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QUESTION

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

The oxidation number of polyatomic ions is equal to the charge on the ion. The oxidation number of "Mo" in "MoO"_4^(2-) will be +6

The of all the atoms in a compound must add up to the charge of that compound.

"O"= -2

"Mo"= x

The charge of the molybdate anion, "MoO"_4^(2-), is 2-, therefore

 x + 4* (-2) = -2

x-8 = -2 implies x = -2 + 8 = +6

Therefore, "Mo" has an oxidation number of +6 in the molybdate anion.