# Metallic and magnetic nature of transition metal oxides

Which of the following oxides is both metallic and ferromagnetic:
$\ce{VO2}$, $\ce{CrO2}$, $\ce{MnO2}$, $\ce{TiO2}$?

This is how I went about solving it:

Since ferromagnetic materials also show paramagnetism, so the condition for paramagnetism must be checked.

In $\ce{TiO2}$, $\ce{Ti}$ has no unpaired electrons, it cannot be paramagnetic and as a consequence it cannot be ferromagnetic.

$\ce{CrO2}$ is amphoteric. A metallic oxide would show basic properties in most cases. So $\ce{CrO2}$ may not be the right answer.

$\ce{VO2}$ is basic. All three except $\ce{TiO2}$ are paramagnetic.

Then I googled, I didn't get an exact answer. But I read that $\ce{CrO2}$ is termed as a "half metal" which is related to its magnetic properties, most part of which I didn't understand.

Then I also found articles about half metallic ferromagnetism of $\ce{TiO2}$. But that goes against my assumption.

So my questions are:

1. Explain where I'm wrong.
2. Why $\ce{CrO2}$ is the answer? (Found that when I finally referred to the solutions of the question paper. However the solutions given can have mistakes. So it is possible that $\ce{CrO2}$ may not be the true answer).
3. If you think $\ce{CrO2}$ is not the answer, then why?
• Basic or amphoteric nature of a compound is not related to ferromagnetism whatsoever. – Ivan Neretin Apr 18 '16 at 11:22
• What the heck is a ‘metallic oxide’, under the assumption that not all of those oxides are metallic because the counterion is metallic? – Jan Jul 29 '16 at 22:24
• @Jan ok, i will take care next time. – Nilay Ghosh Aug 24 '16 at 10:53
• @Jan 'metallic oxide' here probably means 'oxide with metallic conductivity' – permeakra Sep 21 '16 at 20:55
• @permeakra Most metal oxides have band gaps, so I'm not sure if they can be truly metallic – CoffeeIsLife Sep 22 '16 at 4:36

This question should be cross-posted to Physics.SE, it is not a simple chemistry question. I'll give short rationalization of the fact, but beware: it is a rationalisation, not a deduction. It is also given in layman's terms.

We know the crystal structure of $\ce{TiO2}$, it is stereotypical and quite common. It contains octahedrons $\ce{TiO6}$ with oxygens shared between three octahedrons. Since the four compounds have similar composition and the cations have similar size, we may expect them to have similar structures. This is, however, a mere guess, but lets' go with it.

For a compound to have metallic conductivity, it must have mobile electrons, i.e. partially occupied group of orbitals. Thus, $\ce{TiO2}$ is obviously out.

It is known, that $d$-orbitals of a metal in octahedral neighbourhood split: three are lower in energy then two other.

In $\ce{MnO2}$ there are three electrons, so all three lower orbitals are occupied. This makes electrons immobile: for an electron to move to a neighboring atom, it has to pair with one of the already present electrons, which requires energy. I do not expect, however, that $\ce{MnO2}$ shall be a great insulator: the pairing energy is quite low, and furthermore, any impurity shall create charge carriers. So, $\ce{MnO2}$ is probably a semiconductor.

To choose from remaining two, we should recall that ferromagnetism is common among compounds with a lot of half-filled d-orbitals. This suggests that $\ce{CrO2}$ has higher chances, than $\ce{VO2}$ Still, the latter should be paramagnetic and conducting under right conditions.

Again, you should cross-post the question on Physics.SE and probably read a good solid state physiscs book.

$\ce{TiO2}$ is diamagnetic. $\ce{VO2}$ has only 1 unpaired electron, definitely not ferromagnetic. But I am still confused between $\ce{MnO2}$ and $\ce{CrO2}$. Both are ferromagnetic. Metallic character is decided by the strength of the Metal-Oxygen bond. In case the metal is in a very high O.S. (e.g. $\ce{MnO4-}$, $\ce{Cr2O7^2-}$) the M-O bond is more covalent and cannot be ruptured by water to give a base.

In this case however, both $\ce{MnO2}$ and $\ce{CrO2}$ seem appropriate.