Oxidation States Help?

I'm doing some transition work for the A-level chem course and am slightly confused. So for calculating the oxidation states for specific elements in a compound, why aren't group numbers always considered. E.g. V2O5 I did this as it has to equal 0 :
V^x x 2 and O^2- x 5
x=+5 -10

So my query is firstly is this correct and secondly why is the Vanadium oxidation state not +3 like its group number? Is there an exception to this element? When most of my previous answers have related to the elements group number.

(Also I should probably double-check this but are oxidation state and oxidation number the same thing?)

Any help is appreciated, thanks!

This is a very common query! Basically, most elements will have the same oxidation state as their group, but the transition metals as a whole are an exception. Transition metals have variable oxidation states - they can take on different oxidation states in different compounds. For example, Vanadium can have oxidation states of +2, +3, +4 or +5! And yes, oxidation state and number are the same thing.

I must admit, it’s quite difficult to expect a new chemistry A-Level student to work out that by themselves - probably intended to prompt a discussion, so hopefully I haven’t ruined that!

Thank you so much! I've been trying to watch some youtube videos to understand it but your response has cleared up my confusion! Another question, if you don't mind answering it, when the compound is charged does that just mean the oxidation state has to equal that instead of 0? For example Cr^2O^7^2- the oxidation states would just have to add to equal -2???

Yes, it does! So for example, in the Cr2O7 2- ion, the overall charge is -2 and the oxidation state of O is -2, so 2*(oxidation of Cr)+7*-2=-2, so 2*(oxidation state of Cr) = +12, so each Cr ions has an oxidation state of +6.