Most salts in the solid state are electrical insulators, since the ions are not mobile (e.g. sodium chloride). However, solid titanium(II) oxide, $\ce{TiO (s)}$, is a conductor. How is this so?


The electron configuration of $\ce{Ti^{2+}}$ is $\mathrm{[Ar]~3d^2~4s^0}$.

$\ce{TiO}$, like e.g. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$, adopts the rock salt ($\ce{NaCl}$) structure.

Due to the lower nuclear charge of $\ce{Ti}$ its 3d orbitals are less contracted than that of $\ce{Fe}$, $\ce{Co}$ and $\ce{Ni}$ which allows for overlapping of the 3d orbitals and the formation of a metal d band in $\ce{TiO}$. The d electrons of $\ce{Ti^{2+}}$ partially fill this band and lead to electrical conductivity. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$ on the other hand are electrical insulators.

Note that in this case it is electrons that act as the mobile charge carriers, not the ions.

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    $\begingroup$ I always thought it was related to defect structures. $\ce{FeO}$ is a fairly good conductor on that account IIRC. $\endgroup$ Aug 25 '17 at 23:13
  • $\begingroup$ As far as I know conductivity of TiO is not related to defect structures. However, $\ce{FeO_{1-x}}$ might not be the best example for an insulator since it rather has the conductivity of a semiconductor than an insulator at normal temperature. Its conductivity is indeed due to $\ce{Fe^{3+}}$ holes hopping between Fe sites in the lattice. $\endgroup$
    – aventurin
    Aug 26 '17 at 20:46
  • $\begingroup$ $\ce{Fe_{1-x}O}$ maybe? Much of my job involves dealing with oxide scales, hence I see the q through wustite-colored glasses. $\endgroup$ Aug 26 '17 at 22:15
  • $\begingroup$ TiO has a defect NaCl structure (1/6 lattice points are vacant); I've read that this permits a more contracted structure and hence better overlap between 3d orbitals on neighbouring Ti. Disclaimer, I have only very superficial knowledge of solid state chem. $\endgroup$
    – orthocresol
    Feb 18 '18 at 1:43

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