Section 4d Metallic bonding / Page 1
Metals have several characteristic properties, for example: i] high melting points and boiling points ii] high electrical conductivity iii] high thermal conductivity iv] ductile, malleable and hard
Therefore the type of bonding in metals must be able to explain the above properties.
A simple model to explain the bonding in metal is that the metal consist of cations
(from the metal atoms) with their charge balanced by a sea of mobile valence electrons
(delocalized electrons). The mobile valence electrons distributes uniformly throughout the lattice attracts the positively charged cation and thus bonds them all together. At the same time, the cations attract the electrons cloud and prevent it from dispersing.
Since a strong electrostatic attraction was established between the mobile electrons and the cations , this account for the high melting point and boiling point of metals.
The presence of a sea of delocalized electrons account for the high thermal and electrical conductivity of metals. For example, the high conductivity of metal can be explained by the presence of this mobile electrons. Under the influence of an external potential difference, the valence electrons can be made to flow in one direction. As one electron enters the metal from outside, an electron is displaced from the metal on another side of the metal crystal thus result in smooth flowing of electrons through the metallic crystal. force before 'slip' after 'slip'
The mobile nature of metallic bonding between atoms makes possible the deformation of metals without fracture. When a crystal of a metal change shape, layers of atoms acting if they were perfect sphere, will slip over each other along definite gliding planes with little loss of energy. This account for the malleability and ductility of metals.
.
1. The following diagram shows a simple model of part of a metallic structure:
+ + + + + +
+ + + + +
+ + + + + +
(a) Which electrons make up the 'sea of electrons' ?
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(b) Why are these electrons referred to as 'mobile' or 'delocalized' ?
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(c) How does a metal conduct electricity ?
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(d) In what 2 ways is metallic bonding similar to both ionic and covalent bonding ?
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Section 4d Metallic bonding / Page 2
The metal crystals can be imagined to consist of identical spheres packed together as close as possible. And almost all of the metals crystallize into one or more of these systems:
Hexagonal close packed (hcp) examples: magnesium, zinc
Face-centred cubic (fcc)
[ Cubic close packed (ccp) ] examples: aluminium, copper, lead, gold, platinum
Body-centred cubic (bcc) examples: iron, manganese, alkali metals
N.B. There is no clear relationship between the structure of a metal and its position in the periodic table.
In each close-packed layer, each atom is in contact with 6 others.
In building the second layer, the atoms can occupy the holes labelled ' b ' or ' c '.
If the holes 'b' are used then those marked 'c' cannot be used.
And the structures obtained by placing the second layers on top of holes 'b' or 'c' are identical.
c
b
b
c
c
b
c
b
b
c
c
b
Therefore the sphere ' a ' touches 12 spheres in these close-packed arrangement, 6 in the same layer, 3 in the layer above and 3 in the layer below. Hence the coordination number is 12.
(Coordination number is the number of nearest neighboring atoms or ions in contact with a particular atom or ions in the structure.)
The difference between hexagonal close  packed (hcp) and cubic close  packed or face centred cubic (fcc) lies in the arrangement of the third layer.
The third layer of atoms may be added in two different ways. The atoms may occupy the holes
'a', i.e. identical to the first layer, this is called an abab structure. The abab structure is the hexagonal close  packed (hcp) structure.
b
c
b a
b
c
b
c
b b
C
b
C
b
c
b
Hexagonal close-packed structure
(hcp)
Cubic close-packed structure (ccp)
Face centred cubic structure (fcc)
But if the spheres in the third layer occupy hole 'c', i.e. the third layer is different from both the first and second layer, this is called an abcabc structure. This arrangement is known as cubic
close-packed(ccp) or face-centred cubic (fcc).
This is not an close-packed structure. This structure is basically cubic with an atom at the

Section 4d Metallic bonding / Page 3 centre of each cube.
1st and 2nd layers of body-centred cubic structure
A body-centred cubic unit cell
The coordination number is 8 and the packing is more open than the close-packed structures.
Example: Iron crystallizes in a body-centred cubic lattice with a unit-cell edge of 0.2866 nm. Find the apparent radius of an iron atom. [ 0.124 nm]
1. Unit Cell
The unit cell is the simplest arrangement of spheres which when repeated will produce the whole structure.
2. Tetrahedral hole
A tetrahedral hole is formed between three spheres of one layer and one from the next layer a tetrahedral hole
.
3. Octahedral hole
An octahedral hole is formed by six spheres in two layers, three in each layer. It can also be viewed as four spheres, s, in a plane and two others, o, one above and one below, giving a regular octahedron. o o s s s s s s s o s an octahedral hole o
Octahedral holes are larger than tetrahedral holes. In a close-packed structure, the number of tetrahedral holes is greater than that of octahedral holes.
END OF SECTION EXERCISE
- 1. Compare and contrast the two close packing atomic arrangements in metals.

Section 4d Metallic bonding / Page 4
(3 marks) (90IA2.c))
2. The arrangement of atoms in metals can be described by the close packing of spheres. i) Which close-packed structure does abcabcabc... describe ?
Indicate on the diagram below one tetrahedral hole (marking it T ) and one octahedral hole (marking it O). ii) Describe the bonding in metallic crystals. iii) Of the three energy ranges in kJ mol - 1 given below:
5-100
200-700
800-1500 ,
Which is the most likely energy range for the change M(s)  M(g) , where
M is a metal ?
(4 marks) (92IA2.d))
3. (i) Iron has a body-centred cubic structure. Draw a unit cell representation of iron.
(ii) The relative atomic mass of iron is 55.8 and the atomic radius of iron in a bodycentred structure is 0.124 nm. Calculate the density of iron in g cm -3 .
(Avogadro constant is 6.02
 10 23 mol -1 )
(iii) Explain why the electrical conductivities of iron and of caesium chloride are different in the solid state.
(5 marks) (91IA2d))
4. A metal X has a body-centred cubic structure. Its density is found to be 3.5 g cm -
3 . From x-ray diffraction analysis, the length of its unit cell is 0.507 nm. Find the relative molecular mass of the metal. [ 137.3 ]
5. The atoms in metallic nickel pack in the face-centred cubic pattern.
(a) What is the percentage of the total volume occupied by the atoms?
(b) Given that there are 9.1  1022 nickel atoms per cm3 of metal, calculate the apparent radius of the nickel atom.
[ 74.05% ; 0.1248nm]
Making Scheme to Section Exercise
1. Both close packing arrangements have the same coordination number of
12 / same packing efficiency.
One close - packing have abab arrangement in hexagonal close packing.
The other close packing have abcabc arrangement in cubic close packing.
(3 marks)

2. (i)
Section 4d Metallic bonding / Page 5
Cubic close packed structure (ccp) or
Face centred close packed structure (fcc)
T O O T
1
3. (i)
(ii) “Free electron” model or “Mobile electron” description.
(iii) The best answer out of the three is 200-700 kJmol -1 .
4r a
1
1
1
Iron atom
(ii) 3a 2 = 4r
4 a = r
3 density =
( 4

( 1
8
8 1 )

55.8 g

7 cm )
3  23
3
= 7.89 g cm -3 2
(iii) In CsCl, the ions are held together by strong coulombic forces and cannot be easily displaced. Because of lack of mobility of ions, conductivity is low. 1
In iron, the iron atoms occupy the b.c.c. sites and electrons are delocalized. Conductivity is high because electrons can move. 1
4. Let M be the molecular mass of the metal.
Each b.c.c. unit cell contains ( 1
 
) = 2 atoms.
8
2 M
23

( .
1

7 3
)
= 3.5

M

137.296
2
5. (a)
(4r) 2 = 2a 2  a = 8 r
Volume of unit cell = a 3 = 8
Each f.c.c. unit cell has ( 1
8
8 r 3
8
2
6 )

4 atoms
Volume occupied by 4 atoms = 4 ( 4
3
 r
3
)
% of total volume occupied by the atoms
=
4 ( 4
3
 r
8 8 r
3
3
)

100% = 74.05 %
(b) Volume occupied by Ni atoms per cm 3
= 1 cm 3 (74.05%) = 0.7405 cm 3 = (9.1x10
22 ) ( 4
3
 r
3 )
r = 1.248 x 10 -8 cm = 0.1248 nm
2
2