How do you find the number of valence electrons in a ion?

Let's take examples from Group 15 (nitrogen), Group 16 (oxygen), and Group 17 (fluorine). All of these are non-metals, and they tend to be (STRONGLY) oxidizing; that is they accept electron density to give ions that are isoelectronic with the next Noble Gas. Nitrogen thus forms a #N^(3-)# ion, oxygen an #O^(2-)# ion, and, fluorine, by reason of its electron count the most strongly oxidizing element on the table, a #F^(-)# ion.

All of these ions have a full valence shell, and are isoelectronic with #Ne#, which possesses a full valence configuration of #1s^(2)2s^(2)2p^(6)#; this is the origin of the electron shell idea of #2:8:8# etc. that we learnt in lower high school.

Anyway, I don't know which level you need to understand. Of course, the topic can be further elaborated. The practical way of deciding the number of valence electrons is to use the Periodic Table. Reducing main-group metals tend to lose electrons to give the electronic configuration of the last Noble Gas (to form a cation), and oxidizing non-metals tend to gain electrons to give the electronic configuration of the NEXT Noble Gas (to form a cation). Can you see where I am going?