# why is the chemical formula for vanadium (v) oxide V2O5?

if you were given the problem how would you know there is a 2 after the v

### 6 Answers

- Old Science GuyLv 710 years agoFavourite answer
V+5 and O-2

criss-cross the numbers without the signs

V2O5

reduce to lowest possible whole number ration (already there)

this is the empirical formula

- Log in to reply to the answers

- Anonymous10 years ago
The (v) means that vanadium has an oxidation number of 5. (V is the roman numeral for 5).

Oxygen always has an oxidation number of -2.

The oxidation number of a compound always equals the charge of the compound. Here, the compound has no overall charge so it equals 0.

If there was only one vanadium, then the oxidation number would be 5 and you wouldn't be able to balance that out with oxygens because the 2 oxygens would make the overall oxidation number -1 and 3 oxygens would make the overall oxidation number +1. We want it to be 0.

Therefore we have two vanadium atoms, which have an oxidation number of +10 altogether.

We want the overall oxidation number to be 0 so we will have 5 oxygens, each at -2, to get -10 altogether.

- Log in to reply to the answers

- Dr.ALv 710 years ago
Vanadium (V) oxide means that vanadium has oxidation number +5

because oxygen has oxidation number -2 and for neutral species the sum of oxidation numbers times the number of each kind of atoms adds up to zero the formula is V2O5

check ; [ 5 ( +2)] + [5 (-2)]= 0

- Log in to reply to the answers

- 5 years ago
This Site Might Help You.

RE:

why is the chemical formula for vanadium (v) oxide V2O5?

if you were given the problem how would you know there is a 2 after the v

Source(s): chemical formula vanadium oxide v2o5: https://biturl.im/8on93- Log in to reply to the answers

- What do you think of the answers? You can sign in to give your opinion on the answer.
- uriasLv 44 years ago
Vanadium V Oxide

Source(s): https://shrink.im/a8RVp- Log in to reply to the answers

- ?Lv 44 years ago
Chemical Symbol Vanadium

Source(s): https://snipurl.im/aYSEc- Log in to reply to the answers