Transcript 幻灯片 1

INORGANIC 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]
Sept. 2005
For Science Class 2
INORGANIC
CHEMISTRY
Contents
Contents
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
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Contents
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
INORGANIC
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Introduction
Introduction
1. Emergence and Development of Chemistry
Pride of Chemist
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
Set up theoretic
system of
modern
chemistry and
industry
INORGANIC
CHEMISTRY
Introduction
2. Brach of Chemistry
CHEMISTRY
Inorganic Chemistry
Coordinary Chemistry
Organic Chemistry
Polymer Chemistry
Analytical Chemistry
Organometallic Chemistry
Electro-Chemistry
Physical Chemistry
Structural Chemistry
……
Biochemistry
Geochemistry
Spacechemistry
Material Chemistry
Nano-Chemistry
etc.
INORGANIC
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Introduction
3. Ways of Learning Chemistry
Study hard
Learn cleverly
Do it yourself
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Introduction
4. Some basic conception in Chemistry
Nucleus reaction
Protons + Neutrons
= Nucleus
Breaking
Atoms
Molecules
Forming
Get or Lost
Electrons
Chemical reaction
INORGANIC
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Atom
Introduction
Nucleus
+
Electrons
Removal (addition) of
Protons or Neutrons
Removal (addition)
of electrons
Isotopes, New Element
Ions, Free radical
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
INORGANIC
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Introduction
Gas
Species discussed
in this course
Liquid, Solution
Solid
n-----mol
Molar mass-----gmol-1
Molar volume---- m3mol-1 lmol-1, dm3mol-1
SI units:
Pressure, P-----Pa, KPa, atm
Temperature, T-----K, ℃
Energy----- J, or kJ, cal, kcal
Gas
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1. Ideal-gas Law
No intermolecular interaction
Ideal gas
The volume of the molecules in gas
would be considered as zero.
GAS
Real gas
This content would be introduced in Physics and Physical Chemistry .
(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 (gmol-1)
PV= n R T
d
RT

P
M
(i) If SI units would be used, R = 8.314 Jmol-1K-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
INORGANIC
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Introduction
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。
PV=nRT
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？(gdm-3)
d
RT

P
M
d ＝ 1.16 gdm-3
M ＝28×79％＋32×21％＝ 28.8
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Introduction
(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.
If V=0 and  T=0,
PT   Pi  P1  P2  P3  .... 
nT   ni  n1  n2  n3  ....
RT
V
n
PiV  ni RT
PT V  nT RT
xi 
ni
nT
 xi  1
volumefraction 
Mole fraction
Vi
VT
i
partial pressure
of one kind of
component
Gas
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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。
PT   Pi  P1  P2  P3  .... 
RT
V
n
i
PiV  ni RT
PN2= 175mmHg, PO2=240mmHg, PH2=134mmHg， PT = 549mmHg
(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)
nB
pB 
p  xB p
n
gas
INORGANIC
CHEMISTRY
(3) Diffusion Law of gas
u1
d1
M2


u2
d2
M1
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
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Section A
A1
Section A1
Atomic structure
THE NUCLEAR ATOM
Electron 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
Section A1
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Nuclear 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
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Section A1
Isotopes
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;
Tin (Sn): no fewer than 10.
12C, 13C, 14C;
etc.
Relative atomic mass (RAM) is equal to molar mass of element
RAM of an element:
RAM   ( f i  Mri )
fi: the abundance of the isotopes in the nature;
Mri: the relative atomic number of the isotopes
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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
Nuclides (核素):
Isotopes (同位素):
Element (元素):
Atom (原子):
isotone (同中素):
Isobar (同量异位素):
Section A1
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Section A1
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:
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A2
Section A2
ATOMIC ORBITALS
Wavefunctions
The quantum theory is necessary to describe electrons. It predicts discrete
allowed energy levels and wavefunction, which give probability distributions for
electrons. Wave functions for elections in atoms are called atomic orbitals.
Quantum mechanics
Quantized energy
Probability distributions
Schrödinger’s wave equation
Atomic orbitals
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Quantum numbers
and nomenclature
A3
Section A2
A3
Atomic orbitals are labeled by three quantum numbers n,l
and m. Orbitals are called s,p,d or f according to the value of
1; there
are respectively one, three,five and seven different
THE NUCLEAR
ATOM
possible m values for these orbitals.
Electron and Nuclei
Principal
Quantumofnumbers
An atom consists
a very small positively charged nucleus,
n=1, 2, 3…..
surrounded
by negative electrons held by electrostatic
Angular momentum (or azimuthal) Quantum numbers
attraction. The motion of electrons changes when chemical
l=0, 1, 2, …, (n-1)
bonds are formed, nuclei being unaltered.
Magnetic Quantum numbers
Nuclear structure
m=0,
±1,
±2, …,
±(n-1)protons and uncharged neutrons.
Nuclei
contain
positive
Spin
Quantum
The number
of numbers
protons is the atomic number (Z) of an
ms=+1/2, -1/2
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.
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Angular functions:
‘shapes’
A3
Section A2
A3
s orbitals are spherical. p orbitals have two directional lobes,
which can point in three possible directions. d and f oritals
have
coorespondingly greater numbers of directional lobes.
THE NUCLEAR
ATOM
Electron and Nuclei
Nuclear structure
Radial
An atomwavefunction
consists of a very small positively charged nucleus,
surrounded by negative electrons held by electrostatic
Angular
attraction.wavefunction
The motion of electrons changes when chemical
bonds are formed, nuclei being unaltered.
Polar diagrams
Nuclei contain positive protons and uncharged neutrons.
The number of protons is the atomic number (Z) of an
Boundary surface
element. The attractive strong interaction between protons
and neutrons is opposed by electrostatic repulsion between
Nodal plane
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
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Section A2
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Radial distributions
A3
Section A2
A3
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.
THE NUCLEAR
ATOM
Electron and Nuclei
Radial
probability
distributions
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.
Nuclear structure
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.
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Section A2
INORGANIC
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Energies in hydrogen
A3
Section A2
A3
The allowed energies in hydrogen depend on n only. They
can be compared with experimental line spectra and the
ionization
THE NUCLEAR
ATOMenergy.
En= -R/n2
An
atom consists
of a very
smalleVpositively
Rydberg
constant:
R=13.595
per atom charged nucleus,
Electron and Nuclei
surrounded
heldlimit
by electrostatic
When
n=∞, Eby
which is electrons
the ionization
∞=0negative
attraction.
The
motion Eof∞-E
electrons
changes when chemical
Ionization
energy:
1=R
bonds
are formed,
nuclei
being
unaltered.
Degenerate:
Orbitals
with
the same
n but different values of l
and m have the same energy.
Nuclear structure
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.
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Hydrogenic ions
A3
Section A2
Increasing nuclear charge in a one-electron ion leads to
contraction of the orbital and an increase in binding energy
of the
electron.
THE NUCLEAR
ATOM
For example: He+, Li2+
Electron and Nuclei
<r>=
n2aconsists
An atom
of a very small positively charged nucleus,
0/Z
Where
a0 is Bohr
radius, theelectrons
average radius
of aelectrostatic
1S orbital in
surrounded
by negative
held by
hydrogen.
attraction. The motion of electrons changes when chemical
Z: atomic number
bonds are formed, nuclei being unaltered.
En= -Z2R/n2
Nuclear structure
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.
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A3
Section A3
MANY-ELECTRON ATOMS
The orbital
approximation
Putting electrons into orbitals similar to those in the
hydrogen atom gives a useful way of approximating the
wavefunction of a many-electron atom. The electron
configuration specifies the occupancy of orbitals, each of
which has an associated energy.
Orbital approximation
Electron configuration
Orbital energy
INORGANIC
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Electron spin
A3
Section A3
Electrons have an intrinsic rotation called spin, which may
point in only two possible directions, specified by a quantum
number
ms. Two electrons in the same orbital with opposite
THE NUCLEAR
ATOM
Electron and Nuclei
spin are paired. Unpaired electrons give rise to
paramagnetism.
An atom consists of a very small positively charged nucleus,
Spin
Quantum
numbers electrons held by electrostatic
surrounded
by negative
m
s=+1/2, -1/2
attraction.
The motion of electrons changes when chemical
bonds are formed, nuclei being unaltered.
Nuclear structure
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.
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Pauli exclusion
principle
A3
Section A3
When the spin quantum number m, is included no two
electrons in an atom may have the same set of quantum
numbers.
THE NUCLEAR
ATOMThus a maximum of two electrons can occupy any
orbital.
Electron and Nuclei
Nuclear structure
An atom consists of a very small positively charged nucleus,
Li
: (1s)2 (2s)
surrounded
by1 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.
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Effective nuclear
charge
A3
Section A3
The electrostatic repulsion between electrons weakens their
binding in an atom, this is known as screening or shielding.
TheATOM
combined effect of attraction to the nucleus and
THE NUCLEAR
Electron and Nuclei
Nuclear structure
repulsion from other electrons is incorporated into an
effective
nuclear charge.
An atom consists
of a very small positively charged nucleus,
surrounded by negative electrons held by electrostatic
2R/n2
E
attraction.
motion of electrons changes when chemical
n= -Zeff The
bonds are formed, nuclei being unaltered.
σ = Z-Zeff
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.
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Screening and
penetration
A3
Section A3
An orbital is screened more effectively if its radial
distribution does not penetrate those of other electrons. For a
given
n, s orbitals are least screened and have the lowest
THE NUCLEAR
ATOM
energy; p, d,... orbitals have successively higher energy.
Electron and Nuclei
Nuclear structure
An atom consists of a very small positively charged nucleus,
Degenerate
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons changes when chemical
Penetrating
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.
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Hund’s first rule
A3
Section A3
When filling orbitals with l>0, the lowest energy state is
formed by putting electrons so far as possible in orbitals with
different
m values, and with parallel spin.
THE NUCLEAR
ATOM
Electron and Nuclei
Nuclear structure
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.
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A4
History
Section A4
THE PERIODIC TABLE
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
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Section A4
Building 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.
A3
THE NUCLEAR ATOM
Elements in the same group have the same configurati0n of outer electrons. The
way different orbitals are filled is controlled by their energies (and hence their
An atom consists of a very small positively charged nucleus,
Electron and Nuclei
different screening by other electrons) and by the Pauli exclusion principle.
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons changes when chemical
According to the aufbau principle, the ground-state electron configuration of
bonds are formed, nuclei being unaltered.
an atom can be found by putting electrons in orbitals, starting with that of
Nuclei
contain positive
protons
and uncharged neutrons.
lowest energy
and moving
progressively
to higher
energy.
Nuclear
structure
（1） the lowest energyThe number of protons is the atomic number (Z) of an
（2）Pauling exclusionelement.
principleThe attractive strong interaction between protons
and neutrons is opposed by electrostatic repulsion between
（3）Hund’s rule
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
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A3
Section A4
Penetration effect
THE NUCLEAR ATOM
Electron and Nuclei
Nuclear structure
Screening effect.
Intervening
of
An atom consists of a very small positively charged
nucleus,
energy level
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.
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Electron configuration:
The outmost electron configuration:
Valence electron configuration:
Section A4
INORGANIC
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Block structure
A3
Section A4
The 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.
THE NUCLEAR ATOM
Electron and Nuclei
Nuclear structure
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.
INORGANIC
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Section A4
Group numbers and
names
Modern group numbering runs from 1 to 18, with the j bl0cks being subsumed into
groupTHE
3. Older
(and contradictory)
A3
NUCLEAR
ATOM numbering systems are still found. Some groups
of elements are conventionally given names, the most commonly used being alkali
metals (group l), alkaline
halogens
(17) small
and noble
gases charged
(18). nucleus,
Anearths
atom (2),
consists
of a very
positively
Electron and Nuclei
Alkali metal :
Alkaline metal:
Transition metal:
Nuclear
structure
Coinage
metal:
Chalcogen:
Halogen:
Noble gas:
surrounded by negative electrons held by electrostatic
attraction.rare
Theearth:
motion of electrons changes when chemical
bonds arelanthanide:
formed, nuclei being unaltered.
actinide:
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.
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Section A5
A5 TRENDS IN ATOMIC PROPERTIES
Energy and sizes
Trends in orbital energy and size reflect changes in the
principal quantum number and effective nuclear charge.
They are seen experimentally in trends in ionization energy
(IE) and apparent 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
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Horizontal trends
A3
Section A5
Increasing nuclear charge causes a general increase of IE and a
decrease of radius across any period. Breaks in the IE bend are
found following the complete or half filling of any set of orbitals.
THE NUCLEAR ATOM
Electron and Nuclei
Nuclear structure
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.
INORGANIC
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Section A5
Vertical trends
A3
THEANUCLEAR
ATOM
general increase
of radius and decrease in IE down most
groups is dominated by the increasing principal quantum
An atom
consistsEffective
of a very nuclear
small positively
nucleus,
Electron and number
Nuclei of outer
orbitals.
charge charged
also
by to
negative
electrons
heldIEbytrends.
electrostatic
increases, andsurrounded
can give rise
irregularities
in the
attraction. The motion of electrons changes when chemical
bonds are formed, nuclei being unaltered.
Nuclear structure
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.
INORGANIC
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Section A5
States of ionization
A3
IEs for positive
ions always increase with the charge.
THE NUCLEAR
ATOM
M2+, M3+, …., second, third,….
consists of a very small positively charged nucleus,
Electron and NucleiElementsAn atom
IE
IE
IE
1
2
3
surrounded by negative electrons held by electrostatic
Be (1s22s2)
9.3ev 18.2ev
154ev
attraction. The motion of electrons changes when chemical
bonds 6.1ev
are formed,
nuclei51ev
being unaltered.
Ca (Ar)4s2
11.9ev
Nuclear structure
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.
INORGANIC
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Section A5
Electron affinity of an atom: defined as the ionization energy of the negative
ion, thus the energy input in the process:
M-  M + e-
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Section B
Section B1
Introduction to inorganic substance
B1 ELECTRONEGATIVITY AND BOND TYPE
Definition
Electronegativity is the power of an atom to attract electrons to itself in a
chemical bond. Elements of low electronegativity are called electropositive.
Pauling Electronegativity: based on bond energies
Mulliken Electronegativity: the average of the first ionization energy and the
electron affinity of an atom.
Allred-Rochow Electronegativity: proportional to Zeff/r2.
INORGANIC
CHEMISTRY
Section B1
Different numerical estimates agree on qualitative trends:
Electronegativity increases from left to right along a period, and generally
decreases groups in the periodic table.
INORGANIC
CHEMISTRY
Section A3
B1
The bonding triangle
Electronegativity elements form meta1lic solids. Electronegative elements form
or polymericATOM
solids with covalent bonds. Elements of very different
A3molecules
THE NUCLEAR
electronegativity combine to form solids that can be described by the ionic model.
An atom consists of a very small positively charged nucleus,
Delocalization of electrons:
surrounded by negative electrons held by electrostatic
Covalent compounds:
attraction. The motion of electrons changes when chemical
Giant covalent lattices
(Polymeric
solids):
bonds
are formed,
nuclei being unaltered.
Localized bonds:
Nuclei contain positive protons and uncharged neutrons.
Nuclear
structure
Cation：
The number of protons is the atomic number (Z) of an
Anion：
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.
Electron and Nuclei
INORGANIC
CHEMISTRY
Section B1
INORGANIC
CHEMISTRY
Bond polarity
A3
Section A3
B1
The polarity of a bond arises from the unequal sharing of electrons
between atoms with different electronegativities. There is no sharp
dividing
line between polar covalent and ionic substances.
THE NUCLEAR
ATOM
Electron and Nuclei
Nuclear structure
Homopolar
(A covalent between
two atoms
of the same ELs)
An atom consists of a very
small positively
charged
nucleus,
Bond
polarity
surrounded
by negative electrons held by electrostatic
Heteropolar
(A covalent when
between two
atoms of the different ELs
attraction. The motion of
electrons changes
chemical
bonds are formed, nuclei being unaltered.
Electric
dipolepositive
momentprotons and uncharged neutrons.
Nuclei contain
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.
INORGANIC
CHEMISTRY
B2
Section B2
CHEMICAL PERIODICITY
Introduction
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.
INORGANIC
CHEMISTRY
Section A3
B2
Metallic and nonmetallic elements
A3
THE NUCLEAR ATOM
Metallic elements
are
electropositive,
form
electrically
An
atom
consists
of
a
very
small
positively
charged nucleus,
Electron and Nuclei
conducting solids
and have
Non-metallic
surrounded
by cationic
negative chemistry.
electrons held
by electrostatic
elements, found
in the upper
right-hand
portion ofchanges
the periodic
attraction.
The motion
of electrons
when chemical
table, have predominantly
cova1ent
andbeing
anionic
chemistry. The
bonds are formed,
nuclei
unaltered.
chemical trend is continuous and elements on the borderline
Nuclei
contain positive protons and uncharged neutrons.
show intermediate
characteristics.
Nuclear structure
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.
INORGANIC
CHEMISTRY
Horizontal trends
A3
Section A3
B2
Moving to the right in the periodic table, bonding character
changes as electronegativity increases. The increasing number
of ATOM
electrons in the valence shell also gives rise to changes in
THE NUCLEAR
Electron and Nuclei
Nuclear structure
the stoichiometry and structure of compounds. Similar trends
operate
the d block.
An atominconsists
of a very small positively charged nucleus,
surrounded by negative electrons held by electrostatic
Closed
shell
attraction.
The motion of electrons changes when chemical
Lattice
energy
bonds are
formed, nuclei being unaltered.
Octet rule
Nuclei contain positive protons and uncharged neutrons.
Non-bonding electrons
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.
INORGANIC
CHEMISTRY
Section B3
B3 STABILITY AND REACTIVITY
Introduction
Stability and reactivity can be controlled by thermodynamic
factors (depending only on the initial and final states and not
on the reaction pathway) or kinetic ones (very dependent on
the reaction pathway). Both factors depend on the conditions,
and on the possibility of different routes to decomposition or
reaction.
Thermodynamic stable：stable or unstable
Kinetic stable ：reactive or unreactive
INORGANIC
CHEMISTRY
Section B3
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
INORGANIC
CHEMISTRY
Section B3
(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.
U   En
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.
INORGANIC
CHEMISTRY
Section B3
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
Hess’ Law
Section B3
Enthalpy 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
A3 THE NUCLEAR
ATOM
values for reactions can be calculated from tabulated enthalpies of
formation.
An atom consists of a very small positively charged nucleus,
Electron and Nuclei
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons changes when chemical
bonds are formed, nuclei being unaltered.
Nuclear structure
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.
INORGANIC
CHEMISTRY
Section B3
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
A3 THE
NUCLEAR
substance
at the standard
state. ATOM
C (graphite) + O2(g)  CO2 (g) Hf0 (CO2)
An atom consists of a very small positively charged nucleus,
Electron and
C (diamond)
+ Nuclei
O2(g)  CO2 (g)  Hf0 (CO2)
surrounded by negative electrons held by electrostatic
0
By definition, Hf is zero for any elements in its stable (standard) states.
attraction. The motion of electrons changes when chemical
Application: (Hess’ Law) bonds are formed, nuclei being unaltered.
The standard enthalpy change (H0) in any reaction would be calculated by using
Nuclei contain positive protons and uncharged neutrons.
structure
theNuclear
standard
enthalpy of formation.
The number of protons is the atomic number (Z) of an
For example, for this reaction
element. The attractive strong interaction between protons
a A (state) + b B (state) and
c C neutrons
(state) + is
dD
(state) by
(thermochemical
equation)，
opposed
electrostatic repulsion
between
protons.
Repulsion
dominates
H 0   i  H  ( formation
)
k  H  (reac
tan ts) as Z increases and there is
f
f
only
a limited number
of stable elements.



  [c  H f (C ) d  H
f ( D)]   [ a  H f ( A) b  H f ( B )]
INORGANIC
CHEMISTRY
Section B3
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 , kJmol-1) of these compound
are listed in Table.
NO2 (g) 33.18 kJmol-1
H2O(l) -285.83 kJmol-1
HNO3(l) -173.23 kJmol-1
NO(g) 90.25 kJmol-1
Applying Hess’ law:
rHm = ifHm (product)- ifHm (reactant)
rHm = [ 2  (-173.23) + 90.25] – [3  33.18 + (-285.83)]
= -69.92 kJmol-1.
INORGANIC
CHEMISTRY
Section B3
Entropy
1. Entropy(熵) is a measure of molecular disorder or more precisely ‘the number of
microscopic arrangements of energy possible in a macroscopic sample’.
A3
THE NUCLEAR
(1) Entropy
increases withATOM
rise in temperature and depends strongly on the state.
(2) S(gas) >>S (liquid) > S (solid)
Entropy
change (△S)
are invariably
for reactions
generate
gas
An atom
consists ofpositive
a very small
positively that
charged
nucleus,
Electron
and Nuclei
molecules.
surrounded by negative electrons held by electrostatic
(3) molecule mass or itsattraction.
structure: The motion of electrons changes when chemical
2. Entropy:——state function
bonds are formed, nuclei being unaltered.
Applying the Hess’ Law
contain positive protons and uncharged neutrons.
△Sreaction
= ∑S(products) Nuclei
- ∑S(reactants)
Nuclear
structure
number change
of protons
is the
(Z)direction
of an for
For an isolated system, The
the entropy
would
be atomic
used tonumber
judge the
spontaneous reaction. element. The attractive strong interaction between protons
If △S >0, the forward reaction would be spontaneous.
and neutrons is opposed by electrostatic repulsion between
= 0, the reaction would keep the balance state.
as Z increases and there is
< 0, the backwardprotons.
reactionRepulsion
would bedominates
spontaneous.
a limited
of stable elements.
———— theonly
second
law ofnumber
thermodynamics
INORGANIC
CHEMISTRY
Section B3
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,，S0m) 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)
 r H m   i  H f ( products)   k  H f (reac tan ts)
  [c  H f (C ) d  H f ( D)]   [a  H f ( A) b  H f ( B)]
INORGANIC
CHEMISTRY
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.
Section B3
INORGANIC
CHEMISTRY
Section B3
△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,
G 0   i  G f ( products)   k  G f (reac tan ts)
  [c  G f (C ) d  G f ( D)]   [a  G f ( A) b  G f ( B)]
INORGANIC
CHEMISTRY
Section B3
For example:
CaCO3 (s)  CaO(s) + CO2 (g)
(1) At the Standard state, G0 = +130kJmol-1.
However, (2) at 1273K, 100kPa, G = -25 kJmol-1
(3) at 1273K, 0.1kPa, G = -174 kJmol-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):
 r Gm (T )   r Gm (T )  RT ln J
J  (
Pi  i
) (气体反应体系)
P
or
J  (
Ci  i
) (溶液反应体系)

C
P. T
G
INORGANIC
CHEMISTRY
Section B3
(2) 吉布斯－亥姆霍兹方程
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)
 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
INORGANIC
CHEMISTRY
Section B3
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: rHm0 = -161
kJmol-1, rSm0 = -420 Jmol-1K-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.
INORGANIC
CHEMISTRY
Chemical 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
Section B3
INORGANIC
CHEMISTRY
Section B3
(2) Equilibrium constants
For a general reaction such as
a A (state) + b B (state)  c C (state) + d D (state)
c
d
[C ] [ D]
K
a
b
[ A] [ B]
(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
Section 3
The relationship between Equilibrium constants and rGm
 r Gm (T )   r Gm (T )  RT ln J
J  (
Pi  i
) (气体反应体系)

P
A3 orTHE NUCLEAR ATOM
Ci  i
) (溶液反应体系
An atom consists) of a very small positively charged nucleus,
Electron and Nuclei
C
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons changes when chemical
If at balanced state, △rG
m= 0.are formed, nuclei being unaltered.
bonds
△rG0m= －RT ln J = —RT ln K = △rH0m－T △rS0m
Nuclei)contain positive protons and uncharged neutrons.
J = K (structure
equilibrium constant
Nuclear
Theofnumber
of is
protons
thethe
atomic
number
(Z)free
of an
The equilibrium constant
reaction
relatedis to
standard
Gibbs
element.
The attractive
interaction between
energy change. Equilibrium
constants
change strong
with temperature
in a wayprotons
that depends on △H forand
the neutrons
reaction. is opposed by electrostatic repulsion between
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
J  (
INORGANIC
CHEMISTRY
Section B3
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) + H
Effect 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
Section B3
Reaction rates
(1) Average velocity
It is defined as the increment of the products or decrement of reactant
in unitNUCLEAR
time.
A3 THE
ATOM
n B
r
t
n
r
 t
An atom consists of a very small B
positively charged nucleus,
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons
B changes when chemical
bonds are formed, nuclei being unaltered.
B－－the coefficient of B
in protons
reactionand
equation.
Nuclei contain positive
uncharged neutrons.
Nuclear structure
The number of protons is the atomic number (Z) of an
(a) concentration:
element. The attractive strong interaction between protons
(b) temperature: Reaction
generally
increase
rise in temperature
andrates
neutrons
is opposed
bywith
electrostatic
repulsion between
(c) Catalysis provide alternative
reaction pathways
of lower
energy. Aand
truethere
catalyst
protons. Repulsion
dominates
as Z increases
is
by definition can be removed
after the
And so does not alter the
only unchanged
a limited number
of reaction.
stable elements.
thermodynamics or the position of equilibrium.
Electron and Nuclei
INORGANIC
CHEMISTRY
Section B3
(2) Rare equation: the relationship between the rate and the
concentration of the reactant.
a A + b B  products
A3
n [B]m
THE NUCLEAR
ATOM
Rate= k[A]
rate constant: k
consists
of amechanism
very small and
positively
charged nucleus,
Electron
and Nuclei
Orders
of reaction: An
m, atom
n. depend
on the
not necessarily
surrounded
by negative
equal to the stoichiometric
coefficients
of theelectrons
reactants.held by electrostatic
attraction. The motion of electrons changes when chemical
For example:
nuclei being unaltered.
(1) H2＋ I2 ＝＝2HI bonds
r＝k are
c(Hformed,
2) c(I2)
1
(2) H2＋ Br2 ＝＝2HBr
Nuclei contain positive protons
k c( Hand
)c 2uncharged
( Br ) neutrons.
Nuclear structure
r
1
2
2
The number of protons is the atomic
number
k2c( HBr
) (Z) of an
1

(3) H2＋ Cl2 ＝＝2HCl
element. The attractive strong interaction
c( Br2 ) between protons
and neutrons is opposed by electrostatic repulsion between
1
protons. Repulsion
dominates as Z increases and there is
2
r  kconly
( Ha2 limited
)c (Cl
)
2
number of stable elements.
INORGANIC
CHEMISTRY
Section B3
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
divided 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 = k
mol 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
Rate = k c3
mol-2 dm3 s-1
3 (third order reaction)
Another means: by designing a lots of favorable experiments.
INORGANIC
CHEMISTRY
Section B3
2 NOg   2H 2 g   N 2 g   2H 2 Og 
1073k
1 1
1
-1

/(
mol

L
s )


c
H
/(
mol

L
)


c
NO
/(
mol

L
)
试验编号
2
1
2
3
4
5
0.0060
0.0060
0.0060
0.0030
0.0015
Rate = k [c(NO)]n [c(H2)]m
n=2 , m = 1. k= ?
0.0010
0.0020
0.0040
0.0040
0.0040
7.9  10 7
3.2 10 6
1.3 10 5
6
6.4  10
3.2 10 6
INORGANIC
CHEMISTRY
Section B3
(3) Arrhenius equation——the relationship between the rate constant and the
temperature.
k  Ae

Ea
RT
K------the rate constant.
Ea-------the activation energy (J.mol-1)
Ea
ln k  ln A 
RT
T-temperature （K）
A-------the constant
Ea
lg k  lg A 
2.303  R  T
lnk (or lgk)---1/T ： linear relationship
the intercept（截距）: lnA
the slope（斜率）: －Ea / R
INORGANIC
CHEMISTRY
Section B3
Ea
ln k  ln A 
RT
(1) If these data T1,T2, and k1, k2 were given,
Ea (T1  T2 )
k1 Ea 1 1
ln

(  )

k2
R T2 T1
R
T2T1
(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.
INORGANIC
CHEMISTRY
Section B3
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
4.8710-3
1.5010-3
4.9810-4
1.3510-4
3.4610-5
7.8710-7
Solution 3.:
Applying the Arrhenius equation ( k = A e–Ea/RT),
lg k = -Ea / (2.303 RT) + 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 =
 2.50  (5.73)
3


5
.
38

10
K
3
3
3.0  10  3.6  10

Ea = 5.38 103  2.303  8.314
= 103.0  103 Jmol-1 = 103 kJmol-1
INORGANIC
CHEMISTRY
Section B4
B4 OXIDATION AND REDUCTION
Definitions
(1) Oxidation means combination. with a more electronegative element or the
removal 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
thus 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
Section B4
Oxidation states
Oxidation states of atoms in a compound are calculated by assigning electrons in a
bond to the more electronegative element. In simple ionic compounds they are the
same as the ionic charges. In any redox reaction the oxidation states of some
elements change.
Calculation rules of Oxidation states:
An element
atom consists
ofincluded.
a very small
positively
nucleus,
(a) Bonds between the same
are not
Elements
havecharged
oxidation
state
2－ theelectrons
by O
negative
zero. In an ion such surrounded
as peroxide
electronsheld
in by
theelectrostatic
O－O bond are
2
attraction.
distributed equally, making
O-1. The motion of electrons changes when chemical
2 －formed,
nucleielectronegative
being unaltered.and electropositive
(b) Except in cases suchbonds
as Oare
the most
2
elements in a compd have
an oxidation
state equal
to the
ionic neutrons.
charge.
Nuclei
contain positive
protons
andnormal
uncharged
(c) The sum of the oxidation
states must
equalisthe
on the species,
The number
of protons
the charge
atomic number
(Z) of anand is
therefore zero in a neutral
compd.
thisstrong
rule and
(b) above,
we have
H1 in
element.
The Using
attractive
interaction
between
protons
H2O, H-1 in CaH2.
and neutrons is opposed by electrostatic repulsion between
(d) Complex formation, and
donor-acceptor
interactionas in
general don’t
alteris the
protons.
Repulsion dominates
Z increases
and there
oxidation state.
only a limited number of stable elements.
A redox reaction is any reaction involving changes of oxidation state.
INORGANIC
CHEMISTRY
Section B4
Balancing redox
reactions
In complete redox reactions the overall changes in oxidation state must
A3
THE NUCLEAR ATOM
balance. When reactions involve ions in water it is convenient to split the
overall reaction into two half reactions. To balance these it may also be
An atom consists of a very small positively charged nucleus,
Electron
and Nuclei
necessary
to provide water and H+ or OH-.
surrounded by negative electrons held by electrostatic
attraction. The motion of electrons changes when chemical
Half reactions:
bonds are formed, nuclei being unaltered.
Fe2+  Fe3+ + e
Nuclei
contain
positive protons and uncharged neutrons.
(1) Electrons were included
in half
reaction.
Nuclear structure
Thebalance
numberinof
protonsand
is the
atomic number (Z) of an
(2) The half reaction must
charges
elements.
strong interaction between protons
MnO4－ ＋ 5 e  Mn2＋element. The attractive
（×）
2＋ ＋ 4H
and
is2O
opposed
MnO4－ ＋ 5 e ＋ 8H＋
Mnneutrons
(  ) by electrostatic repulsion between
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
INORGANIC
CHEMISTRY
Section B4
Example 1
Problem 1. Balance the equation H2SO3 + MnO4- SO42- + Mn2+ (in
solution).
A3acidic
THE
NUCLEAR ATOM
Solution.: (the method of half-reactions)
Step 1. Write the two skeletal half-reactions.
An atom consists of a very small positively charged nucleus,
Electron and Nuclei2H2SO3  SO4 (the reducing agent, H2SO3, is oxidized to SO42-)
surrounded by negative electrons held by electrostatic
2+
MnO4  Mn (the oxidizing agent, MnO4-, is reduced to Mn2+)；
attraction. The motion of electrons changes when chemical
Step 2. Balance each half-reaction.
bonds are formed, nuclei being unaltered.
(i) To balance the O, add water. H2SO3 + H2O  SO422+
Nuclei contain positive protons and uncharged neutrons.
MnOstructure
4  Mn + 4H2O
Nuclear
+.
2+
number
of protons
is the atomic
number
(Z) of an
(ii) To balance the H, The
add H
H2SO
3 + H2O  SO4 + 4 H
MnO4- + 8 H+ Mn2+element.
+ 4H2O The attractive strong interaction between protons
and neutrons
is opposed by electrostatic repulsion between
(iii) To balance the charge,
add electrons.
dominates as Z increases and there is
H2SO3 + H2O  SO42-protons.
+ 4 H+ +Repulsion
2 ea limited
MnO4- + 8 H+ + 5 e-only
Mn2+
+ 4H2Onumber of stable elements.
INORGANIC
CHEMISTRY
Section B4
Example 1
H2SO3 + H2O  SO42- + 4 H+ + 2 e-
A3
（× 5）
THE NUCLEAR ATOM
MnO4- + 8 H+ + 5 e- Mn2+ + 4H2O
（× 2）
An atom consists of a very small positively charged nucleus,
surrounded by negative
electrons held
by electrostatic
Step 3. Add the two half-reactions.
Since 2 electrons
are lost
in the first
of electrons
changesmultiple
when chemical
while 5e are gained in attraction.
the second,The
andmotion
since the
lowest common
of
bonds the
are formed,
unaltered.
2 and 5 is 10, we multiply
first by nuclei
5 and being
the second
by 2. When the
Electron and Nuclei
resulting equations areNuclei
added,contain
the 10epositive
on the two
sidesand
of the
equations
will
protons
uncharged
neutrons.
Nuclear structure +
cancel. Also 16H and The
5 H2number
O will cancel
from iseach
of protons
the side.
atomic number (Z) of an
The total reaction:
element. The attractive strong interaction between protons
22+ + 4 H+ + 3 H O
5 H2SO3 + 2 MnO4- 5
4 + 2Mn
2
andSO
neutrons
is opposed
by electrostatic
repulsion between
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
INORGANIC
CHEMISTRY
Example 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
Section B4
INORGANIC
CHEMISTRY
Section B4
Extraction of the
element
A3
THE NUCLEAR ATOM
Redox reactions are used in the extraction of nearly all elements from natura1ly
occurring compounds. Carbon
is consists
used to of
reduce
some
metal
oxides,
but many
An
atom
a
very
small
positively
charged
nucleus,
Electron and Nuclei
elements require strongersurrounded
reducing agents,
or theelectrons
use of electrolysis.
by negative
held by electrostatic
The methods shown as following
common
attraction.are
Theinmotion
of use.
electrons changes when chemical
（1） carbon
bonds are formed, nuclei being unaltered.
（2）reduction by using H2 or CO.
Nuclei contain
positive
protonsmetals
and uncharged
neutrons.
（3）the
replacement reaction
by using
the reactive
M such as
alkali
Nuclear
structure
The number of protons is the atomic number (Z) of an
metals or alkaline metals.
element. The attractive strong interaction between protons
（4）electrolysis
and neutrons is opposed by electrostatic repulsion between
protons. Repulsion dominates as Z increases and there is
only a limited number of stable elements.
INORGANIC
CHEMISTRY
Section B5
B5 Describing inorganic compounds
formulae
Stoichiometric (empirical) formulae describe only the
relative numbers of atoms present.
Molecular formu1ae and/or representations giving
structural information should be used when they are
appropriate. The physical state of a substance is often
specified.
INORGANIC
CHEMISTRY
Section B5
Name
Systematic nomenclature can be based on three systems, binary, substitutive
(similar to that in organic chemistry) or coordination. Many nonsystematic or
trivial(习惯） names are used.
A3
THE NUCLEAR ATOM
Systematic nomenclature is based on three systems.
Binary names:
An atom consists of a very small positively charged nucleus,
ElectronNaCl:
and Nuclei
sodium Chloride
PCl3: phosphorous trichloride
surrounded by negative electrons held by electrostatic
N2O4: dinitrogen tetroxide
MnO2: Manganese (IV) dioxide
attraction. The motion of electrons changes when chemical
F－：Fluoride; F2: fluorine.
bonds are formed, nuclei being unaltered.
Substitutive names:
SiH4: silane; SiCl4Nuclei
: tetrachloride
contain silane
positive protons and uncharged neutrons.
Nuclear structure
NH3: amine; NH2OH:
hydroxyamine
The number
of protons is the atomic number (Z) of an
Coordination names: element. The attractive strong interaction between protons
[Cu(NH3)4]2＋ ：tetraaminecopper(2＋)；tetraaminecopper
ion between
and neutrons is opposed by electrostatic(II)
repulsion
[CuCl4]2－：tetrachlorocuprate(2－)
or tetrachlorocuprate
(II) and there is
protons. Repulsion
dominates as Z increases
[Ni(NH3)6]Br2：hexaaminenickel
only a limiteddibromide
number of stable elements.
K[PF6]: potassium hexafluorophosphate (V)
INORGANIC
CHEMISTRY
A3
Section B5
THE NUCLEAR ATOM
Electron and Nuclei
Nuclear structure
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.
INORGANIC
CHEMISTRY
Section B5
Structure and bonding
A3
THE
NUCLEARnumber
ATOMand geometry of an atom describe the
The coordination
number of bonded atoms and their arrangement in space. Oxidation
An atom
consists
a very small
positively
charged nucleus,
states
rather than
valencies
are ofgenerally
used
for describing
Electron and
Nuclei
by negative electrons held by electrostatic
different possiblesurrounded
stoichiometries.
Bond lengths: attraction. The motion of electrons changes when chemical
Bond angles: bonds are formed, nuclei being unaltered.
Coordinate number（CN）
Nuclear structure
INORGANIC
CHEMISTRY
Section C
Section C Structure and bonding in molecules
C1 Electron Pair Bonds
Lewis 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.
INORGANIC
CHEMISTRY
:O
£º
H O£º
£º
£º
H
H C H
H
Section C-1
O: :N
N:
H
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]－
INORGANIC
CHEMISTRY
Section C-1
Octets and
‘hypervalence’
In most stable molecules and ions of the elements C-F, each of these atoms has eight
electrons (an octet) in its valence shell. Expansion of the octet and increased valency is
possible 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
£º
F
F:
£º
£º
£º
F:
£º £º
B
F:
:S
F
F
F
INORGANIC
CHEMISTRY
Section C-1
Resonance
£º
£º
2£-
:O£º
:O£º
£º
£º
:O £º
£º
:O£º
2£-
:O C
:O C
£º
:O C
:O
£º
:O£º
£º
2£-
£º
£º
£º
When several alternative valence structures are possible, the
bonding may be described in terms of resonance between them.
(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)
INORGANIC
CHEMISTRY
Section C-1
Formal Charges:
Formal charges are assigned by apportioning bonding electrons equally between
the two atoms involved. They can be useful to rationalize apparent anomalies in
bonding, 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
covalent bonds were broken, with the electrons being assigned equally to the atoms
involved.
Formal charge = (No. of valence electrons in neutral atom)－ (No. of
£º
£º
£º
:O£º
:O£º
£º
£º
:O £º
£º
:O£º
2£-
:O C
:O C
£º
:O C
:O
£º
:O£º
2£-
£º
£º
£º
nonbonding electrons) －1/2 (No. of electrons in bonds formed.)
2£-
INORGANIC
CHEMISTRY
Section C-1
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.
INORGANIC
CHEMISTRY
Section C-1
Limitations
Many covalent molecules and ions cannot be understood in terms of
electron pair bonds between two atoms. They include electrondeficient boron hydrides and transition metal compounds.
Two-center bonds: localized bonds
Three-center bonds: delocalized bonds
Molecular Orbital theory: （MOT）
INORGANIC
CHEMISTRY
Section C-1
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: 10928
Pauling: hybrid orbital theory:
2s
2p
ground state of C atom
excited state
4 sp3 hybrid orbitals
Atom 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；
INORGANIC
CHEMISTRY
Section C-1
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
BeCl2
CO2
BF3
NO3－
CO32－
CH4
CCl4
CHCl3
PdCl42－
Complex
PCl5
SF6
SiF62－
等性杂化： CH4 （sp3）
不等性杂化: (H2O) （sp3）
INORGANIC
CHEMISTRY
C2
Section C-2
Molecular shapes: VSEPR
VSEPR principles
The valence shell electron pair repulsion (VSEPR) model is based on the
observation that the geometrical arrangement of bonds around an atom is
influenced by nonbonding e1ectrons present. It is assumed that electronic pairs –
whether bonding or nonbonding - repel each other and adopt a geometrical
arrangement that maximizes the distances between them.
VSEPR principles have widely applied in predicting the possible steric
configuration of covalent molecules for non-transition elements.
————shapes and bond angles of the molecules
INORGANIC
CHEMISTRY
Section C-2
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.
A
A
A
2. linear
sp
3. trigonal planar
sp2
A
A
5. trigonal bipyramidal
sp3d
4. tetrahedral
sp3
A
6. Octahedral 7. pentagonal bipyramidal
sp3d2
Basic VESPR geometries with 2-7 electron pairs
INORGANIC
CHEMISTRY
Section C-2
(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.
INORGANIC
CHEMISTRY
Section C-2
Using VESEPR
It is first necessary to decide which atoms are bonded together. Drawing a valence
structure gives the total number of electron pairs around an atom, sometimes known as
its steric 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
not apply to hydrides or organic groups.
Step2: Calculate the steric number (SN) of the central atom, equal to the number of
bonded 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
INORGANIC
CHEMISTRY
Section C-2
Steric numbers 2－4：
Linear species （SN＝2）：BeH2，HgCl2，CO2，N3＋， NO2+
Trigonal planar species: （SN＝3）：BF3, NO3Tetrahedral 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
INORGANIC
CHEMISTRY
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－
Section C-2
INORGANIC
CHEMISTRY
Section C-2
Extensions, difficulties
and exceptions
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
INORGANIC
CHEMISTRY
C3
Section C-3
Molecular orbitals: Homonuclear diatomics
Bonding and antibonding orbitals
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= c11 + c2 2—— bonding MO（成键分子轨道）
2= c11 - 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
has 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
the internuclear region, and internuclear repulsion being “deshielded”.
INORGANIC
CHEMISTRY
Section C-3
MO diagrams
An MO diagram is a representation of the energies of bonding and antibonding
MOs formed from atomic orbitals. Electrons are assigned to MOs in accordance
with the exclusion principle, leading to the same building-up procedure as for the
periodic table of elements.
(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，1sg能量升高， 1su能量降低，
分子的总能量降低。
1su
(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。
H 1s AO
H 1s AO
(a) Exclusion principle;
(b) Lowest energy principle
1sg
(c) Hund’s first rule
INORGANIC
CHEMISTRY
Section C-3
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。
INORGANIC
CHEMISTRY
Section C-3
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+ (1sg)1 BO = (1-0)/2 = 0.5
H2 (1sg)2 BO = (2-0)/2 = 1
He2+ (1sg)2(1su)1 BO = (2-1)/2 = 0.5
1su or 1s *
He2 (1sg)2(1su)2 BO = (2-2)/2 = 0
1s AO
1s AO
1sg or 1s 
INORGANIC
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Section C-3
Second 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：
INORGANIC
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Section C-3
2p atomic orbitals can give rise to both  and  MOs, the former overlapping
along 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
electron-pair bonding models.
O
O2
O
O2 (2sg)2(2su)2(2pg)2(2pu)4(2pg)2
BO = 2
2p
2p
N2 (2sg)2(2su)2(2pg)2(2pu)4
BO = 3
2s
2s
INORGANIC
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Section C-3
用MO描述第二周期的物种及电子的排布情况：P55
Electron configuration
BO
D（kJmol-1）
R (pm)
N2
(2sg)2(2su)2(2pg)2(2pu)4
3
945
110
O2+
(2sg)2(2su)2(2pg)2(2pu)4(2pg)1
2.5
630
112
O2
(2sg)2(2su)2(2pg)2(2pu)4(2pg)2
2
498
121
O2(2sg)2(2su)2(2pg)2(2pu)4(2pg)3
1.5
-
128
F2
(2sg)2(2su)2(2pg)2(2pu)4(2pg)4
1
158
142
Higher predicted BO values correspond to stronger and shorter bonds
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C4
Section C-4
Molecular orbitals: Heteronuclear diatomics
Basic principles
Orbitals from atoms of different elements overlap to give unsymmetrical MOs,
the bonding MO being more concentrated on the more electronegative atom.
The greater the electronegativity difference, the greater the degree of
localization.
The basic principle of forming the MO for the heteronulclear diatomics is
similar 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
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Section C-4
HF and BH
In HF the F 2s orbital is too low in energy to be involved significantly in
bonding, but in BH both 2s and 2p orbitals on B can contribute to MOs with
H 1s. This situation is described as sp hybridization, and leads to bonding
and nonbonding MOs with a spatial localization similar to that assumed in
VSEPR theory.
H
H
HF
3
1s
2p
Orbital energies
Orbital energies
B
F
3
1
HB
1
2
2p
2
1s
1
2s
2s
1
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Section C-4
CO
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
orbital resembling a lone-pair on C the lowest unoccupied MO (LUMO) is an
antibonding orbital.
C
CO
O
4
Very complicated
LUMO:
(最低未被占用的分子轨道)
the Lowest unoccupied MO
Orbital energies
HOMO:
(最高被占用的分子轨道)
the highest occupied MO
2
2p
3
2p
2s
1
2
2s
1
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C5
Molecular orbitals: polyatomics
Localized and delocalized orbitals:
Polyatomics:
Localized orbitals: Two center
Delocalized orbitals: three or more center
Directed Valence
Hybrid Orbital Theroy
VSEPR theory: determine the molecular structure
Section C-5
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Section C-5
Multiple Bonds
 And/or  MO orbitals would be formed by LCAO.
H
H
C
C
C
H
H
O
O
O
C2H4
CO32-
Normal  bonding
大键 46
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Section C-5
Three-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)
H
H
B
H
H
B
H
H
B2H6
Al2(CH3)6 and BeH2
Three-center four-electron bonding provides an explanation of some
hypervalent molecules.
Hydrogen bonds:
F——HF (3c-4e bonds)
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C6
Section C-6
Rings and Clusters(簇合物)
Introduction
Rings and clusters can be formed by using simple two-center bonding models;
in others it is necessary to used delocalized MO models.
S
S
S
S
S
N
P
N
S
P
S
S
S
S
P
Aromatic Rings
Wade’s Rules
P
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C7
Section C-7
Bond Strengths
Bond 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＋
 H1
2H +
½ Ｏ２  H2O
 H2
O
 H3 =
fH0(H2O)
2 B.E. (O-Ｈ)
H1+H2 +2 B.E (O-H) = fH0(H2O)
INORGANIC
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Section C-7
Major 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.
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Section C-7
(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.
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C8
Section C-8
Lewis acids and bases
Definition and scope
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,
and ones such as SiF4 where octet expansion is possible.
A Lewis base (or donor) is any species with a pair of non-bonding electrons available
for donation.
for examples: the molecules such as NH3, and anions such as F-.
The Lewis Acid-base definition
The Brnsted 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.
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Section C-8
Lewis acids and bases may interact to give a donor-acceptor complex.
BF3 + (CH3)3N:
SiF4 + 2 F-
(CH3)3N:BF3
[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.
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Section C-8
Modeles of interaction
Interaction between the orbitals in Fig.1 will be strongest when the energy difference
between 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 most highly charged (positive) acceptor, and (negative) donor atoms.
INORGANIC
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Section C-8
Hard-soft classification
(HSAB classification)
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 donor and acceptor atoms tend to be larger and more polarizable.
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Hard-soft classification
Section C-8
It is found that the relative strength of donors depends on the nature of the acceptor and
vice versa. The hard and soft acid-base (HSAB)classification is often used to rationalize
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
harder base, and the softer acid with the softer base. As a standard for comparison the
prototype 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+,
and 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- ).
INORGANIC
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Section C-8
Hard-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,
CO32-, NO3-, O2-,
SO42-, PO43-, ClO4-,
NO2-, SO32-, Br-,
N3-, N2, C5H5N,
SCN-
H-, R-, CN-, CO,
I-, NCS-, R3P,
R2S, C6H6, S2-
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.
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Section C-8
Polymerization
Formation of dimers and polymeric structure is a manifestation of
donor-acceptor interaction between molecules of the same kind.
2 AlCl3  Al2Cl6
Cl
Cl
Al
Al
Cl
Cl
Cl
Cl
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
INORGANIC
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C9
Section C-9
Molecules in Condensed Phases
Molecular 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
INORGANIC
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Section C-9
Intermolecular 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
沸点、酸性
HF 的某些性质的反常表现。
INORGANIC
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Section C-9
Molecular Polarity
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.
INORGANIC
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Section D-1
Section D
D1 Introduction to solids
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
Cubic
trigonal
tetragonal
hexagonal
orthogonal
Monoclinic
triclinic
Unit cell parameter
a=b=c
a=b=c
a = b≠c
a = b≠c
a≠b≠c
a≠b≠c
a≠b≠c
α=β=γ= 900
α=β=γ≠900
α=β=γ= 900
α=β= 900, γ= 1200
α=β=γ= 900
α=β= 900, γ≠ 900
α≠β≠γ≠ 900
compds
NaCl
Al2O3
SnO2
AgI
HgCl2
KClO3
CuSO4·5H2O
INORGANIC
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Section D-1
Looking at unit cells
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
atom（原子坐标）.
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
as MX). (see below). －－P 81, Fig 1.
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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.
Section D-1
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.
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(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.
Section D-1
(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）.
INORGANIC
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Section D-1
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):
INORGANIC
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Section D-1
Non-stoichiometry (非化学计量比)
非化学计量比化合物， 非整比化合物
Some solids, especially natural minerals and transition metal compounds, are
nonstoichiometric 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 cell)
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
INORGANIC
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Section D-1
Chemical Classification
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
relatively weak intermolecular forces; ——low melting point, low boiling
point, low rigidity, insulator, etc
2) Metallic Solids (金属晶体) have atoms with high coordination numbers,
bound 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
one- or two- three- dimensional network;—— high rigidity, high m.p. and b.p.,
insulators, etc
4) Ionic Solids (离子晶体) are bound by electrostatic attraction between anions
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
INORGANIC
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Section D-1
INORGANIC
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Section D-2
D2 Element structures
Sphere packing
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 closepacked (hcp) and cubic close-packed (ccp, also known as face-centered cubic, (fcc)
structures. The body-centered cubic (bcc) structure is slightly less efficiently close
packed.
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%)
INORGANIC
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Section D-2
Simple cubic close packing: （简单立方密堆积）
C.N.： 8
Efficient Space filling: 52.36 %
2r  a
4 3
r
3
100%  52.36%
3
a
INORGANIC
CHEMISTRY
Section D-2
体心立方堆积：bcc
4r 
配位数：8
空间占有率: 68.02%
2
3a
4 3
r
3
 100%  68.02%
3
a
INORGANIC
CHEMISTRY
Section D-2
b
f
h
h
f
f
h
b
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
面心立方密堆积：fcc
b
b
c  1.633a, b  a
4
2  r 3
3
100%  74.05%
abc sin 120
INORGANIC
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Section D-2
b
f
h
h
f
f
h
b
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
六方密堆积 Hcp
b
b
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Section D-2
Hexagongal close packing
Cubic close packing
fcc
INORGANIC
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Section D-2
Metallic elements
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
metals 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
INORGANIC
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Section D-2
Nonmetallic 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.
INORGANIC
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Section D-2
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
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Section D-3
D3 Binary compounds: simple
structures
Coordination 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)
INORGANIC
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Section D-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 CdI2);
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)
INORGANIC
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Section D-4
D4 Binary compounds: Factors influences
structures
Ionic 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.
INORGANIC
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Section D-4
Table Ionic radii (pm) for six coordination, based on a value of 140 pm for O2Li+
Be2+
27
Na+ 102
Mg2+
72
Al3+
53
K+
138
Ca2+
100
Ga3+
62
Rb+ 149
Sr2+
116
Cs+
Ba2+
149
76
167
H-
146
O2－
140
F-
133
S2－
182
Cl-
167
Br-
182
I-
206
(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).
INORGANIC
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Section D－4
Radius 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 anioncation 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
CN
8
0.4 ~ 0.7
6
For examples:
BeF2: 0.20 CN=4;
MgF2:
0.54
6
CaF2 :
0.75
8
0.2~ 0.4
4
INORGANIC
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Section D-4
Ion 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.
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Section D-4
Covalent 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.
INORGANIC
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Section D-6
D6 Lattice energies
The 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.
INORGANIC
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Section D-6
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)
INORGANIC
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Section D-6
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.
N 0 Az z e 2
1
UL 
(1  )
40 r0
n
(b) Kapustinskii equations:
C z  z 
UL 
r  r
(1) lattice energies increase strongly with increasing charge on the ions.
(2) lattice energies are always larger for smaller ions.
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NaCl 型
离子晶体
NaF
NaCl
NaBr
NaI
MgO
CaO
SrO
BaO
Section D-6
Z1
Z2
r+
/pm
1
1
1
1
2
2
2
2
1
1
1
1
2
2
2
2
95
95
95
95
65
99
113
135
rU
/pm /kJ·mol-1
136
920
181
770
195
733
216
683
140
4147
140
3557
140
3360
140
3091
熔点
/o C
992
801
747
662
2800
2576
2430
1923
硬度
3.2
2.5
<2.5
<2.5
5.5
4.5
3.5
3.3
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Application
Section D-6
INORGANIC
CHEMISTRY
Section D-6
(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)
INORGANIC
CHEMISTRY
Section D-6
(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.
INORGANIC
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Section D-7
D7 Electrical and optical properties of
solids
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.
INORGANIC
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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
Section D-7
INORGANIC
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Section D-7
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.
INORGANIC
CHEMISTRY
Section D-7
Band 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.
INORGANIC
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Section D-7
Dielectric 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)：
INORGANIC
CHEMISTRY
Section D-7
Influence of defects
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.
INORGANIC
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Section D-7
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.
INORGANIC
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Section E1
Section E
E1 Solvent types and properties
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.
INORGANIC
CHEMISTRY
Section E1
Polarity: polar molecular.
(1) The molecules with large dipole moments form polar solvents, such as NH3,
H2O, 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
INORGANIC
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Section E1
INORGANIC
CHEMISTRY
Section E1
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)  [FeCl3Py]
FeCl3 + DMSO {(CH3)2SO}  [FeCl2DMSO4]+ + Cl－
FeCl3 + MeCN  [FeCl2MeCN4]+ + [FeCl4]－
INORGANIC
CHEMISTRY
Section E1
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 + + OH2 NH3  NH4+ + NH2An 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+ + SnF62BrF3 --------- Lewis base to produce F- ions.
SnF4---------- Lewis acid to react with F- ions
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(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
Section E1
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E2
Section E2
Brønsted Acids and Bases
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-/HPO42-
HPO42- / PO43-
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.
INORGANIC
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酸
HAc
H 2 PO
HPO
NH

4
2
4

4
[CH 3 NH 3 ]
[Fe(H 2 O) 6]
[Fe(OH)(H 2 O) 5 ]
3
2
Section E2
H+ +碱


H  Ac

2

H
HPO 4

3

H
PO 4

H  NH 3

H  CH 3 NH 2

H  [Fe(OH)(H 2 O) 5 ]

2

H  [Fe(OH) 2 (H 2 O) 4 ]
INORGANIC
CHEMISTRY
Section E2


c(H3O ) c(OH )
KW 
c
c
pH value:
Water undergoes autoprotolysis (self-ionization):
2 H2O  H3O + + OHThe equilibrium constant: Kw = [H3O+][OH-] = 1×10-14 (25℃)
T ， KW 
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-]
INORGANIC
CHEMISTRY
Section E2
Strong and weak behavior
HA + H2O  H3O + + AAcidity constant or acid dissociation constant of HA:
pKa ＝ －lg Ka
[ H 3O  ][ A ]
Ka 
[ HA]
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, HSO4B + H2O  OH- + HB+
basicity constant
pKb ＝ －lg Kb = 14- pKa
Strong base: pKb < 0
Weak base:
pKb > 0
[ HB  ][OH  ]
Kb 
[ B]
INORGANIC
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Section E2
INORGANIC
CHEMISTRY
Section E2
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 × 
[ H 3O  ][ A ]
[ H 3O  ]2
c 2
Ka 



[ HA]
c  [ H 3O ] 1  
(1) If  < 5%, or c/Ka > 500;
pH = - lg [H+]
(2) If  > 5%,
K a  c 2
[ H 3 O  ]  c  K a c
2

K

K
a
a  4K ac

[H 3 O ] 
2
INORGANIC
CHEMISTRY
Section E2
For weak monoatomic base:
B +H2O  OH- + BH+
[ HB  ][OH  ] [OH  ]2
c 2
Kb 



[ B]
c  [OH ] 1  
(1) If  < 5%, or c/Ka > 500;
(2) If  > 5%,
2

K

K
b
b  4Kbc

[OH ] 
2
pOH = - lg [OH-]
pH = 14- pOH
[OH  ]  K b c
INORGANIC
CHEMISTRY
Section E2
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)
NH4Ac(aq)
NH (aq) + Ac–(aq)

4
If addition of HCl, the same phenomenon would be observed.
INORGANIC
CHEMISTRY
Section E2
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)
If no HCl,
y = 1.33 ×10-3 mol·L-1 ，’ = 1.33×10-3 / 0.10 ×100 ％ ＝ 1.33 ％
pH1 = 2.876
HAc ＋ H2O  H3O＋ ＋ Ac－
Initial state(mol·L-1)
0.10
Balanced (mol·L-1)
0.10 – x
x + 0.1
x
[ H 3O  ][ Ac  ] x  (0.10  x)
Ka 

[ HAc]
0.10  x
x << 0.10 , x = Ka = 1.75×10-5 mol·L-1
pH = 1.000
 = 1.75×10-5 / 0.10 ×100 ％ ＝ 0.018 ％
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。
解：
HAc(aq)+H2O(l)
0.10
ceq / (mol·L-1) 0.10 – x
c0/ (mol·L-1)
x  (0.10  x)
5
 1.8  10
0.10  x
H3O (aq)+Ac－(aq)
+
0
x
0.10
0.10 + x
0.10 ± x ≈ 0.10
x = 1.8×10-5 c(H+) = 1.8×10-5 mol·L-1
pH = 4.74，α = 0.018%
0.10 mol·L-1 HAc溶液：pH = 2.89，α = 1.3%
INORGANIC
CHEMISTRY
Section E2
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:
For weak polyatomic acid or base,
fractional ionization
H 3O  (aq)  HCO－
3 (aq)
第一步：H 2 CO3 (aq)  H 2 O(l)

c(H O )c(HCO )
(H CO ) 
 4.2 10

K a1
3
2
c(H 2CO3 )
3

第二步：HCO (aq)  H 2 O(l)
H 3 O (aq)  CO
－
3
K a1
K a2
2－
3
3
2
3
7
2－
3

c(H O )c(CO )
(H CO ) 
 4.7  10
c(HCO )

K a2
－
3
(aq)
11
－
3
 10 3 溶液中的H 3 O  主要来自于第一步解离反应，
c(H 3 O  )的计算可按一元弱酸的解离平衡
做近似处理。
INORGANIC
CHEMISTRY
Section E2
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 H3O+,
H2CO3, HCO－3，CO32－和OH－的浓度以及溶液的pH值。
H 2 CO 3 (aq)  H 2 O(l)
解：
1
H 3 O  (aq)  HCO－
3 (aq)
0.010 －x
平衡浓度 /( mol  L )
x
K a1 (H 2 CO 3 )  4.2  10
x
7

c(H O )c(HCO )



－
3
3
c(H 2 CO3 )
0.010 － x  0.010

x
0.010－x
x  6.5  10
c(H 3 O )  c(HCO )  6.5  10
－
3
c(H 2 CO 3 )  0.010mol  L－1
－5
2
－5
mol  L
－1
根据第二步解离计算c(CO 32－ )：
H 3 O  (aq)  CO 32－ (aq)
HCO－
3 (aq)  H 2 O(l)
ceq /( mol  L1 ) 6.5  10－5 －y
6.5  10－5  y
K a2 (H 2 CO 3 )  4.7  10

{c(H 3 O )}{c(CO

－
{c(HCO 3 )}
5
2－
3
)}
y
11
5
(6.5  10  y ) y

5
6.5  10  y
5
6.5  10  y  6.5  10 , y  K a2  4.7  10
c(CO
2－
3
1
)  K a2 mol  L  4.7  10
11
11
1
mol  L
INORGANIC
CHEMISTRY
Section E2
pH 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 + OHKh = { [HAc][OH-]} / [Ac-] = Kw/Ka
(II) NH4＋ ＋ 2 H2O  NH3H2O ＋ H3O＋
Kh = { [NH3H2O][H3O+]} / [NH4+] = Kw/Kb
(III) In NH4Ac solution
Kh = Kw / (Ka  Kb)
INORGANIC
CHEMISTRY
Section E2
Problem 14.: Calculate the pH of 0.200 moldm-3 solution of NH4Cl.
Solution.:
Cl-, the conjugate base of a strong acid, does not react with water, but NH4+, the
conjugate acid of NH3, does react:
NH4+ + 2H2O
NH3  H2O + H3O+
[ H 3O  ][ NH 3 ] K w
Kh 

Kb
[ NH 4 ]
Let [H3O+] = x, [NH3] = x, and [NH4+] = 0.200 -x  0.200
The solution of NH4Cl in water is acidic, as expected.
K w 1.0  10 14
x2
Kh 


 5.6  10 10  [ H 3O  ]  x  1.1  10 5
5
0.200 K b 1.8  10
 pH  4.96
INORGANIC
CHEMISTRY
Section E2
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-
CHA
CA-
[ A  ][ H 3O  ]
Ka 
[ HA]
K [ HA] K a C HA
[ H 3O  ]  a 

[A ]
C A
O+
+ H3
pH  pKa  lg
For the weak base- salt system,
[OH  ] 
K bCbase
Csalt
pH  14  pK b  lg
Cbase
Csalt
C acid
C salt
INORGANIC
CHEMISTRY
Section E2
Table I. Some common buffer system
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
10.25
9.25 ~ 11.25
HCO3- –CO32-
INORGANIC
CHEMISTRY
Section E2
Problem 13.: Given 50 mL of 0.20 moldm-3 solution of the acid HA (Ka =
1.010-5) 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 moldm-3 to get this value of x
after dilution.
[ A  ][ H 3O  ] x  1.0  10 4
Ka 

 1.0  10 5  x  1.0  10 2 mol  dm 3
[ HA]
0.10
INORGANIC
CHEMISTRY
Section E2
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。
Cacid
pH  pKa  lg
C salt
5.00  4.75  lg
x  32mL
50  0.10  0.10 x
0.10 x
例题：若在 50.00ml 0.150mol·L-1 NH3 (aq)
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HCl
前后溶液的pH值各为多少？
解：加入 HCl 前：
c(B)
pH  14  pK b  lg

c(BH )
0.150
 14  ( lg 1.8  10 )  lg
0.200
 9.26  (0.12)  9.14
5
加入 HCl 后：
1.00  0.100
c(HCl) 
mol  L1  0.0020mol  L1
50.10
NH4
加HCl前浓
度/(mol·L-1)
加HCl后初始
浓度/(mol·L-1)
NH3(aq) + H2O (l)
0.150
0.150-0.0020
(aq) +OH－(aq)
0.200
0.200+0.0020
平衡浓度
/(mol·L-1)
0.150-0.0020-x 0.200+0.0020+x x
(0.202  x) x
5
 1.8  10
0.148  x
5

5
1
x  1.3  10
c(OH )  1.3  10 mol  L
pOH  4.89
pH  9.11
INORGANIC
CHEMISTRY
Section E2
Solvent Leveling 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.
Differentiating solvent effect
The strong & the weak acid in water (HCl, HAc)
The strong & the weak base （NaOH，NH3H2O）
HI，HBr，HCl ----in H2O, the acidity be leveled; in CH3OH, differentiated
INORGANIC
CHEMISTRY
Section E2
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
INORGANIC
CHEMISTRY
Pauling’ s rules
Section E2
INORGANIC
CHEMISTRY
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]-
Section E2
INORGANIC
CHEMISTRY
Section E3
E3 Complex formation
Introduction of complex
Coordination compounds, complex
A complex in general is any species formed by specific association of molecules or ions by donoracceptor 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:
Monodentate (unidentate); bidentate ; tridentate, tetradentate
For example: en, ethylenediamine, EDTA (ethylenediamine tetraacetic acid)
optical isomerism
INORGANIC
CHEMISTRY
Section E3
en, ethylenediamine
Phen
EDTA
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
INORGANIC
CHEMISTRY
Section E3
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+
INORGANIC
CHEMISTRY
Section E3
Nomenclature 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))
INORGANIC
CHEMISTRY
Section E3
Equilibrium constants
Stepwise formation constant, K1, K2, K3…….(逐级稳定常数)
M + L  ML
ML + L  ML2
[ ML ]
[ ML2 ]
K2 
[ M ][ L ]
[ ML ][ L ]
K1  K 2  K 3    
K1 
Cumulative stability constant, 1, 2, 3……….(累积稳定常数)
M + n L  MLn
 n  K1  K 2  K3    
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
NH3H2O： Na2S2O3，NaCN
AgCl, AgBr, AgI
《近代化学导论》：P343 配离子的不稳定常数——配合物的稳定常数
INORGANIC
CHEMISTRY
Section E3
Hard 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
donor 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
ions unaffected by complex formation.
INORGANIC
CHEMISTRY
Section E4
E4 Solubility of ionic compounds
Introduction 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
K sp  [M m ]n [ X n ]m
Solubility product: (for dilute solutions)
K sp ( PbI 2 )  [ Pb 2 ][ I  ]2
2
K sp ( BaSO 4 )  [ Ba 2 ][ SO4 ]
3
K sp (Ca3 ( PO4 ) 2 )  [Ca 2 ]3 [ PO4 ]2
AgCl Ksp= 1.77 ×10-7
AgCl > AgBr > AgI
AgBr Ksp= 5.35 ×10-13
AgI Ksp= 8.51 ×10-17
INORGANIC
CHEMISTRY
Section E4
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
(1) AgCl ( s )  Ag  (aq )  Cl  (aq )
K sp ( AgCl )  [ Ag  ][Cl  ]  [ Ag  ]2  1.77 10 10
[ Ag  ]  K sp ( AgCl )  1.33 10 5 mol  dm 3
2
(2) Ag 2CrO4 ( s )  2 Ag  (aq )  CrO4 (aq )
2
2
K sp ( Ag 2CrO4 )  [ Ag  ]2 [CrO4 ]  4[CrO4 ]3  1.12 10 12
2
[CrO4 ]  3
K sp ( Ag 2CrO4 )
4
 1.04 10  4 mol  dm 3
Solubility: AgCl < Ag2CrO4
INORGANIC
CHEMISTRY
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
Section E4
INORGANIC
CHEMISTRY
Section E4
Problem .: Determine whether a precipitate form in solution or not when Pb2+ solution
(0.100 moldm-3, 100 mL ) is mixed with Cl- (0.30 moldm-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 moldm-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
exceed that required for equilibrium, and the precipitation will form in the mixture.
INORGANIC
CHEMISTRY
Section E4
The influence of pH value in solution
Problem .:Whether Mg(OH)2 precipitate or not in a solution containing Mg2+ (0.0010
moldm-3) if the OH- concentration of the solution were altered to be (a) 10-5 moldm3 (b) 10-3 moldm-3 ? (K (Mg(OH) ) = 1.2  10-11)
sp
2
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 moldm-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.
Application:
The selectivity separation of metal ions in mixture solution would be
successfully finished by controlling the pH value.
INORGANIC
CHEMISTRY
Section E4
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 moldm-3
M (OH ) n ( s )  M n  (aq )  OH  (aq )
K sp ( M (OH ) n )  [ M
[OH  ]  n
pH  14 
n
K sp
[M n ]
K sp
1
lg
n [M n ]
][OH  ]n
例题：在含有0.10mol·L-1 Fe3+和 0.10mol·L1 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?
解：
K sp 开始沉淀pH  沉淀完全pH 
16
Ni(OH) 2 5.0  10
39
Fe(OH) 3 2.8  10
6.85
Fe3+沉淀完全
2.82
c(Fe3+)≤10-5
2.82
Ni2+开始沉淀
6.85
可将pH值控制在 2.82 ~ 6.85 之间
pH
INORGANIC
CHEMISTRY
Section E4
common ion effect
MnXm(s)  n Mm+ (aq) + n Xn-(aq)
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 moldm-3) solution.
(1) Before adding HCl, [Ag+]=1.33×10-5 moldm-3
AgCl (s)  Ag+(aq) + Cl－(aq)
(2)
After adding HCl,
x
x
x+0.010
K sp ( AgCl )  [ Ag  ][Cl  ]  ( x)( x  0.010)  1.77 10 10
x  0.010mol  dm 3 , x  0.010  0.01mol  dm 3
 x  1.77 10 8 mol  dm 3
INORGANIC
CHEMISTRY
Section E4
Complexing effect
If adding some compound able to complex the insoluble salt, its solubility would
markedly rise.
For example:
Calculate the solubility of AgCl in NH3H2O (6 moldm-3) sultion.
Answer.:
Initial state (moldm-3)
Balanced state(moldm-3)
AgCl (s) + NH3H2O  [Ag(NH3)2]+(aq) + Cl－(aq) + 2 H2O
6
0
0
6 - 2x
x
x
Ksp(AgCl) =1.7×10-10， K{[Ag(NH3)2]+}=1.7×107
[ Ag ( NH 3 ) 2 ] [Cl  ] [ Ag ( NH 3 ) 2 ]
K

 [ Ag  ][Cl  ]
2
2

[ NH 3 ]
[ NH 3 ] [ Ag ]
 K sp ( AgCl )  K {[ Ag ( NH 3 ) 2 ] }  (1.7 10 10 )  (1.6 107 )  2.7 10 3
x2
 2.7 10 3  x  0.283mol  dm 3
2
(6  2 x )
INORGANIC
CHEMISTRY
Section E4
Thermodynamic Analysis
Go =－RT lnKsp
(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.
INORGANIC
CHEMISTRY
Section E4
Major 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 [Bu 4N]+ can be helpful.
(ii) Salts where both ions have multiple charges are less likely to be soluble than ones with
single charges.
CO32- & SO42-, 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.
INORGANIC
CHEMISTRY
Section E4
Influence 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.
MnXm(s)  n Mm+ (aq) + n Xn-(aq)
(a) pH value will therefore influence the solubility in a range where one of the ions has
significant Brnsted 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.
INORGANIC
CHEMISTRY
Section E5
E5 Electrode Potentials
Standard 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+
氧化型

Z e 
Zn 2  /Zn
还原型
， Cu
2  /Cu
() Zn Zn 2 (1.0mol  L1 ) ‖ Cu 2 (1.0mol  L1 ) Cu ()
INORGANIC
CHEMISTRY
Section E5
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。
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 。
③纯液体、固体和气体写在惰性电极一
边用“,”分开。
INORGANIC
CHEMISTRY
Section E5
例：将下列反应设计成原电池并以原电池符号表示。


2Fe 1.0mol  L  Cl2 101325Pa 
2
1
3

1



1

 2Fe 0.1mol  L  2Cl 2.0mol  L

正 极 Cl 2 (g )  2e

2

负 极 Fe (aq ) e
2

1

3


2Cl (aq )
3
Fe (aq )
1
() Pt Fe 1.0mol  L , Fe 0.1mol  L
‖ Cl 2.0mol  L

1



Cl2 101325Pa  ，Pt ()
INORGANIC
CHEMISTRY
Section E5
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 cmol-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+] = 1moldm-3, PH2= 100 kPa)
H+ + e  ½ H2
INORGANIC
CHEMISTRY
Section E5
Only differences in electrode potential are significant, the absolute values having no
meaning except in comparison with H+ / H2 potentials.
INORGANIC
CHEMISTRY
Section E5
Direction 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)
nFE 
ln K 
RT
E 0   0   0
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。
3


Fe
(
aq
)
e
解：
2
Fe (aq)  2e 


O2 (g) 2H (aq)  2e 
2
Fe (aq)
Fe(s)
  0.769V
   0.4089V
H2O2 (aq)   0.6945V
2H2O(l)   1.763V


H2O2 (aq) 2H (aq)  2e 
2
H2O2与 Fe 发生的反应：

2


H2O2 (aq) 2Fe (aq) 2H (aq)
3
2Fe (aq) 2H2O(l)
3
2


MF  (H2O2 / H2O)  (Fe / Fe )
 1.763V 0.769V  0.994V  0.2V
INORGANIC
CHEMISTRY
Section E5
Non-standard conditions
Under nonstandard conditions the electrode potential of a couple can be calculated
from the Nernst equation:
RT
[Ox ]
0.0592 [Ox ]
ln
 0 
lg
nF [Re d ]
n
[Re d ]
RT
[Ox ]
0.0592 [Ox ]
E  E0 
ln
 E0 
lg
nF [Re d ]
n
[Re d ]
  0 
(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.

4




MnO 8H 5e
 MnO /Mn  

4
2
Mn  4H 2 O
2

4

8
0.0592V {c(MnO )}{c(H )}
lg
 (MnO /Mn ) 
2
5
{c(Mn )}

4
2
② 介质的酸碱性

3

例： 已知 EA (ClO /Cl )  1.45V

3

1
求：当c(ClO )  c(Cl )  1.0mol L ，

1




c(H ) 10.0mol L 时，E (ClO3 /Cl )  ?
解：ClO3 (aq)  6H (aq)  6e

3

Cl (aq)  3H2O(l)

A (ClO /Cl )

3

6
0.0592V {c(ClO )}{c(H )}
 A (ClO /Cl ) 
lg

6
{c(Cl )}
0.0592V
6

1.45V
lg10.0  1.51V
6

3

③沉淀的生成对电极电势的影响


例： 已知  (Ag / Ag)  0.799V，若在Ag
和Ag组成的半电池中加入NaCl会产生AgCl s ，

1

当 c(Cl )  1.0mol  L 时， (Ag / Ag)  ? 并求
10


 (AgCl/ Ag) ? ( Ksp (AgCl) 1.8×10 )
Ag
Ag 
c(Cl )  1.0mol  L1



Ag (aq ) Cl (aq )
解：AgCl (s)


{c( Ag )}{ c(Cl )}  Ksp (AgCl)

1

若c(Cl )  1.0mol  L 时 , c( Ag )  Ksp (AgCl)

Ag (aq )  e

Ag(s)

 ( Ag / Ag)


  ( Ag / Ag)  0.0592V lg {c( Ag )}

  ( Ag / Ag)  0.0592V lg Ksp ( AgCl)
 0.799V  0.0592V lg 1.8×10
 0.222V
10
INORGANIC
CHEMISTRY
Section E3
(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.
2
例 ：已知  (Cu /Cu )  0.3394V ，
2
12
2

K f (Cu(NH 3 ) 4 ) 2.30×10 。在 Cu /Cu 半 电
1


池 中，加入氨水，当 c ( NH 3 ) 1.0mol L ,
2
3 4
1
2
c (Cu(NH ) )  1.0mol  L 时，  (Cu /Cu )  ?
2
3 4
并求  (Cu(NH )
/ Cu )  ?
Cu
c( NH3 )  c(Cu(NH 3 ) 24 )  1.0mol  L1
Cu 2
氨水
解：Cu 2 (aq)  4NH 3 (aq)
2
3 4
Cu(NH ) (aq)
2
3 4
{c(Cu(NH ) )}
 Kf
2
4
{c(Cu )}{c( NH 3 )}
2
3 4
1
当 c( NH 3 )  c(Cu(NH ) )  1.0mol  L 时
1
2
c(Cu ) 
2
K f (Cu(NH 3 ) 4 )
2
Cu (aq)  2e

Cu(s)
2
 (Cu / Cu)
0.0592
2
  (Cu / Cu) 
lg{c(Cu )}
2
0.0592
1
2
  (Cu / Cu) 
lg
2
2
K f (Cu(NH 3) 4 )
2
0.0592
1
 0.3394V 
lg
12
2
2.30×10
 0.0265V
2
3 4


Cu(NH ) (aq) 2e
Cu(s)  4NH 3(aq)
1
2



当 c (NH3 ) c(Cu(NH3 ) 4 ) 1.0mol L 时 ,
2
3 4
2
 (Cu(NH ) / Cu)   (Cu /Cu)  0.0265V
2
3 4
即  (Cu(NH ) / Cu)
0.0592V
1
2
  (Cu /Cu) 
lg
2
2
K f ( Cu(NH3) 4 )
2
3 4
2

 (Cu(NH ) / Cu)  (Cu /Cu)
INORGANIC
CHEMISTRY
Section E5
Diagrammatic Representation:
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
1.判断歧化反应能否发生

2
2Cu (aq)
Cu(s)  Cu (aq)
0.1607V
0.5180V

2
Cu
 / V Cu
0.3394V
 

(Cu / Cu)  
2

(Cu / Cu )
 0.5180V  0.1607V
 0.3573 V  0
右  左
发生歧化反应；
右  左
发生歧化逆反应。
Cu
2.计算电对的电极电势 （Hess’ Law）
E2
E1
E3
A
B
C
D
(Z1)
(Z3)
(Z2)
Ex
(Zx)
A  Z1e

B  Z 2e
+）C  Z3e




A Z xe
B
1
 r Gm(1)  Z1 F 1
C
2
 r Gm(2)  Z 2 F 2
D
3
 r Gm(3)  Z 3 F3
D
x
 r Gm(x )  Zx F x
Z x  Z1  Z 2  Z 3
 r Gm( x )   r Gm(1)   r Gm(2)   r Gm(3)
 Z x F   Z 1 F 1  Z 2 F 2  Z 3 F 3
 x  Z1 1  Z 2  2  Z 3 3
Z1 1  Z 2  2  Z 3 3
x 
Zx
INORGANIC
CHEMISTRY
G10  G20  G30
3 0 ( Mn 3 / Mn )   0 ( Mn 3 / Mn 2 )  2 0 ( Mn 2 / Mn )
 0 ( Mn 3 / Mn)  0.29V
Section E5
例题：已知Br的元素电势图如下
E2

3
BrO
E1

BrO
0.4556
Br2
1.0774
Br

E3
0.6126
(1) 求 1 、 2 和 3 。
(2)判断哪些物种可以歧化?
(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 。
E2
解：(1)

3
BrO
E1

BrO
0.4556
Br2
1.0774
Br

E3
0.6126
(0.6126  6  0.4556 1  1.0774 1)V
E1 
 0.5357V
4
(0.4556 1  1.0774 1)V
E2 
 0.7665V
2
(0.6126  6  1.0774 1)V
E3 
 0.5196V
5
(2)
0.7665

3
BrO
0.5357

BrO
0.4556
0.5196

Br2、BrO 可以歧化。
Br2
1.0774
Br


(3) 因为 BrO 能歧化 ，不稳定，
所以 Br 2 (l) 与 NaOH 混合最稳定的产物

3

是 BrO 和 Br 。

3Br 2 ( l )  6OH ( aq )

E MF

3
5Br ( aq )  BrO ( aq )  3H 2 O(aq)


 E (Br 2 /Br )  E (BrO3 /Br 2)
 1 . 0774V  0.5196V  0 . 5578 V
ZE MF
5×0.5578V

 47 . 11
lg K 
0.0592V
0.0592V
47

K
1 . 29×10
INORGANIC
CHEMISTRY
Section E5
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.
INORGANIC
CHEMISTRY
Section E5
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)
INORGANIC
CHEMISTRY
Section E5
Kinetic 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.
f
f
作业：
6, 7, 9, 11, 12.
INORGANIC
CHEMISTRY
Sec. F
Section F
Introduction to Non-metals
Covalent 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
INORGANIC
CHEMISTRY
Section F
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
Many halides and oxides are Lewis acids; BCl3 SO3
The compounds with lone-pairs are Lewis bases;
NH3，H2O
Brnsted acidity is possible in hydrides and oxoacids.
Halide complexes can be also formed by ion transfer.
Redox 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.
INORGANIC
CHEMISTRY
Sec. G
Section G
Introduction to Non-transition metals
Scope
Non-transition metals includes groups 1 to 2 of the s-block elements, group 12 and pblock 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 pretransition metals.
lower oxidation states. (the sixth period, Tl+, Pb2+, Bi3+)
INORGANIC
CHEMISTRY
Section G
Positive ions
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.
Group trends
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 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).
INORGANIC
CHEMISTRY
Sec. H
Section H
Introduction to transition-metals
Scope
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
INORGANIC
CHEMISTRY
Section H
Vertical 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 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.
INORGANIC
CHEMISTRY
Section H
Electron 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.
INORGANIC
CHEMISTRY
Sec. I
Section I
Lanthanum and lanthanides
The 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
INORGANIC
CHEMISTRY
Section I
Oxidation 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 Ln 3+ ions.
INORGANIC
CHEMISTRY
Section I
Other 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) )