Sinclair- Thermodynamic Treatment of Defects and Conduction Models Flashcards
What are the 3 variables for semiconducting oxides (SCO)?
Temperature, x (metal vacancies due to oxygen gain or oxygen loss depending where the x is in the formula), pO2 (partial pressure of oxygen)
Examples of p or n type SCO
Ni1-xO is p-type
TiO2-x is n-type
Kroger-Vink notation in equation for p-type SCO Ni1-xO
1/2O2(g) Oo^x + Vm” + 2h•
In this case sub M is Ni
Equilibrium constant for p-type SCO and simplify
Use equation with Kroger-Vink from before to do normal thing with products concentrations over reactants to powers of coefficients. Not from initial equation [h•]=2[Vm”] and replace [Vm”] in equation for K1 to get:
K1=[Oo^x][h•]^3/2pO2^1/2
Deriving the slope of lnσ vs ln(pO2) for p-type SCO
Rearrange the simplified K1 equation to make [h•] the subject involving a cube root of a fraction times pO2^1/6. Means it is proportional to pO2^1/6. For σ=nqμ, in this case n is [h•]. Means conductivity proportional to pO2^1/6 so the plot of ln of both sides has slope of +1/6. This is only for a fixed temperature
Why might the slope of lnσ vs ln(pO2) graph not be 1/6?
Metal vacancies Vm” and holes can be associated with each other (more common at lower T). This reduces the defective charge in the resulting defect:
Vm”+h•->Vm’ so now equilibrium equation has only one hole and a Vm’. When you rearrange K2 equation find [h•] proportional to pOo^1/4 so slope of +1/4. In extreme cases all holes may become trapped by metal vacancies (defect association)
How to prove you are in thermodynamic equilibrium
Prove σ is not a function of time
Real graphs for log(σ) vs log(pO2) of p-type SCO
Different plots for different temperatures. No simple relationship but gradient close to +1/4 at lower T and higher pressures of O2 and gets nearer +1/6 at higher T and lower pressures of O2.
Equation for ratio of double and singly charged cation vacancies for NiO
k=(VNi”]/[VNi’]=1/2(n-4)(6-n)
n is reciprocal of normal gradient for lnσ vs ln(pO2)
How does percentage of doubly ionised vacancies in NiO vary with T?
Percentage of doubly ionised vacancies increases with increasing T
Where do the holes go in NiO?
Chemically most feasible option is on Ni2+ ions (d8) to form Ni3+ (d7). Otherwise could oxidise O2- ions (p6) to O- (p5). If bonding treated as purely ionic then all holes either on Ni or all on O ions. In reality bonding has some covalence so is possible that some holes on Ni and some on O
Lowest energy type of defect in NiO
Schottky defects VNi”+Vo••
Kroger-Vink notation equation for n-type SCO
Example is TiO2-x
Oo^x -> 1/2O2(g) + Vo•• + 2e’
Deriving the slope of lnσ vs ln(pO2) for n-type SCO
Start with Kroger-Vink notation equation and do normal thing to find equilibrium constant Kred (reduction). Remember the 1/2O2 is written as (pO2)^1/2. Note that [e’]=2[Vo••] and sub in to eliminate [Vo••]. Rearrange to make [e’] the subject involving a cube root time pO2^-1/6. σ is proportional to [e’] and so proportional to pO2^-1/6. Therefore slope is -1/6
Lowest energy defect type for TiO2
Frenkel defect
V(Ti)””+Ti(i)••••
Vacancy and interstitial
