What is the charge on the copper ion in antlerite?
Ok, here is a short answer. Since you already know the chemical formula it's actually quite easy provided you are familiar with the process of determining the oxidation numbers in chemical compound. From the Wikipedia article (and the way the chemical formula is written indicates that too) you can get the information that the $\ce{SO4}$ part of the formula is sulfate and sulfate ions carry a charge of $-2$, i.e. $\ce{SO4^{2-}}$. Furthermore, in most cases $\ce{H}$ gets an oxidation number of $+1$ and $\ce{O}$ gets an oxidation number of $-2$ so that the 4 $\ce{OH}$ units in the compound each have a charge of $1 - 2 = -1$, so you have $\ce{OH^{-}}$, i.e. hydroxyl groups - this is supported by the way those $\ce{O}$'s and $\ce{H}$'s are grouped, if you see "$\ce{OH}$" in a compound it usually means "hydroxyl group". So, you have one sulfate group with a charge of $-2$ and four hydroxyl groups with a charge of $-1$ each, giving you a charge of $-6$ in total. But your compound $\ce{Cu3(SO4)(OH)4}$ is neutral, i.e. has a charge of $0$, so the three $\ce{Cu}$ have to counteract the negative charge exerted by the anions, in other words you have to distribute a charge of $+6$ between the three copper atoms. If you assume that the charge is evenly distributed, i.e. that all the $\ce{Cu}$ are the same (this need not necessarily be the case, you can also have mixed oxides like $\ce{Fe3O4}$ where you have $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ within the same mineral), you get to a charge of $+2$ for each copper ion, i.e. $\ce{Cu^{2+}}$.