Non-stoichiometric cuprous oxide,$Cu_{2}O$ can be prepared in the laboratory. In this oxide,the copper to oxygen ratio is slightly less than $2:1$. Can you account for the fact that this substance is a $p$-type semiconductor?

(N/A) In the cuprous oxide $(Cu_{2}O)$ prepared in the laboratory,the copper to oxygen ratio is slightly less than $2:1$.
This implies that the number of $Cu^{+}$ ions is slightly less than twice the number of $O^{2-}$ ions.
This occurs because some $Cu^{+}$ ions are replaced by $Cu^{2+}$ ions to maintain electrical neutrality.
Since every $Cu^{2+}$ ion replaces two $Cu^{+}$ ions,it creates a cation vacancy or a 'hole'.
These positive holes facilitate the conduction of electricity.
Therefore,the substance acts as a $p$-type semiconductor.