2. 2
We should grasp…
What is a metal?
What is an alloy?
What are the differences between their
properities?
Explain solid solution and intermetallic
compound.
6. 6
What is a Metal?
Q: What is a Metal?
A: Metal is consisted of positive centers
(or ions) sitting in a “gas ” of free-electrons.
It tends to be good electrical conductors.
7. 7
What is an Alloy?
Q: What is an Alloy?
A: An alloy consists of a mixture of a pure
metal and one or more other elements
which can be metals or non-metal.
Q: What are the differences between their
properties?
A:Alloys are usually less malleable and
ductile than pure metals and the tend to
have lower melting points.
8. 8
Solid Solutions
In many cases ,metals are quite soluble in
other metals .For example, solid copper
and solid nickel are fully soluble in each
other.
9. 9
This type of perfect solid solubility is a
side effect of having free electrons. Since
the electrons are free to move, the exact
number of valence electrons possessed
by any given atom shouldn’t matter.
10. 10
So a “solid-solution” is that one metal
serves as the solvent and the other as the
other as the solute, although in a case
like copper and nickel where these are
mutually soluble at all compositions the
terms solvent and solute can be a little
misleading .
11. 11
For example, the
figure in the right
is Au-Ag phase
diagram, below
fusiform area is
solid solutions.
12. 12
Intermetallic Compounds
Not all metals are soluble in other metals,
thus, produced another new phase called
“intermetallic compounds”, such as nickel
will dissolve some aluminum, so that at
low aluminum contents a solid solution is
produced. But if larger amounts of
aluminum are added, then produce
intermetallic compounds.
13. 13
Some of these compounds have a very
well defined composition, such as Ni3Al.
Others have quite a wide range of
composition, such as NiAl.
Nickel Aluminum Alloy
14. 14
Three types of circumstances of
intermetallic compounds forming
The first circumstance: size difference.
The second circumstance: a large
difference in electronegativity.
The third circumstance: certain ratios of
the number of valence electrons to the
number of atoms in a structure.
15. 15
Hume-Rothery Rule
Hume-Rothery Rule 1:Atomic Size Factor (the 15%)
Rule.
Extensive substitutional solid solution occurs only if
the relative difference between the atomic diameters
(radii) of the two species is less than 15%. If the
difference > 15%, the solubility is limited.Comparing
the atomic radii of solids that form solid solutions,
theempirical rule given by Hume-Rothery is given as:
Mismatch= %
15
100
solvent
solvent
solute
r
r
r
16. 16
Hume-Rothery Rule 2:Crystal Structure Rule :
For appreciable solid solubility, the crystal structures
of the two elements must be identical.
Hume-Rothery Rule 3: Valency Rule :
A metal will dissolve a metal of higher valency to a
greater extent than one of lower valency. The solute
and solvent atoms should typically have the same
valence in order to achieve maximum solubility.
Hume-Rothery Rule 4: The Electronegativity Rule :
Electronegativity difference close to 0 gives
maximum solubility. The more electropositive one
element and the more electronegative the other, the
greater is the likelihood that they will form an
intermetallic compound instead of a substitutional solid
solution. The solute and the solvent should lie
relatively close in the electrochemical series.
20. 20
1. Given that many of the compounds
formed in alloys are size factor or electron
compounds,these do not follow the rules
of valency.
在合金中,许多化合物是按比例组合或是
电子化合物,不能很好的遵循价键理论。
21. 21
2.Whereas the chemical formula of an
ionic compound , like NaCl or Al2O3 can
be predicted easily from the compound’s
position in the periodic table ,this is not the
case for many intermetallic compounds.
换句话说,尽管像氯化钠,氧化铝的价键
可以通过他们在元素周期表的位置进行预
测,但却不能解释合金中金属元素的成健
方式。
22. 22
3.Thus,knowing from their position in the
periodic table that Na wants to from Na+
ions and chlorine wants to from Cl- ions
explans why an ionic compound with a
formula NaCl is observed.
从元素周期表可以知道,钠有变成钠离子
的趋势,氯有变成氯离子的趋势,这就解
释了氯化钠的形成。
23. 23
4.In contrast,this dosen’t explain why a carbide
with a formula Cr23C6 is formed in stainless
steels.
然而,这却解释不了Cr23C6在钢铁中的成健方
式。
24. 24
5.The precipitation of Cr23C6 is a big
problem when stainless steels are welded.
这样 Cr23C6 成为钢铁焊接的大问题
高铬高碳冷作钢
25. 25
6.Cr23C6 forms in the “heat affected
zone”around the weld (this region is
heated but not melted during welding).
在“受热区”的Cr23C6遍布在焊接点周围
(这个区域在焊接的时候受热但不熔化)
26. 26
7.The formation of this chromium-rich
phase pulls chromium out of solution in the
surrounding iron.
富铬相的结构,使铬在离子溶液的氛围之外。
27. 27
8. This in turn,prevents the formation of a
protective layer of Cr2O3 on the surface of
the stainless steel and so the stainless
steel is no longer stainless,but instead
suffers from catastrophic localized
corrosion.
这样,就保护了钢铁表面氧化铬保护层的
结构,从而钢铁不再是单纯的铁,而是可
以抗局部严重腐蚀的物质。
28. 28
9.Consider a different example,NiAl is an
electron compound which froms due to
having a “magic” 1.5 valence electrons per
atom (this magic number isn’t really magic ,
but is a results of deviations from the free
electron model).
另外一个不同的例子,NiAl是具有“魔法”
的化合物因为其中每个原子都是1.5价的
(神奇的不是数字本身,而是1.5价不符合
自由电子模型)。
29. 29
10.There is one nickel atom and one
aluminum atom in NiAl and the valency of
nickel is two and that of aluminum is
three.given that (2+3)/2=2.5 this doesn’t
seem like 1.5 valence electons per atom .
一个镍原子和一个铝原子构成了NiAl,从
价键理论看,镍是二价,铝是三价。照这
样,应该平均每个原子有2.5个价电子,与
实际上每个原子1.5个价电子不同。
8 2
:3d 4s
Ni 2 3
:3 3
Al s p
30. 30
11.However ,as a transition metal ,nickel is
able to act as if it does nor have a valency
by “hiding”electrons in the emply states in
the shell.thus , effectively ,there are
(0+3)/2=1.5 valence electrons per atom in
NiAl.
镍作为一种过渡金属,可以将外层电子
“隐藏”在次外电子层的空轨道,看起来
没有价电子。所以,铝镍合金中,每个原
子有1.5个价电子。
32. 32
12.Even in cases where intermetallic formation
does not occur ,there may not be perfect solid-
solubility ,if tow metals have different crystal
structures then at some intermediate
composition there will have to be a change from
the crystal structure of one metal to that of the
other .
即使两种金属形成的晶体没有缺陷,也不
能说明它们能形成完美的固溶体,如果两
种金属有不同的晶格结构,必须在介质中
改变一种金属的晶格结构,使其与另外一
种金属的晶格结构相适应
33. 33
13.In such a case the result would be ,on
gradually changing the composition of an
alloy from pure metal A(froming the a-
phase )to pure metal B (froming the b-
phase)
为了形成这种晶格,逐渐将合金从A(α相)
变到B(β相)。
34. 34
Single-phase a soild-solution;
Two-phase mixture of a soild-solution and
b soild-sulition;
Single-phase b solid-solution.
α单相固溶体
α单相固溶体与β单相固溶体混合
β单相固溶体
35. 35
Think about….
Q: Give examples to explain soild solution
and intermetallic compound?
A: For example, nickel will dissolve some
aluminum,so that at low aluminum
contents a solid solution is
produced.However,if larger amounts of
aluminum are added,then a series of
intermetallic compounds(for example
Ni3Al and NiAl) are produced.