What is the oxidation number of the #"Mo"# in #"MoO"_4^(2-)#?

1 Answer
Jun 21, 2016



Oxygen almost always has an oxidation number of -2, so it is safe to make the assumption that it does here (the exception are certain peroxides and superoxides).

Four oxygens give a total of (4 x -2) = -8.

The Molybdenum must be positive, but if the complex overall has a charge of -2, there are 2 negative charges left over.

To put it mathematically:
#x# -8 = -2
#x# must be +6.