Question

The metal titanium, which is useful in making alloys to resist high temperatures and extremely corrosive environments, adopts two crystal structures – close-packed hexagonal (α-phase) at room temperature and body-centred cubic (β-phase) above 882 °C. Under favourable circumstances, the α phase can dissolve up to 32 atomic % (nearly 15 wt %) of oxygen whereas the β phase can dissolve very little oxygen at the transition temperature (882 °C) and a little more at higher temperatures. Oxygen atoms have a radius of approximately 0.7 x 10-10 m.

a) Given that the “distance of closest approach” between two titanium atoms is approximately 2.92 x 10-10 m, calculate the radii of the octahedral and tetrahedral interstices in both the hexagonal and cubic forms of titanium. .

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b) Use your answers to part a) to explain why the solubility of oxygen is different in the two structures of titanium? .

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c) Can you explain how a metal that is capable of dissolving so much oxygen below 882 °C can simultaneously have excellent oxidation resistance? .

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Answer 1

(a) In body centered cubic form,

                (i) For tetrahedral, Void Radius = 0.29*dist of closest approach

                                                                                = 0.29*2.92*10-¹⁰

                                                                                = 0.8468*10-¹⁰m

                (ii) For octahedral, Void radius = 0.155* dist of closest approach

                                                                                = 0.155*2.92*10-¹⁰

                                                                                = 0.4526*10-¹⁰m

In hexagonal form

                (i) For tetrahedral, Void Radius = 0.225*dist of closest approach

                                                                                = 0.225*2.92*10-¹⁰

                                                                                = 0.657*10-¹⁰m

                (ii) For octahedral, Void radius = 0.414* dist of closest approach

                                                                                = 0.414*2.92*10-¹⁰

                                                                                = 1.21*10-¹⁰m

(b) The voids have a comparable space to co-exist in the alpha phase only. In the beta phase, the space is too low for the oxygen atom to stay stable as compared to the alpha form. It can only occupy the tetrahedral voids in beta phase.

(c) The structure of the Base Centred Cubic crystals means that there are more planes for the alloys lattice to deform. on that temperature, it practically undergoes a heat treatment and has its entire lattice changes. So, it can no longer accomodate oxygen atoms as it udes to do in the hexogonal close packing