INORGANIC CHEMISTRY

INORGANIC CHEMISTRYINORGANIC CHEMISTRY

Teachers: Prof. Zhang Wei-guang ( 章伟光） Tel: 13416250262(HP), E-mail: [email protected]

Dr. Fan Jun （范军） Tel: 85210267(H), 85210875(o); E-mail: [email protected]

Teachers: Prof. Zhang Wei-guang ( 章伟光） Tel: 13416250262(HP), E-mail: [email protected]

Dr. Fan Jun （范军） Tel: 85210267(H), 85210875(o); E-mail: [email protected]

Sept. 2005For Science Class 2

Sept. 2005For Science Class 2

ContentsContentsINORGANIC CHEMISTRY

ContentsContents ContentsContents

Introduction

Section A Atomic structure

Section B Introduction to inorganic substance

Section C Structure and bonding in molecules

Section D Structure and bonding in solids

ContentsContentsINORGANIC CHEMISTRY

Section E Chemistry in solution

Section F Chemistry in nonmetals

Section G Non-transition metals

Section H Chemistry of transition metals

Section I Lanthanides and actinides

Section J Environmental, biological and industrial aspects

IntroductionIntroductionINORGANIC CHEMISTRY

Introduction

1. Emergence and Development of Chemistry

Four Inventions in ancient China and Alchemy in ancient Greece

Nature phenomenon: Burning, Volcano breaking out, and Lightning, etc.

High technology production in modern civilization

Pride of Chemist

Set up theoretic system of modern chemistry and industry


2. Brach of Chemistry

Organic Chemistry

CHEMISTRY

Inorganic Chemistry

Analytical Chemistry

Physical Chemistry

Biochemistry

Geochemistry

Spacechemistry

Material Chemistry

Polymer Chemistry

Organometallic Chemistry

Coordinary Chemistry

……

etc.

Structural Chemistry

Electro-Chemistry

Nano-Chemistry


3. Ways of Learning Chemistry

Study hard

Learn cleverly

Do it yourself


4. Some basic conception in Chemistry

Molecules Atoms

Protons + Neutrons = Nucleus

Electrons

Get or Lost

Nucleus reaction

Chemical reaction

Breaking

Forming


Atom Nucleus Electrons +

Isotopes, New Element Ions, Free radical

Removal (addition) of Protons or Neutrons

Removal (addition) of electrons

Isotopes are atoms with the same atomic number but different numbers of the neutrons for one element, so their mass numbers will be different.

They have the same amount of protons.

Potential application: NMR, dating and origin of rocks,

the studies for the reaction mechanism, etc


Species discussed in this course

Gas

Solid

Liquid, Solution

SI units:

n-----mol

Molar mass-----gmol-1

Molar volume---- m3mol-1 lmol-1, dm3mol-1

Pressure, P-----Pa, KPa, atm

Temperature, T-----K, ℃

Energy----- J, or kJ, cal, kcal

GasGasINORGANIC CHEMISTRY

1. Ideal-gas Law

GAS

Real gas

Ideal gas

This content would be introduced in Physics and Physical Chemistry.

The volume of the molecules in gas would be considered as zero.

No intermolecular interaction

(1) Ideal-gas equation For the mono-component ideal-gas

P——pressure (Pa); V ——volume (m3); n —— moles (mol);

T ——temperature (K) T= t + 273.15; d —— density (gmol-1)

P V = n R T

(i) If SI units would be used, R = 8.314 Jmol-1K-1;

(ii) P—KPa, V—dm3, T—K, n—mol, R = 8.314 J mol-1 K-1;

(ii) P—atm, V—l, T—K, n—mol, R = 0.08206 atm l mol-1 K-1

M

RT

P

d


For example:

(a) 现在一般能得到的最高真空是 10-12 托 (1 托＝ 1 mmHg) 。求 27℃下体积为 1ml 的真空体系中的气体分子数。

P V = n R T n = PV/(RT) = 5×10-20 mol

N ＝ n×6.023×1023 ＝ 3×104 个

(b) 空气的 N2 和 O2 的体积比例分别为 79 ％与 21 ％，计算空气在 25℃和 100KPa下空气的密度是多少？ (gdm-3)

M ＝ 28×79 ％＋ 32×21 ％＝ 28.8

d ＝ 1.16 gdm-3

M

RT

P

d


(2) Law of partial pressure For the Poly-component ideal-gas

If the chemical reaction don’t take place among the component, the total moles would keep constant after mixed them.

T

i

i

T

ii

TT

ii

iT

iiT

V

Vtionvolumefrac

x

n

nx

RTnVP

RTnVP

nnnnn

nV

RTPPPPP

1

....

....

321

321

If V=0 and T=0,

Mole fraction

partial pressure of one kind of component

GasGasINORGANIC CHEMISTRY

For example:

(1) 将 730mmHg 压力的氮气 60ml ， 300mmHg 压力的氧气 200 ml 及 420mmHg 压力的氢气 80ml 压入 250ml 的真空瓶中，求混合后的气体分压力和混合的总压力。

(2) 某容器中含有 NH3 、 O2 、 N2 等气体的混合物。取样分析后，其中 n(NH3)=0.320mol ， n(O2)=0.180mol ， n(N2)=0.700mol 。混合气体的总压 p=133.0kPa。试计算各组分气体的分压。

n= n(NH3)+n(O2)+n(N2)

iiT nV

RTPPPPP ....321

RTnVP ii

PN2= 175mmHg, PO2=240mmHg, PH2=134mmHg ， PT = 549mmHg

pxpnn

p BB

B

gasgasINORGANIC CHEMISTRY

(3) Diffusion Law of gas

1

2

2

1

2

1

M

M

d

d

u

u

u －－ the diffusion rate of the gas molecules

d －－ the density of the gas

M －－ the mole mass of the gas

(4) Real gas

Homework: 1, 3, 7

Section A1Section A1INORGANIC CHEMISTRY

Section A Atomic structure

A1 THE NUCLEAR ATOM

Electron and NucleiElectron and Nuclei An atom consists of a very small positively charged nucleus,

surrounded by negative electrons held by electrostatic attraction.

The motion of electrons changes when chemical bonds are

formed, nuclei being unaltered.

Nucleus: positive charge

Electron: negative charge

Electrostatic attractive force: Electrons are attracted to Nucleus.

Electrostatic repulsion force: among the protons.

From planetary model to quantum theory


Nuclear structureNuclear structure

Nuclei contain positive protons and uncharged neutrons. The number of

protons is the atomic number (Z) of an element.

Electrostatic attractive force: Electrons are attracted to Nucleus

The attractive strong interaction between protons and neutrons is

opposed by electrostatic repulsion between protons. Electrostatic repulsion force: among the protons.

Repulsion dominates as Z increases and there is only a limited

number of stable elements.

Magic numbers:

2 8 20 28 50 82 126 16(8+8)O 208(82+126)Pb


Isotopes are atoms with the same atomic number but different numbers of neutrons. Many

elements consist naturally of mixtures of isotopes, with very similar chemical properties.

Only one stable isotopes: 9Be, 19F, 23Na, 27Al, 31P, Sc, Mn, etc.

total numbers: 20.

Several stable isotopes: 1H, Deuterium: 2H, 3H; 12C, 13C, 14C; etc.

Tin (Sn): no fewer than 10.

Relative atomic mass (RAM) is equal to molar mass of element

RAM of an element:

IsotopesIsotopes

)( ii MrfRAM

fi: the abundance of the isotopes in the nature;

Mri: the relative atomic number of the isotopes


Nuclides ( 核素 ):

Isotopes ( 同位素 ):

Element ( 元素 ):

Atom ( 原子 ):

isotone ( 同中素 ):

Isobar ( 同量异位素 ):

Application: (1) NMR: 2H, 13C, nuclear magnetic resonance (2) Unstable radioactive: medical diagnostics (3) 14C: judge years; (4) investigate the detailed mechanism of chemical reactions


Radioactivity:

Radioactive decay: is a process whereby unstable nuclei change into more stable ones by emitting high-energy particles of different kinds. All elements with Z > 83 are radioactive.

The process involved are as follows:

(a) decay: 4He

(b) decay: electron

(c) decay: high-energy electromagnetic radiation. It often accompanies and decay.

Half-life:


A2 ATOMIC ORBITALS

WavefunctionsWavefunctions

The quantum theory is necessary to describe electrons. It predicts discrete allowe

d energy levels and wavefunction, which give probability distributions for electro

ns. Wave functions for elections in atoms are called atomic orbitals.

Quantum mechanics

Quantized energy

Probability distributions

Schrödinger’s wave equation

Atomic orbitals


A3 THE NUCLEAR ATOM

Electron and NucleiElectron and Nuclei


An atom consists of a very small positively charged nucleus,

surrounded by negative electrons held by electrostatic

attraction. The motion of electrons changes when chemical

bonds are formed, nuclei being unaltered.

Nuclei contain positive protons and uncharged neutrons.

The number of protons is the atomic number (Z) of an

element. The attractive strong interaction between protons

and neutrons is opposed by electrostatic repulsion between

protons. Repulsion dominates as Z increases and there is

only a limited number of stable elements.


Quantum numbers and nomenclature

Quantum numbers and nomenclature

Atomic orbitals are labeled by three quantum numbers n,l an

d m. Orbitals are called s,p,d or f according to the value of 1;

there are respectively one, three,five and seven different pos

sible m values for these orbitals.Principal Quantum numbers

n=1, 2, 3…..

Angular momentum (or azimuthal) Quantum numbers

l=0, 1, 2, …, (n-1)

Magnetic Quantum numbers

m=0, ±1, ±2, …, ±(n-1)

Spin Quantum numbers

ms=+1/2, -1/2


A3 THE NUCLEAR ATOM














Angular functions: ‘shapes’

Angular functions: ‘shapes’

s orbitals are spherical. p orbitals have two directional lobes,

which can point in three possible directions. d and f oritals h

ave coorespondingly greater numbers of directional lobes.

Radial wavefunction

Angular wavefunction

Polar diagrams

Boundary surface

Nodal plane



A3 THE NUCLEAR ATOM














Radial distributionsRadial distributions The radial distribution function shows how far from the

nucleus an election is likely to be found. The major features

depend on n but there is some dependence on l.

Radial probability distributions



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Energies in hydrogenEnergies in hydrogen The allowed energies in hydrogen depend on n only. They ca

n be compared with experimental line spectra and the ionizat

ion energy.

En= -R/n2

Rydberg constant: R=13.595 eV per atom

When n=∞, E∞=0 which is the ionization limit

Ionization energy: E∞-E1=R

Degenerate: Orbitals with the same n but different values of l an

d m have the same energy.

INORGANIC CHEMISTRY

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Hydrogenic ionsHydrogenic ions Increasing nuclear charge in a one-electron ion leads to

contraction of the orbital and an increase in binding energy

of the electron.

For example: He+, Li2+

<r>= n2a0/Z

Where a0 is Bohr radius, the average radius of a 1S orbital in

hydrogen.

Z: atomic number

En= -Z2R/n2


A3 MANY-ELECTRON ATOMS

The orbital approximation

The orbital approximation

Putting electrons into orbitals similar to those in the hydroge

n atom gives a useful way of approximating the wavefunctio

n of a many-electron atom. The electron configuration specif

ies the occupancy of orbitals, each of which has an associate

d energy.

Orbital approximation

Electron configuration

Orbital energy

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Electron spinElectron spin Electrons have an intrinsic rotation called spin, which may p

oint in only two possible directions, specified by a quantum

number ms. Two electrons in the same orbital with opposite

spin are paired. Unpaired electrons give rise to paramagnetis

m.

Spin Quantum numbers

ms=+1/2, -1/2

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Pauli exclusion principle

Pauli exclusion principle

When the spin quantum number m, is included no two

electrons in an atom may have the same set of quantum

numbers. Thus a maximum of two electrons can occupy any

orbital.

Li : (1s)2 (2s)1

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Effective nuclear charge

Effective nuclear charge

The electrostatic repulsion between electrons weakens their

binding in an atom, this is known as screening or shielding.

The combined effect of attraction to the nucleus and repulsio

n from other electrons is incorporated into an effective nucle

ar charge.

En= -Zeff2R/n2

σ = Z-Zeff

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Screening and penetration

Screening and penetration

An orbital is screened more effectively if its radial distributi

on does not penetrate those of other electrons. For a given n,

s orbitals are least screened and have the lowest energy; p,

d,... orbitals have successively higher energy.

Degenerate

Penetrating

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Hund’s first ruleHund’s first rule When filling orbitals with l>0, the lowest energy state is for

med by putting electrons so far as possible in orbitals with di

fferent m values, and with parallel spin.


A4 THE PERIODIC TABLE

HistoryHistory The periodic table - with elements arranged horizontally in

periods and vertically in groups according to their chemical

similarity - was developed in an empirical way in the 19th

century. A more rigorous foundation came, first with the use

of spectroscopy to determine atomic number and, second

with the development of the quantum theory of atomic

structure.

INORGANIC CHEMISTRY

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Building upBuilding up

The 'aufbau' or 'building up' principle gives a systematic method for determining the

electron configurations of atoms and hence the structure of the periodic table. Elem

ents in the same group have the same configurati0n of outer electrons. The way diff

erent orbitals are filled is controlled by their energies (and hence their different scre

ening by other electrons) and by the Pauli exclusion principle.

According to the aufbau principle, the ground-state electron configuration of a

n atom can be found by putting electrons in orbitals, starting with that of lowes

t energy and moving progressively to higher energy.

（ 1 ） the lowest energy

（ 2 ） Pauling exclusion principle

（ 3 ） Hund’s rule

INORGANIC CHEMISTRY

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Penetration effect

Screening effect.

Intervening of energy level


Electron configuration:

The outmost electron configuration:

Valence electron configuration:

INORGANIC CHEMISTRY

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Block structureBlock structureThe table divides naturally into s, p, d and f blocks according

to the outer electron configurations. s and p blocks form the

main groups, the d block the transition elements, and the f

block the lanthanides and actinides.

INORGANIC CHEMISTRY

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Group numbers and names

Group numbers and names

Modern group numbering runs from 1 to 18, with the j bl0cks being subsumed into g

roup 3. Older (and contradictory) numbering systems are still found. Some groups of

elements are conventionally given names, the most commonly used being alkali met

als (group l), alkaline earths (2), halogens (17) and noble gases (18).

Alkali metal : rare earth:

Alkaline metal: lanthanide:

Transition metal: actinide:

Coinage metal:

Chalcogen:

Halogen:

Noble gas:


A5 TRENDS IN ATOMIC PROPERTIES

Energy and sizesEnergy and sizes Trends in orbital energy and size reflect changes in the princ

ipal quantum number and effective nuclear charge. They are

seen experimentally in trends in ionization energy (IE) and a

pparent radius of atoms.

From:

En= -Zeff2R/n2

Ionization energy: E∞-E1=R

To

Average radius <r>=n2a0/Zeff

Metallic radii:

Covalent radii:

Ionic radii:

Van der Waals’ radii:

INORGANIC CHEMISTRY

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Horizontal trendsHorizontal trends Increasing nuclear charge causes a general increase of IE and a d

ecrease of radius across any period. Breaks in the IE bend are fou

nd following the complete or half filling of any set of orbitals.

INORGANIC CHEMISTRY

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Vertical trendsVertical trends

A general increase of radius and decrease in IE down most g

roups is dominated by the increasing principal quantum num

ber of outer orbitals. Effective nuclear charge also increases,

and can give rise to irregularities in the IE trends.

INORGANIC CHEMISTRY

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States of ionizationStates of ionization

IEs for positive ions always increase with the charge.

M2+, M3+, …., second, third,….

Elements IE1 IE2 IE3

Be (1s22s2) 9.3ev 18.2ev 154ev

Ca (Ar)4s2 6.1ev 11.9ev 51ev


Electron affinity of an atom: defined as the ionization energy of the negative ion, thus the energy input in the process:

M- M + e-

Section B1Section B1INORGANIC CHEMISTRY

Section B Introduction to inorganic substance

B1 ELECTRONEGATIVITY AND BOND TYPE

DefinitionDefinition

Electronegativity is the power of an atom to attract electrons to itself in a che

mical bond. Elements of low electronegativity are called electropositive.

Pauling Electronegativity: based on bond energies

Mulliken Electronegativity: the average of the first ionization energy and th

e electron affinity of an atom.

Allred-Rochow Electronegativity: proportional to Zeff/r2.


Different numerical estimates agree on qualitative trends:

Electronegativity increases from left to right along a period, and generally decreases groups in the periodic table.


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The bonding triangleThe bonding triangle

Electronegativity elements form meta1lic solids. Electronegative elements form

molecules or polymeric solids with covalent bonds. Elements of very different ele

ctronegativity combine to form solids that can be described by the ionic model.

Delocalization of electrons:

Covalent compounds:

Giant covalent lattices (Polymeric solids):

Localized bonds:

Cation ：Anion ：



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Bond polarityBond polarity The polarity of a bond arises from the unequal sharing of electrons bet

ween atoms with different electronegativities. There is no sharp dividi

ng line between polar covalent and ionic substances.

Homopolar (A covalent between two atoms of the same ELs)

Bond polarity

Heteropolar (A covalent between two atoms of the different ELs)

Electric dipole moment


B2 CHEMICAL PERIODICITY

IntroductionIntroduction Major chemical trends, horizontally and vertically in the

periodic table, can be understood in terms of changing

atomic properties. This procedure has its limitations and

many details of the chemistry of individual elements cannot

be predicted by simple interpolation from their neighbors.


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Metallic and non-metallic elements

Metallic and non-metallic elements

Metallic elements are electropositive, form electrically

conducting solids and have cationic chemistry. Non-metallic

elements, found in the upper right-hand portion of the periodic

table, have predominantly cova1ent and anionic chemistry. The

chemical trend is continuous and elements on the borderline

show intermediate characteristics.


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Horizontal trendsHorizontal trends Moving to the right in the periodic table, bonding character ch

anges as electronegativity increases. The increasing number of

electrons in the valence shell also gives rise to changes in the s

toichiometry and structure of compounds. Similar trends opera

te in the d block.

Closed shell

Lattice energy

Octet rule

Non-bonding electrons


B3 STABILITY AND REACTIVITY

IntroductionIntroduction

Stability and reactivity can be controlled by thermodynami

c factors (depending only on the initial and final states and n

ot on the reaction pathway) or kinetic ones (very dependent

on the reaction pathway). Both factors depend on the conditi

ons, and on the possibility of different routes to decompositi

on or

reaction.

Thermodynamic stable ： stable or unstable

Kinetic stable ： reactive or unreactive


1. Very important concepts of thermodynamics:

(1) System: the objects in research:

(a) closed, (b) open, (c) isolated system

(2) Surroundings: the others except for the system

(3) State: the total parameters used to describe the physical and chemical properties of the system.

The parameters P, V, T, n would be needed for the state of gas 。

(4) process: the route from the initial state to final state.

isothermal process (T=0); isobaric process (P=0);

isovolumic process (V=0); adiabatic process ( Q=0).

(5) State function:

From state A to state B, some processes were adopted, but the total value for any process would be keep constant.

state A —— state B—— State A, = 0


(6) Heat (Q): the energies related with the change of temperature.

positive Q: the system would absorb the energy from the circumstance;

negative Q: release the energy from the system.

(7) Work (W) : the others energies except for the heat. positive W: negative W: the volume work W = (PV)

(8) Intrinsic energy (U) (state function): the total energy of the object in research.

if from State A (U1)—— state B(U2),

U1 = U2 － U1

if from State B (U2)—— state A(U1), U2 = U1 － U2

U1 ＋ U2 ＝ 0.

nEU


2. The fist law of thermodynamics (the law of conservation of energy )

for a closed system, U = Q － W

3. Enthalpy ( 焓 )

if P ＝ 0 ，

U ＝ U2 － U1 ＝ Q － W = (Q2 － Q1) －（ P2V2 － P1V1 ）

Qp ＝ (U2 ＋ P2V2) －（ U1+P1V1 ） =H2 － H1 ＝ H

defined as U ＋ PV H (enthalpy) －－ state function

So, (i) if P ＝ 0 ,

H ＝ Qp ＋ (positive), endothermic process

－ (negative), exothermic process (ii) if V ＝ 0 , U ＝ QV

Qp ＝ QV ＋ n RT

INORGANIC CHEMISTRY

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Hess’ LawHess’ LawEnthalpy change ( H) is the heat input to a reaction, a useful △

measure of the energy change involved. As H does not depend △

on the reaction pathway (Hess' Law) it is often possib1e to

construct thermodynamic cycles that allow va1ues to be

estimated for processes that are not experimentally accessible.

Overall H values for reactions can be calculated from tabulated △

enthalpies of formation.

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Standard state: 100kPa, 298.15K

The standard enthalpy of formation (Hf0):

H0f of any compounds is defined as the energy change from its most stable elementary

substance at the standard state.

C (graphite) + O2(g) CO2 (g) Hf0 (CO2)

C (diamond) + O2(g) CO2 (g) Hf0 (CO2)

By definition, Hf0 is zero for any elements in its stable (standard) states.

Application: (Hess’ Law)

The standard enthalpy change (H0) in any reaction would be calculated by using the standard enthalpy of formation.

For example, for this reaction

a A (state) + b B (state) c C (state) + d D (state) (thermochemical equation) ，

)]()([)]()([

)tan()(0

BHbAHaDHdCHc

tsreacHkformationHiH

ffff

ff


Example.: Calculate the standard enthalpy change (rHm) in the following reaction, respectively.

3 NO2 (g)+ H2O(l) 2 HNO3 (l) + NO (g)

Answer.:The standard enthalpy of formation (Hf

, kJmol-1) of these compound are listed in Table.

NO2 (g) 33.18 kJmol-1 H2O(l) -285.83 kJmol-1

HNO3(l) -173.23 kJmol-1

NO(g) 90.25 kJmol-1

Applying Hess’ law:rHm = ifHm (product)- ifHm (reactant) rHm = [ 2 (-173.23) + 90.25] – [3 33.18 + (-285.83)] = -69.92 kJmol-1.

INORGANIC CHEMISTRY

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EntropyEntropy

1. Entropy( 熵 ) is a measure of molecular disorder or more precisely ‘the number o

f microscopic arrangements of energy possible in a macroscopic sample’.

(1) Entropy increases with rise in temperature and depends strongly on the state.

(2) S(gas) >>S (liquid) > S (solid)

Entropy change ( S)△ are invariably positive for reactions that generate gas molecu

les.

(3) molecule mass or its structure:

2. Entropy:——state function Applying the Hess’ Law △Sreaction= ∑S(products) - ∑S(reactants)

For an isolated system, the entropy change would be used to judge the direction for spontaneous reaction.If S >0, the forward reaction would be spontaneous.△ = 0, the reaction would keep the balance state. < 0, the backward reaction would be spontaneous. ———— the second law of thermodynamics


S :

by definition: the standard entropy of the ideal crystal for any pure compounds would be considered as zero at 0 K.

——the third law of thermodynamics

S0m (the standard entropy) in standard state:

the related data(△H0f, ， S0

m) would be found in the appendix.

For any reaction, a A (state) + b B (state) c C (state) + d D (state)

△rSm0

(reaction)= ∑Sm0

(products) - ∑Sm0

(reactants)

)]()([)]()([

)tan()(

BHbAHaDHdCHc

tsreacHkproductsHiH

ffff

ffr m


Gibbs free energy （ G ）Gibbs free energy （ G ）

△ S —— the application were be confined for an isolated system.

If P△ ＝ 0 and T=0△ ,

From the first law and the second law of thermodynamics,

G = H － TS

G ： the Gibbs free energy, (state function)

△G = G (final state) △ －△ G (initial state) If G <0, the forward reaction would be spontaneous.△ = 0, the reaction would keep the balance state. > 0, the backward reaction would be spontaneous.


△G0f of any compounds is defined as the Gibbs free energy change from its most

stable elementary substance at the standard state.

By definition, G0f is zero for any elements in its stable (standard) states.

for this reaction

a A (state) + b B (state) c C (state) + d D (state)

Applying the Hess’ Law,

)]()([)]()([

)tan()(0

BGbAGaDGdCGc

tsreacGkproductsGiG

ffff

ff


For example:

CaCO3 (s) CaO(s) + CO2 (g)

(1) At the Standard state, G0 = +130kJmol-1.

However, (2) at 1273K, 100kPa, G = -25 kJmol-1 P. T G

(3) at 1273K, 0.1kPa, G = -174 kJmol-1

G0f （ standard state ）

If under non-standard state, how to obtain G(T) ?

(1) Van’t Holf equation( 范特霍夫等温方程 ):

)()(

)()(

ln)()(

溶液反应体系

气体反应体系

i

i

C

CJ

orP

PJ

JRTTGTG

i

i

mrmr


G = H － TS G = H － T S G (T) = H0 － T S0

G (T) = H0(298K) － T S0(298K)

If G = 0, the reaction would keep the balanced state. Tm （ eq ） = H0(298K) / S0(298K)

(2) 吉布斯－亥姆霍兹方程

H0(298K) S0(298K) G0(298K) Example

－ (exothermic) ＋ (entropy increase)

－ (spontaneous)

H2(g) + F2(g) 2 HF (g)

＋ (endothermic)

－ (entropy decrease)

＋ (not spontaneous)

CO (g) C (s) + O2 (g)

－ (exothermic) －＋ or － NH3 + HCl NH4Cl

＋ (endothermic)

＋＋ or － CaCO3 CaO ＋ CO2


Example. In general, the following course was applied in purification of coarse nickel: first, Ni(CO)4 was synthesized by the reaction of nickel and CO at 323K; then, decomposition of Ni(CO)4(l) would produce the high-purity nickel. Ni(s) + 4 CO (g) Ni(CO)4 (l) Try to analysis the rationality of this process from these data: rHm

0 = -161 kJmol-1, rSm

0 = -420 Jmol-1K-1.Answer. : Applying rGm= rHm – T rSm, the temperature at equilibrium (rGm) would be obtained. rGm (T) = 0 T = rHm / rSm = － 161 103 / ( － 420) = 383 K.if T < 383 K, rGm < 0, the normal reaction will occur successfully.If T > 383K, rGm > 0, the reverse reaction will take place.So, at lower temperature, Ni(CO)4 was synthesized by the reaction of coarse nickel and CO. And, at higher temperature, the preparation of the high-purity nickel would be performed by decomposition of Ni(CO)4.


Chemical EquilibriumChemical Equilibrium

(1) Dynamic equilibrium, or running balance

A (state) B (state)

forward reaction:

Backward reaction:

For example:

Some NaCl was dissolved into the water at room temperature.

Chemical equilibrium:

the object: the reaction in the closed system


(2) Equilibrium constants

For a general reaction such as

a A (state) + b B (state) c C (state) + d D (state) ba

dc

BA

DCK

][][

][][

(1) Where the terms [A], [B], [C],[D] represent activities, but are frequently approximated as concentrations or partial pressures.

(2) Pure liquids (solvent) and solids are not included in an equilibrium constant as they are present in their standard state.

(3) The change of temperature has very important effect on the equilibrium constant.

(4) The form of reaction equation will have effect on the value of the equilibrium constant.

K >>1, indicates a strong thermodynamic tendency to react, so very little of the reactants A and B will remain at equilibrium.

K<<1 indicates very little tendency to react and the reverse reaction will be very favorable.

INORGANIC CHEMISTRY

A3 THE NUCLEAR ATOM













Section 3Section 3INORGANIC CHEMISTRY

The relationship between Equilibrium constants and rGmThe relationship between Equilibrium constants and rGm

If at balanced state, △rGm= 0.

△rG0m= － RT ln J = —RT ln K = rH△ 0

m － T △rS0m

J = K ( equilibrium constant )

The equilibrium constant of reaction is related to the standard Gibbs free

energy change. Equilibrium constants change with temperature in a way th

at depends on H for the reaction. △

)()(

)()(

ln)()(

溶液反应体系

气体反应体系

i

i

C

CJ

orP

PJ

JRTTGTG

i

i

mrmr


Le Chatelier’ s principle （勒沙特列原理 , 吕查德克原理）：

It would be applied to qualitatively judge the direction of chemical reaction.

a A (state) + b B (state) c C (state) + d D (state) + HEffect factor to the equilibrium ：(1) Change of Concentration use the equilibrium constant(2) Change of Pressure use the equilibrium constant(3) The temperature （ H ）

K ： increase with rise in the temperature for an endothermic reaction, and decrease for an exothermic one.

INORGANIC CHEMISTRY

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Reaction ratesReaction rates

(1) Average velocity

It is defined as the increment of the products or decrement of reactant in unit time.

t

nr B

t

nr

B

B

B －－ the coefficient of B in reaction equation.

(a) concentration:

(b) temperature: Reaction rates generally increase with rise in temperature

(c) Catalysis provide alternative reaction pathways of lower energy. A true catalyst by definition can be removed unchanged after the reaction. And so does not alter the thermodynamics or the position of equilibrium.

INORGANIC CHEMISTRY

A3 THE NUCLEAR ATOM














(2) Rare equation: the relationship between the rate and the concent

ration of the reactant.

a A + b B products

Rate= k[A]n [B]m

rate constant: k

Orders of reaction: m, n. depend on the mechanism and not necessarily

equal to the stoichiometric coefficients of the reactants.

For example:

(1) H2 ＋ I2 ＝＝ 2HI r＝ k c(H2) c(I2)

(2) H2 ＋ Br2 ＝＝ 2HBr

(3) H2 ＋ Cl2 ＝＝ 2HCl )()(

1

)()(

2

2

22

1

21

BrcHBrck

BrcHckr

)()( 22

1

2 ClcHkcr


a A + b B products

Rate= k[A]n[B]m orders of reaction: n = a, m=b. （很少的例子） —————— （基元反应， the process need only one-step reaction

from the reactants to products ） for most chemical reactions, n a, m b, because the process would be div

ided into many steps.—————— depend on the mechanism.

n + m == the total orders of a reaction

The order of reactions ( n + m)

The rate equation

The unit of rate constant

0 (zero order reaction) Rate = kmol dm-3 s-1

1 (first order reaction) Rate = k c s-1

2 (second order reaction Rate = k c2 mol-1 dm3 s-1

3 (third order reaction) Rate = k c3 mol-2 dm3 s-1

Another means: by designing a lots of favorable experiments.


gOH2gNgH2gNO2 22k1073

2

试验编号

1

2

3

4

5

)Lmol/(H 12

c )Lmol/(NO 1-c )sLmol/( 11 7109.7 6102.3

5103.1 6104.6 6102.3

Rate = k [c(NO)]n [c(H2)]m

n=2 , m = 1. k= ?


(3) Arrhenius equation——the relationship between the rate constant and the temperature.

TR

EAk

RT

EAk

Aek

a

a

RT

Ea

303.2lglg

lnln

K------the rate constant.

Ea-------the activation energy (J.mol-1)

T-temperature （ K ）A-------the constant

lnk (or lgk)---1/T ： linear relationship

the intercept （截距） : lnA

the slope （斜率） : － Ea / R


RT

EAk alnln

(1) If these data T1,T2, and k1, k2 were given,

12

21

122

1 )()

11(ln

TT

TT

R

E

TTR

E

k

k aa

(2) If many T and k were obtained in the experiment, the method of diagram would be prior to use to calculate the Ea and A.


Problem 3.: Calculate the activity energy of the reaction N2O5 (g)2 NO2 (g)+ 1/2 O2(g) from the relationship (listed in Table) between the temperature and the rate constant.

T/K 338 382 318 308 298 273

k/ s-1 4.8710-3 1.5010-3 4.9810-4 1.3510-4 3.4610-5 7.8710-7

Solution 3.:Applying the Arrhenius equation ( k = A e–Ea/RT), lg k = -Ea / (2.303 RT) + lg A.A linear relationship was shown between lg k and 1/T. So, the slope of this line is –Ea / (2.303 R). The data of 1/T and (lg k/s-1) are given in table III

T-1/10-3K-1 2.96 3.05 3.14 3.25 3.36 3.66

Lg(k/ s-1) -2.31 -2.82 -3.30 -3.87 -4.46 -6.10

The slope =

Ea = 5.38 103 2.303 8.314 = 103.0 103 Jmol-1 = 103 kJmol-1

K333

1038.5106.3100.3

)73.5(50.2


B4 OXIDATION AND REDUCTION

DefinitionsDefinitions

(1) Oxidation means combination. with a more electronegative element or the remo

val of electrons.

Reduction means combination with a less electronegative element or the addition of

electrons.

A complete redox reaction involves both processes.

2H ＋ (aq) ＋ Zn (s) Zn2 ＋ (aq) ＋ H2(g)

(2) A strong oxidizing agent is substance capable of oxidizing many others, and is thu

s itself easily reduced;

A A strong reducing agent is substance capable of reducing many others, and is thus

itself easily oxidized.

(3) A redox reaction is any reaction involving changes of oxidation state.

INORGANIC CHEMISTRY












Oxidation statesOxidation states

Oxidation states of atoms in a compound are calculated by assigning electrons in a b

ond to the more electronegative element. In simple ionic compounds they are the s

ame as the ionic charges. In any redox reaction the oxidation states of some eleme

nts change.

Calculation rules of Oxidation states:

(a) Bonds between the same element are not included. Elements have oxidation state

zero. In an ion such as peroxide O22 － the electrons in the O － O bond are distrib

uted equally, making O-1.

(b) Except in cases such as O22 － the most electronegative and electropositive elemen

ts in a compd have an oxidation state equal to the normal ionic charge.

(c) The sum of the oxidation states must equal the charge on the species, and is theref

ore zero in a neutral compd. Using this rule and (b) above, we have H1 in H2O, H-1

in CaH2.

(d) Complex formation, and donor-acceptor interaction in general don’t alter the oxida

tion state.

A redox reaction is any reaction involving changes of oxidation state.

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Balancing redox reactions

Balancing redox reactions

In complete redox reactions the overall changes in oxidation state must balan

ce. When reactions involve ions in water it is convenient to split the over

all reaction into two half reactions. To balance these it may also be neces

sary to provide water and H+ or OH-.

Half reactions:

Fe2+ Fe3+ + e

(1) Electrons were included in half reaction.

(2) The half reaction must balance in charges and elements.

MnO4－＋ 5 e Mn2 ＋（ × ）

MnO4－＋ 5 e ＋ 8H ＋ Mn2 ＋＋ 4H2O ( )

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Example 1Example 1

Problem 1. Balance the equation H2SO3 + MnO4- SO4

2- + Mn2+ (in acidic

solution).

Solution.: (the method of half-reactions)

Step 1. Write the two skeletal half-reactions.

H2SO3 SO42- (the reducing agent, H2SO3, is oxidized to SO 4

2-)

MnO4- Mn2+ (the oxidizing agent, MnO4

-, is reduced to Mn2+) ；Step 2. Balance each half-reaction.

(i) To balance the O, add water. H2SO3 + H2O SO42-

MnO4- Mn2+ + 4H2O

(ii) To balance the H, add H+. H2SO3 + H2O SO42- + 4 H+

MnO4- + 8 H+ Mn2+ + 4H2O

(iii) To balance the charge, add electrons.

H2SO3 + H2O SO42- + 4 H+ + 2 e-

MnO4- + 8 H+ + 5 e- Mn2+ + 4H2O

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Example 1Example 1

Step 3. Add the two half-reactions. Since 2 electrons are lost in the first

while 5e are gained in the second, and since the lowest common multiple of

2 and 5 is 10, we multiply the first by 5 and the second by 2. When the

resulting equations are added, the 10e on the two sides of the equations will

cancel. Also 16H+ and 5 H2O will cancel from each side.

The total reaction:

5 H2SO3 + 2 MnO4- 5 SO4

2- + 2Mn2+ + 4 H+ + 3 H2O

H2SO3 + H2O SO42- + 4 H+ + 2 e- （ × 5 ）

MnO4- + 8 H+ + 5 e- Mn2+ + 4H2O （ × 2 ）


Example 2Example 2

In acidic solution, it is more appropriate to use H ＋ or H2O.

Yet ， in alkaline solution, it is more appropriate to use OH － or H2O.

(1) Al (s) ＋ 4OH － [Al(OH)4] －＋ 3e －

(2) 2H2O + 2e － 2OH －＋ H2

(3) 2Al （ s ）＋ 2OH －＋ 6H2O 2 [Al(OH)4] －＋ 3H2

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Extraction of the element

Extraction of the element

Redox reactions are used in the extraction of nearly all elements from natura1l

y occurring compounds. Carbon is used to reduce some metal oxides, but many

elements require stronger reducing agents, or the use of electrolysis.

The methods shown as following are in common use.

（ 1 ） carbon

（ 2 ） reduction by using H2 or CO.

（ 3 ） the replacement reaction by using the reactive metals M such as alkali

metals or alkaline metals.

（ 4 ） electrolysis


B5 Describing inorganic

compoundsformulaeformulae Stoichiometric (empirical) formulae describe only the relat

ive numbers of atoms present.

Molecular formu1ae and/or representations giving structura

l information should be used when they are appropriate. The

physical state of a substance is often specified.

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NameName

Systematic nomenclature can be based on three systems, binary, substitutive (simil

ar to that in organic chemistry) or coordination. Many nonsystematic or trivial( 习惯） names are used.

Systematic nomenclature is based on three systems.

Binary names:

NaCl: sodium Chloride PCl3: phosphorous trichloride

N2O4: dinitrogen tetroxide MnO2: Manganese (IV) dioxide

F －： Fluoride; F2: fluorine.

Substitutive names:

SiH4: silane; SiCl4: tetrachloride silane

NH3: amine; NH2OH: hydroxyamine

Coordination names:

[Cu(NH3)4]2 ＋： tetraaminecopper(2 ＋ ) ； tetraaminecopper (II) ion

[CuCl4]2 －： tetrachlorocuprate(2 － ) or tetrachlorocuprate (II)

[Ni(NH3)6]Br2 ： hexaaminenickel dibromide

K[PF6]: potassium hexafluorophosphate (V)

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Structure and bondingStructure and bonding

The coordination number and geometry of an atom describe the nu

mber of bonded atoms and their arrangement in space. Oxidation sta

tes rather than valencies are generally used for describing different p

ossible stoichiometries.

Bond lengths:

Bond angles:

Coordinate number （ CN ）

INORGANIC CHEMISTRY Section CSection C

Section C Structure and bonding in molecules

C1 Electron Pair Bonds

Lewis and valence StructureLewis and valence Structure

A single covalent bond is formed when two atoms share a pair of electrons.

Double and triple bonds can be formed when two or three such pairs are shared.

—— octet rule

A Lewis structure shows the valence electrons in a molecule. Two shared

electrons form a single bond, with correspondingly more for multiple bonds.

Some atoms may a1so have nonbonding electrons (lone-pairs). Valence

structures show the bonds simply as lines.

Section C-1Section C-1INORGANIC CHEMISTRY

C H

H

H

H

O£ºH

H

£º

:O O:£º£º :N N:

Important Notes: Non-bonding electrons or lone-pair electrons localized on one atom would be labeled in Lewis structure.

Isoelectronic principle( 等电子规律 ): the molecules or ions having the same number of valence electrons should have similar valence structure.

[NH4]+ [BH4] －


In most stable molecules and ions of the elements C-F, each of these atoms has eight ele

ctrons (an octet) in its valence shell. Expansion of the octet and increased valency is pos

sible with elements in periods 3 and below.

Octet rule:

Incomplete octet ———— 缺电子分子Octet expansion or hypervalence: —— valence expansion or hybrid orbital theory

Octets and ‘hypervalence’

Octets and ‘hypervalence’

B F:

F:

F:£º

£º

£º£º

£º

£º

:SF

F

F

F


When several alternative valence structures are possible, the

bonding may be described in terms of resonance between them.

ResonanceResonance

C:O:O

:O

£º

£º£º

£º

£º

2£

C:O:O

:O

£º

£º£º

£º £º

2£

C:O:O

:O

£º

£º£º

£º

£º

2£ £º

(1) 分散电荷，达稳定结构。

(2) 使所有原子都能满足 octet rule.

Resonance may also be appropriate with different valence structures that are not equivalent but look equally plausible.

For examples: N2O (12)


Formal charges are assigned by apportioning bonding electrons equally between th

e two atoms involved. They can be useful to rationalize apparent anomalies in bon

ding, and to assess the likely stability of a proposed valence structure.

A formal charge on an atom is essentially the charge that would remain if all cova

lent bonds were broken, with the electrons being assigned equally to the atoms inv

olved.

Formal charge = (No. of valence electrons in neutral atom) － (No. of nonbon

ding electrons) － 1/2 (No. of electrons in bonds formed.)

Formal Charges:Formal Charges:

C:O:O

:O

£º

£º£º

£º

£º

2£

C:O:O

:O

£º

£º£º

£º £º

2£

C:O:O

:O

£º

£º£º

£º

£º

2£

£º


Some general principles for formal charges are:

(1) Structure without formal charges are preferred if possible;

(2) Structures with formal charges outside the range -1 to ＋ 1 are generally unfavorable;

(3) Negative formal charges should preferably be assigned to more electronegative atoms, positive charges to more electropositive atoms.

For example (P44):

CO N2O BF

Formal charge is very different from Oxidation state, which is assigned by apportioning electrons in a bond to the more electronegative atom rather than equally.

Both are artificial assignments, useful in their respective ways, but neither is intended as a realistic judgment of the charges on atoms.


Many covalent molecules and ions cannot be understood in terms of e

lectron pair bonds between two atoms. They include electron-deficien

t boron hydrides and transition metal compounds.

Two-center bonds: localized bonds

Three-center bonds: delocalized bonds

Molecular Orbital theory: （ MOT ）

LimitationsLimitations


hybrid orbital theory—— 解释多原子分子的方向性问题hybrid orbital theory—— 解释多原子分子的方向性问题

For example: CH4 ： Lewis structure: planar

real structure: tetrahedron

bond angles: 90, real: 10928

Pauling: hybrid orbital theory:

2s 2p

ground state of C atom excited state 4 sp3 hybrid orbitals

the foramtion of sp3 hybird orbitals

sp3 hybrid orbitals; sp2 ; sp; sp3d (dsp3), sp3d2(d2sp3), dsp2

意义： (1) 解释了多原子分子结构的方向性问题；

(2) 中心原子的成键能力增加；

Atom orbitals


Types sp sp2 sp3 dsp2 sp3d

dsp3

sp3d2

d2sp3

No. of atom orbitals hybridized

2 3 4 4 5 6

No. of hybrid orbitals 2 3 4 4 5 6

The angle between the hybrid orbitals ()

180 120 109.5 90

180

120

90

180

90

180

Configuration Linear Trigonal

planar

Tetrahedral

Planar square

Trigonal bipyrimdal

Octahedral

Examples BeCl2CO2

BF3

NO3－

CO32 －

CH4

CCl4CHCl3

PdCl42 －

Complex

PCl5 SF6

SiF62 －

等性杂化： CH4 （ sp3 ）

不等性杂化 : (H2O) （ sp3 ）


The valence shell electron pair repulsion (VSEPR) model is based on the obser

vation that the geometrical arrangement of bonds around an atom is influenced by

nonbonding e1ectrons present. It is assumed that electronic pairs – whether bon

ding or nonbonding - repel each other and adopt a geometrical arrangement t

hat maximizes the distances between them.

VSEPR principles have widely applied in predicting the possible steric configurati

on of covalent molecules for non-transition elements.

————shapes and bond angles of the molecules

VSEPR principlesVSEPR principles

C2 Molecular shapes: VSEPR


The basic principles of the model are as follows:

(i) Valence electron pairs round an atom (whether bonding or nonbonding) adopt a geometry that maximizes the distance between them. The basic geometries usually observed with 2-7 pairs are shown in fig 1.

AA A

A A A

2. linearsp

3. trigonal planar

sp24. tetrahedral

sp3

5. trigonal bipyramidal

sp3d

6. Octahedral

sp3d2

7. pentagonal bipyramidal

Basic VESPR geometries with 2-7 electron pairs


(2) Nonbonding electron pairs are closer to the central atom than bonding pairs and have larger repulsions: in fact the order of interaction is:

Nonbonding-nonbonding > nonbonding-bonding > bonding-bonding

(3) If double (or triple) bonds are present the four (or six) electrons involved behave as if they were a single pair, although they exert more repulsion than do the two electrons of a single bond.

(4) As the terminal atoms become more electronegative relative to the central one, bonding electron pairs are drawn away from the central atom and so repel less.


It is first necessary to decide which atoms are bonded together. Drawing a valence struct

ure gives the total number of electron pairs around an atom, sometimes known as its ster

ic number. The basic VSEPR geometry is then used to assign positions for bonding and

nonbonding e1ectrons.

Step1: Designate the central atom: usually the least electronegative one, which does n

ot apply to hydrides or organic groups.

Step2: Calculate the steric number (SN) of the central atom, equal to the number of b

onded atoms, plus the number of non-bonding electron pairs.

Step 3: Give the possible steric configuration of the molecules.

Another process to calculate the SN:

the No. of electron pairs in valence shell = the total No. of valence electrons in central

atom + the No. of the electrons bonded atoms +/- the charge in the ions

e.g. 配位原子： H, X, =1 ；中心原子： X ＝ 7

O ＝ 0 ；中心原子： O ＝ 6

Using VESEPRUsing VESEPR


Steric numbers 2 － 4 ：

Linear species （ SN ＝ 2 ）： BeH2 ， HgCl2 ， CO2 ， N3＋， NO2

+

Trigonal planar species: （ SN ＝ 3 ）： BF3, NO3-

Tetrahedral species （ SN ＝ 4 ） : CH4 , NH4+, SiCl4

AX2 with SN=3 or 4, bent ： NO2- H2O H2S

AX3 with SN =4, pyramidal ： NH3 ， PH3

Steric numbers 5 (AX5) ：—— trigonal bipyramidal

AX4: see-saw with two axial and two equatorial X postions, SF4, XeO2F2

AX3 ： T － shapes ， ClF3

AX2 ： linear ： XeF2


Steric numbers 6 (AX6) ：—— octahedral

AX5: square pyramidal structure, XeOF4

AX4 ： square planar; XeF4

Steric numbers 7 (AX7) ：—— pentagonal bipyramidal IF7

AX5: pentagonal planar ， XeF5－


In spite of a lack of firm theoretical foundation the VSEPR model

is widely applicable to molecular geometries and even to some

solids. Occasionally it fails to predict the correct structure.

(1) H2NOH: N as central atom: pyramidal

O: bent

(2) N(CH3)3 —— pyramidal

N(SiH3)3 —— planar

Extensions, difficulties and exceptions

Extensions, difficulties and exceptions


Molecular Orbitals (MOs) are wavefunctions for electrons in molecules.

Molecular Orbitals are formed by linear combination of AOs.---LCAO approximation.

LCAO: linear combination of atomic orbitals ( 原子轨道的线性组合形成分子轨道 )

Overlapping atomic orbitals can give bonding and antibonding MOs

1= c11 + c2 2—— bonding MO （成键分子轨道）

2= c11 - c2 2 —— antibonding MO （反键分子轨道）…… ..Fig 1.

Nonbonding MO( 非键分子轨道 ) ： An electron in the bonding MO has lower energy than in the AO of an isolated atom, as it h

as enhanced probability of being close to both nuclei simultaneously.

In an antibonding MO an electron has higher energy, with a reduced probability of being in t

he internuclear region, and internuclear repulsion being “deshielded”.

Bonding and antibonding orbitalsBonding and antibonding orbitals

C3 Molecular orbitals: Homonuclear diatomics


An MO diagram is a representation of the energies of bonding and antibonding MO

s formed from atomic orbitals. Electrons are assigned to MOs in accordance with t

he exclusion principle, leading to the same building-up procedure as for the period

ic table of elements.

MO diagramsMO diagrams

1sg

1su

H 1s AOH 1s AO

(1) 能量相近原则；

(2) 最大重叠原则；

(3) 对称性匹配原则；

(4) 轨道总数目不改变；

(5) 轨道通过线性组合后，轨道的能量发生分裂， 1sg 能量升高， 1su 能量降低，分子的总能量降低。

(6) 电子填充进入分子轨道的原子与电子在AO 中填充的规则相同。

(a) Exclusion principle;

(b) Lowest energy principle

(c) Hund’s first rule


The types of forming MOs

（重叠方式来分）

沿键轴方向 ———— 键（ head － to － head ，头碰头形式） :

the overlapping of s orbitals or p orbitals

or overlapping of s orbital and p orbital.

垂直于键轴方向 —— 键（ shoulder － to － shoulder ，肩并肩形式） :

the overlapping of p orbitals.

the overlapping of p orbital and d orbitals.

Very important Notes ：

（ 1 ）相同符号的原子轨道才能重叠；

(2) 两个原子间只有一个键，其他成键均为键。


The definition of Bond Order (BO) in MO theory recognizes that a normal single bond is formed by two electrons.

BO = (1/2) [(No. of electrons in bonding MOs) - (No. of electrons in bonding MOs)]

Fractional values are possible, as in the molecular ions H2+ or He2

+, which are both bonded but more weakly than H2.

H2+ (1sg)1 BO = (1-0)/2 = 0.5

H2 (1sg)2 BO = (2-0)/2 = 1

He2+ (1sg)2(1su)1 BO = (2-1)/2 = 0.5

He2 (1sg)2(1su)2 BO = (2-2)/2 = 01s AO 1s AO

1sg or 1s

1su or 1s *


Second period diatomicsSecond period diatomics

In extending the MO theory to second period elements two additional principles need to be considered.

(1) Only valence-shell are shown in MO diagrams as inner shells are too tightly bound to be involved in bonding, and don’t overlap significantly.

(2) Both 2s and 2p valence orbitals need to be included.

Two fundamental rules in the LCAO MO model are important:

(1) The number of MOs formed is equal to the total number of starting AOs.

(2) Only AOs of the same symmetry type combine to make MOs.

(3) Exclusion principle ：

Lowest energy principle ： Hund’s first rule ：


2p atomic orbitals can give rise to both and MOs, the former overlapping al

ong the molecular axis and the latter perpendicular to it. Multiple bonding in O2

and N2 arises from as well as bonding.

bonding < bonding < antibonding < antibonding

Trends in bond strengths and lengths follow predicted bond orders.

The O2 molecule has two unpaired electrons, a fact not predicted by simple electr

on-pair bonding models.O OO2

O2 (2sg)2(2su)2(2pg)2(2pu)4(2pg)2

BO = 2

N2 (2sg)2(2su)2(2pg)2(2pu)4

BO = 3

2p

2s

2p

2s


Electron configuration BO D （ kJmol-

1 ）R (pm)

N2

(2sg)2(2su)2(2pg)2(2pu)4

3 945 110

O2+

(2sg)2(2su)2(2pg)2(2pu)4(2pg)1

2.5 630 112

O2

(2sg)2(2su)2(2pg)2(2pu)4(2pg)2

2 498 121

O2-

(2sg)2(2su)2(2pg)2(2pu)4(2pg)3

1.5 - 128

F2

(2sg)2(2su)2(2pg)2(2pu)4(2pg)4

1 158 142

Higher predicted BO values correspond to stronger and shorter bonds

用 MO 描述第二周期的物种及电子的排布情况： P55


Orbitals from atoms of different elements overlap to give unsymmetrical MOs, t

he bonding MO being more concentrated on the more electronegative atom.

The greater the electronegativity difference, the greater the degree of localizatio

n.

The basic principle of forming the MO for the heteronulclear diatomics is simila

r to that of the homonuclear diatomics.

（ 1 ） LCAO among the AOs having the approximate energy.

（ 2 ） The number of MOs formed is equal to the total number of starting AOs.

（ 3 ） the principles of filling the electrons to the MOs

Basic principlesBasic principles

C4 Molecular orbitals: Heteronuclear diatomics


In HF the F 2s orbital is too low in energy to be involved significantly in bon

ding, but in BH both 2s and 2p orbitals on B can contribute to MOs with H 1

s. This situation is described as sp hybridization, and leads to bonding and no

nbonding MOs with a spatial localization similar to that assumed in VSEPR

theory.

HF and BHHF and BH

H FHF

2s

2p

1s

Orb

ital e

nerg

ies

1

2

1

3

H BHB

2s

2p

1s

1

2

1

3


When sp hybridization occurs on both atoms in a diatomic the order of MOs can

be hard to predict In CO the highest occupied MO (HOMO) is a nonbonding u or

bital resembling a lone-pair on C the lowest unoccupied MO (LUMO) is an antib

onding orbital.

Very complicated

HOMO:

( 最高被占用的分子轨道 )the highest occupied MO

LUMO:( 最低未被占用的分子轨道 )the Lowest unoccupied MO

COCO

C OCO

2s

2p

2s

Orb

ital e

nerg

ies

2p

1

2

1

3

4

2


C5 Molecular orbitals: polyatomics

Localized and delocalized orbitals:Localized and delocalized orbitals:

Polyatomics:

Localized orbitals: Two center

Delocalized orbitals: three or more center

Directed ValenceDirected Valence

Hybrid Orbital Theroy

VSEPR theory: determine the molecular structure


Multiple BondsMultiple Bonds

And/or MO orbitals would be formed by LCAO.

C C

HH

HH

C2H4

Normal bonding

C O

O

O

CO32-

大键 46


Three-center BondingThree-center Bonding

Three-center bonds are necessary for the description of some molecules.

Bridge bonds are of the three-center two-electron type;

For example: diborane B——H——B (3c-2e bonds)

Al2(CH3)6 and BeH2

Three-center four-electron bonding provides an explanation of some hypervalent molecules.

Hydrogen bonds:

F——HF (3c-4e bonds)

B B

H

H

H

H

H

H

B2H6


C6 Rings and Clusters( 簇合物 )

IntroductionIntroduction

Rings and clusters can be formed by using simple two-center bonding models;

in others it is necessary to used delocalized MO models.

P

P

PP

S

N

N

S

S

S

SS

S S

S S

Aromatic Rings

Wade’s Rules


C7 Bond Strengths

Bond enthalpiesBond enthalpies

Bond energies:

Mean bond enthalpies are defined as the enthalpy involved in breaking bonds in molecules.

H2O (g) ２Ｈ　＋　Ｏ

They may be determined from the thermo-chemical cycles using Hess’ Law.

　Ｈ２＋　 ½ Ｏ２　　 H2O fH0(H2O)

2 H + O

H1 H2 H3 = 2 B.E. (O- Ｈ )

H1+H2 +2 B.E (O-H) = fH0(H2O)


Major TrendsMajor Trends

Stronger bonds are generally formed with lighter elements in a group, when multiple bonding is present, and when there is a large electronegativity difference between the two elements.

Some important trends are summarized below:

(1) Bond energies often become smaller on descending a main group. This is expected as electrons in the overlap regions of a bond are less strongly attracted to larger atoms.

(2) Bond energies increase with bond order, although the extent to which B(A=B) is larger than B(A-B) depends greatly on A and B, the largest differences occurring with elements from the set C ， N ， O 。 Strong multiple bonding involving these elements may be attributed to the very efficient overlap of 2p orbitals compared with that of larger orbitals in low orbitals.

(3) In compounds ABn, with the same elements but different n values, B(A-B) decreases as n increases. The differences are generally less for larger A, more electronegative B.


(4) Bonds are stronger between elements with a large electronegativity difference.

(5) Single A-B bonds where A and B are both from the set N, O, F are weaker than expected from group comparisons. This is often attributed to a repulsion between nonbonding electrons.

(6) Other expectations to rule (1) above occur with A-O and A-X (halogen) bonds, which generally increase in strength between periods 2 and 3.

This may be partly due to the increased electronegativity difference when A is period 3, but repulsion between lone-pairs electrons on nonbonded atoms may also play a role.


A Lewis acid (or acceptor) is any species capable of accepting a pair of electrons.

for examples: H+, and metal cations, molecules such as BF3 with incomplete octets, a

nd ones such as SiF4 where octet expansion is possible.

A Lewis base (or donor) is any species with a pair of non-bonding electrons availabl

e for donation.

for examples: the molecules such as NH3, and anions such as F-.

The Lewis Acid-base definition The Brnsted Acid-bases defination

Brǿnsted bases are also Lewis bases, and H+ is a Lewis acid, but Bronsted acids such as

HCl are not Lewis acids.

Definition and scopeDefinition and scope

C8 Lewis acids and bases


Lewis acids and bases may interact to give a donor-acceptor complex.

BF3 + (CH3)3N: (CH3)3N:BF3

SiF4 + 2 F- [SiF6]2-

(the bond of N-B or Si-F is the dative bond )

covalent bond-------dative bond

The formation of solvated ions, complexes in solution and coordination compounds are the examples of this types of interactions.


Interaction between the orbitals in Fig.1 will be strongest when the energy difference bet

ween the acceptor LUMO and the donor HOMO is least. In this model the best acceptors

will have empty orbitals at low energies, the best donors filled orbitals at high energies. By

contrast, the strongest electrostatic interactions with take place between the smallest and m

ost highly charged (positive) acceptor, and (negative) donor atoms.

Modeles of interactionModeles of interaction


Hard donors interact more strongly with hard acceptors, soft donors with soft accept

ors ( 硬亲硬，软亲软，软和硬的反应不容易发生 ).

Hard acids tend to be more electropositive, and harder bases more electronegative.

Softer donor and acceptor atoms tend to be larger and more polarizable.

Hard-soft classification

(HSAB classification)

Hard-soft classification

(HSAB classification)


It is found that the relative strength of donors depends on the nature of the acceptor an

d vice versa. The hard and soft acid-base (HSAB)classification is often used to rationaliz

e some of the differences. When two acids (A1 and A2) and in competition for two bases

(B1 and B2) the equilibrium.

A1:B1 + A2:B2 Al:B2 + A2:B1

will lie in the direction where the harder of the two acids is in combination with the harde

r base, and the softer acid with the softer base. As a standard for comparison the prototyp

e hard acid H+ and soft acid [(CH,)Hg]+ are often used. Thus the equilibrium

[B:H]+ + [(CH,)Hg]+ H+ + [(CH,)Hg:B]+

will lie to the left or right according to the degree of hardness of the base B.

Examples of hard acids are H+, cations of very electropositve metals such as Mg2+, a

nd nonmetal fluorides such as BF3.

Soft acids include cations of late transition and post-transition metal such as Cu+, Pd2+ and Hg2+.

The hardness of bases increase with the number of the donor atom (e.g. NH3 < H2O

< F-) and decreases down any group (e.g. NH3 > PH3, and F- > Cl- > Br- > l- ).

Hard-soft classificationHard-soft classification


Hard Hard-soft Soft

Acids H+, Li+, Na+,

Be2+, Mg2+, Ca2+, Cr3+, RE3+, BF3, SO3

Fe2+, Co2+, Ni2+, Cu2+, Zn2+, Pb2+, SO2, BBr3

Cu+, Ag+, Ti+, Hg+, Pd2+, Cd2+, Pt2+, Hg2+, BH3

Bases F-, OH-, H2O, NH3, CO3

2-, NO3-, O2-,

SO42-, PO4

3-, ClO4-,

NO2-, SO3

2-, Br-, N3

-, N2, C5H5N, SCN-

H-, R-, CN-, CO, I-, NCS-, R3P, R2S, C6H6, S2-

Hard-soft classificationHard-soft classification

Hard-hard interactions have a greater electrostatic component and soft-soft ones depend more on orbitals interactions. But many other factors may be involved.

Soft acceptor and donor atoms are often large and Van der Waals’ force may contribute to the bonding; some soft bases such as CO also show - acceptor behavior is defined in a competitive situation.


PolymerizationPolymerization

Formation of dimers and polymeric structure is a manifestation of donor-acceptor interaction between molecules of the same kind.

2 AlCl3 Al2Cl6

Al

Cl

Cl

ClCl

ClCl

Al

Hydrogen Bonding can also be regarded as a donor-acceptor interaction in which the acceptor LUMO is the unoccupied antibonding orbital of hydrogen bonded to an electronegative element.

F—H---F


C9 Molecules in Condensed Phases

Molecular solids and liquidsMolecular solids and liquids

Intremolecular forces cause molecular substances to condense to form solids and liquids.

Enthalpies of fusion (melting, 熔解热 ) and vaporization ( 汽化热 )

Molecular solids: molecular retain their identity with geometries similar to those in the gas phase.

with highly unsymmetrical

the directional nature of intermolecular forces

Molecular liquids:

more disorganized than molecular solids


Intermolecular forcesIntermolecular forces

With polar molecules, the force between permanent electric dipoles is the dominant one.

When polarity is absent, the force arises from the interaction between instantaneous dipoles. ------- the London dispersion or Van der Waals’ force

Fig 1.

Polarizability generally increases with the sizes atoms, and the sequence of boiling points He< Ne <Ar < Kr.

The boiling point increase down the group in most series of nonpolar molecules, for examples, CH4 < SiH4 < GeH4.

CF4 < CCl4 < CBr4 < CI4

Hydrogen bonding: （ P77, Fig 1 ）

NH3 H2O HF 的某些性质的反常表现。

沸点、酸性


Dipole moment (, Debye, 德拜 ): = q (imagine charge)× d (distance)

Polarity is a measure of charge separation in bonds and can often be related to the electronegativtiy difference between the atoms.

HF (6.4), HCl (3.6), HBr (2.7), HI (1.4)

(1) Lone-pair electrons also have dipole moments.

(2) In polyatomic molecules the dipoles associated with each bond add vectorially to give a resultant that depends on the bond angles. The highly symmetrical molecules such as BF3, CF4, PF5, the net dipole moment is zero even though the individual bonds may be strongly polar.

(3) Dielectric constant: molecular dipole moments

(4) In nonpolar substances the dielectric constant arises from the molecular polarizability, and is generally much smaller than with polar molecules.

Molecular PolarityMolecular Polarity

Section D-1Section D-1INORGANIC CHEMISTRY

Section D

D1 Introduction to solids

Crystals and glasses Crystals and glasses

A crystalline solid is characterized by a unit cell containing an arrangement of atoms

repeated indefinitely. Long-range order: in crystal

For a crystal it is only necessary to give the details of one unit cell.

The substances are said to have the same structure if the arrangement of atoms within

a unit cell are different. ------X-ray structure analysis method or X-ray diffraction

Polymorphism ( 同质异晶现象 ) ： TiO2: rutile ( 金红石 ) ， brookite ( 板钛矿 ) ， anatase （锐钛矿）Non-crystalline or glassy solids do not have a unit cell.

Short-range order: local chemical bonding arrangement

Short-range order resulting from the local bonding of atoms may, however, be similar

in crystals and glasses.

Unit cell （晶胞） :

晶体的最小重复单元，通过晶胞在空间平移无隙地堆砌而成晶体

Unit cell parameter( 晶胞参数 ):

a ， b ， c ， α ， β ， γ;

Crystal system

Unit cell parameter compds

Cubic a = b = c α =β =γ = 900 NaCl trigonal a = b = c α =β =γ ≠900 Al2O3

tetragonal a = b≠c α =β =γ = 900 SnO2 hexagonal a = b≠c α =β = 900, γ = 1200 AgI orthogonal a≠ b≠ c α =β =γ = 900 HgCl2 Monoclinic a≠ b≠ c α =β = 900, γ ≠ 900 KClO3

triclinic a≠ b≠ c α ≠β ≠γ ≠ 900 CuSO4· 5H2O


Different representations of unit cel1s are possible.

It is important to understand how to use them to determine the stoichiometry

of the compound （化学计量比，即组成） and the coordination of each at

om （原子坐标） .

Looking at unit cellsLooking at unit cells

To specify the complete structure of a crystalline solid it is only necessary to

show one unit cell, but interpreting these Pictures requires practice.

Figure 1 shows some views of the cesium chloride structure (CsCl, depicted a

s MX). (see below). －－ P 81, Fig 1.


透射图 ( 斜射投影图 )

(a) Figure 1a is a perspective view ( 透视图 ) (more correctly known as a clinographic projection ( 斜射投影图 ), which is the most common way of showing a unit cell.

轴向投影图(b) Figure 1b shows a projection down one axis of the cell ( 轴向投影图 ). The Position of an atom on the hidden axis is given by a specifying a fractional coordinate (e.g. 0.5 for the central atom showing it is halfway up). No coordinate is given for atoms at the base of the cell.


(c) Figure 1c shows the atoms shifted relative to the unit cell, and emphasizes the fact that what is important about a unit cell is its size and shape ( 晶体尺寸与形状 ); its origin is arbitrary because of the way in which it is repeated to fill space.

(d) In Figure.1d the drawing has been extended to show some repeated positions of the central atom. This helps in seeing the coordination of the corner atom （给出离子的配位情况） .


Stoichiometry ( 计量比 ): the relative numbers of different types of atoms.

counting all the atoms depicted, and taking account of those that are shared with neighboring cells.

Any atom at a corner of a unit cell is shared between eight cells, any at an edge between four, and any on a face between two.

the Coordination of each atom ( 配位情况 ):


Some solids, especially natural minerals and transition metal compounds, are nonstoichio

metric with variable composition.

1) Defects （晶格缺陷） Vacanies ( 空位， atoms missing from their expected sites.)

Interstitials （间隙，填隙原子， extra atoms in sites normally vacant in the unit cel

l)

An imbalance of defects involving different elements can introduce nonstoichiometry.

for example: sodium tungsten bronzes NaxWO3 (x = 0 ~ 0.9 )

2) Replacement （同晶置换） :

aluminosilicate feldspars (Na, Ca) (Al, Si)4O8

Non-stoichiometry ( 非化学计量比 )

非化学计量比化合物，非整比化合物

Non-stoichiometry ( 非化学计量比 )

非化学计量比化合物，非整比化合物


Solids are often classified according to their chemical bonding, structures and

properties:

1) Molecular solids( 分子晶体 ) contain discrete molecular units held by relativ

ely weak intermolecular forces; ——low melting point, low boiling point, lo

w rigidity, insulator, etc

2) Metallic Solids ( 金属晶体 ) have atoms with high coordination numbers, bo

und by delocalized electrons that give metallic conduction; －－ thermal and

electrical conduction.

3) Covalent or polymeric solids ( 原子晶体 ) have atoms bound by directional

covalent bonds giving relatively low coordination numbers in a continuous on

e- or two- three- dimensional network;—— high rigidity, high m.p. and b.p., i

nsulators, etc

4) Ionic Solids ( 离子晶体 ) are bound by electrostatic attraction between anion

s and cations with structures where every anions is surrounded by cations and

vice versa. high m.p. and b.p., brittleness, insulator in solid state, etc

Chemical ClassificationChemical Classification



Element structures where chemical bonding is non-directional are best introduced

by considering the packing of equal spheres.

———— Close-packed structures （球密堆积形成晶体）Spheres of equal size may be packed in three dimensions to give hexagonal close-pa

cked (hcp) and cubic close-packed (ccp, also known as face-centered cubic, (fcc) str

uctures. The body-centered cubic (bcc) structure is slightly less efficiently close pac

ked.

ABABABABABAB….. Hexagongal close packing (hcp, 六方密堆积 , 74.05%)

ABCABCABCABC……Cubic close packing (ccp, 立方密堆积 , 52.36%)

body-centered cubic (bcc) structure ： ( 体心立方密堆积， 68.02%)

Sphere packingSphere packing

D2 Element structures


Simple cubic close packing: （简单立方密堆积）

C.N. ： 8

Efficient Space filling: 52.36 %

%36.52%10034

2

3

3

a

r

ar


体心立方堆积： bcc配位数： 8空间占有率 : 68.02% %02.68%1003

42

34

3

3

a

r

ar


面心立方密堆积： fcc

第三层与第一层有错位，以 ABCABC… 方式排列空间占有率 : 74.05%

b b

b b b

b b

f

f fh

h h

%05.74%100120sin

34

2

,633.1

3

abc

r

abac


六方密堆积 Hcp

第三层与第一层对齐，产生 ABAB… 方式

配位数： 12 ；空间占有率： 74.05%

b b

b b b

b b

f

f fh

h h


Hexagongal close packing

Cubic close packing

fcc


Many metallic elements have hcp, fcc or bcc structures. There are some clear

group trends in structure, although there are exceptions to these and some met

als have less regular structures, especially in the p block.

Some factors: depend on temperature and/or pressure

some regularities （ the numbers of valence electrons ）The commonest stable structures according to group numbers are ：Group 1 ( IA) ： bcc

Group 2 (IIA): varied

Group 3,4 (IIIB, IVB): hcp

Group 5, 6 (VB, VIB): ccp

Group 7, 8 (VIIB, VIII): hcp

Group 9-11 (VIII, IB): fcc

Metallic elementsMetallic elements


Nonmetallic elementsNonmetallic elements

Single-bonded structures where each element achieves an octet lead to the following predictions.

Group 14 (IVA) ： four tetrahedral bonds as shown in the diamond structure of C, Si, Ge, and Sn.

Group 15 (VA): three bonds in a pyramidal (non planar ) geometry, which can give rise to P4 molecules (white phosphorus) or a variety of polymeric structures shown by P and As. Phosphorus has several allotropes, some with apparently complex structures, but all are based on the same local bonding.

Group 16 (VIA): Two bonds, noncolinear, as found in S8 rings and in spiral chains with Se and Te. The different allotropes of sulfur all have this bonding.

Group 17 (VIIA): one bonds, giving diatomic molecular structures shown by all the halogens.

Group 18(0 group): no bonds, leading to monatomic structures with atoms held only by van der Waals’ forces. The normal solid structure of the noble gas elements is fcc.


The structural chemistry of the period 2 elements C, N, and O shows a greater tendency to multiple bonding than in lower periods.

Carbon ： graphite, diamond, fullerence

nitrogen: N N,

oxygen: O =O


D3 Binary compounds: simple

structuresCoordination number and geometryCoordination number and geometry

Binary compounds: ones with two elements present

Simple crystal structures: ones in which each atom (or ions) is surrounded in a regularly way by atoms (or ions ) of the other kind.

Coordination number( 配位数 ):

In regularity binary solids, the ratio of CN values must reflect the stiochiometry.

For example: in AB, both A and B have the same CN. zinc blende(4:4); rocksalt (6 : 6);

in AB2, the CN of A must be twice that of B. Rutile (6:3), CdI2, (6:3)


Coordination geometry ( 配位构型 ):

In the structures shown many of the atoms have a regular coordination geometry.

CN = 2, linear (B in ReO3 );

CN = 3, planar (B in rutile ( 金红石 )) ；

CN ＝ 4 ， tetrahedral (A and B in zinc blende( 闪锌矿 ), B in fluorite( 萤石 ) ；

CN ＝ 6 ， octahedral (A and B in rocksalt ( 岩盐矿 ) ， A in NiAs, rutile, and CdI

2);

CN= 8, cubic (A and B in CsCl, A in fluorite)

Other geometries are sometimes found, especially for the nonmetal B atom:

CN =2, bent, (SiO2 structures);

CN=3, pyramidal (in CdI2);

CN=6, trigonal prismatic (in NiAs)


D4 Binary compounds: Factors

influences structuresIonic RadiiIonic Radii

Ionic radii （离子半径） are derived from a somewhat arbitrary division of the observed anion-cation distances.

Sets of ionic radii are all based ultimately on somewhat arbitrary assumptions.

Four kinds of ionic radius:

(a) Goldschmidt ionic radius: standard: O2-: 132pm. F-: 133 pm

(b) Pauling ionic radius: standard: O2-: 140pm. F-: 136 pm

(c) Shannon and Prewitt ionic radius, based on the assumed radius of 140 pm for O2- in six coordination. （有效离子半径）

(d) Yatsmirskii ionic radius: ( 热化学离子半径 )

Any consistent set of radii should be able to give estimates of the total anion-cation distance and hence the unit cell dimensions if the structures is known.It is essential not to mix values from different sets.Although the values may differ, all sets show the same trends.


H- 146

Li+ 76 Be2+ 27 O2 － 140 F- 133

Na+ 102 Mg2+ 72 Al3+ 53 S2 － 182 Cl- 167

K+ 138 Ca2+ 100 Ga3+ 62 Br- 182

Rb+ 149 Sr2+ 116 I- 206

Cs+ 167 Ba2+ 149

Table Ionic radii (pm) for six coordination, based on a value of 140 pm for O2-

(i) For isoelectronic ions, radii decrease with increasing positive charge (Na+ > Mg2+ > Al3+) or decreasing negative charge.

(ii) Radii increase down each group;

(iii) For elements with variable oxidation state, radius decreases with increasing positive charge;

(iv) Most anions are larges than most cations.

(v) Ionic radii increase with coordination number (CN). For example, the Shannon and Prewitt radii for K+ with different CN are 138(6), 151 (8), 159 (10), 160 (12).

Section D－4

Section D－4

INORGANIC CHEMISTRY

Radius ratiosRadius ratios

An ionic solid should achieve maximum electrostatic stability, when (i) each ion is surrounded by as many as possible ions of opposites charges; (ii) the anion-cation distance is as short as possible.

The radius ratio rules( 半径比规则 ): predict the likely CN for the smaller ion (usually the cation) in terms of the ration r>/r< where r< is the smaller and r> the larger of the two radii, the approximate radius ratios for different CN are:

r>/r< > 0.7 0.4 ~ 0.7 0.2~ 0.4

CN 8 6 4

For examples:

BeF2: 0.20 CN=4;

MgF2: 0.54 6

CaF2 : 0.75 8


Ion PolarizabilityIon Polarizability

Polarizabilty of the ion: refers to the ability of an applied electric field to distort the electron cloud and so induce an electric dipole moment.

The most polarizable ions are large ones, especially anions from later periods （ S2 －， I －， Br －） .

In layer and chain structure, the anions are generally in asymmetric environments and experiences a strong net electric field from neighboring ions. Polarization lowers the energy of an ion in this situation , giving a stabilizing effect not possible when the coordination is symmerical.

Layer structures occur frequently with disulfides and dichlorides (and with heavier anions lower in the same groups), but almost never with difluorides and dioxides:

TiO2 and FeF2; TiS2 and FeI2.


Covalent bondingCovalent bonding

Purely electronstatics forces between ions are nondirectional, but with increasing covalent character the directional properties of valence orbitals become more important.

Compounds between nonmetallic elements have predominantly covalent bonding and the structures can often be rationalized from the expected CN and bonding geometry of the atom presents.


D6 Lattice energies

The Born-Haber CycleThe Born-Haber Cycle

The lattice energy UL of a solid compound is defined as the energies required to transform it into gas-phase ions:

(1) NaCl (s) Na+ (g) + Cl- (g) UL (positive value)

(2) Na+ (g) + Cl- (g) NaCl (s)

The reverse process ------- UL (negative value)

Lattice energies may be estimated from a thermodynamic cycle knows as a Born-Haber cycle, which makes uses of Hess’ Law.


Lattice enthalpy HL= -fH (NaCl) + atH(Na) + ½ B(Cl2) + I (Na) – A (Cl)

the enthalpy of formation of NaCl: fH (NaCl)

the enthalpy of atomization of Na solid: atH (Na)

the bond enthalpy of Cl2: B(Cl2)

the ionization energy of Na: I (Na)

the electron affinity of Cl: A (Cl)


1) Thermodynamic data —— Hess’ Law

2) The other methods for lattice energy (theoretical estimates):

(a) Born- Mayer equation

The long-range Coulomb interaction between charges.

(b) Kapustinskii equations:

(1) lattice energies increase strongly with increasing charge on the ions.

(2) lattice energies are always larger for smaller ions.

)1

1(4 00

20

nr

ezAzNU L

rr

zzCU L


NaCl型离子晶体

Z1 Z2 r+

/pmr-

/pmU

/kJ· mol-1熔点/oC

硬度

NaFNaClNaBrNaI

MgOCaOSrOBaO

11112222

11112222

959595956599113135

136181195216140140140140

9207707336834147355733603091

9928017476622800257624301923

3.22.5

<2.5<2.55.54.53.53.3


ApplicationApplication


(ii) Stabilization of high and low oxidation states

When an element has variable oxidation states, it is often found that the highest value is obtained with oxide and/or fluoride. The ionic model again suggests that a balance between IE and lattice energy is important.

Small and/or highly charged ions provide the highest lattice energies according to equation 1. and increase in lattice energy with higher oxidation state is more likely to compensate for the high IE.

A large ion with low charge such as I- is more likely stabilize a low oxidation state, as the smaller lattice energy may no longer compensate for high IE input.

CuF2 CuCl2, CuBr2 （）

CuF (×) or CuX (X=Cl, Br I)

(III) Stabilization of large anions or cations

(a) Large cations stabilize large anions.

(b) Oxoanions salts such as carbonates are harder to decompose thermally when combined with large cations.

(c) Solids where both ions are larger are generally less soluble in water ones with a large ion and a small one.

A satisfactory explanation depends on a balance of energies.



D7 Electrical and optical

properties of solidsThe band model ( 能带模型 )The band model ( 能带模型 )

The band model: an extension of the molecular orbital method

The overlap of atomic orbitals in an extended solid gives rise to continuous bands of electronic energy levels associated with different degrees of bonding. In a simple monatomic solid the bottom of the band is made up of orbitals bonding between all neighboring atoms; orbitals at the top of the band are antibonding, and levels in the middle have an intermediate bonding character.

Metallic and nonmetallic behavior results from a band partially occupied by electrons, so that there is no energy gap between the top filled level and the lowest empty one.

On the other hand, a nonmetallic solid has bandgap between a completely filled band (the valence band) and a completely empty one (the conduction band)

So, different atomic orbitals can, in principle, give rise to different bands.


the distinction between nonmetallic and metallic solids.

Band gap （带隙，能隙）：

Valence band ( 价带 ) ： a completely filled band

Conduction band ( 导带 ): a completely empty one


Nonmetallic solids include ionic and covalent compounds.

In the former case, the VB is made up of the top filled anion levels and the CB of the lowest empty cation levels.

In covalent solids, such as diamond the VB consists of bonding orbitals and the CB of antibonding orbitals.

Simple metallic solids are elements or alloys with close-packed structures where the large number of interatomic overlaps gives rise to wide bands with no gaps between levels from different atomic orbitals.


Band gapsBand gaps

Band edges determine the optical absorption of nonmetallic solid and the possibility of semiconduction. Band gaps in binary solids decrease with decreasing electronegativity difference between the elements. In both ionic and covalent solids bandgaps are smaller with elements in lower periods.

A solid with a small bandgap is a semiconductor with a conductivity that (unlike the case with a metal) increases as temperature is raised. The bandgap also determines the minimum photo energy required to excite an electron from the VB to CB, and hence the thresold for optical absorption by a solid.

In covalent solid the bandgap is related to the energy splitting between bonding and antibonding ortibals. And thus to the strength of bonding,

In an ionic solid the bandgap is determined by the energy required to transfer an electron back from the anion to cation, which is related to the lattice energy.

Some systematic trends:

(a) In a series of isoelectronic solids, such as CuBr-ZnSe-GaAs-Ge the bandgap decrease with decreasing electronegativity difference between the two elements.

(b) In series such as C-Si-Ge- or LiF-NaF-KF the bandgap decreases as the group is descended and atoms or ions become larger.


Dielectric propertiesDielectric properties

The dielectric constant of a medium is a measure of the electrostatic polarization, which reduces the forces between charges.

the static dielectric constant: depend on the displacement of ions from their regular positions in an applied electric field. It is applicable for static fields, or frequencies of electromagnetic radiation up into the microwave range.

The high-frequency dielectric constant: is measured at frequencies faster than the vibrational motion of ions. ----applicable in the visible region of the spectrum and determines the refractive index( 折射率 ).

——depend on electronic polarizability.

Large ions contribute to this ：

Glass containing Pb2+used for lenses——a high refractive index.

TiO2( 钛白粉 ) ：


Defects including impurities have a major influence on the electrical properties of nonmetallic solids. They can provide extra electrons or holes, which enhance semiconduction, and they can also facilitate conduction by ions.

All solids contain defects where the regularity of the ideal periodic lattice is broken.

Line and plane defects: for mechanical properties;

Point defects: for electrical properties;

They include

(a) Vacancies or atoms missing from regular lattice positions;

(b) Interstitials or atoms in positions not normally occupied;

(c) Impurities either accidentally present or introduced as deliberate doping.

Defects that introduce extra electrons, or that give missing electrons or “holes”, have a large influence on electronic conduction in nonmetallic solids.

Influence of defectsInfluence of defects


Providing the electrons:

N-type semiconductor:

doping Si with P replaces some tetrahedral bonded Si atoms in the diamond lattice with P. Each replacement provides one extra valence electron, which requires only a small energy to escape into the CB of silicon.

P-type semiconductor:

replacing an Si atom with Al gives a missing electron of hole.

Other examples:

the slight reduction pf TiO2:

add some electrons---------a n-type semiconductor.

Oxidation of NiO:

decrease some electrons-------a p-type semiconductor

Ionic conduction: atoms in defect sites may themselves be mobile

these would be correlated with the number of defects, and may be especially large in disordered solids.

Section E1Section E1INORGANIC CHEMISTRY

Section E

E1 Solvent types and properties

Polarity and solvation Polarity and solvation

A solvent is a liquid medium in which dissolved substances are known as solutes.

application:

(a) storing substances that would otherwise be in inconvenient states (gases, eg. NH3).

(b) facilitating reactions that would otherwise be hard to carry out (solids and liquids).

Important: controlling what substances dissolve easily and what types of reactions

may be performed.

The chemical as well as the physical state of solutes may be altered by interaction with the solvent.


Polarity: polar molecular.

(1) The molecules with large dipole moments form polar solvents, such as NH3, H2

O, etc;

dielectric constant (r):

solvation ( 溶剂化 ): involves donor-acceptor interactions

solvent molecules are ordered round the solute, not only in the primary solvation sphere but affecting more distant molecules.

(2) That with little or no dipole moments form nonpolar solvents, such as CO2, hexane, benzene, etc.

they are better at dissolving mompolar molecules and for carrying out reactions where no ions are involved.

Between nonpolar solvent and solute: Van der Waals’ force

Nonpolar solvent are generally poor solvents for polar molecules.

Nonpolar molecules cannot compete with the strong intermolecular forces in a polar solvent. —— 相似相溶原理



Donor and acceptor properties Donor and acceptor properties

(1) Donor or Lewis Base: most polar solvents : lone-pair electrons

H2O, NH3, C5H5N （ Py ） , etc.

solvating cations and other Lewis acids.:

(2) Acceptor or Lewis Acids: empty of from hydrogen bonding

solvating anions.

(3) Solvolysis ( 溶剂解反应 ):

OPCl3 + 6 H2O OP(OH)3 + 3 H3O ＋＋ 3 Cl －

OPCl3 + 6 NH3 OP(OH)3 + 3 NH4＋＋ 3 Cl －

FeCl3 in different solvents: for the difference of polarity and the solvation of small ions.

FeCl3 + pyridine (C5H5N) [FeCl3Py]

FeCl3 + DMSO {(CH3)2SO} [FeCl2DMSO4]+ + Cl －

FeCl3 + MeCN [FeCl2MeCN4]+ + [FeCl4] －


Ion － Transfer solvents Ion － Transfer solvents

(1) Protic solvents ( 质子溶剂 ): have transferable H+

water ， NH3 ， C2H5OH ， CH3OH ， etc

Autoprotolysis ( 自质子解 ) ：

2 H2O H3O + + OH-

2 NH3 NH4+ + NH2

-

An acid is the positive species (NH4+ ) formed or any solute that gives rise to it.

A base is the negative species (NH2-) formed or anything producing it in solution.

(2) Aprotic Solvents ( 非质子溶剂 ) ： don’t have transferable H+, but some other ion such as halide or oxide can be involved. autoionization: ion transfer

2 BrF3 + SnF4 2 BrF2+ + SnF6

2-

BrF3 --------- Lewis base to produce F- ions.

SnF4---------- Lewis acid to react with F- ions


(3) Lux-Flood acid/base definition

in oxide melts the solvent system.

An oxide donor-------- a base ; An oxide acceptor -----an acid

CaO (base) + SiO2 (acid) CaSiO3


E2 Brønsted Acids and Bases

Definitions Definitions

Brønsted acids: a proton donor ; Brønsted bases: a proton acceptor

an acid-base reaction involves protolysis HA + B HB+ + A-

Conjugate Acid- Base relationship( 共扼酸碱关系 ):

Conjugate Acid- Base pair ( 共扼酸碱对 ):

For example: HCl / Cl- ， H2O/OH-, H3O+ / H2O ， NH4+/NH3

H2PO4-/HPO4

2- HPO42- / PO4

3- H2SO4/HSO4-

This definition of acids and bases should not be confused with Lewis acid-base definition. Brønsted bases are also Lewis bases.

H+ _____ Lewis acid and Brønsted acid, BF3 _____ Lewis acid, not Brønsted acid

Brønsted Acidity is solvent dependent. Substances such as HCl are covalent molecules that undergo protolysis only in solvents polar enough to solvate ions and when a base is present.


AcHHAc 2

442 HPOHPOH 3

424 POHHPO

34 NHHNH

2333 NHCHHNHCH 2

523

62 O)Fe(OH)(HHO)Fe(H 422

252 O)(HFe(OH)HO)Fe(OH)(H

酸 H+ +碱


pH value: pH value:

Water undergoes autoprotolysis (self-ionization):

2 H2O H3O + + OH-

The equilibrium constant: Kw = [H3O+][OH-] = 1×10-14 (25 )℃

In pure water and in solutions that don’t provide any additional source of H3O+ or OH- both ions have molar concentrations equal to 10-7.

pH scale: pH = -lg [H+]

At 25 ℃，

Neutral water ： pH = 7

pH < 7 , acidic solution [H+] > [OH-]

pH > 7 , basic solution (alkaline) [H+] < [OH-]

)OH()OH( 3

c

cc

cKW

T ， KW


Strong and weak behavior Strong and weak behavior

HA + H2O H3O + + A-

Acidity constant or acid dissociation constant of HA:

pKa ＝－ lg Ka

a larger Ka value corresponds to a smaller pKa.

pKa < 0, strong acid, the equilibrium lies strongly to the right, H2SO4, HCl, etc.

pKa > 0, weak acid, and undergo only partial protolysis. HF, HSO4-

][

]][[ 3

HA

AOHKa

B + H2O OH- + HB+

basicity constant

pKb ＝－ lg Kb = 14- pKa

Strong base: pKb < 0

Weak base: pKb > 0

][

]][[

B

OHHBKb



the degree of ionization: （电离度）

Degree of electrolytic dissociation:

HA + H2O H3O + + A- (for weak monoatomic acid ( 一元弱酸 ))

C1 （ initial state ）

C2 （ the balanced state ）

＝ C2/C1 × 100 ％

[H3O+] = C1 ×

1][

][

][

]][[ 2

3

233 c

OHc

OH

HA

AOHKa

cKcOH

cK

a

a

][ 3

2

(1) If < 5%, or c/Ka > 500;

pH = - lg [H+]

(2) If > 5%,

2

4][

2

3

cKKKOH aaa


For weak monoatomic base:

B +H2O OH- + BH+

1][

][

][

]][[ 22 c

OHc

OH

B

OHHBKb

cKOH b ][(1) If < 5%, or c/Ka > 500;

(2) If > 5%,

2

4][

2 cKKKOH bbb

pOH = - lg [OH-]

pH = 14- pOH


Common ion effect （同离子效应）：

The degree of ionization would decrease for being the common ions in solution.

HAc(aq) + H2O (l) H3O+ (aq) + Ac–(aq)

Ac–(aq) 4NHNH4Ac(aq) (aq) +

If addition of HCl, the same phenomenon would be observed.


For example:

Calculate the pH value and the degree ionization of the mixture of HCl (0.10 mol·L-1) and HAc (0.10mol·L-1). (Ka = 1.75×10-5)

HAc ＋ H2O H3O ＋＋ Ac －

Initial state(mol·L-1) 0.10

Balanced (mol·L-1) 0.10 – x x + 0.1 x

x

xx

HAc

AcOHKa

10.0

)10.0(

][

]][[ 3

x << 0.10 , x = Ka = 1.75×10-5 mol·L-1

pH = 1.000 = 1.75×10-5 / 0.10 ×100 ％＝ 0.018 ％

If no HCl, y = 1.33 ×10-3 mol·L-1 ，’ = 1.33×10-3 / 0.10 ×100 ％＝ 1.33 ％ pH1 = 2.876

　　例：在 0.10 mol·L-1 的 HAc 溶液中，加入 NH4Ac (s) ，使 NH4Ac 的浓度为 0.10 mol·L-1 ，计算该溶液的 pH值和 HAc 的解离度。

5108.110.0

)10.0( x

xx

x = 1.8×10-5 c(H+) = 1.8×10-5 mol·L-1

0.10 ± x ≈ 0.10

0.10 mol·L-1 HAc 溶液： pH = 2.89 ， α = 1.3%

　　解： HAc(aq)+H2O(l) H3O+(aq)+Ac － (aq)

ceq / (mol·L-1) 0.10 – x x 0.10 + x

c0/ (mol·L-1) 　 0.10 0 0.10

pH = 4.74 ， α = 0.018%


Salt effect Salt effect

The degree of ionization () of weak electrolyte would increase if adding some kind of salt into the solution.

Owing to increasing of electrostatic attraction.

Common ion effect:

(aq)HCO(aq)OH O(l)H(aq)COH 33232－第一步：

(aq)CO(aq)OH O(l)H(aq)HCO 23323－－第二步：

7

32

3332a1 102.4

)COH(

)HCO()OH()COH(

c

ccK

－

11

3

233

32a2 107.4)HCO(

)CO()OH()COH(

－

－

c

ccK

做近似处理。

解离平衡的计算可按一元弱酸的

反应，主要来自于第一步解离溶液中的

)OH(

OH 10

3

33

a2

a1

c

KK

For weak polyatomic acid or base, fractional ionization


For weak polyatomic acid or base,

fractional ionization

H2S + H2O HS- + H3O+ Ka1 = 5.7 ×10-8

HS- + H2O S2- + H3O+ Ka2 = 1.5×10-15

例题：计算 0.010 mol·L-1 H2CO3 溶液中的 H3

O+,

H2CO3, 　　，　和 OH －的浓度以及溶液的 pH值。

－23CO－

3HCO

(aq)HCO(aq)OH O(l)H(aq)COH －33232 解：

5106.5010.0 0.010 －－ xx15

33 Lmol105.6)HCO()OH( －－－ cc1

32 Lmol010.0)COH( －c

x

x

c

cc

－

－

0.010)COH(

)HCO()OH( 2

32

33

732a1 102.4)COH( K

0.010 )Lmol/(1 xxx－平衡浓度

y

yy

c

cc

5

5

3

233

105.6

)105.6(

)}HCO({

)}CO()}{OH({－

－

：根据第二步解离计算－)CO( 23c

10 .56 10 .56 )Lmol/(

(aq)CO (aq)OH O(l)H(aq)HCO 551

eq

23323

yyyc

－－

－－

－

1132a2 107.4)COH( K

11a2

55 107.4 , 106.5106.5 Kyy

1111a2

23 Lmol107.4Lmol )CO( Kc －


pH value of salt solutionpH value of salt solution

(1) Strong acid + strong base compounds, neutral, pH =7.

(2) weak acid + strong base compounds ，

the hydrolysis of the anion occurs.

For example, (i) in NaAc solution

Ac- + H2O HAc + OH-

Kh = { [HAc][OH-]} / [Ac-] = Kw/Ka

(II) NH4＋＋ 2 H2O NH3H2O ＋ H3O ＋

Kh = { [NH3H2O][H3O+]} / [NH4+] = Kw/Kb

(III) In NH4Ac solution

Kh = Kw / (Ka Kb)


Problem 14.: Calculate the pH of 0.200 moldm-3 solution of NH4Cl.

Solution.:Cl-, the conjugate base of a strong acid, does not react with water, but NH4

+, the conj

ugate acid of NH3, does react:

NH4+ + 2H2O NH3 H2O + H3O+

Let [H3O+] = x, [NH3] = x, and [NH4

+] = 0.200 -x 0.200

The solution of NH4Cl in water is acidic, as expected.

b

wh K

K

NH

NHOHK

][

]][[

4

33

96.4

101.1][106.5108.1

100.1

200.05

310

5

142

pH

xOHK

KxK

b

wh


Buffer solution Buffer solution

The pH value would keep stable if adding a little acid or base into the solution.

The buffer solution consist of the conjugate acid-base pair.

for example: NaAc- HAc, NH3H2O-NH4Cl, Na2HPO4-NaH2PO4, etc.

In buffer solution (weak acid (HA)-salt (A-),

HA + H2O A- + H3O+

CHA CA-

For the weak base- salt system,

salt

acid

A

HAaa

a

C

CpKapH

C

CK

A

HAKOH

HA

OHAK

lg

][

][][

][

]][[

3

3

salt

baseb

salt

baseb

C

CpKpH

C

CKOH

lg14

][


The agents The content of buffer solution

pKa The range

HCOOH-NaOH HCOOH-HCOO- 3.75 2.75 ~ 4.75

HAc-NaAc CH3COOH-CH3COO- 4.75 3.75 ~ 5.75

Na2B4O7 ~ HCl H3BO3 _ B(OH)4

- 9.14 8.14 ~ 10.14

NH3 H2O-NH4Cl

NH4+ - NH3 9.25 8.25 ~ 10.25

NaHCO3 – Na2CO3

HCO3- –CO3

2- 10.25 9.25 ~ 11.25

Table I. Some common buffer system


Problem 13.: Given 50 mL of 0.20 moldm-3 solution of the acid HA (Ka = 1.010-5) and 50 mL of an NaA solution. What should the concentration of the NaA solution be to make a buffer solution with pH = 4.00?Solution.:

HA + H2O A- + H3O+

The addition of the two solutions halves the concentration of each solute assuming no reaction at all. The original NaA solution must be 2.0 10-2 moldm-3 to get this value of x after dilution.

3254

3 100.1100.110.0

100.1

][

]][[

dmmolxx

HA

OHAK a


想配制 pH ＝ 5.00 的缓冲溶液，需在 50ml 0.10 mol·L-1 的 HAc 溶液中加入 0.10 mol·L-1 的 NaOH 多少毫升？

选择合适的酸，以 pKa 接近所要求的 pH值的酸最为适合。

mLxx

x

C

CpKapH

salt

acid

3210.0

10.010.050lg75.400.5

lg

　　例题：若在 50.00ml 0.150mol·L-1 NH3 (a

q) 和 0.200 mol·L-1 NH4Cl 组成的缓冲溶液中，加入 0.100ml 1.00 mol·L-1 的 HCl ，求加入 H

Cl前后溶液的 pH值各为多少？解：加入 HCl 前：

14.9)12.0(26.9 200.0

150.0lg)108.1lg(14 5

)BH(

)(Blg p14pH b

c

cK

加入 HCl 后：11 Lmol0020.0Lmol

10.50100.000.1

)HCl( c

加 HCl前浓度 /(mol·L-

1)

0.150-0.0020-x 0.200+0.0020+x x

0.150 0.200

加 HCl 后初始浓度 /(mol·L-1)

0.150-0.0020 0.200+0.0020

平衡浓度 /(mol·L-1)

5108.10.148

)202.0(

x

xx4 NH NH3(aq) + H2O (l) (aq) +OH － (aq)

9.11pH 4.89pOH

155 Lmol101.3)OH( 101.3 cx


Solvent Leveling effect Solvent Leveling effect

Differentiating solvent effectDifferentiating solvent effect

Because the strong acid completely protolyzed, it is impossible to study this species itself in water. So the pH value would be the same for the strong acid with the same concentration and H3O+ is effectively the strongest acid possible there and any stronger acid is said to be leveled.

Strong base would be leveled in water. HI, HCl, HBr, HNO3, HClO4

Solvent leveling limits the range of acid-base behavior that can be observed in given solvent, and is one reason for using other solvents with different leveling ranges. Liquid ammonia is very basic compared with water, and H2SO4 is very acidic.

The strong & the weak acid in water (HCl, HAc)

The strong & the weak base （ NaOH ， NH3H2O ）

HI ， HBr ， HCl ----in H2O, the acidity be leveled; in CH3OH, differentiated


Trend in pK value Trend in pK value

(1) AHn compounds, acid strength increases from left to right in the periodic table, for example, CH4 < NH3 < H2O << HF . This trend is most simply to related to the electronegativity increase of the element attached to hydrogen, which gives more bond polarity in the direction A- -H+

(2) Acid strength also increases down the group, HF < HCl < HBr < HI; because of the deceasing H-X bond strength down the group.

(3) Oxides of nonmetallic elements, acids —— oxoacids in water

(4) Oxides and hydroxides of metal, bases

(5) Metals in high oxidation states can also form oxoacids


Pauling’ s rulesPauling’ s rules


Aqua cations Aqua cations

Many metal cations are acidic in water.

Salt containing these ions Al3+, Fe3+ form acidic solutions, and if the pH is increased successive protolysis may lead to polymerization and precipitation of an soluble oxide or hydroxide such as Al(OH)3.

Some compounds show amphoteric behavior and dissolve in alkaline solution to give oxoanions.

Al(OH)3 —— [Al(OH)4]-


E3 Complex formation

Introduction of complex Introduction of complex

A complex in general is any species formed by specific association of molecules or ions by donor-acceptor interactions.

A metal cation (Lewis acid)—— ligands. (Lewis base, ions and neutral molecules)

The ligand acts as a donor and replaces one or more water molecules from the primary solvation sphere.

Important terms:

Coordination spheres: （内界）： Counterions(抗衡离子，外界 ) ：

Coordination bond, dative bond: steric isomerism

Coordination numbers: coordination configuration isomerism( 异构现象 )

Chelate(螯合，螯合物 ): Chelate ligands: optical isomerism

Monodentate (unidentate); bidentate ; tridentate, tetradentate

For example: en, ethylenediamine, EDTA (ethylenediamine tetraacetic acid)

Coordination compounds, complex


en, ethylenediamine EDTA

Phen 2, 2’- bpy

Chelating ligands generally form stronger complexes than unidentate ones with similar donor properties.

application ： complexmetric titration ， removing the toxic metals, etc.

The most stable complexes generally being formed with four or five atoms, so that with the metal ion a five- or six- membered ring is formed


Macrocyclic compounds: macrocyclic effect

Crown ether, & open-chain crown ether

Cryptand ether, & open-chain ones

Size selectivity: corresponding to different cavity sizes.

e.g. Benzo-15-C-5, K+ & 6 + Na+


Nomenclature of the complexesNomenclature of the complexes

IUPAC nomenclature

Follow the normal nomenclature of some inorganic compounds,

(a) If the counterion is the simple anions, 某化某

(b) If the counterion is the complicate anions, 某酸某

(阴离子配体－中性配体－“合”－中心离子 )

For example:

[Cu(NH3)4]SO4 ： tetraaminecopper(II) sulfate ；硫酸四氨合铜 (II)

K2[CuCl4] ： potassium tetrachlorocuprate(II) (四氯合铜 (II) 酸钾 )

[Ni(NH3)6]Br2 ： hexaaminenickel(II) dibromide (溴化六氨合镍 (II))

K[PF6]: potassium hexafluorophosphate (V) (六氟磷酸 (V)钾 )

[Co(H2O)(NH3)3(Cl)2]Cl: dichloro-triamine-hydrate cobalt(III) chloride

(氯化二氯三氨一水合钴 (III))


Stepwise formation constant, K1, K2, K3…….(逐级稳定常数 )

M + L ML

ML + L ML2

Cumulative stability constant, 1, 2, 3……….(累积稳定常数 )

M + n L MLn

The bigger the equilibrium constants is, the higher the stability of the complex are.

[Ag(NH3)2]+: 1.1×107 [Ag(CN)2] － : 1.3×1021 (for the same types ions)

[Ag(SCN)2] － : 3.7×107 [Ag(S2O3)2]3 － : 1.1×1013

NH3H2O ： Na2S2O3 ， NaCN

AgCl, AgBr, AgI

321

221 ]][[

][

]][[

][

KKK

LML

MLK

LM

MLK

321 KKKn

Equilibrium constantsEquilibrium constants

《近代化学导论》： P343 配离子的不稳定常数——配合物的稳定常数


Hard and soft behaviorHard and soft behavior

Hard donors interact more strongly with hard acceptors, soft donors with soft acceptors ( 硬亲硬，软亲软，软和硬的反应不容易发生 ).

Hard acids tend to be more electropositive, and harder bases more electronegative. Softer dono

r and acceptor atoms tend to be larger and more polarizable.

Class A(hard acid): the cations formed metals in early groups in periodic table

Na+, K+, Mg2 ＋， Al3 ＋， etc.

Class B(soft acid): some later transition metals and many post-transition metals

Hg2+, Pb2+, Pt2+, Cd2+, etc.

Class b forms strong complexes with ligands such as ammonia.

But ， Class a ions don’t complex with such ligands appreciably in water.

Hard cations with low charge/size ratio, such as alkali ions, form very weak complexes with all

ligands, except macrocycles.

Some polyatomic ions such as NO3-, ClO4

-, BF4-, PF6

-, etc have very low complexing power to

either class a or class b. ---- counterions for studying the thermodynamic properties of metal i

ons unaffected by complex formation.


E4 Solubility of ionic compounds

Introduction of solubility ProductIntroduction of solubility Product

When the ion compounds were dissolved in water, the following equilibrium will be found,

MnXm(s) n Mm+ (aq) + n Xn-(aq)

Equilibrium constant: the activities, not concentration

Solubility product: (for dilute solutions)mnnm

sp XMK ][][

234

32243

24

24

222

][][))((

]][[)(

]][[)(

POCaPOCaK

SOBaBaSOK

IPbPbIK

sp

sp

sp

AgCl Ksp= 1.77 ×10-7 AgBr Ksp= 5.35 ×10-13 AgI Ksp= 8.51 ×10-17

AgCl > AgBr > AgI


For the ionic compounds with different type, calculate the solubility from the solubility product.

AgCl: Ksp= 1.77 ×10-7 Ag2CrO4 Ksp= 1.12 ×10-12

343 4224

12324

24

242

2442

35

102

1004.14

)(][

1012.1][4][][)(

)()(2)()2(

1033.1)(][

1077.1][]][[)(

)()()()1(

dmmolCrOAgK

CrO

CrOCrOAgCrOAgK

aqCrOaqAgsCrOAg

dmmolAgClKAg

AgClAgAgClK

aqClaqAgsAgCl

sp

sp

sp

sp

Solubility: AgCl < Ag2CrO4


Solubility product Principle ( 溶度积原理 )

It was widely applied to judge the trend and the extent for forming the precipitation from the mixture solution containing the Mm+ and Xn － ions.

Firstly, calculate the J value (it is defined to be [Mm+]n[Xn-]m ) from the concentration of ions;

Secondly, compare the relationship between the J value and the Ksp.

If Ksp < J ， the precipitation will be formed in solution and the equilibrium shifts to the direction of forming the precipitation.

If Ksp > J ， the precipitation will be dissolved and the equilibrium shifts to the direction of dissolving the precipitation to the solution.

If Ksp = J ， the equilibrium keep the balanced states.

the rate of precipitating = the rate of dissolving


Problem .: Determine whether a precipitate form in solution or not when Pb2+ solution (0.100 moldm-3, 100 mL ) is mixed with Cl- (0.30 moldm-3) in 100 mL solution. (Ksp(PbCl2)

= 1.7 10-4 )

Solution.:

Assuming that no precipitate form and the final volume keep to be 200 mL, the lead ion and chloride ion concentrations would be 0.005 and 0.15 moldm-3, respectively.

PbCl2 Pb2+ + Cl-

Ksp(PbCl2) = [Pb2+][Cl-]2 = 1.7 10-4

But if there were no precipitation, [Pb2+][Cl-]2 = (0.005)(0.15)2 = 1.1 10-3.

For this product is much greater than the value of Ksp(PbCl2), the concentration of ions ex

ceed that required for equilibrium, and the precipitation will form in the mixture.


Problem .:Whether Mg(OH)2 precipitate or not in a solution containing Mg2+ (0.0010 mo

ldm-3) if the OH- concentration of the solution were altered to be (a) 10-5 moldm-3 (b) 10-3 moldm-3 ? (Ksp(Mg(OH)2) = 1.2 10-11)

Answer.:

Ksp = [Mg 2+][OH-]2 = 1.2 10-11

[OH-]2 = 1.2 10-11/ [Mg 2+] = 1.2 10-8 [OH-]sat = 1.1 10-4 moldm-3

(a) [OH-] does not exceed [OH-]sat, so no precipitation will be observed.

(b) [OH-] exceeds [OH-]sat, so Mg(OH)2 will precipitate from the solution.

The influence of pH value in solution

Application:

The selectivity separation of metal ions in mixture solution would be successfully finished by controlling the pH value.


The influence of pH value in solution

The basis of control the pH value of solution to separate the metal ions:

(a) the condition of forming the precipitation in solution: J > Ksp

(b) the condition for judging whether the metal ion precipitate completely or not:

[Mn+] < 10-5 moldm-3

][lg

114

][][

]][[))((

)()()()(

n

sp

nn

sp

nnsp

nn

M

K

npH

M

KOH

OHMOHMK

aqOHaqMsOHMn

2.82 6.85 pH

可将 pH值控制在 2.82 ~ 6.85 之间

Fe3+沉淀完全 Ni2+开始沉淀

例题：在含有 0.10mol·L-1 Fe3+ 和 0.10mol·L-1 Ni2+ 的溶液中，欲除掉 Fe3+ ，使 Ni2+仍留在溶液中，应控制 pH值为多少 ?

2.82 102.8 Fe(OH) 393

6.85 105.0 Ni(OH) 16

2

pH pH sp 沉淀完全开始沉淀K解：

c(Fe3+)≤10-5


common ion effect


Adding one of the ions Mm+ or Xn- will shift the equilibrium to the left and so reduce the solubility of the salt.

For example,

calculate the solubility of AgCl(s) in HCl (1.0×10-2 moldm-3) solution.

(1) Before adding HCl, [Ag+]=1.33×10-5 moldm-3

(2) AgCl (s) Ag+(aq) + Cl － (aq)

After adding HCl, x x x+0.010

38

33

10

1077.1

01.0010.0,010.0

1077.1)010.0)((]][[)(

dmmolx

dmmolxdmmolx

xxClAgAgClK sp


Complexing effectComplexing effect

If adding some compound able to complex the insoluble salt, its solubility would markedly rise.

For example:

Calculate the solubility of AgCl in NH3H2O (6 moldm-3) sultion.

Answer.:

AgCl (s) + NH3H2O [Ag(NH3)2]+(aq) + Cl － (aq) + 2 H2OInitial state (moldm-3) 6 0 0Balanced state(moldm-3) 6 - 2x x x

Ksp(AgCl) =1.7×10-10 ， K{[Ag(NH3)2]+}=1.7×107

332

2

371023

23

232

3

23

283.0107.2)26(

107.2)106.1()107.1(}])({[)(

]][[][][

])([

][

][])([

dmmolxx

x

NHAgKAgClK

ClAgAgNH

NHAg

NH

ClNHAgK

sp


Go = － RT lnKsp

Thermodynamic AnalysisThermodynamic Analysis

(a) the formation of gas-phase ions; (b) their subsequent solvation

Total energy: (a) lattice energy of the solid (b) solvation enthalpies of the ions

With ions of very different size, solvation is relatively favored and solubility tends to be larger than with ions of similar size.


Major trends in WaterMajor trends in Water

(i) Soluble salts are more often found when ions are of very different size rather than similar size.

(a) For example: different alkali metal cations: OH- & F- + Li+ -------least soluble

Cl- & NO3- —— soluble

(b) precipitate a large complex ions, a large cation such as [Bu4N]+ can be helpful.

(ii) Salts where both ions have multiple charges are less likely to be soluble than ones with single charges.

CO32- & SO4

2-, insoluble

(iii) With ions of different charges, especially insoluble compounds result when the lower charged one is smaller.

M3+ + fluorides and hydroxides: very insoluble

+ heavier halide or nitrates: very soluble.

(iv) Lower solubility result from covalent contributions to the lattice energy.

—— with ions of less electropositive metals in combination with more polarizable cations.

Late transition and post-transition elements + S2- —— insoluble; X- —— insoluble.


Influence of pH and complexingInfluence of pH and complexing

Any substance in solution that reacts with one of the ions formed the following reaction will shift the equilibrium to the right and hence increase the solubility of the solid.


(a) pH value will therefore influence the solubility in a range where one of the ions has significant Brnsted acid or base properties.

The anion is the conjugate base of the weak acid solubility will increase at low pH.

MxOy, MCO3 ， MS,

M(OH)n.—— dissolve in acid solution, and conversely such solids may be precipitated from a solution containing a metal ions as the pH is increased.

Amphoteric:

(b) The conjugate acid of the anion is volatile, or decomposes to form a gas.

MCO3 & MS

(c) Any ligand that complexes with metal ion will also increase solubility.


E5 Electrode Potentials

Standard potentialsStandard potentials

Electrochemical cell:

Two Electrodes: anode (reduction reaction ) and cathode (oxidation reaction)

half reaction:

Zn + Cu2+ = Cu + Zn2+ -- cell reaction

(+) Cu2+ + 2e Cu ----half reaction

(-) Zn - 2e Zn2+

/CuCu，/ZnZn 22

还原型 e 氧化型 Z

)( Cu )L(1.0molCu )L(1.0mol Zn Zn )( 1212 ‖


书写原电池符号的规则： ①负极“ -” 在左边，正极“ +” 在右边，盐桥用“‖”表示。

③纯液体、固体和气体写在惰性电极一边用“ ,” 分开。

②半电池中两相界面用“ ”分开，同相不同物种用“ ,” 分开，溶液、气体要注明cB ， pB 。


113

212

L2.0mol2ClL0.1mol2Fe

101325PaClL1.0mol2Fe

例：将下列反应设计成原电池并以原电池符号表示。

‖

)(Pt ， 101325PaCl L2.0molCl

L0.1molFe , L1.0molFe Pt )(

21

1312

)(aq2Cl 2e)g(Cl 极正 2

)(aqFe e)(aqFe 极负 32


In a electrochemical cell, a redox reaction occurs in two halves.

Electrons are liberated by the oxidation half reaction at one electrode and pass through an electrical circuit to another electrode where they are used for the reduction.

Cell potential: is the potential difference between the two electrodes required to balance the thermodynamic tendency for reaction, so that the cell is in equilibrium and no electrical current flows.

Go = - nFE = - RT lnK (F---- the Farady constant, 96500 cmol-1, n--- the number of moles of electrons passed per mole of reaction)

--------- measurable

Electrode potential: ----- not measurable

Standard electrode potential.:

The potential of standard hydrogen electrode

is defined as zero at standard state.

([H+] = 1moldm-3, PH2= 100 kPa)

H+ + e ½ H2


Only differences in electrode potential are significant, the absolute values having no meaning except in comparison with H+ / H2 potentials.


Direction of reactionDirection of reaction

Comparing two couples Ox-Red, a more positive means that the corresponding species Ox is a stronger oxidizing agent.

0 (electrode potential)

Ox1 + ne Red 1 big

Ox2 + ne Red 2 samll

Ox1 + Red 2 Red 1 + Ox2

the positive reaction-------- very easy

the reverse reaction--------- hard

Go = - nFE = - RT lnK (K ---equilibrium constant)

000

ln

E

RT

EnFK

　　例：判断在酸性溶液中 H2O2 与 Fe2+ 混合时，能否发生氧化还原反应？若能反应，写出反应方程式。

0.2V V994.0 0.769V1.763V

Fe OH 222

发生的反应：与

解：

)aq(OH 2e)aq(2H)g(O 222 V6945.0

)l(O2H 2e)aq(2H)aq(OH 222 V763.1

)aq(Fe e)aq(Fe 23 0.769V

)s(Fe 2e)aq(Fe2 0.4089V

)l(O2H)aq(Fe2)aq(2H )aq(Fe2)aq(OH 232

22

)Fe/Fe( )OH/OH( 23222MF


Non-standard conditionsNon-standard conditions

Under nonstandard conditions the electrode potential of a couple can be calculated from the Nernst equation:

][Re

][lg

0592.0

][Re

][ln

][Re

][lg

0592.0

][Re

][ln

00

00

d

Ox

nE

d

Ox

nF

RTEE

d

Ox

nd

Ox

nF

RT

/MnMnO 24

O4HMn 5e8HMnO 22

4

4

)}Mn({

)}H()}{MnO({lg

5

V0592.0)/MnMnO( 2

82

4

c

cc

(1) When a reaction involves H+ or OH- ions, these must be included in the Nernst equation to predict the pH dependence of the couple.

)/ClClO( 3A

?)/ClClO( 时 L10.0mol)H( 31

Ec ，

L1.0mol)Cl()ClO( 13

cc ，求：当

② 介质的酸碱性V45.1)/ClClO( 3A

E已知例：

3

)}Cl({

)}H()}{ClO({lg

6

0.0592V)/ClClO(

6

3A

c

cc

)l(O3H)aq(Cl 6e)aq(6H)aq(ClO 解：23

V51.10.10lg6

V0592.0 6 1.45V

，

Ag

1L1.0mol)Cl( c

Ag

③沉淀的生成对电极电势的影响

)108.1)AgCl(( ?Ag)/(AgCl

?Ag)/(Ag Lmol0.1)Cl(

s AgClNaCl Ag

AgV799.0Ag)/(Ag

10sp

1

Kc

并求时，当会产生加入电池中组成的半和

，若在已知例：

0.222V108.1lgV0592.00.799V 10

)Ag/Ag(

)aq(Cl)aq(Ag (s) AgCl 解：

Ag(s) e)aq(Ag

(AgCl))}Cl( )}{Ag({ spKcc

(AgCl))Ag( , Lmol0.1)Cl( sp1 时若 Kcc

)}Ag({ lgV0592.0)Ag/Ag( c

AgCl)(lgV0592.0)Ag/Ag( sp K


(2) Complex formation:

In general, any ligand that complexes more strongly with the higher oxidation state will reduce the potential.

Conversely, the potential increases if the lower oxidation state is more strongly complexed.

?)/CuCu(L1.0mol))Cu(NH( 21243 时，c

,L1.0mol)NH( 中，加入氨池 13 水，当 c

Cu

氨水2Cu

12433 Lmol0.1))Cu(NH()NH( cc

，0.3394V)/CuCu( 2 ：已知例电半 Cu/Cu 。在1030.2))Cu(NH( 2122

43f K

?)Cu/)Cu(NH( 243 并求

解：

时 Lmol01))Cu(NH()NH( 12433

当 .c c

)aq()Cu(NH )aq(4NH)aq(Cu 2433

2 2

)}NH()}{Cu({

)})Cu(NH({f4

32

43 K

cc

c

))Cu(NH(

1)Cu( 2

43f

2

K

c

0.0265V1030.2

1lg

2

0592.00.3394V 12

)Cu /Cu( 2

)}Cu(lg{2

0592.0)Cu /Cu( 22

c

)s(Cu 2e)aq(Cu 2

))Cu(NH(

1lg

2

0592.0)Cu /Cu( 2

43f

2

K

, Lmol0.1))Cu(NH()(NH 12433 时当 cc

)aq(4NH)s(Cu 2e)aq()Cu(NH 3243

V0265.0/Cu)Cu(Cu)/)(Cu(NH 2243

Cu)/)(Cu(NH 243即

/Cu)Cu( Cu)/)(Cu(NH 2243

))Cu(NH (

1lg

2

V0592.0/Cu)Cu( 2

43f

2

K


Diagrammatic Representation:

(a) Latimer diagram: shows the standard electrode potentials associated with the different oxidation states of an element. （元素电极电势图）

(b) Frost or oxidation state diagram

Diagrammatic Representation:Diagrammatic Representation:

1. 判断歧化反应能否发生

0 V 0.3573 0.1607V0.5180V

Cu 0.5180V

Cu 0.1607V Cu2

0.3394V

V/

发生歧化反应；左右

发生歧化逆反应。左右

Cu/(Cu Cu) / Cu( 2 )

)aq(Cu )s(Cu )aq(2Cu 2

+）

2. 计算电对的电极电势（ Hess’ Law ）

FZG Z B eA 11m(1)r11

FZG Z C e B 22m(2)r22

FZG Z D e C 33m(3)r33

FZG Z xxxxx DeA )m(r

(Z1) (Z2) (Z3)A B C D

E1 E2 E3

(Zx)Ex

ZZZZ x 321

GGGG x m(3)rm(2)rm(1)r)m(r

FZFZFZFZ x 332211

ZZZx 332211

Z xx ZZZ 332211


VMnMn

MnMnMnMnMnMn

GGG

29.0)/(

)/(2)/()/(330

2023030

03

02

01

(2)判断哪些物种可以歧化 ?

例题：已知 Br 的元素电势图如下

Br

1.0774Br

0.4556BrO

BrO 23

0.6126

E2

E3

E1

(1) 321 。和、求

(3) Br2(l) 和 NaOH(aq) 混合最稳定的产物是什么？写出反应方程式并求其。K

解：(1)

Br

1.0774Br

0.4556BrO

BrO 23

0.6126

E2

E3

E1

V5357.04

1)V1.077410.45566(0.6126 1

E

V7665.02

V)10774.114556.0( 2

E

V5196.05

V)10774.166126.0(3

E

(2)

可以歧化。、 BrOBr2

0.5196

0.7665

Br

1.0774Br

0.4556BrO

BrO 23

0.5357

V5578.00.5196V0774V.1

2 )aq(6OH)l(3Br

3 Br 和 BrO 是。2 混合最稳定的产物NaOH 与(l)Br所以

能歧化 BrO (3) ，不稳定，因为

23 O(aq)3H)aq(BrO)aq(5Br

MF 11.470.0592V

0.5578V5

0.0592Vlg ZE

K

471029.1 K

232MF /Br(BrO/Br(Br EEE ) )


Frost or oxidation state diagram

The potential 0 in volts between any two oxidation states is equal to the slope of the line between the points in this diagram. ]

Steep positive slopes show strong oxidizing agents, steep negative slopes strong reducing agents.


Application:

(a) Display the comparative redox properties of elements.

(b) Provide a visual guide to when disproportionation of a species is expected.

pH=0,

Mn3+ (aq) + 2 H2O MnO2 (s) + Mn2+ (aq) + 4 H+ (aq)

Mn(V) Mn(VI)


ff

Kinetic limitationsKinetic limitations

Electrode potentials are thermodynamic quantities and show nothing about how fast a redox reaction can take place.

Simple electron transfer reactions are expected to be rapid, but redox reactions where covalent bonds are made or broken may be much slower.

MnO4-/MnO2 + H2O/ O2

Kinetic problems can also affect redox reactions at electrodes when covalent substances are involved.

作业：

6, 7, 9, 11, 12.

Section FSection FINORGANIC CHEMISTRY

Sec. F Introduction to Non-metals

Covalent chemistryCovalent chemistry

(1) Hydrogen and boron stand out in their chemistry.

Hydrogen —— one covalent bond.

Boron —— electron deficiency

(2) In the other elements, valence states depend on the electron configuration and on the possibility of octet expansion which occurs in period 3 onwards.

octet rules—— octet expansion

(3) Multiple bonds are common in period 2 but are often replaced by polymerized structures with heavier elements.

C, N ， O

Section FSection FINORGANIC CHEMISTRY

Ionic chemistry Ionic chemistry

Simple anionic chemistry is limited to oxygen and the halogens, although polyanions and polycations can be formed by many elements.

Acid-base chemistry Acid-base chemistry

Many halides and oxides are Lewis acids; BCl3 SO3

The compounds with lone-pairs are Lewis bases; NH3 ， H2O

Brnsted acidity is possible in hydrides and oxoacids.

Halide complexes can be also formed by ion transfer.

Redox ChemistryRedox Chemistry

The oxidizing power of elements and their oxides increases with group number. Vertical trends show an alternation in the stability of the highest oxidation states.

O, F, Br, Cl.

Section GSection GINORGANIC CHEMISTRY

Sec. G Introduction to Non-transition metals

ScopeScope

Non-transition metals includes groups 1 to 2 of the s-block elements, group 12 and p-block elements in lower periods.

(a) Aluminum and the elements of groups 1 and 2 are classed as pre-transition metals,

—— typical metals, very electropositive in character and almost invariably found oxidation states expected for ions in a noble-gas configuration.

—— found widely in silicate minerals

(b) The remaining ones (metallic elements from periods 4-6 in groups following the transition series) as post-transition metals.

—— less electropositive than the pre-transition metals and are typically found nature as sulfides rather than silicates.

in solution, post-transition metals less ionic in character than corresponding compounds of pre-transition metals; In solution, post-transition metals form stronger complexes than pre-transition metals.

lower oxidation states. (the sixth period, Tl+, Pb2+, Bi3+)

Section GSection GINORGANIC CHEMISTRY

Positive ionsPositive ions

Group trendsGroup trends

Formation of compounds with positive ions depends on a balance between ionization energies and lattice or solvation energies. Post-transition metals have higher ionization energies and are less electropositive than pre-transition metals.

Trends down groups 1 and 2 are dominated by increasing ionic size. In later groups the structural and bonding are less regular and there is an increased tendency to lower oxidation states, especially in period 6.

Non － cationic chemistryNon － cationic chemistry

Many of the elements can form anionic species. Compounds with covalent bonding are also known: these include organometallic compounds, and compounds containing metal-metal bonds( especially with post-transition metals).

Section HSection HINORGANIC CHEMISTRY

Sec. H Introduction to transition-metals

ScopeScope

Transition elements form groups 3-11 in the d block.

They have distinct chemical characteristic resulting from the progressive filling of the d shells. These includes the occurrence of variable oxidation states, and compounds with structures and physical properties resulting from partially filled d orbitals.

Typical transition metal characteristics include:

(a) the possibility of variable oxidation states;

(b) Compounds with spectroscopic, magnetic or structural features resulting from partially occupied d orbital.

(c) An extensive range of complexes and organometallic compounds including ones with very low oxidation state (zero or even negative)

(d) Useful catalytic properties shown by metals and by solid or molecular compounds


Vertical trendsVertical trends

Elements of the 3d series are chemically very different from those in the 4d and 5d series, showing weaker metallic and covalent bonding, stronger oxidizing properties in high oxidation states and the occurrence of many more compounds with unpaired electrons.

Horizontal trendsHorizontal trends

Electropositive character declines towards the right of each series. Elements become less reactive and their compounds show a tendency towards softer behavior. Later elements in the 4d and 5d series are relatively more inert.


Electron configurationElectron configuration

Neutral atoms have both s and d valence electrons, but in chemically important states are often regarded as having purely d configuration.

In the 3d series most atoms have the configuration (3d)n(4s)2, where n increase from one to 10; chromium and copper are however exceptions with (3d)5(4s)1 and (3d)10(4s)1 respectively. The configurations depend on a balance of two factors:

(a) 3d orbitals are progressively stabilized relative to 4s across the series;

(b) Repulsion between electrons is large in the small 3d orbitals, and so minimum energy in the neutral atom is achieved in spite of (a) by putting one or two electrons in the 4s orbitals.

In forming positive ions, 4s electrons are always removed first.

For M2+ ions and ones of higher charge, outer shell electrons are left only in the 3d series.

Section ISection IINORGANIC CHEMISTRY

Sec. I Lanthanum and lanthanides

The elementsThe elements

(1) The elements (sometimes called rare earths) are found together in nature and are electropositive metals. —— with the filling of seven orbitals of the 4f shells.

Ln (lanthanide, 镧系元素 ) ——La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

RE(rare earth) —— La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu ＋ Sc, Y

(2) Chemistry is dominated by ＋ 3 state with ions in (4f)n configuration, and is similar for all elements.

Ligand field and chemical bonding effects associated with incomplete 4f orbitals are very small and hardly detectable in chemical trends.

The chemistry of all Ln3+ ions is very similar and differentiated only by the gradual contraction in radius associated with increasing nuclear charge.

The lanthanide contraction( 镧系收缩效应 ) ：

+2, +4


Oxidation States ＋ 3Oxidation States ＋ 3

A wide range of +3 compounds is formed as well as aqua ions. The ionic radius decrease gradually across the series, leading to changes in solid structures, and an increase in stability of complexes in solution. Organometallic compounds are more ionic than in the d block.

Ln3+ : halides, LnX3, oxides, Ln2O3

this relatively large size for 3+ ions is associated with correspondingly high coordination numbers in solid compounds.

LnF3 （ CN ＝ 9 ） , Ln2O3. （ CN ＝ 7 ） for earlier elements.

For later Ln elements the decrease in radius leads to changes in structure with reduction in coordination.

The aqua Ln3+ ions show slightly acidity, which increases from La to Lu as the radius decrease but still much less than for Al3+.

Strong complexes are formed with hard oxygen donor ligands, and especially chelating ones such as EDTA, or － diketones. Complexes strengths generally increase across the series as the radius decreases, and this may be used to separate a mixture of Ln3+ ions.


Other Oxidation StatesOther Oxidation States

Sm, Eu and Yb form many compounds in the +2 oxidation state.

With the other elements, compounds in this state are formed only with large anions and are often metallic.

LnS, LnI2

Ce, and to a lesser extent Pr, and Tb show the ＋ 4 state.

CeO2, Pr6O11, Tb4O7 ( Tb(IV) )

Documents

INORGANIC CHEMISTRY