How is solid titanium(II) oxide an electrical conductor?

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.
• I always thought it was related to defect structures. $\ce{FeO}$ is a fairly good conductor on that account IIRC. – Oscar Lanzi Aug 25 '17 at 23:13
• 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. – aventurin Aug 26 '17 at 20:46
• $\ce{Fe_{1-x}O}$ maybe? Much of my job involves dealing with oxide scales, hence I see the q through wustite-colored glasses. – Oscar Lanzi Aug 26 '17 at 22:15