe]6s1

56 Ba [Xe]6s2

57 La [Xe]6s25d 1

58 Ce [Xe]4f 16s25d 1

59 Pr [Xe]4f 36s2

60 Nd [Xe]4f 46s2

61 Pm [Xe]4f 56s2

62 Sm [Xe]4f 66s2

63 Eu [Xe]4f 76s2

64 Gd [Xe]4f 76s25d 1

65 Tb [Xe]4f 96s2

66 Dy [Xe]4f 106s2

67 Ho [Xe]4f 116s2

68 Er [Xe]4f 126s2

69 Tm [Xe]4f 136s2

70 Yb [Xe]4f 146s2

71 Lu [Xe]4f 146s25d 1

72 Hf [Xe]4f 146s25d 2

73 Ta [Xe]4f 146s25d 3

74 W [Xe]4f 146s25d 4

75 Re [Xe]4f 146s25d 5

76 Os [Xe]4f 146s25d 6

77 Ir [Xe]4f 146s25d 7

78 Pt [Xe]4f 146s15d 9

79 Au [Xe]4f 146s15d 10

80 Hg [Xe]4f 146s25d 10

81 Tl [Xe]4f 146s25d 106p1

82 Pb [Xe]4f 146s25d 106p2

83 Bi [Xe]4f 146s25d 106p3

84 Po [Xe]4f 146s25d 106p4

85 At [Xe]4f 146s25d 106p5

86 Rn [Xe]4f 146s25d 106p6= [Rn]

87 Fr [Rn]7s1

88 Ra [Rn]7s2

89 Ac [Rn]6d 17s2

90 Th [Rn]6d 27s2

91 Pa [Rn]5f 27s26d 1

92 U [Rn]5f 37s26d 1

93 Np [Rn]5f 47s26d 1

94 Pu [Rn]5f 67s2

95 Am [Rn]5f 77s2

96 Cm [Rn]5f 77s26d 1

97 Bk [Rn]5f 97s2

98 Cf [Rn]5f 107s2

99 Es [Rn]5f 117s2

100 Fm [Rn]5f 127s2

101 Md [Rn]5f 137s2

102 No [Rn]5f 147s2

103 Lr [Rn]5f 147s26d 1

18 Chapter 1 . Some basic concepts

adding to 5f (5fn7s26d1)

adding to 3d (5s03d10)

adding to 4d (5s14dn)

adding to 5f (5fn7s2)

Alk

ali

met

als

Alk

ali

eart

h m

etal

s

d-block elements

Pnic

togen

s

Chal

cogen

s

Hal

ogen

s

Noble

gas

es25

Hund’s Rule

Step 1 Step 2 Step 3

&c

&e

total pairing energy=&=&c +&e

Oxygen

Or

&=3&e +&c &=2&c +2&e

&e add stability &c add instability

When two electrons (e.g. Helium atom) occupy the same part of the space around an atom, they repel each other because of their

mutual negative charges, with a Coulombic energy of repulsion, &c.

In addition, there is an exchange energy, &e, which arises from purely quantum mechanical considerations; the number of

electrons at the same energy that could be exchanged while retaining the same overall spins. see Box 1.8

26

In this atom, the electrons of 1s are the core electrons (lower energy quantum levels)and those of 2s, 2p are valence electrons (outer)

2s and 2p ...etc penetrate the 1s2s more penetrating than 2p

The core electrons shield the valence electrons from the nuclear charge, i.e. 1s shield the nucleus and the atomic number will replace with effective charge Zeff

Zeff can be calculated by Slater’s rules, which is based on experimental data.

Effective charge(Z*)=number of protons(Z)-Shielding(S)

1. The electronic structure of the atom is written in groupings as follows: (1s) (2s, 2p) (3s,

3p) (3d) (4s, 4p) (4d, 4f) (5s, 5p), etc.

2. Electrons in higher groups (to the right in list above) don not shield those in lower groups.

3. For ns or np valence electrons:

a. Electrons in the same ns, np group contribute 0.35, except the 1s, where 0.30 works better.

b. Electrons in the n-1 group contribute 0.85.

c. Electrons in the n-2 or lower groups contribute 1.00.

4. For nd and nf valence electrons:

a. Electrons in the same nd or nf group contribute 0.35.

b. Electrons in groups to the left contribute 1.00.

Black plate (19,1)

Fig. 1.12 Radial distribution functions, 4!r2RðrÞ2, for the 1s, 2s and 2p atomic orbitals of the hydrogen atom.

CHEMICAL AND THEORETICAL BACKGROUND

Box 1.6 Effective nuclear charge and Slater’s rules

Slater’s rules

Effective nuclear charges, Zeff, experienced by electrons indifferent atomic orbitals may be estimated using Slater’srules. These rules are based on experimental data forelectron promotion and ionization energies, and Zeff isdetermined from the equation:

Zeff ¼ Z $ S

where Z ¼ nuclear charge, Zeff ¼ effective nuclear charge,S ¼ screening (or shielding) constant.Values of S may be estimated as follows:

1. Write out the electronic configuration of the element inthe following order and groupings: (1s), (2s, 2p),(3s, 3p), (3d ), (4s, 4p), (4d ), (4f ), (5s, 5p) etc.

2. Electrons in any group higher in this sequence than theelectron under consideration contribute nothing to S.

3. Consider a particular electron in an ns or np orbital:(i) Each of the other electrons in the (ns, np) group

contributes S = 0.35.(ii) Each of the electrons in the ðn$ 1Þ shell contri-

butes S ¼ 0:85.(iii) Each of the electrons in the ðn$ 2Þ or lower shells

contributes S ¼ 1:00.4. Consider a particular electron in an nd or nf orbital:

(i) Each of the other electrons in the (nd, nf ) groupcontributes S ¼ 0:35.

(ii) Each of the electrons in a lower group than the onebeing considered contributes S ¼ 1:00.

An example of how to apply Slater’s rules

Question: Confirm that the experimentally observed electro-nic configuration of K, 1s22s22p63s23p64s1, is energeticallymore stable than the configuration 1s22s22p63s23p63d1.

For K, Z ¼ 19:

Applying Slater’s rules, the effective nuclear charge

experienced by the 4s electron for the configuration

1s22s22p63s23p64s1 is:

Zeff ¼ Z $ S

¼ 19$ ½ð8& 0:85Þ þ ð10& 1:00Þ(

¼ 2:20

The effective nuclear charge experienced by the 3d electronfor the configuration 1s22s22p63s23p63d1 is:

Zeff ¼ Z $ S

¼ 19$ ð18& 1:00Þ

¼ 1:00

Thus, an electron in the 4s (rather than the 3d) atomic orbitalis under the influence of a greater effective nuclear charge andin the ground state of potassium, it is the 4s atomic orbitalthat is occupied.

Slater versus Clementi and Raimondi values of Zeff

Slater’s rules have been used to estimate ionization energies,ionic radii and electronegativities. More accurate effectivenuclear charges have been calculated by Clementi andRaimondi by using self-consistent field (SCF) methods, andindicate much higher Zeff values for the d electrons.However, the simplicity of Slater’s approach makes thisan attractive method for ‘back-of-the-envelope’ estimationsof Zeff. "

Chapter 1 . Many-electron atoms 19

At maximum 1s, 2s>2p,

and so the electron will

insert in the orbital that is

closer to the nucleus.

Oxygen (1s2) (2s2p4) Z=8For outmost electron Z*=Z-S=8-2*0.85-5*0.35=4.55Z (1s) (2s,2p)

4.55/8*100%=57% force expected for 8 nucleus and -1 electron

Shielding (S) (Screening)

27

Nickel The electron configuration (1s2) (2s2p6) (3s23p6) (3d8) (4s2)

For a 3d electron

Z*=Z-S

=28-[18x(1.00)]-[7x(0.35)]=7.55

(1s,2s,2p,3s,3p) (3d)

The 18 electrons in 1s, 2s, 2p, 3p, levels contribute 1.00 each, the other 7 in 3d contribute 0.35, and the 4s

contribute nothing. The total shielding constant is S=20.45 and Z*=7.55 for the last 3d electron.

For the 4s electron,

Z*=Z-S

=28-[10x(1.00)]-[16x(0.85)]-[1x(0.35)]=4.05

(1s,2s,2p) (3s,3p,3d) (4s)

The ten 1s, 2s and 2p electrons each contribute 1.00, the sixteen 3s, 3p and 3d electrons each contribute 0.85,

and the other 4s electron contributes 0.35, for a total S=23.95 and Z*=4.05, considerably smaller than the value

for the 3d electron above. The 4s electron is held less tightly than the 3d and should therefore the first removed

in ionization. This is consistent with experimental observations on nickel compounds, Ni2+, the most common

oxidation state of nickel, has an electron configuration of [Ar]3d8 (rather than [Ar]3d64s2), corresponding to loss

of the 4s electrons from nickel atoms. All the transition metals follow this same pattern of losing ns electrons

more readily than (n-1)d electrons.

Electron Configuration and Shielding: Nickel as an Example

28

Ionization energy The general increases in IE1 across a given period is a consequence of an increase

in Zeff. A group 15 element has a ground state electronic configuration ns2np3 and the np level is half-

occupied. A certain stability is associated with such configurations and it is more difficult to ionize a group

15 element than its group 16 neighbor. In going from Be (group 2) to B (group 13), there is a marked

decrease in IE1 and this may be attributed to the relative stability of the filed shell 2s2 configuration

compared with the 2s22p1 arrangement; similarly, in going from Zn (group 12) to Ga (group 13), we need to

consider the difference between 4s23d10 and 4s23d104p1 configurations.

Electron Affinity The attachment of an electron to an atom is usually exothermic. Two electrostatic

forces oppose one another: the repulsion between the valence shell electrons and the additional electron, and

the attraction between the nucleus and the incoming electron. In contrast, repulsive interactions are dominant

when an electron is added to an anion and the process is endothermic.A(g)+ e

-! A

"(g) Electron affinity=EA=--U

An+(g)! A

(n+1)+(g)+ e

-Ionization energy=-U

Ionization energy and Energy Affinity

Exceptions: B Be, N O, Mg Al, P S ... etc

29

Ion Protons Electrons Radius(pm)

O2- 8 10 126

F- 9 10 119

Na+ 11 10 116

Mg2+ 12 10 86

Ion Protons Electrons Radius(pm)

O2- 8 10 126

S2- 16 18 170

Se2- 34 36 184

Te2- 52 54 207

Ion Protons Electrons Radius(pm)

Ti2+ 22 20 100

Ti2+ 22 19 81

Ti2+ 22 18 75

+++ -

- - -+++

- - -+

Atomic and Ionic Radii Trends

1A 2A 4A 5A 6A 7A 8A3A

Ion Cation

30

Orbital is like ghost, we cannot see it but some believe

that it exists

Section 4-4 Atomic Units 109

of the set of associated Laguerre polynomials. Like the Legendre functions, theseare mathematically well characterized. A few of the low-index associated Laguerrepolynomials are

L11(ρ) = 1, L1

2(ρ) = 2ρ − 4,

L13(ρ) = −3ρ2 + 18ρ − 18, L3

3(ρ) = −6(4-47)

4-4 Atomic Units

It is convenient to define a system of units that is more natural for working with atomsand molecules. The commonly accepted system of atomic units for some importantquantities is summarized in Table 4-1. [Note: the symbol h (“h-cross or h-bar”) isoften used in place of h/2π .] Additional data on values of physical quantities, units,and conversion factors can be found in Appendix 10.

In terms of these units, Schrodinger’s equation and its resulting eigenfunctions andeigenvalues for the hydrogenlike ion become much simpler to write down. Thus, the

TABLE 4-1 ! Atomic Units

Quantity Atomic unit in cgs or other units

Values of some atomicproperties in atomic units

(a.u.)

Mass me = 9.109534 × 10−28 g Mass of electron = 1 a.u.Length a0 = 4πε0h2/mee

2 Most probable distance of 1s= 0.52917706 × 10−10 m electron from nucleus of H

(= 1 bohr) atom = 1 a.uTime τ0 = a0h/e2 Time for 1s electron in H

= 2.4189 × 10−17 s atom to travel one bohr= 1 a.u.

Charge e= 4.803242 × 10−10 esu Charge of electron =−1 a.u.= 1.6021892

×10−19 coulombEnergy e2/4πε0a0 = 4.359814 × 10−18 J Total energy of 1s electron in

(= 27.21161 eV ≡ 1 hartree) H atom = −1/2 a.u.Angular h= h/2π Angular momentum for

momentum = 1.0545887 × 10−34 J s particle in ring = 0, 1,2, . . . a.u.

Electric fieldstrength

e/a20 = 5.1423 × 109 V/cm Electric field strength at

distance of 1 bohr fromproton = 1 a.u.

Atomic units

Conclusions and Appendices

con lo que podemos comprobar que se cumple [X, Px] = ih. Y en general :

pm =h

i

∂

∂qm

Tenemos ya una regla para construir los operadores:

a) Expresar cualquier variable dinamica en funcion de las coordenadas y los mo-mentos.

b) Sustituir los momentos por el operador asociado al momento.

Para acabar con postulados, nos queda el ultimo, que nos relaciona la funcion deestado con el tiempo.

6) La funcion de estado viene ligada con el tiempo por la relacion siguiente:

− h

i

∂Ψ

∂t= HΨ ihΨ = HΨ

∂Ψ

∂t=

1

ihHΨ = − i

hHΨ

Es la ecuacion de Schrodinger dependiente del tiempo, que nos da la evolucion delestado de un sistema con el tiempo.

Erwin Rudolf Josef Alexander Schrodinger.Erdberg, Vienna, Austria(12-VII-1887) -Vienna, Austria(4-I-1961)

Es el primer postulado en que aparece un operador concreto, el operador H.

Existen ciertos sistemas en los que H no depende del tiempo, estos sistemas se llamanestacionarios.

Dado que la energıa cinetica no depende del t, el potencial tampoco debe depender det, por lo que solo sera funcion de las coordenadas, y las funciones de onda de cualquierestado de un sistema estacionario Ψ(q, t) se puede escribir como el producto de unafuncion dependiente de las coordenadas y otra que dependa del tiempo.

Ψ(q, t) = Ψ0(q)ϕ(t) ihΨ = HΨ

HΨ = HΨ0(q)ϕ(t) = ϕ(t)HΨ0(q)

ihΨ = ih∂Ψ∂t = ihΨ0(q)

∂ϕ(t)∂t

ih

ϕ(t)

∂ϕ(t)

∂t=

1

Ψ0(q)HΨ0(q) = W

3

Old Quantum Theory Spectroscopy and Quantum Number

angle relative to the direction of the sunlight, and when thedielectric constant is decreased by adding less polar sol-vents like alcohols. The quinine in tonic water is excited bythe ultraviolet light from the sun. Upon return to the groundstate the quinine emits blue light with a wavelength near450 nm. The first observation of fluorescence from a qui-nine solution in sunlight was reported by Sir John FrederickWilliam Herschel (Figure 1.2) in 1845.1 The following is anexcerpt from this early report:

On a case of superficial colour presented by a homo-geneous liquid internally colourless. By Sir John

Frederick William Herschel, Philosophical Translationof the Royal Society of London (1845) 135:143–145.

Received January 28, 1845 — Read February 13, 1845.

"The sulphate of quinine is well known to be ofextremely sparing solubility in water. It is howevereasily and copiously soluble in tartaric acid. Equalweights of the sulphate and of crystallised tartaricacid, rubbed up together with addition of a very littlewater, dissolve entirely and immediately. It is thissolution, largely diluted, which exhibits the opticalphenomenon in question. Though perfectly transparentand colourless when held between the eye and the

light, or a white object, it yet exhibits in certainaspects, and under certain incidences of the light, anextremely vivid and beautiful celestial blue colour,which, from the circumstances of its occurrence,would seem to originate in those strata which the lightfirst penetrates in entering the liquid, and which, if notstrictly superficial, at least exert their peculiar powerof analysing the incident rays and dispersing thosewhich compose the tint in question, only through avery small depth within the medium.

To see the colour in question to advantage, allthat is requisite is to dissolve the two ingredientsabove mentioned in equal proportions, in about ahundred times their joint weight of water, andhaving filtered the solution, pour it into a tall nar-row cylindrical glass vessel or test tube, which isto be set upright on a dark coloured substancebefore an open window exposed to strong day-light or sunshine, but with no cross lights, or anystrong reflected light from behind. If we lookdown perpendicularly into the vessel so that thevisual ray shall graze the internal surface of theglass through a great part of its depth, the wholeof that surface of the liquid on which the lightfirst strikes will appear of a lively blue, ...

If the liquid be poured out into another vessel,the descending stream gleams internally from all

2 INTRODUCTION TO FLUORESCENCE

Figure 1.1. Structures of typical fluorescent substances.

Figure 1.2. Sir John Fredrich William Herschel, March 7, 1792 toMay 11, 1871. Reproduced courtesy of the Library and InformationCentre, Royal Society of Chemistry.

Wavemecanics

Atomic Orbitals

31

Bonding model

Homonuclear diatomic molecules

The octet rule isoelectronic species

Molecular shape and the VSEPR model

Molecular shape: Stereoisomerism

Electronegativity values

Dipole moments

Chapter 2: Basic Concepts: molecules

32

A cartoon rendition of Irving Langmuir and G.N. Lewis ; Source: http://www.minerva.unito.it/humor/Langmuir%20&%20Lewis.htm

Lewis, Kossel, and Langmuir made several important proposals on bonding which lead to the development of Lewis Bonding Theory

Elements of the theory: Valence electrons play a fundamental role in chemical bonding. Ionic bonding involves the transfer of one or more electrons from one atom to another. Covalent bonding involves sharing electrons between atoms. Electrons are transferred or shared such that each atom gains an electron configuration of a noble gas

(ns2np6), i.e. having 8 outer shell (valence) electrons. This arrangement is called the octet rule. Exceptions to the octet rule do exist and will be explored later.

Lewis Bonding Theory

33

Lewis Symbols represent the resulting structures that accommodate the octet rule.

In a Lewis symbol, an element is surrounded by up to 8 dots, where elemental symbol represents the nucleus and the dots represent the valence electrons.

1A(1)ns1

2A(2)ns2

3A(13)ns2np1

4A(14)ns2np2

5A(15)ns2np3

6A(16)ns2np4

7A(17)ns2np5

8A(18)ns2np6

Li Be B

Na Mg Al

C

Si

N O F Ne

P S Cl Ar

2

3

Octet Rule

34

Ionic bond: e.g. Na ! Na- +

!+

Cl Cl-

Cl-

Na

+

metal nonmetal

small ion large ionLoss Gain

Covalent bond: e.g. ClShare

Cl Cl Cl!

nonmetal nonmetal

O

H

H !

Lewis formula

O

H

H

Kekulé formula

"

"

How can we write Mg3N2???

35

Homonuclear covalent bondHomonuclear molecule

Homonuclear covalent bondHomonuclear molecule

Homonuclear covalent bondHeteronuclear molecule

Homonuclear covalent bondHeteronuclear molecule

H ! H! H ! H!

F!

!!

!!

!!

F!!!

!!

!!

F!

!!

!!

!!

F!!!

!!

!!

O

!!

!!

!!

C!

!

!

!

2 C!

!O

!!

!!

!!!! O

!!

!!

!!

!

! !

!!

! !

C!

!!

!

C!

!

!

!

F !

! !

!!

! ! !

!!

!!

!!

!

!!

!!

!!!

!!

! !

!! F

F

F

F

+

+

+

+

Covalent Bond

36

Unequally shared: polar covalent bond

Bond order: Single, Double, Triple, Quadruple, Quintuple, Sextuple

Single e.g. ClShare

Cl Cl Cl! O

H

HShare!

O

H

H

Equally shared: nonpolar covalent bond

""

"

Double e.g. O O""

Triple e.g. N N"""

Quadruple e.g. Cl4Re ReCl4""""

Quintuple e.g. RCr CrR

"""""

;R=large organic group

Sextuple e.g. W W

"""""

"

Zeruple e.g.The carbon atom in C(PPh3)2 (where Ph stands for a benzene ring) has

two lone pair electrons, but no electrons connected to the bonded groups

2005 Philip Power, U of California, Davis

1964 F. Albert Cotton, Texas A&M University

2006 (Roos, Gagliardi, Borin, U of São Paulo)

37

Xe 3+O 1-

Expanded Shells

1. Elements in the third or higher shells can have >8 valence electrons

2. Empty d-orbitals are used to hold excess electrons (s2p6d10=18 electrons max)

S

N

F F

F

S

O

Cl O

Cl

Xe

O

O O

S

O

O O

O

S

O

O O

S

N

F F

F

S

O

Cl O

Cl

S

O

O O

O

S

O

O O

I

O

F

FF

FF

I

O

F

FF

FF

••••

••

••

••

••

••

••••••

•• ••

••

•• ••

••

••

••

••

••

••

••

••••••

••

••

••

••

••

••

••

••••

••

••

••

••

••

••

•• ••

••

••

••

••

•• ••

••

•• ••

••

••••

••

••

••

••

••

••

••

•• ••

••

••

••

••

••

••

••

••••

••

••

••

••

••

•• •• ••

••

••

••

••

••

••

Xe

O

O O

••

••••

••

••••

••

••••

••

•• ••

••

••••

••

•• ••

••

••••

••

••••

••

•• ••

••••

••

••

•• ••

S 2+N 2-

S 2+O 1-

S 2+O 1-

S 1+O 1-

S 0N 0

S 0O 0

Xe 0O 0

S 0O 0,1-

S 0O 0,1-

O 1+I 1-

O 0I 0

Anion Formal Charge Anion Formal ChargeMolecule Expanded to

SNF3

SO2Cl2

XeO3

SO42-

SO32-

IOF5

No octet rule

12

12

14

12

10

12 or 14Or

Decreasing the formal charge

No octet mean, it is not possible to draw the molecule with 8 electrons around the central atom

38

Resonance and Formal Charge

1. Ability to draw multiple Lewis structures for the same molecule

S

O

O O

S

O

O O

S

O

OO

a. Taken as a set to represent the true structures

b. Interconverted by movement of e- only

c. Separated by double headed arrows

More resonance structures means a lower energy for the compound; Spreading the electrons overe more atoms lets them occypy more space.

Isoelectronic=resonance structures with identical electronic structuresa) SO3 and CO3

2-

Resonance structures may be electronically different as well amides

CH3C

O

NH2 CH3C

O

NH2

distance in pm S-C C-N

SCN- 165 117

HSCN 156 122

Single bond 181 147

double bond 155 ~128

triple bond 116

Data from A. F. Wells, structural Inorganic Chemistry, 5th ed. OXford University Press, New York,

1984, pp. 807, 926, 933-934

distance in pm O-C C-N

OCN- 113 121

HNCO 118 120

Single bond 143 147

double bond 116 ~128

triple bond 113 116

Thiocyanate SCN-

Cyanate OCN-

Fluminate CNO-

S=C=N S!C"N S"C!N

O=C=N O!C"N O"C!N

C=N=O C!N"O C"N!O

1- 1- 2-1+

1- 1- 2-1+2-

1+ 1+ 1+3- 1+ 1+1-

Formal charge=valence electrons in Neutral atom-(unshared valence electrons+half os shared electrons)

39

Could not explain:

a. Why metals conduct electricity

b. How a semiconductor works

c. Why liquid oxygen is attracted into the magnetic field of a large magnet

d. Be and B compounds1. BeX2 and BX3 compounds have <8 electrons around Be/B if single bonds2. Be/B=X double bonds create large charge separation3. Solids tend to have extended structures relieving charge separation/octet4. Monomers are reactive as Lewis acids (electron pair acceptors), even though the small atom size allow less than an octet around Be/B

Shortcomings of Lewis theory

Be

F

F F

F

Be

Be

FF

F

F

F

F

Be FF Be FF

B F

F

F

B F

F

F

B F

F

F

B F

F

F

Predicted: The B-F bond length is 131 pm the calculated single-bond length is 152 pm

Predicted

Actual salt

40

Valence Shell Electron Pair Repulsion Theory

! 1.! All paires of electrons, both bonding pairs and lone pairs, are important in determining the shape of a

molecule.

! 2.! Bonding pairs are smaller than lone pairs because there are 2 positively charged nuclei pulling them in.

! 3.! Single bonds are smaller than double bonds and double bonds are smaller than triple bonds.

! 4.! If a central atom (A) is surrounded by different atoms (B and C) in the molecule ABxCy, the relative sizes

of B and C can affect the structure of the molecule.

Determine the Valence#Determine the valency#Draw Lewis#Count the lone pairs

The true bond angles will usually be distorted from the idealized angles in the pictures above

because all bonds and non-bonding electron pairs don't have the same "size".

lone pairs > triple bond > double bond > single bond

41

Valence Shell Electron Pair Repulsion Theory: Coordination Geometry

AX2

Linear

BeCl2 , CO2 , HCN

AX3

Trigonal planar

SO3, BeCl3

AX2E

Bent

O3, SnCl2

42

AX4

AX5

Valence Shell Electron Pair Repulsion Theory: Coordination Geometry

AX3E AX2E2

Tetrahedral Trigonal pyramidal Bent

SO4 , POCl3 , SiF4 (H3O)+ , NH3 , PCl3 H2O , Cl2O, (NH2)-

AX4E AX3E2 AX2E3

Trigonal bipyramidal Seesaw T-shaped Linear

SOF4 , PCl5 XeO2 , IF2O2 , SiF4 ClF3 I3-

43

AX7 AX8

AX6

Valence Shell Electron Pair Repulsion Theory: Coordination Geometry

AX5E AX4E2 AX3E3

Octahedral Square pyramidal Square planar T-shaped

IOF5 , SF6 XeOF4 , BrF5 ICl4-

[Ln(H2O)8]3+

Square antiprismatictrigonal bipyramidal

44

Valence Shell Electron Pair Repulsion Theory: Second order effects

Influence of the nature of electron pairs on the angles between bonds

It's logical to admit that the non-bonding pairs, less confined in the internuclear space than bonding pairs, occupy a

higher volume. And so, we see a diminution of the angles between the bonds when we observe in the order CH4, NH3

and H2O.

109.5º 107º 104.5

[NH4]+ NH3 [NH2]-

Influence of the volume of the multiple bonds

The geometric form depends only on s bond. So we can class the molecules with p bonds in the same groups that

these with only s bonds. But the volume occupied by the electrons depends on the number of p bonds, and then we

observe a diminution of the opposite angle of the p bond.

109.5º 102º 98º

SiF4 POF3 NSF3

45

Black plate (49,1)

If an octahedral species has the general formula EX3Y3,then the X groups (and also the Y groups) may be arrangedso as to define one face of the octahedron or may lie in aplane that also contains the central atom E (Figure 1.31).These geometrical isomers are labelled fac (facial) andmer (meridional) respectively. In [PCl4][PCl3F3], the[PCl3F3]

! anion exists as both mer- and fac-isomers (1.28and 1.29).

––

PF

F Cl

Cl

F

Cl

PF

F F

Cl

Cl

Cl

fac-isomer mer-isomer

(1.28) (1.29)

An octahedral species containing two identical groups (e.g. oftype EX2Y4) may possess cis- and trans-arrangements ofthese groups. An octahedral species containing threeidentical groups (e.g. of type EX3Y3) may possess fac- andmer-isomers.

Trigonal bipyramidal species

In trigonal bipyramidal EX5, there are two types of X atom:axial and equatorial. This leads to the possibility of geome-trical isomerism when more than one type of substituent isattached to the central atom. Iron pentacarbonyl, Fe(CO)5,is trigonal bipyramidal and when one CO is exchanged forPPh3, two geometrical isomers are possible depending onwhether the PPh3 ligand is axially (1.30) or equatorially(1.31) sited.

OC Fe

CO

CO

CO

PPh3

Ph3P Fe

CO

CO

CO

CO

(1.30) (1.31)

For trigonal bipyramidal EX2Y3, three geometrical iso-mers (1.32 to 1.34) are possible depending on the relative

positions of the X atoms. Steric factors may dictate whichisomer is preferred for a given species; e.g. in the staticstructure of PCl3F2, the F atoms occupy the two axialsites, and the larger Cl atoms reside in the equatorial plane.

Y E

Y

Y

X

X

X E

X

Y

Y

Y

X E

Y

Y

Y

X

(1.32) (1.33) (1.34)

In a trigonal bipyramidal species, geometrical isomerismarises because of the presence of axial and equatorial sites.

High coordination numbers

The presence of axial and equatorial sites in a pentagonalbipyramidal molecule leads to geometrical isomerism in asimilar manner to that in a trigonal bipyramidal species. Ina square antiprismatic molecule EX8, each X atom isidentical (Figure 1.29). Once two or more different atomsor groups are present, e.g. EX6Y2, geometrical isomers arepossible. As an exercise, draw out the four possibilities forsquare antiprismatic EX6Y2.

Double bonds

In contrast to a single (!) bond where free rotation is gener-ally assumed, rotation about a double bond is not a lowenergy process. The presence of a double bond may thereforelead to geometrical isomerism as is observed for N2F2. EachN atom carries a lone pair as well as forming one N!F singlebond and an N¼N double bond. Structures 1.35 and 1.36show the trans- and cis-isomers† respectively of N2F2.

N N

F

F

(1.35)

N N

F F

(1.36)

Further reading

First-year chemistry: basic principlesC.E. Housecroft and E.C. Constable (2002) Chemistry, 2nd edn,

Prentice Hall, Harlow – A readable text covering fundamentalaspects of inorganic, organic and physical chemistry whichgives detailed background of all material that is takenas assumed knowledge in this book. An accompanyingmultiple-choice test bank and Solutions Manual can befound through www.pearsoned.co.uk/housecroft

Fig. 1.31 The origin of the names fac- and mer-isomers. Forclarity, the central atom is not shown.

† In organic chemistry, IUPAC nomenclature uses the prefix (E)- for atrans-arrangement of groups and (Z)- for a cis-arrangement, but forinorganic compounds, the terms trans- and cis- remain in use.

Chapter 1 . Further reading 49

equatorialaxial

M

Y

X

X

Y M

Y

X

Y

X

M

XX

X Y

Y

Y

M

XX

X X

Y

Y

M

XX

X Y

Y

Y

M

YX

X Y

X

Y

MY

Y

Y

X

X

MX

Y

X

Y

Y

MX

Y

Y

Y

X

In trigonal bipyramidal species, geometrical isomersim arises because of the presence of axial and equatorial sites

Square planar species of the general form MX2Y2 or MX2YZ may possess cis- and trans-isomers

cis-isomer trans-isomer

cis-isomer trans-isomer

mer-isomerfac-isomer

mer-isomerfac-isomer

Isomerism

46

Bond Polarity

• .

0 0.4 2.0 4.0Electronegativity Difference

covalent ionic

polarnon

polar

3.0-3.0

= 0.04.0-2.1

= 1.9

3.0-0.9

= 2.1

Pauling electronegativity values $P

bond enthalpy=D(XY)=!(D(XX)+D(YY))

"#= #P(Y)-#P(X)=$"D

Mulliken electronegativity values $M

#M=!(Ionization energy+Electron affinity) eV

Allerd-Rochow electronegativity values $AR

#AR=(3590xZeff/rcov2)+0.744; rcov in pm

Eleectronegativity difference

The symmetrical electron distribution in the bond of a homonuclear diatomic renders the bond non-polar. In a heteronuclear

diatomic, the electron withdrawing powers of the two atoms may be different, and the bonding electrons are drawn closer

towards the more electronegative atom. The bond is polar and possesses an electric dipole moment (%)

!="ede charge of the electron, d bond distance % electric dipole moment" point charge

Measure of the pull an atom has on bonding electrons

Increases across period (left to right) highest at F

Decreases down group (top to bottom)

Large difference in electronegativity, more polar the bond

Negative end toward more electronegative atom

47

System operations and symmetry elements

Successive operations

Point groups

Introduction to character tables

An introduction to Molecular SymmetryChapter 4

48

Symmetry operations are geometrical operations applied to molecules (or better, molecular models). The

application of a symmetry operation to a particular molecular geometry produces a spatial orientation

which is indistinguishable from the initial orientation.

The five basic symmetry operations that need to be considered are identity, rotation, reflection (mirroring),

inversion, and rotary reflection.

For each symmetry operation there is a corresponding symmetry element, with respect to which the

symmetry operation is applied. For example, the rotation (a symmetry operation) is carried out around an

axis which is the corresponding symmetry element. Other symmetry elements are planes (for reflections)

and points (for inversions).

Different (valid) combinations of symmetry operations that leave at least a single common point (the

molecular center) unchanged give rise to the so called point groups (not all arbitrary combinations of

symmetry operations and elements are possible, and some combinations of symmetry operations implicitly

imply others, see below).

Symmetry

49

The basic symmetry operations are:

Identity Operation E The identity operation does nothing and leaves any molecule unchanged, the corresponding symmetry element

is the entire object (molecule) itself. The reason for including this symmetry operation is that some molecules have only this symmetry

element and no other symmetry properties. Another reason is the logical completeness of the mathematical description of group theory.

Inversion i The inversion (the symmetry operation) through a center of inversion (the symmetry element, which must be identical to the

center of geometry of the molecule) takes any point in the molecule, moves its to the center, and then moves it out the same distance on

the other side again (sometimes called point reflection). The benzene molecule, a cube, and spheres do have a center of inversion,

whereas a tetrahedron does not.

Reflection ! The reflection (the symmetry operation) in a plane of symmetry or mirror plane ! (the corresponding symmetry element)

produces a mirror image geometry of the molecule. The mirror plane bisects the molecule and must include its center of geometry. If this

plane is parallel to the principal axis (and includes it, see below), it is called a vertical mirror plane denoted !v, if it is perpendicular to the

principal axis (and bisects it in the molecular center of geometry), it is denoted a horizontal mirror plane !h. Vertical mirror planes

bisecting the angle between two Cn axes are called dihedral mirror planes !d (these mirror planes all include principal axis and intersect

in it).

n-Fold Rotation Cn The n-fold rotation (the symmetry operation) about a n-fold axis of symmetry (the corresponding symmetry

element) produces molecular orientations indistinguishable from the initial for each rotation of 360°/n (clockwise and counter-

clockwise). A water molecule has a single C2 axis bisecting the H-O-H bond angle, and benzene has one C6 axis (amongst one C3 axis

and seven C2 axes of which the C3 and one C2 axis coincide with the C6 axis). Linear molecules display a C" axis (any infinitely small

rotation about this axis produces unchanged orientations), and perfect spheres posses an infinite number of symmetry axes along any

diameter with all possible integral values of n. If a molecule has one (or more) rotation axes Cn or Sn (see below), the axis with the

greatest n is called the principal axis.

n-Fold Rotary Reflection Sn The n-fold rotary reflection (or n-fold improper rotation, the symmetry operation) about an n-fold rotary

reflection axis (or n-fold axis of improper rotation) is composed of two successive geometry transformations: first, a rotation through

360°/n about the axis of that rotation, and second, reflection through a plane perpendicular (and through the molecular center of

geometry) to that axis. Neither of these two operations (rotation or reflection) alone is a valid symmetry operation, but only the outcome

of the combination of both transformations.

50

"C6

C6 S6

"

C2

"

C6!v

!v

C2

C2

!d

!v

C2! !h

Between two C2Parallel to the principal

rotation axes (!v)

Perpendicular to principal

rotation axes (!h)Mirror plane (!)

Identity (E) Inversion (i)

n-fold rotation axis (Cn) n-fold improper axis (Sn)

The basic symmetry operations are:

51

only E only ! only i

Symmetry operations examples:

x

y

(x,y)

(x,y,z)#(x,y,z)

E

No change Mirror inversion

x

y

(x,y)

(x,y,z)#(-x,y,z)

!(yz)

(-x,y)

x

y

(x,y,z)

(x,y,z)#(-x,-y,-z)

i

(-x,-y,-z)

52

Chiral?

All molecules (or molecular conformations)

belonging to the C1, Cn, Dn, or the (very rare)

T, O, or I point groups MUST be chiral. All

other point groups include symmetry elements

that (any Sn axis, including mirror planes ! = S1

or inversion centers i = S2) require a molecular

geometry to be superimposable to its mirror

image geometry, and therefore these molecules

MUST be achiral

Polar molecules must have a permanent

electric dipole moment. Only molecules

belonging to the point groups Cn (including

C1), Cnv, and Cs, may have a permanent dipole

moment, which in the case of Cn and Cnv must

lie along the rotation axis. All other point

groups posses symmetry operations that include

Cn axes (n > 1, and thus no dipole moment

perpendicular to these axes), and symmetry

operations which take one end of the molecule

into the other (therefore no dipole moment

along the principal axis can be observed).

Permanent electrical dipole moment?

yes

yes yes

yesyesyes

yes

yes

yes yes

yes

yes

yes

Yes

Yes

yes yes

no

no

no no no no

nono

no

no

no

no

no

yes

no

no

no

no

linear? i

C$%

D$h

Ih

Dnh

DndCnh

Cn%

S2n

Ci

Cs

Td

Th Oh

! or iCn&5Cn&4Q1

! i

Q2Cn

!

i

!h

n!v

S2nCnC1

n!d

!h

T O

I! or i

Chiral point groups

Achiral point groups

Q1 Two or more Cn with n&3

Q2 Select Cn with highest n, then

are nC2 perpendicular to Cn?

Molecular Geometry

D

S1= !S2= i

For this reason, if the molecule has S1 or S2, we use ! and i.

53

cont.

EXAMPLES

54

cont.

EXAMPLES

55

cont.

C"v

EXAMPLES

56

cont.

D"h

EXAMPLES

57

cont.

EXAMPLES

58

EXAMPLES

59

Chemists and physicists use a special convention for representing character tables which is

applied especially to the so-called point groups, which are the 32 finite symmetry groups possible

in a lattice. In the following example:

Character Table

1. The symbol used to represent the group in question (in this case C3v).

2. The conjugacy classes, indicated by number and symbol, where the sum of the coefficients gives the

group order of the group.

3. Mulliken symbols, one for each irreducible representation.

4. An array of the group characters of the irreducible representation of the group, with one column for

each conjugacy class, and one row for each irreducible representation.

5. Combinations of the symbols x, y, z, Rx, Ry and Rz the first three of which represent the coordinates

x,y and z and the last three of which stand for rotations about these axes. These are related to

transformation properties and basis representations of the group.

6. All square and binary products of coordinates according to their transformation properties.

C3v E 2C3 3!v

A1 1 1 1 z x2+y2,z2

A2 1 -1 1 Rz

E 2 -1 0 (x,y)(Rx,Ry) (x2-y2,xy) ,(xz, yz)

! "

# $ % &

!!!!

! !

60

3. Mulliken Symbols

The character tables contain various information, for instance the number and kind of irreducible

representations for a specific point group. Usually these irreducible representations are denoted using

a scheme suggested by Robert S. Mulliken (1896-1986, awarded with the Nobel prize in 1966) in the

early 1930s. Here, an overview on the meaning of Mulliken's symbols is given.

1. The dimension of characters are denoted with one of the following capital letters:

dimension Mulliken symbol

1 A and B

2 E

3 T threefold degeneracy

twofold degeneracy

not degenerated

one electron system

a and b

e

t

2. If Cn represents rotation about the principal axis, the one-dimensional characters are denoted dependent of the

value obtained for !(Cn).

!(Cn) denoted as

+1 A

-1 B

cont.

symmetric

antisymmetric

61

3. Mulliken Symbols

3. Indices reflect an additional classification of symmetry. If the molecule possesses an axis C2 or a plane of reflection "

or "d perpendicular to the principal axis Cn, the values for function # change or keep their sign and are therefore

regarded as symmetric and antisymmetric respectively.

function # Index

Sign unaffected 1

Change sign 2

4. Dependent on the effect of inversion i, Mulliken's symbols take the indices g for gerade and u for ungerade.

5. The way reflection on a horizontal plane affects a function is denoted by primed or doubly primed symbols.

!(i) Index

Sign unaffected g

Change sign u

!("h) Index

Sign unaffected ‘

Change sign “

62

4. Orbital symmetry (C2v as an example)

#(E) #(C2(z)) #(!v(xz)) #(!v(yz))z

x

y

pz

z

x

y

py

+1 +1 +1 +1

+1 -1 -1 +1

x

y

px

+1 -1 +1 -1

z

63

4. Translational motion

z

y

x

OH H #(E) #(C2(z)) #(!v(xz)) #(!v(yz))

+1 -1 -1 +1

doesn’t translate translate translate doesn’t translateH translation:

+1 +1 -1 -1

doesn’t rotate rotate rotateH rotation along z : doesn’t translate

64

Infrared (IR) active vibrations

IR : one dimensional

Raman: two dimensional

There are two types of spectroscopy that involve vibrational transitions. infrared

spectroscopy. During infrared spectroscopy experiments we observe transitions between

vibrational energy levels of a molecule induced by the absorption of infrared (IR) radiation.

The second type of vibrational spectroscopy is Raman spectroscopy. In Raman

spectroscopy, vibrational transitions occur during the scattering of light by molecules.

At room temperature almost all molecules are in their lowest vibrational energy levels with

quantum number n = 0. For each normal mode, the most probable vibrational transition is

from this level to the next highest level (n = 0 -> 1). The strong IR or Raman bands resulting

from these transitiions are called fundamental bands. Other transitions to higher excited

states (n = 0 -> 2, for instance) result in overtone bands. Overtone bands are much weaker

than fundamental bands

65

Valence bond (VB) theory: hybridization of atomic orbitals

Valence bond (VB) theory: multiple bonding in polyatomic molecules

Bonding in Polymeric Molecules

Chapter 5 - Part 1

66

Valence Bond Theory and Hybrid Orbitals

The Linear Combination of Atomic Orbitals (LCAO) is a mathematical method to construct a new set of orbitals -

the hybrid orbitals - by mixing appropriate atomic orbitals (AOs)

Valence bond theory describes a chemical bond as the overlap of atomic orbitals.

repulsion occurs

some overlap attraction begins

maximum overlap attraction begins

Potential Energy

r74 pm

-436 kJ/mol

0

H•+•H!H:H

!i= c

i,1"1+ c

i,2"2+ ...+ c

i,n"n

Orthogonalityci, j2

j

! =1!!and !!! ci, j ck , jj"k

! = 0

67

2sp-orbitals

e.g. BeH2 Be

1s 2s 2p

1s 2s 2p

1s 2sp 2p

Be

Be

BeH2

1s 2sp 2pH H

px px

+ =

+ = - = !1=1 2"

2s+1 2"

2 px

!2=1 2"

2s#1 2"

2 px

68

2sp2-orbitals

e.g. BH3 B

1s 2s 2p

1s 2s 2p

1s 2sp2 2p

B

B

H H

px px

1s 2sp2 2p

BH3

H

BH3

+ + =

+ +

+ +

=

=

!1=1 3"

2s+1 6"

2 px+1 2"

2 py

!2=1 3"

2s+ 2(#1 3"

2 px)

!3=1 3"

2s+1 6"

2 px+ (#1 2"

2 py)

69

2sp3-orbitals

e.g. CH4 C

1s 2s 2p

1s 2s 2p

1s 2sp3 2p

C

C

H H

px px

1s 2sp3

CH4

H H

CH4

+ +

++

--

--

x

y

+-

+z-z

!1=1 4 "

2s+1 4 "

2 px+1 4 "

2 py+1 4 "

2 pz

!2=1 4 "

2s+1 4 "

2 px#1 4 "

2 py#1 4 "

2 pz

!3=1 4 "

2s#1 4 "

2 px#1 4 "

2 py+1 4 "

2 pz

!4=1 4 "

2s#1 4 "

2 px+1 4 "

2 py#1 4 "

2 pz

70

CH3!CH3

CH2=CH2

CH"CH

sp3-sp3

sp2-sp2

sp-sp

1s 2sp3

CH3

1s 2sp2

CH2

1s 2sp

CH

2p

2p

4#

3# and $

2# and 2$

Multiple bonding in polyatomic molecules

71

H!C"N

C [He]2s22p2

N [He]2s22p3

H 1s2

H

1s 2s 2p

H NC

2$

##

H

72

N [He]2s22p3

1s 2s 2p

sp3 lone pair1s

O [He]2s22p2

1s 2s 2p

1s sp2 two lone pairs

Examples: lone pairs in ammonia and water

73

Resonance in conjugated polymers

px px px px px px px px

$

# # # ## # #

$ $ $

px px px px px px px px

$

# # # ## # #

$ $

px px px px px px px px

# # # ## # # C ...C ...C ...C ...C ...C ...C ...C

C=C-C=C-C=C-C=C

C-C=C-C=C-C=C-C

$ $ $ $ $ $ $

74

Hybridization vs. Shape

sp s, px 2 linear

sp2 s, px,y 3 trigonal-planar

sp3 s, px,y,z 4 tetrahedral

sd3 s, dxy, xz, yz 4 tetrahedral

sp2d s, px,y, dxy, xz, yz 4 square-planar

sp3d s, px,y,z, dxy, xz, yz 5 square-pyramidal

sp3d s, px,y, z, dx2-y2 5 trigonal-bipyramidal

sp3d2 s, px,y, z, dx2-y2,z2 6 octahedral

sp3d3 7 pentagonal-bipyramidal

sp3d4 8 square-antiprismatic

CO2

CO32-

BF4-

CrO42-

[PtIICl4]2-

Ni(PR3)2Br3

PF5

SF6

IF7

TaF83-

Hybrid Orbital used n Orientation Example

75

Group theory and Hybridization

Determine the shape of the molecule by VSEPR techniques and consider each sigma bond to the

central atom and each lone pair on the central atom to be a vector pointing out from the center.

Find the reducible representation for the vectors, using the appropriate group and character table,

and find the irreducible representations that combine to form the reducible representation.

The atomic orbitals that fit the irreducible representations are those used in the hybrid orbitals.

e.g. BH3

1. VSEPR x

!BH

Black plate (870,1)

C5v E 2C5 2C25 5!v

A1 1 1 1 1 z x2 þ y2; z2

A2 1 1 1 "1 Rz

E1 2 2 cos 728 2 cos 1448 0 ðx; yÞðRx;RyÞ ðxz; yzÞ

E2 2 2 cos 1448 2 cos 728 0 ðx2 " y2; xyÞ

D2 E C2ðzÞ C2ðyÞ C2ðxÞ

A 1 1 1 1 x2; y2; z2

B1 1 1 "1 "1 z;Rz xy

B2 1 "1 1 "1 y;Ry xz

B3 1 "1 "1 1 x;Rx yz

D3 E 2C3 3C2

A1 1 1 1 x2 þ y2, z2

A2 1 1 "1 z, Rz

E 2 "1 0 (x; y) (Rx, Ry) (x2 " y2, xy) (xz, yz)

D2h E C2ðzÞ C2ðyÞ C2ðxÞ i !ðxyÞ !ðxzÞ !ðyzÞ

Ag 1 1 1 1 1 1 1 1 x2; y2; z2

B1g 1 1 "1 "1 1 1 "1 "1 Rz xy

B2g 1 "1 1 "1 1 "1 1 "1 Ry xz

B3g 1 "1 "1 1 1 "1 "1 1 Rx yz

Au 1 1 1 1 "1 "1 "1 "1

B1u 1 1 "1 "1 "1 "1 1 1 z

B2u 1 "1 1 "1 "1 1 "1 1 y

B3u 1 "1 "1 1 "1 1 1 "1 x

D3h E 2C3 3C2 !h 2S3 3!v

A1’ 1 1 1 1 1 1 x2 þ y2, z2

A2’ 1 1 "1 1 1 "1 Rz

E’ 2 "1 0 2 "1 0 (x; y) (x2 " y2, xy)

A1’’ 1 1 1 "1 "1 "1

A2’’ 1 1 "1 "1 "1 1 z

E’’ 2 "1 0 "2 1 0 (Rx;Ry) (xz, yz)

870 Appendix 3 . Selected character tables

B

H

H

H

y

Black plate (870,1)

C5v E 2C5 2C25 5!v

A1 1 1 1 1 z x2 þ y2; z2

A2 1 1 1 "1 Rz

E1 2 2 cos 728 2 cos 1448 0 ðx; yÞðRx;RyÞ ðxz; yzÞ

E2 2 2 cos 1448 2 cos 728 0 ðx2 " y2; xyÞ

D2 E C2ðzÞ C2ðyÞ C2ðxÞ

A 1 1 1 1 x2; y2; z2

B1 1 1 "1 "1 z;Rz xy

B2 1 "1 1 "1 y;Ry xz

B3 1 "1 "1 1 x;Rx yz

D3 E 2C3 3C2

A1 1 1 1 x2 þ y2, z2

A2 1 1 "1 z, Rz

E 2 "1 0 (x; y) (Rx, Ry) (x2 " y2, xy) (xz, yz)

D2h E C2ðzÞ C2ðyÞ C2ðxÞ i !ðxyÞ !ðxzÞ !ðyzÞ

Ag 1 1 1 1 1 1 1 1 x2; y2; z2

B1g 1 1 "1 "1 1 1 "1 "1 Rz xy

B2g 1 "1 1 "1 1 "1 1 "1 Ry xz

B3g 1 "1 "1 1 1 "1 "1 1 Rx yz

Au 1 1 1 1 "1 "1 "1 "1

B1u 1 1 "1 "1 "1 "1 1 1 z

B2u 1 "1 1 "1 "1 1 "1 1 y

B3u 1 "1 "1 1 "1 1 1 "1 x

D3h E 2C3 3C2 !h 2S3 3!v

A1’ 1 1 1 1 1 1 x2 þ y2, z2

A2’ 1 1 "1 1 1 "1 Rz

E’ 2 "1 0 2 "1 0 (x; y) (x2 " y2, xy)

A1’’ 1 1 1 "1 "1 "1

A2’’ 1 1 "1 "1 "1 1 z

E’’ 2 "1 0 "2 1 0 (Rx;Ry) (xz, yz)

870 Appendix 3 . Selected character tables

How many H-B bonds have been changed after applying the operation?

3 0 1 3 0 1

Summation of =s s=12

76

Now we reduce the reducible presentation to irreducible ones:

!BH 3 0 1 3 0 1

Black plate (870,1)

C5v E 2C5 2C25 5!v

A1 1 1 1 1 z x2 þ y2; z2

A2 1 1 1 "1 Rz

E1 2 2 cos 728 2 cos 1448 0 ðx; yÞðRx;RyÞ ðxz; yzÞ

E2 2 2 cos 1448 2 cos 728 0 ðx2 " y2; xyÞ

D2 E C2ðzÞ C2ðyÞ C2ðxÞ

A 1 1 1 1 x2; y2; z2

B1 1 1 "1 "1 z;Rz xy

B2 1 "1 1 "1 y;Ry xz

B3 1 "1 "1 1 x;Rx yz

D3 E 2C3 3C2

A1 1 1 1 x2 þ y2, z2

A2 1 1 "1 z, Rz

E 2 "1 0 (x; y) (Rx, Ry) (x2 " y2, xy) (xz, yz)

D2h E C2ðzÞ C2ðyÞ C2ðxÞ i !ðxyÞ !ðxzÞ !ðyzÞ

Ag 1 1 1 1 1 1 1 1 x2; y2; z2

B1g 1 1 "1 "1 1 1 "1 "1 Rz xy

B2g 1 "1 1 "1 1 "1 1 "1 Ry xz

B3g 1 "1 "1 1 1 "1 "1 1 Rx yz

Au 1 1 1 1 "1 "1 "1 "1

B1u 1 1 "1 "1 "1 "1 1 1 z

B2u 1 "1 1 "1 "1 1 "1 1 y

B3u 1 "1 "1 1 "1 1 1 "1 x

D3h E 2C3 3C2 !h 2S3 3!v

A1’ 1 1 1 1 1 1 x2 þ y2, z2

A2’ 1 1 "1 1 1 "1 Rz

E’ 2 "1 0 2 "1 0 (x; y) (x2 " y2, xy)

A1’’ 1 1 1 "1 "1 "1

A2’’ 1 1 "1 "1 "1 1 z

E’’ 2 "1 0 "2 1 0 (Rx;Ry) (xz, yz)

870 Appendix 3 . Selected character tables

1 2 3 1 2 3

1 1 1 1 1 1

!BH 3 0 1 3 0 1

Black plate (870,1)

C5v E 2C5 2C25 5!v

A1 1 1 1 1 z x2 þ y2; z2

A2 1 1 1 "1 Rz

E1 2 2 cos 728 2 cos 1448 0 ðx; yÞðRx;RyÞ ðxz; yzÞ

E2 2 2 cos 1448 2 cos 728 0 ðx2 " y2; xyÞ

D2 E C2ðzÞ C2ðyÞ C2ðxÞ

A 1 1 1 1 x2; y2; z2

B1 1 1 "1 "1 z;Rz xy

B2 1 "1 1 "1 y;Ry xz

B3 1 "1 "1 1 x;Rx yz

D3 E 2C3 3C2

A1 1 1 1 x2 þ y2, z2

A2 1 1 "1 z, Rz

E 2 "1 0 (x; y) (Rx, Ry) (x2 " y2, xy) (xz, yz)

D2h E C2ðzÞ C2ðyÞ C2ðxÞ i !ðxyÞ !ðxzÞ !ðyzÞ

Ag 1 1 1 1 1 1 1 1 x2; y2; z2

B1g 1 1 "1 "1 1 1 "1 "1 Rz xy

B2g 1 "1 1 "1 1 "1 1 "1 Ry xz

B3g 1 "1 "1 1 1 "1 "1 1 Rx yz

Au 1 1 1 1 "1 "1 "1 "1

B1u 1 1 "1 "1 "1 "1 1 1 z

B2u 1 "1 1 "1 "1 1 "1 1 y

B3u 1 "1 "1 1 "1 1 1 "1 x

D3h E 2C3 3C2 !h 2S3 3!v

A1’ 1 1 1 1 1 1 x2 þ y2, z2

A2’ 1 1 "1 1 1 "1 Rz

E’ 2 "1 0 2 "1 0 (x; y) (x2 " y2, xy)

A1’’ 1 1 1 "1 "1 "1

A2’’ 1 1 "1 "1 "1 1 z

E’’ 2 "1 0 "2 1 0 (Rx;Ry) (xz, yz)

870 Appendix 3 . Selected character tables

1 2 3 1 2 3

1 1 -1 1 1 -1

"

"

3 0 3 3 0 3

3 0 -3 3 0 -3

(#)/s = 12/s=12/12=1

0 A2`

1 A1`

(#)/s = 0/s=0/12=0

Finally after testing all the irreducible representation we obtain= A1`+E`

Related orbitals s from A1` px py or dxy dx2-y2 from E` Hybridization sp2 or sd2

According to VSEPR of trigonal planar p is less energetic than d so the hybridization is sp2

77

Molecular orbital (MO) theory : Homonucelar diatomic molecules: (chapter 2)

Molecular orbital (MO) theory : Heteronuclear diatomic molecules (chapter 2)

Molecular orbital (MO) theory: the ligand group orbital approach and application to triatomic molecules

Molecular orbital (MO) theory applied to the polyatomic molecules BH3, NH3 and CH4

Chapter 5 - Part 2

78

In molecular orbital (MO) theory, we begin by placing the nuclei of a given molecule

in their equilibrium positions and then calculate the molecular orbitals (i.e. regions of

space spread over the entire molecule) that a single electron might occupy. Each MO

arises from interactions between orbitals of atomic centers in the molecule, and such

interactions are:

. Allowed if the symmetries of the atomic orbitals are compatible with one another;

. Efficient if the region of overlap between the two atomic orbitals is significant;

. Efficient if the atomic orbitals are relatively close in energy.

An important ground-rule of MO theory is that the number of MOs that can be

formed must equal the number of atomic orbitals of the constituent atoms.

You have to apply aufbau principles on MO’ as we did for atomic ones

MOLECULAR ORBITAL THEORY

79

Hydrogen

!1s

*

!1s

!1s

*

!1s

Antibonding (LUMO)

bonding (HOMO)

Bond order = !(bonding-antibonding)

The symbol (*) instead of antibonding

Bond order= $(2-0)=1 Bond order= $(1-0)= !

H2 H2+

Bond energy=270 kJ/mol Bond energy=452 kJ/mol

2H+

H2+

H2 -452

-270

0 kJ

1s 1s1s1s

H!:!1s

1

Bond length=74 pm Bond length=105 pm

80

!1s

*

!1s

!1s

*

!1s

Antibonding

bonding

Bond order= $(2-2)=0 Bond order= $(2-1)= !

He2 He2+

Bond energy=0.05 kJ/mol Bond energy=301 kJ/mol

1s 1s1s1s

Helium

Molecule not observed

THE MORE BOND ENERGY, THE MORE BOND STABILITY

81

px py pz

Other Homonuclear Molecules

1s1s

!2 p

*

!2s

*

!2s

!2 p

!2 p

*

!2 p

px py pz

1s

!2s

*

!2s

px py pz

B2 , C2 , N2 O2 , F282

+

!

!

Linear Combination of Atomic Orbitals LCAOAO AO

#p

+

!

+

+

+

+

!

!

!

!+

+ !

2px

2py

2py

2px

1s

1s

#s*

#s

2!p*

#p*

2!p

Orbital

Potential

Energy

+-

The s...s and p...p Overlaps

83

px py pz

1s1s

!2 p

*

!2s

*

!2s

!2 p

!2 p

*

!2 p

px py pz

O2

PARAMAGNETIC:

Unpaired electrons

Diradical

Magnetism in Oxygen

http://sol.sci.uop.edu/%7Ejfalward/magneticforcesfields/magneticforcesfields.html

84

The variation in orbital energies for Period 2 homonuclear diatomic molecules as far as F2

O2 F2 Li2 Be2 B2 C2 N2

1!u

1!g

2!u

2!g

1!u

1!g

!g!,!"

u! crossing

Bond order 1 220 11 3Bond distance (pm)

Bond energy (kJ/mol)267 121124- 141159 110

110 498607- 159297 945

85

134 Chapter 5 Molecular Orbitals

5-5 HETERONUCLEAR

The actual energies of molecular orbitals for diatomic molecules are intermediate

between the extremes of this diagram, approximately in the region set off by the verti-

cal lines. Toward the right within this region, closer to the separated atoms, the energy

sequence is the "normal" one of O2 and F2; further to the left, the order of molecular or-

bitals is that of B2, C2 and N2, with u g ( 2 p ) above n U ( 2 p ) .

DIATOMIC MOLECULES

5-3-1 POLAR BONDS

Heteronuclear diatomic molecules follow the same general bonding pattern as the

homonuclear molecules described previously, but a greater nuclear charge on one of

the atoms lowers its atomic energy levels and shifts the resulting molecular orbital lev-

els. In dealing with heteronuclear molecules, it is necessary to have a way to estimate

the energies of the atomic orbitals that may interact. For this purpose, the orbital poten-

tial energies, given in Table 5-1 and Figure 5-13, are useful. These potential energies are

negative, because they represent attraction between valence electrons and atomic nu-

clei. The values are the average energies for all electrons in the same level (for example,

all 3 p electrons), and are weighted averages of all the energy states possible. These

TABLE 5-1 Orbital Potential Energies

Orbital Potential Energy (eV)

Atomic Number Element I s 2s 2~ 3s

SOURCE: J . B. Mann, T. L. Meek, and L. C. Allen, J. Am. Chem. Soc., 2000,122,2780.

NOTE: All energies are negative, representing average attractive potentials between the electrons and the nucleus for all terms of the specified orbitals.

86

MO’s of Heteronuclear MoleculesThe 2s orbital of the F atom has an energy about 27eV lower than that of the

hydrogen 1s, so there is very little interaction between them. The F orbital retains a

pair of electrons. The F 2pz, orbital and the H 1s, on the other hand, have similar

energies and matching A, symmetries, allowing them to combine into bonding ! and

antibonding !* orbitals. The F 2px and 2py orbitals have B1 and B2 symmetries and

remain nonbonding, each with a pair of electrons. Overall, there is one bonding pair

of electrons and three lone pairs.

"

!

!

"

!*

1s

2p

2s

a1

b1 b2

a1

-13.6eV

-14.05eV

-40.17eV

H HF F

H F

H

H

#

#

#*

$

HF

87

Bond order of CO=%(bonding-antibondig)=%(8-2)=3

C"O- +

98 Fundamentals of Bonding Theory

Fig. 3.3.5.Energy level diagram of a heteronucleardiatomic molecule.

X Y

2 ∗σ

σ

σ

σ

π

π

1 ∗

1

1 ∗

1

2

2p

2p

2s

2s

E

XY

Fig. 3.3.6.Energy level diagrams of CO and NO.

–14.0

(a) (b)–9.2

–14.6–15.2

–23.4

–40.4

–16.5

E (e

V)

E (e

V)–19.7

–39.8 1σ 1σ

2σ 2σ

1π∗1π∗

1π 1π1σ∗

1σ∗

CO NO

bond energy of BN is 385 kJ mol−1, suspiciously low compared with 602kJ mol−1 for C2. Clearly more experimental work is required in this case.

(b) BO, CN, and CO+ (with nine valence electrons): The electronic configura-tion for these molecules is 1σ 21σ*21π42σ 1, with bond order 21/2. They allhave bond lengths shorter than BN (or C2), 120 pm for BO, 117 pm for CN,and 112 pm for CO+. Also, they have fairly similar bond energies: 800,787, and 805 kJ mol−1 for BO, CN, and CO+, respectively, all of whichare greater than that of C2.

(c) NO+, CO, and CN− (with ten valence electrons): Here the electronic con-figuration is 1σ 21σ*21π42σ 2, with a bond order of 3. They have similarbond lengths: 106, 113, and 114 pm for NO+, CO, and CN−, respectively.The bond energy of carbon monoxide (1070 kJ mol−1) is slightly greaterthan that of N2 (941 kJ mol−1). The energy level diagram for CO is shownin Fig. 3.3.6(a).

(d) NO (with eleven valence electrons): The ground electronic configuration is1σ 22σ*21π42σ 2

z 1π*1 and the bond order is 21/2. The bond length of NO,115 pm, is longer than those of both CO and NO+. Its bond dissociationenergy, 627.5 kJ mol−1, is considerably less than those of CO and N2. Theimportance of NO in both chemistry and biochemistry will be discussed

Bond order of NO =%(bonding-antibondig)=%(8-3)=2.5

•N=O

!ss

!ss*

!pp

!pp*

"pp

"pp

"pp*

"pp*

CO and NO

88

2nodes

3nodes

1node

4nodes

2nodes

5nodes

3nodes

7nodes

3nodes

7nodes

3nodes

nonbonding

nonbonding

bonding

bonding

bonding

bonding

bonding

bonding

antibonding

antibonding

antibonding

antibonding

antibonding

antibonding

0 nodes

CO2

O-"-O O=C=O89

144 Chapter 5 Molecular Orbitals

Oxygen Orbitals Used Group Orbitals

i i i -0---c-0-z

J J J Y Y Y 2pZ A6 -0- 0-

3 4

FIGURE 5-19 Group Orbital Symnetry in COz.

The group orbitals of the oxygen atoms are the same as those for the fluorine

atoms shown in Figure 5-16. To determine which atomic orbitals of carbon are of cor-

rect symmetry to interact with the group orbitals, we will consider each of the group or-

bitals in turn. The combinations are shown again in Figure 5-19 and the carbon atomic

orbitals are shown in Figure 5-20 with their symmetry labels for the DZh point group.

FIGURE 5-20 Symmetry of the Carbon Atomic Orbitals in the D2h

Point Group.

2s group orbitals

Group orbitals 1 and 2 in Figure 5-21, formed by adding and subtracting the oxy-

gen 2s orbitals, have Ag and B1,, symmetry, respectively. Group orbital 1 is of appropri-

ate symmetry to interact with the 2s orbital of carbon (both have Ag symmetry), and

group orbital 2 is of appropriate symmetry to interact with the 2pZ orbital of carbon

(both have B1, symmetry).

0 C 0 .cm O C O 0.0 0 C 0 .C. O C O 0.0

FIGURE 5-21 Group Orbitals 1 and 2 for COz.

Group orbitals 3 and 4 in Figure 5-22, formed by adding and subtracting the oxy- ; gen 2pz orbitals, have the same Ap and B1, symmetries. As in the first two, group

orbital 3 can interact with the 2s of carbon and group orbital 4 can interact with the

carbon 2p,.

5-4 Molecular Orbitals for Larger Molecules 145

2p7 group orbitals

\ \

1

\ I

\\ - 0 0 0 1 ,

O C O .C. O C O

\ / '

\

\\ COCOeO /'

O C O .C. O C O 0 .0

FIGURE 5-22 Group Orbitals 3 and 4 for C 0 2 .

The 2s and 2p, orbitals of carbon, therefore, have two possible sets of group or-

bitals with which they may interact. In other words, all four interactions in Figures 5-21

and 5-22 occur, and all four are symmetry allowed. It is then necessary to estimate

which interactions can be expected to be the strongest from the potential energies of the

2s and 2p orbitals of carbon and oxygen given in Figure 5-23.

Interactions are strongest for orbitals having similar energies. Both group orbital

1, from the 2s orbitals of the oxygen, and group orbital 3, from the 2pz orbitals, have the

proper symmetry to interact with the 2s orbital of carbon. However, the energy match

between group orbital 3 and the 2s orbital of carbon is much better (a difference of

3.6 eV) than the energy match between group orbital 1 and the 2s of carbon (a difference

of 12.9 eV); therefore, the primary interaction is between the 2pz orbitals of oxygen and

Orbital - 2s 2l2

Carbon -19.4 eV -10.7 eV

Oxygen -32.4 eV E -20 -

-15.9 eV

-30 t 2s -

0

Orbital

potential energies

CO2

90

146 Chapter 5 Molecular Orbitals

2pJ, group orbitals 8@ -

6 - ------- - 2py

OCO - C . OCO 0.0

FIGURE 5-24 Group Orbitals 5 and 6 for COz.

the 2s orbital of carbon. Group orbital 2 also has energy too low for strong interaction

with the carbon p, (a difference of 21.7 eV), so the final molecular orbital diagram

(Figure 5-26) shows no interaction with carbon orbitals for group orbitals 1 and 2.

2px group orbitals

I EXERCISE 5-5

Using orbital potential energies, show that group orbital 4 is more likely than group orbital 2

with the 2p, orbital of carbon.

The 2py orbital of carbon has BZu symmetry and interacts with group orbital 5 (Figure 5-24). The result is the formation of two IT molecular orbitals, one bonding and

one antibonding. However, there is no orbital on carbon with B3g symmetry to interact

with group orbital 6, formed by combining 2p, orbitals of oxygen. Therefore, group or-

bital 6 is nonbonding.

Interactions of the 2p, orbitals are similar to those of the 2py orbitals. Group or-

bital 7, with B2, symmetry, interacts with the 2p, orbital of carbon to form IT bonding

and antibonding orbitals, whereas group orbital 8 is nonbonding (Figure 5-25).

The overall molecular orbital diagram of C02 is shown in Figure 5-26. The 16 va-

lence electrons occupy, from the bottom, two essentially nonbonding a orbitals, two

bonding o orbitals, two bonding .rr orbitals, and two nonbonding IT orbitals. In other

words, two of the bonding electron pairs are in o orbitals and two are in n orbitals, and

there are four bonds in the molecule, as expected. As in the FHF- case, all the occupied

molecular orbitals are 3-center, 2-electron orbitals and all are more stable (have lower

energy) than 2-center orbitals.

The molecular orbital picture of other linear triatomic species, such as N3-, CS2,

and OCN-, can be determined similarly. Likewise, the molecular orbitals of longer

FIGURE 5-25 Group Orbitals 7 and 8 for COz

146 Chapter 5 Molecular Orbitals

2pJ, group orbitals 8@ -

6 - ------- - 2py

OCO - C . OCO 0.0

FIGURE 5-24 Group Orbitals 5 and 6 for COz.

the 2s orbital of carbon. Group orbital 2 also has energy too low for strong interaction

with the carbon p, (a difference of 21.7 eV), so the final molecular orbital diagram

(Figure 5-26) shows no interaction with carbon orbitals for group orbitals 1 and 2.

2px group orbitals

I EXERCISE 5-5

Using orbital potential energies, show that group orbital 4 is more likely than group orbital 2

with the 2p, orbital of carbon.

The 2py orbital of carbon has BZu symmetry and interacts with group orbital 5 (Figure 5-24). The result is the formation of two IT molecular orbitals, one bonding and

one antibonding. However, there is no orbital on carbon with B3g symmetry to interact

with group orbital 6, formed by combining 2p, orbitals of oxygen. Therefore, group or-

bital 6 is nonbonding.

Interactions of the 2p, orbitals are similar to those of the 2py orbitals. Group or-

bital 7, with B2, symmetry, interacts with the 2p, orbital of carbon to form IT bonding

and antibonding orbitals, whereas group orbital 8 is nonbonding (Figure 5-25).

The overall molecular orbital diagram of C02 is shown in Figure 5-26. The 16 va-

lence electrons occupy, from the bottom, two essentially nonbonding a orbitals, two

bonding o orbitals, two bonding .rr orbitals, and two nonbonding IT orbitals. In other

words, two of the bonding electron pairs are in o orbitals and two are in n orbitals, and

there are four bonds in the molecule, as expected. As in the FHF- case, all the occupied

molecular orbitals are 3-center, 2-electron orbitals and all are more stable (have lower

energy) than 2-center orbitals.

The molecular orbital picture of other linear triatomic species, such as N3-, CS2,

and OCN-, can be determined similarly. Likewise, the molecular orbitals of longer

FIGURE 5-25 Group Orbitals 7 and 8 for COz

CO2

91

! - Bonding in CO2

Black plate (120,1)

Self-study exercise

Work out a qualitative MO description for the !-bonding in CO2

and show that this picture is consistent with leaving eight electronsto occupy the "-type MOs shown in Figure 4.24.

[NO3]!

In worked example 4.2, we considered the bonding in [NO3]!

using a VB approach. Three resonance structures (one ofwhich is 4.4) are needed to account for the equivalence ofthe N!O bonds, in which the net bond order per N!Obond is 1.33. Molecular orbital theory allows us to representthe N!O !-system in terms of delocalized interactions.

O

N–O O–

(4.4)

The [NO3]! ion hasD3h symmetry and the z axis is defined

to coincide with the C3 axis. The valence orbitals of each Nand O atom are 2s and 2p orbitals. The !-bonding in [NO3]

!

can be described in terms of the interactions of the N 2pzorbital with appropriate LGOs of the O3 fragment. UnderD3h symmetry, the N 2pz orbital has a2’’ symmetry (seeTable 4.1). The LGOs that can be constructed from O 2pzorbitals are shown in Figure 4.25 along with their sym-metries; the method of derivation is identical to that forthe corresponding LGOs for the F3 fragment in BF3

(equations 4.23–4.25). The partial MO diagram shown inFigure 4.25 can be constructed by symmetry-matching ofthe orbitals. The MOs that result have !-bonding (a2’’),non-bonding (e’’) and !-antibonding (a2’’

") character; thea2’’ and a2’’

" MOs are illustrated at the right-hand side ofFigure 4.25. Six electrons occupy the a2’’ and e’’ MOs. Thisnumber of electrons can be deduced by considering that ofthe 24 valence electrons in [NO3]

!, six occupy "-bondingMOs, 12 occupy oxygen-centred MOs with essentially non-bonding character, leaving six electrons for the !-typeMOs (see problem 4.18 at the end of chapter).

Molecular orbital theory therefore gives a picture of [NO3]!

in which there is one occupiedMOwith !-character and this isdelocalized over all four atoms giving anN!O !-bond order of13. This is in agreement with the valence bond picture, but it isperhaps easier to visualize the delocalized bonding schemethan the resonance between three contributing forms of thetype of structure 4.4. The bonding in the isoelectronic species[CO3]

2! and [BO3]3! (both D3h) can be treated in a similar

manner.

SF6Sulfur hexafluoride (4.5) provides an example of a so-calledhypervalent molecule, i.e. one in which the central atomappears to expand its octet of valence electrons. However,a valence bond picture of the bonding in SF6 involvingresonance structures such as 4.6 shows that the S atomobeys the octet rule. A set of resonance structures isneeded to rationalize the observed equivalence of the sixS!F bonds. Other examples of ‘hypervalent’ species of thep-block elements are PF5, POCl3, AsF5 and [SeCl6]

2!. The

Fig. 4.24 A partial MO diagram that illustrates the formation of delocalized C!O !-bonds using the ligand group orbitalapproach. The CO2 molecule is defined as lying on the z axis. The characters of the !g and !u MOs are shown in the diagrams atthe top of the figure.

120 Chapter 4 . Bonding in polyatomic molecules

92

CO2

"u

!u*

!g*

"g

"u

!u*

!g*

2p

2s

2p

2OC CO2

Energy

93

Molecular orbitals of nonlinear molecules can be determined by the same procedures. Water will

be used as an example, and the steps of the previous section will be used.

1. Water is a simple triatomic bent molecule with a C2 axis through the oxygen and two mirror

planes that intersect in this axis. The point group is therefore C2v.

2. The C2 axis is chosen as the z axis and the xz plane as the plane of the molecule. Because the

hydrogen 1s orbitals have no directionality, it is not necessary to assign axes to the hydrogens.

148 Chapter 5 Molecular Orbitals

FIGURE 5-27 Symmetry of the Water Molecule.

Molecular orbitals of nonlinear molecules can be determined by the same procedures.

Water will be used as an example, and the steps of the previous section will be used.

1. Water is a simple triatomic bent molecule with a C2 axis through the oxygen and

two mirror planes that intersect in this axis, as shown in Figure 5-27. The point

group is therefore C2,.

2. The C2 axis is chosen as the z axis and the xz plane as the plane of the molecule.21

Because the hydrogen 1s orbitals have no directionality, it is not necessary to as-

sign axes to the hydrogens.

3. Because the hydrogen atoms determine the symmetry of the molecule, we will

use their orbitals as a starting point. The characters for each operation for the Is

orbitals of the hydrogen atoms can be obtained easily. The sum or thc contribu-

tions to the character ( I , 0, or - 1, as described previously) for each symmetry op-

eration is the character for that operation, and the complete list for all operations

of the group is the reducible representation for the atomic orbitals. The identity

operation leaves both hydrogen orbitals unchanged, with a character of 2.

Twofold rotation interchanges the orbitals, so each contributes 0, for a total char-

acter of 0. Reflection in the plane of the molecule (a,) leaves both hydrogens un-

changed, for a character of 2; reflection perpendicular to the plane of the

molecule (a,') switches the two orbitals, for a character of 0, as in Table 5-2.

TABLE 5-2 Representations for Ch Symmetry Op,erationr fol ~ydrogen &oms ,in Wker

Cz, Character Table

[:::I = [: :I[:;] for the identity operation

[:::I = [ y b][::] for the C2,, operation

[EL:] = [b :][:,] for the G,, reflection (xz plane)

[;;:I = [ y b][:,] for the D,,' reflection ( y z plane)

The reducible representation F = A 1 i- BI :

2 1 ~ o m e sources use the yz plane as the plane of the molecule. This convention results in r = A, + B2 and switches the hl and b2 labels of the molecular orbitals.

3. Because the hydrogen atoms determine the

symmetry of the molecule, we will use their

orbitals as a starting point. The characters for each

operation for the Is orbitals of the hydrogen atoms

can be obtained easily. The sum of the

contributions to the character (I, 0, or - 1) for each

symmetry operation is the character for that

operation, and the complete list for all operations of

the group is the reducible representation for the

atomic orbitals. The identity operation leaves both

hydrogen orbitals unchanged, with a character of 2.

Twofold rotation interchanges the orbitals, so each

contributes 0, for a total character of 0. Reflection

in the plane of the molecule (!v) leaves both

hydrogens unchanged, for a character of 2;

reflection perpendicular to the plane of the

molecule (!v') switches the two orbitals, for a

character of 0.

148 Chapter 5 Molecular Orbitals

FIGURE 5-27 Symmetry of the Water Molecule.

Molecular orbitals of nonlinear molecules can be determined by the same procedures.

Water will be used as an example, and the steps of the previous section will be used.

1. Water is a simple triatomic bent molecule with a C2 axis through the oxygen and

two mirror planes that intersect in this axis, as shown in Figure 5-27. The point

group is therefore C2,.

2. The C2 axis is chosen as the z axis and the xz plane as the plane of the molecule.21

Because the hydrogen 1s orbitals have no directionality, it is not necessary to as-

sign axes to the hydrogens.

3. Because the hydrogen atoms determine the symmetry of the molecule, we will

use their orbitals as a starting point. The characters for each operation for the Is

orbitals of the hydrogen atoms can be obtained easily. The sum or thc contribu-

tions to the character ( I , 0, or - 1, as described previously) for each symmetry op-

eration is the character for that operation, and the complete list for all operations

of the group is the reducible representation for the atomic orbitals. The identity

operation leaves both hydrogen orbitals unchanged, with a character of 2.

Twofold rotation interchanges the orbitals, so each contributes 0, for a total char-

acter of 0. Reflection in the plane of the molecule (a,) leaves both hydrogens un-

changed, for a character of 2; reflection perpendicular to the plane of the

molecule (a,') switches the two orbitals, for a character of 0, as in Table 5-2.

TABLE 5-2 Representations for Ch Symmetry Op,erationr fol ~ydrogen &oms ,in Wker

Cz, Character Table

[:::I = [: :I[:;] for the identity operation

[:::I = [ y b][::] for the C2,, operation

[EL:] = [b :][:,] for the G,, reflection (xz plane)

[;;:I = [ y b][:,] for the D,,' reflection ( y z plane)

The reducible representation F = A 1 i- BI :

2 1 ~ o m e sources use the yz plane as the plane of the molecule. This convention results in r = A, + B2 and switches the hl and b2 labels of the molecular orbitals.

H2O

94

4. The representation # can be reduced to the irreducible representations A1 +B1, representing the symmetries of the

group orbitals. These group orbitals can now be matched with orbitals of matching symmetries on oxygen. Both 2s

and 2pz orbitals have A1 symmetry, and the 2px orbital has B1 symmetry. In finding molecular orbitals, the first step is

to combine the two hydrogen 1s orbitals. The sum of the two, 1/$2[%(Ha)+%(Hb)], has symmetry A1 and the

difference, 1/$2[%(Ha)-%(Hb)], has symmetry B1. These group orbitals, or symmetry-adapted linear combinations,

are each treated as if they were atomic orbitals, so they contribute equally to the group orbitals. The normalizing

factor is 1/$2. In general, the normalizing factor for a group orbitals is N=1/$(&ci2), where ci=the coefficients on the

atomic orbitals. Each group orbital is treated as a single orbital in combining with the oxygen orbitals.

5. The same type of analysis can be applied character when a p orbital changes sign. Each orbital can be treated

independently.

The s orbital is unchanged by all the operations, so it has A1 symmetry.

The px orbital has the B1 symmetry of the x axis

The py orbital has the B2 symmetry of the y axis

The pz orbital has the A1 symmetry of the z axis

The x, y, and z variables and the more complex functions in the character table assists in assigning representations to

the atomic orbitals.

150 Chapter 5 Molecular Orbitals

Molecular Oxygen Atomic Group Orbitals Symmetry Orbitals Orbitals from Hydrogen Atoms Description

BI T6 - c 9 NPX) + .lo [ $ ( H a ) - W b ) ] antibonding (clo is negative) -

A 1 9 5 - C7 $($I + cx [$(Hn) + *(Hb)l antibonding (c8 is negative) -

B2 9 4 - ~ ( P Y ) nonbonding -

A1 9 3 - c 5 ~ ( P z ) + C6 [@(Ha) + d ~ ( ~ b ) l nearly nonbonding (slightly -

bonding; c6 is very small)

B1 9 2 - C3 N P X ) + C4 [$(Ha) - w b ) l bonding (q is positive) -

A I 9 I - C i C(s) + cz [$(Ha) + W b ) ] bonding (c2 is positive) -

The s orbital is unchanged by all the operations, so it has Al symmetry.

The p, orbital has the B1 symmetry of the x axis.

The py orbital has the B2 symmetry of they axis.

The p, orbital has the Al symmetry of the z axis.

The x, y, and z variables and the more complex functions in the character tables assist in assigning representations to the atomic orbitals.

6. The atomic and group orbitals with the same symmetry are combined into molec-

ular orbitals, as listed in Table 5-3 and shown in Figure 5-29. They are numbered q, through q6 in order of their energy, with 1 the lowest and 6 the highest.

The A , group orbital combines with the s and p, orbitals of the oxygen to form

three molecular orbitals: one bonding, one nearly nonbonding (slightly bonding), and one antibonding (three atomic or group orbitals forming three molecular orbitals, q 1 ,

!P3, and q 5 ) . The oxygen p, has only minor contributions from the other orbitals in the

weakly bonding q3 orbital, and the oxygen s and the hydrogen group orbitals combine

weakly to form bonding and antibonding !PI and q5 orbitals that are changed only slightly from the atomic orbital energies.

The hydrogen B1 group orbital combines with the oxygen p, orbital to form two

MOs, one bonding and one antibonding (Y2 and !P6). The oxygen py ( q 4 , with B2

symmetry) does not have the same symmetry as any of the hydrogen 1s group orbitals, and is a nonbonding orbital. Overall, there are two bonding orbitals, two nonbonding or

nearly nonbonding orbitals, and two antibonding orbitals. The oxygen 2s orbital is near-

ly 20 eV below the hydrogen orbitals in energy, so it has very little interaction with them. The oxygen 2p orbitals are a good match for the hydrogen 1s energy, allowing

formation of the bonding bl and a1 molecular orbitals. When the eight valence electrons available are added, there are two pairs in bond-

ing orbitals and two pairs in nonbonding orbitals, which are equivalent to the two bonds and two lone pairs of the Lewis electron-dot structure. The lone pairs are in molecular

orbitals, one b2 from the py of the oxygen, the other a l from a combination of s and p, of the oxygen and the two hydrogen Is orbitals. The resulting molecular orbital diagram is shown in Figure 5-29.

The molecular orbital picture differs from the common conception of the water molecule as having two equivalent lone electron pairs and two equivalent 0 - H bonds.

In the MO picture, the highest energy electron pair is truly nonbonding, occupying the 2py orbital perpendicular to the plane of the molecule. The next two pairs are bonding .:

pairs, .resulting from overlap of the 2p, and 2p, orbital with the 1s orbitals of the hy- j

drogens. The lowest energy pair is a lone pair in the essentially unchanged 2s orbital of j the oxygen. Here, all four occupied molecular orbitals are different.

6. The atomic group orbital combines with the same symmetry are combined into molecular orbitals. They are

numbered %1 through %6 in order of their energy, with 1 the lowest and 6 the highest

H2O

95

Hydrogen orbitals

E C2 !v !v’

+1 +1 +1 +1

+1 -1 +1 -1

148 Chapter 5 Molecular Orbitals

FIGURE 5-27 Symmetry of the Water Molecule.

Molecular orbitals of nonlinear molecules can be determined by the same procedures.

Water will be used as an example, and the steps of the previous section will be used.

1. Water is a simple triatomic bent molecule with a C2 axis through the oxygen and

two mirror planes that intersect in this axis, as shown in Figure 5-27. The point

group is therefore C2,.

2. The C2 axis is chosen as the z axis and the xz plane as the plane of the molecule.21

Because the hydrogen 1s orbitals have no directionality, it is not necessary to as-

sign axes to the hydrogens.

3. Because the hydrogen atoms determine the symmetry of the molecule, we will

use their orbitals as a starting point. The characters for each operation for the Is

orbitals of the hydrogen atoms can be obtained easily. The sum or thc contribu-

tions to the character ( I , 0, or - 1, as described previously) for each symmetry op-

eration is the character for that operation, and the complete list for all operations

of the group is the reducible representation for the atomic orbitals. The identity

operation leaves both hydrogen orbitals unchanged, with a character of 2.

Twofold rotation interchanges the orbitals, so each contributes 0, for a total char-

acter of 0. Reflection in the plane of the molecule (a,) leaves both hydrogens un-

changed, for a character of 2; reflection perpendicular to the plane of the

molecule (a,') switches the two orbitals, for a character of 0, as in Table 5-2.

TABLE 5-2 Representations for Ch Symmetry Op,erationr fol ~ydrogen &oms ,in Wker

Cz, Character Table

[:::I = [: :I[:;] for the identity operation

[:::I = [ y b][::] for the C2,, operation

[EL:] = [b :][:,] for the G,, reflection (xz plane)

[;;:I = [ y b][:,] for the D,,' reflection ( y z plane)

The reducible representation F = A 1 i- BI :

2 1 ~ o m e sources use the yz plane as the plane of the molecule. This convention results in r = A, + B2 and switches the hl and b2 labels of the molecular orbitals.

A1

B1

H+H

H-H

H2O

96

Oxygen orbitals

E C2 !v !v’

148 Chapter 5 Molecular Orbitals

FIGURE 5-27 Symmetry of the Water Molecule.

Molecular orbitals of nonlinear molecules can be determined by the same procedures.

Water will be used as an example, and the steps of the previous section will be used.

1. Water is a simple triatomic bent molecule with a C2 axis through the oxygen and

two mirror planes that intersect in this axis, as shown in Figure 5-27. The point

group is therefore C2,.

2. The C2 axis is chosen as the z axis and the xz plane as the plane of the molecule.21

Because the hydrogen 1s orbitals have no directionality, it is not necessary to as-

sign axes to the hydrogens.

3. Because the hydrogen atoms determine the symmetry of the molecule, we will

use their orbitals as a starting point. The characters for each operation for the Is

orbitals of the hydrogen atoms can be obtained easily. The sum or thc contribu-

tions to the character ( I , 0, or - 1, as described previously) for each symmetry op-

eration is the character for that operation, and the complete list for all operations

of the group is the reducible representation for the atomic orbitals. The identity

operation leaves both hydrogen orbitals unchanged, with a character of 2.

Twofold rotation interchanges the orbitals, so each contributes 0, for a total char-

acter of 0. Reflection in the plane of the molecule (a,) leaves both hydrogens un-

changed, for a character of 2; reflection perpendicular to the plane of the

molecule (a,') switches the two orbitals, for a character of 0, as in Table 5-2.

TABLE 5-2 Representations for Ch Symmetry Op,erationr fol ~ydrogen &oms ,in Wker

Cz, Character Table

[:::I = [: :I[:;] for the identity operation

[:::I = [ y b][::] for the C2,, operation

[EL:] = [b :][:,] for the G,, reflection (xz plane)

[;;:I = [ y b][:,] for the D,,' reflection ( y z plane)

The reducible representation F = A 1 i- BI :

2 1 ~ o m e sources use the yz plane as the plane of the molecule. This convention results in r = A, + B2 and switches the hl and b2 labels of the molecular orbitals.

+1 -1 +1 -1

+1 +1 +1 +1

+1 -1 -1 +1

+1 +1 +1 +1A1s

B2px

A1pz

B1py

H2O

97

C2v E C2 !v !v’

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

Unchanged bonds E C2 !v !v’

#O-H 2 0 2 0

#O-H=A1+B1 means bonds (O of symmetry a1) and (H of symmetry a1)

H2O

98

H2O

Symmetry-adapted orbitals in several MLn complexes

Figure 6.5. Construction of the MOs forH2O from AO on oxygen and thesymmetry-adapted orbitals on thehydrogen atoms.

a1

a1

a1

b2

b1

b2

1b2

2b2

1a1

2a1

3a1

1b1

symmetry, three molecular orbitals are formed: 1a1 (bonding), 2a1 (non-bonding), and 3a1 (antibonding). The atomic orbital with B1 symmetry(2px) cannot, by symmetry, interact with any other, and so it staysunchanged in shape and in energy (1b1, nonbonding).

6.6. Symmetry-adapted orbitals in severalMLn complexes

The aim of this section is to construct the symmetry-adapted orbitalsfor the principal ligand fields, making use of the reduction (6.5) andprojection (6.10) formulae. In what follows, with the exception of §6.6.6, we shall only consider a single orbital on each ligand, the onethat is used to create the σ bond with the metal (Chapter 1, § 1.5.1).The orbital on ligand Li will be written σi. We shall suppose that all theligands, and thus all the orbitals σi, are identical.

M

!1

!2!3

!4

6-23

C4,C2, S4z

x

y

C2a!

C2b!

C2b"

C2a"

ML4 L1

L2L3

6-24

6.6.1. Square-planar ML4 complexes

Consider a complex in which the metallic atom is surrounded by fourligands that are placed at the corners of a square (6-23). The sym-metry elements of this system are characteristic of the D4h point group.The axes are shown in 6-24. The planes of symmetry are xy (σh), xz

(σda), and yz (σdb), respectively, together with the planes that bisect xz

and yz and each contain two M−−L bonds (σva and σvb, respectively).The inversion centre is of course at the origin, coincident with thecentral atom.

99

D3h E 2C3 3C2 !h 2S3 3!v

A1` 1 1 1 1 1 1

A2 ` 1 1 -1 1 1 -1

E ` 2 -1 0 2 -1 0

A1`` 1 1 1 -1 -1 -1

A2 `` 1 1 -1 -1 -1 1

E `` 2 -1 0 -2 1 0

#B-H=A1`+E`Unchanged bonds E 2C3 3C2 !h 2S3 3!v

#N-H 3 0 1 3 0 1

x

y

BH3

E C3 C32 C2(1) C2(2) C2(3) !h S3 S3

2 !v (1) !v (2) !v (3)

"1 "1 "2 "3 "1 "3 "2 "1 "2 "3 "1 "3 "2

"1

"2 "3

LGO(1)="1+"2+"3+"1+"3+"2+"1+"2+"3+"1+"3+"2=4"1+4"2+4"3=1/#3("1+"2+"3)

LGO(2)=1/#6(2"1-"2-"3)

LGO(3)=1/#2("2-"3)

100

2s orbital has a1` symmetryE 2C3 3C2 !h 2S3 3!v

1 1 1 1 1 1

2pz orbital has a2` symmetry1 1 -1 1 1 -1

2 -1 0 2 -1 0 2px and 2py orbital has e` symmetry

B

BH3

LGO(1) orbital has a1` symmetryE 2C3 3C2 !h 2S3 3!v

1 1 1 1 1 1

2 -1 0 2 -1 0LGO(2) and LGO(3) orbital has e`

symmetry

nonbonding

doesn’t belong to a1`+e`

101

Black plate (113,1)

schematically as the in-phase combination of 1s orbitalsshown as LGO(1) in Figure 4.17.

ða1’Þ ¼ 1 þ 2 þ 3 þ 1 þ 3 þ 2 þ 1 þ 2 þ 3

þ 1 þ 3 þ 2

¼ 4 1 þ 4 2 þ 4 3 ð4:14Þ

ða1’Þ ¼1ffiffiffi3

p ð 1 þ 2 þ 3Þ ð4:15Þ

A similar procedure can be used to deduce that equation4.16 describes the normalized wavefunction for one of thedegenerate e’ orbitals. Schematically, this is represented asLGO(2) in Figure 4.17; the orbital contains one nodal plane.

ðe’Þ1 ¼1ffiffiffi6

p ð2 1 % 2 % 3Þ ð4:16Þ

Each e’ orbital must contain a nodal plane, and the planesin the two orbitals are orthogonal to one another. Thus, wecan write equation 4.17 to describe the second e’ orbital; thenodal plane passes through atom H(1) and the 1s orbital onthis atom makes no contribution to the LGO. This isrepresented as LGO(3) in Figure 4.17.

ðe’Þ2 ¼1ffiffiffi2

p ð 2 % 3Þ ð4:17Þ

The MO diagram for BH3 can now be constructed byallowing orbitals of the same symmetry to interact. The 2pz

orbital on the B atom has a2’’ symmetry and no symmetrymatch can be found with an LGO of the H3 fragment.Thus, the 2pz orbital is non-bonding in BH3. The MOapproach describes the bonding in BH3 in terms of threeMOs of a1’ and e’ symmetries. The a1 orbital possesses !-bonding character which is delocalized over all four atoms.The e’ orbitals also exhibit delocalized character, and thebonding in BH3 is described by considering a combinationof all three bonding MOs.

NH3

The NH3 molecule has C3v symmetry (Figure 4.18) and abonding scheme can be derived by considering the interactionbetween the atomic orbitals of the N atom and the ligandgroup orbitals of an appropriate H3 fragment. An appropri-ate axis set has the z axis coincident with the C3 axis of NH3

(see worked example 3.2); the x and y axes are directed asshown in Figure 4.19. Table 4.2 shows part of theC3v charac-ter table. By seeing how each symmetry operation affects eachorbital of the N atom in NH3, the orbital symmetries areassigned as follows:

. each of the 2s and 2pz orbitals has a1 symmetry;

. the 2px and 2py orbitals are degenerate and the orbital sethas e symmetry.

Fig. 4.17 A qualitative MO diagram for the formation of BH3 using the ligand group orbital approach. The three H atoms in theH3 fragment are out of bonding range with each other, their positions being analogous to those in the BH3 molecule. OrbitalsLGO(2) and LGO(3) form a degenerate pair (e’ symmetry), although for clarity, the lines marking their orbital energies are drawnapart; similarly for the three 2p atomic orbitals of boron. [Exercise: where do the nodal planes lie in LGO(2) and LGO(3)?]

Chapter 4 . Molecular orbital theory applied to the polyatomic molecules BH3, NH3 and CH4 113

102

C3v E 2C3 3C2

A1 1 1 1

A2 1 1 -1

E 2 -1 0

#N-H=A1+EUnchanged bonds E 2C3 3C2

#N-H 3 0 1

NH3

103

NH32pz and 2s orbital has a1 symmetry

2 -1 0 2px and 2py orbital has e symmetry

LGO(1) orbital has a1 symmetry

LGO(2) and LGO(3) orbital has e symmetry

x

y

E 2C3 3C2

1 1 1

1 1 1

C3v E 2C3 3C2

1 1 1

1 1 1

104

Construction of MO: H2O as an example

Figure 6.3. Symmetry-adapted orbitals forNH3 (C3v point group) and orbitals of thesame symmetry on the central atom.

N

N

N

H 1

H3

H2H1

H3

H2H1

H1 H2

H1 H2

H3

H3

2pz2s

2px

2py

A1

E!E (1) = ex

!E (2) = ey

!A1

These orbitals are shown in Figure 6.3, together with the orbitals onthe central atom whose symmetry is given in the character table for theC3v point group (Table 6.6).

The set of orbitals that we have found for the E representations,φE(1) and φE(2), is not unique. Any pair of independent linear combin-ations of these two orbitals also constitutes a basis for this representation,and the same is true for the 2px and 2py orbitals on the central atom.The φE(1) and φE(2) orbitals shown here are, however, the set that isused most frequently. The first is transformed in the same way as 2px byall the symmetry operations of the C3v point group and the second like2py. They are therefore often written ex and ey, respectively.

6.5. Construction of MO: H2O as an example

The problem of allowing AO, or SALCO of these orbitals, to interact toform molecular orbitals (MO), is simplified considerably if the symmetryproperties of the system are taken into account. The use of symmetryallows us to identify rapidly those interactions which are exactly zero,and therefore to consider only those which really do contribute to theformation of the MO.

6.5.1. Symmetry and overlap

Two orbitals φ1 and φ2 interact if their overlap is non-zero (Chapter 1, §1.3). This overlap is equal to the integral over all space of the product ofthe functions φ1 and φ2:

S12 =∫

spaceφ∗

1 φ2dτ (6.13)

It can be shown that this integral is non-zero if the product functionis a basis for the totally symmetric representation of the group (or of a

105

Black plate (114,1)

To determine the nature of the ligand group orbitals, weconsider how many H 1s orbitals are left unchanged byeach symmetry operation in the C3v point group (Figure4.18). The result is represented by the row of characters:

E C3 !v

3 0 1

It follows that the three ligand group orbitals have a1 and esymmetries. Although the symmetry labels of the LGOs of

the H3 fragments in NH3 and BH3 differ because themolecules belong to different point groups, the normalizedwavefunctions for the LGOs are the same (equations 4.15–4.17). Schematic representations of the LGOs are shown inFigure 4.19.

Self-study exercises

1. Give a full explanation of how one derives the symmetries of theLGOs of the H3 fragment in NH3.

2. By following the same procedure as we did for BH3, deriveequations for the normalized wavefunctions that describe theLGOs shown schematically in Figure 4.19.

The qualitative MO diagram shown in Figure 4.19 isconstructed by allowing interactions between orbitals ofthe same symmetries. Because the nitrogen 2s and 2pz

Fig. 4.18 The NH3 molecule has C3v symmetry.

Ene

rgy

N NH3

H H

x

y

H

a1

e

LGO(1)

LGO(2) LGO(3)

node

a1

a1

e*

a1*

e

2pz (a1)2px 2py (e)

2s (a1)

C3v

Representation of the HOMO (a1)

Representation of one of the e MOs

Representation of the lowest lying a1 MO

Fig. 4.19 A qualitative MO diagram for the formation of NH3 using the ligand group orbital approach. For clarity, the linesmarking degenerate orbital energies are drawn apart. The diagrams on the right-hand side show representations of three of theoccupied MOs; the orientation of the NH3 molecule in each diagram is the same as in the structure at the bottom of the figure.

Table 4.2 Part of the C3v character table; the complete table isgiven in Appendix 3.

C3v E 2C3 3!v

A1 1 1 1A2 1 1 !1E 2 !1 0

114 Chapter 4 . Bonding in polyatomic molecules

NH3

106

Packing of spheres

The packing-of-spheres model applied to the structures of elements

Polymorphism in metals

Metallic radii

Melting points and standard enthalpies of atomization of metals

Alloys and intermetallic compounds

Bonding in metals and semiconductors

Semiconductors

Sizes of ions

Ionic lattices

Crystal structures of semiconductors

Lattice energy: The Born-Haber cycle

Lattice energy: ‘calculated’ versus experimental values

Applications of lattice energy

Defects in solid state lattices

Structures and energetics of metallic and ionic solidsChapter 6

107

Structures of the 14 possible crystal structures (Bravais lattices)

Bravais Lattices

108

x

y

x

z

A AB ABA

A AB ABC

x

z

x

y

Packing of Spheres

Hexagonal close packing (hcp)

Cubic close (ccp) or

face-centered (fcc) packing

Cube

Octahedral hole

Tetrahedral

hole

Simple Cubic non-close Packing Body-Centered Cubic non-close Packing

109

The Packing-of-Spheres Model Applied to the Structures of Elements

Group 18 Elements ccp (except He)

H2 and F2hcp,packing-of-spheres model to the crystalline structures of such

molecules is only valid because they contain freely rotating molecules

Metallic elements

h hexagonal closet-packing

!cubic closet-packing

" body-centered cubic packing

# structures type of its own

* slightly distorted

If a substance exists in more than one crystalline form, it is polymorphic.

*

*

*

*

*

110

ra

ra

c

b

ra

b

scc

a=2r

bcc

b2=a

2+a

2

c2=a

2+b

2= 3a

2

c= 3a=4r

a=4r

3

fcc

b2=a

2+a

2

16r2=2a

2

a= 8r

Relationship between Atomic Radius and the Edge Length in Three Different Unit Cells

111

Gold (Au) crystallizes in a cubic close-packed structure (the face-centered cubic unit cell) and has a density of

19.3 g/cm3. Calculate the atomic radius of gold in picometers and its percentage occupancy.

density of

unit cell

Volume of

unit cell

Edge of

unit cell

Radius of

Au atom

•

Edge

•

• Face

Corner

Body (Center)

Corner=1⁄8 atom

Edge=1⁄4 atom

Face=1⁄2 atom

Center=1 atom

Atoms per unit cell

Atoms per unit cells=how many atoms can be situated inside

a unit cell (fractions are permitted)

8*(1/8)+6*(1/2)=4 atoms per unit cell

Mass of the unit cell=4!atoms

1!unit!cell!

1!mol

6.022 !1023 !atoms!197.0g!Au

1!mol!Au=1.31!10"31g/unit!cell

Volume of the unit cell=m

d=1.31!10"31

19.3!g/cm3= 6.79 !10"23cm3

a = V3 = 6.79 !10

"23cm

33= 4.08 !10

"8cm

r =a

8

=4.08 !10

"8cm

8

=1.44 !10"8cm=144pm

8r = a

( 8 )3r3= a

3! 22.63r

3

(Atoms!per!unit!cell) * 43!r3 =16

3!r3 " 16.76r3

Unit cell edge length=

Unit cell volume=

The unit cell contains 4 spheres, and the volume of a sphere=4/3!r3

Volume of the unit cell occupied=

Percentage occupancy=Volume of the unit cell occupied/Unit cell volume*100%=16.76/22.63*100%=74%

Atomic Radii and Percentage occupancy

Holes occupancy=1-74%=26%

112

Pauling considered a series of alkali metal halides, each member of which contained

isoelectronic ions (NaF, KCl, RbBr, CsI). In order to partition the ionic radii, he assumed that the

radius of each ion was inversely proportional to its actual nuclear charge less an amount due to

screening effects (can be calculated using Slater's rules).

For example, the difference in anion-cation distance between the halides NaX and KX is close to

36 pm irrespective of the anion X, and it is natural to attribute this to the difference in radii

between Na+ and K+. To separate the observed distances into the sum of two ionic radii is,

however, difficult to do in an entirely satisfactory way. One procedure is to look for the

minimum value in the electron density distribution between neighboring ions, but apart from the

experimental difficulties involved such measurements do not really support the assumption of

constant radius. Sets of ionic radii are therefore all based ultimately on somewhat arbitrary

assumptions. Several different sets have been derived, the most widely used being those of

Shannon and Prewitt, based on the assumed radius of 140 pm for O2 in six-coordination.

Goldschmidt and, more recently, Shannon and Prewitt, concentrated on the analysis of

experimental data (mostly fluorides and oxides) with the aim of obtaining a set of onic radii

which, when combined in pairs, reproduced the observed internuclear distances, they found that

rion for a given ion increases slightly with an increase in coordination number.

Cation Li+ Na+ K+ Rb+ Cs+

r+/pm 76 102 138 149 170

Anion F- Cl - Br - I -

r-/pm 133 181 196 220

Ionic Radii

2rx-

d=2rx-+2rx+

d

Li+

Cl-

113

7.1 Radius Ratios 53

Fig. 7.1The three most

important structuretypes for ioniccompounds of

composition MX.Compared to theireffective sizes, the

ions have beendrawn to a smaller

scaleCsCl

Pm3mNaCl

F m3m

zinc blende (sphalerite, ZnS)F 43m

one eighth of the unit cell

a

a 2

a 3!

12 a 2

12 a

!!

!!

face of theunit cell

12 a

rX

rX 3

2rX

12 a 2 rX 2 1

2 a rX 2 14 a 2 rX

12 rX 6 1

4 a 12 rX 2

!

!

!

!

!!

!

!

!

Fig. 7.2Calculation of the

limiting radius ratiosrM rX

CsCl type

rM rX rX 3rM rX 3 1

0 732

NaCl type

rM rX rX 2rM rX 2 1

0 414

zinc blende type

rM rX rX12 6

rM rX12 6 10 225

complicated. If we only take into account the electrostatic part of the lattice energy, thenthe relevant magnitude in equation (5.4) is the ratio A R (A MADELUNG constant, Rshortest cation–anion distance). Fig. 7.3 shows how the electrostatic part of the lattice en-ergy depends on the radius ratio for chlorides. The transition from the NaCl type to thezinc blende type is to be expected at the crossing point of the curves at rM rX 0 3 insteadof rM rX 0 414. The transition from the NaCl type to the CsCl type is to be expectedat rM rX 0 71. The curves were calculated assuming hard Cl ions with rCl = 181pm. If, in addition, we take into account the increase of the ionic radius for an increasedcoordination number, then we obtain the dotted line in Fig. 7.3 for the CsCl type. As aconsequence, the CsCl type should not occur at all, as the dotted line always runs below

Ionic Radii Ratio: Why NaCl is Cubic?

For LiF, this corresponds to an NaCl type lattice, in

agreement with the observed structure. In fact each of the

group 1 halides (except CsCl, CsBr and CsI) at 298K and

1 bar pressure adopts the NaCl type structure; CsCl, CsBr

and CsI adopt the CsCl type lattice. Radius ratio rules

predict the correct structures in only some cases; they

predict tetrahedral coordination for the cations in LiBr

and LiI, and cubic coordination in NaF, KF, KCl, RbF,

RbCl, RbBr and CsF (in addition to CsCl, CsBr and CsI).

rM/rX

Predicted

coordination

number of cation

Predicted

coordination

geometry of cation

<0.15 2 Linear

0.15-0.22 3 Trigonal planar

0.22-0.41 4 Tetrahedral

0.41-0.73 6 Octahedral

>0.74 8 Cubic

rM/rXCoordination number

and polyhedronStructure type

>0.732 8 cube CsCl

0.414-0,732 6 octahedron NaCl

<0.414 4 tetrahedron Zinc blende

CsCl NaCl ZnS

114

Metallic Radii

CN=how many atoms can make bond with the cation/anion* Inorganic Chemistry, Houescroft and Sharpe page 153

2rmetal

Coordination number

Relative Radius

12 8 6 4

1.00 0.97 0.96 0.88

For example in the case of K (potassium) the lattice system is

cubic body centered=K with 8 other K around.

r12coordiante

r8coordinate

=1

0.97

From!Table!6.2 *!! r8coordinate = 0.97 * r12coordinate = 0.97 *235 = 228pm

115

Binary compounds (compounds consisting of two elements) may have very simple crystal structures and

can be described in several different ways. If the larger ions (usually the anions) are in close-packed

structures, ions of the opposite charge occupy these holes, depending primarily on two factors:

1. The relative sizes of the atoms or ions. The radius ratio (usually r+/r- but sometimes r-/r+, where r+

is the radius of the cation and r- is the radius of the anion) is generally used to measure this. Small

cations will fit in the tetrahedral or octahedral holes of a close-packed anion lattice. Somewhat larger

cations will fit in the octahedral holes, but not in tetrahedral holes, of the same lattice. Still larger cations

force a change in structure.

2. The relative numbers of cations and anions. For example, a formula of M2X will not allow a close-

packed anion lattice and occupancy of all of the octahedral holes by the cations because there are too

many cations. The structure must either have the cations in tetrahedral holes, have many vacancies in

the anion lattice, or have an anion lattice that is not close-packed

Binary compounds

116

NaCl is made up of face-centered cubes (fcc) of sodium ions and face-centered cubes of chloride ions combined, but offset by

half a unit cell length in one direction so that the sodium ions are centered in the edges of the chloride lattice (and vice versa). If

all the ions were identical, the NaCl unit cell would be made up of eight simple cubic unit cells. Many alkali halides and other

simple compounds share this same structure. For these crystals, the ions tend to have quite different sizes, usually with the anions

larger than the cations. Each sodium ion is surrounded by six nearest-neighbor chloride ions, and each chloride ion is surrounded

by six nearest-neighbor sodium ions.

Na+

Cl-

Corner

Edge

Face

Center

Total

Na+ Cl-

8 0

0

6

0

12

0

1

8*1/8=1 Na+

12*1/4=3 Cl-

6*1/2=3 Na+

1*1=1 Cl-

4 Na+ and 4 Cl-

CN 6 6

Cation is small

Anion is large

117

Corner

Edge

Face

Center

Total

Cs+ Cl-

8 0

0

0

0

0

0

1

8*1/8=1 Cs+

0*1/4=3 Cl-

0*1/2=3 Cs+

1*1=1 Cl-

1 Cs+ and 1 Cl-

CN 8 8

As mentioned previously, a sphere of radius 0.73r will fit exactly in the center of a cubic

structure. Although the fit is not perfect, this is what happens in CsCl where the chloride

ions form simple cubes with cesium ions in the centers. In the same way, the cesium ions

form simple cubes with chloride ions in the centers. The average chloride ion radius is 0.83

times as large as the cesium ion (167 pm and 202 pm, respectively), but the inter-ionic

distance in CsCl is 356 pm, about 3.5% smaller than the sum of the average ionic radii.

Only CsCl, CsBr, Csl, TlCl, TlBr, TII, and CsSH have this structure at ordinary

temperatures and pressures, although some other alkali halides have this structure at high

pressure and high temperature. The cesium salts can also be made to crystallize in the NaCl

lattice on NaCl or KBr substrates, and CsCl converts to the NaCl lattice at about 469ºC.

Cation is large

Anion is large

118

The wurtzite form of ZnS is much rarer than zinc blende, and is formed at higher temperatures

than zinc blende. It also has zinc and sulfide each in a tetrahedral hole of the other lattice, but

each type of ion forms a hexagonal close-packed lattice. As in zinc blende, half of the tetrahedral

holes in each lattice are occupied.S

Zn

One sulfide layer and one zinc layer of wurtzite. The

third layer contains sulfide ions, directly above the

zinc ions. The fourth layer is zinc ions, directly

above the sulfides of the first layer.

ZnS has two common crystalline forms, both with coordination

number 4. Zinc blende is the most common zinc ore and has

essentially the same geometry as diamond, with alternating layers of

Zn and S. It can also be described as having zinc ions and sulfide ions,

each in face-centered lattices, so that each ion is in a tetrahedral hole

of the other lattice. The stoichiometry requires half of these tetrahedral

holes to be occupied, with alternating occupied and vacant sites.diamondZnS blend

ZnS wurtzite

119

The fluorite structure can be described as having the calcium ions in a cubic close-packed lattice,

with eight fluoride ions surrounding each one and occupying all of the tetrahedral holes. An

alternative description of the same structure has the fluoride ions in a simple cubic array, with

calcium ions in alternate body centers. The ionic radii are nearly perfect fits for this geometry. There

is also an antifluorite structure in which the cation-anion stoichiometry is reversed. This structure is

found in all the oxides and sulfides of Li, Na, K, and Rb, and in Li2Te and Be2C. In the antifluorite

structure, every tetrahedral hole in the anion lattice is occupied by a cation, in contrast to the ZnS

structures, in which half the tetrahedral holes of the sulfide ion lattice are occupied by zinc ions.

Fluorite shown as Ca2+ in a cubic close packed lattice,

each surrounded by eight F- in the tetrahedral holes.

120

Ni

As

TiO2 in the rutile structure has distorted TiO6 octahedra that form columns by sharing edges, resulting

in coordination numbers of 6 and 3 for titanium and oxygen, respectively. Adjacent columns are

connected by sharing corners of the octahedra. The oxide ions have three nearest-neighbor titanium

ions in a planar configuration, one at a slightly greater distance than the other two. The unit cell has

titanium ions at the corners and in the body center, two oxygens in opposite quadrants of the bottom

face, two oxygens directly above the first two in the top face, and two oxygens in the plane with the

body-centered titanium forming the final two positions of the oxide octahedron. The same geometry is

found for MgF2, ZnF2, and some transition metal fluorides. Compounds that contain larger metal ions

adopt the fluorite structure with coordination numbers of 8 and 4.

The nickel arsenide structure has arsenic atoms in identical close-packed layers stacked directly

over each other, with nickel atoms in all the octahedral holes. The larger arsenic atoms are in the

center of trigonal prisms of nickel atoms. Both types of atoms have coordination number 6, with

layers of nickel atoms close enough that each nickel can also be considered as bonded to two others.

An alternate description is that the nickel atoms occupy all the octahedral holes of a hexagonal

close-packed arsenic lattice. This structure is also adopted by many MX compounds, where M is a

transition metal and X is from Groups 14, 15, or 16 (Sn, As, Sb, Bi, S, Se, or Te). This structure is

easily changed to allow for larger amounts of the nonmetal to be incorporated into

nonstoichiometric materials.

Ti

O

NiAs - a semiconductor

Rutile

121

Perovskite is an example of a double oxide; it does not, as the formula might imply, contain[TiO3]2- ions, but is a mixed Ca(II) and Ti

(IV) oxide.

Many double oxides or fluorides such as BaTiO3, SrFeO3, NaNbO3, KMgF3 and KZnF3 crystallize with a perovskite lattice.

Deformations of the lattice may be caused as a consequence of the relative sizes of the ions, e.g. in BaTiO3, the Ba2+ ion is relatively

large (rBa2+=142pm compared with rCa2+=100 pm) and causes a displacement of each Ti(IV) centre such that there is one short TiO

contact. This leads to BaTiO3 possessing ferroelectric properties

The structures of some high-temperature superconductors are also related to that of perovskite. Another mixed oxide lattice is that of

spinel, (complicated structure and its structure will not be drawn here) MgAl2O4.

Corner

Edge

Face

Center

Ba2+ O2-

0 0

0

0

1

0

6

0

1*1=1 Ba2+

6*1/2=3 O2-

8*1/8=1 Ti+

CN 6 2

Perovskite Ti+

8

0

0

0

8

ccp (fcc)

BaTiO3 a ferroelectric complex

122

CdI2 Graphite

123

SiO32-

Si2O52-

SiO32-

Si4O116-

Si6O1812-Si3O9

6-

Common Silicate Structures

Quartz

=

Consider the first tetrahedron in the chains of SiO32- to have four oxygen atoms, or SiO4. Extending the

chain by adding SiO3 units with the fourth position sharing an oxygen atom of the previous tetrahedron

results in an infinite chain with the formula SiO3. The charge can be calculated based on Si4+ and O2-.

Si2O52- can be described similarly. Beginning with one Si2O7 unit can start the chain. Adding Si2O5 units

(two tetrahedra sharing one corner and each with a vacant comer for sharing with the previous unit) can

continue the chain indefinitely. Again, the charge can be calculated from the formula.

SiO2: $-Cristobalite and "-Quartz

ß-cristoballite

124

Alloys and Intermetallic Compounds

Substitutional alloys:

In a substitutional alloy, atoms of the solute occupy sites in the lattice of solvent metal, e.g. Sterling Silver (92.5% Ag and 7.5% Cu)

1. Both metals must have the same lattice (ccp, hcp...etc)

2. radius of the first metal is approximately equal to the other.

Interstitial alloys:

In a interstitial alloy, atoms with small radius occupy the

octahedral holes, e.g. Stainless steel Cr in Fe and Carbon

steel (see Box 6.2 page 159, of Inorganic chemistry

Housecroft and Sharpe)

Intermetallic compounds:

When melts of some metal mixtures solidify, the alloy formed may possess a definite structure type that is different from those of the

pure metals. e.g. CuZn (ß-Brass)

Replace (an) atom(s) from

the lattice by different one

Insert (an) atom(s) in the

lattice.

125

Metals

Explain why metals are usually lustrous (mirror-like)

In the band model there is a continuum of empty energy levels rather than discrete energy levels. Thus light quanta of all

energies within a wide range of wavelengths will be absorbed equally, and then the energized electrons will re-emit the light

when they fall back into their ground-state orbitals. This is the mechanism for reflection of light of all frequencies, which we

call "luster".

Predict how the Group II metals differ from group I in density, melting point, and mechanical strength.

In any given period the Group II ions are smaller and may thus approach each other more closely. At the same time twice as

many electrons are present sea. The closer approach and the much greater electrostatic interactions between the 2+ ions and the

sea of high negative charge density lead to greater density and much greater bonding energy, which in turn leads to much higher

melting point and hardness. Actually going from Group I to Group II increases the density by a factor of about 2 and raises the

melting point by hundreds of degrees.

Metals usually feel cool to the touch compared with other materials because they are such good conductors of heat. How

can we explain this unusual thermal conductivity?

In most materials heat is conducted by atom-to-atom transfer of vibrational motion from the hot end to the cold end. In metals

the thermal energy is transferred primarily by the motion of the electron sea's free electrons, which are very mobile.

126

Conductors

Li Li

Li Li Li

Li Li Li Li

Li Li.....

6.022!1023 Li (metallic solid)

Li2

Li3

Li4

LiAvogadro

empty orbitals

filled orbitals

In any solid, metallic, ionic or covalent the valence

orbitals of atoms and molecules combine to form

molecular orbitals. The number of molecular orbitals

is equal to the number of atomic orbitals, and for any

material, this is a very large number but the energy

difference between highest and lowest doesn't change

by much. Many orbitals separated by very small

energy gaps give rise to bands of orbitals.

For lithium metal, the band is made up of molecular

orbitals formed from the combination of 2s atomic

orbitals and about half of the orbitals are filled. There

is only a very small energy gap between the highest

occupied molecular orbital (HOMO) and the lowest

unoccupied molecular orbital (LUMO). When an

electron goes to an unoccupied orbital with available

thermal energy, it is free to travel throughout the

metal. The "hole" left in the HOMO can travel too.

127

Band gap

Bandgap Fermi Level!

Ele

ctro

nic

(orb

ital

ener

gy)

nonmetallic band picture metallic solid band picture

electron sea are in blue color

The band gap in a nonmetallic solid is important for electrical

and optical properties. A solid with a small band gap is a

semiconductor with a conductivity that (unlike the case with a

metal) increases as temperature is raised. The band gap also

determines the minimum photon energy required to excite an

electron from the valence band (VB) to the conduction band

(CB), and hence the threshold for optical absorption by a solid.

VB

CB

In silica (SiO2) there are covalent bonds between each silicon and 4 oxygens around it that are

combinations of Si 3s and 3p atomic orbitals and O 2s and 2p atomic orbitals. We can think of the

molecular orbitals forming similar bands but there is a large energy separation or band gap between the

occupied and unoccupied levels. The electrons in the orbitals of silica are not free to travel and the

material is an insulator.

In pure, crystalline silicon, there is also a band gap between the occupied and unoccupied molecular

orbitals but it is smaller. Electrons in the filled valence band can be promoted to the conduction

band by thermal energy or when the material absorbs light. The electrons would then be free to

travel through the material and holes (electron vacancies) would travel through the valence band.

128

Semiconductors

Intrinsic semiconductors: pure semiconductors e.g. C, Si, Ge and a-Sn

Extrinsic (n- and p-type) semiconductors: extrinsic semiconductors contain dopants; a dopant is an impurity introduced into a semiconductor in minute amounts to enhance its electrical conductivity, e.g. Ge-doped Si. Doping involves the introduction of only a minutely small proportion of dopant atoms, less than 1 in 106, and extremely pure Si or Ge must first be produced.

Acceptor level

Small band gap

p-type

Small band gap

Donor level

n-type

In a p-type semiconductor (e.g. Ga-doped Si), electrical conductivity arises from thermal population of an acceptor level which leaves vacancies (positive holes) in the lower band.

In an n-type semiconductor (e.g. As-doped Si), a donor level is close in energy to the conduction band.

Carbon in diamond form is an

insulator with extremely high

resistivity. But in graphite form its

interatomic spacing is larger, making the

band gap small enough to support some

electrical conduction

Used as p-dopants

to produce p-type

semiconductors

Tin can be considered to be a

semiconductor with a very small

band gap, but at room temperature

it supports metallic conduction

Silicon and germanium are the

intrinsic semiconductors employed in

solid state electronics.

Used as n-dopants

to produce n-type

semiconductors.

The bands overlap in lead,

making it metallic conductor.

BC

2p2N

AlSi

3p2P

GaGe

4p2As

InSn

5p2Sb

TlPb

6p2Bi

129

g / Georg Simon Ohm (1787-

1854).

3.4 Resistance

Resistance

So far we have simply presented it as an observed fact that abattery-and-bulb circuit quickly settles down to a steady flow, butwhy should it? Newton’s second law, a = F/m, would seem topredict that the steady forces on the charged particles should makethem whip around the circuit faster and faster. The answer is that ascharged particles move through matter, there are always forces, anal-ogous to frictional forces, that resist the motion. These forces needto be included in Newton’s second law, which is really a = Ftotal/m,not a = F/m. If, by analogy, you push a crate across the floor atconstant speed, i.e., with zero acceleration, the total force on it mustbe zero. After you get the crate going, the floor’s frictional force isexactly canceling out your force. The chemical energy stored inyour body is being transformed into heat in the crate and the floor,and no longer into an increase in the crate’s kinetic energy. Simi-larly, the battery’s internal chemical energy is converted into heat,not into perpetually increasing the charged particles’ kinetic energy.Changing energy into heat may be a nuisance in some circuits, suchas a computer chip, but it is vital in a lightbulb, which must get hotenough to glow. Whether we like it or not, this kind of heating effectis going to occur any time charged particles move through matter.

What determines the amount of heating? One flashlight bulbdesigned to work with a 9-volt battery might be labeled 1.0 watts,another 5.0. How does this work? Even without knowing the detailsof this type of friction at the atomic level, we can relate the heatdissipation to the amount of current that flows via the equationP = I∆V. If the two flashlight bulbs can have two different valuesof P when used with a battery that maintains the same ∆V , itmust be that the 5.0-watt bulb allows five times more current toflow through it.

For many substances, including the tungsten from which light-bulb filaments are made, experiments show that the amount of cur-rent that will flow through it is directly proportional to the voltagedifference placed across it. For an object made of such a substance,we define its electrical resistance as follows:

definition of resistanceIf an object inserted in a circuit displays a current flow pro-portional to the voltage difference across it, then we define itsresistance as the constant ratio

R = ∆V/I

The units of resistance are volts/ampere, usually abbreviated asohms, symbolized with the capital Greek letter omega, Ω.

88 Chapter 3 Circuits, Part 1

The electrical conductivity of a metal decreases with temperature

A metal is characterized by the fact that its electrical resistivity increases as the temperature increases, i.e. its electrical conductivity decreases as the temperature increases.

The electrical conductivity of a semiconductor increases with temperature.

Resistivity: Conductors vs. Semiconductors

The degree to which a substance or device opposes the passage of an

electric current, causing energy dissipation. Ohm's law resistance

(measured in ohms) is equal to the voltage divided by the current.

ohms

130

PerovskitesMgB2

Cu

Y

Ba

Mg

B

YBa2Cu3O7

The conductivity of metals and of semiconductors changes with temperature. Why? It's because the

electrons flowing through a material will always pass close by positively charged nuclei. This impedes

flow. As temperature increases, there is more thermal motion of the nuclei and this resistance increases.

Resistivity is a property of the metal or semiconductor. In superconductors, the resistivity goes to zero at

some temperature.

Some of the recently discovered high temperature superconductors are based on perovskites. You drew the

lattice of CaTiO3 last week and most of you drew a cubic lattice of Ca with Ti in the cubic hole and

oxygen atoms on the edges. Another perovskite with the same structure is BaCuO3 You could also draw

the same lattice as a cubic array of copper atoms. Now imagine that some of the oxygen atoms have been

removed and 1/3 of the barium atoms have been exchanged for yttrium. You would have the basic

structure of a 1-2-3 material or YBa2Cu3O7.

Superconductors

131

Lattice Energy

The energy of the crystal lattice of an ionic compound is the energy released

when ions come together from infinite separation to form a crystal:

M(g) ++X(g)

-"MX(s)

In the crystal lattice there are more interactions. The summation of all of these geometrical

interactions is known as the Madelung constant, A. The energy then can be expressed as following:

From Coulomb’s law, the attraction is obtained as following:

Ec=Z+Z!

4"#0r

The energy increases as the interionic distance decreases.

It is common to express Z+ and Z- as multiples of electronic charge:

Ec=e2Z+Z!

4"#0r

Ec=Ae

2Z+Z!

4"#0r

132

Madelung Constant

a distance r

a distance #2 r

a distance #3 r

a distance #4 r=2r

6Xz!

12Mz+

8Xz!

6Mz+

!U = "e2

4#$0[(6

rz+ z" )" (

12

2rz+ z+ )+ (

8

3rz+ z" )" (

6

4rz+ z+ )...]

!U = "e2

4#$0[(6

rz+ z" )" (

12

2rz+

2)+ (

8

3rz+ z" )" (

6

4rz+

2)...]

!U = "z+ z" e

2

4#$0r[6" (

12 z+

2 z"

)+ (8

3)" (3

z+

z"

)...]...

...

133

Madelung Constant

!U = "z+ z" e

2

4#$0r[6" (

12 z+

2 z"

)+ (8

3)" (3

z+

z"

)...]z+

z!

=1!for!NaCl

!U = "z+ z" e

2

4#$0r[6"

12

2+8

3" 3...]

!U = "N

AvogadroA z+ z" e

2

4#$0r

A Madelung constant.

NAvogadro Avogadro’s number=6.0223x1023 atom/mol

|z+| e.g. NaCl; Na has a charge of +1 then |+1|=1

|z-| e.g. NaCl; Cl has a charge of -1 then |-1|=1

e 1.602x10-18 C

!0 permittivity of vacuum=8.854x10-12 Fm-1

r the ionic radius

Lattice type A

Sodium chloride (NaCl) 1.7476

Caesium chloride (CsCl) 1.7627

Wurtzite ($-ZnS) 1.6413

Zinc blend (ß-ZnS) 1.6381

Fluorite (CaF2) 2.5194

Rutile (Ti2O) 2.408

Cadmium iodide (CdI2) 2.355

Lattice energy

increases with increasing |z+| and |z-|

decreases with increasing r

increases with increasing the number of ions

134

Lattice Energy: Some Approximation Formulas I

The Coulomb equation is a simple form for the lattice energy. Beside

Coulombic interactions there are other type of forces, nucleus-nucleus

repulsion are one of them. These forces are called Born forces,.

Mathematically, one can add the following term for such forces: !U =N

AB

rn

B repulsion coefficient

n Born exponent

Electronic configuration of the

ions in an ionic compound MXn

[He][He] 1.7476

[Ne][Ne] 1.7627

[Ar][Ar] or [3d10][Ar] 1.6413

[Kr][Kr] or [4d10][Kr] 1.6381

[Xe][Xe] or [5d10][Xe] 2.5194

n Born exponent For LiCl=Li+Cl-=

([He]+[Ne])/2=(5+9)/2=7

!U(0K) =AN

Ae2z+z"

4#$0r

+N

AB

rn

!"U(0K)

!rr=r0

= 0

B =AN

Ae2z+z!r0

n!1

4"#0n

Born-Landè equation

rr0

E

n

e

r

g

y

!U(0K) = "AN

Ae2z+z"

4#$0r0

(1"1

n)

135

Lattice Energy: Some Approximation Formulas II

Further approximations to the lattice energy equation may be done by

replacing the 1/rn term by e-%/r. This will lead to Born-Mayer equation !U(0K) = "AN

Az+z"e2

4#$0r0

(1"%

r0

)

An example of these approximations among other are NaCl. In NaCl, the contributions to the total

lattice energ (766 kJ mol) made by electrostatic attractions, electrostatic and Born repulsions,

dispersion energy and zero-point energy are 860, -99, 12 and -7kJmol1 respectively. In fact, the

error introduced by neglecting the last two terms (which always tend to compensate each other) is

very small

From the other hand, approximations can be done to simplify the lattice energy equations. Attempts

have been made to use the fact that Madelung constants for MX and MX2 lattice types are in an

approximate ratio of 2:3. In 1956, Kapustinskii derived what has become the best known general

expression for estimating lattice energies, and one form of this is given in equation:

!U(0K) = "(1.07#10

5)$ z+ z"

r+ + r"

Gross approximation and has to be treated with caution

Kapustinskii equation ! Number of ions; e.g. CaF2 ! = 3

136

Born-Haber Cycle: Experimental Determination of the Lattice Energy

K(s) K+(g)

K(g)

&Cl2(g) Cl-(g)

Cl(g)

"fHº(K+,g)

"fHº(K,s) IE1(K,g)

"fHº(Cl-,g)

"EAHº(Cl,g)"aHº(Cl,g)

"fHº(K,s)+ IE1(K,g)= "fHº(K+,g)

"aHº(Cl,g)+ "EAHº(Cl,g)= "fHº(Cl-,g)

K(s)+&Cl2(g)!KCl(s) "fHº(KCl,s)

K+(g)+Cl-(g)!KCl(s) "latticeHº(KCl,s)

K(s)!K(g)

&Cl2(g)!Cl(g)

Cl(g)+e!Cl-(g)

K(g)!K+(g)+e

"aHº(K,g)

"aHº(Cl,g)

IE(K,g)

EA(Cl,g)

K(g)+Cl(g)!K+(g)+Cl-(g) IE(K,g)+EA(Cl,g)

K(s)+&Cl2(g)!K(g)+Cl(g) "aHº(K,g)+"aHº(Cl,g) III

IV

III+IV

K(s)+&Cl2(g)!K+(g)+Cl-(g) "aHº(K,g)+"aHº(Cl,g)+IE(K,g)+EA(Cl,g)

I

II

KCl(s)!K(s)+&Cl2(g) -"fHº(KCl,s) I

KCl(s)!K+(g)+Cl-(g) "aHº(K,g)+"aHº(Cl,g)+IE(K,g)+EA(Cl,g)-"fHº(KCl,s)

"latticeHº(KCl,s)=

"fHº(KCl,s)-("aHº(K,g)+"aHº(Cl,g)+IE(K,g)+EA(Cl,g))

K+(g)+Cl-(g)!KCl(s)

137

Defect in crystal lattice: Schottky and Frenkel Defect

Removing an anion/cation from the lattice Moving an anion/cation in the lattice

Schottky Frenkel

138