What is the oxidation state of the vanadium atom in (VO4)-?
Well I thought I had this down because I assumed O's oxidation was -8 but then I noticed the negative so.... help?
Well I thought I had this down because I assumed O's oxidation was -8 but then I noticed the negative so.... help?
phuong
( VO4) -1
oxidation of V + 4 ( -2 ) = -1 => oxidation number of V = -1 + 8 = -7
Anonymous
Assuming the VO4- has a charge of -1 it's 7. If the ion has a charge of -1 there must be one more negative charge than positive.
Oxygen ions are -2 so 4 oxygens are -8. To make the whole thing -1 the vanadium must be +7.
7 positives cancel out 7 negatives leaving the 1 negative on the oxygen to give the molecular ion a charge of -1 overall.