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Can anyone explain to me why most articles describe chromium as an acceptor in titanium dioxide? In TiO2, titanium has the charge state Ti$^{4+}$ and oxygen has the charge state O$^{2-}$. When Cr substitutes for Ti, it does so as Cr$^{3+}$. 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, Cr$^{3+}$ actually has three more valence electrons than the Ti$^{4+}$ ([Ar]$3d^34s^0$ vs [Ar]$3d^04s^0$). It should therefore be a donor, right? 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?

My understanding has always been simply this: more electrons than host $\Rightarrow$ donor, fewer electrons than host (more holes) $\Rightarrow$ acceptor. The position of the impurity level, to my (perhaps incorrect) knowledge, does not determine whether or not the impurity is actually a donor or acceptor, but rather whether it is a recombination center or trap. We can have localized states near the middle of the bandgap that are technically donors/acceptors but function as recombination centers, so I'm not sure what I'm missing here.

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Is it a well-known fact that Cr is substituting Ti? (Not substituting O, not interstitial, not defect complex, not a combination of the above?) –  Steve B May 11 '13 at 13:38
    
Yes, absolutely. Not only does it substitute Ti, but interstitial Cr is energetically unfavorable. TiO$_2$ is described in terms of chains of distorted TiO$_6$ octahedra, so it does form a coordination complex. Cr$^{3+}$ just replaces Ti$^{4+}$ as the central ion. –  SMPAT May 11 '13 at 13:53
    
Can you provide a reference? –  Vasiliy Aug 10 '13 at 9:13
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A 3+ ion substituting for a 4+ ion is called p-type doping. Since it only contributes 3 electrons there is one missing, called a hole. This hole will readily accept an electron, so it is an acceptor. As you say this will also give an impurity band just above the valence band. That Cr gives fewer electrons to the lattice than Ti has to do with splitting of the bands in the lattice.

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I understand what you're saying about the charge state indicating one fewer electron, but what about the total number of valence electrons? This is what has me confused... does the charge state or number of valence electrons determine the donor/acceptor designation? I remember one Phys. Rev. paper that refers to Cr as a "double donor", and another that refers to Cr as "n-type." This gets pretty confusing. –  SMPAT May 12 '13 at 12:42
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For heavier atoms determing the number of valence electrons is more complex than just counting down to Argon. Also, in an environment the orbitals can split (ligand field). Cr could exist as 3+ or 5+ or perhaps other states depending on environment. What matters is that in TiO2 it is 3+ and therefore readily accepts an electron in the missing spot. In some other environment it could be n-type and donate electrons. –  Groda.eu May 12 '13 at 15:39
    
So what happens to the traditional ideas taught in solid state, where, for example, P is a donor in Si because it has one more electron (higher group)? Do we just disregard the traditional way of thinking in this regard? I guess what I'm asking is what happens to the additional $d^3$ electrons introduced by Cr? –  SMPAT May 12 '13 at 15:55
    
That works very well for low n-quantum numbers (for instance P and Si) since they are far apart in energy. Further down the periodic table the energy levels are closer and small energy shifts (due to oxidation states, ligand fields or whatever) can shift the order of orbitals or even make the normal orbital basis set less useful. –  Groda.eu May 12 '13 at 18:13
    
I'm starting to think this is the way to describe ionic semiconductors, while we usually think of covalent SCs like Si or Ge. In that case, we assume $\text{P}_\text{Si}$ provides an additional electron not involved in bonding. In this case, we seem to look at things differently. If that's the case, what role do valence electrons play? Cr$^{3+}$ is $d^3$, so shouldn't the three $d^3$ electrons have consequences (compared to the $d^0$)? –  SMPAT May 12 '13 at 18:30
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