Test 1 METB113 Sem 1 2015/2016

METB113

MID -TERM TEST

SEMESTER 1 2015/2016

TIME: 10.00 – 11.00 AM

DATE: 12 JULY 2015

NAME :

ID NO :

SECTION :

LECTURER :

INSTRUCTION: ANSWER ALL QUESTIONS IN THESE QUESTION PAPERS (ON BOTH PAGES).

Marks:

Question Marks Obtained Marks Allocated

1 10

2 10

3 10

Total 30

Question 1:

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Test 1 METB113 Sem 1 2015/2016

a) Explain the differences between primary bond and secondary bond in materials.

Primary bond arises from either transfer or sharing of electrons (valence electrons)

between the atoms in materials.

It is strong interatomic forces that can vary from very high energy (ionic bond) in

ceramic to very low energy (covalent bond) in polymer.

Can be directional (covalent) or non-directional (ionic, metallic).

Secondary bond (van der Waals) is intermolecular forces.

Secondary bonding arises from dipole interactions between the molecules.

Small energy at 10-42 kJ/mol.

Directional only due to the dipole interactions.

[4 marks]

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Test 1 METB113 Sem 1 2015/2016

b) For Figure 1 below;

Figure 1

i) What are the type of materials that are formed between positive ions and negative

ions as shown in Figure 1? [1 Mark]

Ionic material

ii) Explain the bonding properties in Figure 1. [5 Marks]

Ionic bond are formed by strong electrostatic force (Coulombic force) between the

cation and anion. Lattice energies and melting temperature of ionically bonded

solids are high. Metal oxides especially MgO has the highest bonding energies at

ca. 3932 kJ. Ionic solids are hard (do not dent), rigid (do not bend or stretch),

strong (hard to break, brittle (deform little before fracture) due to strong

electrostatic force that holds the ions together.

Question 2 (10 Marks)

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Test 1 METB113 Sem 1 2015/2016

Consider a Face Centered Cubic (FCC) unit cell.

a) Identify the coordination number of FCC. [1 M]

Coordination Number = 12 [1Mark]

b) Prove the relationship between the length of the cube side a and the atomic radius R is

a=4 R√2

[2M].

a

a

c

c=4 R−−Eq (1 ) [0.5 Mark]

c=√2a – Eq (2) [0.5 Mark]

Eq (1 )=Eq (2 ) [0.5 Mark]

4 R=√2a

∴a=4 R√ 2

[0.5 Mark]

c) Calculate the Atomic packing factor of FCC [2M]. What is the significant of this number? [1M]

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Test 1 METB113 Sem 1 2015/2016

[0.5 Mark]

[0.5 Mark]

[0.5 Mark]

[0.5 Mark]

74% of the unit cell is occupied by atoms. [1 Mark]

d) Draw the close packed plane in FCC. [1M].

Plane (111) – *or (111) family of plane [1 Mark]

e) Identify the position of atoms in plane Question (1d) [3M].

(1,0,0) ; (0,1,0); (0,0,1); ( ½ ,0, ½); ( ½ , ½ , 0) and (0, ½, ½ ) [0.5 Mark x 6 = 3 Marks] *

[or based on (111) family of plane drawn]

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Test 1 METB113 Sem 1 2015/2016

Question 3 (10 Marks)

a) Describe and illustrate the solidification process of a pure metal in terms of the

nucleation and growth of crystals. [4Marks]

Answer

Steps of solidification:

Nucleation: Formation of stable nuclei [1M]

Growth of nuclei: Formation of grain structure [1M]

[2M]

b) What are the 2 mechanisms in formation of stable nuclei? [1Mark]

Answer

Homogeneous Nucleation [0.5M]

Heterogeneous Nucleaction [0.5M]

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Test 1 METB113 Sem 1 2015/2016

c) What is a solid solution? [1Mark]

Answer

Solid solution is a simple type of alloy in which elements are dispersed in a single

phase.

2 types of solid solutions: substitutional and interstitial [1 M]

d) Explain the differences between a substitutional solid solution and an interstitial solid

solution. [4Marks]

Answer

Substitutional:

• Solute atoms substitute for parent solvent atom in a crystal lattice.

• The structure remains unchanged.

[2M]

Interstitial:

• Solute atoms fit in between the voids (interstices) of solvent atoms.

• Solvent atoms in this case should be much larger than solute atoms.

[2M]

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Test 1 METB113 Sem 1 2015/2016

-END OF QUESTION-

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