In ammonium-phosphomolybdate, the oxidation state of Mo is +______

We need to find the oxidation state of Mo in ammonium phosphomolybdate.

The chemical formula of ammonium phosphomolybdate is $$(NH_4)_3PO_4 \cdot 12MoO_3$$, also written as $$(NH_4)_3[PMo_{12}O_{40}]$$.

Let us find the oxidation state of Mo using the second formula $$(NH_4)_3[PMo_{12}O_{40}]$$.

The ammonium ion $$NH_4^+$$ has a charge of $$+1$$, and there are 3 of them, so the total positive charge from cations is $$+3$$.

Therefore, the anion $$[PMo_{12}O_{40}]$$ has an overall charge of $$-3$$.

In the anion $$[PMo_{12}O_{40}]^{3-}$$:

- Phosphorus (P) has an oxidation state of $$+5$$

- Oxygen (O) has an oxidation state of $$-2$$

- Let the oxidation state of Mo be $$x$$

Writing the charge balance equation:

$$(+5) + 12(x) + 40(-2) = -3$$

$$5 + 12x - 80 = -3$$

$$12x - 75 = -3$$

$$12x = 72$$

$$x = +6$$

Therefore, the oxidation state of Mo in ammonium phosphomolybdate is $$+6$$.

The answer is 6.