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?
Answer
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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