Sinclair- The Case of TiO and BaTiO3

Question 1

Useful properties of TiO2

Answer 1

Photo-catalysis, electrolysis of H2O, pigment, dielectric

Question 2

Why should TiO2 be written as TiO2+-x?

Answer 2

Because of the existence of both n and p type behaviour observed under 1000°C.

Question 3

What influences the value and distribution of x?

Answer 3

Purity, sintering atmosphere and cooling conditions

Question 4

Graph of resistance vs -log(pO2)

Answer 4

Resistance on a log scale, -log(pO2) goes from 30 to 0 left to right. Shape is a dome. The dome moves lower and the peak moves right for increasing temperature. Left of the peak is O loss so n-type TiO2-x and right of peak is O gain so p-type TiO2+x. At the peaks of resistance there is stoichiometric TiO2

Question 5

What are the intrinsic point defects in rutile (TiO2)?

Answer 5

Oxygen vacancies, Ti vacancies, interstitial Ti ions

Question 6

What are trivalent Ti and monovalent O ions equivalent to

Answer 6

Trivalent Ri (Ti3+, d1) equivalent to quasi free electron
Ti(Ti)’=e’
Monovalent O ions (O-, p5) equivalent to quasi free hole
Oo•=h•

Question 7

The 5 main defect equations (descriptions) that dominate the defect chemistry

Answer 7

Conventional O loss with partial reduction of lattice Ti to Ti3+.
Interstitial Ti4+ ions with O loss resulting in partial reduction of lattice Ti to Ti3+.
Interstitial Ti3+ ions with O loss resulting in further reduction of lattice Ti to Ti3+.
Intrinsic generation of charge carriers and no change in O content.
Oxygen gain on surface creating metal vacancies with partial oxidation of some lattice O2- to O-

Question 8

Conventional O loss with partial reduction of lattice Ti to Ti3+.

Answer 8

Oo^x -> 1/2O2(g) + Vo•• + 2e’
2e’ + 2Ti(Ti)^x -> 2Ti(Ti)’
Slope of -1/6

Question 9

Interstitial Ti4+ ions with O loss resulting in partial reduction of lattice Ti to Ti3+.

Answer 9

Ti(Ti)^x + 2Oo^x -> VTi”” + Ti(i)•••• + 2Vo•• + O2(g) + 4e’
4e’ + 4Ti(Ti)^x -> 4Ti(Ti)’
Slope of -1/5

Question 10

Interstitial Ti3+ ions with O loss resulting in further reduction of lattice Ti to Ti3+.

Answer 10

Ti(Ti)^x + 2Oo^x -> VTi”” + Ti(i)••• + 2Vo•• + O2(g) + 3e’
3e’ + 3Ti(Ti)^x -> 3Ti(Ti)’
Slope of -1/4

Question 11

Intrinsic generation of charge carriers and no change in O content.

Answer 11

nil -> e’ + h•

pO2 independent

Question 12

Oxygen gain on surface creating metal vacancies with partial oxidation of some lattice O2- to O-

Answer 12

O2(g) -> 2Oo^x + VTi”” + 4h•
4h• + 4Oo^x -> 4Oo•
Slope of +1/5

Question 13

The real formula of Ti2+-x with 8 species

Answer 13

(Ti(Ti)^x)a(Ti(Ti)’)b(VTi””)c(Ti(i)••••)d(Ti(i)•••)e(Oo^x)f(Oo•)g(Vo••)h

Question 14

Which defect is useful for electrolysis of H2O?

Answer 14

The VTi”” has been identified as a surface active site able to provide holes to adsorb H2O and enable splitting into H2 and O2

Question 15

3 main conditions for predominant defects

Answer 15

x<0: O vacancies, Ti interstitials, electrons, n-type TiO2-x.
x=0: intrinsic TiO2 (Eg roughly 3.2eV)
x>0: excess O, Ti vacancies, holes, p-type TiO2+x

Question 16

Graph of logσ vs log(pO2) for undies BaTiO3

Answer 16

Roughly diagonally down left to right then reaches minimum and starts to curve back up. Higher temperatures move the curve up and shift minimum right. Left of the min is n-type and right is p-type

Question 17

Problem with BaTiO3

Answer 17

There is an n-p transition but you would always expect n-type behaviour due to O loss and reduction of Ti4+ to Ti3+. Don’t expect p-type as O gain not expected and can’t oxidise Ba2+ or Ti4+ and not favourable to create both Ba and Ti vacancies on surface of a crystal.

Question 18

n-type region equations for BaTiO3

Answer 18

Oo^x -> 1/2O2(g) + Vo•• + 2e’
2e’ + 2Ti(Ti)^x -> 2Ti(Ti)’
Slope of -1/6 in low pO2 for O loss.
Electron can enter the CB created by Ti ions

Question 19

p-type region equations for BaTiO3

Answer 19

Vo•• + 1/2O2(g) -> Oo^x + 2h•
2h• + 2Oo^x -> 2Oo•
Slope of +1/4 in high pO2 for O gain.
Electron holes can enter VB created by the O ions

Question 20

What is the origin of Vo•• in undoped BaTiO3 for p-type behaviour?

Answer 20

When making it using TiO2 this can have acceptor impurities in it if dirty (<99 to 99.999% pure). Fe3+ and Al3+ ions can occupy the oct sites in TiO2 so Ti1-x(Al,Fe)xO(2-x/2). This is:
Ti(Ti)^x + 2Oo^x -> (Fe,Al)sub(Ti)’ + +3/2Oo^x + 1/2Vo••
This is the source of Vo•• in undoped BaTiO3 which is
BaTi1-x(Al,Fe)xO(3-x/2).
Controversial as could be acceptor impurities and/or metal vacancies or due to O- ion. Any way holes must be created