What are the molecular, complete ionic, and net ionic equations for the reaction between CuO and HCl?

Hint: The molecular equation is simply the balanced chemical reaction we write. In this we consider all the ionic compounds or acids as neutral molecules. The state of each reactant is shown in brackets after the chemical formula of the compound.

Complete answer:
The complete ionic equation gives us information about all the ions that are present in the solution during the reaction whereas the molecular equation tells us about the ionic compounds that were the sources of the individual ions present in the solution.
A net ionic equation is the one that shows only the chemical species that are involved in that reaction, whereas a complete ionic equation also includes the spectator ions. The ions that are unchanged during the reaction are known as the spectator ions. These species appear on both sides of the equation. We can cancel them out in a similar manner to cancel out equal terms on either side of a mathematical equation.
We are given a reaction between CuO and HCl. This is a double displacement neutralisation reaction that takes place between the base CuO and acid HCl. CuO is known as cuprous oxide which is black in color and used to make nanoparticles.
The general/molecular reaction can be given as:
$CuO(aq) + 2HCl(aq) \to CuC{l_2}(aq) + {H_2}O(l)$
This is a balanced chemical equation between Copper (II) Oxide and HCl gives us Cuprous chloride (salt) and water.
The complete ionic equation will include all the ions present in the molecular equation, including the spectator ions. The reaction goes as:
$C{u^{ + 2}}(aq) + {O^{2 - }}(aq) + 2{H^ + }(aq) + 2C{l^ - }(aq) \to C{u^{ + 2}}(aq) + 2C{l^ - }(aq) + {H_2}O(l)$
The net ionic equation can be given by cancelling out equal terms in the complete ionic equation. The net ionic equation can hence be given as:
${O^{2 - }}(aq) + 2{H^ + }(aq) \to {H_2}O(l)$.

Note:
A net ionic equation is mostly written with the smallest integer value of the stoichiometric coefficients. We might have to divide the stoichiometric coefficients by a common divisible number to get the final net ionic equation.