Non-stoichiometric cuprous oxide, Cu2O can be prepared in 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? 


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In the cuprous oxide (Cu2O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu+ ions is slightly less than twice the number of O2− ions. This is because some Cu+ ions have been replaced by Cu2+ ions. Every Cu2+ ion replaces two Cu+ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor. 

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The ratio less than 2:1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p-type semiconductor.

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