Sinclair- The Case of TiO and BaTiO3 Flashcards

1
Q

Useful properties of TiO2

A

Photo-catalysis, electrolysis of H2O, pigment, dielectric

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2
Q

Why should TiO2 be written as TiO2+-x?

A

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

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3
Q

What influences the value and distribution of x?

A

Purity, sintering atmosphere and cooling conditions

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4
Q

Graph of resistance vs -log(pO2)

A

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

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5
Q

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

A

Oxygen vacancies, Ti vacancies, interstitial Ti ions

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6
Q

What are trivalent Ti and monovalent O ions equivalent to

A

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

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7
Q

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

A

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-

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8
Q

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

A

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

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9
Q

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

A

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

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10
Q

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

A

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

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11
Q

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

A

nil -> e’ + h•

pO2 independent

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12
Q

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

A

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

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13
Q

The real formula of Ti2+-x with 8 species

A

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

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14
Q

Which defect is useful for electrolysis of H2O?

A

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

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15
Q

3 main conditions for predominant defects

A

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

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16
Q

Graph of logσ vs log(pO2) for undies BaTiO3

A

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

17
Q

Problem with BaTiO3

A

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.

18
Q

n-type region equations for BaTiO3

A

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

19
Q

p-type region equations for BaTiO3

A

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

20
Q

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

A

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