Non-stoichiometric cuprous oxide, $\mathrm{Cu}_2 \mathrm{O}$ can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor?

Question

Non-stoichiometric cuprous oxide, $\mathrm{Cu}_2 \mathrm{O}$ can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor?,

MARKS App · Accepted Answer

The stoichiometric ratio slightly less than $2: 1$ in $\mathrm{Cu}_2 \mathrm{O}$ shows that some cuprous $\left(\mathrm{Cu}^{+}\right)$ions have been replaced by cupric $\left(\mathrm{Cu}^{2+}\right)$ ions. In order to maintain electroneutrality, every two $\left(\mathrm{Cu}^{+}\right)$ions will be replaced by one $\mathrm{Cu}^{2+}$ ion thereby, creating a hole. As the conduction will be due to the presence of these positive holes, hence it is a $p$-type semiconductor.

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