Donors/Acceptors Metal Oxides

[citw]

Can anyone explain to me why most articles describe chromium as an acceptor in TiO2? In TiO2, titanium has the charge state Ti⁴⁺ and oxygen has the charge state O^2-. When Cr substitutes for Ti, it does so as Cr³⁺. Now, at first glance, Cr has atomic number 24 and Ti 22. Cr therefore has two more valence electrons and is a donor. In TiO2, Cr³⁺ actually has three more valence electrons than Ti⁴⁺ ([Ar]3d³4s⁰ vs [Ar]3d⁰4s⁰). It should therefore be a donor. The thing is, it forms a deep impurity level near the valence band. TiO2 has an energy gap of around 3.2 eV, and the impurity state is about 1.0 eV from the valence band maximum. To me, that makes it a deep donor. For some reason, journals almost always describe it as an acceptor. Can someone help me make sense of this?

 

[citw]

It has been explained to me that Cr is an acceptor because its oxidation state indicates one fewer electron is given to the lattice than with Ti (Cr³⁺ vs. Ti⁴⁺). I understand this point, but Cr³⁺ still has more valence electrons than Ti⁴⁺, so can someone help me out here?

 

[Useful nucleus]

It is not clear to me what do you mean by "valence electrons" for "ions". One can talk about valence electrons of atoms but not ions. So for Cr you can say it has 3 valence electrons but the same language would not be convenient in describing Cr+3.

The interpretation that you were given is correct, Cr+3 has an effective charge of -1 with respect to Ti+4 and hence can be regarded as an acceptor.

 

[Useful nucleus]

I thought again about what you wrote in your initial post. It seems you think that since Cr3+ still has 3 electrons in its d shell, it can donate them and become Cr+6 and hence a donor. Well, it would be really a donor if it can do this but it seems this is unfavorable energetically and it will remain Cr+3 and hence an acceptor.

 

[citw]

It is not clear to me what do you mean by "valence electrons" for "ions". One can talk about valence electrons of atoms but not ions. So for Cr you can say it has 3 valence electrons but the same language would not be convenient in describing Cr+3.

The interpretation that you were given is correct, Cr+3 has an effective charge of -1 with respect to Ti+4 and hence can be regarded as an acceptor.

This is exactly what I'm having trouble with. In solid state, you're taught that when P substitutes for Si, it's a donor because it has five valence electrons (3s²3p³) and Si has four (3s²3p²) (see page 50 of http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px432/px432.pdf).

In this case, like I said in the beginning, Cr has six valence electrons (3d⁵4s¹) vs. Ti with four (3d²4s²) (e.g. neutral Cr has two more). Considering the charge state (which I've never seen described before in terms of donors/acceptors), Cr³⁺ has three more valence electrons than Ti⁴⁺ (3d³ vs. 3d⁰). However, from what I'm being told, we don't care! All we care about now is the charge state -- that the Cr³⁺ ion has -1 lower in oxidation state than Ti⁴⁺.

How do we reconcile this with the Si/P example? Do you understand the source of my confusion when I compare my line(s) of thinking?

 

[Useful nucleus]

Alright, so I will give it another try :-)

The example you brought from basic semiconductor physics does not capture all the complexity especially when it comes to transition metals like Cr that can have multiple oxidation states. Transition metals can potentially have a multitude of oxidation states when they are present as impurity in a material with a band gap (just like TiO2). The predominant charge state will in general depend on the chemical potential of electrons (or the so-called Fermi level). What we really care about is the charge states that are proven either experimentally or through quantum mechanical calculations to be the dominant ones. As far as I know there is no unified explanation for why the transition metal X is 2+ in the material Z, while X is usually 3+ in material Y.

However, for some "simple" impurities like Hydrogen, there is exists somehow a line of thinking that explains why it takes certain charge state but no the other in semiconductors. See for example:
http://www.nature.com/nature/journal/v423/n6940/abs/nature01665.htmlTo appreciate the complexity of transition metal impurities, you can have a look at the following comprehensive study for these impurities in silicon:
http://prb.aps.org/abstract/PRB/v41/i3/p1603_1

 

[citw]

Alright, so I will give it another try :-)

The example you brought from basic semiconductor physics does not capture all the complexity especially when it comes to transition metals like Cr that can have multiple oxidation states. Transition metals can potentially have a multitude of oxidation states when they are present as impurity in a material with a band gap (just like TiO2). The predominant charge state will in general depend on the chemical potential of electrons (or the so-called Fermi level). What we really care about is the charge states that are proven either experimentally or through quantum mechanical calculations to be the dominant ones. As far as I know there is no unified explanation for why the transition metal X is 2+ in the material Z, while X is usually 3+ in material Y.

OK, thanks for sticking with me... you clearly understand what's going on here, so hopefully you can clear this up for me once and for all. Is the example that I described to you technically wrong? In other words, do the number of valence electrons not really matter, just the charge state? It sounds like you're saying the whole "Group V in Group IV/ five valence electrons v. four valence electrons" is just an approximation for "simple" materials, but in reality this is not how it works. Am I on the right track?

 

[Useful nucleus]

All what I'm saying is that the notion of a fixed number of valence electrons does not work for all elements in the periodic table. Yes, silicon in all its compounds , assumes a charge state of 4+ and hence we can safely say that it has 4 valence electrons.

But how about Mn , for example? Mn can have charge states from 2+ all the way to 7+ . All what we can say about Mn atom is that it has two electrons in its outer S-shell and 5 electrons in its outer d-shell. Will it lose all of them when it is present as an impurity in TiO2? We do not know a priori. If it loses only 3 then it will be Mn3+ and hence an acceptor (with respect to TiO2). If it loses 4, then it will be Mn4+ and hence electrically inactive. If it so happens and it loses 5 electrons , then it is a donor (again with respect to Ti+4). But in any case I cannot predict a priori what will happen, unless I do the calculations (or the experiment).

 

[citw]

All what I'm saying is that the notion of a fixed number of valence electrons does not work for all elements in the periodic table. Yes, silicon in all its compounds , assumes a charge state of 4+ and hence we can safely say that it has 4 valence electrons.

But how about Mn , for example? Mn can have charge states from 2+ all the way to 7+ . All what we can say about Mn atom is that it has two electrons in its outer S-shell and 5 electrons in its outer d-shell. Will it lose all of them when it is present as an impurity in TiO2? We do not know a priori. If it loses only 3 then it will be Mn3+ and hence an acceptor (with respect to TiO2). If it loses 4, then it will be Mn4+ and hence electrically inactive. If it so happens and it loses 5 electrons , then it is a donor (again with respect to Ti+4). But in any case I cannot predict a priori what will happen, unless I do the calculations (or the experiment).

In looking back at your previous post, you said "One can talk about valence electrons of atoms but not ions." I think you're saying that for solids with (at least some) ionic character (like TiO₂), we can't think in terms of valence electrons (by which I mean unpaired electrons outside of the noble gas core) and we are only concerned with charge state. This is a pretty new concept for me, so my first question would be this: what is the consequence of the additional electrons outside of the core? As you pointed out, Cr⁶⁺ is not energetically favorable, so we have Cr³⁺ with three more electrons outside of its core. What happens to these extra electrons?

Secondly, for covalent elemental semiconductors like Si or Ge, what is the story here? Si forms four covalent bonds with its 3s²3p² electrons, but since the electrons are shared equally, we assume an oxidation number of 0 and in this case we're just discussing atoms. If this statement is correct, it seems like for covalent solids, the donor/acceptor designation is determined by the number of valence electrons, while in ionic solids it's determined by the charge (oxidation) state. If this is the case, this clears up a lot. Is this correct?

 

[Useful nucleus]

we can't think in terms of valence electrons (by which I mean unpaired electrons outside of the noble gas core)

I think your definition of valence electrons does not always work. Try to apply it to Ga which we know it has one unpaired electron but in spite of that it usually donates 3 electrons. Honestly, the notion of valence electrons is not something that has a clear definition. It is just a convenient way to introduce the notion of the chemical bond in introductory discussions. But even the "chemical bond" is not something that has a clear cut definition!

what is the consequence of the additional electrons outside of the core? As you pointed out, Cr⁶⁺ is not energetically favorable, so we have Cr³⁺ with three more electrons outside of its core. What happens to these extra electrons?

This is a hard question. I'm currently doing a literature survey for this topic. People invoke the so-called crystal field theory (or the refined ligand field theory) to study the impact of the neighboring ions (like oxygen in TiO2) on these remaining electrons. May be somebody with better expertise can share a concise answer here.

but since the electrons are shared equally, we assume an oxidation number of 0 and in this case we're just discussing atoms.

I believe the notion of oxidation state is more relevant to ionic materials where complete charge transfer takes place from the cations to the anions.

 

[citw]

I think your definition of valence electrons does not always work. Try to apply it to Ga which we know it has one unpaired electron but in spite of that it usually donates 3 electrons. Honestly, the notion of valence electrons is not something that has a clear definition. It is just a convenient way to introduce the notion of the chemical bond in introductory discussions. But even the "chemical bond" is not something that has a clear cut definition!

This is a difficult concept to accept, but I understand your point.

This is a hard question. I'm currently doing a literature survey for this topic. People invoke the so-called crystal field theory (or the refined ligand field theory) to study the impact of the neighboring ions (like oxygen in TiO2) on these remaining electrons. May be somebody with better expertise can share a concise answer here.

I'm really glad you mentioned that, because I had completely dismissed crystal field theory's relevance with respect to the extra electrons, and now I remember you're right. With respect to CFT, you can find diagrams/explanations for the splitting of d¹ through d³ in chapter 21, pp 666-668 of Houscroft's Inorganic Chemistry, 3rd edition. I don't want to post anything from it here since it'll probably be a violation of of forum rules. This book is absolutely fantastic re CFT, and provides the most clear explanations I've seen thus far.

I believe the notion of oxidation state is more relevant to ionic materials where complete charge transfer takes place from the cations to the anions.

Thank you! I figured as much.

Now, so you can understand why I've had such confusion with this and why it continues to be a headache, have a look at this article on codoping:

http://prl.aps.org/abstract/PRL/v103/i22/e226401

Here, they mention that Cr-N codoping is "net n-type" and Cr-C is "compensated". There is absolutely no discussion of charge state, unfortunately. Now, I'm still trying to find the charge state for C in TiO₂, but I can tell you that when N substitutes for O it does so as N^3-. In other words, in comparison to O^2-, one more electron is taken from the lattice and we have an acceptor. So with Cr_Ti and N_O, we have two acceptors, which is definitely not n-type! (Note that they're saying the net doping scheme is net n-type, not TiO₂, which is usually n-type). I don't think there's any other way to reach this conclusion than the simplified "valence counting" that I was doing earlier, whereby they would say that Cr has two greater valence than Ti, and N has one less than O, making it net n-type. This would also explain the notion of Cr-C being compensated, since C is two elements below O and Cr is two above Ti.

Indeed, if you have a look at this citing article,

http://apl.aip.org/resource/1/applab/v102/i17/p172108_s1?isAuthorized=no

these authors also refer to N(+1) and Cr(-2)! This most certainly does not refer to the oxidation state, but the net "valence electron" difference from the substituting constituent (as I mentioned directly above in the previous paragraph). They say "If Cr doping transfers 6 electrons and replaces Ti⁴⁺, the net gain is 2 electrons", which contradicts what you (and others) have explained about the notion of valence electrons being a clear concept. I'm actually not sure what they mean by this. They also point out that in reality, Cr is incorporated as Cr³⁺, but say this introduces "an extra electron per formula unit". Huh?? As we've discussed, Cr³⁺ introduces one fewer electron to the lattice! Am I missing something obvious here?

 

[Useful nucleus]

With respect to CFT, you can find diagrams/explanations for the splitting of d¹ through d³ in chapter 21, pp 666-668 of Houscroft's Inorganic Chemistry, 3rd edition.

Thank you for recommending this book! I will borrow a copy form the library and check the pages you suggested.

Indeed, if you have a look at this citing article,

http://apl.aip.org/resource/1/applab/v102/i17/p172108_s1?isAuthorized=no

these authors also refer to N(+1) and Cr(-2)! This most certainly does not refer to the oxidation state, but the net "valence electron" difference from the substituting constituent (as I mentioned directly above in the previous paragraph). They say "If Cr doping transfers 6 electrons and replaces Ti⁴⁺, the net gain is 2 electrons", which contradicts what you (and others) have explained about the notion of valence electrons being a clear concept. I'm actually not sure what they mean by this. They also point out that in reality, Cr is incorporated as Cr³⁺, but say this introduces "an extra electron per formula unit". Huh?? As we've discussed, Cr³⁺ introduces one fewer electron to the lattice! Am I missing something obvious here?

I had a quick look at the paper, in particular page 3, the pargraph on the top of the right column. I think there is no inconsistency between what the paper says and what we discussed. It is just not well-written.

To clarify a bit let's take Cr+6 as an example. If it substitutes Ti+4, then we will have a positively charged defect with an "effective charge " of +2 (That is +6 referrenced to +4). Now in the bulk of a semiconductor charge neutrality has to be maintained. One way to achieve this is to charge-balance this doubly positive defect by creating two free electrons in the conduction band. This is why those authors mention that Cr+6 introduces two free electrons.
Same argument applies to N-3 setting on O-2 site and hence introduced a hole in the valence band.

I will diverge here a bit to say that this compensation by electronic charge carriers dominates the thinking in the semiconductors community. However, there are other ways of compensation. For example a doubly negative oxygen interstitial or a positively charged oxygen vacancy or in general the native defects of the material.

 

[citw]

I had a quick look at the paper, in particular page 3, the pargraph on the top of the right column. I think there is no inconsistency between what the paper says and what we discussed. It is just not well-written.

To clarify a bit let's take Cr+6 as an example. If it substitutes Ti+4, then we will have a positively charged defect with an "effective charge " of +2 (That is +6 referrenced to +4). Now in the bulk of a semiconductor charge neutrality has to be maintained. One way to achieve this is to charge-balance this doubly positive defect by creating two free electrons in the conduction band. This is why those authors mention that Cr+6 introduces two free electrons.
Same argument applies to N-3 setting on O-2 site and hence introduced a hole in the valence band.

But what about Cr³⁺ introducing "an extra electron per formula unit"? This can't be correct, can it? And the notion of the Cr-N being a "net n-type" doping scheme, if Cr is Cr³⁺ and N is N^3-, this can't be correct either, can it?

 

[Useful nucleus]

But what about Cr³⁺ introducing "an extra electron per formula unit"? This can't be correct, can it? And the notion of the Cr-N being a "net n-type" doping scheme, if Cr is Cr³⁺ and N is N^3-, this can't be correct either, can it?

Well, after spending more time on this paper, I really believe it is not well-written at all. In addition it is not conclusive in their interpretation of the results (But this may be ok for a letter!).

I think what they mean is: Cr+3 is a ngeative defect with respect to Ti+4. Of course this meana that it has to be compensated with something posititve (free holes or oxygen vacancies). But Cr+3 itself carries a "localized" electron or polaron that can hop and contribute to conductivity. This ,however, means that it can change valence to +4 (when the polaron departs) and then can change back again to +3 when it happens to be in the way of a migrating polaron. [One can even go far in thiking and imagines that Cr+3 is actually Cr+6 and three localized polarons!]

As I said this paper is not conlcusive and they do not know what is the defetc equilibrium state in their samples. They "conjecture" from the negative seeback coefficient that all their samples are dominated by electrons from the creation of oxygen vacancies.

 

[citw]

Well, after spending more time on this paper, I really believe it is not well-written at all. In addition it is not conclusive in their interpretation of the results (But this may be ok for a letter!).

I think what they mean is: Cr+3 is a ngeative defect with respect to Ti+4. Of course this meana that it has to be compensated with something posititve (free holes or oxygen vacancies). But Cr+3 itself carries a "localized" electron or polaron that can hop and contribute to conductivity. This ,however, means that it can change valence to +4 (when the polaron departs) and then can change back again to +3 when it happens to be in the way of a migrating polaron. [One can even go far in thiking and imagines that Cr+3 is actually Cr+6 and three localized polarons!]

As I said this paper is not conlcusive and they do not know what is the defetc equilibrium state in their samples. They "conjecture" from the negative seeback coefficient that all their samples are dominated by electrons from the creation of oxygen vacancies.

After our discussion, I question statements made in both if the papers I linked. The implication of these (most explicitly the first) papers appears to be that the doping scheme itself (directly, not by vacancy compensation, etc) is net n-type. Given the charge states, and the points you've made, the idea of Cr-N must be p-type. The outcome (whether it is experimentally found to be p- or n-type) is another matter. Correct?

 

[Useful nucleus]

After our discussion, I question statements made in both if the papers I linked. The implication of these (most explicitly the first) papers appears to be that the doping scheme itself (directly, not by vacancy compensation, etc) is net n-type. Given the charge states, and the points you've made, the idea of Cr-N must be p-type. The outcome (whether it is experimentally found to be p- or n-type) is another matter. Correct?

I did not read the PRL, but according to the APL what you said is true. N is -3 and negative with respect to O-2. Also, Cr+3 is negative with respect to Ti+4. In total Cr-N doped TiO2 is anticipated "naively" to be p-type since these negative defects requires free holes to charge-balance them. Surprisingly, the experiment measured n-type conductivity in this system.

 

[madhusoodan]

Hi all
Can you please let me know why almost all metal oxide semicondductors are n type?

 

[Useful nucleus]

Hi all
Can you please let me know why almost all metal oxide semicondductors are n type?

This is not true some metal oxides are p-type (ex: some copper and iron oxides if I remember correctly). The point defects are the deciding factors.